BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF 1891 Jf..iiZ.Uo.^^ a. 3513-1 VM605 .369 •" """"'"""' ^'"""^ 'ISimMlfiiiiiMlllS.'ij!?,!.^ scienc olin 3 1924 030 902 930 DATE urn 4a S 1 '3B Cornell University Library The original of tiiis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924030902930 MECHANICAL AND Marine Engineering SCIENCE (Bssays, Problems, Demonstrations). SPECIALLY WRITTEN AS A HANDBOOK TO THE BOARD OF TRADE EXAMINATIONS FOR EXTRA-FIRST-CLASS ENGINEERS. A. N. SOMERSCALES, GOLD MEDALLIST IN STEAM SCIENCE AND ART DEPARTMENT, SOOTH KENSINGTON; HONORARY MEMBER AND PAST PRESIDENT (HULL BRANCH) MARINE ENGINEERS' ASSOCIATION. Entered at ^^ / l ^ Stationer^ Sail. JAMES MUNRO & CO., LTD., Engineering and Nautical Publishers. Soniion : SIMPKIN, MARSHALL, HAMILTON, KENT, & CO., LTD. MDCCCCIX. GEO. C. MACKAY, Printer, 39 Bernard Street, Leith. PREFACE. ■'T'HE present work is an attempt to exhibit in some kind of *■ order and sequence a few out of the many applications of scientific principles to Mechanical and Marine Engineering- matters. In the choice of subjects the Author has been guided more particularly by the requirements of the Marine Engineer, but with A few exceptions the subjects dealt with are not exclusively marine. The selection has arisen out of the Author's experience in assisting -candidates, preparing for the Board of Trade Examination, for Extra-First-Class Engineers. The knowledge required fpr these ■examinations covers an interesting and useful portion of the ground usually included in Mechanical and Marine Engineering ; and upon these lines the present work has been written, in the hope that it may be acceptable to readers, who are not necessarily aspiring to Examination Honours, as well as to those who desire a Handbook for the Extra Examination. liuLL, February igop. CONTENTS. FAGS Preface, 3 Board of Trade Regulations, ... g Part I.— Short Essays on Physical and Engineering Subjects— I. — Earth's Atmosphere, 15 2. — The Mercurial Barometer, ... ... ... ... ... ... 18 3. — The Aneroid Barometer, 21 4. — Sea Water, 23 5. — The Electric Current, ... ... ... ... ... ... ... 26 6. — A Candle Flame and Bunsen Burner, 29 7. — Life Buoys with Inextinguishable Lights, 33 8. — The Absolute Zero of Temperature, 35 9. — Pyrometers, 38 10. — Calorimeters 40 II. — The First and Second Laws of Thermo-Dynamics, 43 12. — The Steam Hygrometer, 45 13.— What is Meant by "Entropy?" 48 14. — Entropy Diagrams, 51 15. — Use of Entropy — Efficiency of Engine, 56 t6. — Efficiency of a Refrigerating Plant, 59 17. — Lime and Cement, 61 Portland Cement, 62 l8. — Iron Ores, 64 ig. — The Coal Measures, 66 Composition of Coal and Coke, ... ... ... ... ... 67 Coke, 68 20.— The Blast Furnace, 70 2i.^-The Composition and Properties of Pig Iron 74 22.^The Manufacture of Wrought Iron, 77 23.^The Bessemer Acid Process of Making Steel, 80 24.— The Basic Process (Bessemer Steel), 83 25.— Siemens-Martin Process for Open-Hearth Steel, ... ... ... 85 26.^The Cementation Process for Making Tool Steel, 89 27. — Welding of Tubes and Furnaces, 91 28. — Lubricating Oils, 94 29. — Corrosion in Boilers 98 30. — Remedies for Corrosion in Boilers ... 103 31. ^Softening Water for Steam Boilers, 105 Clark's Test for Hardness of Water, ic8 page; 32. — Liquid Fuel for Steam Vessels, no- 33. — The Balancing of a Marine Engine, ii5- Paet II. — Solutions of Questions in Mensuration, etc. — I. — Mensuration Questions, I29' 2. — Specific Gravity and Weights, ... 138 Specific Gravity of Liquids, 140 3. — Moments, 146- 4. — The Parallelogram of Forces, 152- The Parallelogram of Velocities, ... ... ... ... IS3- 5. — Work and Horse Power, ... ... ... ... ... ... 163. 6. — Riveted Joints, 172 7. — Stress and Strain, 182: 8.— Hydraulics 191 9. — Centrifugal Force, 201 10. — Shafts 207 Hollow Shafts, 213. Combined Twisting and Bending, ... ... ... ... 216- Coupling Bolts, 217 Horse Power Transmitted by Shafts, 219- Rigidity of Shafts, 220 II. — Beams, ... ... ... ... ... ... ... ... ... 224 Bending Moments (See Part III), 226. Moments of Resistance (See Part III), ... ... ... 226 Curvature of Beams, ... ... ... ... ... ... 240- Deflection of Beams, ... ... ... ... ... ... 242 12. — Stability of Ships. 247 More Advanced Principles, ... ... ... ... ... 254. RoUing of Ships, 255 13. — Exercises in Stability 257- 14. — Heat, 270- 15.— Efficiency and Horse-Power, 286- 16. — Ship's Performance, ... ... ... ... ... ... ... 290 17. — The Valve Diagram, ... 295. 18. — The Piston and Crank Diagram, 302- 19. — Indicator Diagrams, 2o8- Part III. — Proof of Rules and Formula — — Common Logarithms, ... ... ... ... ... „, •321 — Hyperbolic Logarithms, ... ... ... ... ... ... 326" — The Trigometrical Ratios and Angular Measures, ... ... 329. — Acceleration, ... ... ... ... ... ... ... .., 33c — Kinetic Energy, 338. — Cubical Expansion, ... ... ... ... ... ... ._ 34Q — Centre of Gravity of a Triangle, 342 — Centre of Gravity of a Semi-Circle, .. 34^ — Centre of Gravity of a Pyramid 3^5. — Moment of Inertia of a Rectangle, 347 — Radius of Gyration, ^cq. — Moment of Inertia of a Circle, 352- — Moment of Resistance of Rectangular Beams, 354 — Moment of Resistance of a Shaft, ... ... ... ... ... ^cg, ^Moment of Resistance of any Beam, ... iQi PAGE 16.— Bending Moments in Beam Problems, 363 17- — Radius of Curvature, 366 18. — Bending Moment and Curvature 368 19- — Deflection of Beams, ... ... ... ... ... ... ... 370 20— ^Centrifugal Force, 373 21.— Fly Wheels 376 22. — Strength of a Cylindrical Pipe or Boiler 378 23. — Longitudinal Strength of a Boiler 380 24. — Compressive Stress on Boiler Tube Plates, 382 25. — Stress allowed in B.T. Formula for Girders, 384 26. — Safety Valve SpringSj 387 27. — Angle of Twist in a Shaft 389 28.— Compression of a Safety Valve Spring, 391 29. — Thrust of a Propeller, ... ... ... ... ... ... ... 393 30. — Alteration of Pitch of Screw, 395 31. — Speed Against a Current, 396 32. — Horse Power from Wetted Surface, 398 Second Method — Kirk's Analysis, ... ... ... ... 399 33 — Tank Experiments for Predicting the Speed and Ilorse-Power of Vessels, ... ... ... ... .. ... ... ... 403 34.— Cut-off by Slide Valve, 407 35.— Differeiice in Cut-off, ... ... ... ... ... ... ... 409 36. — Formula connecting the Pressure, Temperature, and Volume of Air, 411 37, — Steam Consumption, ... ... ... ... ... ... ... 412 38. — Loss of Fuel by Blowing Boiler Down, ... ... ... ... 414 39! — Best Temperature of Funnel for most Effective Draught, ... 416 40. — Work done during Expansion, 419 41. — Mean Pressure of Steam, 421 42. — ^Work done during Isothermal Expansion, 423 43. — Linear Velocity of Escaping Steam, 425 44. — The Action of Giffard's Injector, 427 Appendix. — Tables and Examination Papers — Chemical Symbols and Atomic Weights of Common Elements, ... 432 Specific Gravity and Weight of Solids and Liquids, 433 Specific Gravity and Weight of Various Gases 434 Specific Heat of Substances, 43S Melting Point, of Metals, etc 436' Temperature and Weight of Steam, 437 Breaking Strength of Materials, 438 Elastic Strength of Materials in Tension, 439' Factors of Safety (from Unwin), 4.39' Young's Modulus of Elasticity (E), 440' Modulus of R'gidity (K), 440 Questions set at the B.T. Examination held in January 1909, for Extra- First-Class Engineers 44* Board of Trade Regulations. The following extract from the Board of Trade Regulations applies to the Extra First-Class Engineer's Examination. Extra First- Class. 24. Extra First-Class Engineer. — This examination is voluntary, and is intended for such persons as wish to prove their superior quahfications, and are desirous of having Certificates for the highest grade granted by the Board of Trade. Extra Examina- tions, when held 25. The extra examinations are held once in every three months, and at ports where examinations of engineers are ordinarily conducted. The examination extends over several days. According to present arrangements these examinations will begin on the second Tuesday in January, April, July and October in each year. Notice required. 26. Notice must be given by the candidate at least one week before the day on which the examination begins, and these notices must be immediately reported by the Examiners to the Chief Examiner in London. Qualifica- tions required. 27. A candidate for an Extra First-Class Engineer's Certificate must possess a First-Class Engineer's Certificate and have served an apprenticeship of five years at least or its equivalent, viz., three years' apprenticeship and three years at an approved technical school; or he must have served for one year at sea as senior of a watch on the main engines or boilers while holding a First-Class Engineer's Certificate. BOARD OF TRADE REGULATIONS. 9 {a) He must be able to write g'ood English. {b) He must possess a thorough knowledge of the con- struction and working of the different forms of marine engines and propellers in all their parts, and be so far acquainted with the elements of theoretical mechanics as to comprehend the general principles on which the machine works, and to illustrate his knowledge of these principles by numerical examples. {c) He must possess a knowledge of the theory of strain and stress sufficient to be able to deduce the ordinary rules for the bending of rectangular bars and for the twisting and bending of round bars. ^d) He must be acquainted with the principles of expan- sion and the modern theory of heat, and be able to solve, with the assistance of his own books or without books, according as the examination papers may be set, questions in economy, and duty in connection with engines and boilers. {e) He must understand how to apply the indicator, and to draw the proper conclusion from the diagrams, and to construct the approximate diagrams for any given data. {/) He must be able to produce, without a copy, a fair working drawing of any part of the machinery, with figured dimensions fit to work from. (^ ) He must understand the principles of the action of the screw propeller and the paddle-wheel, and must be able to estimate numerically the effect in speed of ship and consumption of fuel due to any alteration in pitch, diameter, revolutions, etc., etc. lO BOARD OF TRADE REGULATIONS. (A) He must be able to give a description of boilers and the method of staying them, and must show that he possesses a knowledge of the theoretical prin- ciples which regulate their construction, and that he is able to calculate the strength of the boiler shell, stays, and riveting. (z) He must understand the general nature of the strains- and stresses produced by the steam pressure, and by the expansions due to unequal temperatures ia boiler shells. '^) He must have a knowledge of safety-valve construc- tion, and the principles involved in determining the size of a safety-valve, and the construction of spring-loaded and deadweight valves. ,(/) He must, possess a thorough knowledge of the theory of combustion ; the chemical composition of fuels ^ the evaporative duty of fuels of given composition ;. the production of draught; the effect in regard to- economy, safety, and wear and tear, pf increasing or of diminishing the proportion of heating surface, of grate bar surface, of area of section of air passages, of area of water surface, of steam space capacity, and water capacity. (m) He must be able to explain the formation of scale and the precipitation of salt, and the precautionary means adopted in respect thereto, with jet con- densers, and with surface condensers. (71) He must understand the general principles involved in the construction of the barometer, thermometer, salinometer, and steam and vacuum gauges. (p) He must be familiar with the general results obtained from past experience in relation to corrosion,. BOARD OF TRADS REGULATIONS. II pitting^, and galvanic action in boilers, and the use of zinc and of soda in boilers. (p) He must give a variety of illustrations of how defects have arisen from accident, imperfect construction^ or deterioration, and how these defects might have been prevented, and the best way of repairing such defects. {g) He must possess an intelligent knowledge of the properties of the lubricants, boiler cements, and india-rubber in general use in steamers. (r) He must understand the causes of spontaneous com- bustion and the formation of explosive gases in coal holds, and the precautionary measures proper to prevent accidents from these causes. (s) He must be able to explain the construction and working of the refrigerating machinery in use on board ship, the electric lighting plant, the steering engines, hydraulic and pneumatic engines, the pumps, and all other auxiliary machinery placed under the Chief Engineer's control. (/!) In order to intelligently deal with ballast tanks, the cocks, valves, and pumps of which are under the Chief Engineer's control, and to co-operate the more readily with the master in keeping the vessel in a safe condition, especially when she is light, and when coaling operations are proceeding, candidates are expected to possess an elementary knowledge of the stability of floating bodies. (m) If the candidate does not obtain 67 per cent, of the total number of marks allotted for the papers, he will be declared to have failed. The papers will be founded chiefly on the foregoing sub-paragraphs. FART SHORT ESSAYS ON PHYSICAL AND ENGINEERING SUBJECTS. 1. — The Earth's Atmosphere. The atmosphere has been aptly described as an ocean or ■"deep" of air surrounding- the earth. What the depth of this aerial ocean may be does not appear to be known ; it has been variously estimated at from 45 to 200 miles. The atmosphere resembles the ocean in many ways. It is subject to the attraction of gravity, as water is, and hence it presses upon the surface of the earth and ocean in the same wav and for the same reason that the water of the ocean presses upon the ocean floor; but of course with much less intensity. At the ' deepest parts of the ocean the pressure amounts to over three tons on each square inch, whereas the weight of the atmosphere only amounts to i4"7 lbs. per square inch. Then the atmosphere is subject to currents, due to heat and cold, like the ocean, and very probably to tides also. It is the abode of life, like the ocean, and one of its constituents, viz., oxygen, is also a principal constituent of the ocean water. On the other hand, the atmosphere is a gas — or rather a mixture of g^ases — and is, therefore, endowed with properties quite different, and in some ways opposed, to the properties of a liquid such as the water of the ocean. One such property is the remark- able elasticity or compressibility of air under pressure. Now if we ' realise that upon the surface of the earth we are living, at the ' bottom of an ocean of air, extending for many miles overhead, and ' that what we call the pressure of the atmosphere is merely? the ' weight of the superincumbent air, we shall realise that the air near the ground must be in a permanent state of compression, and its particles pressed very much closer together than the air, say, at ' thie top of a mountain. This compression is so great that, • although the atmosphere extends to over forty-fiye miles high, ^it l6 THE earth's atmosphere. can be proved that one half the entire ocean of air is contained within three miles of the earth's surface. We are also led to enquire — what is the actual state of com- pression or density of the air at the sea level? Well, a cubic foot of air v^eighs i '2 ounces (527 grains) when subject to the usual pressure of i4'7 lbs. per square inch ; and it is a simple application of Boyle's law to state, that if we ascend a mountain until the atmospheric pressure is reduced to one half, the same quantity of air, viz., i'2 ounces, will now occupy two cubic feet, and the density will be just one half that at the sea level, viz., '6 ounce per cubic foot. Thus the higher we ascend in the atmosphere the more the air is expanded, and it is difKcult to say why there should be any limit to the height at which some very rarefied air may exist. The total quantity of atmospheric air surrounding the globe is better realised if we make a simple calculation of how high it would extend supposing its density remained the same as at the sea level. If we use the weight of the atmosphere upon a square foot of the earth's surface, which is i^"j x 144 = 2ii61bs., the calculation is easy, for every cubic foot weighs V2 ounces, and i"2 ounces is contained in 21 16 lbs. about 28,000 times, which is the height in feet of the homogeneous atmosphere, as it is called. All the earth's atmosphere, therefore, would fall within five and a half miles of the surface, if everywhere of the surface density. The actual composition of air is 20 "8 volumes of oxygen, and 79"2 volumes of nitrogen, in 100 volumes of air ; or expressed l^y weight, the proportions are 23 of oxygen and 77 of nitrogen. There is also present in the atmosphere a varying amount of water vapour (HjO), quite invisible, some carbonic dioxide (CO^), and a trace of ammonia (NH3). In the air of towns sulphurous acid is found (SO2), and also sulphuretted hydrogen (SHj). These various gases are not chemically combined (as in the case of oxygen and hydrogen forming water by chemical combination), they are merely mixed mechanically, and each gas retains the same properties that it has in a pure state. The oxygen, for instance, does not lose its property of supporting combustion by being diluted with four times THE earth's atmosphere. If its volume of nitrog-en, but it acts with less energy; and the nitro- gen does not gain any new property by being mixed with oxygen. Each retains its own nature and gives nothing to the others. All animal and vegetable life is dependent upon the atmos- phere — even the fishes breathe the air which is dissolved in the water of rivers, lakes, and seas — and while animals absorb oxygen and exhale carbonic acid; plants, under the influence of sunshine, absorb carbonic acid and exhale oxygen gas. In this way, it is said, the composition of the atmosphere is maintained constant. To say that air is dissolved in the water of rivers, lakes, and seas, is merely stating a chemical fact which is difficult of explana- tion. That the air is not contained in the water in the form of bubbles of gas, is evident from observation ; but until Professor Dewar liquefied air it was only known in the gaseous form. It seems also unlikely that air exists as a liquid in water at ordinary temperatures. However, the fact remains, that water at 32°' absorbs air to the extent of about five per cent, of its volume, and less at higher temperatures. Moreover it is found that oxygen is^ absorbed more readily than nitrogen, with the result that the air dissolved in water has not by any means the composition of atmospheric air, its composition being about two volumes of" nitrogen to one of oxygen. Now whatever gases are absorbed by water when at a low temperature, are partly given up when the water is heated, and completely thrown out upon boiling. Hence the large proportion of oxygen in the air absorbed by water has an important bearing upon the internal corrosion of boilers, as well as propeller blades and ships' plates. Oxygen being the essential agent in all forms of oxidation, the fact that the air liberated from feed water contains twice the normal quantity of oxygen, must render this action far more energetic than under ordinary circumstances. Those who- hold that air is the principal cause of internal corrosion in boilers, obtain considerable support from the above facts ; and there is more reason to expect benefit from depriving the feed water of air before it enters the boiler. 2. — The Mercurial Barometer. fn\ yacuLim spite c The simplest barometer may be constructed by taking a gflass tube about 33 inches long, closed at one end and open at the other. After pouring in enough mercury to fill the tube, the finger is placed over the open end and the tube is inverted, the lower end with the finger is then lowered carefully into a small cup of mer- cury, and when well below the surface the finger is removed. The tube being upright some mercury will be found to run out, but only until the level in the tube has fallen to about 30 inches above the level of the mercury in the cup (as in fig. i), the space left being a vacuum. A balance is then established, and by inclining the tube a little the liquid can be made to oscillate up and down quite freely, as there is perfect continuity between the mercury in the tube and that in the cup (shown also in sketch). Such a barometer will measure, in a rough way, the atmospheric pressure, for it is the pressure of the atmosphere upon the mercury in the cup which holds up the mercury in the tube ; and if we sup- pose the tube to have a section of one square inch, «very two inches in length of the tube will contain two cubic inches, or very nearly a pound of mer- cury. So 30 inches will indicate that the pressure of the atmosphere is 15 lbs. per square inch. A tube of smaller bore would, of course, contain a less weight of mercury, but there would also be a proportionately smaller area at the orifice for the atmospheric pressure to act upon, and hence the size of the tube has no influence upon the reading of such a barometer. The defects of this simple barometer are : — (first) the vacuum is far from perfect owing to air bubbles, which cling to the sides THE MERCURIAL BAROMETER. 19 of the tube and ultimately rise to the top vacuum space. To cute this defect barometers are carefully heated, after filling, to the boil- ing point of mercury, which expels all air bubbles, so that when inverted a perfect Torricellian vacuum is obtained (called after the first maker of a barometer); and (second) the barometer jiist described could not easily be moved about without spilling the mercury in the cup. In the marine type of barometer the cup is closed in (as shown in fig. 2). ^ is a wooden bottle into the neck of which the barometer tube is cemented. The bottle is screwed into the wooden case of the instrument at B, and it is furnished with a screwed bottom C, having a hole covered with a flexible piece of leather. A brass cap, Z), screws over alh The tube is filled with mercury when the barometer is upside down, and the bottle is also filled about two-thirds full. The cap being screwed on, the barometer can be righted, when the flexible leathei* will allow the atmospheric pressure to be communicated to the air inside, and so act upon the mercury in the bottle as if it were open to the atmosphere. Another feature of the marine baro- meter is a contracted place in the tube, shown at E. The object of this is to ■check any tendency of the mercury to oscillate in the tube (called pumping) when the ship is pitching or rolling. Fig. 2. Sometimes the tube of a barometer is bent round at the lower end and formed into a glass cup (as in fig. 3), and in another kind there is-no cup but a simple U-tube (fig. 4). In this kind there is a float placed in the open tube, which actuates a small pulley, carrying an index like a pressure gauge, by means THE MERCURIAL BAROMETER. of a silk cord. The barometer may be said to be read off from the lower end when this arrangement is adopted. The scale of a barometer is merely a scale of inches divided very accurately into loths or 2oths and furnished with a vernier. The inches must be measured from the level of the mercury in the cup at the bottom, and inasmuch as this level rises and falls with every change in the height of the column, the scale must either be moveable or a correction must be made, in order to get the true height. In one form of standard barometer the cup at the bottom (fig. 2) is made of glass, with a scratched line at the level from which the inches on the top scale are measured. Before reading off the height of the mercury, the screw S is turned until the mercury in the cup is exactly up to the scratched line or zero mark 00, There is also a correction necessary for capillarity. The fact is, when any liquid is contained in a tube, a certain attraction or repulsion is exerted (called capillary attraction) between the tube and liquid, which either lifts it or depresses it from its natural hydrostatic level. In the case of mercury im a glass tube, the liquid is depressed — ^thus giving a lower reading. The error increases with the smallness of the bore at the surface, level, so that it is usual to swell the tube out at the upper part where the readings are taken. In order to compare barometric readings at different places and different times, it is necessary to bring them all to some standard temperature (usually 60° Fahrenheit). This introduces a correction for temperature also. 3« — The Aneroid Barometer. One form of aneroid (no liquid) barometer is constructed with a Bourdon tube, almost exactly like the vacuum gauge of a steam engine. It is indeed obvious that if the Bourdon tube of a vacuum gauge were completely exhausted of air and then sealed up, the gauge would always indicate a "perfect vacuum." Any variation in its reading would be due, not to the "vacuum," but to the variations in the pressure of the atmosphere acting upon the out- side of the tube. fe^/->///y/y//////////////^!^/^^ ^ t23-^rZ> Fig. 5- The usual construction, however, of the aneroid is upon a different mechanical principle. A vacuum box is made with corrugated sides — a round flat box about 3 in. or 4 in. diameter — (A) in Fig. 5. The under side of the box at its centre is securely bolted down to a stout iron plate forming the back of the barometer case. The ■upper side is similarly bolted to the end of a strong spring of flat steel (S) bent over from a frame F. Now there being a vacuum in 32 THE ANEROID BAROMETER. the box the atmospheric pressure tends to force the two corrugated sides together, but this tendency is resisted by the steel spring. One force is acting against the other. As the spring remains practically unaffected by the weather, any movement of the centre pin up or down must be due to variations in the atmospheric pressure; The movement is very minute at any time, and quite imperceptible to the eye. But the rest of the mechanism of the instrument (and the most delicate part) is for the express purpose of magnifying the movement of the top of the box, so as to make it manifest. A stiffish arm (R) is attached to the end of the spring, and its extremity is connected by a link to a bell-crank lever L, whose long arm reaches nearly to the glass cover. A piece of watch chain now completes the connection with the hand or index of the dial. The chain passes round a small pulley on the spindle of the index, which, of course, works in a light frame. A hair spring, also upon the spindle, takes up any back lash and keeps the chain taut. The Aneroid barometer is of no use as a standard barometer; it must be adjusted from time to time by comparison with a mercurial standard ; but for recording changes of atmospheric pressure it is very convenient and remarkably sensitive. The adjustment is made by the small screw D, which can be reached from the back of the case. 4- — ^^Sea Water. Sea water differs from the water of springs, rivers, and most lakes in the large amount of common salt which it contains. Spring water often contains considerable quantities of magnesia and lime, although not to such an extent as sea water (if we except the carbonate of lime or chalk of which sea water contains remark- ably little). But the amount of salt in sea water is so large and so far above the salt in any known spring or river water, that the term " salt water" is often used as equivalent to "sea water." The actual amount of solid matter dissolved in the water ot the ocean varies in different parts of the world. During the voyage of the Challenger, in 1873 and following years, it was found that the densest water occurred in the centre of the N. Atlantic Ocean, and also opposite the coast of Brazil, the amount of solid matter in these two localities being 3 "73 per cent. The next densest was found in the centre portions of the South Pacific and Indian Oceans (3 '6 per cent.), which is also the density of the N. Atlantic, west of the British Isles. The Arctic and Antarctic Oceans have only 3^4 per cent, of solid matter. The Challenger did not penetrate into the Mediterranean Sea, but it is known that the Mediterranean is Salter than the Atlantic Ocean, and the Red Sea Salter than either. The effect of the large amount of salt in the sea is to increase its specific gravity above that of pure water. Thus an amount of 3^ per cent, gives a specific gravity of 1027.* The composition of 'sea salt' or in other words the exact nature and amount of the various ingredients in sea water, has been often ascertained. In the Challenger Reports the analysis is given thus : — * If the excess of the specific gravity of salt water over IQOO is multiplied by "13 the result is the percentage of salt. 24 SEA WATER. Soda 41*43 Magnesia 6-21 Potash - - - I '33 Lime - - . 1-69 Muriatic acid 42-92 Sulphuric acid 6-42 1 00 00 But a more useful statement for the marine engineer is the following : — * In a gallon of sea water there is: — 1 87 1 grains of Chloride of Sodium (common salt) 262 grains of Chloride of Magnesium 160 grains of Sulphate of Magnesium 113 grains of Sulphate of Calcium or lime (plaster of Paris) 9 grains of Carbonate of Calcium (chalk) 2415 Total solid matter In round numbers it is easy to remember that each gallon of sea water contains : — 4 to 5 ounces of common salt ; I ounce of salts of magnesia ; and ^ ounce of sulphate of lime. When sea water is heated in the open air it boils at 2i3'2 Fah. {Bar. at 30"), and if allowed to become more dense through evapora- tion its boiling point increases by i "2 degrees for every addition of ^ by the salinometer. When the density has increased to || the temperature of boiling is 2i2+(i2 x i*2) = 226 degs. Fah. After this no increase of density occurs, because salt is deposited from the liquid and the water is now saturated. But long before the water is saturated with the common salt "(at about -^^) a separation of lime occurs, which is shown by the water suddenly becoming milky, * From Babcock & Wilcox's Engineer's Reference Book. SEA WATER. 25 If common salt is deposited very slowly from a saturated •solution of sea water, as might occur if hot water were allowed to •cool down, the crystals formed are cubical and transparent, large or small according to the time occupied and perhaps to other ■conditions. The effect of sea water upon the plates and tubes of a boiler will be considered in the essay on Corrosion (No. 29). -The Eledtric Current. The term "galvanic action" is an old name for the effect which occurs when two different metals are in contact and exposed to some corroding influence such as acid or saline water. But there are in reality itvo results immediately observable rather than one, whenever such a combination exists. There is a current of electricity through the two metals and the exciting fluid, as well as the more obvious corrosion of the positive metal going on at the same time. The term "galvanic action," as generally used, relates more to the corrosion that occurs than to the existence of the current which is produced; for Galvani, an Italian physician, was the first who observed the effects of an electric current upon animal tissue. The current is generally called a "voltaic current," after Volta, another Italian, who was the first to arrange a battery of metal plates to produce it. To detect the existence of a current, it is convenient to connect the plates solely by a long copper wire, so as to compel the current to pass through the wire. If the wire is then laid parallel to a compass needle, the latter is deflected from its position by the current. Upon this principle, galvanometers and voltmeters are often constructed. If the common metals including carbon are arranged in the- following order, viz. : — carbon, copper, iron, lead, tin, zinc, we have a rough guide as to the effect when any two are connected in a. voltaic circuit. The rule is: — The metal v]\\\Qh follows the other in the list is the one which suffers corrosion, and is said to be positive- with respect to the preceding one. Hence it is that copper becomes- dangerous near a ship's skin if exposed to bilge water. -THE ELECTRIC CURRENT. 27 Nor is it absolutely necessary for the generation of a voltaic current that two different metals should be used. Two pieces of the same metal providing they are at different temperatures, or their surfaces differently prepared, will generate a current, although perhaps a weak one. To such weak currents has been attributed a considerable amount of the corrosion in certain parts of marine boilers. A very fair voltaic current can be got by combining a plate of steel and a plate of iron ; or a piece of cast iron and a piece of wrought iron. The rule for the direction of the cur- rent is also simple. The current flows in the direction in which the metals appear in the list, that is from the copper to the iron at their junction, or from iron to zinc, if these metals are in contact. Moreover, by reversing the current, or rather by starting a more powerful current in the opposite direction (by some other agency) and making this: current pass, say, from iron to copper, a protection against corrosion is obtained Fig. 6. A combination of metal plates or materials for the purpose of producing electric currents is called a "Primary Battery" (or if .28 THE ELECTRIC CURRENT, only one pair of elements is used it is called a "Voltaic Cell." A very common primary battery is that made up of Leclanch^ cells. In figure 6, C is a plate or block of carbon (gas carbon from old gas retorts of the gas works) which is placed in a cylindrical pot P, of -porous earthenware, and packed tightly round with crushed carbon peroxide of manganese. The porous pot stands inside a glass or earthenware jar G, containing a solution of sal ammoniac (ammonium chloride NH^ CI). A rod of zinc is placed in the jar also, and forms the positive element, whilst the carbon is the negative element. When the terminal screws on the carbon and zinc are joined by a wire, the current flows from the carbon to the zinc through the wire, and from the zinc to the carbon through the liquid ■and porous pot, as shown by the arrows in the sketch. The carbon terminal is therefore called the positive pole. The electro motive force or pressure of a Leclanchd cell is about i\ volts. To test the direction of a current is a simple matter. Lay the wire carrying the current over a pocket compass so that the wire is parallel to the compass needle ; then if the N end of the needle deflects to the East the current is running from North to South, and if it deflects to the West the current is running from South to North. As an aid to memory it has been suggested to use the word SNOW, in which the letter S stands for 'South,' the letter N for 'North,' the letter O for 'Over,' and the letter W for 'West.' The word Snow then stands for the sentence : — A current running from South to North, Over a compass, deflects the needle to the West. The chemical change that occurs in a Leclanch^ cell, con- sists in zinc decomposing the ammonium chloride and becoming •chloride of zinc which forms a white incrustation. Hydrogen and ammonia gases are set free, but the hydrogen combines with some •of the oxygen of the peroxide of manganese, and reduces it to sexquioxide, water being formed at the same time. Thus begin- ning with Zn, aCNH^Cl) and aCMnOj), the result is Zn C\, aCNHg), MnsOg and HjO. 6. — A Candle Flame and Bunsen Flame. Flame is nothing more than a portion of some gas (or mixture of gases) at a temperature sufficiently high to make the gas- luminous; and in-as-much as our experience of gases at a high temperature is usually confined to instances of combustion (as in the case of a coal-fire or lamp flame) we are in the habit of thinking- of flame as always the result of something burning, which is wrong. For example, the flame produced when a high tension electric current is switched off' in an electric-light station, is merely a portion, of atmospheric air highly heated by a discharge of electricity, and. may not be accompanied by the burning of anything. As this hot and slightly luminous air — a blue flame — rises to the ceiling it gradually cools and ceases to be luminous. The same air may circulate round again and produce a flame any number of times, without change. A similar confusion between the nature of incandescence and of combustion is often made with respect tO' solid bodies as well as gases. For instance, when electric glow lamps were first introduced a misapprehension arose in some minds as to how the filament could burn so brightly without burning away ! We have now become accustomed to the fact that light is not always a result of combustion or burning, but only an accom- paniment (and not always that) of intense Iieat. Hence, if the substance of the filament can be made hot enough without burning, we get light without combustion. The foregoing remarks are rather off the line as regards the: subject of this essay, because in a candle flame we have an instance of combustion as well as heat and light, but they may lead us to recognise more clearly what is the function performed by different parts of the flame. .30 A CANDLE FLAME AND BUNSEN FLAME. A candle flame has exactly the same structure as a lamp flame 'Or a gas light. But a candle when burning is a gas manufactory as well as a gas burner, for the heat of the lighted wick first melts the tallow or wax of the candle, then draws it up the wick by ■capillary attraction, and then decomposes it into gases. It is these -hydro-carbon gases which burn in the flame. Referring to figure 7, the transparent and dark-looking central ■part of the flame (a) consists of unburnt and comparatively cool gas, which is evolved from the tallow or wax, and has not yet got air enough to burn it. Next to this portion comes the luminous part of the flame (b) which consists of gas and air more or less chemically combined. Combustion is here .going on, and one result of the combustion is the throwing out from the hydro-carbon gas of minute iparticles of solid carbon, which raised to a white heat are the cause of the great and useful luminosity of this part of the flame. (Some chemists explain the iluminosity differently, they say it is due simply to certain gases in a state of incandescence). It will be noticed that the air needed for combustion can only reach this part of the flame after passing through the • outer zone or mantle (c). The combustion, therefore, ■is not completed in the zone (b) but in (c) where the surrounding air is plentiful. Even the solid particles of carbon are consumed in the mantle (c) with the •result that this portion is very slightly luminous, and can only be ■seen well by cutting out a piece of cardboard to the shape and size •of the luminous flame, and holding it up so as to screen the eye from its glare. Fig- 7- Since the gas which burns in a candle flame contains the two ■elements, carbon and hydrogen, the combustion of the gas gives rise to two diff"erent products, viz. : — Carbonic-acid gas (COj) and .«team (HjO). It is the greater affinity of oxygen for hydrogen over Us affinity for carbon which causes the carbon to be thrown out in the first stage of the combustion. But the final product of the flame, and what forms the mantle is a mixture of COa'and HjO with a.. A CANDLE FLAME AND BUNSEN FLAME. 31 large excess of nitrogen and some oxygen. Indeed very much the same mixture as is exhausted after each stroke of a gas engine. A Bunsen Flame or an atmosptieric burner has a simpler structure than the candle flame, see figure 8. As the gas issues from the top of the tube it is already mixed with about 2^ times its volume of air which has been drawn in through apertures at the base of the tube. The space (a) in the figure is the unburnt mixture of gas and air. Now for coal gas to burn completely it must have about ten times its volume of air, but by mixing the gas with a certain quantity of air at the bottom of the tube a less amount of additional air is required at the top ; the result is that the combustion is completed much quicker and more perfectly. There is no throwing out of solid particles of carbon (or formation of denser hydro-carbon, according to some investigators) nor is there any smoke. But there is greater heat. The Bunsen flame consists of two portions only, viz. : — The dark central space of cool gas and air (a), and the envelope (c), of gas and air in the process of burning. But by shutting off the supply of air a little white luminous tip is produced at (b), and by shutting off" the air entirely this tip of white develops into a complete luminous zone as in the ordinary gas jet. Again, by admitting too much air to mix with the gas, the phenomenon of "firing back" is produced, because with so much air the gas is able to burn without the extra supply at the top of the tube, and, therefore, it burns in the tube instead at the top. Fig. 8. Considering the zone of combustion of any flame whether from a candle or a gas jet or a Bunsen burner, the chemists distinguish the inner boundary from the outer boundary by their special reaction. The outer boundary (y) is called the oxidising flame, because it is here that the oxygen is abundant, and if a small portion of copper {for example) is held in this part, it will be rapidly oxidised, that is made into black scale. The inner boundary (x), on the other hand, is known as the reducing flame because there is here a deficiency of 32 A CANDLE FLAME AND BUNSEN FLAME. oxygen, and the heated carbon or hydrogen will take oxygen from oxides of metals when inserted in this part of the flame and sO' 'reduce' them to the metallic state. T. — Life Buoys with Inextinguishable I^ights. The common life-buoy is a ring of cork, built up in sections, and covered with canvas well painted. The diameter is about 32 inches, with a width of 6 inches and a thickness of 4 inches. There are lines fixed round the outer edge for more readily seizing hold of in the water, About 12 lbs. of cork is used in the construc- tion of a life-buoy, which enables it to support three or four persons in the water at one time. The Board of Trade require thai the Life Buoys on board ship shall be capable of supporting a weight of 32 pounds of iron for a period of 24 hours without sinking. A life-buoy naturally floats very flat upon the surface of the water and unless in a smooth sea, is not easily seen, even in the day time, by the person for whose rescue it may have been thrown overboard. And especially at night, in the darkness, a person may be drowning within a few yards of a life-buoy without knowing of its existence, and, therefore, quite unable to save himself by means of it, unless the buoy can indicate its position to the drowning person, by a light or other signal. The self-lighting life-buoy accomplishes this object in a satisfactory manner as follows : — Attached to the buoy by a short line is a tin canister, the upper half of which furnishes the necessary buoyancy for supporting the weight of 16 ounces of phosphide of calcium which fills the lower half of the canister. Before throwing the life-buoy overboard it is necessary to pierce the top and bottom of the canister with a sharp instrument provided for the purpose and attached to the apparatus. When, upon throwing the buoy overboard, the canister reaches the water, it floats with the bottom hole open to the sea, which immediately commences to leak in c 34 LIFE BUOYS WITH INEXTINGUISHABLE LIGHTS. amongst the phosphide of calcium. The contact of water then evolves a gas called phosphuretted hydrogen, which has the peculiar property of spontaneously taking fire upon contact with the atmos- phere. The gas escapes by the hole in the top and rushes out as a flame, a foot or more in height, thus indicating the position of the life-buoy to the person in the water. Phospide of Calcium (P Ca) is a brownish stone-like mass made by exposing quick lime (Ca O) at a red heat to the vapour of phos- phorus, for some hours. It is broken up into small fragments and sealed up in the tins as soon as made. The action of water upon the compound consists in it giving up its hydrogen to the phos- phorus forming phosphuretted hydrogen (PHj), and its oxygen to the calcium forming lime again (Ca O). But it appears that another gaseous compound of phosphorus and hydrogen (Pj HJ is also evolved at the same time to which the property of spontaneous combustion is due. Not only has the gas evolved the property of spontaneously taking fire, but the flame is also practically inextinguishable ; for even if the whole apparatus should happen to be submerged for a time, the gas would still find its way through the water to the air and continue to burn ; and if smothered for a moment it would re-light itself immediately. Spontaneous combustion always implies that the chemical .affinity of two bodies even when at ordinary temperatures is so strong and develops heat so quickly, that there is not time for the heat to be dissipated as fast as it is formed: with the result, that the temperature quickly rises to the ignition point, when the affin- ities are so enormously increased that flame or incandescence is the result, according to whether the product of the chemical union is a gas or a solid. The part played by the gas Pj H^ is upon this view, the part of a match which lights the P H3. The former only having the necessary chemical affinity for atmospheric air to give .rise to spontaneous combustion. 8, — The Absolute Zero of Temperature. On the Centigrade scale the freezing temperature of water is marked o or zero. On the Fahrenheit scale the zero is 32° below freezing point. Neither of these arbitrary points is a real zero, for temperatures occur even in the atmosphere much below both these zeros; and temperatures are obtained every day in refrigerating machines 50 and 60 degrees below the zero of Fahrenheit. Scientific men have thus been led to speculate upon what is the real zero of temperature on Centigrade or Fahrenheit scale, in other words, to find how many degrees a body would have to be cooled down to extract all heat from it, and leave it in a condition perfectly devoid of that particular form of energy called Heat. One such enquiry is based upon the property of air and other g-ases. When a so-called permanent gas is cooled lower and lower in temperature, it is found to contract in volume at a steady and definite rate, and to remain in the gaseous state — if not at the lowest temperature obtainable — still at extremely low temperatures. The rate at which gases contract by cold and expand by heat is known with great accuracy. The law was discovered by Charles and Guy Lussac, and may be stated as follows : — Suppose we have just 493 cubic inches of air (or other permanent gas) at 32 degrees Fah., and under the atmospheric pressure (the tube of an air thermometer might be supposed to hold just 493 cubic inches). Then for every fall of one degree in temperature, the volume of air is reduced by one cubic inch ; and for every rise of one degree the volume is increased one cubic inch, the pressure being kept constant in both cases. More generally, we may state the law thus: — "All gases expand ^^ of their 36 THE ABSOLUTE ZERO OF TEMPERATURE. volume at 32 degs. Fah. for every rise of one degree in tempera- ture." It must be noted that although the amount of expansion is the same for every degree, the fraction ^^ only applies to the volume that the gas possesses at freezing point. This is curious, but it is the law. Reckoning in this manner it follows that at a temperature of 493 degs. above freezing point the volume of our original 493 cubic inches of air will be exactly doubled, for the expansion' will have amounted to 493 cubic inches. But what is to be said about a/a// of temperature of 493 degs. below freezing point? D ^^ ..... ^_,^ •n 1 ^.^"'^'^ •« 1 B^^ •i -& u. - ' lFo~ •4 ^^""^"^"^ ") ^^"''^ * ^ ^^0^""^ "^ ir ^ — ^ «NJ ^^^^"""^ "N* C 1 .^-^^""''^ ^ I c t--. --493 32 ISO 210. — > Fig. 9. If the rate of contraction really holds good it follows that there will be nothing left of the air whatever, for we cannot take away 493 cubic inches from 493 cubic inches and leave anything over. The law probably fails at very low temperatures. However, the point is that Charles' law, applied to the air thermometer, leads to an absolute zero at 493 degs. below freezing point or 461 degs. below zero on Fahrenheit's scale, and this is equivalent to 273 degs. below zero on Centigrade. Another and graphic representation is the following (Figure 9). — Starting with the experimental fact that a cubic foot of air at 32 degs. becomes i"365 cu. ft. at 212 degs., we set off to scale a line AB = i and a line CD = 1-365. Produce the line CA until it meets BD produced in the point X. THE ABSOLUTE ZERO OF TEMPERATURE. 37 Now XC is a scale of temperature, and the point A is freezing point, and the point C is boiling point. AB and CD represent the respective volumes of the air, and it is clear that if the rate of contraction between 212 and 32 follows a "linear law" it can be represented by a straight line, and the same continues to hold good, the point X is the limit of contraction possible, and is called the absolute zero. To find AX, first draw BN parallel to AC, so that ND = -365, then we have by similar triangles NDB and ABX — ND : AB = BN : AX •365 : I = 180° : AX which works out to A X = 493 degs. Or if AC = 100 as on Centigrade scale •365 : I = 100 : AX AX = 273 degs. g. — Pyrometers. When temperatures have to be measured higher than the limit of a mercurial thermometer (which is fixed by the fact that mercury- boils at 660 degs. Fah. ) the instrument used is called a ' ' Pyrometer," no matter what its construction may be. Many kinds of pyrometers have been invented, depending upon very different principles. a.tCfO.T^eJf^GCJ'^ ^W///A\ p. Y / / / / /// i Fig. 10. Wedgevvood used an instrument depending upon the contrac- tion ot a baked clay plug when heated. Others have depended upon the expansion of a metal rod such as platinum. The pressure of the steam formed in a miniature boiler of sufficient strength, when placed in the furnace whose temperature is to be measured, has been successfully used. Also an air thermometer with a hard porcelain bulb. Siemen's electrical pyrometer depends upon increased electrical resistance of a platinum wire as its temperature rises, and, lastly, there is the Le Chatelier pyrometer, which consists of a thermo- electrical couple, and is essentially a thermo-pile with only one pair of elements. The two metals made use of are platinum and an alloy of platinum and 10 per cent, of rhodium. PYROMETERS. 39 In figure lo, we have two wires, P and R, of platinum and the alloy respectively. These are twisted together (and sometimes fused as well at H) to form the hot junction which is exposed to the heat of the furnace whose temperature is to be measured. The other ends of the wires are kept apart at Tj^ and T^ in a wooden case. But copper wire conductors are taken from Tj and Tg to a sensitive galvanometer, so that there is a complete circuit through the two wires across the hot junction and through the coil in the galvanometer. In using the instrument it is essential that the two points in- the circuit, T^ and T^, should be kept at an equal temperature, which must be known. The instrument is graduated by exposing- the pyrometer to the heat of fluids whose temperatures are known such as boiling water, melting tin, etc. The reading of the galvanameter being noted at the same time, a scale of correspond- ing temperatures and galvanometer readings can be made. lo. — Calorimeters. A calorimeter is an instrument, or rather an apparatus, for measuring the quantity of heat lost or gained by a body. The object of such measurement is generally to ascertain, once for all, and with as much accuracy as possible, the specific heat of a substance, which is the quantity of heat required to raise the temperature of one pound of the substance exactly one degree in temperature. The principle of the apparatus used by Regnault is that a known weight of the substance under investigation is heated in a hot chamber to a certain temperature and then quickly plunged into a known weight of water, at a temperature a little below 60 degrees. The water is thereby raised to a temperature a little above 60 degs., and the exact rise of temperature is accurately measured by a delicate thermometer. The number of pounds of water multiplied by its observed rise of temperature, gives the quantity of heat g^atned by the water, and, therefore, also the quantity of heat /ost by the substance in cooling from the hot chamber temperature to that of the water at the finish. In figure 11, H is the hot chamber, which has a steam jacket round it, supplied from a small boiler, C is the fine wire cage in which the substance is placed, and W is the vessel of water placed immediately below a hole in the bottom of the hot chamber, the hole being opened and closed by a slide S. The calorimeter vessel W is made of thin polished brass so as to radiate as little heat as possible to surrounding bodies, and is also protected from the radia-' tion of the hot chamber and of the boiler by screens containing cold water. Loss or gain of heat by conduction is prevented by making the vessel rest upon three pointed feet. CALORIMETERS. 41 Let us suppose that the specific heat of iron 'is under investiga- tion. A definite quantity (say just a pound) of iron in small frag- ments would be placed in the wire cage C, and lowered into the Fig. II. chamber H through the hole in the top, and allowed to remain until it had acquired the same temperature as the chamber (say 212 degs.), as tested by a thermometer in the same wire cage. The 42 CALORIMETERS. temperature of the water in W would now be read off and the slide S opened, so as to enable the cage to be quickly lowered into the water. After stirring- well to bring all the water to an equal temperature, the thermometer is again read off. We will suppose the water to have risen from 58 degs. to 62 degs., the difference being 4 degs., and that the weight of water in W is 4 lbs. It is evident that the water in this case would have gained exactly 4x4 = 16 units of heat, which is also the amount of heat lost by the pound of iron in cooling from 212 degs. to 62 degs. or through 150 degs. If X is the specific heat of iron ixXxi5o=i6 .'. X=-j/t= 'lo?- Practically it is necessary to take into consideration the cap- acity for heat of the wire cage and of the metal of the vessel W, besides applying corrections for the unavoidable loss and gain of heat during the stirring of the water, etc. But supposing all these corrections made, we should have found in the supposed case just taken, the average specific heat of iron between 62 degs. and 212 degs. In the case of liquids Regnault enclosed them in thin glass tubes, hermetically sealed, while they were under investigation. In the case of gases a different kind of apparatus is used altogether. 11. — The First and Second La-ws of Thermo-Dynamics. The first law states that "Heat and mechanical work are^ mutually interchangeable, one unit of heat (British Thermal Unit) being equivalent to 774 units of work or foot pounds." The value of this constant was first obtained satisfactorily by Dr Joule, of Manchester. Joule expended mechanical work so as- to produce heat. His apparatus consisted of a kind of churn driven by falling weights. The paddles of the churn agitated the water, or other liquid, and thereby raised its temperature. By carefully measuring the rise of temperature after a certain interval of time, and also recording the amount of mechanical work expended during the same time, and then making numerous and elaborate correc- tions. Joule divided the number of foot pounds of work by the number of units of heat produced, and obtained the number 772. Other experimenters, notably the late Professor Rowland, of America, have, since Joule's time, verified and slightly altered the value to 774. The reverse experiment, viz., converting heat into mechanical work, is what is done every day by a steam engine, but the difficulties in getting an accurate value for the mechanical equivalent of heat by experimenting with steam engines is greater than in the other case. On the Continent, Hirn attacked the problem and was more or less completely successful ; but the accuracy of the number 774 rests upon the first kind of investigation. The second law is expressed in different words by different men. It attempts to formulate the conditions under which heat can be converted into work, and is, therefore, the fundamental principle of all steam, gas, and oil engines. 44 THE FIRST AND SECOND LAWS OF THERMO-DYNAMICS. The most important corollary or consequence of the second law is the well-known formula for the efficiency of a perfect heat engine, T - 1 viz., efficiency = where T and t are the absolute temperatures at which the heat is received and rejected respectively, in other words, the temperature of the boiler and the temperature of the condenser. The pioof of the formula will be given in the article on Entropy. la. — The Steam Hygrometer. The terms "wet" and "dry" applied to steam have a different meaning than when applied to the air. "Wet" steam is steam which contains minute drops of water held mechanically in suspen- sion, due to their small size, but still being water in the liquid state. "Dry" steam contains no water in the liquid state whatever. A steam hygrometer is an instrument for measuring the amount of water in the liquid state, which may exist in any sample of steam. It is usual to express the wetness or dryness of steam by means of a fraction, which denotes how much of the weight of the "stuff" is really in the gaseous state The fraction is then called the "Dryness Fraction" of the steam. For example, if a pound of water passes out of a boiler containing one tenth of a pound of water in the liquid slate, and only nine-tenths of a pound in the gaseous state, the mixture of steam and water would be called steam whose dryness fraction is "g. Peabody's Steam Hygrometer consists of a small vessel K (figure 12), well lagged with hair felt and wood. There is a con- nection to it from the main steam pipe controlled by a valve E, and an outlet into the atmosphere controlled by a valve N. Also pressure gauges, fixed as shown, and a thermometer to show the temperature in K. To use the apparatus, valves E and N are opened; steam, therefore, flows from the steam pipe into the vessel K, and out through N. The orifice or nozzle through which the steam enters being only small, compared with the outlet, the steam expands -46 THE STEAM HYGROMETER. considerably in the vessel K, and having an excess of heat for its reduced pressure, not only evaporates its watery particles, but produces some few degrees of superheat in the low pressure steam in K. If the two pressure gauges and the thermometer are read off after a few minutes' action, the readings afford sufficient data for calculating the wetness of the steam taken from the steam pipe. Thus: — Let the pressure in the steam pipe (gauge C) be P, and that in the vessel K (gauge B) be p, and let the temperature in K be t. By calculation, or from tables, the latent heat of the steam at pressure P is known = L, and also its tem- perature =T. Similarly, the latent heat of common steam at pressure p is known = L, and its temperature Now, put X for the dryness fraction of the steam in the steam pipe, so that in each pound of stuff which comes through E, there is X pounds of steam and i - X pounds of water. Then units of heat per lb. of stufT= X x L+ 1 x T, but as no work has been taken out of the steam in entering K, the steam in K must have the same total heat as in the pipe. But the steam in K is superheated (t - TJ degrees. .-. units of heat per lb. = (Lj + Tj) + -48 (t - Tj). Equating these two values for the units of heat, we have — (X X L) + T = (Li + Tj) + -48 (t-TO V - (Li +T,)+ •48(t-T,)-T L THE STEAM HYGROMETER. 47 Perhaps the result is better expressed in words as follows : — To the units of superheat in K add the total heat of the low pressure steam, and subtract the temperature of the high-pressure steam. Then divide by the latent heat of the high-pressure steam. 13. — What is meant by "Entropy?" Some practical working- notion of the meaning of "Entropy" may be acquired by making a few simple calculations, thus : — 1. Suppose a mass of water equal to 1500 lbs. is heated from 60 degs. to 62 degs. in temperature, the amount of heat put into, the water is 1500 x 2 = 3000 units. Now let us divide this number of units of heat by the absolute temperature of the water which is 460 + 60 at the commencement, and 460 + 62 at the finish, say> 460 + 61 = 521 average temp., the result is 3 = ^-y^g. 521 We express this result by saying that the entropy of the water has been increased by 5 "758. 2. Take another example. Suppose that i lb. of water at 212 degs. is made into steam at atmospheric pressure, what amount of entropy has the water gained ? Here, the temperature has remained the same all the time that heat has been put into the water, the absolute temperature being 460+212 = 672 F. Also the amount of heat added is just the latent heat of atmospheric steam, viz., 966-6 units. Hence the gain of entropy =?g— =1-438. 3. Again suppose that 10 lb. of water has been cooled from 212 degs, to 100 degs., the heat lost is 1120 units, and the average temperature is 460 + 156 = 616, we can find approximately the loss of entropy by dividing 1 120 by 616, 1^4?= i '82. This result is only 616 ■' WHAT IS MEANT BY ENTROPY? 49 approximate, because the temperature has changed too much to allow us to use the tiverag-e temperature without error. It would be more correct to calculate the entropy in several stages, say every lo degrees, but leaving this point for the present, the first idea to get hold of is that entropy is calculated by dividing the number of units of heat by the absolute teriiperatute 4. As another example, let a quantity of steam, whose weight is 5 lbs., be superheated from 250 degs. Fah. to 300 degs. Fah.,: what gain of entropy does this represent? The specific heat of steam being taken as -48, the heat added = 5 X -48 X 50= 120 units. The average temperature (absolute) = 460 + 275 = 735, hence the gain of entropy = ='i63. 735 This result, like No. 3, is only approximate, owing to the change of temperature being so great. The correct entropy can only be calculated in such cases by logarithms, the formula being — T Entropy = H x loge. — where H is heat added for every degree rise of temperature, and T and t, the two temperatures absolute. H = 5 X -48= 2-4 T = 760 logt. =6-6333 t = 710 loge. =6-5653 •0680 H = 2-4 2720 1360 Gain of Entropy = •16320 This shows that where the difference of temperature is not very great, the average temperature gives a very close approximation. D 50 WHAT IS MEANT BY ENTROPY? Before leaving these calculations of Entropy, it should be mentioned that, when not otherwise stated, all calculations of entropy are understood to be made with one pound weight of the substance, whatever it may be, whether water, steam, or air for instance. Thus the result found in example 2, viz., the entropy gained by i lb. of water when evaporated at atmospheric pressure, will be found quoted in any table of entropy; because in this example the standard quantity of the substance (i lb.) has been used. 14* — Entropy Diagrams. An "Entropy" diagram — called a "Th eta-phi" diagram, from the two Greek letters, 6 and i^, which are used to designate the temperature and entropy respectively — is set out after the manner of an ideal indicator diagram. But the vertical scale, instead of representing pressures as in the indicator diagram, now represents yoo J •=0 BOO. t % A- 0, « 5 0. I 0. ahsolzzTo. &12 ■3he area of I'h'is rectanale rehTasents the anrtaunt of heat = 672 x i: VO >« 2 0. rl NO ir II f^ «l J. 100. ci <3 <.— -/IS-— = +-•103-. ;_ .Of3-» a. b c d &ntroht4 Scale iH- Fig. 14. AD da, the total entropy is •115 + "103+ "093= •311, and the total area= 180 units of heat. It should be noticed that the point D could have been plotted at one step, if we had calculated by logarithms the gain of entropy due to the rise of temperature from 32 to 212. Or quicker still, if we had taken the value of the entropy from a published table. 54 ENTROPY DIAGRAMS. We will now represent on an entropy diagram, figure 15, the total amount of heat put into the feed-water of a boiler until it becomes superheated steam. The temperature of the feed may be taken as 160 degs. F., the temperature of the boiler gs 300 degs., and the steam may be superheated to 400 degs. The absolute temperatures are: — Feed i 3 e I 00- 000 a S60 yeo 630 io^a? enlro, 1/,^. I-^S. f ao Fig. 15. v. cL vs 620 degs., boiler 760 degs., superheater 860 degs. Approximately 140 heat units are put into the feed-water at an average temperature of '— =690. Entropy = ^=-203. Thus we get the point B, and the line AB which may be drawn slightly curved as shown. Again, the latent heat of steam at 300 degs. is 902 '8. .'. Entropy = " =i'i88. Setting the extra entropy off from B to C gives us the point C. ENTROPY DIAGRAMS. 55 To find the entropy required to superheat the steam from 760 to 860, we have heat added = loox '48 = 48, average temperature 760+860 Q T, . 48 ' =810. Entropy = -'^ = "059. This gives the point D, and the line CD, which should also b^ slightly curved. ,, The , total amount of heat given to each pound of water or steam is now the area of the irregular figure aABCDd, and it is made up of — Area aABb = 140 units. ,, b B C c = 902-8 ,, cCDd = 48 Total heat =1090 '8 15.— Use of Entropy — Efficiency of Engine. To understand the utility of Entropy or of the entropy diagram requires a knowledge of what is meant by a "cycle of operations." Def. : — Whenever a quantity of any substance such as water, steam, or air is made to change its condition as regards temper- ature, volume, or pressure (or all three) and ultimately to come back to the same condition as at first, it is said to undergo a "cycle of operations." In the working of a condensing steam engine the feed-water goes through just such a cycle of changes. It is heated to the boiler temperature, then evaporated and passed through the engine doing work, and after being condensed to water it is pumped into the hot well to become feed-water again, at the same temperature and in exactly the same condition as before. The value of considering the working fluid of a steam-engine as passing through a "cycle" consists in this: — If a substance returns to the same condition as regards temperature, volume, and pressure, any property of the substance which depends upon these conditions will not be altered. Now the entropy of a body is such a property. We cannot suppose a pound of hot-well water at say 120 degs. temp, to contain more heat at one time than at another, the temperature being the same, hence, whatever entropy it may have gained during one part of its cycle of changes, it must have lost during the remaining part. USE OF ENTROPY — EFFICIENCY OF ENGINE. 57 We will take the same data as used for figure (15) viz..: Feed temp. i6o degs. =620 degs. absolute; Boiler temperature 300 degs. =760 degs. absolute, and superheat 100 degs. =860 degs. absolute. For simplicity we may also assume that the steam is- ■expanded down to the feed temperature before release. Now, we have already drawn the Theta-Phi diagram as far as the point D, in Fig. 15, and we know that the total amount of heat supplied to the feed water =iogo'8 units and the total increase of entropy = i "45. The next step is to complete the diagram' so as to represent the complete cycle of changes until the steam returns to feed water again. During this part of the cycle the steam first does work upon the piston and afterwards gives up its remaining heat to the circulating water in the condenser. Now, in a perfectly non-conducting cylinder the steam would neither receive heat nor give up heat, while doing its work, although its temperature would fall all the time. As a consequence, //le entropy ■of /he steam does 7iot change owing to performing work. This is represented hy the vertical line DX and the point X is the lower limit of temperature at the end of the expansion, in this case, 620, the same as the feed water (by supposition). The line XA represents the loss of entropy as the steam is condensed after the release. The temperature is supposed to remain the same during this period, but Ave cannot calculate the loss of heat owing to our ignorance at present of how much of the steam at the end of the expansion is steam and how much water. Nor do we need to make such a calculation. If the significance of a cycle of changes has been understood it will be seen that the loss of entropy during a cycle must always equal the gain. Now, the gain of entropy amounted to 1-45, hence the loss is also equal to 1-45. This is our justifica- tion for returning to the point A, when condensation is completed and the water has returned to the same condition in every respect as at the commencement of the cycle. Moreover, knowing the loss of entropy during condensation (after release) we can calculate the amount of heat rejected by the steam during this period for loss of entropy x absol. temp. = heat 58 USE OF ENTROPY EFFICIENCY OF ENGINE. rejected, which is the area of the fig-are aAXd. I "45 X 620 = 8gg units of heat. We can now arrive at a useful result, for we can find the thermal efficiency of our engine. The amount of heat given to a pound of feed water=i40 + 902-8 + 48= iogo-8. This is represented by the area aABCDd. And the amount of heat rejected by the same pound of water or steam = 899 which is represented by the area aAXd. The difference between these two amounts is the amount of heat converted into worli viz: — 1090 '8 -899= 191-8 which is represented by the area ABCDX. The thermal efficiency of the engine is, therefore, -^ — 5 =-176. logoS ' i6. — Efficiency of a Refrigerating Plant. If we trace the action of a refrigerating plant, such as the CO^ machine, we find that, as far as the heat which is dealt with is concerned, what happens is as follows : — I St. In the evaporator the brine is continually giving up its kea^ to the gas in the coils, because, although the brine is very cold, it is not so cold as the gas which has just been produced by release ot pressure. 2nd. In the compressor, worA is done upon the gas, which considerably increases its temperature, but not by direct addition of heat, only through the conversion of work into heat. 3rd. In the cooler or condenser, the gas gives heat to the circulating water, and returns to the liquid state. 4th. As the liquefied gas escapes through the regulator valve into the evaporator coils it expands adiabatically and falls in temperature. We may represent these changes upon a Theta-phi diagram, figure 16, XA is the absolute temperature of the evaporator coils, or rather of the gas when first admitted. The area XABY represents the quantity of heat absorbed by the gas from the brine. The temperature during this period only rising slightly (as shown by the line AB) because the heat received is mainly the latent heat of evaporation. At B compression commences and the tempera- ture rises without gain of heat or entropy. This brings us to C, the highest temperature reached. ■6o EFFICIENCY OF A REFRIGERATING PLANT. Condensation of the g;as now commences with some fall of temperature, but chiefly with loss of latent heat to the circulating- -water. The heat rejected is represented by the area XDCY. At D the liquid re-evaporates and falls in temperature to A without loss or gain of heat. If we take the difference between the heat received and the heat rejected, we get the area ADCB, and as the heat rejected is greater than the heat received this area must represent the work done upon the gas. Now the object of the machine is to take heat from the brine, and the ' "" ^^,*' B ■efficiency of the machine is measured by the amount <3f this heat, compared with the work spent in S oo- 4>oo- •t oo~ etfecting it. Efficiency ■- ABYX ADCB 300 - I oo £>TvtT Fig. I 6. So far, if the diagram were drawn correctly, the efficiency could only be ' worked out accurately by measuring the respective areas. But if we consider an ideal case we may suppose the lines AB and DC to be parallel to the base line XY. This would require all the heat received by the gas to be received at one temperature XA, and all the heat rejected to be rejected at one temperature XD. We can now give a simple measure of the efficiency, for ^^^^=__, which is the lowest temperature divided by the difference of the temperatures! In the usual nomenclature — Efficiency = T - t 17> — I/itne and Cement. Lime. — There is a metal of a whitish colour, very light and ductile, called calcium. The metal can only be obtained with great difficulty from compounds which contain it, and is on that account rather a chemical curiosity. But when combined with oxygen we have the familiar substance known as lime. Lime has the same relation to the rare metal calcium that iron rust has to the metal iron, that is we might call lime the rtist of calcium, being the oxide of calcium, its chemical formula is CaO. Lime is obtained by heating limestone, chalk, or any substance, which is composed of lime and carbonic acid, in kilns, so as to drive off the carbonic acid gas. The lime obtained is unaff"ected by even the highest tempera- ture, and is taken out of the kilns in lumps, which easily break u;> and form a white powder. Obtained in this way it is called quick- lime, and its composition is CaO plus impurities. The Lime-light so much used in lantern exhibitions is obtained by directing the flame of an oxy-hydrogen blow-pipe — that is, a gas flame fed with pure oxygen — upon a lump of quick-lime. The lime becomes dazzlingly luminous. When water is thrown upon a lump or upon a heap of quick-lime, a very violent chemical action is set up which consists in the union of the lime with a definite quantity of water. Great heat is produced causing the mass to swell very much and crack and the surplus water to boil and give off steam. The operation is a verv common one during building operations, and results in a fine wliite powder called slaked lime whosi chemical formula is C&O^H^ or (CaO -(- OHj). Mixed with fresh water, sand and water, slaked lime forms the builder's mortar (2 or 3 of sand to r of lime). Lime dissolves in water to a slight extent (800 of water to i of lime). The clear liquid is lime-water, but if more lime is added than the above proportion and the whole kept agitated, a 62 LIME AND CEMENT. white opaque fluid is obtained, called milk of lime from which the surplus lime will gradually settle, leaving the clear lime-water above. Lime enters as a base into many natural rocks, for example, as carbonate of lime it occurs as marble, limestone, and chalk. As sulphate of lime it forms gypsum — -a substance that also contains water. When gypsum is heated it gives up its water and falls to a powder which has the property of re-absorbing water again, when mixed with it to the consistency of thick cream. The mass quickly hardens or sets, and is thus used for taking casts of objects, and iilso for putting a finishing layer of Plaster of Paris upon the walls of buildings. In steam boilers, especially marine, it is sulphate of lime chiefly which torms the adhesive scale. Portland Cement. Common builder's mortar consists of 2 or 3 parts clean fresh silica sand and i part freshly-slaked lime. When mixed with water to a convenient consistency it remains moist and soft for a long time. But it ultimately dries and hardens. The chemical change that occurs, consists in a gradual absorption of carbonic acid from the air (especially on the outer surface), which converts the lime in the mortar into carbonate of lime, which cements the grains of sand together to form an artificial stone. In Portland Cement the ingredients are different. It is manu- factured as follows : — Lime is used as in builder's mortar, but in place of sand, clay is added. Now clay is composed of two principal ingredients, viz., silica and alumina, in very fine particles, so that when mixed with lime there are thred ingredients, viz. : — Lime - - - in powder. Silica - - - in very small grains. Alumina - Do. LIME AND CEMENT. 03 Hence, viechanically, cement is not so coarse as mortar, and, chemically, it has an additional ingredient, viz., alumina. The limestone and clay are ground up with water and thoroughly mixed together. Then the paste is made into balls and dried. Afterwards the balls are calcined in a special furnace, and finally ground up to a very fine powder. The setting of cement when mixed with water is much more rapid than in the case of mortar, as it does not depend upon the action of the air. The ingredients are so intimately mixed together, and in the right proportions, that upon the addition of water they combine chemically. The hardened cement is said to be a complex silicate of alumina and lime. Many natural limestones contain all the elements for making cement, all that is necessary to do is to burn them in kilns, so as to convert the carbonate of lime in them into quick-lime. After being ground up fine, it is ready for use, by mixing with water as above. i8. — Iron Ores. Iron is found in the earth combined with many different elements^ as oxygen, sulphur, carbon, etc., as well as in a metallic state (meteoric iron for example), but the compounds from which all the iron of commerce is obtained are the oxides and the carbonate of iron ; these compounds being thus distinguished from the others as- "iron ores." Taking the ores of iron in order of their richness and purity we have : — ist. Magnetic oxide (Fe304), which, if perfectly pure, would contain 72 per cent, of metallic iron.* This ore is black, often very lustrous, and sometimes crystallized. It obeys the magnet, and when it is found, as a magnet itself, it is the natural loadstone of the ancients. Geologically it occurs in the oldest rocks, and is not found largely in Great Britain. Its chief mines are in Sweden, Norway, and Lapland. The famous mine of Dannemora being of this ore. Considerable quantities are imported into Great Britain from Gelllvara, in Sweden, and also from Marbella, in Spain. The next ore in richness is the Ferric oxide (Fe^Oj), which,, when pure, contains 70 per cent, of iron. When found crystallinef as in the island of Elba, it is called "specular iron ore," but its ordinary commercial name is "red hematite." A large deposit of this ore occurs in Lancashire, near Ulverston, and is smelted in the iron works there. Red and brown hematite are imported in large quantities from Bilbao, Elba, Santander, Algeria, Greece, and Russia. * Fe = 56, O = 16, hence (56 x 3) + (16 x 4) - 168 + 64 - 232 .•. 232 : 168 : : 100 : 72 per cent IRON ORES. 65 The next is the hydrated ferric oxide (FeaOj + Hfi). This is known commercially as "brown hematite." The old iron works in the Forest of Dean smelted this ore. It is also the ore of the Northampton district in England. But the chief iron ore found and worked in England is the carbonate of iron (Fe €03) = 49% of iron. This ore is known as *'clay iron stone" from its association with the ancient clay or mud of its geological period, which is more recent than the period of hematite ores. These ores are very impure in consequence, con- taining lime, magnesia, silica, and phosphorus, the latter from the remains of vertibrate animals. The percentage of iron may thus be as low as only 20%. Their colour is yellow or brown. Clay iron stone is a sedimentary rock itself. Carbonate of iron, however, sometimes occurs in a crystallized state, especially in Westphalia, and in several localities in England (Weardale being one). It is then called "spathic iron ore" or "iron spar," and contains considerable quantities of manganese in its composition. Another impure carbonate of iron is the "black band iron stone" of the coal fields. Its composition is carbonate of iron mixed with a bituminous shale coal to the extent of 20%. A large deposit occurs at Airdrie, near Glasgow, and also in N. Staffordshire, and in the Cleveland district. ig. — The Coal Measures. In Britain the coal occurs in layers or strata, called "seams," of various thicknesses, alternating with strata of sandstone, lime- stone, or shale. The whole series of strata being known as the carboniferous or coal-bearing system. Each coal seam represents a long period of time, especially so in the case of the seams 6 feet or more in thickness, for a coal seam is the remains of a swampy 'growth of vegetation, which, after accumulating, during many centuries, to a great thickness has been submerged by the subsidence of the land to some depth below the surface of the ocean. Another long period of time has then elapsed, during which, the materials of the sandstone and limestone rock have been gradually deposited upon the top of the old vegetation. Then owing to pressure and heat, and perhaps other causes, the mass of vegetation has lost all its water and become a solid mass of carbon and bituminous matter. In the oldest seams the change being almost entirely into carbon. When several seams of coal occur one below another, separated by strata of sedementary rock, it proves that the land has been up and down more than once, each elevation to the sea level being followed by the accumulation of the kind of vegetation that makes coal. In any case, the present condition of the coal fields of Great Britain is due to the last lifting up of the land, with its coal-bearing strata, out of the ocean to its present level. And this final elevation has occurred many thousands of years ago, so that much of the strata, originally over-lying the coal seams, has been removed by denudation. THE COAL MEASURES. 67 Under-lying each coal seam is found a clay bed in which can still be traced the fossil remains of the roots of the coal plants, and in the shale and sandstone over-lying a coal seam are found prints of the foliage of the coal plants, often resembling huge ferns. The coal fields of Great Britain occur in a broad belt which crosses the island in a direction from N.E. to S.W. The northern limit is a line drawn from the Tay to the Clyde, and the southern limit a line drawn from the Humber to the Bristol Channel. Within this area some of the principal coal fields are, as follows : — ist. South Wales, having an area of 900 square miles. Here, at a depth of over 3000 feet, is mined the famous Welsh steam coal. 2nd. South Yorkshire, Nottingham, and Derbyshire. 3rd. Northumberland and Durham. The coal raised in this district is generally known as Newcastle coal. It has about the same area as that of South Wales. 4th. The Coal Basin of the Clyde, which yields 15 million tons annually, is remarkable for the proximity of the coal seams to the surface. Several seams often occur within 300 or 400 feet of the surface in this district. 5th. The Lancashire Coal Field, situated almost surround- ing Manchester. The coal is used chiefly in the surrounding district. Composition of Coal and Coke. It is usual to distinguish broadly three varieties of coal, viz., Anthracite, semi-bituminous, and bituminous. I. Anthracite coal is the oldest coal, geologically speaking, and is found in the deepest workings. It consists almost entirely 68 THE COAL MEASURES. of fixed carbon, hence it burns with scarcely any flame. It is difficult to ignite, but it gives out intense heat. About 90 per cent, of anthracite is pure carbon. 2. Semi-bituminous coal is a steam coal containing from 70 to 80 per cent, fixed carbon. Its advantage for steam boilers Consists in it burning freely with a good flame and little tendency to form a cake. 3. Bituminous coal contains still less fixed carbon, but a considerable quantity of bituminous or tarry matter which readily melts and causes the coal to cake on the bars. It also produces much smoke and bright flame. House coal is usually a bituminous coal. For purposes of calculating the evaporative power of fuels their ultimate composition is required. This can be obtained from published analyses of the coal from the various pits, but an average composition of good Welsh, Newcastle, and Lancashire steam coal may be stated in round numbers to be as follows : — Welsh. Newcastle. Lancashire, Carbon, - - - - 85 81 79 Hydrogen, 5 5 5 Oxygen, - - - - 4 6 8 Sulphur, - I I I Ash (including nitrogen). 5 7 7 Inferior coals have less carbon and more oxygen and ash. Coke. When coal is heated without being allowed to combine with oxygen or "burn," the volatile portions escape as gases, and what THE COAL MEASURES. 69 is left is the "fixed carbon" mentioned in the previous descriptions, together with the earthy matter or ash. This fixed carbon with its impurities is the commercial "coke." The oldest method of making coke consists in piling the coal in a heap, and after plastering it over with mud and qoal dust, setting fire to the heap. By controlling the admission of air and renewing the covering as it burns through, the greater part of the coal is converted into coke. The operation is completed when all the gases have escaped. In coke ovens, specially built for the purpose, the gases ar_e not wasted> but utilised for heating purposes. Certain bye-products are alse extracted. A "coke oven" is a circular brick structure about ii feet diameter and 7 feet high inside. It is built with a double thick- ness ; the space between the two walls being filled with sand or loam. There is a flue at the top to take away the gases, which are used for raising steam for the colliery engine. A number of sucih coke ovens (called Beehive ovens) are built together forming a complete plant. The small coal and slack is used for converting intp coke. The process consists in charging several tons of coal into each oven, which is then lighted and allowed to burn for a few days with a limited supply of air; by which only a portion of the coal is con- sumed, the heat from which drives off' all volatile matter from the rest, and leaves a mass of hard dense coke, which, after slaking, is withdrawn, and the oven charged again. The weight of coke obtained from a given weight of coal depends upon the amount of fixed carbon in the coal. From 6q to 70 per cent, is usual, the remaining 40 or 30 per cent, passing away as gases. The actual composition of coke is about go pure carbon, and the rest sulphur and ash. 20. — The Blast Furnace. The smelting of iron is carried out in the blast furnace, which is a structure varying from 50 to 100 feet in height and from 12 to 25 feet in diameter (figure 17). The interior shape is shown in the diagram ; the exterior is generally cylindrical. It is built of brick or masonry, and has a lining or "shirt" of fire bricks. The exterior tO ^ stare, ^or ha at mil fl H! ^O f^m^mm Fig. 17. surface is, generally, now formed of iron plates, and the whole structure is bound round with angle iron rings. The orifice at the top is closed by an inverted cone, which can be lowered to admit the charges of ore and fuel. Just below the THE BLAST FURNACE. 7 1 cone is a lateral opening for taking away the gases through an external pipe to the stove for heating the blast (unless cold blast iron is being produced). The widest part of the furnace is called the "boshes," below which the furnace contracts until the crucible or "hearth" is reached. Here the blast tuyeres enter, and at one side of the hearth is fixed the dam plate. The "blowing in" or starting ot a blast furnace is a gradual operation, extending over several weeks, but when in full action the method of working the furnace is briefly as follows : — Successive charges of ore, coal, and limestone are fed in at intervals through the cone at the top. The blast being always on, as the materials settle down to the hotter parts of the furnace, they undergo, first, a roasting action which drives off water, coal gas, and carbonic acid (if the ore is a carbonate of iron) and in any case render the ore more porous, and more susceptible to reduction. As the mass settles down to the red hot zone, it meets the upward current of carbonic oxide gas (due to the partial combustion of the coal lower down) the real smelting commences. The CO gas combines with the oxygen in the ore and sets free the iron ; carbonic acid gas being formed — Fe203 + 3(CO) = Fe2 + 3(C02). when first separated from the ore or "reduced" — as the separation of iron from the ore is called — the iron is not fused, but in a spongy mass, mixed with the earthy impurities of the ore and the limestone, or rather quick-lime, for the limestone has by this time lost its carbonic acid. The coal also has been "coked" in the first stages of its descent, and forms a part of the spongy semi-fused mass. As the charge descends still lower and reaches a higher tem- perature, recombination take place, the iron combines to some extent with sulphur and carbon from the coal ; also with silicon, and with phosphorus, when the latter is present in the ore. These compounds are easily fusible, and when melted have the power of dissolving the remainder of the spongy iron. The result 72 THE BLAST FURNACE. is that molten cast iron trickles down through the incandescent fuef on to the hearth, where it accumulates. Meanwhile the lime has also to form its own combinations. Its duty is to combine with the remaining' large amounts of silica^ alumina, etc. The resulting compounds are also fusible as the heat increases. So these trickle down also, and being lighter substances than cast iron they float upon the top of the molten metal. We have thus upon the hearth of the furnace molten iron at the bottom, and melted slag or "cinder," as it is called, at the top. The tapping of the furnace may be done every 6 or 12 hours. The cinder is run off through a notch in the top of the dam plate, and the metal through a hole lower down in the hearth. The origin of the name "pig iron" is due to the method of running the metal from the furnace. In front of the furnace a large area of sand is prepared, called the "pig-bed." Channels are cut in the sand for the metal to flow along, and branching out from the main channels are smaller cavities which form the moulds for the pigs. The appearance then presented has some resemblance to a sow and a litter of pigs. • The pigs are broken off from the sow when cold, and carefully assorted into the various grades or numbers, according to the appearance of the fracture, which to an expfi-ienced eye is a sufficient guide. The broad distinctions are grey, mottled, and white, but the numbers run from i to 6 or 8. The air blast for a furnace is a very powerful one, the pressure being several pounds per sq. inch. Also the temperature is very high, viz., from 1000 to 1500 Fah. The air is treated in regenerative furnaces or stoves, in which the waste gases from the top of the blast furnace are burnt. A blow^ing engine is used to force the air through the tuyeres into the blast furnace, after the manner of a double-acting force pump. A large Cleveland furnace may require as much as 50,000 cubic feet of air per minute, and its output of pig iron may amount to 1000 tons a week. THE BLAST FURNACE. 73 Before the year 1828 all Blast furnaces were worked with a cold blast, and cold blast iron is still produced to a limited extent as at Blenavon. The introduction of hot blast cheapened the production considerably but without improving the quality, hence, in spite of its higher price, cold blast iron is still used for special mixtures in the foundry — cylinder castings for example. The management of a blast furnace is the iron master's business. There must be a nice adjustment of the relative amounts of "mine," and fuel and flux from time to time, as well as the air pressure and temperature. Once blown in, the furnace has to run for years without stopping. Everything that goes in at the top as solid must come out at the bottom as liquid, unless it escapes as gas. An obstruction called a "scaffold," caused by anything that cannot be fused, would entail immense loss in having to blow the furnace out and start afresh. 31. — The Composition and Properties of Pig Iron. The same blast furnace using the same ore and fuel yet pro- duces many grades of pig iron. The various grades are known by numbers, but a rough classification makes three very distinctive grades, viz. , Grey, Mottled, and White. The broad facts about these three grades ot pig iron are : — I St. Grey iron has nearly all its carbon uncombined and in the form of graphite scales dispersed throughout the mass. Whereas, white iron has nearly all its carbon in chemical combination, and mottled iron is intermediate between the other two. 2nd. The amount of silicon is always greatest in the grey iron, amounting often to three-fourths of the amount of carbon. And its effect is similar to that of combined carbon, viz., making the iron more brittle and hard. 3rd. The amount of manganese varies very much and does not affect the quality of the pig iron to any great extent. 4th. Sulphur and Phosphorus are very objectionable in pig iron which is intended for conversion into steel, or into wrought iron, for they are difficult impurities to remove and if not got rid of, the iron or steel produced is unworkable, being either 'cold-short' or 'red- short.' But pig-iron containing much phosphorus is remarkably ■fluid when melted and is consequently used for ornamental castings where strength is not important. THE COMPOSITION AND PROPERTIES OF PIG IRON. 75 The following- three analyses, taken from Prof. Greenwood's work, show the relative amounts of combined and uncombined carbon ; also the large amount of silicon in the grey iron, and the very great range in the amounts of sulphur, phosphorus, and manganese that may occur: — Grey. Mottled. White. Graphite carbon 3"i 1-99 •87 Combined carbon •04 2-78 2-46 Silicon 2-i6 71 I-I2 Sulphur - •II — 2-52 Phosphorus ■63 1-23 •91 Manganese ■5 — 2-72 Metallic iron - - 94 '56 93-29 89-40 1 00 '00 lOO'OO 100-00 The grades of iron from i to 3 are used in the foundry, owing to their fluidity when melted as well as to their toughness. Mixtures are made of several different brands and some scrap, so as to obtain the desired quality as regards strength, closeness of grain, freedom from blow holes, etc. Mr Seaton, in his "Manual of Marine Engineering," gives for a good cylinder metal a mixture of one-third cold blast Blenavon pig, one-third No. 3 Scotch pig, and one-third picked scrap. White iron is only used for conversion into wrought or into steel. The higher grades 5, 6, etc., are, therefore, known as forge iron. Cast iron whatever its composition is always much stronger in compression than in tension. An average number for com- pressive strength is 40 tons per sq. inch, and for tensile strength 7 or 8 tons, For a given load or stress, cast iron stretches more than either wrought iron or steel. In other words, its modulus of elasticity is only about ^ that of wrought iron. Its ductility is very small, so 76 THE COMPOSITION AND PROPERTIES OF PIG IRON. that it breaks suddenly without warning and with little reduction of area or permanent elongation. But the brittleness of ordinary cast iron can be removed by a process exactly the reverse of case hardening; and resulting in "malleable castings." The castings to be treated in this way are generally small and are run from white or mottled hematite iron. The process consists in packing the articles in cast-iron boxes surrounded with powdered hematite ore (which is oxide of iron). When sealed up, the boxes are placed in a furnace and kept at a red heat for several days. By this means the carbon and some of the impurities are removed from the iron to a greater or less depth dependijig on the time the heat is continued. 32. — The Manufacture of "Wrought Iron. Ths conversion of cast iron (pig iron) into malleable wrought iron is known as "puddling," because during the process the iron is kept in a pasty condition or like puddle. When the pig iron is allowed to become thoroughly fluid during the process, it is called "pig boiling" instead of puddling, from the appearance of boiling due to the escape of gases. In all the methods for converting cast iron into -wrought, the object is the same, namely, to make the carbon and other impurities combine with oxygen and thus pass away, either as CO gas, or form a slag which can be run out of the tapping hole. In figure 18, F is the furnace for burning the fuel, A the ash pit, and C is the bed upon which the charge of pig iron is placed. The top of the furnace is shaped so as to deflect the flame and gases down upon the charge, before escaping to the chimney W. In this furnace the purity of the fuel is of less importance, as its impurities cannot affect the iron. In dry puddling only white cast iron is used. The reason being that white cast iron assumes, before melting, the pasty condition which is so favourable to the decarburizing action of the air, whereas grey cast iron melts all at once to a liquid. In dry puddling, as in the open hearth process, the purifying of the cast iron is chiefly affected by the action of the air, and only assisted to a small extent by the pressure of "fettling," that is, oxides and slag thrown into the furnace. 78 THE MANUJACTURE OF WROUGHT IRON. "Wet Puddling" or "Pig Boiling" is the most extensively used of all processes now for the production of wrougnt iron. Grey cast iron is used, and the furnace is the same as for dry puddling. The bottom of the furnace, however, must have a basic lining (not sand as for dry puddling). It consists of slag, tap cinder, hammer scale, etc., all of which consist largely of oxide of iron. The lining is plastered over the iron bed when semifused and the furnace is then ready for the charge. 5'(a ^ f ///////////////////////"////"/// 7T. STcvatto'^' ^m W777777777777777777777777^77777Zf7777P7?!yZ^ ^///////////////Jm? • Fig, 1 8. The charge may amount to 4 or 5 cwt. The first stage consists in melting down the pigs, and it is during this period that most of the silicon and manganese are burnt out. About half-an- hour completes the melting down. The next stage consists in adding fettling material — oxide oi iron, etc. — and gathering the metal from the sides of the furnace to the centre and mixing all together. This occupies about 7 minutes. The third stage requires a much higher temperature, and it is during this stage that the carbon is removed from the iron. This THE MANUFACTURE OF WROUGHT IRON. 79 is effected by the furnace lining and slag rich in iron oxides. The metal appears to boil as the CO formed escapes — phosphorus and sulphur are removed also during the "boil." Time about half- an-hour. Fourth stage consists in collecting the charge (which is now in a pasty condition) into balls of 60 or 80 lbs. apiece, rolling them to the hottest part of the furnace to obtain a welding heat, and then removing them one at a time to the steam hammer, to be pressed into blooms. The first rolling out results in only a very unsatisfactory kind of iron, called puddled bars. These are broken up into short lengths, piled up cross and across, put into a furnace to raise them to a welding heat, and taken again to the steam hammer, and so to the roller. This gives merchant bars. A second repetition of the piling up, hammering, and rolling, gives best iron, and so on to best best, and treble best iron. 33. — The Bessemer Acid Process for Making Steel. The pig iron which is used for conversion into Bessemer steel is known as Bessemer pig. It is smelted from the pure hematite ores of Cumberland and Lancashire, or those from Spain, etc., because such pig iron contains only a very small proportion of phosphorus. The Bessemer process (acid) cannot burn out the phosphorus, and if it remains in the steel produced the steel is rendered hard and brittle. The conversion of Besse- mer pig into Bessemer steel is effected in the converter, which is a wr ought-iron vessel mounted on trunnions, as shown in fig. ig. There is a kind of spout or mouth at the top and a number of air holes or twyers at the bottom. The inside is lined to a con- siderable thickness with a mineral called "Gannister," which is chiefly composed of silica or sand. After the converter has been treated by burning charcoal in it, assisted by an air blast, a charge of molten pig is run in (perhaps 5 or more tons). The con- verter is charged while laid with the mouth nearly horizontal, fig. 20, so that the metal may not run into the air holes at the bottom. The blast is then turned on and the converter brought to the upright position. The effect of thus blowing air through the mass of molten metal is very violent. Various coloured flames are discharged from the mouth accompanied by showers of sparks. What happens inside the converter is that the air blast burns out ^ VTTT/TTTTTTT? Fig. 19. THE BESSEMER ACID PROCESS FOR MAKING STEEL. 8l of the pig- iron its carbon, silicon, and manganese, converting the two latter into slag. Some of the iron also is oxidised. The "blow" is continued for about 20 minutes during which the temperature rises very considerably. At the end of this time the flame from the mouth suddenly diminishes, which is the sigO that the carbon has all been eliminated. The converter is then quickly laid horizontal again and the blast shut off. At this stage the metal is almost pure iron in a molten condition. It is now necessary to add just the necessary amount of carbon to produce the kind of steel required. This is done by running into the converter a definite quantity of a special kind of cast iron from a cupola. From 7 to 10 per cent, of molten spiegeleisen will answer the. purpose, or a smaller quan- tity of ferro - manganese. The "blow" can be put on Kg. 20. again for a short time in order to mix them thoroughly, but this is not always necessary. The only remaining operation is to empty the steel into ladles and pour into moulds so as to have ingots suitable for rolling into rails, etc., when sufficiently cooled. The changes which occur can be studied from the following tables, which are taken from Prof. Greenwood's work :— Melted pig charged into Converter. Carbon uncombined - 2-07 percent. Carbon combined - - - i'2 ,, Silicon - - - - I '95 >> Sulphur ... - -oi „ Phosphorus - - - - '05 ,, Manganese - - - "OQ i> 82 THE BESSEMER ACID PROCESS FOR MAKING STEEL. At end of Blow. Carbon Silicon - Phosphorus Sulphur Manganese •I per cent. •02 ,, •06 „ trace only trace only After adding; Spiegeleisen. Carbon - - - - 'sSS per cent. Silicon - - - '03 ,, Phosphorus - - - - "05 ,, Manganese - - - - "31 ,, Copper - - - - '04 ,, It appears that Mr Bessemer originally expected that he would be able to stop the "blow" at a certain stage of the process, if advisable, and leave the metal with just as much carbon as he wanted for the particular kind of steel required. But it was soon found that this was not practicable, not because it was impossible to leave the desired amount of carbon in the steel, but because other impurities were left in also. In order, indeed, to remove the last traces of silicon, it is necessary to continue the blow until the iron itself begins to be oxidised or burnt. This is proved by the rotten nature of the metal, if a sample taken at the end of the blow is cooled down and treated mechanically. Hence, it is considered that the addition of the spiegeleisen does more than supply a due amount of carbon. The late Mr Mattieu Williams considered that the manganese in the spiegeleisen acts as a purifier, by combining" with both the oxygen of the burnt iron and the remaining silicon, to make a silicate of manganese slag, which forms a scum on the: top of the metal. 34» — The Basic Process (Bessemer Steel). As stated in the description of the Acid Bessemer process, pig- iron containing' much phosphorus, cannot be used for conversion into steel by that process. But Messrs Thomas and Gilchrist have brought out what is known as the basic process, which will eliminate phosphorus from molten pig-iron as well as carbon, silicon, and manganese. The difference consists entirely in the lining material of the converter; whereas in the original Bessemer and acid process the lining is acid (gannister or silicic acid), in the basic process the lining is a magnesian limestone called dolomite, which is a base instead of an acid. This lining is either built up of hard burnt dolomite bricks set in tar or asphalte, or the dolomite, after being burnt and ground up, is mixed with tar and rammed in soft. The thickness at the bottom of the converter should be 2 feet. The converter is first charged with a ton or so of lime, and some small coke breeze. This enables the temperature to be raised (using the blast) to a bright glow. The charge of pig iron is now run in (7 to 15 tons), and the blow is turned on as in the acid process. After some twenty minutes all the carbon is burnt out, but the blow is still kept on in this process for another three minutes or 30 ; and it is during this after blow, when the temperature is very high, that the phosphorus is expelled, and forms, with the lime added, a slag the same as the silicon. 84 THE BASIC PROCESS (BESSEMER STEEL). It is necessary to take samples of metal from the converter, to ascertain if all the phosphorus has been eliminated. This is judg-ed of by casting a small ingot, cooling it, and breaking it across; large bright crj'stals show, that the process is incomplete, and the blow in that case is repeated for a further minute or so. The completion of the process is the same as in the acid process namely, the addition of sufficient spiegeleisen or ferro-manganese to bring up the percentage of carbon to what is required. The following table, taken from Prof. Greenwood's work, shows the composition of basic steel at different stages in the converter: — The numbers are percentages. Original After blowing After 2 min. After addition pig iron. 10 min. over-l)low. of Spiegel. Carbon - 3-276 0-590 0-026 0-302 Silicon - 0-476 •222 0-002 0-016 Phosphorus - 2-6 2-064 0-062 0-092 Manganese - 1-131 •122 •197 0-540 Sulphur - 0-062 •139 •051 0-040 25* — Siemens-Martin Process for Open- Mearth Steel. In the "open-hearth" process, by which the steel for boiler and ship plates is made, a furnace is employed to melt the steel and to keep it molten during the process. A Siemens furnace is essentially a gas furnace, fig. 21. Gas and air are brought into the furnace through separate apertures or ports (L and L^), say at the left-hand side of furnace in diagram, and after combustion pass out on the right-hand side. But it differs from any ordinary gas fire in that both gas and air are heated to a very high temperature previous to entering the furnace. This is efi^ected withoutany additional expenditure of fuel in a most ingenious manner. The waste gases (which in a furnace for melting steel are at a very high temperature) are made to pass downwards through a great pile — or rather through two piles — of looselj'-stacked fire- bricks. By this means the bricks under Bj Bjj in diagram are heated up during perhaps thirty minutes, when the upper tiers of bricks will have become almost as hot as the furnace itself, and the lower tiers less hot, the waste gases passing away at only a mode- rate temperature — say 300 deg. Fah. — through outlet, 0. At the end of this time the draught valves are altered so as to bring the supply of gas and air from D and H on their way to the furnace through these heated stacks of firebrick, the air through the stack Bj^ and the gas through the stack Bjj. By this means the incoming supply of gas and air picks up in a manner the heat which was left behind by the waste gases during the half-hour previously. The waste gases now pass downwards through the bricks on the left- hand side and heat these up in turn, and so every half-hour or so the stacks on the right and left of the furnace change duties — one pair are receiving heat from the waste gases while the other pair 86 SIEMENS-MERTIN PROCESS FOR OPEN-HEARTH STEEL. are giving up heat to the unburnt gases. The stacks of firebrick are called regenerators, and the furnace is called a regenerative furnace. In the production of steel by this furnace, it must first be lined with a good thickness (15 inches) of sand pat on in layers while the Fig. 21. furnace is hot, and the surface of each layer partly fused before the next layer is rammed in. The charge is now introduced; this consists of about 30 per cent, of Bessemer pig iron (not melted), and the rest of steel scrap, consisting of shearings, broken rails, etc. The cast iron melts first. And when the scrap has reached a white heat it is rolled over into the molten metal. When all is completely fused, a further addition of hjEmatite iron ore is made at intervals. The ore is thrown in SIEMENS-MARTIN PROCESS FOR OPEN-HEARTH STEEL. 87 in lumps, and the chemical action of the ore is as follows: — The ore being nearly pure oxide of iron does not introduce impurities, such as silicon, phosphorus, etc. — but its oxygen combines with the large amount of carbon in the pig iron, forming CO (carbonic oxide gas), this gas escaping from the body of the molten metal causes an appearance like boiling — ;and this action goes on until the percentage of carbon is reduced to i per cent., when the boiling Meanwhile the oxidizing flame of the gas has been at work, converting whatever manganese and silicon the pig iron contained into slag. At this stage samples are taken and tested, and if satisfactory the proper amount of ferro-manganese is added and the metal tapped. The time occupied may be some 12 or 13 hours. Ferro-manganese is a particular kind of cast iron containing over 20 per cent, of the metal manganese combined with the iron as an alloy, as well as the usual carbon. Being very free from the usual impurities found in cast iron, its addition to the steel has the €fFect of giving the desired amount of carbon and manganese with- out adding injurious elements. The effect of manganese in steel seems to be beneficial, chiefly by neutralizing the bad effects of phosphorus, which is so difficult to get rid of in the Siemens furnace unless a different lining is used. Now even this scanty description of steel-making shows us one or two vital facts about the steel for boiler plates and steel castings. We see at once why there is no grain or fibre in mild steel, for it has all been produced in a molten state like, an iron casting, and when required in the form of plates or bars it is first run into a rectangular mould to form an ingot of weight somewhat greater than the weight of the plate required, and when sufficiently cool is rolled out into plate or bar form. We also see the practicability in such a process — where the metal is kept for several hours upon an open-hearth in a molten state — of controlling the chemical composition with far more certainty than in any of the older processes for either iron or steel. 88 SIEMENS-MARTIN PROCESS FOR OPEN-HEARTH STEEL. The following table is an extract from Greenwood's work on steel and iron : — Analyses of Steel. Soft Siemens- Siemens Crucible Hard In 100 parts ot Martin Steel Steel for Tool Steel. Plates. Forgings. Steel. Carbon - - -167 •21 •36 1-144 Manganese •044 •36 •30 •104 Silicon - •023 ■047 •02 •166 Sulphur - - -013 •052 •02 — Phosphorus •062 •03s ■03 — Copper •076 trace — 36. — The Cementation Process for Making Tool Steel. The cementation furnace, fig. 22, consists of an outer casing- terminating in a chimney C. Inside the outer casing is a furnace F, covered by a fire brick arch D, and having several small vents or flues c, c, leading into the main chimney; p, p, are two fire-brick troughs 8 or 12 feet long, and 2^ feet square, open at the top. These are called "pots," and are to contain the bars which it is intended to convert into steel. The wrought iron bars are generally of the best Swedish iron, 3 in. by f in., and the length of the pots. To charge the furnace, a layer of small lumps of charcoal is laid upon the bottom of the pot, next comes a layer of bars with a space between each bar ; upon these a second layer of charcoal, and then another Pig. 22. layer of bars, and so on, until the pot is nearly full. The last layer being of charcoal, the pot is sealed up by covering over with grindstone mud. The fire is now lit, and the whole gradually raised to a white heat, during the course of a couple of days, and kept at that tem- perature for 8 or 10 days longer. What happens is that the iron absorbs carbon to an extent depending upon the time given. ^o THE CEMENTATION PROCESS FOR MAKING TOOL STEEL. From time to time a bar is withdrawn through holes left for the purpose in the pots and furnace walls, and after cooling it down it is broken to examine the fracture. The depth to which the carbon has penetrated is known by the crystalline nature of the fracture extending so far inwards from the outside. The various stages are distinguished by numbers : — No. I , being spring heat. No. 4, double shear heat. No. 6, melting heat, etc. When the desired result is obtained, the fire is allowed to die out, and the furnace to cool down. It is then entered and the steel withdrawn. In this condition it is called blister steel, from the blistered appearance of the surface. To produce cast steel suitable for cutting tools, the bars are broken up and melted in crucibles, and the molten homogeneous steel poured out to form ingots, or run into other forms for use. Whilst in the crucibles also various additions can be made to alter the composition as desired. Steel thus made is the purest of all steel, and contains the most carbon, sometimes exceeding one per cent. The cementation process for the making of steel has been practised in England for 150 or 180 years. And the melting of steel in crucibles is said to have been first accomplished by a Sheffield man named Huntsman, in the year 1760. The above description of the cementation process is given, although now rather out of date for steel making; but it is important as being the only process for a long and eventful period in our history, as well as being in principle identical with the engineer's procedure for case-hardening wrought-iron finished articles. What is taking its place in the making of tool steels is the direct fusion of wrought iron with charcoal and other elements in crucibles, the temperature necessary for such fusion being now more easilv obtained than formerlv. 37.— Welding of Tubes and Furnaces. Two operations are generally necessary in the making of a w&lded tube. The strip or plate of metal to form the tube has first to be bent into the circular form, when it is called a "skelp," and afterwards it is reheated and the weld effected. The method -of performing each operation varies with the size and purpose of the tube. In the case of small gas, water, and steam pipes, both operations can be done by drawing the strip or skelp through a ■die having a funnel-shaped aperture. The draw bench, fig. 23, DTLUtrtq ^ rrheei FTg. 23. is fixed immediately opposite the furnace mouth, and as each strip is pulled out of the furnace and its extremity passed through the ■die, it is clamped to the draw chain, and the two halves of the die being closed, the strip of metal is drawn through the die and becomes a tube. If a butt weld only is intended the tube can be formed and welded at one operation, because sufficient pressure •can be put upon the tube by the die to form the weld without the assistance of a core ; but for pipes of I inch bore and upwards, lap welds are commonly made. 92 WELDING OF TUBES AND FURNACES. The skelp is made to take the form shown in figure 24, and in order to make the weld it is necessary to draw the tube at a weld- ing- heat over a mandrel, which consists of a rod with a bulb or plug at the end, equal to the bore of the tube. Grooved rolls are generally used for this purpose. The bulb of the mandrel is held in position exactly between the rolls, and the tube is drawn by the rolls over the mandrel, which have semicircular grooves of the exact radius of the outside of the finished tube. Fourthly, the tube is straightened in a reeling mill. Larger sizes of steam pipes are welded by some makers upon a "beak" extending the full length of the pipe. This method is similar to the plan adopted for welding boiler furnaces by gas — as follows : — Boiler furnaces are welded piecemeal, either before the corru- gations are formed or after. The first plan is adopted in the manufacture of Fox's furnaces. The plate to form the furnace is rolled in the plate-bending rolls to the correct radius, so as to be ready for welding. It is then mounted upon a sliding gantry, with the joint to be welded lying horizontally at the top, fig. 25. A double jet of water gas, fed by a blow-pipe with air, is arranged to play upon the joint both inside and outside, to obtain a welding heat ; and a small pneumatic hammer is used, with an WELDING OF TUBES AND FURNACES. 93 anvil, or beak, reaching inside the tube. The weld is made by running the furnace along the gantry until the part of the seam to be welded is exposed to the gas fire ; the blast is turned on and the heat obtained. Then the furnace is run back to the position of the hammer, when a few rapid blows unites that portion of the seam. In this way, about 8 inches is welded at a time. The furnace is corrugated afterwards by raising it to a red heat and rolling it Fig. 25- between corrugated rolls, one of which has to be placed in position inside the tube as soon as the furnace comes out of the fire on to the rolling mill. Holmes' furnaces are also welded before the corrugations are formed. But the particular form ot the corrugations is produced, bit by bit, by means of dies actuated by hydraulic pressure. A portable coke fire is used to obtain the welding heat in these furnaces, and the hammer is moved into position to suit the part to be welded, instead of the furnace being moved to suit the position of the hammer. 28. — Lubricating Oils. There are three natural sources from which oils for lubricating^ machinery may be obtained. These are : — 1st. The fat of animals. 2nd. The seeds and other parts of plants. 3rd. Crude petroleum from the earth. With respect to animal oils, there is Neafs Foot Oil, which, when cold-drawn from the bullock's teet, used to be much thought of, as it never gums to clog a bearing. Sperm Oil — taken from the head of the sperm whale — is also a fine lubricating oil for light machinery, but both sperm and neat's foot oil become solid near the freezing point of water, and they are both rather high priced. Lard Oil is obtained from hog's lard by pressure. Other animaj oils, such as common Whale Oil, are not suitable for lubrication as they are subject to gumming. The vegetable oils chiefly used for lubrication are Rape (colza), Olive, Castor, and Cotton oils. Of these, olive oil is the dearest, and cotton seed oil the cheapest. Linseed oil is never used for lubrication owing to its marked tendency to thicken and become gummy (hence its value as a paint oil). Blown oils are those vegetable oils which have been thickened' by blowing air through them when heated. Rape and cotton seed oil are sometimes treated in this way, and afterwards mixed with a mineral oil to make a better lubricating oil ; for it appears that such a "compound oil" makes a better "lather" when used with water upon a marine engine bearing. LUBRICATING OILS. 95, The third source is from petroleum. Crude Russian petroleum is heated in retorts, and at the lowest heat applied naptha is given off, and at a somewhat higher tem- perature illuminating oils pass off, leaving a residue called Astatki. Further evaporation is promoted by admitting superheated steam under the surface of the liquid. The vapours now given off condense and form lubricating oils. When the temperature is- about 320 F., the lightest lubricating oil is obtained, sp. gr. "85, and at higher temperatures heavier oils of a darker colour are distilled, until at a temperature ot 550 degs. F. a cylinder oil of a red colour, and sp. gr. "92 is obtained. The final residue, called goudron, is brown black in colour, and only fit for liquid fuel. When used for lubricating the external parts of steam and gas- engines, the chief qualities or properties of the various oils which are of consequence are briefly, ist, their degree oi viscosity; 2nd, their freedom irom gumming ; 3rd, their freedom or otherwise from acids. A vegetable or animal oil can always be selected of the required! degree of viscosity, and of very slight tendency to gum, but free acid is generally present in these oils, and sometimes to a large extent, rendering them liable to corrode the surfaces they lubricate. Mineral oils are nearly perfectly free from acids and do not gum or thicken when exposed to the air. The viscosity of the light oils is as low as the most fluid sperm oil, and the dark, heavier,^ oils have a greater viscosity even than castor oil. There is plenty of range, but mineral oils lose viscosity (that is become more fluid)' when heated, at a much greater rate than the vegetable oils do ; which has to be considered if a bearing is likely to run warm. Turning now to internal lubrication the case is very different for the three classes of oil. At a certain temperature above the boiling point of water all animal and vegetable oils give off inflammable vapours, the same as mineral oils, but there is this difference : — The vapour given off by g6 tUBRICATING OILS. mineral oil is the oil itself in a state of vapour. Condense it and we get the oil back again ; whereas, when a vegetable or animal oil vaporises, the oil is being decomposed. It is oil no longer, for it resolves itself into its elements. Instead of oil we get a substance called glyceryl ; and an acid which is stearic acid in the case of animal oil ; and oleic or palmitic acid in the case of the vegetable oils. Solid carbon or soot is also formed. For internal lubrication, therefore, the mineral oils alone should be used ; and if the oil is not to be vaporised at the working temperature the flashing point of the oil should be sufficiently high. What is called the "open test" for the flash point is very easily made as follows : — A little oil is poured into an open cup, and a small high temperature thermometer placed in the liquid. Heat is now applied, and, as the temperature rises, from time to time the flame from a lighted taper is passed at a distance of ^ in. over the surface. When the flash point is reached, a pale blue flame will be obtained. The temperature of the oil by the thermometer is the flash point. The "closed test" is made with a metal box having a sliding lid, which, when opened, depresses a small flame into the air space of the box above the heated oil. A lower flash point is obtained with the closed test than with the open test, and it is said to be not so uncertain. The flash point of cylinder oil varies from 300 to 600 degrees. Archbutt and Deeley, in their work on lubricating oils (from which most of the above facts have been taken), say that cylinder oil, if kept at a temperature of 370 degs. in a current of air, should not lose more than i per cent, of its weight in an hour. The evil effects resulting from vegetable oils being used for internal lubrication are pretty obvious. Firstly, the internal parts are very soon deprived of lubrication. Secondly, the hard portions LUBRICATING OILS. 97 of the decomposed oil cut and injure the rubbing surfaces. Thirdly, passages are blocked up. Fourthly, the acids attack the metal. Fifthly, when the acids are carried through the condenser to the boiler, the water is made corrosive, and the boiler suffers. More- over, any oil or grease getting into a boiler is liable to mix mechanically with the fine particles of lime or magnesia floating in the water, and perhaps settle on the furnace crowns. Being almost a perfect non-conductor of heat, the furnace crowns are most likely brought down in consequence. 29' — Corrosion in Boilers. The wasting away of the interior surfaces of the plates in a boiler has been attributed to various causes ; these are chiefly : — (a) Mechanical straining. (6) Galvanic action. (c) Acids from the oil brought in with the feed. (d) The salts in sea water, especially magnesia chloride. (e) Air in the feed water and carbonic-acid gas. (a) The first mentioned is not so much a cause of wasting as an action that determines where the wasting shall be most acutely localized. A common place to see the effects attributed to this action is on the crown plate of a vertical boiler's fire box, just out- side the flange of the uptake. The expansion and contraction of the fire box and uptake causes the crown plate to be alternately depressed and lifted, with consequent straining of the metal. The skin of the plate is thereby cracked or fretted, just outside the rigid double thickness of the flanged joint. Any corrosive tendency existing in the vicinity attacks these minutes cracks, and ultimately works a groove more or less completely round the flange. A similar effect may occur next the circular seams of a main boiler at the lower part, either outside or inside the shell. (b) Considering now the reputed causes of corrosion, we may first take galvanic action as being one of the earliest theories put forward. There does not appear to be any evidence to support this theory as a complete explanation of general wasting. But locally corrosion is probably intensified by electric currents in places where brass fittings are attached to the boiler, or where copper pipes are brought sufficiently near to the iron or steel plates. CORROSION IN BOILERS. 99 (c) Next with respect to the oil and grease brought in with the feed water in cases where these are not arrested by filtering. All vegetable and animal oils are decomposed by heat into glycerine and a vegetable acid. Thus a greasy mass lodging upon a heating- surface will be speedily decomposed, and the acids set free will proceed to attack the plate with which they are in contact. It is said that the water in a boiler is seldom, if ever, found to be acid throug-h the decomposition of g-rease or salts, but this does not prove that corrosion is not due to acids. For if the decomposition takes place, as above stated, the acids will only exist in close proximity to the plate, and will almost immediately be killed, as a necessary consequence of their combination with the metal. (d) With respect to the action of sea water upon iron or steel, 5t is a matter of common observation that salt water rusts iron more rapidly than fresh water, at all events when exposed to the atmosphere. Air, indeed, seems to be necessary in the formation ■of ordinary rust, for if a bit of bright i'roii is immersed in distilled ■water free from air it will not rust.* But the action of sea water inside a boiler is probably different to ordinary rusting. Messrs Babcock & Wilcox in their Eng-ineer's Reference Book give the following explanation. Chloride of anagnesia (of which sea water contains nearly one ounce per gallon) is decomposed by contact with heated iron, the result being- ;magnesia and hydro-chloric acid. Mg Cl2+H20 = MgO + 2 H CI. Now the natural effect of hydro-chloric acid upon iron is to produce •chloride of iron, which is soluble in water: — Fe -{- 2 H ei = Fe Clj -f- Hj. We should thus have chloride of iron dissolved in the boiler water instead of chloride of magnesia, whilst the hydrogen from the decomposed acid escapes as a gas. But it appears that chloride of iron in presence of magnesia produces chloride of magnesia again, Ihus : — Fe CI2 H- Mg O = Fe O + Mg Cl^. * The writer has kept bits of iron and steel bright for six years now in this way. 1,00 CORROSION IN BOILERS. We have now left upon the boiler plate a deposit of ferrous oxide (Fe O), and the chloride of magnesia may begin the action over again and deposit more ferrous oxide. In this view of the action of sea water it will be seen that the- oxygen to form the rust or scale has been taken from the water of the boiler (see the first equation).* (e) Now when air is admitted partly dissolved in feed water to a boiler the action is much simpler, for oxygen is here in a free- state and diluted with only twice its volume of nitrogen (see essay on air No. i). The feed water only requires heating to a higher temperature to give up its dissolved oxygen. It is reasonable tO' suppose that this will occur to a considerable extent when the water through circulation flows round the furnace sides and reaches- suddenly the great heat at the line of fire bars. The oxygen, adheres to the furnace sides in small bubbles which commence their injurious action immediately. Ferrous oxide is thus formed direct from the air, and not from the water as in the previous case. Fe + O = Fe O. The pitting of furnace sides is accounted for in this way by Mr Stromeyer in his work on boilers. But whether the first film of ferrous oxide is formed in one way or the other, such a film affords no protection to the plates. The ferrous oxide becomes in a remarkable wa}' a carrier of oxygen to- the plate beneath, which accounts for the tendency always observed. for iron rust to grow and extend from a nucleus. Hence the little pits first formed tend to deepen in the same place and spread. Iron rust or scale may consist of several oxides of iron, and its colour may be black, red, or brown. The brown rust is known as a hydrated ferric oxide (FeaOg combined with water). The scale from red hot iron is a mixture of FcaOg with Fe O, and the dark scale formed inside the boiler has a similar composition. * The chemical equalions are added by the present writer. CORUOSiOX IN BOil.lilJi. lOI It is a well-known chemical experiment to pass steam through a. red-hot iron tube, the iron is oxidised and hydrogen gas set free, which passes out at the other end of the tube. The oxide formed is similar to smith's scale. Now if this action takes place below a red heat, there is good reason to refer the corrosion in the steam space of a boiler to the action of the steam. The front plate of a boiler is particularly liable to become heated (through its proximity to the uptake) to a higher temperature than the steam itself, and the same thing may be said of the main stays near the front plate. Lastly, when iron is exposed to water containing carbonic acid in the solution, or when a moist surface of iron is exposed to •carbonic-acid gas, rust is formed very rapidly (a freshly-filed surface breathed upon shows a yellow stain almost immediately). It is ■said that carbonate of iron is first formed : — Fe -1- CO2 + H2O = Fe CO3 -f- Hj ■which is subsequently converted into oxide by the escape of the •carbonic acid. The question now remains as to which of these possible ■chemical actions is responsible for the corrosion of boiler plates, tubes, and stays? Or are they all responsible? The question is complicated by the fact that all boilers do not corrode in the same places, even when worked under identical conditions. Just one or two points may be noticed. In the first place a very local injury may sometimes be traced to its immediate source, local galvanic action has been mentioned already. There is also the wasting of a ■C.C. plate, where the feed water has unwisely been permitted to impinge. Again, the fact that corrosion occurs more below the water level than in the steam space seems to favour the decomposi- tion of sea water and grease, and the same at the water line. In connection with this point, the following average analysis of the "mud" from the bottom of a marine boiler is significant.* * From a paper by Mr I. B. Dodds, North-East Coast Institution of Engineers and Shipbuilders. CORROSION IN BOILERS. Ferric oxide ------ 65 •oo Calcic sulphate ----- 9-02 Calcic oxide. ------ 75 Magnesic hydrate - - - - - io"i2 Zinc oxide ------ "75 Sand, etc. 170 Oily organic acids combined with ■. the ferric, calcic, and magnesic l - io'65 oxides - - _ - ; Free uncombined oil - - - - 2 "00 A similar composition was obtained from the froth which lodges on the upper plates and stays. 30' — Remedies for Corrosion in Boilers. Beginning with a new boiler, it is a very good practice of some firms to thoroughly clean the inside of their boilers with soda, so as to remove effectually all grease and dirt that may have accumulated during construction. The boiler, when put on board the vessel, has then a fair start, and any subsequent trouble is more easily traced to its true origin. A new boiler if first worked with very pure water is liable to show the effects of ordinary rusting all over the interior; and the joints to bleed very much. In order, therefore, to encourage the formation of a thin lime scale, which is a protection against ordinary rusting, it is recommended to lime-wash the inside of the boiler, or to brush it over with thin cement and water before it is put to work. Also common soda added to the boiler water is a good preventative of rusting. (The Admiralty Regulations provide for boilers out of use being filled up completely to the top with fresh water, containing i part of soda crystals to loo parts of water, to prevent rusting). Sea water is sometimes used in new boilers until a thin lime scale is formed, but probably the use of quick lime dissolved in water, as recommended by Messrs Babcock & Wilcox in their reference book, is a safer plan and equally effective. They advise 5 lbs. of lime per looo H. P. to be put into a new boiler, and a further addition of i or 2 lbs. per 1000 H.P. daily, until a thin coating is obtained. Zinc slabs are generally fitted now in new boilers. They should be fixed on studs screwed into the sides and backs of combustion chambers, so as to be in metallic contact with the plates. Slabs may also be hung in stirrups between the tubes ; or clamped on to main stays for special protection of those parts. Also on the furnace sides. I04 REMEDIES FOR CORROSION IN BOILERS. The Admiralty rule is one slab for every 50 square feet of heating surface. The theory of the action of zinc in a boiler is that it forms with the iron an electric couple, after the manner of a galvanic battery. Zinc, being electro-positive with respect to iron, is alone attacked by the salt water, and thus the iron is preserved. But, apart from any galvanic action, there is no doubt that zinc perishes in boiling salt water, and, therefore, draws in a manner the corrosive influence to itself, which might otherwise spend itself upon the iron. Mr Stromeyer further states that the zinc salts thus formed in the boiler water have a protective action upon iron — the same as common soda — and in marked contrast to other salts, such as contain tin, for instance. Local pitting and wasting being generally attributed to injurious ingredients in the feed water — such as acidity, due to decomposed lubricants, or decomposition of sea water in the evaporator, as well as air from the hot well — the more the feed-water can be purified before it reaches the boiler the better ; prevention being better than cure. But should corrosion show itself in spite of every precaution, the parts affected should be carefully cleaned and the oxide of iron removed from the pit holes until the bare iron is exposed, after Avhich a thin wash of Portland cement should be applied to the iron, and the parts examined closely at every opportunity. The boiler water should never be allowed to become acid. If it is ever found to turn blue litmus paper pink, the acidity should be counteracted by putting soda into the feed-water, until the water is neutral or slightly alkaline. 31. — Softening "Water for Steam Boilers. What is known as "hardness of water" is due to mineral substances dissolved in the water obtained from rivers, lakes, or -spring's, and which it is advisable to remove before the water is used for steam boilers. The substances which cause "hardness" in water are the salts of lime and magnesia, viz. , bi-carbonate of lime, ,bi-carbonate of magnesia, sulphate of lime and sulphate of magnesia, with perhaps some chlorides. The reason that it is advisable to remove these substances is that if admitted into the boiler they are sure to form deposits upon the heating surfaces and endanger its safety, besides causing trouble in having to clean the boiler and remove the scale at frequent intervals. In softening these hard waters certain properties of the above substances can be taken advantage of : — Thus if the water contains the bi-carbonate of lime or magnesia, and it is boiled for a short time, the bi-carbonates are reduced to carbonates through the -escape of one half their carbonic acid. And as the simple carbonates -are very slightly soluble in water they are precipitated, and can be removed by filtering or by settling. The bi-carbonates are, there- fore, said to give only temporary hardness to water, since they are removed by boiling. Another method often used to remove temporary hardness consists in adding slaked lime in just sufficient •quantity to combine with the surplus carbonic acid, forming carbonate of lime, which, with that reduced from the bi-carbonate, all settles to the bottom of the softening tank. This is known as ■Clark's process and it is very commonly used. To remove sulphates (which give permanent hardness) it is usual to add carbonate of soda (common washing soda) and some- times caustic soda. The carbonate of soda effects a double Io6 SOFTENING WATER FOR STEAM BOILERS. decomposition ; that is, the carbonic acid leaves the soda and com- bines with the lime, whilst the sulphuric acid leaves the lime and combines with the soda. The result is carbonate of lime and sulphate of soda. The former settles and the latter remains in solution. Ca S04 + Naj C03 = Ca C03 + Nag S04 calcic sulphate + sodic carbonate = calcic carbonate +sodic sulphate. If permanent hardness is due to sulphate of magnesia the best chemical to use is caustic soda. The hydrate of magnesia is then precipitated. Mg S04 + Nag Hg 02 = i\Ig H3O2 + Nag So^ magnesic sulphate + sodic hydrate = magnesic hydrate + sodic sulphate. It should be noticed in connection with the methods of removing temporary hardness, that, although in both cases, the lime and' magnesia are deposited in the filtering tank yet the water is not left in the same condition in the two cases. With temporary hard- ness, if the water is boiled to precipitate the lime, or if slaked lime is added for the same purpose, the water is left practically free from anything in solution. Whereas when carbonate of soda is added to precipitate the lime in the case of permanent hardness, an. equivalent of sulphate of soda remains in solution after all precipita- tion has taken place. So that in the latter case although the water is softened it is not purified. And as only pure water is evaporated in the boiler, the sulphate of soda will accumulate and make the- water gradually more dense. Messrs Babcock & Wilcox fit a water-softening plant, in whiclv the various tanks are arranged as in figure 26. The soda tank. A, is charged every morning with a solution of carbonate of soda, sufficient to last the day. From the soda tank a cock (a) admits a certain quantity into the reaction tank B, to mix with the hard water brought in by pipe (b). A steam injection is fitted to B, in order to boil the water in the reaction tank and so- precipitate the bi-carbonates. The softened water flows out througb SOFTENING WATER FOR STEAM BOILERS. 107- outlet (c) into the filtering tank C, and again overflows through vertical pipe (d) into the feed tank, from which it is pumped into the- boiler. The admission of hard water and soda solution to the reaction tank is regulated by valves or cocks worked by a float in the feed tank D, and the position of the valves and cocks is made to suit this object, but not shown in the sketch. Also the pro- portion of soda solution to hard water can be regulated by a separate cock on the soda tank. An excess of soda being indicated by the water in the feed tank giving an alkaline reaction with a chemical test, such as red litmus paper. ^ q JnJet fi Soda, Tanli Rcaclicn Taiik B ^ 5) Ou.l-tel.- Iilicriny; Ta'nh. Sojtencd. yialer _ Fig. 25. When slaked lime is used to precipitate the carbonates which give temporary hardness to water, it is necessary to have two mixing tanks, one of which is being used from, while the other is being treated with the lime ; and after thorough mixing, allowed to settle. Each tank may contain about 400 gallons. From the mixing tank the water is run through a filtering tank, and from thence into the feed tank, as already described. The carbonate of soda or caustic soda to precipitate the sulphates is also put into the mixing tanks with the lime. The chemical equation for the- action of lime is as follows : — Ca (003)2 + Ca O2 H3 = 2 (Ca CO3) + 2 U^O. ■lo8 SOFTENING WATER FOR STEAM BOILERS. With respect to the quantity of slaked lime required to remove the temporary hardness of water, the calculation is the following-: — Looking at the last equation quoted, we see that each molecule of calcic bi-carbonate requires a molecule of slaked lime. Now the hardness of water is expressed by the grains of calcic carbonate per gallon, so we must compare the relative weights of Ca CO 3 and •Ca O2 Hj, using atomic weights : — Ca CO3 = 40 + 12 + 48 = TOO. Ca O2 Hg = 40 + 32 + 2 = 74. This relation is very nearly as 4 to 3. Taking a case where the temporary hardness is equal to 16 grains per gallon, we should require 16 x f = 12 grains of slaked lime per gallon. To soften 400 gallons of such water would, therefore, require 400 x 12 = 4800 grains of lime, equal to lO'i ounces. The objection to putting in more lime than necessar}' is that water dissolves lime to the extent of 75 grains in each gallon of water which is partly precipitated on boiling, hence it is important that there should be no uncombined lime left after the reaction shown by the equation has taken place. Clark's Test for Kardness of Water. The testing solution is made by dissolving curd soap in proof -spirit, 1 20 grains of soap to a gallon of spirit is the standard solution. The test is carried out as follows : — Pour 1000 grains of the water to be tested into a stoppered bottle, ■which will hold twice that amount (so that the bottle will be about ihalf-full). Next add 10 grains of the soap solution to the bottle and shake well for about a minute. Lay the bottle upon its side and observe if a lather remains for at least three minutes. If not, add -another 10 grains of soap solution and try again. When a lather SOFTENING WATER FOR STEAM BOILERS. 1 09 remains for three minutes or more it shows that all the lime and magnesia (the substances which cause hardness) have been con- verted into stearates, leaving a slight excess of soap uncombined in, the water which forms the lather. According to the number of measures of soap solution which- have been added, Dr Clark compiled a table showing the amount of Ca CO 3 in the water. Thus — I grain of chalk per gallon requires 32 grains of soap solution.. ^ M M M 54 » J n 3 » » >> >» 7 ' » ' » 16 ,, ,, ,, 320 ,, ,, 32.— Iriquid Fuel for Steam Vessels. When crude petroleum is refined by distillation in the oil 'district of Russia, the first vapour or spirit which is given off and condensed is Benzine or naptha; then at a somewhat higher temperature comes Kerosene or lamp oil; and at a still higher ■ temperature are distilled over lubricating oils. The residue after these more or less volatile parts have thus been separated is called "Astatki." Astatki is a hydrocarbon or mixture of hydrocarbons of very -dark brown colour, and having an ultimate composition of: — ■ Carbon - - 88 per cent. Hydrogen - 1075 ,, Oxygen - - 1-25 ,, The flash point may vary from 150 degs. Fah. to 300 degs. Fah. , and its heat value is equal to about 20,000 British thermal units per pound ; which is half as much again as good coal, if reckoned by weight ; but if reckoned by stowage room, 36 cubic feet of liquid fuel are equivalent, in heat value, to 67 cubic feet of coal as usually stowed. As a liquid fuel Astatki has been used by steam vessels on the •Caspian sea since 1870, and its use was confined chiefly to that sea for a good many years, owing, it is asserted, to restrictions upon exportation. Now, however, supplies of liquid fuel can be got from Burmah and Borneo in the east, and from Texas and California in the west. Stores of liquid fuel have now been established at many ports, at London, Hamburg, Bombay, Alexandria, Melbourne, Zanzibar, etc. LIQUID FUEL FOR STEAM VESSELS. 11? The use of oil for fuel, on board ship, may be considered under two heads, viz. : — Storage and furnace arrangements. Oil fuel lends itself very well to storage on board ship, for it can be stored in the water-ballast tanks and in the extreme end •compartments of a vessel, without encroaching upon space that i^ available for cargo. It thus sets free certain bunker spaces for cargo, which, ordinarily, are required for bunker coal It is necessary, however, to have two large tanks in the "between decks, called "settling tanks," into which the oil is pumped from the storage places. Each settling tank should hold 12 hours* ■consumption of fuel, and while one tank is being used from, the other is first pumped full, and then allowed to "settle," by which ■any water which may have been pumped from the ballast tanks along with the oil, settles to the bottom, due to its slightly greater density. A steam coil is fitted in each tank for heating the oil before it is used. From the settling tanks the oil is taken by suitable pipes and connections to the burners at tlie mouth of the furnaces. Fig. 27. The nozzle of one form of burner, patented by Rusden and Eeles, is shown in figure 27. It will be seen that the burner con- sists of a central spindle which terminates as a conical valve. Twa concentric tubes are fitted over the spindle, the annular ■ spape between the spindle and the inner tube being supplied with steamy; and the annular space between the inner and outer tubes being supplied with oil. There is also a steam jacket in the outer tube., The steam and oil are supplied to the burner through trunnions in, the middle of its length, each having its own supply pipe f^pm the ii: LIQUID FUEL FOR STEAM VESSELS. boiler and settling tanks respectively. The inner tube and spindle are adjustable endwise, so as to regulate the amount of steam and oil at the nozzle. The function of the steam jet is to pulverise the oil as it issues from the outer annular orifice and to project it in a powerful jet inside the furnace. The arrangement of the pipe connections is shown, in figure 28, for one furnace. Two burners are shown whose nozzles pass through holes in the fire door. They can be swung back out of the way when it iv necessary to open the fire door, or to examine the burners themselves. As regards the interior of the furnaces different plans are adopted, partly to suit different burners and partly as alternative plans. The simplest arrangement, and one that enables a return to coal burning to be made most readily at any time, is to keep the fire-bars in the furnaces, but to cover them with fragments of broken fire-bricks. The oil burners spray the oil just over the layer of broken bricks, which soon become heated and thus vaporise any oil that may fall from the spray upon them, besides absorbing some portion of the intense heat of the spray and radiating it to the LIQUID FUEL FOR STEAM VESSELS. I Ij crown. It is also necessary, as a further protection, to build a brick arch over the bridge, and a brick lining to the back of the combustion chamber. With this arrangement fitted in s.s. Trocas, the change from going full speed with oil to going full speed with coal was effected in 28 minutes. Another plan of furnace arrangement is to remove the fire-bars entirely and build a fire-brick stack about half way along the furnace, figure 29, and another structure at the end of the furnace where the bridge would be. The burners project their sprays right against the centre stack, which baffles the flame, at the same time as it absorbs some of the heat, becoming white hot in consequence. Fig. 29. The advantages claimed for oil fuel, are : — I St. A great saving in labour of stoking and trimming, as when once lighted the furnaces need no further attention. 2nd. A saving of both weight and space. The weight saved IS due to the higher calorific power of the oil compared with coal, but the space saved is much more than this, owing to the storage of the oil in the ballast tanks, etc. 3rd. There is no cleaning of fires necessary, with its accom- panying waste and loss of pressure. H 1 14 "LIQUID FUEL FOR STEAM VESSELS. There are disadvantages with oil fuel, one of which is thfe greater complexity and cost of the outfit ; with its pipes and pump- ing arrangements, its settling tanks and burners, etc. Also there is a direct loss consequent upon the use of steam for pulverising the oil in the burners. There is also still an uncertainty as to the •supply of liquid fuel nt convenient ports. 33« — The Balancing of a Marine Engine. What is generally meant by "balancing an engine" is arranging the moving parts, such as cranks, rods, pistons, etc., together with special weights fixed upon the crank webs or shaft, so that the engine will run under steam with the least possible ■vibration. Now the vibration which is set up when an unbalanced engine IS running, may be traced to several causes : — I St. The turning eifort upon the shaft not being uniform all round, such as occurs more especially with a single crank engine. and. The shaft with its cranks and crank pin brasses, etc., mot being balanced to run without vibration even by itself, due to centrifugal force. 3rd. The great inertia of the reciprocating parts of the engine, .-such as pistons, crossheads, etc. 4th. The swing and inertia of the connecting rods as they .follow the motion of the crank. With respect to the first cause in the above list, it may be as well to point out that, as far as the steam pressure is concerned, .every engine is balanced naturally. The engine frame provides for •that. This can easily be proved. Consider a marine engine. The pressure upon the piston on the down stroke is exactly equal to the pressure upon the cylinder cover. The pressure upon the cover is itransmitted to the columns, which transmit it to the bed plate as aa. Il6 THE BALANCING OF A MARINE ENGINE. upward pull. But this upward pull is met by the downward thrust of the connecting rod forcing the shaft upon the bottom brass of the main bearings, making a balance of force. There is another point to notice. If we follow the screw shaft through the stern bush to the propeller, we seem to meet with a difficulty. Since action and reaction are always equal and opposite, the questions occurs, "Where is the reaction which enables the propeller to overcome the great resistance to revolving which it meets with in the water?" Surely, upon the above principle of action and reaction there must be a reaction to revolving somewhere. As a matter of fact, we shall find this reaction in the engine room. If it were not for the holding-down bolts of the bed plate (probably the weight of the whole engine suffices) it would tilt up and the engine would fall over to the port side when turning a right hand screw propeller. In other words, the engine tends to revolve round the shaft in the opposite direction to the propeller, and is only prevented by its attachment to the ship's hull. It is thus proved that a ship takes a slight list to port whenever she is being propelled by a right-handed single screw and vice versa. We can easily imagine that a "cigar ship" might thus revolve completely round her own shaft when being propelled. And in the Whitehead torpedo this would actually occur were it not that the torpedo is driven by two screws attached to concentric shafts, right and left handed, each of which takes the reaction of the other (also in propellers for air ships). Now, it is in the reaction of the turning- effort exerted upon the shaft that vibration may, and indeed must, arise if the turning effort is not uniform, which it never is absolutely. But in marine engines the vibration due to this cause is probably- inappreciable. From considerations such as these, when carefully thought out on mechanical principles, we arrive at the conclusion that as regards the pressure of steam and the power exerted by the engine, every- thing is fairly balanced with the work it has to. do, and if only the- moving parts could be made without weight there would be no need. THE BALANCING OF A MARINE ENGINE. 117 for any further balance. It is the inertia of the moving parts which •destroys the otherwise perfect balance of the machinery in most ■eng-ines. Taking up next the second cause of vibration — centrifugal force in the shaft itself — it is necessary to distinguish carefully between a static balance and a dynamic balance. A very simple experiment will make this clear. Let us' turn up a wood spindle and mount it between V centres upon a board, as in figure 30. Next bore two holes, one near each end and exactly opposite each other radially ; fit convenient set-screws in these holes and adjust them so that the spindle, although free to turn, will remain at rest in any position. Fig- 3°' We have now a crude model of a two-throw crank shaft, with -cranks at 180 degs., and the model is balanced statically, as the real shaft would be in the lathe. Now wind a piece of string round the spindle and spin it by this means, when it will be found to make the board vibrate most violently, showing that there is no balance -when revolving — no dynamic balance. In order to prove that the •vibration is not due to imperfect adjustment of the weights, it is as -well to shift one screw as B to a position Bj, exactly opposite to A and at the same end of the spindle. The balance when revolving will now be found perfect. The explanation is very simple. A revolving weight develops a centrifugal force which acts, as its name implies, from the centre Il8 THE BALANCING OF A MARINE ENGINE. outwards, hence it is always changing the direction of its pull as it rerolves, and so causes the observed vibration. Now a single forc& can only be counteracted by an equal force acting along the same line in space, like the tension of the rope in a "tug-of-war," or like two walking sticks pushing at each other, point against point. Well, it is quite obvious that the set-screws A and B, in figure 30,. are not opposite each other in this sense, although they are opposite- each other radially in the end view of the spindle. Indeed, the result of centrifugal force in such a shaft is simply what in. mechanics is called a "couple," viz., two parallel and equal forces acting upon a body at a certain distance from each other, called the "arm" of the couple. This action upon a shaft we may call a "rocking action," and we may have to allude to it again. /«44- JL 2 5 Fig- 31- It is for a similar reason that a three-throw crank shaft, with cranks at 120 degs., is not a balanced shaft dynamically, although it may be perfectly balanced statically when swung between the lathe centres. (The model just described can be readily fitted with set-screws to test the truth of what is now being stated, which is- more convincing than any amount of theory.) The least number of cranks of equal weight which make a dynamic balance is four, and then only as arranged in figure 31, all at 180 degs., which is the- shaft used in the four-cylinder motor car and petrol launch. Six cranks also make a perfect balance if placed in pairs at 180 degs., I and 4 being opposite, 2 and 5 being opposite, and 3 and 6 being- opposite, figure 32. This, however, is not the usual arrangement for a six-crank shaft for a marine engine or motor car. It is usual to have i and 6 on the dead centre together, 2 and 5 at 1 20 degs. on one side, and 3 and 4 at 120 degs. on the other side. The arrangement makes a perfect balance. THE BALANCING OF A, MARINE ENGINE. 119 But it is safer to try the model to make sure that the arrange- ment is correct. Owing to the absence of a natural balance in ordinary marine engine shafts, it is necessary, if they are to be balanced, to fijf; weights upon the crank webs. Each crank requires a weight upon each web or we get the "rocking action." A three-throw shaft can be balanced perfectly by weights upon one web of each of the number i and 3 cranks alone if the position is carefully worked out and the weights are smaller in that case, but as a set off a certain amount of bending strain is induced in the shaft. We come next to the fourth cause of vibration. Fig. 32- It is the reciprocating parts of an engine — the piston, piston rod, and connecting rod — which cause the most trouble when unbalanced. The need for balancing these parts becomes evident when we consider that the rapid motion of the piston and rods at half stroke has to be arrested by the time the stroke is finished. Now whatever it may be that arrests this movement — whether cushion steam in the cylinder or the crank pin of the shaft — the framing of the engine has to receive the shock just as an anvil has to receive and absorb the shock due to the descent of a steam hammer. Not that the shock is as violent as that of a steam hammer, for the motion of a piston is arrested more gradually. But it is shock all the same and causes a more or less severe vibration up and down in a vertical engine, which might result in an engine actually jumping up when the piston reached the top of its stroke if it were not for the holding-down bolts. 120 THE BALANCING OF A MARINE ENGINE. Now, just to show one arrangement of cylinders and cranks which would result in a natural balance, very nearly in all respects perfect, figure 33, is drawn. In this engine no balance weights are required. We have here four cylinders and four cranks; the cylinders are in pairs ; A and B form one pair, placed on opposite sides of the shaft, with their cranks at 180 degs., C and D are another similar pair. The two cranks of B and C are pointing in the same direction and also the cranks of A and D. All that is necessary to make this engine balanced is to ensure that the valve Fig. 33- gears of the engines are symmetrically placed with respect to the middle line like the cranks, and that the reciprocating parts of each engine are of the same weight. The points to notice are, first, that ■ the crank shaft is balanced of itself, both statically and dj'namically as already explained. And secondly, that notwithstanding the obliquity of the connecting rods and eccentric rods, the reciprocating parts upon opposite sides of the shaft are at every instant moving ■with the same velocity, and with the same acceleration of velocity; THE BALANCING OF A MARINE ENGINE. 121 for the pistons will pass corresponding positions in their respective -cylinders at absolutely the same instant. But these equal velocities and accelerations are in exactly opposite directions. And since the mass of the weights moving from right to left is exactly equal to the mass moving from left to right, the shock that would otherwise tend to move the frame of the engine one way is counteracted by an ■equal tendency to move the frame the other way. To get a clearer notion of what has been done by the arrange- ,ment just described, we may consider the following modification. Let there be only one pair of engine, A and B, instead of both pairs (one cylinder upon each side of the shaft) ; and in this case too, let 1;he weight of the corresponding parts of each engine be exactly ■equal. We have now as before, equal weights moving with equal velocities and accelerations in opposite directions. It may, there- fore, be asked, why is not the balance as good as with both pairs? The answer is because the cranks cannot be placed opposite each ■other in plan as well as in elevation. Looking at the plan of the engine, a shock in the line Aa cannot be counteracted perfectly by Sl shock in the line Bb without causing a rocking action upon the frame, due to the distance between the crank centres. But by using four cranks, as sketched, this rocking action is counteracted by the second pair. There is, however, one defect in the original plan, which is -apparent when we consider the movement of parts in an up and .doTiin direction. The shaft is perfectly balanced for centrifugal force and need not be again considered, but the connecting rods have a swinging motion up and down, which is arrested and started again twice in a revolution. In the position sketched, revolving from right to left, the rods are approaching the limit of their swing and, therefore, are being checked in their upward and downward motions respectively. Now although there is the same momentum up, as down, and, therefore, the same shock received by the engine frame, these two effects cannot perfectly counteract each other owing to the distance (x y) between their lines of action. The two forces form, in fact, a couple in a vertical plane and the result is a slight tendency to make the frame rock in a vertical plane. 122 THE BALANCING OF A MARINE ENGINF. It will be noticed, too, that the particular effect just considered is much greater in the position shown in sketch than it will be in iinpther half revolution, when the connecting- rods overlap each other, thus bringing their centres of gravity nearly one over the other. The "period" of this vibration would, therefore, be chiefly once every revolution instead of twice. Referring again to the balanced engine of figure 33, let us see what can be done to make something like a marine engine of it and still keep an approximate balance. In the first place, then, we cannot A ■* 33 fis- 34. have cylinders and connecting rods below the shaft as well as above, so when we make an inverted marine engine of it, we will swing all the four cylinders up until they are directly over the shaft, figure 34. The balance of the reciprocating parts is now, unfortunately, not sa perfect as before, and the reason is not difficult to see. If we keep the cranks at the same angles, the sliaft is still balanced as well as before, and the reciprocating parts are also still moving in each pair of cylinders in opposite directions at every instant. This is not th& difficulty. The reason that the balance is worse is because the acceleration of velocity is now different on the up and down strokes of each pair of engines. THE BALANCING OF A MARINE ENGINE. 12^ It is well known that the obliquity of the connecting rod makes- the acceleration of the piston's motion much greater at the top' centre than at the bottom. With connecting rods equal to 4 cranks- in length, the ratio is about 5 to 3, hence the stopping and starting of, say, the B engine at the top centre, causes a shock nearly twicfr as great (5 to 3) as th,e simultaneous shock due to the A engine at the bottom centre. These shocks do not neutralize each other, therefore, as before ; and as well as a rocking moment due to A and B there is an unbalanced vertical force as these two cranks- approach and leave the dead centres. Of course, if the cranks of the C and D cylinders could be allowed to come on to their dead centres at the same instant as A and B, the rocking moment would. be neutralized. But other considerations, such as the necessity for handiness and a more uniform turning effect, make it unadvisable that alt four cranks should be on their dead centres together. The defective balance rocking moment cannot, therefore, be altogether avoided in such an engine without sacrificing more important functions. Something, however, can be done by setting the cranks at certain angles with each other to suit the relative weights of the moving parts of the several engines. This leads to the Yarrow-Schlick- Tweedy system of balancing. The essential points in this system are, ist, the engine must have at least four cranks; 2nd, counting- from forward, the first and second cranks are nearly at 1 80 degs. , likewise the third and fourth cranks ; 3rd, the two middle cranks- are not far from 90 degs. apart; 4th, no balance weights are used,, but an approximate balance is obtained by the nice adjustment of the positions of the four cranks. Another system, which is also used for marine engines, and ■which has the advantage of being equally applicable to i, 2, 3, or 4-crank engines, may now be mentioned. In its crudest form this system consists in fixing to the crank webs of each engine weights- which would balance statically the entire weight of the reciprocating- parts of that engine if hung from the crank pin. Thus, in figure 35, Wa = Pb. The action of such a balance weight is as follows: — With respect to motion in a -vertical direction, the balance weight is- 124 THE BALANCING OF A MARINE ENGINE. -always moving in the opposite direction to the piston, and since the motion of the piston is controlled by the circular motion of the •crank pin, and the balance weight is fixed opposite the crank pin, we may say, that as far as vertical movement is concerned (and neglecting the obliquity of the connecting rod), the shocks, as the ■dead centres are approached, are neutralizec". When, however, we consider the horizontal, side to side, swing ■of these huge balance weights, there is nothing but the bottom end of the connecting rod and crank webs to counteract their inertia, and prevent a horizontal vibration being set up almost as bad as the vertical vibration which they are intended to cure. A compromise, however, is often made by only attempting to balance one half the reciprocating weights of the engine, over and above the parts which revolve with the crank pin. The engine then runs with an ■over-balance horizontally, but with a. very much less vertical vibration than if not balanced at all. Locomotive engines are balanced upon the same principle, the balance weights' being placed inside the rims Fig. 35. of the driving wheels, and the centri- fugal force of these balance weights causes excessive hammering upon the rails, which cannot be avoided. As regards the slide valves and their gear, of course, a similar method could be adopted, as with the main engine. But it is inconvenient to fix balance weights upon eccentrics, neither is it necessary. By altering the position and amount of two of the balance weights upon different cranks, the same effect can be ■obtained, as if balance weight were fixed to the eccentric itself. THE BALANCING OF A MARINE ENGINE. 125 Neither is it necessary to actually fix weig-hts to the centre crank of i a three-crank engine, for the weights upon the No. 1 and 3, cranks: can be adjusted to balance the centre one, with the additional advantage that the balance weights are then of less amount than if each engine had been treated separately. In order to demonstrate the principles discussed in the present essay, the writer constructed the vibration model illustrated in figures 36 and 37. It consists of a shaft with six cranks which can> be set to any angle ; each crank can be fitted with balance weights, easily detached. The crossheads can be loaded to represent weights 126 THE BALANCING OF A MARINE ENGINE. •of pistons and rods, etc., and the model is driven by an electric rnotor and counter shaft. Fig- 37- FART SOLUTION OF QUESTIONS IN MENSURATION ETC. 1. — Mensuration Questions. 1. The sides of a rectangular tank measure 10 feet, 8 feet^ and 6 feet respectively. It is required to give the dimen- sions of a tank of the same proportions, but exactly one -half the capacity. The answer to the above is most readily obtained by dividing- each given dimension by the cube root of 2. But in order to prove the rule Let X = the greatest dimension of the proposed tank. then — ^ = next dimension; and — j*; = least dimension. lO lO Content = x y. — x x — x = i_ a;* ID lO loo which is required to be cu. ft. • 48^_iox8x6. ^^ _ lo . 3_iooo , , X^ f ■ i X ■ lOO 2 lOO 2 2 X = ^J 500 or ^ = 10 -=- ^J2 The cube root of 2 = i "26 nearly. The dimensions work out to Tg2>Ji 6"35, and 4-762 feet. — Ans. 2. Find the dimensions of a tank holding only one-fourth the one given in the previous example. 6-3> 5 "04, 2,lS.— Ans. 3. A straightedge, 1 foot long, laid inside a furnace tube at right angles to the length, is found to be 1 inch from the furnace plate at the middle of its length. Find the diameter of the furnace. Put X for the distance from the straightedge to the top of the furnace tube. Then by the property of a circle, figure 38. I30 MENSURATION QUESTIONS. :r X I = 6 X 6. AT = 36 and the diameter of the furnace = 37 in. — Ans. i. A boiler is 10 feet diameter, and a line drawn across one end forming a chord to the circle measures 6 feet. How far is the line from the centre of the boiler end ? 4 feet. — Ans. Fig. 38. 5. The chord of an arc of the inside circle of a fly-wheel measures 48 in. The versine of one half the arc measures 12 in. The thickness of the rim is 5 in. Find the. outside diameter. Inside diam. 60 in. ; outside diam. 70 in. — Atis. 6. A cylindrical boiler is 10 ft. 6 in. long and 13 ft. diam. The water level is 3 ft. i in. from the crown. Find the area of water level. II ft. 4'2 in. X 10 tt. 6. in. 7. Draw a semi-circle with a 5 in. radius. From one extremity of the diameter draw a chord making an angle of 60 degs. with the diameter. Complete the triangle in the semi- circle and calculate its height. (Figure 39.) The angle ACB being an angle in a semi-circle is a right angle, therefore angle CB A = 30 degs. Such a triangle has its shortest side exactly one half its AB longest side, hence AC = — = 5 in. 2 Fig- 39- Also the small triangle ACP has the same property, hence AP AC = — = 2'5. The perpendicular CP can now be found by squaring AC and AP. 4'33 in. — Afis. MENSURATION QUESTIONS. I3I S. The altitude of a ship's f annel measured from a horizontal line level with the deck is 30 degs., and 50 feet farther away from the ship the altitude is 15 degs. What is the height of the funnel above the deck ? (Figure 40.) The angles given here are such that the solution is possible without tables. Since .angle ABF = 30 degs., angle CBF=i5o degs., and as the three angles of the tri- angle CBF must amount to 180, the angle CFB = i5. Therefore the triangle is isosceles and BF = BC which is 50 ft. Again in the right-angled triangle ABF, one angle being 30 degs., the other angle = 60 degs. The triangle is just the half of an equilateral triangle and AF = -|- BF. The height of the funnel is, therefore, 25 ft. — Ans. 9. Find the ratio between the measurement of a hexagonal nut over the angles and the measurement over the flats. I to -866. — Ans. iO. If the angle between the shank of a spanner and the jaws is any of the following angles, the spanner by turning over will move the nut only 30 degs. at a time. The angles are 15 degs., 45 degs., 75 degs. Prove this. 11. Calculate the steam space in a cylindrical boiler 15 ft. diameter and 18 ft. long (inside measure) if the distance from water level to crown of boiler is i ft. 6 in., and the arch of the crown plate = 17"3 ft. in length, (Figure 41.) The principle to adopt here is to subtract the area of the tri- angle ACB from the area of the sector AVBC, thus leaving the .^e.ment .AVBD. Now DC = 3 ft. and CB = 7-5 ft. and DB by 47th prop, will be found = 6-87. The area of triangle ABC =^6-87 x 3 = 20-62 sq. ft. 132 MENSURATION QUESTIONS. For the sector we have the length of the arc AVB = 1 7 "3 and as the whole circumference = 47 • 1 24 ft. the area of the 1 7 "3 .7-3 •-. sector IS of the area of the whole 47-124 circle, which works out to 64'875 sq. ft. Subtracting the triangle from the sector gives us 64 '875 - 2o'62 = 44'2S5 area of segment. Therefore steam space = 44*255 X i8=7g6'59 cu..ft. — Ans. Fig. 41. Nbie. — In the above example it is not necessary to be given the- length of the arch of the boiler crown, for it would be easy to find the angle of the sector from the other measures, but it would need Trigonometrical Tables to work this out. 12. In a circle of diam. 2 in., draw the largest equilateral triangle and find: — (a) The length of a side; (b) The area of the triangle ; (c) The distance of the centre of gravity from one side; (d) Bisect the triangle by a^ line parallel to one side. In figure 42, ACB is the equilateral triangle, and O the centre of the circle. OA, OB, and OC are each = i in. Now it is readily seen that AOP is a right - angled triangle, with angles 30 degs. and 60 degs., respectively, hence OP = -5andAP2^i2--52=-75 AP = -866, AB = 1732 in. i OA / \ ^ \ y^>\- 'S&& > p y Fig. 42. Next we must note that CP = I "5 and since area of equi- lateral triangle = AP X CP, area= •866 x i"5= i'2g9 sq. in. centre of gravity is in the point O, "5 in. from each side. The To bisect the triangle we must draw a line as XY such that the- triangle cut off, viz., CXY shall be one-half the whole triangle. MENSURATION QUESTIONS. 133 Now the areas of similar figures are proportional to the squares of their corresponding dimensions. We take the perpendicular height CP of the whole triangle and the height CZ of the triangle cut off and square them, which must give the ratio of the whole to the half, thus : — The whole : the half : : CP^ : CZ^ i: i :: {isf : CZ^ This works out to CZ = i "0606 and ZP = i "5 - i 'oS = '44 in. Side, 1732 in.; Area, i'299 sq. in.; c. g. "5 in. from each side; bisection, '44 in. from side. — Ans. 13. Find the weight of a triangular piece of boiler plate 1| in. thick, if the length of the sides are respectively 12 feet, 11 feet, and 7 ft. The area of a triangle may be found by the following rule : — The sides being named a, b, and c, and half the sum of the sides being s. Area = Js {s- a) {s -b) (s- c) The area being found to be 38 square feet nearly, the weight of the plate works out to about 1900 lbs. — Ans. a. A triangular plate of iron has one angle=100 degs., another angle = iS degs., and the side between them != 20 inches. Find the weight of the plate if its thickness is 1 in. B The perpendicular BD of the triangle, figure 43, is readily found from the property of the right-angled triangle BDC, for angle C being 45 degs. , angle B is also 45 degs. and BD = DC. .-. 2 BD2 = BC2=400. BD= ^400^ 2=14-14 in. But to find the length of the base AC we must use a principle in trigonometry, viz. : — Fig. 43- 134 MENSURATION QUESTIONS. The sides of any triangle are proportional to the sines of ther opposite angles. (If any angle is over 90 degs. subtract from iSo- degs.) The three angles of the triangle being equal to 180 degs. and B and C making 145 degs., there is only left 35 degs. for angle A» The proportion runs : — Sine A : Sine B : : BC : AC Sine 35 degs. : Sine 100 degs. : : 20 : AC •5736 : -9848 : : 20 : AC from which AC is found = 34*34 in. The area of the triangle is AC X BD_34*34 X 14' 14, equal to 2 2 measure of the i in. plate. ■242 "78 which is also the cubic 2 A 2 *'70 The weight is, therefore, — / =67 -44 lbs. — Ans. 3-6 15. A piece of steel plate 1 in. thick, in the form of an equilateral triangle, weighs 1470 lbs., find the length of each side. 9 ft. yX ir -Ans. 16. A piece of sheet metal, in the form of a trapezium, has the length of its parallel sides 1^ in. and 10 in. respec- tively and the perpendicalar distance between them 8 in. Find its area and the distance of the centre of gravity from each parallel side. U 10 ->i^ — 4- -"— > A I P ' Let ABDC, figure 44, be the trapezium. The area of the figure presents no diffi- culty, for area = —^ — x 8 2 = 96 square inches. But to find the position of the centre of gravity let us draw DP parallel to CA, so as to divide the figure into a parallelogram and a triangle. Fig. 44. .>ID MENSURATION QUESTIONS. 135 We can now take moments about the line CD as an axis. The area of the parallelogram APDC is= 10 in. x 8 in. =80 and its c. g. being 4 in. above the axis, the moment 13 = 80x4 = 320. Also the area of the triangle DBF = i^= 16 and its c. g. being f of 8 from the axis, the moment = 16 x (f x 8) = 85 '33. Adding the two moments together and dividing by the total area, height of c. g. _ 320+85 "33 96 !=4"222 m. The centre of gravity is, therefore, 4f in. from the short side and 3!^ in. from the long side. 17. Given length of connecting rod ii ft., stroke 6 ft. Find angle between crank and rod when piston has moved '6 of the stroke. Also find how far piston is below half stroke when crank is horizontal. Taking the second half of the problem first, ■we have, in figure 45, a right-angled triangle in which AB= 14 and BC = 3. By the 47th prop. AC= VI?^=i3-67- Now at half stroke the distance of the cross- head A from the shaft centre C, is equal to the length of the connecting rod, 14 ft. It results that the present position is 14- 13-67= -33 feet below the half stroke, or more accurately 4 -08 inches. .^.jiJitrokg) Fig. 45- Again, making use of the same figure for the first part of the question, the distance AC is evidently 14 -(Jg- of 6), for the crosshead is to be one-tenth of the stroke below the half stroke /. AC= 13-4 feet. The triangle ABC is not right-angled in this position and to find angle B we must use a trigonometrical formula. A useful one is: — Cosine B = BA2+BC2-I-AC2 2 BAxHC t36 Cosine B MENSURATION QUESTIONS 14^ + 3^- (13-4)^ =.3028 2 X 14 X 3 By the tables the angle = 72 degs., 22 min. 18. A fragment of an exploded boiler is thrown up to a height of 55 feet above the ground. The angle at the instant of projection is 20 degs. from the yertioal. Required the horizontal distance covered. Apart from the influence of the air resistance the path of a projectile is a parabola as AVB, in figure 46. And it is a property of a para- bolic curve that a tangent to the curve at any point (as AC) cuts the axis of the parabola at a point C such that VC = VD. Applying this principle to the example, we have A as the seat of the explosion and AC the first direction of the fragment, which is immediately deflected by gravity and constrained to follow the parabolic path AVB. The highest point reached is V, and, there- fore, VD= 55 ft. But since VD=VC, we know that CD = no. In the triangle ACD the angle A being 70 degs., and CD no. AD = no X cotan. 70 degs. AD = no X "364 = 40 "04. The whole distance AB is, therefore, 8o'o8 feet. — Ans. 19. A cylindrical boiler 15 ft. diam. and 16 ft. long. If 30 per cent, of the whole capacity is occupied by furnaces, etc.; and 62 per cent, of the water space is steam space; how much water is there in the boiler? The total capacity = 2463 cu. ft. .'. furnaces, etc., occupy 738"9 cu. ft. Now if .» = water space, "62 ^ = steam space. Adding all together 738'9-f-^+ "62 x is the whole capacity = 2463 from which jt;=io64'2 cu. ft. — A7ts. MENSURATION QUESTIONS. 137 20. Some combustion chamber stays are pitched 8^ inches apart between centres. The nuts measure 2| inches across the flats. Find the area of plate exposed between each four stays. The nut occupies 5 '41 2 sq. in. Area left = 65"838 sq in. — Ans. 21. A block of iron 20 in. diameter, cylindrical, 12 in. high, but with the top corner rounded off all round to a 6- in. radius. Find cubic inches in block. -/3'Ocf ■ 20 Fig. 46a. 7^ It is evident, from figure 46a, that the block may be cut up into three portions, viz., ist, a cylindrical portion 20 in. diam. and 6 in. high; 2nd, a ■cylindrical portion 8 in. diam. and 6 in. high ; and 3rd, a ring having a section the form of a quadrant of a circle. Now the volume of a ring is found by multiplying the sectional area by the circumference measured at the ■centre of gravity of the section, and it is proved in Part III. that the e.g. of a quadrant is at "424 of the radius from either side. The •diameter of the ring for the purpose is, therefore, 8 in. +(-424 of 6 in. X 2)= 13 '088 in., and the circumference = i3'o88 x tt. Volume of ring = area of quadrant x circum. 6^ X — ) X 1 3 -088 X TT The total volume of the block works out to i8S4"95+30i"59-|- Ii62'5 = 3349 cu. inchej. 3.— Specific Gravity and Weights. When we say that the specific gravity of wrought iron is 7 7,, we mean that the weight of wrought iron is 77 times the weight of an equal bulk of pure water at some standard temperature. It is generally more useful to know the actual weight of a cubic inch or of a cubic foot of the material we are dealing with, but specific gravities are important because in original investigations it is always easier to covipare two things than it is to obtain absolute measurements. For example, in order to find directly the weight of a cubic inch of iron, we should have to measure very accurately the size of a piece of iron which was of some regular shape so as to be able to calcalute its volume, and after very accurate weighing with chemists' weights, we should divide the weight of the sample by the number of cubic inches in its volume and so obtain .the weight of a cubic inch. The specific gravity is obtained much more readily as follows : — The sample (which may be of any form whatever, so long as it is not hollow) is first weighed in a chemist's balance in the ordinary way. This is called its weight in air. It is then hung from one arm of the balance by a fine thread and allowed to immerse itself in a vessel of distilled water at the temperature of 39 degs., Fah. (at this temperature water is at its greatest density and expands whether heated or cooled, but very slightly indeed, for a degree or two). The sample is now weighed whilst entirely immersed in the water. This weight is less than the former, due to the buoyancy of the water. The actual difference betnoeen the "weight in air and the weight in water gives the exact weight of an equal bulk of water to the sample. To find the specific gravity we must divide the greater of the two weights by the difference between them. SPECIFIC GRAVITY AND WEIGHTS, 1 3^/ 1. A piece of iron was found to weigh 1521 grains in air and 1324'5 grains in water. What was its specific gravity? Weight of displaced water= 1521 - i324'5= i96'5 Sp. gr. = 1521^196-5 ., =774.— ^»^. 2. A piece of metal weighing 652 grains in air is tied to a piece of cork which weighs 56 grains in air. The two together weigh il2 grains in water and the metal alone weighs 587 grains in water. What is the specific gravity of the metal and of the cork? Firstly, take the metal alone: — Displaced water = 652 - 587 = 65. Sp. gr. of metal = 652 -4-65 = io'03. Secondly, take both together : — Displaced water = 708 -412 = 296. Now subtract water displaced by metal, viz., 65, and we getr Water displaced by cork = 296- 65 = 231, but cork weighs in air 56 grains. .'. Specific gravity = 56 -^ 23 1 ,, = '2424. — Ans. 3. To find the sp. gr. of a lump of tallow which would not sink in water by itself an iron bolt was attached to it, and it was found that the combined weigat in water was 18 oz. The tallow weighed 32 oz. and the iron bolt 23 oz. in air. If the sp. gr. of the iron is 7*7 what is the sp. gr. of the tallow ? Taking both together we find the displaced water = 55 - 18 = 37 oz. and since the iron is 77 times heavier than water, the bolt will displace 23-4-77 = 2 '987 oz. Subtracting this from the total dis- placement leaves 34 '013 for the displacement of the tallow; its sp.. gr. is thus 32^34-013 = -94 approx.— ^M. 140 SPECIFIC GRAVITY AND WEIGHTS. -4. A piece of wood and a piece of metal weigh 20 lbs. and iO lbs. respectively in air. The two together weigh 28 lbs. in water. The metal alone weighs 36 lbs. in water. Find the sp. gr. of the wood and of the metal. Wood '714, metal 10. — Ans. Specific Gravity of Liquids. Remembering that specific gravity always means the weight of a body compared with an equal bulk of pure water, it is obvious that a liquid will lend itself very readily to such a comparison, inasmuch as a definite volume of any liquid can be measured by the simple operation of filling a bottle with it. The specific gravity bottle is a small bottle holding about two or three ozs. of water, •fitted with a glass stopper having a small hole through the centre. It is usual to adjust the bottle and stopper so that it will hold exactly 1000 grains of distilled water at 39 degs. or 60 degs. Fab.; but this is not essential, it merely saves a double weighing and any stoppered bottle may be used for determining the specific gravity of a liquid. ■For example, suppose the sp. gr. of a sample of sea water is required. Proceed as follows: — Let the sample of sea water and also a vessel of distilled water remain together in the room until they have acquired an equal •temperature (39 degs. preferred). First place the empty sp. gr. bottle in one pan of the balance and counterpoise it with dry sand •or shot. Next fill the bottle with distilled water, replace it on the scale and weigh it. Suppose that it weighs 1520 grains. Now ■empty out the distilled water and fill the bottle with sea water, place it on the scale and weigh again. Suppose the weight to be 1561 grains. The specific gravity = -5 — = i -027. 1520 Instead of counterpoising the empty bottle as above described, the empty bottle may be weighed in the balance, and its weight subtracted from the weight of the full bottle each time. SPECIFIC GRAVITY AND WEIGHTS. 141" 6. A sp. gr. bottle weighs when empty 509 grains. When filled with distilled water it weighs 1412 gr. and when filled with petrol it weighs 1205 gr. What is the sp.. gr. of the petrol? ■793-— ^'«- 7. The density of oxygen gas is 1-10565 times the density of air at the same temperature. At what temperature will oxygen have the same density as air at 32 degs. F. The law of Charles states that all gases expand by heat in- proportion to their absolute temperature. Therefore under the same pressure the density will decrease in the inverse ratio of their temperatures. ■10565 X : (460 + 32) X — 544 degs. absolute or 84 Fah. — Ans. 8. A salinometer has a space equal to 6 in. between the fresh water mark and the i/32 mark. Find the correct position for the 1/32 mark. In solving this problem it is convenient to consider the salinometer as a uniform stem with no bulb, figure 47, but a capacity equal to it. We proceed to find the total length of the instrument (L). The sp. gr. of sea water at ^^ = i '028 and the sp. gr. at ^ = 1 • II 2. Now the weight of displaced water is the same, whatever its density, for it equals the weight of the instrument, and since with a uniform section for the instrument the displacement is proportional to the length Lx I =(L-6) X I-II2 from which L = 59'5 in. ^, whose distance from zero we call x. :h- -X I M 32 A similar equation will give us the position for Fig. 47. 59-5 X I =(S9'5-^)x I '028 from which ^=i'62 in. This proves (what is always noticed on a hydrometer stem) that the- divisions are smaller towards the bulb. Equal divisions would.. 142 SPECIFIC GRAVITY AND WEIGHTS. result in the ^V mark being one-fourth of the distance to ^^ or 1-5 9. A steamer's displacement is 5000 tons. The area of water line is 10,300 sq. ft. and she draws 19 ft. in fresh water. Required her dri.ft in sea water of specific gravity 1-028. Assuming that 36 cu. ft. of fresh water weighs i ton, the displacement of the vessel in cu. ft. of fresh water is 5000 x 36 = 180,000 cu. ft. And since the wsighi of water displaced is the same, whether the vessel is floating in fresh water or in salt, it follows that the product of volume by specific gravity is constant, hence, cu. ft. sea water X I '028= 180,000 X I. r^ . 180,000 CU. It. sea water = = 17?, 100. I -028 '^ The difference between 180,000 and 175,100 is the amount which the vessel lifts at the water line and when divided by the area of water line gives the difference in draft, viz. : — -3^ — = '466 feet = 5'6 in. 10500 The draft in sea water is 18 ft. 6-4 in. — Ans. 10. A steamer leaves a salt-water port, and, after burning 150 tons of coal, she anchors in a fresh-water harbour. The draught is found to be exactly the same. Find the original displacement of the steamer, having given that the sp. gr. of the sea water was 1-028 and of the fresh water 1-005. Let X be the original displacement in tons. The two specific gravities may be taken to mean that a cubic foot of sea water weighs 1028 ounces and a cubic foot of fresh water 1005 ounces. Now if the displacements are reckoned in ounces also, it is obvious that displacement in cubic feet will be obtained by dividing the ^number of ounces by 1028 in one case and by 1005 in the other. Hence — a. j- 1 , xx. 2240 x 16 cu. ft. displacement = ^ — Also „ „ ^(-v-i5o)x.g.2_4oxi6 1005 SPECIFIC GRAVITY AND WEIGHTS. 1 43 Since the draft is the same, the cu. ft. of displacements are equal. X _X— ISf) (after canceUing common factors). 1028 1005 1005 X = 1028 X - 154200 23 jc = 154200 and X = 6704 tons.^ — Ans. 11. Find the weight in pounds and kilograms of a pipe 60 in. long, the outside diameter being 15 cm. and the inside diameter 11 cm. Sp. gr of metal 7-1^. It will be best to work in the metric system here, and since a •c.m. = '3937 inches, length of pipe=6o-T- "3937= i52"4 cm. Volume = (i5^ - 11^) X — xi52'4. Volume = 12448 cu. cm. 4 Now a cu. cm. of water weighs exactly i gram in the metric system, but the sp. gr. of the pipe being 7"i4, each cu. cm. weighs .7'i4 grams, and the total weight in kilograms = 1244S x 7'i4-h 1000 = 88-88 kilos. To express the weight in English pounds we must multiply by .2"204, which gives i95'9 lbs. nearly. — Ans. 12. Find the weight of a hollow sphere 10 in. outside diameter and thickness ^ in., which is filled with liquid. Th e sp. gr. of the sphere = 78 ; the sp. gr. of the liquid = VS12. Note that the volume of a sphere = fi?^ x _. Also a cu. ft. of water 6 ■weighs 62*32 lbs., hence a cu. ft. of the metal weighs 62'32X7'8, ^nd a cu. ft. of the liquid weighs 62-32 x -8. Weight = 51 -04 lbs. — Ans. 13. The inside diameter of a hollow globe is 7/10 of the out- side diameter. The globe is of cast iron (sp. gr. 7) and weighs 100 lbs. Find its outside diameter. Let D = outside diam. in feet. ^^^ D = inside diam in feet. )3 ^ . . - = volume of metal. But smce weight of cu. ft. 6 of iron = 62-32 x 7 = 436-24 lbs. Volume of metal = 100 -i- 436-24, D3T-f7D 6 \ 10 / 144 SPECIFIC GRAVITY AND WEIGHTS. The equation is therefore — ■ 6 \ 10/ 6 436'24 \6 1000 6/ 436'24 This works out to D= "87 feet (approx.) or io'4 inches. — Ans, \i. A cast iron ball 6 in. diameter is coated with lead \ in> thick all over. What is the area of the surface of th& lead? Also what is the combined weight? Use '23 lbs> for a cu. inch of iron and -^ lbs. for a en. inch of lead. Note. — The surface of a sphere is equal to the rounded surface of a cylinder of the same diameter as the sphere, and heigfht equal to its diameter. Surface, 153 '93 sq. ins. Weight = 26-6+28'27 = 54'87 lbs. — A71S. 15. A metal pipe 3 metres long, 15 cm. ontside diameter^ and 11'2 cm. inside diam.; sp. gr. of metal 7*2. Find the weight in kilos and in pounds- (1 kilogram = 2-20'i6 lbs.) 168-9 kilos, 372"37 lbs. — Ans. 16. There is a dead-weighted safety valve 4 in. diam., loaded to 60 lbs. pressure. The valve and spindle weigh 16 lbs. The weights are 12 in. diameter with a 1^ in., hole and their total height is 25 in. One weight is lead, the rest cast iron. Find the thickness of the lead weight. Given iron = -26 lbs. per cu. inch; lead =4 lbs. per cu. inch. If we put X for the thickness of the lead weight, 24 - a; is th& total thickness of the iron weights. We have now simply to make an equation between the total pressure of the steam on the 4 in. valve, and the total amount of the load upon the valve, and the value of X is found to work out correctly by solving the equation. The equation, however, is rather complicated and it is perhaps easier to first find how much the load is deficient upon the supposition that all weights are of iron, thus: — Total weight (if all iron) works out to 694-69+ 16 = 710-69 lbs. And steam pressure on valve =753-96 lbs., showing a deficiency of 43-27 lbs. to be made up by changing: SPECIFIC GRAVITY AND WEIGHTS. 145 some iron into lead. Now for every cubic inch thus changed there is a gain of "4- '26= -14 lbs., dividing this into the deficiency gives us 309 cubic inches and dividing again by the area of weights gives us the thickness = 2 '77 inches. — Ans. 3- — Moments. If a bar of any material is first balanced upon a centre, axis, or fulcrum, so as to remain in a horizontal position, and weights are afterwards hung from the bar at various points in its length, we have the simplest application of the "principle of moments," sometimes called the "principle of the lever." If only two weights are acting, one at each side of the axis, as for example, 12 lbs. at 4 feet from the axis and 16 lbs. at 3 feet from the axis on the opposite side, the bar will still be balanced, because — 12 X 4= 16 X 3 Each of these products is called a moment and the bar is said to be in equilibrium because the moments are equal. If there are more than two weights, equilibrium will still be established, so long as the sum of the moments acting upon one side of the fulcrum is equal to the sum. of the moments on the other side. For example, if in addition to the 12 lbs. at 4 feet from the axis, there was 3 lbs. at 5 feet and 7 lbs. at 3 feet, all upon one side, the total moment would be— {i2X4)+(3X5)+(7X3)=84 To counteract this would require an equal moment upon the other side such as — (i6x3) + (9x4) = 84. It remains to add, that so long as only one weight or one distance is unknown, it can be found by making an equation of the moments upon opposite sides of the fulcrum. A more difficult application of the principle occurs when the bar or lever is not itself balanced upon its centre of gravity, but has its axis at some other point in its length. In this case the whole ■weight of the bar must be considered as acting at its centre of gravity. MOMENTS. 147 By multiplying the weight of the bar by the distance of its centre of gravity from the axis, we obtain a moment, which is an addition to the other moments and perfectly satisfies the conditions of equilibrium. Again, forces may be acting upon a lever in an upward direction as in a safety valve lever, due to the steam pressure under the valve. In such cases it is necessary to consider the direction 0/ ?otation due to each force, and equate the moments tending to rotate the lever "clockwise," to the moments tending to rotate the lever anti-clock- wise. For example, let the centre of gravity of a safety valve Jeve/ be 15 in. from the fulcrum ; its weight 20 lbs ; let the upward steam pressure upon the valve amount to 1000 lbs. and act upon the lever at 5 in. from the fulcrum; let the weight at the end of the lever be 25 in. from the fulcrum. To find the weight W, the several moments are — 20x15 moment of lever. 1000 X 5 moment of steam force. W X 25 moment of weight. It is evident that the last moment is assisted by the first. .". (W X 25)-J-(20X 15)= 1000 X 5 25 W = 5000 - 300 and W = 4700 -i- 25 = 1 88 lbs. Just one point more. The principle of moments is not restricted to moments taken about an actual working centre or axis, for so ■Jong as all the forces acting upon a body (including the reaction of its supports) are taken into consideration, any point whatever may be used as an axis. For example, in figure 51, let the weight of the beam be 400 lbs., which will put 200 lbs. upon each of the supports A and B. Let us take moments about the extreme right-, hand end of the beam. There is an upward force of B = 200 lbs., acting at a distance of 8 ft., assisted by a force A =200 lbs., acting at a distance of 28 ft. These tend to produce a clock-wise rotation (if the end is fixed). Now acting in the opposite direction there is -the weight of the beam, 400 lbs., whose centre of gravity is 18 feet from the assumed axis. . 148 MOMENTS. Equating the moments — (200 X 8) +(200 X 28) = 400 X 18 1 600 + 5600 = 7200 7200 = 7200 In a similar way it can be shown that the principle of moments is true about any other assumed axis. The example, No. 9, is a practical application of the same principle. 1. A bar of uniform section weighs 8 lbs. It balances on a point 7 in. from one end when 56 lbs. hangs from the same end, and 12 lbs. hangs from the other end. Find, the length of the bar. Put X for the total leng-th of the bar as in figure 48. Now the centre of gravity of the bar being in the middle of its length, its weight is re- 7 ->!< Jr — 7 • — i — «rF --*- acting at a distance = — - 7 -^'S- 48- © 2 from the fulcrum to the right. The moment of the beam is, there- fore, assisting the moment of the 1 2 lbs. hanging weight, and botb are counteracted by the moment of the 56 lbs. Equating these moments — f--7j x8+{x-y)y. 12 = 56x7 4;c-56+i2 ^-84 = 392 i6;«r=392 + 56+84 a; =33^ inches. 2, A deck beam is being lifted by a chain attached at a point 2 feet from the middle of the beam. A man weighing 168 lbs. stands at a point 3 feet from the shorter end of the beam and just makes the beam balance horizon- tally. The beam is 34 feet long. Find its weight. 1008 lbs. — Atis. MOMENTS. 149 3. A uniform bar balances about a point 12 inches from one end when 120 lbs. is hung from the same end. If now the point of support be shifted i inches nearer the centre it is found to balance with only 80 lbs. hanging from the same end. Find the length of the bar and its weight. If L is put for the length of the bar and W for its weight, two equations of moments may be written : — ist. 120X i2 = W X (■!■ L— 12) and. 8oxi6 = Wx(iL-i6) .-. 1440 = 1 W L-12W 1280 = 1 WL- 16 W Subtracting bottom from top, -^ W L goes out and 160 = 4 W. •'• W = 40 lbs. Then substituting this value of W in the first equation — 1440=40 X (-^ L- 12) 1440 = 20 L-480 20 L= 1920 L = 96 inches. <■ i. In a balance with unequal arms (a false balance) if a body W is balanced by A pounds when placed in one scale and by B pounds when placed in the other scale, prove that the correct weight of W is equal to VA x B, Let the lengths of the arms of the balance be, respectively, tn and n inches. Then by the principle of the lever Wx7w = Ax n and also Wx w=B xm. multiplying the corresponding sides of the equations. W^ xtnn = A'Bxtnn cancelling the factor m n occurring on both sides. W2 = ABand W= VAB^ ISO MOMENTS. A plate ABC triangular shape lies on a table, as in fignrift i9. It weighs 12 lbs., and the corner C overhangs the edge of the table a distance CS = SD. Find the force at C to upset the plate. Note that the centre of gravity of a triangle is at l/3rd the distance from the base ta the vertex. In figure 49 let CD be divided into 6 parts, then CS = 3 parts = SD and the eg-. = i part from the edge of the table. Taking moments about the edge of the table, 12 X I =x X 3 .'. ^ =4 lbs. 6. A circular plate 12 in. diam. weighs i lbs. and lies on a square table. The centre of the plate is 1 io. from the edge of table. Find the force required to upset the plate when applied at the outer edge. •8 lbs. — Ans. Weston's differential pulley of 2 in. and 2 l/8th in. radius. If 1200 lbs. is lifted and the efSciency is i8 per cent., find the pull upon the chain by the principle of moments. Neglecting efficiency and putting x for pull required, see figure 50. {x X 2^)+ (600 X ?) = 600 X 2i •»=35"3 lbs. And allowing for small efficiency — pull = 35 '3 X 48 = 73 '5 lbs. 6,00 \ 6ir' Find the relation between the power and the weight in Weston's pulley block having given diameters 8 in. and 8^ in., and efQciency 'i2. P _ I W 2772' -Ans. MOMENTS. 151 9. A uniform beam, 36 ft. long, is supported at points 8 ft. from each end. The beam is merely loaded by its own weight. Find the position of two points in the length of the beam where the bending moment is nil. (See figure 51.) Jfc • ->' C . 1 1 1 Let one position of no bending moment be at x distance from one end as in figure. Also let W be I ^ " B "• the weight of the beam per foot .■*-»*«-- io >*»-»: run. Load at A = ^ total =18 W. Fig. ^i. Taking moments about the position C we have 18 W x (jc — 8) upwards and x W x- downwards. 2 Equating — 18 Wx(^-8)=;«rWx- 2 x^ 18 ^- 144 = — or 2 x^-2€>x= -288 completing the square, etc. x-i8=±6 x = 2^ or 12 feet. These are the two positions of no bending moment measured from the end of beam. 4. — The Parallelogram of Forces. It is an obvious principle that if two forces are acting upon a body at a point, both in the same direction, the effect upon the body is the same as if a single force were acting equal to the sum of the other two forces. Also we may have two forces acting upon a body in exactly opposite directions — as in a steam cylinder where the back pressure is opposing the steam pressure — the effect now is equal to the difference of the two forces. When, however, two forces are acting neither exactly in the same direction nor exactly in the opposite direction, the principle to apply is "The Parallelogram of Forces." Let there be two forces acting upon a body at A. (Figure 52). Let one force act in the direction AB and the other in the direction AC. To construct the parallelogram of forces we must make the ■length of the line AB repre- sent the magnitude of that force to some scale, and the length of the line AC must be made to represent the magnitude of the other force to the same scale. Thus if the forces are 2 and 3 tons respectively, we can make AB = 2 in. and AC = 3 in. The parallelogram is now completed by drawing BD parallel to AC and CD parallel to AB. The diagonal AD is now measured (say = 4 in). AD is called the resultant of the two forces and it shows both in magnitude and direction the equivalent single force upon the body at A. THE PARALLELOGRAM OF FORCES. '53 Figure 53 shows a practical application of the principle under consideration. A piece of machinery is being lifted by a sling, the angle between the two legs of which =120 degs. Suppose the weight lifted is 4 tons. The vertical part AX evidently carries the entire weight, hence the tension in AX is equivalent to, or is the resultant of, the tensions on the two legs AB and AC. If we now produce XA downwards and make Ad = 4 in. and complete the parallelogram by drawing db parallel to AC and dc parallel to AB, we can measure the length of Ab or Ac and thus obtain the tension in those parts of the sling. It will be found that with 120 degs, as the angle, each leg of the sling carries 4 tons. Fig. 53 The Parallelograxa of Velocities* The principle of the parallelogram of velocities enables us to combine two velocities in the same way as we combine two forces. As an illustration, suppose a cricket ball is moving from one wicket to the other at a speed of 20 ft. per second. This is velocity No. I . Now suppose the ball is struck in such a manner that if it had been at rest when struck it would have been given a velocity of ID ft. a second at right angles to the line of the wickets. This is velocity No. 2. In figure 54, A is the position of the ball when hit, having been travelling as per dotted arrow. If allowed to continue untouched it will arrive at B, 20 feet further on, at the end of one second. Also if the ball were struck from A while at rest, it would arrive at C, 10 '54 THE PARALLELOGRAM OF FORCES. feet off the line, at the end of one second. By completing' the parallelogram, we find that the ball really arrives -at D, at the end of one second, having- travelled on the diagonal AD. Since AD measures Js°° ~ Fig- 54- 22-36, the resulting velocity is proved to be 22-36 feet per second'. 1. A derrick polo on board ship is 28 ft. long. A stay from the head is secured to the mast 18 ft. above the foot of the derrick, and the stay is 16 ft. long. When lift- ing 5 tons from the head of the derrick, what are the forces on the pole and stay and what is the bending moment on the mast? The head of the derrick A, in figure 55, is a point where three forces act, as shown by arrows. Complete the parallelogram ABCD by drawing CD parallel to BA. The ratio between the sides and the diagonal of this figure gives the ratio between the forces, hence: — AD : AB : : 5 tons : force on 18 : ' : 16 : : 5 tons : 4^ tons. Also AD : AC : : 5 tons : force on pole, 18 : 28 : : 5 tons : 7I tons. Fig. SS. For the bending moment on the mast we must measure the horizontal distance from the foot C to the line in which the weight hangs. This is found to be 15 feet. The bending moment is 5 x 15 = 75 ton feet. THE PARALLELOGRAM OF FORCES. 155 2, A ball welshing 100 lbs. hangs by a cord from a hook in a vertical smooth wall. The length of the cord from the hook to the eyebolt in the ball is 19 inches and the ball is ii inches diameter. Find the tension on the cord. (Figure 56). Complete the parallelogram as in figure 56, and calculate length CA by 47th prop. This line represents the weight of the ball, and CB the tension on the cord. 103-8 lbs. — Ans. A shaft weighing 6 tons is lifted by a chain sling. Each chain is 6-4 feet long, and the ends are secured to the shaft at points 8 ft. apart in a horizontal plane. Find the tension on the chains. Consider the point A where three forces meet, as shown in the figure 57. By completing the parallelogram ABDC, we see that AD represents the weight of 6 tons, and AB and AC the tensions on the two parts of the sling. Now AE= ^AB2-BE2= y' AB2- (^Y and AD = 2AE .'.2 . /AB2-('15)^AB::W:tension required WxAB Tension = Vab.-(b^)' This works out to 3*84 tons. — Ans. 156 THE PARALLELOGRAM OF FORCES. 4. Two forces of 2 and 3 lbs. respectiYcly act at an angle of 60 degs. to each other, find their resultant. In figure 58, the resul- tant is, of course, AD and if the figure is drawn to scale it can be measured off'. But the angle of 60 degs. happens to be one of the angles which can be dealt with arithmetically. Thus draw AE perpendicular to AC. Angle EAB = 30 degs. and KB = -|^AB = i. .-. AE= n/22-i2= 1732 ED = EB+BD= 1+3 = 4. In the triangle AED, AD = VaE^+ED^ which comes out 4-36 — Ans. 5. A tnnnel shaft weighing. 6 tons is lifted by two chains secured near the ends of the shaft and passing through an opening in the top of the tunnel. One chain makes an angle of 105 degs. and the other an angle of 110 degs. with the shaft which is horizontal. Find the tension on the chains to three decimal places. The first principle to apply in this problem is that when three forces keep a body in equilibrium the direc- tions of the three forces must all meet in the same point. Figure 59. By producing the directions of the two chains downwards imtil they meet in A this point is found. The third force, which is ofravity, must act vertically through A, thus proving that the centre of gravity of the shaft is at B. Now complete the parallelogram ACBD and we have AB representing the weight of Fig 59- THE PARALLELOGRAM OF FORCES. 157- 6 tons, and AC and AD the tensions on the chains. A measure- ment to scale would probably not be accurate enough in this- problem, so we will use trigonometry. First let us mark all the angles of the figure with their degrees. This should not be difficult, since the three angles of every triangle- must amount to 180 degs. Now in every triangle the sines of the angles are proportional to the sides opposite to them, and in the triangle ABD, AB = 6 tons. Sine 145 : Sine 20 : : 6 : AD. In the table of natural sines 145 degs. is entered as its supplement,. viz., 35 degs. •5735 •• "3420 : : 6 : AD AD = 3 "578 tons tension on right-hand chain. In the triangle ACB Sine 145 : Sine 15 : : 6 : AC •5735 : -2588 : : 6 : AC. AC = 2 •7076 tons tension on left-hand chain. An engine has a stroke of iO in. and a connecting rod of 70 in. If the steam is cut off at half stroke and the pressure on the piston is then 23 tons, what is the pressure on the guide? An approximate answer to this question is easily obtained if we assume the crank to be horizontal at half stroke, for drawing the parallelogram of the three forces acting at the cross head, as figure 60, we have the distance from cross head to shaft = 70 in. , which repre- sents the pressure on the piston and AB = CP = 20 in., representing the pressure on the guide. Then 70 : 20 : : 23 tons : x. \ Hence x = 20 X 23 70 6^ tons. Fig 60 ,t8 THE PARALLELOGRAM OF FORCES. The correct solution is as follow*^ :— A is the cross head and S the shaft. The cross being at half stroke, AS = 70 in. the same as AC the connecting rod. The parallelogram is now ABPC, in which AP the diagonal is less than 70 inches. To find the length of AP is a little troublesome and we shall require to use trigonometry. In the triangle ACS the cosine of angle A= — ^ ^C x AS — ^^^^^ usefuj property of every triangle) taking the given dimensions. Cos. A = 75i+7^!^^.959.. 2 X 70 X 70 AP u How in the right-angled triangle ACP Cos. A = ^, and as both 'iP AP J values of Cos. A must be the same -;-=,= "QSQ^- . .—— = '9592 ana AC 70 AP = 67-14 in. Also CP2 = AC2-AP2 which works out to CP= 19-804 in. =AB. The parallelogram of forces now applied gives us : — 67-14 : 19-804 : : 23 : 6-784 tons. J\roie. — ^The above method is applicable to any position of the ■piston on its stroke if we first find the distance from cross head to shaft, as in the next exercise. 7. Using the same data as in the previous example, find the pressure on the guide where the piston has moved i stroke from the top. 5-433 tons. — Aits. 8. A Vfeight of 6000 lbs. is lowered into a ship's hold by means of a span rigged from main-mast to mizzen-mast, which are 120 feet apart. The span is secured to the masts at an equal height on each, and when the load is hanging to the span 40 ft. from the mizzen-mast and 80 ft. from the main-mast, the dip of the span is 30 feet. Find the strain on each part of the span and the force pulling the masts together. Having sketched' the arrangement to scale, a graphic solution is readily obtained by setting up a distance representing 6000 lbs. on the vertical line of force and completing the parallelogram. The tension on the short part is 6650 lbs. and on the, long part 5680 lbs. THE PARALLELOGRAM OF FORCES. 159 The horizontal pull is obtained by makings the oblique pull upon either span a diagonal of another parallelogram with rectangular components. The pull = 5300 lbs. 9. A pair of shear legs, 60 ft. long, are 25 ft. apart at tlieir heels. The back stay is 85 ft. long. Find the forces on the poles and back stay, when lifting a 40-ton boiler, which overhangs 30 ft. * Fig. 61.' In figure 61, first find the vertical height AD when upright, •which = 58 -68. Next take a profile view in which (figure 62) AE = 85^ AD = 58-68, PP = 30. If the figure is drawn to scale arid the parallelogram ACDB drawn for the three forces at A, we shall have the proportion: — AB : AC : : 40 tons : x the force on the back stay ■which works out to 55 "3 tons. " From the same parallelogram we also get AB : AD : 40 : the force on AD which comes to 88-3 tons. But as this force is taken by the two shear poles it must be resolved, as shown in the first figure, that is 88"3 tons in direction AD is equal to 45 'i tons alorjgAB and AC. On each pole 4S'i tons, on back stay 55"3 tons. — Ans, , n ' l6o THE PARALLELOGRAM OF FORCES. 10. Two forces at an angle of 60 degs. have a resultant=2$ lbs. When the same two forces act at an angle of 90 degs. they have a resultant equal to 22 lbs. Find the two forces. Referring back to figure 58, we may consider AB to be one force = :» and AC to be the other =jy, when the angle is 60 degs. the resultant is AD. Draw AE perpendicular to AC, so that angle E AB = 30 degs. Then EB = - and ED = 5+^ Also, 2 2 AE2 = *2-^5y = 3ff Now in the triangle A ED, AD2 = AE2+ED2 which becomes 24^ 4 2^^=x^+xy+j/^ (l.) It is obvious that when the forces act at go degs. AD2 = AB2+BD2 or 22^ = x^+y^ (2.) Equations (i) and (2) solve the problem by well-known Algebra, jf = 4'34 lbs., j/=2i*58 lbs. — A?is. 11. A weight of iOO lbs. rests upon a horizontal plane. The co-efScient of friction is -5. Find the force which acting at an angle of 30 degs. to the horizontal will move the weight along the plane. The reason that the answer to this question is twi 200 lbs. (400 X . 5) is two-fold. First, an oblique force has not the same effect as a direct one in overcoming any resistance, and secondly, because the oblique force tends to lift the weight off the plane and so reduces the amount of friction to be overcome. When it is stated that the co-efficient of friction is •<;, it means that the friction is -5 of the pressure between the weight and the plane, whatever that pressure may happen to be. THE PARALLELOGRAM OF FORCESJ i6i • Let A, figure 63, be the position of the weig-ht and AB the direction of the oblique force. Set up-AC to represent the pressure between the weight and the plane, and also AD = -^ AC to represent the friction due to the pressure AC. Now when the weight is being moved in the direction AX, the plane is acting in two ways upon it. Fig. 63, First, it is pushing upwards with a force equal to the pressure of the weight downwards, represented by AC, and secondly, the friction of the plane is opposing the movement by a force exactly ^ of AC, represented by AD. These two forces have a resultant AF, whose direction is the same for any value we take for AC. It is usual to naine AF the "the reaction of the ptane. " The direction of the reaction can always be found, as above, if only the co-efficient of friction is known. F_ : _ __c y^ ( V iV' ^ \ " \ V \^ ^ ^ b ^ A H X The problem is now simplified to one of three forces acting at a point upon the weight. They are :^The force of gravity acting downwards, equal tp 400 lbs. =AK. The reactioq of the plane acting in the direction AF, and the required force AB. A parallelo- gram of forces is easily drawn by setting up 400 lbs. to K, and drawing KL parallel to AB and KM parallel tp AF., Bj/' measure- ment AM= 1 79 "2 lbs. which is the force required. \%^ A derrick pole, 30 ft, long, stepped 2X the foot of a mast and supported by a homontal ti? to V&& same mast. The head of derrick overhangs 24 ft. A single chain passeEf OYer pulley at the head, and leads down the pole to a winck, double-geared, pinions 12 teeth and. 10; wheels 72 and 86 teeth. Barrel 15 in. diameter. Handles 18 in. long. Find force on handles to lift 3 tons on the chain. The question so far is simply one of the "principle of work." If we suppose the barrel to make one revolution, the handle will L 102 THE PARALLELOGRAM OF FORCES; make — X — = 5 1 '6 revolutions. ' Now let P = force on handles; 12 lO , ,'i 'i i work put into machine = Px36 ;:;x5i"6; and work accomplished^, 3 X 2240 X 15 5r. Equating these, P comes out = 54'26 lbs. 13r , In the same derrick find tension on tie, compression on the pole, and bending moment on the mast. , , As a resolution offeree question, the tie will be found to carry 4 tons and the pole 5 tons, but we must not overlook the extra compression put on the pole by the chain being led down the same to the winch barrel. The actual compression is 5+3 = 8 tons. The bending moment on the mast = 3 x 24 = 72 ton-feet. II. In the same derrick the mast is hollow, 12 in. outside diameter and 6 in. inside diameter. Find the stress in the material. The rule for the moment of resistance of a hollow cantilever of round section is — (D* - d*) X Stress D X IO*2 Equating this to the bending moment in inch tons we have — (i2*x6*) X Stress > — = 3 X 24 X 12. 12 X 10"2 From this equation the stress works out to 5 •44 tons per sq. inch. 15. If a balance weight is fitted to the mast, consisting of a weight of 4 tons at 10 feet leverage. What will the stress amount to when lifting a load of 3 tons? 2'4i8 tons per sq. inch. — Ans, 5^^ — Work and Horse Power. 1. The force on a punch is 25 tons, which is exerted for one- half the plate thickness. The plate is ^ in. thick, and the holes are f in. diameter. If i holes are punched per minute, what horse power is being exerted? Also what is the shearing stress offered by the plate and the compression stress upon the punch? The force exerted being given, w$ have only to multiply by the distance for which the resistance is overcome to obtain the work done per hole. Using pounds and inches we have : — - Work done per hole = 56000 X "25 ,, ,, = 14000 inch lbs. Work per minute = -2 J =4666-6 ft. lbs. 12 H.P. =4666-6^. 33000 Next, to obtain the shearing stress, we must consider the area of the fractured surface, which is the circumference times the thick- ness ; and as all this surface is available to withstand the pressure of the punch. Shearing stress = — ^ = 47480 lbs. For the compression stress on the f-in. punch we also have :^ Stress = 2 = 127000 lbs. nly. 2. Assuming the shear strength of plates at 47000 lbs. per square inch. Find horse power required to punch l-in. holes in a |-in. plate at the rate of 10 holes per minute, the force being exerted for i the plate thickness. •524 H.P. — Ans. 164 WORK AND HORSE POWER. 8. A hammer weighing 5 tons falls 5 ft. on to a steel bloonn at a white heat and compresses it 3/16ths in. Find the force of the blow. If the accumulated work in the 5-ton hammer is all spent in. compressing the steel we can write the equation 60 in. X 5 = y\ in. x Force exerted. Force = 1600 tons. 4. A steam hammer falls 4-5 feet. Its weight is 3 tons and the steam acts during the down stroke with a total pressure of S'7 tons. If the mean resistance of the steel bloom is 950 tons, what amount will it be com- pressed by the blow? •498 in. — Ans. 5. A Prony brake is used for measuring the horse power- passing through a shaft. Sketch the apparatus and calculate H.P. when shaft is i in. diam., running at 75 revolutions per minute. The lever is i'9 feet long and supports weight at end of 550 lbs. (Figure 64.) Fig. 64. What the Prony brake tells us is that the power passing through the shaft is capable of overcoming a force of 550 lbs. act- . ing at 4'g ft. leverage. It would, therefore, be able to wind" up a rope, round a drum 9 '8 ft. diameter, with a load' upon the rope of 550 lbs. All we have to do, therefore, is to find the distance the load would be raised in one minute and multiply by the load itself., \ . Work per minute = 9'8 X ff X 75 X 550. H.P. - 9"8^x75><55o ' 33000 H.P. = 38-48. WORK AND HORSE POWER. 165 6. In a Prony brake, the shaft is 7 in. diam., the revolutions per minute 100, the leverage 60 in., and the load 590 lbs. IShat is the H.P.? 56-1 n.P.—Ans. 1. Steamer of 5000 I.H.P., speed IS knots. If iO per cent, of the I.H.P. is efTective in driving the ship, find the area of thrust collars, allowing 40 lbs. per square inch of surface. Find foot pounds of work available per minute and divide by ■number of feet ship goes in same time. This gives force on thrust. Thrust 43421 lbs., area 1085 sq. in. — Ans. 8. A vessel is steaming at the rate of 15 knots. The actual horse power exerted at the thrust is 509. The bearing surface of the thrust collars being 150 sq. inches, what is the pressure per sq. inch upon them? 72 '3 lbs. — Ans. 9. A vessel is steaming at the rate of 15 knots. Displace- ment 7000 tons. Find the retarding force necessary to stop the vessel in 1000 feet. Kinetic ienergy is -^ The weight (W) or displacement 64 may be reckoned in pounds or in tons, but the velocity must be in feet per second. Now 15 knots per hour expressed in feet per , . ik x6ooo ,1 second is -^—p =25*^. 3600 The energy, therefore, is '-. — . } "^ ^^' ■ „ „ = 70193 ft- tons. The retarding' force being x ATX: 1000= 701 93 jc= 70 "193 tons. ;t66 WORK AND HORSE POWER. 10. S ship of 1400 tons displacement, and steaming 20 knots,, is suddenly required to stop, and is reversed. If there is a constant retarding force of 30000 lbs., how far will she go before stopping? i860 ft. — A?is. Hi Fihd the horse power necessary to work a feed pump for an engine of 4000 indicated horse power. Given pressure in pump 170 lbs. per sq. in., and 16 lbs. of steam used per horse per hour. Efficiency of pump '6.. The quantity of water dealt with by the feed pump works out to 29568 cu. inches per minute. Now to force 29568 cu. inches of fluid against a pressure of 170 lbs. per sq. in., we may assume a ram just one square inch in area and movingf at the speed of 29568- inches per minute. The work done is ^^ i— foot pounds, 12 which, divided by 33000 gives 12 "7 horse power, and dividing by the efficiency '6 gives 21 "i gross power for the pump. 12. The displacement of a vessel is 10,000 tons and the I.H.P. of the engines is 9000. The ship is steaming at the rate of ii knots. Find in what time the work developed by the engines is just equal to the kinetic energy of the vessel. Having found the kinetic energy of the ship- to be i"95,67o,ooo foot pounds, it only remains to divide by the foot pounds of work done by the engines per minnte, viz. : — gooo x 33000 to obtain the time requii-ed in minutes. "6588 minutes. — A?is. 13. A piece of plate is blown out of the shell of a boiler with a velocity of 700 feet per second. The weight is 10 lbs. Find its kinetic energy after rising 100 feet above the boiler. Kinetic energy when first projected = '°° ^ '° = 76562 -5 ft. lbs. 64 Work done against graivity= 10 x ioo=iooo ft. lbs. ' Balance left = 75562 -5 ft. l,bs. WORK AND HORSE POWER. 167 ' '''Nbie.—^The'eS&yt of the resistance of the air; in 'checking the velocity of the fragment of plate in this example, wdUld be consider- able, tut is entirely omitted. 14. A fly-wheel i^ tons weight has a radias of gyration of 10 feet. Speed 58 revolutions per minute. Axle 9 in. diameter. Find the number of reYolntions it will make before stopping, if the co-e£9cient of friction on the^ axle is '06. Also find the time of stepping. ' The kinetic energy of the wheel is the first thing to find, and for this purpose we require' the linear speed of the rim at its radius- of gyration. -.r , .. J 20irx «8 e,8 X Velocity per second = — , !i_ = J . , 6o 3 Kinetic energy =^_§ — i5 L foot tons. 64 X 9 Next we find the work absorbed by friction each revolution of the axle. The load upon the axle bearings being 4*5 tons, the drag of friction is 4*5 x '06= '27 tons. ! Also circum. of axle = < feet, 12 The work absorbed each revolution is, therefore, 4*5x*o6x "-^. It is now only a question of how many times is contained I2 4 in the energy of the wheel already found. (By leaving each expression in the at^ove form the x cancels partially out.) Thus — ''■^■. ■ Timc- ''-'^ ^ ^5^ ^y H- ^'5 ^''^ ^^- ' 64 X 9 ' 12 ,.-s • becomes.^ ^ ^ —-r^ — . 64 X 9 "OO X 9 i.r This works out to 407 "7 revolutions. For the time of stopping we may assume the average speed of revolutions to be 29 per minute =^ time in minutes =■ '7 = i4'o6. 1 68 WORK AND HORSE POWER. 15. A gun is mounted so that it can Fecoil (after firing) against a constfint resistance of 4000 lbs. Tlie weight which recoils, includiiig the gan, is 10 tons. If a 60-lb. shell is fired with a muzzle velocity of 2600 feet per second, how far will the gun recoil? First apply the principle of action and reaction being equal and opposite in the case of momentum. Thus momentum in shell =. and if the same momentum is possessed by the gun at the instant of firing 22400 X Velocity of recoil _ 60 x 2600 ? g Velocity of recoil = 7 ft. per second. The next step is to find the accumulated work or kinetic energy of the gun at this velocity. Work = ?£4^iil!= 17150 ft. lbs. 64 Now this work has to be absorbed by the resistance to recoil = 4000 lbs. Distance = ^7i5° — ..ggy. fgg). 4000 ^ ' •' 16. A bullet weighing 2 ozs. is fired at a suspended target, whose weight is 40 lbs. If the velocity of the bullet is 1000 ft. per second, find the velocity given to the target. Equate the momentums as above; 3^ ft. per sec. — Ans. 17. Explain fully how the energy of the bullet is spent in the exercise above. The kinetic energy of the bullet weighing ^ lb. moving at 1000 ft. per second is = ^211522.= 1^53 ft. lbs. All this energy ought to be accounted for in one form or another. WORK AND HORSE POWER. 1 69 Now the kinetic energy given to the target accounts for very little, viz. :— 4^ 3|^ = 6-104 ft- lbs. 64 The balance of 1947 foot lbs. remains to be accounted for, and the greater part of it is to be found in the heat developed in the bullet itself, for suppose the temperature of the bullet raised only 500 degs. Fah., which is still below its fusing point, the units of heat will be -g- x "03 x 500 where '03 is the specific heat of lead. In units of work this is ■I X -03 X 500 X 774= 1 45 1 ft. lbs. Also the target will be heated to some extent as well as the bullet, and this may easily account for the other 496 units of work. 18. A thin hollow cylinder rolls down an incline, whose vertical height is 100 feet, find its velocity at the bottom. Also the velocity if sliding down. No body can acquire more kinetic enefgy (accumulated work) due to falling a given height than W x h. In the example this amounts to 100 W where W is the weight of the cylinder. Now if the cylinder merely slid down the incline its accumulated work due to its velocity would be simply — — . .•. ^^^ =100 W and V=8o ft. 64 But a cylinder rolling at a certain velocity has more energy than , for it has energy due to its rotation as well as that due to its movement over the ground. If the cylinder is thin its radius of g-yration is practically the same as its full radius, arid the linear velocity of its mass round its axis is, therefore, the same as its rate of progression over the ground. Energy due to rotation = —^ — . WV2- „ „ progression = -g—. 170 WORK AND HORSE POWER.' These two energies only amount to 100 W as before. 64 Y= ^-452 = 56*5 feet,per sec. 'J ' ' '] 19. Ttffo heights of 10 lbs. and 3 lbs. respectively, are con-i nected by a string which passes over a pulley. Find, the velocity of the weights after 1^. seconds from the commencement of the motion. Working from first principles, we must, first obtain the acceleration of velocity per second. Now when a body is acteii ,upon by a force equal to its own weight (as in the case of a falling weight) the acceleration is at the rate of 32 ft. per second for every second it falls. In the given case we have a total weight of 10+3 = 13 lbs. set in motion by a force equal to (10 — 3) = 7 lbs. only. The acceleration is, therefore, less than 32 ft., or Fig. 64a. 'A' I 13 ; 7 :: 32 : 17 '2 feet per sec. This gives the velocity after falling for one second, hut in i-^ seconds the velocity will be i7'2 x i-5 = 25'8 feet. Of course, one weight will be rising as the other falls. 20. A door is formed of ah iron plate. It is 7 ft. high an^ 3 ft. wide. The weight, is 280 lbs. The door is hinged at one side. Now suppose a force of 20 lbs. to be exerted at the side opposite the hinge for exactly f of a second of time. Required the kinetic energy of the ' door when the force is removed. Friction is neglected. ■ '^1' ■■' •' ■ > •--'. .■■',:\y. . . . - ,■ I The general principle; tp;^pply in this , question is that all th^ •work exerted to mov? the door and make it swing open must appear as kinetic energy when the force ceasesto acti, To apply the principle we require to knoW-^not the time during which the force acts — but the distance through which it acts, for work is equal to force multiplied ty distance! WORK AND HORSE POWER. Ift Also in calculating the kinetic energy or accumulated work of a revolving or oscillating body we require to use the velocity of its WV2 centre of gyration in the formula 64 We may begin by putting V for the final linear velocity at the radius of gyration (which, in the case of a plane revolving on one edge, is proved in Part III. to be "577 of the width from the axis). It follows that the linear velocity of the outer edge of the door is greater than V in the proportion of -577 to I. Velocity of outer edge = V -e- '577. Now this is the j^nai velocity afte'r a constant force of 20 lbs. h^s been acting for f of a V J second. The average velocity is onerhalf ,:the final or •577 X 2 the distance moved in "75 seconds is x '75. We can now ■577x2 express the work done as equal to 20 lbs. x x -75 ■577 ^ 2 ' ' =13 V foot lbs. Equating this work to the formula for accumulated work we have — 286 xV2 64 13 V from which,V=?-2— — ^=2'q'ji. 280 And the work= 13 x 2-971 =38'623 foot pounds. 6. — Riveted Joints. t. Design a riveted joint for boiler plates | in. thick of steel. The joint to be a double riveted zig-zag lap with f-in. rivets. Make a hand-sketch dimensioned, and state percentage of strength compared with solid plate. The principle to adopt in all riveted joints is to endeavour to ■obtain equal strength in the various lines of possible fracture. Now if the pitch of the rivets is first considered, it is obvious that the greater the pitch is made the stronger the plate will be that is left between the holes ; and on the other hand, the less the pitch is made the stronger the rivets will be in proportion. When the plates and rivets are of iron it is usual to consider that equality of strength between rivets and plate is sufficiently obtained by making the areas of fracture equal (although the rivets al-e in shear and the plate in tension). But when plates and rivets are of mild steel the Board of Trade rules require us to credit the rivets with 23 tons per square inch, and the plate with 28 tons per square inch to obtain equality of strength. Consider a portion of the seam lying between AB and CD, figure 65, namely a length of the seam equal to a pitch space =^. If the plate fractures, the line of fracture will probably be along /, and the area of metaj fractured will he (p-d)xi. If the rivets shear off, as there are two in the pitch-space considered, the area of metal fractured will be d^ x 7854 x 2. Equating the resistance to frac- fi2y^>^2'/^ Plate •^' RIVETED JOINTS. 173-, ture of these two lines we have {p - d) y. i -k 28 = d^ y. 7854 x 2 x 23, This is the equation for a double riveted lap of steel plates. Referring' to the example, it becomes 28 X (^ - -75) X -5 = -4417 X 2 X 23 (/-75) = i-4S p — 2'2 inches. Referring- again to the Board of Trade rules, the diagonal pitch' of rivets in such a joint should be = &p-\-/i,d 10 which, is equivalent to making the distance between the two rows- of rivets = v/(ii/+4^)x(;>+4^) 10 Working this out gives us I'lg in. as the distance between rows,. say i^ in. There is only the distance of outer row from edge of plate now to fix, and this should not be less than i-^ times diam. of rivet = i|-, but is often made greater than what this rule gives, more especially in butt joints. The sketch is dimensioned approximately. to the above calculations. For the percentage of strength use the formula for plate- strength as it is the simplest, viz. : — (j>-^)xroo^(2;2-f)ioo^^g per cent. nly. p 2-2 ^ ^ Note. — The distance between the rows is deduced from the diagonal pitch as follows (figure 66), CD= VAC^ - AD2 in which AC is the diagonal pitch, AD is one-half the ordinary pitch, and CD is the distance between the rows, substituting values cD=y(5^^'-(l)' but the difference of two squares is . always equal to the sftni and difference multiplied together, therefore, J 74 RIVETED JOIl>fTS; V V lO .2/ \ ' lU; • 2/ "0 V \ lo ; lo ' , / ■ r2. Design a treble riveted lap joint for the middle circnlar seam of a steel boiler. The rivets also of steel and 1^ in. diatD. The shell plates being ii thick. Also find the efficiency of joint. The pitch works out to 5 in. nearly, and the efficiency to 70 per •cent. Distance between rows = 2*6 nearly. — Ans. 3. Given thickness of plates i^ in. Joint to be treble riveted with double butt straps. Pitch 6 in. Rivets allowed 1| times single shear strength. Find diameter of rivets to give equal strength with plate when rivet metal is 23 tons per sq. in. and plate metal 28 tons. The only change needed, in the argument used in example 1, is to substitute 3 for 2 in the nnmber of rivets in one pitch-space, and to credit the rivets with if times their real area due to being- 'in double shear. Strength of plate between holes in tons = 28x(6-rf)xi^ = 210-35 d (i). Strength of rivet metal in one pitch = 23x^2^ -7854x3x11 = 94-837^2 (2j_ Equating (i) and (2) gives us a quadratic: — 94 '8 ^2 = 210-35 ^ or d-^+Jld=^ 94-8 94-8 This works out to d= 1*315 in. "The percentage of strength is (6- i"'5i5)x 100 o ^ - ^-^ = 78 per cent. RIVETED JOINTS. *?§ 4. Design a riveted double butt joint tor plates If in. thick. The pitch of rivets to be 8 in* Put five rows of rivets at each side of butt. Other conditions the same as in the last example. / Diaip. of rivets i'28 in.; .distance between rows 3^ in. nearly; efficiency 84 per. cent. — Ans. Note.—\\. is not usual to have the diameter of rivet less than the thickness of plate in practice. 5. Design a longitudinal seam for a steel boiler 16 feet diameter. The seam is to be a treble-riveted double- strapped joint. EfBciency 80 per cent. The plates are 1^ in. thick. Also find the working pressure of the boiler. Here we have to find both pitch of rivets and their diameter, Begip by writing the equation for efficiency of joint referred to the plate viz. : — {p-d)x i oo_gQ P from this we get 100 / - 100 ^= 80 / 20 p= 100 d P = Sd (i). Next write the equation for efficiency of joint referred to the rivets, viz. : — 100 -nd^-x. 7854 X 3 X if X 23 _ g^ / X i^ X 28 Now substitute ior p in this equation its value from (i). looxd^y. 7854 X 3 X if X 23 _ g^ 5 - -0- - -0- - A -(^y^rfltoi^- . Q. h Q . ; . 1 '•« q) 1 _^ Fig. 67. 6. Sketch a treble-riveted lap joint for the longitudinal seams of a steel boiler. The plates are | in. thick and the joint is to have a percentage of 72 compared with the solid plate. Use 23 and 28 tons. Also having given diameter of boiler 13^ feet find working pressure and i state how thick shell should be if of iron and carrying, the same pressure. Factor of safety 5. , . -, Rivets I in. nearly; pitch 3 "56 in.; working pressure 83-6 lbs.; thickness of iron shell i in. — Ans. 1. Design a riveted joint for shell plates 1 in. thick. Treble- riveted, double-strapped. Tensile strength of plates 35 tons. Shearing strength of rivets 23 tons. Factor of safety 5-5. , Diameter of boiler 10 feet. Also find working pressure for boiler. EfSciency of joint -TS. Pitdfi' 4'42S in.; diam. I'loy in.; working pressure 217-7 lbs. RIVETED JOINTS. 177 8. Design a riveted joint for longitudinal seam of a boiler. The plates are If in. thick of steel. The pitch is not to exceed 10 inches. The efficiency is to he as great as practicable with the usual numbers 23 and 28 for shearing and tension respectively. As a preliminary calculation we will assume that the pitch is 10 in. and the rivets if in., which will give an efficiency of plate = 10- if „, , ^x 100 = 86^ per cent. The question now is, how many rivets shall we require in each pitch-space to equal the strength of the plate? o , 6 , — "-o . O , O '-0 -G-- 0600- o- i o o ' G -a '- o^TvP^-^ Fig. 68. Plate strength = (io- if) x if x 28 = 332 tons. Now each rivet has an area = i '484, and allowing for double shear each rivet has a shearing strength = i -484 x if x 23 = 59-7 tons The result of the preliminary estimate is that it will take between 5 and 6 rivets (each if in.) to equal the strength of the plate in one pitch-space. Must we, therefore, design a joint with five or six rows of rivets ? We might certainly do so, but there is a better plan which is always adopted now for longitudinal seams of high efficiency The joint to design is the one sketched in figure 68, in which it will M 178 RIVETED JOINTS. be noticed that the rivets in the outer row are placed at twice the pitch of the inner rows. By this device Jive rivets are got in the wide pitch-space of the outer row, and althoug'h it might appear that the plate is weak across the middle row of rivets, we shall probably find that the combined strength of the plate in this row together with the shearing strength of the rivets in the outer row (which must shear simultaneously with the plate fracture) is quite up to the percentage of the rest. It remains now to fix the diameter of rivets and design the joint accordingly. We will fix the pitch at the maximum allowed, since that will give us the highest efficiency, and proceed to find the diameter of rivet. To do this rigorously leads, as in a previous example, to a quadratic equation, but as we know the diameter will come out only slightly in excess of if in., we may use without appreciable error the plate strength already found, viz., 332 tons, and equate it to the rivet strength. Let diam. of rivet = d. ^2x-78s4X5x if X 23 = 332 from which d= i "449 inches or say i^^^ in. for rivet diameter. The efficiency is now i 1^2 x 100 =85-6 per cent, if the fracture occurs through the outer row of holes. And if it occurs through the middle row the length of plate fractured is 2 rivets less than the pitch (instead of i rivet less), but there is in addition the shearing strength of one rivet in the outer row. It is easy to balance the loss against the gain here, thus : — Loss of strength in plate section = if x ly^^ x 28 = 55 tons. Gain of one rivet in double shear = area x if x 23 = 65*5 tons. The gain is more than the loss, which shows that we need not trouble about the middle row any further. It remains to find the distance between the rows and the thick- ness of butt straps. RIVETED JOINTS. 179 In the Board of Trade Rules there is a special rule for the distance between the outer and middle rows in a joint of the description we are taking (known as the J joint in the rules), the formula is : — v/(5"5 + iTir) (•5+iiV) = 3'66in. Note. — In finding distance between middle and inner rows, use formula given in the ist example and work with actual pitch in those rows. Last we come to the butt straps, and upon referring to the sketch it will be noticed that the most probable line of fracture for the butt straps is through the row of rivet holes nearest to the butt of the shell plates, and here there are two holes in each wide pitch- space. The metal left is, therefore {p-2d) as against {p~d) in the shell at outer row. On this account the minmum thickness of -each strap is increased from i the shell to ^2 x '^t, which in our p — 2d example w^orks out to i"02 in., say i|- in. The sketch has been dimensioned accordingly. As the type of joint just discussed is very important, the following exercises are given. 5. Find the efficiency of the actual joint described below. Take the fracture, 1st, through the outer row of holes in the shell ; 2nd, through the middle row ; 3rd, through the two butt straps; 5th, as a shear of the riYets. Compare with the solid shell plate in each case. Use 23 tons for the rivets and 28 tons for the plates. A J joint of shell plates 1^ in. thick; straps \\ in.; pitch of rivets in outer rows 10 in.; pitch for inner and middle rows 5 in.; diameter of rivet holes 1^ in. I St, 85 per cent.; 2nd, 87 per cent, nearly; 3rd, 105 per cent.; 4th, 84-6 per cent, efficiency. — Ans. The efficiency is, of course, the last result as it is the lowest. i8o RIVETED JOINTS. 10. A treble -riyeted double butt strap joint for a steel boiler. Every alternate rivet in outer row is left out Pitch of rivets in outer row 7| in., and in inner row 3f in. Diam. of rivets i^ in. Thickness of plates l^V in. Find percentage of strength of plate in outer row and per cent, of rivet strength. Also the combined strength in each of the inner rows. State the advan- tage of this joint. Plate 85-4 per cent. ; rivets 88-9 ; middle row 88-6; inner row 124" I. — A}is. 11. Same data as in the previous exercise, but every alter- nate rivet is now left out in both inner and outer rows. Find strength of plate and also of the rivets. Plate 85'4 per cent.; rivets yi'i per cent. — Ans. 12. A treble-riveted lap joint of steel plates 1 in. thick. Pitch of rivets in outer rows 6 in., and in the middle row 3 in. Find the diameter of rivets to obtain the greatest strength; having regard to the fracture through the middle row as well as in other ways. (Figure 69.) K- 7^ - >l<_~6»rr;p I O O I cp (!) I ©■ '0' Being a lap joint there is no double shear of rivets and in that case the fracture through the middle row is generally the weakest. We must make the rivet strength equal to the plate strength taken through the middle Fig. 69. row in this joint. If d= diam. of rivet, since there are 4 rivets in the 6 in. pitch- space, d^ X "7854 X 4 X 23 = shear strength of rivets Also (6 - 2d) X I X 28 = tensile strength of plate RIVETED JOINTS. l8l adding shear of one rivet in outer row total fracture through middle row = (6 - 2d) X 28+d^ X "7854 X 23. Equating this to shear strength of all the rivets leads to : — d^ X '7854 X 3 X 23 = (6 - 2d) X 28 from this equation d= i '32 in. The percentages of strength works out as follows : — Rivet strength = 74 "4 per cent. Outer rows = 78 per cent. Middle row = 74'6. 13. In the J joint the B.T. rule for outer diagonal pitch is: — ^p+d 10 Prove that the distance between rows is : — (See note to ist example.) 14. Design a J joint for shell plates 1^ in. thick, having an. efficiency of '8i. Use 23/28. Pitch = 6 J times diameter gives '84 efficiency in the plate and equality of plate and rivet is obtained with diam. i 'le in. .". pitch = 7*25 in. Distance between rows i'9i7 in. and 2"8 in. Butt straps I in. thick. — Ans. 15. Find the working pressure for a steel boiler shell 15 ft. diameter under a factor of safety = i-5. The plates are 1| in. and the rivets If in. of steel. Pitch of rivets' 10 in. The joint is a J joint douhle-strapped, and the ratio of shear to tension 23 : 28. Plate 83*75 P®r cent. ; .rivets 91 '"7 per cent.; pressure 2iolbs.. ■ I 7. — Stress and Strain. 1. A steel bar 7 in. broad, 1^ in, thick, and 17 ft. 6 in. long is loaded in tension with 72 tons. Find the elongation and state if permanent. Modulus 30,000,000. Hooke's Law of Elasticity is that the elong-ation is directly- proportional to the load. A simple illustration occurs in a spring- balance used for weighing goods, in which it will be observed that the index moves an equal distance for every additional pound upon the balance. If the elongation of the spring were not proportional to the load the divisions upon the scale would be unequal. Now in order to apply this principle to the calculation of the elongation in any case of a metal bar, it is convenient to make use of two terms which have a very definite and restricted meaning in this connexion. The first is the term "stress." Stress is the amount of force reckoned in pounds which comes upon every square inch in the cross section of the bar. It is the load in pounds divided by the sectional area. The second is the term "strain." Strain is the amount of elongation expressed as a fraction of the length. It is the amount of elongation measured in inches divided by the length of the bar also measured in inches. In practice the "stress" generally amounts to some few thousands of pounds and the "strain" to a very small fraction — less than xothF' STRESS AND STRAIN. 183 When the stress is divided by the strain a very large number is obtained which is constant for each material tested — a result which also follows from Hooke's law. In the case of wrought iron the number obtained is 29,000,000; which is known as Young's- Modulus of elasticity (symbol E). The formula is ^ = Modulus (E). strain In the example we have : — stress = '- 3_ = I c^Go lbs. 7x1-5 Applying the formula ^^ . =30,000,000, we get stram Strain = 1 5360 -^ 30,000,000= "000512 and multiplying by the length of the bar elongation = 210 x •000512 = •10^5 in., which is the answer. The elongation is never perriianent when Hook's law is appli- cable. If the stress exceeds about 30,000 lbs. in the case of steel, and 20,000 lbs. in the case of wrought iron, the limit of elasticity is reached and a "permanent set" produced, which cannot easily be made the subject of calculation. 2. What is the modulus of elasticity? If its amount is 26,SOO,000 what would be the elongation of a bar of 9 sq. inches sectional area under a load of 90,000 lbs.? The bar being 30 feet long. •1358 in. — Ans. 3. A wire \ in. diam. and 9 feet long is found to stretch \ in. with a load of 900 lbs. What is the value of the modulus of elasticity? 28, 180,000. — Ans. 184 STRESS AND STRAIN.' 4. A steel rod ^ in. diam. and 10 ft. long is Stretched -12 in. by a load of 7000 lbs. Find the yalue of E and the resilience of the bar. E = 35,660,006 ; resilience 35 ft. lbs. — Ans. 5. Find the amount of energy (resilience) stored up in a steel bar 2 in. diam. and 20 feet long when strained to its elastic limit, which may be assumed to occur at a stress of 16 tons per sq. inch. Modulus of elasticity 30,000,000. First we must find the strain and elongation corresponding to strsss 16 tons stress by the equation :— = Modulus. strain 16 X 2240 3L = 30,0000,000 strain Strain = "oo II 95 Elongation = 20 X ■ooii9="0239 feet. ■ Now- the load — arSa x 16 x 2240 „ =112600 lbs. , r ■ In multiplying the force by the distance to find the work expended^ we must notice that the load cannot be put upon the bar all at once, and as the load comes on so does the bar stretch. This leads ^o the rule: — ,< . Take the average load, which is one-half the final load, in calculating the work. ,jj , I 12600 X •02'3Q ,^ ,, Woi:k = ^=1345 ft. lbs. 6. A bar of steel 40 ft. long and 2 sq. inches sectional area ^ ' - stretch^ ^^ invby a load of 20,000 lbs. Find Young's modulus, and the resilience^ • - . •'• ''£ = 25,600,000; work = 156-2 ft. lbs. — Ans. STRESS AND STRAIN, 185 7. A bar of iron weighing 200 lbs. is stretched y^ of its length by a load. The modulus £=29,000,000. Find the work expended, or resilience, and express it in heat units. It may appear at first as if the dimensions of the bar should be known, but it will be found that any bar which weighs 200 lbs. will have the same resilience. The simplest bar to take is one of a sq. inch section, which will be 720 inches long. ' ' -'' 870 ft. lbs. = i"i24 heat units.— Ans. 8. A spring is loaded with a weight of 1600 lbs., which compresses it 1 in. The load is removed and this spring allowed to recover its original extension. A weight of 112 lbs. is next allowed to drop i feet on to the spring. How much will the spring be compressed? An approximate answer is easily obtained as follows : — Work due to falling weight =f 112 x 4 = 448 foot lbs. Let X be the amount of compression in inches XX i6go = maximum force, X X 1600 j- = average tprce on spring X :*; in, = work done in inch lbs. 2 Equating this to the work due to the falling weight also in inch lbs. x^x 800 = 448 X r2 x = z'6 in. nearly. A more correct result is obtained by taking into consideration the fact that the weight falls 4 ft + ^ before it comes to rest.' The equation then stands as follows: — X- X 800 = (48+^) X 112 , 112 , _ or jc-*- - — ^j[r = D'72. 800 ; This quadratic works out to' :t=.2736 inches; ''whlcih' is- the correct answer. 2 XX 1600 1 86 STRESS AND STRAIN, 9. A straight copper pipe. 10 feet long and 5 inches diametep outside, and -2 in. thick, just fits between rigid plates when at 60 degs. Fah. temp. Find the force to push the plates apart when steam is in the pipe raising its temperature to 370 degs. F. Given co-efficient of expan- sion for 1 deg. C. = '00000175 and Young's modulus 13,500,000. If the natural and free elongation of the pipe is calculated for the increase of temperature it is evident that when the pipe is not allowed to take this expansion due to the rigid distance apart of the plates, it is the same in effect as if the pipe had been compressed the amount which has been calculated for its elongation. The problem is, therefore, first an expansion question due to heat, and next a stress and strain question with a given modulus. The rise of temp, expressed in Centigrade degrees = f of 310 = i72'2 degs. C. Elongation = "00000175 x 172 '2 ,, = "0003014 of the length. This fraction is now taken as a compressive strain upon the material of the pipe ; hence stress = 13,500,000 •0003014 stress = 4o68"7 lbs. The area of the pipe section will be found to be 3 "016 sq. inches. Total force = 4068 "7 x 3"oi6 ,, =12271 lbs. — Ans. Note. — It will be noticed that the stress produced depends upon, the rise of temperature alone, and is the same way for any length ot pipe. STRESS AND STRAIN. 187- 10. A steam cylinder 20 in. diam. has 26 studs in cover, IJ effective diam. The strain due to nipping up is -OOOl. If steam is used of 160 lbs. pressure by gauge, find maximum tensile stress on studs. And also taking area of joint as 138 sq. inches, find pressure on jointing material when under steam. In solving this problem we may assume that the jointing material is sufficiently elastic to keep up its reaction upon the cover- and studs, even if the latter are elongated by the extra force coming upon them by the steam. Steam pressure on cover = area x i6o = 50265 lbs. dividing by total area of 26 studs. Stress on stud metal = 1576 lbs. Now the initial stress due to nipping up is calculated from the strain^ stress = modulus x strain, stress = 29000000 X 'oooi = 2900 lbs. Total stress = 1576 + 2900 = 4476 lbs. Upon our assumption that the jointing material is elastic, the compression upon it will be slightly less when the steam is turned. on than before, but very slightly. Force exerted by 26 studs when nipped up with stress found. above is : — • I "227 X 2900 X 26 = 92515 lbs. and dividing by area of joint pressure per sq. inch = 670 lbs. 11. A cylindrical pipe of mild steel is i in. diameter inside and ^ in. thick, ultimate strength of the steel 30 tons per sq. inch, and elastic limit 14 tons. Assume that every ton stress above the elastic limit stretches the metal 1*6 per cent, permanently. Find the new diameter and thickness after a pressure of 2240 lbs. per sq. inch has been pumped in. Also using a factor of safety of 14, find the working pressure for the pipe. The rule for fluid pressure in a pipe is the same as for a. cylindrical boiler, viz : — p _ 2 T X Stress l88 STRESS AND STRAIN. Let US first try the effect of P = 2240 lbs. = i ton per sq. inch. _ 2 X ^ X S 4 S = f = 16 tons. ■4 The Stress at the outset is thus 2 tons above the elastic limit, and the permanent stretch is, therefore, i'6 x 2 = 3*2 per cent. But the effect of any amount of stretch is to increase the diameter the same percentage ; and at the same time to reduce the thickness an equal percentage. The stress is, therefore, increased and consequently more stretch occurs, accompanied by a further increase of diameter, etc. To allow for all this leads to more mathematics than is pro- ibably wanted, so we will stop at 3*2 per cent, stretch. The diam. is increased by 4 in. x =3 — = -128 in. and reduces the 100 thickness by i in. x ? — = -004 in. •^ ® 100 ^ The new diameter is, therefore, 4"i28 in., and the new thick- iness •121 in. Applying the rule for working pressure we have 30 tons = 67200 lbs. .". safe stress allowed as per question 67200 o ,, = -i— =4800 lbs. 14 Pressure = ^ T x Stress ^ 2 x -121x4800 diam. 4-128 ,, =281 lbs. Note. — A more rigorous method of solving this question leads to a pressure of 262 lbs. STRESS AND STRAIN. 1891 12. The formula used by the B.T. for copper steam pipes is. p^6000x(T-T-V) jjjj, brazed pipes. Where P = steam pressure, T = thickness, and D = inside diameter. If the stress in the material is 2500 lbs. per sq. inch, what thickness would the pipe be ? Using the fundamental formula for the pressure in a pipe wheUi the stress is 2500 lbs, we have — p_2 TX2500 D Equating the two values for P, we write — 6000 X (T - y^y) _ 2 T X 2500 D b Cancelling the D's and multiplying out gives T = f in. — Ans. 13. A boiler stay is swelled at the end, and a cotter hole cut through the swelled part to receive a cotter, whose depth is four times its thickness. The body of the stay is 2^ in. diam. Find the correct diameter for the swelled part. We will assume that the strength of the cotter in double shear is if times the single shear, and the shearing and tensile stresses are the same. Let T = thickness of cotter 4T = depth of cotter Then 4T x T x if = 2-5 x 2-5 x 7854 7 T2 = 4-908 T = -837 The next thing is to make the section of the stay through the- cotter hole equal to the section through the body. Let D = diameter of swelled part. D^ X - = full area. 4 D X "837 = section lost (approximately). remainder = D^ - - "837 D. 4 iigo STRESS AND STRAIN. Equating this to the area of body (4 "908), D^ - - "837 D = 4-908 4 D2 - i-o6D = 6-25 D - -53 = ± V6-25 + (-53)2 D = -53 + 2-55 = 3 '09 inches. a. A chain cable, studded, composed of links made of bar iron 2 in- diam. requires a ring fitting to it 12 in. diameter (mean). What should be the diameter of the bar iron for the ring so as to have equal strength with the links of the chain? JVoie. — Assume that the strength of a ring varies directly as the section of the bar and inversely as the mean diam. of the ring. The smallest ring into which a 2-in. bar could pass, if made of 2 in. bar iron, is one of 4 in. mean diameter. This ring, not being able to collapse, would have the same strength as the links of the ■cable. — Ans. Working from this 4 in. ring by the rule given in the question. 4:12 : : 2^ : d^ ,, 4 X 12 d^ = 5 = 12 4 d = 3 '46 in. the required thickness. 8. — Hydraulics. 1. Find the velocity of water through the tubes of a condenser having given diam. | in., length 9 ft. The cooling surface is at the rate of If sq. ft. per I.H.P., and 450 lbs. of water circulates per I.H.P. per hour. The water goes twice through the condenser. The surface of one tube being- 2 -06 17 sq. ft., if we take 10 I.H.P., giving i7'5 sq. ft. total surface, we have: — number of tubes = — LJL-. = 8 "488 2 •061 7 each box has .'. 4*244, and sectional area = "601 x 4-244 = 2"55 sq. in. Again, 10 I.H.P. requires 45CO lbs. water per hour, this equals 34 •657 cu. in. per second. Velocity = -''— — — = i3'6 in. per sec. or 68 feet per minute. -■55 2. Prove that velocity through condenser tubes is given by the formula: — r p T Y = ,i7r-ir- in feet per minute. 80 SU *^ '3. A ship's draught being 21 -S feet, find how much water will enter the ship, in three hours, through a hole in her bottom 1^ in. diam. Neglect the effect of increased draught and the effect of water rising in bilges. The theoretical velocity of water issuing through an orifice xinder any "head" is equal to that of a body falling from a height equal to the "head" of water. Velocity = 1^2 gh = 8 \/2i -5 = 37 'i ft. per sec. Area of aperture = i'227sq. in. , 192 HYDRAULICS. Cubic inches per sec. — 1*227 ^ 37'i ^ 12. _ . , 1-227 X '?7'i X 12 X ■^600 X -i Tons in -? hours = '- ^ 1728 X 35 ,, = 97 '546 tons. Practically the quantity admitted wonld be '62 of the above approximately. i, A tank, whose capacity is 1000 gallons, is fixed 20 feet above the level of the sea. A pipe of 2 in. diameter of bore connects the tank with the sea. If a perfect vacuum is made in the tank how long will it be in filling? If the atmospheric pressure is taken as equal to 33 feet of sea water, the available head in this case is 33 - 20 = 13 feet. From this head, the velocity can be calculated as in previous example. Next, the quantity delivered per second and finally the time. Friction is neglected. 4 m. 15 sec. — Aiis. 5. A tank, measuring ^ ft. x 4 ft. x 2 ft. deep, stands in a ship so that the top of the tank is 1^ ft. above the level of water in the bilges. A pipe 1^ in. bore is connected at the top of the tank and leads down into the bilges. Suppose now that a vacuum is created in the tank, find the time it will be in filling with water. 2 m. 7 sec. — Ans. Atmosphere taken as equal to 34 ft. of water. 6. There are two syphons each having a short leg 8 in. long, but one having a long leg 2i in. and the other 72 in. The short syphon empties a tank in 5 minutes, find the time for the long one. Apart from the influence of friction in the action of a syphon, the velocity of the fluid (as in the last examples) is proportional to- the square root of the "head." HYDRAULICS. 1 93, In this example we may assume that the "head" is (24 - 8^ in the first and (72 - 8) in the second syphon ; hence \/64 : \/i6 : : 5 m : x 8 : 4 : : s m : 2^. 2^ minutes. — Ans. 7. £ cast-iron diving bell, with parallel sides, open at the bottom, is lowered into the sea until the water level inside the bell is 50 ft. below the sea level. Find the depth of air space in the beU if the bell is 10 ft. deep inside. Assuming^ that 33 ft. of sea water equals one atmosphere of 15 lbs., we have 33 ft. and 50 + 33 = 83 as the relative pressures of air inside the bell before and after it is lowered into the sea, so by Boyle's law 10 X 33 = ^ X 83 where x is the air space. 3 "976 ft. nly. — Ans. 8. A tube of even bore 10 ft. long, with its npper end closed air-tight, is lowered in the sea until its lower end is 100 feet below the surface. If the water barometer stands at 3^ ft., find how far the water will rise up in the tube. If X is put for the height of water in the tube in feet, \o — x = air space and 100 — x '\s the head, to which if we add one atmosphere, and state Boyle's law we get (lOo - X -\- 34) X (10 - jc) = 34 X 10. This result is a quadratic equation, the solution of which gives X = 7-32 ft. 9. Given temperature of funnel 600 degs. Fah. and of external air 60 degs. F. Height of funnel 80 ft. above fire-bars. Find the draft measured in inches of water. What is wanted here is the difference in weight between a column of hot air in the funnel and a column of cold air outside. N 194 HYDRAULICS. Starting with the fact that a cubic foot of air at 32 degs. F. weighs "08 lbs.; for a cubic foot of air at 60 degs. we have the proportion : — 460 + 60 degs. : 460 + 32 degs. : : "08 : "0757 and for a cubic foot at 600 degs. : — 460 + 600 degs. : 460 + 32 degs. : : '08 : '0371 The difference per cu. ft. being -0386 lbs. A column 80 ft. high will have a difference = -0386 x 80 = 3 '088 lbs. This is the pressure per square /ooi csiusing drsift ; and since 12 inches of water causes a pressure of 62*3 lbs. per sq. foot. 62"3 : 3'o88 : : 12 in. : x X = '59 inches. — Atis. 10. Given height of fannel 55 feet, diameter 8 feet. Temp. of funnel 700 degs. F. and of the external air ^0 degs. F. Calculate draught in inches of water pressure. A cu. ft. of funnel gas at 32 degs. weighs '0807 lbs. the same as air. •477 inches. — A?is. 11. If forced draught is used equal to | in. of water, how high would the above funnel require to be to give the same as forced draught? 1 01 ft. nly. — Ans. 12. If 2i tons of coal are burnt per day with the above funnel and 22 lbs. of air are used per lb. of coal^ what is the velocity of the gases in the funnel? The cubic feet of air used per second divided by the sectional area of the funnel gives the velocity required. 7*964 ft. per sec. — Afis. HYDRAULICS. 195 13. A tank is filled up with fresh water and a hole made in one end If in. diam., situated i feet below the water level. Find the force exerted to move the tank when the water is issuing freely &om the orifice. This is a question of momentum and is of great importance in the theory of jet propulsion. The principle to apply is Newton's third law applied to momentum, viz. : — Action and reaction are equal and in opposite directions. We must first find the velocity of the issuing water, which is S V4 = 16 ft. per second. Next we calculate the weight of water ■escaping per second which is : — Area x 16 ^ ^ ,, . X 62-3 = 14 36 lbs. 144 (We are neglecting here the co-efficient of discharge which is '62). Now consider just what occurs : — Every second of time a ■weight of water equal to i4"36 lbs. is given a velocity of 16 feet a second as it escapes through the orifice. The question is what force is there behind the orifice to account for this? Well, we know that a falling body acquires a velocity of 32 ft. per second after it •has fallen freely for one second of time, due to the force of gravity •acting upon the body. This shows that a force equal to the weight ■of a body acting for one second of time will always give a velocity •of 32 ft. per second. We may safely work from this fact when similar problems occur. Stated plainly it comes to this : — If i4*36 lbs. force had been acting for i second upon 14 "36 lbs. of water the •velocity generated would have been 32 ft. What is the force when the velocity is only 16 feet? 32 : 16 : : i4'36 : x -which gives the force required = 7"i8 lbs. It only remains to add that the reaction upon the tank will be the same as the force behind the water, namely 7*18 lbs. Note. — If the static pressure of the water upon an area equal to the i|-in. hole is calculated at 4 ft. below the surface of water it will be found to be 3.58 lbs. only. And it seems strange that the igfi HYDRAULICS. reaction upon the tank, when the water is allowed to escape, can amount (as found above) to twice this static pressure. But this merely shows how misleading it is to assume that forces, which are acting- when everything is at rest, will continue to act without any modification when things are put into motion. a. A plate 2 ft. by 1 ft. 6 in. is moved at the rate of 6 ft. per second through water in a direction at right angles to its plane, find the resistance offered. It is easier to follow the reasoning adopted in working this example, if we suppose the plate fixed and the water running past it, at the same rate as given. To simplify the solution we must also suppose every particle of water intercepted by the plate to be absolutely stopped in its course before it escapes round the edges to continue its course. The weight of water whose motion is thus arrested by the plate per second is : — 2 X i'5 X 6 X 62"3 = 1 121 '4 lbs. The question now remains, what force must be exerted to bring- to rest IT2I lbs. of water each second, which has a velocity of 6 ft. per second ? As in the last example we can reason from the effect of gravity. The force of gravity will bring a stone to rest in one second if the- stone is thrown vertically upwards with a velocity of 32 ft. per sec. ( =g). And in the case of gravity the force is equal to the weight. .'. 32 : 6 : : 1121 : force. Force or resistance = 210^ lbs. Wx V Note. — The formula (which, it will be noticed, represents- the final operation in the above) is known as momentum. Hence i6 is as well to remember that momentum per second = force. HYDRAULICS. 197 15. The area of a vessel's rudder under water is 28 sq. ft., and the centre of effort is at 20 in. from the gudgeons horizontally. The diameter of the rudder head is 6 in. Let the rudder be put over to an angle of i2 degs., and find the stress upon the rudder head when the speed of the vessel is 16 knots. We will first find the pressure against a surface of 28 sq. feet placed at right angles to the direction of its motion. tj , .^ 16 X 6080 Velocity per sec = =27 "02 3600 pounds of water arrested per second = 28 x 27 x 64 = 48384 lbs. Momentum = =t_J— z 7 ,, = 40824 lbs. which is the force. Now the fact that the rudder is placed at an oblique angle with the direction of motion has a two-fold eifect upon the pressure. In the first place less water is intercepted than if at right angles; and secondly, the water striking obliquely has not the same pressure as when striking at right angles. Thus, in figure 70, XY is the centre line of the vessel, CB is the rudder when at an angle of 42 degs. CA is the same rudder at 90 degs.. From B draw BD parallel to XY. Then CD = CB x sine 42 degs. Again set off Bd to represent the force against a unit area when at right angles to the current. Resolve this force into Bb and bd, of which Bb is alone effective in turning the rudder, and Bb = Bd X sine 42 degs. The conclusion is that we must multi- ply the force already found, viz., 40,824 lbs. by sine 42 degs. to correct it for the Fig. 70. igS HYDRAULICS. quantity of water arrested being less, and again by sine 42 degs. to correct it, for the action upon the rudder being oblique. (Sine 42 degs. = '669). Effective force = 40824 x -669 x -669 ,, = 18271 lbs. The problem is now purely a shaft problem ; for the leverage with which the above force acts is given in the question as 20 in. .'. Twisting moment on rudder head = 18271 x 20 and equating this to the moment of resistance of the 6 in. bar „ 6^ X stress 18271 X 20 = 5-1 Stress comes out = 8628 lbs. 16. A pipe of uneven bore laid horizontal is 1 in. diam. for part of the length and 3 in. diam. at another pajt. The pipe is running full of water, the velocity through the 3-in. portion being ^ foot per second and pressure in the same portion 5 lbs. per sq. inch. Find the velocity and pressure in the 1-in. portion of the pipe. Since the same quantity of water must pass through every part of a pipe in a given time when the pipe is running full, the velocity is inversely as the sectional area, hence 1^ : 3^ : : '5 ; Velocity required which comes out to 4*5 ft. per sec. To obtain the pressure in the small pipe is more diflScult. The principle to use is that a certain amount of the energy of the water due to its pressure is converted into an equivalent amount of energy in the form of velocity, the total energy of the water remaining the same. Now take a pound of water. The velocity being V in feet per I X V^ second, its kmetic energy is by the well-known rule = 64 • This gives its energy due to velocity. To obtain an expression for the energy due to its pressure p, consider what work a pound of water could do if used in a small hydraulic cylinder. Let the piston HYDRAULICS. I Q^ or ram be i sq. inch in area, so that the force upon the piston = p lbs. As soon as the piston moves, more water must be admitted as there is no expansion. A pound of water is 27 "727 cubic inches, therefore, a pound of water can only drive the piston 27 "727 inches along- the cylinder, for as soon as the supply is shut off the piston stops. The work done by the pound of water is, therefore, p X 27727 f. ,. ■'- '_£_< ft. lbs. = 2"'JI p. 12 ^ ^ Combining these two forms of energy, it is clear that the total energy of a pound of water is expressed as ^ + 2"^\ p which is constant. Applying this principle to the example ^ + 2-3. X 5=1^ + 2-31/. From this equation^ = 4*866 lbs. — Ans. 17. A pipe 1\ in. diam. swells out to 6 in. diam. for part of its length. If the water is running at the rate of 24 feet per second in the 2^-in. portion, where the pressure is 20 lbs. per sq. inch. IKhat is the velocity and pressure in the 6-in. portion ? Velocity 4*13 ft.; pressure 23'8 lbs. — Ans. 18. A hydraulic water pipe, 6 in. inside diameter, has to withstand a water pressure of 1000 lbs. per sq. inch. If the stress on the metal is not to exceed 3000 lbs. per sq. inch, find thickness of metal and outside diameter. There are several theories for the stresses in the material of a thick cylinder, but Professor Perry gives the following rule. The outside and inside diameters being D and d, (D^ + d^) : (D^ - d^) : : stress (at inner surface) : fluid pressure The only unknown quantity is D D^ + 36 : D^ - 36 : : 3000 : 1000 200 HYDRAULICS. .'. looo D^ + 36000 = 3000 D^ '•^ 108000 2000 D' = 144000 D = Vyi = 8-48 in. The thickness is, therefore, i'24 in. g. — Centrifugal Force. 1. A seaman taking soundings with a lead weighing 12 lbs., swings the lead at the rate of 20 revolutions a minute. His shoulder may be taken as the centre of the circle, and the radius of the circle as 9 feet. Find the maximum and minimum pull upon the line. Also state the best position for letting go. The most convenient rule for centrifugal force is : — r WRN2, J ■ r. , TIT X c.f. = • (proved in Part III.) 2950 W = weight, N = revol. per min., and R = radius in feet. Applying this rule r 12 X Q X 20^ , „ c.f. = 2 = 14-64 lbs. 2950 We must now observe that when the lead, as it swings, comes vertically below the centre of motion there is an additional pull on the line due to the force of gravity or deadweight of the lead (12 lbs.) making the maximum pull = 26 "64 lbs. Similarly, if the lead is swung right round at the same rate, as it passes over the upper centre, gravity acts against the centri- fugal force and the pull on the line is only 2 '64 lbs. The position for letting go is when the line is inclined 45 ^._ degs. to the vertical, as in figure 71. The lead will then reach the furthest forward before entering the water. Fig. 71 202 CENTRIFUGAL FORCE. 2. A Watt's governor is revolYing at 50 revs, per minute. Find the height of the cone formed by the arms which support the balls. In Watt's governor each ball takes its position due to its own weight and its centrifugal force, so we need only consider one ball as A, in figure 72. The three forces acting upon the ball are, ist, its weight W; 2nd, the centri- fugal force F ; 3rd, the pull of the arm AV. By completing the parallelogram of forces we obtain a diagram in which it is seen that F : W : : radius : heisfht of cone Now F, the centrifugal force, is equal to . Wr N2 W X r X N2 2950 2950 and h = — -, W _ W r X 2950 WrN2 h (feet) 2Q^O N^ Applying this rule to the question height of cone = -?5_ •= fiSft. Note. — ^The foregoing investigation and result shows that no matter what the length of the arm may be which supports the weight, the height of the cone is the same so long as the number of revolutions per minute is unaltered. 3. The arms of the above governor being 18 in. long, what amount in inches will the balls rise when the speed increases from 50 to 55 revs, per minute. The length of the arms is immaterial; balls rise 2'46 in. — Ans, CENTRIFUGAL FORCE. 203, i. A fly-wheel 10 feet mean radius (to e.g. of rim) 30 tons. Find the diameter of each of four bolts which hold the two halves of the wheel together, if the stress on them is 5000 lbs., when the wheel is revolving at the rate of 60 per minute. If we are to work from first principles in solving this problem, ■we must first find the centrifugal force of some definite portion of the rim, such as one foot of length or one inch of length. We will take a foot length measured at the centre of gravity of a rim section, which we may assume to be at the given radius of 10 feet. Circumference = 20 x 3"i4i6 = Sz'S^z ft. Weight of one foot = 5_^ — -- — ^ = 1069 lbs. ^ 62-832 ^ We now find the c.f. of this mass, which is io6q X 10 X 60^ ,, 2 = i-3ot;2 lbs. 2950 The principle to apply now is a very important one, it is the principle also used in finding the strength of a boiler shell, and of a pipe. Generally it may be stated as follows : — The total effect of a number of equal radial forces acting upon each unit in the length of a semi-circular arc, -when resolved in a direction at right angles to the diameter is equal to a single force multiplied by the num,ber of units in the diameter. Figure 73. In the case of a fly-wheel the principle means that the c.f. of one section of the rim must be multiplied by the number of such sections con- tained in the diameter. The resolved force of one- half the wheel is, therefore, equal to : — - 13052 X 20 = 261040 lbs. Fig. 73. 204 CENTRIFUGAL FORCE. and each bolt must carry one-fourth of this force, which, at 5000 lbs. per square inch, requires Area of each bolt = _5£_£ = 13-052 sq. in. 5000 and diameter of bolts = 4*07 inches. 5. Assuming the sectional area of the rim of wheel in above example to be 76 sq. inches, find the stress per square inch upon the material, due to centrifugal force, at the given speed. We have only to divide the total force as obtained above by the sectional area of two sides of the rim. Stress = 1717 lbs. — Ans. 6. A fly-wheel, $ tons weight, making 75 revolutions per minute. The rim is 10 feet outside diameter and 8 feet inside diam., the section of the rim being rectangular. You are required to find the centrifugal force tending to separate the wheel into two halves. (Neglect the effect of the arms ) Workings upon the same lines as in the previous example, it will be noticed that the radius to the centre of gravity of a section may be taken as the mean of the inner and outer radii, which is 4 ft. 6 in. Total force = io"925 tons. — Ans. 7. A fiat-rimmed pulley of cast iron has a mean diameter = 30 in. Find the stress upon the metal of the rim when the pulley is making 1090 revolutions per minute. It might seem that we ought to have given the sectional area of the rim in this problem, but it is proved, in Part III., that the stress upon the rim is the same whatever amount of metal is put into the rim of a wheel. It is convenient to assume a rim i in. thick and i in. broad, and to take a i in. length of the rim, circumferentially, to work CENTRIFUGAL FORCE. 205. with. We have now just a cubic inch of cast iron weighing, say,. ■26 lbs. , for which we must find the centrifugal force. r •26 X I'2q X 1000^ „ c.f. = >> = iio"i7 lbs. 2950 Now multiplying this radial force per inch of circumference by the number of inches in the diameter, in accordance with the principle enunciated previously. Total force = 110T7 x 30 = 3305 lbs. To carry the total force we have only the section of the rim at the two opposite sides, which equals 2 sq. inches. The stress. per sq. inch is, therefore, 5i2— 2 = 1652-5 lbs. — Ans. 8. A cast iron pulley is running with a linear velocity at the centre of gravity of the rim of 200 feet per second. Find the stress in the metal of the rim. Referring to Part III., the reader will notice that there is a rule for centrifugal force depending upon linear velocity, which is :. c.f. = where V = velocity per second in feet. g = force of gravity 32'i6. r = radius in feet. We may assume any radius we like, say i foot ^, - - . , -26 X 200^ the c.f or a cu. inch = r- 32"i6x I ft. multiplying by the diameter in inches 24 '26 X 200 total force = 7 x 24 32-16 which only requires to be divided by twice the sectional areai (2 sq. in.) to give the stress. 3828 lbs. — Ans. .2o6 CENTRIFUGAL FORCE. 9. If the tensile strength of cast iron is taken as 15,000 lbs. per square inch, what would be the bursting speed of the pulley in example 7? Having already found the stress (1652 lbs.) upon the material Tvhen the wheel is running at 1000 revolutions, we can work by .proportion, for taking care to square the revolutions, 1652 : 15000 : : (1000)^ : R^ n /l^OOO X lOOQOOO , R = /y/ -5 g;^ = 3013 nearly. The wheel would, therefore, burst asunder at a speed of about 3000 revolutions per minute or 50 per second. Another method would be to work for the bursting speed from the beginning, which would be the better way in most cases. The point to notice is that the total force separating the two halves must be equated to the total resisting force of the metal at the two sides. Thus if R is the required number of revolutions, T, . , j: -26 X I -21; X R^ Total force = ^ x lo. 2950 -and total resistance = 2 x 15000. Equating these gives R = 3012*7 as before 10. Required the stress per square inch in the rim of a rail- way truck wheel, when the speed of the train is 1 mile per minute. A cu. inch = •28 lbs. Stress = = 741 lbs. — A?is. 10-45 ill. Find the bursting speed of a fly-wheel 20 feet mean diameter; weight, 18 tons; material, oast iron. The bolts connecting the segments of the rim having a resistance equal to 50 per cent, of the rim itself. 267 revol. — Alts. lo. — Shafts. 1. The maximum twisting moment of a two-throw crank shaft is T, when the cranks are at 90 degs; what is the maximum twisting moment when the cranks are at 120 degs.? All other particulars are the same and the steam is carried full stroke. In figure 74, if W is the force acting along each of the connecting rods (which are supposed to have no obliquity) the total moment to turn the shaft is (W x ac) + (W X cV). This is true for any angle of cranks whatever. Now when the cranks cA and cB make angles of 45 degs. with the vertical, ac = «A = & = 6B. Fig 74- ac = Ji. The total moment is W ^ x 2 = W x i"4i4 (i). Let the length of each crank = i then {acY + («A)2 = i^, and 2 {acy = i Again, when the angle, which each crank makes with the vertical, is 60 degs., ah. = -^ Ac, or putting Kc — 1 «A = i = Vi Jt :. ac = \li' - (1)'^ = The total moment now = W ^fx2 = Wx 1732 (2). Comparing (i) and (2) the ratio is i '414 to i 732 or I to I "225 If the moment in the first case is called T, the moment in the ■second case is i "225 T. 2o8 SHAFTS. 2. If the maximum twisting moment of a two-throw crank is T when the cranks are at 90 degs. What is the maximum moment when they are at 180 degs.? I "414 X T. — Ans, 3. Find the ratio of the maximum to the minimum twisting moment, 1st, with cranks at 90 degs., and 2nd, with cranks at 120 degs., assuming 2 cranks and equality of power in each cylinder. ist ratio i'4i4 to i ; 2nd ratio 2 is i. — Ans. i. Find the ratio of maximum to minimum twisting moment in the case of three cranks at 120 degs. JVbie. — The least moment occurs when there is one crank on the dead centre ; and the greatest when one crank is horizontal. Ratio i'i54to i. — Ans. 5. Find the diameter of an iron shaft to work under a twisting moment of 85,000 pound-feet, with a factor of safety of 6. The formula is ™ . D^ X Stress i moment = — (all dimensions in inches and lbs.) D3 X S 85000 X 12 = 5-1 Now taking the shearing resistance of iron as 20 tons per sq^ inch and allowing a factor of safety of 6. 04. 20 X 2240 , ,, Stress = — — - — 3_ = 7467 lbs. D^ X 7467 . . 1020000 = L3_£. T-)3 — 1020000 X 5"I 7467 ' D = 8-865 in- SHAFTS. 209 6. If the correct diameter of shaft is 18 in. when the steam pressure is 160 lbs. and the stroke 5 ft. 6 in., what should be the diameter when the pressure is 185 lbs. and the stroke 6 feat; everything else bging the same ? The cube of the diameter varies as the product of gross pressure and length of crank. This works out to I9'36 inches for the required diameter. — Ans. 7. There are two ship's engines developing the same power. One is a quadruple-expansion with two cranks at 90 degs. The other is a triple-expansion with three cranks at 120 degs. If the latter ship has a 16-in. shaft, what size of shaft should the former have? Revolutions the same in each. If the force upon each of the triple engines' cranks is unity, the force upon each crank when there are only two must be f to give the same power. Now taking the position of maximum twist in each case, we have for the triple (one crank horizontal) Total moment = (1 x i) + (i x '5) + (i x '5) „ = 2-0 and for the quadruple (see example 1.) hence Total moment = (f x '707) + (f x 707) ,, = 2'I2I. 2'o : 2 "1 21 : : 16^ : D^ D = i6'3i in. — Ans. 8. A shaft is reduced by corrosion uniformly all round and thereby loses 16 per cent, of its strength, find what per cent, it hasjost of its diameter. ^ Putting D for- the old diameter, and remembering that, other thing-? being the same, the , strengtlj of a shaft varies directly with the cube of its diameter. , . , SHAFTS. D^ represents the original strength, and ■84 (D^) „ new strength which may also be represented by the new diameter cubed, say jc*. .-. -84 D3 = x^ :» = D X V^ X = T) y. -943 The loss of diam. is, therefore, '057 or 5 '7 per cent. 9. An engine with two cranks at 90 degs. and equal power on each crank, requires a 9-in. shaft. Required the diameter of shaft for the same engine but with 2 cranks at 120 degs. Also with 2 cranks at 180 degs. 9-63 in. for 120 degs.; 10-103 '"• for 180 degs. — Ans. dO. The total effective load upon a piston is 5^,000 lbs. when the crank has moved 45 degs. from the top centre. The length of connecting rod is 10 feet and stroke 5 feet. Find the twisting moment at that angle and the stress upon the material of the shaft, which is ±i in. diameter. When the length of the connecting rod is taken into account in estimating the twisting moment upon a shaft, we may refer to figure 75. CA is the crank making an angle •of 45 degs. with CB, the centre line of the engine, which is vertical. Now if the connecting rod BA is produced to (a) where it meets the horizontal Ca, drawn from the centre of the shaft C, the distance Ca must first be found, for it can be proved that the twisting moment in this position is equal to Ca multiplied by the load upon piston. - — '-th -^C \ (See part III.) \ * Draw AD parallel to aC, then in the triangle ADC, we have Fig- 75- SHAFTS. 211 AD2 + DC2 = (2i)2 =. 6A and since AD = DC, each = J^ = 1 767. Also in triangle ABD, BD2 = AB2 - AD2 = 100 - 3^ = 96I .-. BD = 9-842 The whole distance BC is now found to be 9-842 + I'J^y = 1 1 '609. We can now make the proportional statement BD : DA :: BC : C« or 9*842 : I -767 :: 11 "609 : Ca from which Ca works out to 2 '084 feet. Of course the length of Ca can readily be obtained by drawing the diagram to scale, if that method is preferred. Reduced to inches Ca = 25 in. almost exactly. Next, to find the stress upon the shaft. The formula may be •written approximately as : — .... , d^ X stress twistmg moment = or accurately as : — twisting moment = d^ x stress x — inserting known terms — 54000 X 25 = 14'* X stress x ^ — ^ Stress = 54000 X 25 X 16 2744 X 3-1416 Stress = 2505 lbs. per sq. inch. jgia SHAFTS. 11. Work out by the B.T. rule the diameter of crank shaft for a triple-expansion engine, with cylinders 33 in., 52 in., and 86 in., diam., and 60 in. stroke. The boiler pressure 150 lbs., and the cranks at 120 degs. The B.T. rule is : — / "C X P X D^ 77 d- where C = length of crank in inches. P = absolute pressure of steam. D = diam. of l.p. cylinder. d = ,, h.p. ,, / = mo (constant for 1 20 degs.) S = diam. of shaft. Diam. iS"54 in. — Ans. 12. Having given the B.T. formula for crank shafts^ P = L^^ X ^2 + ?^) ; find what fraction of the gross. steam pressure in the boiler would be sufScient, if exerted in the low-pressure cylinder, to produce by itself a stress of 8000 lbs. in the metal of the crank shaft. The value of / may be taken as 1265 and The B.T. rule with the given numbers is : — (2 + 7) (i). 1265 S3 C D-^ Now the fundamental rule for the moment of resistance of a. S ^ Stl'SSS X TT • shaft being , which equals the twisting moment imposed by the load on the low-pressure engine. .". putting x for the unknown steam pressure on the l.p. piston. ^ X D25 X C ^ S^ ^ 8000 ;r 4 16 S^ X 8000 C X D2 X 4'' .(2). SHAFTS. 3113 The question asks for the ratio of x to P. X _ S^Sooo ^ 1265 S^ P C D2 X 4 • C D2 ^ ^ all the letters cancel out and the result is XI. — = — - — . — Ans. P 5-692 13. The B.T. formula for diam. of shaft being:— C P D^ S 3 = /„ D =*\ f where / = 1110 lbs. ^^ + ^-.>/ and the mean pressure referred to the low-pressure p cylinder being — - where R is the total ratio of 1 + '3 K expansion. You are required to find what stress per sq. inch the shaft is working under. The cut-off in the h.p. cylinder is at '6. In tackling this question the student is advised to assume a ratio of cylinders of i : 8, although it is immaterial what ratio is assumed. The total ratio of expansion will then be 8 -H "6 = R. After simplifying the expression for referred mean pressure, the investigation is very similar to the last example. Two expres- sions for S^ are obtained and equated. Stress = 8/ = 8880 Vos.— Ans. Hollow Shafts. 14. You are required to substitute a hollow for a solid shaft. The solid one is 10 in. diameter and the hole in the new shaft is | the outside diam. of the shaft. Both shafts are of the same material, and both are to have the same resistance to torsion. Equating the formula for moment of resistance in the' two cases : — ' ' ' \ (D* - d^) S ^ 108 X S .0x5-1 5-1 a 14 HOLLOW SHAFTS. Now writing — for d, and cancelling common factors- 2 \ 16/ = 1000. D 16 D* - D* — — = 1000. 16 D IS D3 -:2 ■ = 1000. 16 T-, q 1 6000 J T-« D^ = and D = io"2i7 in. 15 15. A hollow shaft 15 in, outside diam. and 5 in. inside, is taken out and a solid shaft put in its place. What must be the diameter of the new shaft so that its resistance to torsion shall be \\ times greater than the old one, both being made of the same material? The equation is the same as before, except that the formula for the hollow shaft is multiplied by i "5. i7'ii in. — Ans. 16. It is proposed to substitute a hollow shaft for a solid one. The strength to be 1.^ times that of the solid shaft taken out. The hole in the hollow shaft is to be *4 of the outside diam. Express the outside diameter in terms of the diameter of the solid shaft. D = I -096 of solid. — Ans. 17. A and B are two shafts. A is solid, B is hollow, with a hole ^ its diameter. A is 10 in. diameter. Find diameter of B if its material is 10 per cent, stronger than A's material, but its moment of resistance & per cent, weaker. The formula for the hollow shaft must be written with a stress = yJ^ S, and equated to -^^ of the solid. Outside 9'6o5; inside 3 '842. — Atis^ HOLLOW SHAFTS. 2 1 J 18. Compare the strengths of two shafts, one hollow with a hole through it of one-half its diameter, and the other shaft solid, but of the same weight per foot of length. As -| : ^-^~ or 1*443 • '• — Ans. 19. A hollow shaft is required to take the place of a solid one 12 in. diameter. The hole in the hollow shaft is to be 7 in. diameter. Find the outside diameter so thai the two shafts may be of equal strength. If we equate the two formulae for the moment of resistanceof a solid and a hollow shaft, we get — (D^ - 7*) S ^ 123 X S 5-1 D 5-1 and cancelling the factors common to both D D3 - ?gi = ,728 This equation is difficult to solve, but by assuming a reasonable value for D in the term -^-— (say 12^ in.) it becomes practicable and the error involved is inappreciable, thus — D^ - 192 = 1728 D^ = 1920 Extracting the cube root either by logarithms or arithmetic. D = 1 2 '43 in. IVbie. — By using the value obtained for D in place of the assumed value, t2^ in., and working the equation out again, a still nearer approach to an accurate result is obtained. This method may be useful in other equations which we are unable to solve rigorously. 20. A hollow shaft is required to be as strong as a solid one, 12 in. diameter, of the same material. If the hole is 8 in. diameter, what should the outside diameter be? i2"7i in. — Ans. :Sl6 SHAFTS. Combined T-wisting and Bending. 21. A shaft is subject to a twisting moment of 120,000 pound-feet, and at the same time to a bending moment of 160,000 pound-feet. Determine its diameter, if the combined stress is not to exceed 10,000 lbs. ' The rule for combining' a twisting moment and a simultaneous bending moment is as follows : — Equivalent T.M. = J(T^+W) + B where T is the imposed twisting moment, and B the imposed bend- ing moment. This works out to 200,000 + 160,000 = 360,000 pound-feet or 4,320,000 pound-inches. To find the diameter of shaft necessary to transmit this moment the usual formula is used, viz. : — ^— = 4.320,000 ID _, /4,-J20,000 X 16 «= \/- ' = i3"oi V 3'i4i6 X 10,000 Find the equivalent twisting moment on a steamer's tail shaft, having given: — Weight of propeller 12 tons, leverage 28 in. Engines produce a twisting moment of 5,500,000 reckoned in pounds and inches. T.M. = 6,305,700 nly. — Ans. 23. Find diameter of tail shaft for above vessel, using a maximum stress of 6000 lbs. i7"49 in. — Atis. SHAFTS. 217 Coupling Bolts. 2i. Given diam. of shaft 12 in. Number of bolts in coupling 8. Radius of bolt circle 9 in. Find diameter of bolts. Whilst the shaft as a whole is subject to a twisting- moment round its centre, the coupling' bolts are subject to a simple shear stress, which is uniform over the sectional area of each bolt. The total resistance to shear of all the bolts in a coupling is, therefore, found by multiplying the total area by the stress allowed per .sq. inch. Thus — Let d = diameter of each bolt. S (^2 ^ \ ^ Stress = resistance to shear; multiplying now by the leverage or radius of the bolt circle. The moment of resistance = 8fi?2 X - S X 9 (i.) 4 We have now to make this equal to the moment of resistance ■of the shaft itself, which may be written — 12^ S IT Moment = 16 . 8d^7rS X 9 ^ 12^ S IT "4 16 <:ancelling the common factors 72 d^ _ 1728 4 t6 from this we find d = J6 = 2*45 in. 25. Make a formula for the diameter of coupling bolts for a hollow shaft, so as to have equal strength with the shaft itself. V 4^D diam. = a./— — ^^: — ■ — Ansi n 2l8 COUPLING BOLTS, 26. Shaft 15 in. diam.; 8 bolts in coupling at 12-in. radias from centre of shaft. Find diameter of bolts, if their material is 12 per cent, weaker than the material of the shaft, but their actual strength is ii per cent, greater than the shaft. 88 S Proceed as in the last example, but use for the stress in ICO the doli formula, and multiply the shaft formula by — - when writ- lOO ing the equation Diam. 3f in. or 3*374 inches. — A?is. 27. A tunnel shaft has a twisting moment of 385,000 Ib.-feet. The coupling bolts are 3 in. diameter, and 8 in number^ at a radius of 12 in. Find the shearing stress on the bolts. 6808 lbs. per sq. in. — Ans. 28. In the previous example, suppose that the couplings are separated a distance of 1^ in. and the bolts fit tight in the holes in each coupling, what stress due to bending win there be upon each bolt? P u Fig. 76. Looking' at the sketch, figure 76, in which the distance apart of the couplings is exaggerated, it will be seen that the bolts will take a double curvature if they are tight in both couplings. There will consequently be no bending at a section half-way between the couplings, this is at xy. Hence the total shearing force at this section will be acting to bend the bolt as a cantilever pure and simple. Now the total shearing force upon one bolt is 48,125 lbs., and the length of the cantilever from the mid section to the face of the coupling is J in. The bending moment is, therefore, 48125 X "75 = 36093 Ib.-inches. COUPLING BOLTS. 2 19' Equating- this to the moment of resistance of a beam of round section which is ^^ ^ Stress ^ ^^ ^^^^ 32 , Q^ X TT X Stress 36093 = -^ 32 .". Stress = ^ — ?^ 2_ = I -5627 lbs. 27 X 3 1416 2Vb^. — The stress just found is tensile at one side of the bolt and compressive at the other side. It exists, moreover, simul- taneously with the shear stress of exercise 27. Horse Power Transmitted by Shafts. 29. Find the diameter of a steel shaft to transmit 1350 H.P. at SO revolutions per minute, the maximum stress being ^500 lbs. per sq. inch. A troublesome thing- which meets one in working an example like the above from first principles, is in converting horse power into twisting moment. We have to remember that "horse power" means so vasLay foot-pounds of work in a minute ; whereas, twisting: moment has nothing to do with time, as well as being reckoned in the shaft formula in ^oxm&.-inches. It is best to assume a crank a foot long to start with. We have then the distance travelled by crank pin each minute = 2 TT X 50 = 100 TT = 3 14" 16 feet. Bringing the H.P. to foot-pounds and dividing by "feet travelled" gives us the mean force to turn the crank, which is 135° ^ 330°° _ i^iSSolbs. The twisting moment is now obtained 314 by multiplying the length of the crank in inches — T moment = 141880 x 12 320 HORSE POWER TRANSMITTED BY SHAFTS. It only remains to equate this to the moment of resistance of the shaft as in previous examples. D' X 4SOO oo ^ — = 141880 X 12 from which D = 12 '45 in. 30. A hollow shaft (diam. of hole = | diam. of shaft) is required for an engine of 6100 I.H.P., running at 75 revolutions per minute. The stress is not to exceed 8000 lbs., and the ratio of maximum to mean twist is 1*6. Find the diameter of shaft. Proceed, as in the last example, as far as the mean twisting moment. Then multiply by i '6 and equate the result to the formula for a hollow shaft. Diam. 17 '73 in. — Ans. 31. Oiven I.H.P. iiSO. Bevol. 66. Stress 8000 lbs. Ratio of maximum to mean twist 1-5. Find diameter of shaft if solid. 16 in. — Ans. Rigidity of Shafts. The first thing is to understand the meaning of the "co-efBcient •of rigidity." Let us experiment with a block of india-rubber as being easier to stretch or deform in any way than the metals. Let us cement the rubber block down to a table, figure 77, and also cement a flat piece of wood upon the upper face. Now if a spring balance be hooked on to the top piece of wood and _^ _ pulled horizontally, we shall '~^'^^^y77777?777?77?;77?:^7^ be putting a shear stress Fig. 77. upon the india-rubber upon planes parallel to the table. And acconipanying the shear stress there will be a shear strain. It is the shear strain that we are first concerned with. The dotted lines show the outline of the RIGIDITY OF SHAFTS. 2211 rubber when a certain shear force is exerted by the spring balance, say 80 lbs. Now the distance Cc or Bd is the distortion or siram produced by the above force ; but just as in the case of the stretch- ing of a bar where we estimate the "strain" as a fraction of the length, so in the present case the strain Cc is considered as a Cc fraction of AC. That is, we call the strain -— . It should be AC noticed that Cc is at right angles to AC when we are measuring rigidity. Also it will be obvious that when the amount of distortion- Cc becomes equal to AC, the value of the fraction becomes unity (This corresponds to the stretch of a bar when it becomes as much as the original length.) In the experiment -^-^ might amount to xjj- or •!. Next we must notice the amount of shear stress which has- been exerted. This is found by dividing the force of 80 lbs. by the area of the plane of shear. If the area of the plane CD is 4 sq.. inches, shear stress is exactly 20 lbs. per sq. inch. We have now obtained both stress and strain, from which the co-efEcient or modulus of rigidity is found, as in Young's modulus — 5^1?^^ = Modulus = £2 = 200 lbs. Strain "i The modulus of rigidity for india-rubber is thus 200 lbs. The Modulus for cast iron = 6,000,000 lbs. ,, ,, wro't iron = 12,000,000 ,, ,, ,, steel = 13,000,000 ,, 32. If a block of oast iron, 2 in. square and ^ in. thick, were treated like the rubber in the previous example, what force would be necessary to produce a displacement of ■0002 inches? ^, • ■ 1 "0002 . Stress ^ ^ The stram is here = = -0004, . . = 6,000,000, •5 ^ -0004 and stress = 2400 lbs., but as there are 4 square inches of shear surface, fotce = 9600 lbs. — Ans. RIGIDITY OF SHAFTS. ^3. Find the angle of twist in a shaft, 20 ft. long and 16 in. diameter, under a twisting moment of 6,000,000 inch lbs. The first thing- is to apply the ordinary formula for the streng-th •of a shaft and find the stress in the metal due to the twisting moment 1 6' X Stress x tt 6,oco,coo Stress = 7460 lbs. 16 Now Strain Strain Strain Stress -^ Modulus 7460 -e- 12,000,000 •000622. 20 Jeel'- In figure 78, DC is a longitudinal line drawn upon the top of "the shaft before the twisting force is applied. The effect of the ■force is to twist the shaft slightly all the way from the fixed end D to the other end, where the twist- ing force is sup- posed to be applied. The line DC, therefore, takes a spiral direction Dc, and the point C is -displaced to c. Remembering that the displacement is compared with the dimension at right angles to itself when measuring rigidity, Cc -the strain is expressed as ^ , which must have the value already jfound for it, viz., "000622. Fig. 78. Now DC = 20 ft. 240 m. 240 and Cc •000622 = "000622 X 240 = •149 m. RIGIDITY OF SHAFTS. 223 Finally upon the end view of the shaft, in figure 78, we can set off a circumferential displacement, ab equal to Cu = '149 in. And as the diameter of the shaft is 16 in., its circumference is ep'oh in. .'. 50 '26 : '149 :: 360 degs. : angle of twist which works out to i -067 degs. = i deg. 4 min. the required answer. Z\. A formula for the twisting of a shaft is: — ~ _ 140 X (f * 6* T = twisting moment in lb. -feet. L = length of shaft in feet. Q =■■ angle of twist in degrees. Use this formula and find the angle of twist of a shaft 30 feet long and 12 in. diameter, when subject to a twisting moment of 300,000 lb. -feet. 3*099 degs. — Ans. 35. Working from first principles, as in example 33, anfl assuming above angle of twist to be correct in practice, find co-efl9cient of rigidity. 11,788,8^8.— ^WJ. 36. Rudder head 6 in. diam. and i feet 2 in. in its parallel part below the tiller. The total twisting moment upon the rudder head is 126 inch-tons. The length of tiller lis 2 ft. 6 in. Find the load on chains and angle of twist of head. Modulus of rigidity 10,500,000. Load 4-2 tons; stress 3 tons; angle "607 degrees. — Ans. ,' 11. — Beams. In the equilibrium of every beam there will be found an opposition of moments. The load upon the beam tends to deflect the beam and in so doing exerts a bending -moment. Then the tensile and compressive stresses in the material — especially at the upper and lower surfaces — tend to resist this bendingf, and so exert a mrnnent of resistance. These two moments are always equal and opposite at every position in the length of the beam. Although every beam is subject to a bending moment producing stresses in the material, there is no such thing as bending stress. Every stress must either be a shear stress, a tensile stress, or a compressive stress. In a beam all three are generally present at every section. mw The mechanical model, illustrated in figure 79, is for the purpose of showing these three stresses in a simple cantilever beam. It consists of a light block of wood A, connected to an upright support B, by means of two spiral springs T and C, each secured to the block at one end, and to the sup- port at the other. The springs are so stiff that the block is held projecting horizontally without any appreciable deflection due to its own weight. A cord is fastened at the inner end, passes over a pulley, and terminates in a hook, to- which weights may be attached. I Fig. 79- BEAMS. 22^- Now if a weight, W, is hung from the end of the cantilever, two movements are noticeable. Firstly, the beam is tipped down at the outer end, so as to compress the lower spring and stretch the upper one. And secondly, the beam is pulled down bodily so as to make both springs take an oblique position. Next let weights, Wj^, be attached to the hook upon the cord until the beam is raised to its original position, or rather until the springs become again horizontal. It will be found that this occurs when Wj = W. The model, figure 80, now illustrates the three forces or stresses which exist in a loaded beam. If we concieve the springs to be part of the cantilever, the upper one shows the amount of tensile force at the upper surface. Fig. 80. the lower one shows the amount of compressive force at the bottom surface, and Wj, measures the shearing force at the vertical section TC. Moreover it will be noticed that the model can be made to rock when slightly disturbed round the point F, situated between T and C, hence F is a point in the neutral axis, that is where there is neither tension nor compression. As regards the equilibrium of the beam, and neglecting the weight of the beam itself, it is obvious that the bending moment round the axis F caused by the weight W is = W x L. Also that this moment is resisted by the two springs T and C, each acting with a leverage equal to one-half the depth of the beam. If, therefore, S is the force upon each spring, the moment of resistance is = 2x(Sx-) = Sfif. The equation of equilibrium for this model is, therefore, WL = S^. 226 BEAMS. Now this equation is typical of the equation for every beam. The bendingf moment must always be equated to the moment of resistance as in the above equation. It will be seen from the above that the study of beam problems leads in two directions ; one is the question of bending moment due to different kinds of loading; and the other is the question of resistance due to the form and material of the beam. Bending Moments. (See Part III.) Cantilever, load at end, - WL WL 2 WL 4 WL Cantilever, load distributed, ■Supports at the ends, load in centre. Supports at the ends, load distributed Fixed at the ends, load in centre. Fixed at the ends, load distributed. 8 WL 8 WL 12 Moments of Resistance. (See Part III.) S = tensile or compressive stress. b = breadth of beam. d = depth of beam (or diam. if round). I = moment of inertia of cross section. y = distance from neutral axis to where stress (S) is acting, M.R. = moment of resistance. For rectangular section M.R. = ♦ 6 d^ S For circular section M. R. = IO*2 BEAMS. 227 For symetrical hollow section M.R. = (BD^ - M^) S D X 6 !For round hollow section M.R. = (D^ " ^*) S D X 10"2 c iFor any section whatever M.R. = I x — y The following definitions are important : — I St. — The shearing force upon any vertical section of a beam is ithe algebraic sum of all the forces acting on the beam (including the reaction of any support) to the right or left of the section, but not both. 2nd. — The bending irumient at any section in the length of a beam is the algebraic sum of the moments of all the forces acting on the team (including' the re action of any support) to the right or left of the section (but not both) the axis for each mmnent being at the given section. 1. A steel bar 6 in. deep and \.\ in. thick forms a beam. It is 17 feet long, and the ends are securely riveted to supports. If the maximum stress allowed is 5 tons per sq. inch, find the safe load for the beam when distri- buted. Also give the shearing stress when so loaded. Using the formula for resistance to bending of a rectangular ■Dr\2 c , viz., moment = — - — , we at once find the ^nce to be = i-^S x 3^ x 5 = 37.5 ton-inches. ■Dr\2 c fcar, viz., moment = , we at once find the moment of resist- Also it is known that the maximum bending moment of a . WL *iniformly loaded beam, with fixed ends, is Equating the two moments we get ^ ^ ^°4 = 37-5, from which 12 "W = ^-205 tons. 228 BEAMS. For the shearing force we must notice that one-half the load is supported by each end of the beam where the sectional area = 6 in. X i^ in. = 7'5 sq. inches. The shearing stress is therefore : — I'lOZC X 2240 o 11- — 3 3^ = 329*20 lbs. per sq, An air-pump lever is made up of two plates, whose section^ is 16 in. deep and f in. thick, with a hole 8 in. diam. where the centre gadgeon passes through. The length of the cylinder arm is 3i in., and of the pump arm is 26 in. Find what weight can be suspended from the air-pump end to produce a stress of 7000 lbs. in tha lever. The rule for the moment of resistance of a hollow rectangular beam is „ ^ (BD8 - bd^) X S 6D where B and D are the outside dimensions and b and d the inside- dimensions. M = (75 X r6^ - 75 x 8^) x 7000 6 X 16 And this must be doubled for the two plates. The formula works out to M = 196,000, therefore, for two* plates, the moment of resistance is 392,000. We have now to equate this value to the bending moment — ' Let W = weight on pump end W X 26 = 392,000. W = 15077 lbs. 673 tons. — Ans.. BEAMS, 229 3. A box girder of cast iron is 9 in. sqaare in section out- side measure and 7 in. square inside. Its length between supports is 16 feet. Find the amount of load which it will carry in the centre of the span if the stress is limited to 1 ton per sq. inch. (Figure 81.) Using the same rule for moment of resistance as in the last example, M.R. = (9 X 9^ - 7 X 7^) X I ton 6x9 X — 77 '03' •'^'^'i when this is equated to the bending moment ( — - j the load works out to i '604 tons, the Tig. 81. answer. But we will also work this ■example by the principle of "moments of inertia," so as to show the application of the principle to a beam. First fix the position of the neutral axis of the beam, which always passes through the centre of gravity of the section. In the present case the neutral axis XY divides the section equally. We have now to find the moment of inertia (written I) of the section about the axis XY. Take the upper half first. It is proved, in Part III., that a rectangle swinging about one' edge has its I = , where b is the breadth parallel to the axis, and A is the 3 dimension perpendicular to the axis. I of outside rectangle = 2 '■^ •" . = 273*3 I of inner rectangle = '- ^_3L = loo'o 3 Subtracting these results I of half section = 173 "3 and, therefore, of the whole section is twice this or 346-6. , ; 1 330 BEAMS. Now to change "moment of inertia" into "moment of resist^ ance," all we have to do is to multiply by , where S is the stress 2Xy distance from the neutral axis, hence Moment of resistance = 21 "^ ' °" = 77-02 ton-inches. 4-5 The problem is now finished as usual by equating- moment of resistance to bending moment ( for a central loadj. W X 192 in. 4 W = I '604 tons. 77-02 ton-inches. Note. — Since the moment of inertia of a rectangle about an . b h^ . axis through its centre is this formula might have been used 12 in the above beam with advantage. Thus — I = IJL^ - UUl = 346-6 12 12 -^^ as found above. i. K wrougtat-iFon beam of Z section as in sketch, is loaded with 450 lbs. per foot of effective length, which is 16 feet. The ends are rigidly fixed to the supports. Find the maximam stress on the material. (Figure 82.) If the neutral axis is drawn through the centre of gravity of the section, it will be seen that the same methods can be adopted as for the hollow beam of the last example. The outside rectangle is 3'5 in. broad and 9 in. high, and the inside rectangle to be taken away is -45 in. less in breadth and it in. less in height. 1 /r-5'S ^1^ -3/2 ->: •4-5 Fig. 82. Having obtained the I of the beam multiply by — to convert 4"5 I into moment of resistance. BEAMS. 231 The bending moment being gives an equation to find the- stress (S). I = 87*31 ; S = 5937 lbs.— ^mj. 5. A beam is built up of two cbannel irons riveted back to back, as in sketch. The beam is merely supported at the ends, and carries a load, in the centre, of 5 tons. Find the stress at the upper and lower surfaces. The span is 12 feet. (Figure 83.) I = 269"i4; stress 7491 lbs. — Ans. ■•'[^^^^s ■6" g^ i each. :— 8 - Fig. 83. Find the central load for a girder of cast iron 20 foot span. The top flange is 6 in. x 1^ in. The bottom flange 14 in. x If in, and the web 12 in. x 1^ in. (between flanges). Allow 2 tons maximum stress in tension, and 8 tons in compression. (Figure 84.) We have first to find the position of the neutral axis. Now the centre of gravity is readily found by taking moments of the several areas about the base as an axis, and then dividing by the total area. Thus : — Top flange- = (6 x ii) x 14^ = 130-5 Web - - = (12 X li) X 7f = 139-5 Bott. flange = (14 x if) x § = 21-44 ^//////////A Fig. 84. The sum is 291*44 and the total area is 51*5. 5I-S X X = 291*44 X — 5-66 inches. Hence — The neutral axis may, therefore, be taken as 5^ inches from the bottom and 9I from the top surface. 232 BEAMS. The moments of inertia of the parts of the section above and below the neutral axis are now calculated separately and added tog-ether. The total I will work out to 1567-12. Now if we work with the tensile stress of 2 tons at the lower surface of the beam the moment of resistance _ 1567 X 2 tons ii but by using the compressive stress given as 8 tons at the top surface the result is 1567 X 8 tons It is readily seen that the first moment is the lesser of the two and must, therefore, be used for finding the load. The bending moment being i- , the equation is 4 W X 240 I ^G"! y. 2 c 1-1 i- = - - ' from which 4 5'5 W = 9"497 tons. Note. — It must be confessed that the above is a tedious and intricate calculation to make in an examination room, even with the help of logarithms ; but it may be pointed out that a check upon the result can readily be obtained by using the approximate method for a flanged beam, viz. : — Moment of resistance = strength of weakest flange x distance between flanges. In the above case (14 X If X 2) X 12 = WJLJ40 4 W = 9-8 tons. BEAMS. 233 7. A cast-iron flanged beam has a total depth of 20 in. and a length of 2i feet. The weight of the beam is iSO lbs. per cubic foot of material. The top flange measures 8 in. x 2 in., the bottom flange IS in. x 3 in., and the web IS in. x 3 in. Find the maximum stresses when loaded with 80 tons distributed over its length. No/e. — The weight of the beam itself merely increases the tota* distributed load ; and the bending moment for a distributed load is ^^* L stress — ^ — Also when multiplying the I of the section by '— it must o _y be noted that y may be the distance from the neutral axis to the top surface of the beam, or to the bottom surface, and which ever value is used will give the corresponding stress. Weight of beam 7950 lbs.; e.g. of section 12 in. from top surface; I = 5000, stress at top 16170 lbs.; stress at bottom 10780 lbs. — Ans. 3. Find the distributed load for a bulb iron beam fixed at the ends. The length is 20 feet, and the section as in sketch. Allow 5 tons maximum stress both in tension and compression. (Figure 85.) <—6' W/^/Z/Z/^Li!^^ By taking moments about the vpper surface as an axis, the centre of gravity of the section will be found to be 2*35 in. from the top, and, therefore, 4*275 in. from the extreme lowest point. This fixes the neutral axis. The I of the portion above the neutral axis presents no difference to a previous example ; but the portion below the axis consisting partly of a circular section must be treated specially. A .yery near approximation for the I of the hulb consists in multiplying its area by the square of its distance from the axis. Thu;s :— ;- \'/2. diam . Fig. 85. I of bulb = 1767 X (3'525)^ = 21-956. 234 The I of the portion of the vertical web below the neutral axis = -5 X 2775)^ = 3.561 Above the neutral axis we have I _ 6 x' (2'35)^ _ 5'5 X (1725)^ = 15.64 3 3 Adding the moments up — ' Total I = 41 "I w and moment of resistance = 41 '157 x 2 ^ ^' 4'275 Equating this to the bending moment at centre of girder Load X 240 ^ 41 '157 X 5 12 4-275 Load works out to ;j'4o7 tons, which includes the weight of the- beam itself. 9. A girder H section of wrought iron can be placed in twO' positions, viz., H and T. If the dimensions are 6 in. both in height and width, and the thickness of metal is ^ in. all over, compare the strength of the girder in the two positions. In the 1st position we have two vertical plates and a central horizontal plate with the neutral axis running through it. ' For the vertical plates : — I = 12L^ = 18 12 and for the middle portion T 5 X (-5)3 I = '' L-JZ_ = •OC2 12 ^ Total I (list position) = 18.052 (i.) In the and position we have outside and inside rectangles to work out ; I=6j^_ SI^JLi! = S0.7X (2.) 12 12 ' COMPARISON OF BEAMS. 235. Now since the multiplier is the same in both cases, we y need not use it, but simply compare the moments of inertia found in (i) and (2). Ratio = 18 "05 2 : 50-7 I : 2"8i nearly. 10. Find the relative strength in the two positions of an H girder when the flanges are 10 in. broad and 1 in. thick,, and the web 3 in. between flanges and 1 in. thick, Note, in this case, the multiplier must be used, but the stress will cancel out as it is the same for both. Ratio 33-4 S : 33-5 S I : I '003. — Ans. 11. Given flanges 6 in. x 1 in., web | in. Full depth 10 in. Ratio I : 4'5. — Ans. 12. A piece of timber of uniform section 12 in. by 15 in. is used as a beam to support a central load. How does its strength compare, 1st, when placed with 15 in. vertically, and, 2nd, with 12 in. vertically. B D^ S Using the formula — - — , the ratio is as 5 to 4. — Ans. 13. Find the position of the neutral axis and the moment of inertia of a T iron beam, the top flange being 6 in. x f in. thick, and the central web also 6 in. x f in. thick, the total depth being 6| in. Neutral axis 2yV in. from top; I = 39'34. — Ans. -336 SHEARING FORCE OF BEAMS. 14. A rectangular iron beam 10 in. deep, li in. thick, and 20 feet long, is loaded with a weight of 5 tons at one- fourth the length from one end. Draw the bending moment and shearing force diagrams, and find the maximum stress. (The weight of the beam itself may be neglected.) The first step is to find the weight which is taken by the end ■supports. In figure 86, taking moments about B Weight at A X 20 = 5 tons x 15 ' i A = 3f tons, •and taking moments about A Weight at B X 20 = 5 tons x 5 B = i^ tons. fOMS _ „° I Now the shearing force at •any section of a beam is really the force or tendency that ■exists to cause the one portion of the beam to slide past the other portion. The shearing force is always greatest close to the supports, and it is best realized by imagining what would happen if the beam were sawn across at any part, for if there is a + shearing force at that particular part, the right-hand portion would drop below the left. If there is no shearing force (as. at the 15 Fig. 86. •centre of together. a beam uniformly loaded) both portions would drop a For the shearing force diagram, figure 87, we will take the forces to the left of the position C. There is only one force acting, namely, the reaction of the support, which is 3f tons (+). This is set up to •(c) to any scale, and as there is no additional force between A and C, the line is drawn horizontally from c to a. Next take a position as D. The forces to the left of D are the weight of 5 tons acting Fig. 87. mic iL DIAGRAM OF BENDING MOMENTS. 237 downwards ( - ) at C, and the reaction at A 3! tons acting upwards ( + ). The algebraic sum is - 1^ tons, which is, therefore, set ofl" downwards to (d). Also, since there are only these two forces to the left of all positions between C and B, the shearing force is constant for this distance along the beam, and the diagram line is drawn through d horizontally. The shearing force diagram is a c e d b. Next as regards bending moments. Take the position C where- the load is applied. The bending moment at C can be written as 3f tons X AC or as ij tons x CB according to the end we are- working from— Moment at C = i^ x 15 = i8| ton-feet. Also at any intermediate point as at D 7^ from B — Moment at D = i;^ x 7^ = 9! ton-feet. These moments set off to scale, as in figure 88, enables us to draw the diagram AVB, in which it will be noticed that the greatest bending moment is at the position of the load, from whence it decreases Fig. 88. uniformly each way to the ends of the beam where it is nil. To find the maximum stress we- must- equate the bending moment, which is i8f ton-feet, to the- B D^ S moment of resistance of the beam section, viz., . 6 , i'?xiooxS o hence —^ ^ = is'7^ x 12 6 '^ all dimensions being now in inches, from which 8 = 9 tons exactly. 15. A beam 12 feet long has two loads, one of i tons at 3 ft. from one end, and one of 6 tons at 8 ft. from the- same end. Draw shearing-force and bending-moment . diagrams. If the definitions on page 227 are carefully studied, this problem, will not present any great difficulty. 338 BOAT DAVITS. i6. A boat davit 18 ft. high from foot to head has an ovep- hang of 5 feet. Find the stress on the davit at the eye bracket, which is half way up when lifting a boat weighing 30 cwt. and 30 passengers at HO lbs. each. The diameter of davit at bracket is 6 in. A davit is a combined cantilever and column. There is both TDcnding' and crushing- forces upon it, and the actual stress upon the material is found by adding the stresses due to each cause tog^ether. In the first place if we divide the total load between the two davits at head and stern of the boat we shall obtain the load carried by each. This works out to 3780 lbs. apiece. Now as a crushing load upon the davit at the 6 in. diameter, we have crushing stress = ^ — = i-j-; lbs. *" area ■'•' Next consider the davit as a cantilever. The overhang being 5 feet the bending moment is merely 3780 lbs. x 60 in., and since the moment of resistance of a round bar to bending is — - we have IO-2 the equation: — 62 X S = 3780 X 60 IO-2 .'. S = 10710 lbs. per sq. in. This stress, of course, is tensile at one side of the davit and com- pressive at the other, and, therefore, when it is combined with the dead weight compression of 133 lbs. per sq. in., the result is a total compressive stress of 10843 lbs. and a net tensile stress of 10577 lbs. -17. With the same davit as in the last example and sup- porting the same load, let the vessel heel over IS degs towards the boat side, and find the new stress upon the material of the davit. Also the pull upon the eye bracket. Figure 89, shows what forces are acting upon the davit in its ■inclined position. BOAT DAVITS., 239 It will be noticed that the foot- step carries the full load of 3780 lbs. which hangs from the head of the davit, and, therefore, the load and this vertical reaction form a couple with an arm or leverage = AC, so that the moment to tip the davit over-board is 3780 x AC. But this moment is counteracted by the pressure at the eye bracket and the reaction at the foot-step, parallel to the deck, which form another couple. li x = these forces we have : — X X 108 in. = 3780 X AC. As the calculation of AC is troublesome, we have drawn the figure to scale and measured it. (AC = 114 in.) X X 108 in. = 3780 X 114 in. X = 3990 lbs. wliich is the pull upon the bracket. Next for the stress on the metal. The moment of resistance of the bar is the same as before, but the bending moment at the position of the eye bracket (where it is greatest) is now 3780 ibs. X DF and DF being measured = 85 in. .'. the equation is: — ^LJLA = 3780 X 86 10-2 ■^' S = 15351 lbs. If we now add the dead weight compression already used in the previous example, 133 lbs. (although strictly it should be first multiplied by the cosine of 15 degs.) we get — Total stress (compressive) = 15484 lbs. 240 CURVATURE OF BEAMS. Curvature of Beams. 18. A cast-iron girder i in. deep has its neutral plane 3 in. from the upper surface. Find the stress, top and bottom, when the girder is bent to a radius of 100 ft. Take E = 16,000,000. There is a very simple law connecting- the radius of curvature with the stress in the case of a loaded beam. The proof is given in Part III., the law is: — Stress _ jf •^ where R is Modulus R the radius of curvature at the neutral plane and y the distance from the neutral plane to the upper or lower surface where the stress is required. Applying this to the example where R = 1200 inches. Stress T 16,000,000 1200 C-. ■? X 16,000,000 ,, btress = -i '- '- = 40,000 lbs. 1200 Similarly for the bottom surface Oi IX 16,000,000 ,, ^^^^ " IW^ " ^^'333 ^tis. 19. Find the radius of curvature of a wrought-iron girder 10 in. deep with equal flanges top and bottom, when loaded so that the stress in tension and compression is 11,000 lbs. (E = 29,000,000 and neutral plane is in the centre). Radius = 1099 ft. = 13182 inches. — Ans. 20, A 12-in. steel rule, 1^ in. broad and ^ in. thick is bent to a radius of 8 in, Find the force exerted to bend it. We will first find the stress Stress _ I 29,000,000 64 -I- 8 r,. 29,000,000 Stress = ^' ' - = 56640 lbs. 512 CURVATURE OF BEAMS. 241 The question now arises, how is it bent? That is, how is the bit of steel considered as a beam to be loaded? If the curvature is the same from end to end the only way would be to press the rule against a mould of the exact radius required, viz., 8 in., for no kind of loading of an ordinary beam will produce a uniform curvature for the entire length. The nearest approach to it is when a beam is supported at the ends and loaded, as shown in fijfure 90, with two equal weights at equal distances from the ends. It is easily proved that between the points c and rf the beam will take a uniform curvature — or in other words — bend to a circular arc. The bending moment in such a case is, of course, W x ac or W X db, and the bending moment is constant for any point between c and d. To prove this take any intermediate point as x. The bending moment at x is : — Wxa;c-Wxc^ = Wx (ax — ex) = W X <7c which is the same as at c. To find the value of the couple W x «c in the case of the 12- in. steel rule we must use the ordinary formula for a rectangular beam and find the moment of resistance when the stress is 56640 lbs. Moment = ^'^5 x [if x 56640 ^ „ 6 The interpretation of this result is that if the rule is held by the finger and thumb at each end, and pressure exerted to bend the rule, a force of ii-J- lbs. must be exerted when the distance between finger and thumb of each hand is i in. Or a force of 5f lbs. when the distance is 2 in., etc. The rule will then take a curvature equal to a circle of 8 in. radius. Q ^42 CURVATURE OF BEAMS. 21. A packing ring for a piston is turned to a radius of 14 in. It is then out and sprung into a cylinder of 13f in. radius. If the thickness of the ring is 1 in. what stress will there be in the cast iron of the ring? E = 16,000,000. The method of working this example is a little peculiar, as intermediate results are obtained which have no real existence. Nevertheless the method is sound. First we calculate the stress in the ring upon the assumption that it has been bent from a straight bar to a radius of 14 inches. Or more correctly to a radius of 13.J in. reckoning to the neutral plane. Thus StrGss * z = — ^ from which stress = 592,592 lbs. 16,000,000 1 3 "5 Again the stress is calculated as if the straight bar were bent to a radius of 13I in. or 13 J in. reckoning to the neutral plane again Stress _ "5 - from which stress = 603,773 lbs. 16,000,000 13 '25 The difference is 11,181 lbs. which is the actual stress in the ring. Deflection of Beams. In the following rules for deflection S — maximum deflection. W = load. L = length of beam. ' ' E = Youngs Modulus of elasticity. ' ' I = moment of inertia of beam section (which is ' supposed uniform from end to end. ' All measurements are in inches and pounds.- For proof of rules see Part III. • i . DEFLECTION OF BEAMS. 243 Cantilever, load at end, - - - S = 3EI Cantilever, load uniform, - - - S = ' 8EI JBeam, load in centre, - - - - 8 = 48 E I Beam, uniform load, - - - - 8 = 5 ^ ^ 384 EI Beam, fixed at ends, load in centre, - S = ^ 384 El Beam, fixed at ends, load uniform, - - S = 384 E I 22. A rectangular beam of wrought iron, 20 feet between supports, 10 in. deep, and 2 in. broad, is loaded with 2 tons weight in the centre of the span. Find the deflection in the centre. It is necessary to remember the formulse for the deflection of beams and simply apply the formula which is applicable to the particular case. Now when the section is rectangular, T BD* 2 X 1000 ,, , 1 = = . = i66-6 12 12 and using E = 29,000,000 lbs. = 12950 tons. 2 X 24.0 Deflection = — ; -^ ^^ = -2669 in. 48 X lOO'D X 12950 Deflection = "2669 in. — A?is. 23. A bar of wrought iron, 1 in. square and 1 foot long, is fixed at one end and loaded with -18 tons weight at the other end. Find the amount of dip at the loaded end. The formula to use is : — dip = 3 E I I = ]^; deflection = -0928 in. — Ans. 244 DEFLECTION OF BEAMS. 24. An engine room girder is 30 ft. long. The top and bottom flanges a e equal and 6 in. x f in. The web is 9 in. X j in. (measured between flanges). Find th& deflection when 3 tons is swung &om the middle of the length. The calculation of I will give a little more trouble, but in other respects it is similar to Example 22. I = 260 nearly.; deflection = "866 in.; if fixed at ends^ deflection = '216. — Ans. 25. You are required to flnd what camber should be given the girder in the previous exercise, so that when, loaded with a distributed load of 10 tons it would become practically straight. Use formula 2_ — _-_. 1-804 'n. — Ans. 384 E I 26. A rectangular beam of wood, whose modulus of elasticity is 250,000, was loaded in the centre of a span of 3 ft, with 2500 lbs. The depth of the beam was 3 times its thickness, and the deflection in the centre was ■214. Find the dimensions of the beam. If we begin with the formula for deflection for a beam loaded in the centre, viz. : — deflection = -g^ we readily find the moment of inertia for deflection x 48 x E which works out to 45 very nearly. Now let t = thickness and 3 / = depth. n ^* = 12 X 45 I = i X (3^)" = 45 and t^ = i^i^S = 20. 27 ' » 2'ii in., and liepth = 6*33 in. — Ans.- DEFLECTION OF BEAMS. 245 27. Find the dimensions of the strongest reotangulaf beam which can be cut out of a circular log of timber 18 in. diameter. If this question is to be worked from first principles it is an example in the differential calculus. In fig-ure 91, A B C D is the section of the required beam, of which the depth is d, and the breadth b. We have to find the relation of d to b, so that the beam may be the strongest possible. Now the strength of a beam depends upon the depth square multiplied by the breadth. (The length does not enter into the present question). This amounts to saying that rf^ x 6 is to be a w- maximum. Fig. 91. The triangle ABD being right-angled and DB being 18 in. d"^ = \& - b^, and substituting this value of d^ in the statement (i). (18^ - 6^) x 6 is to be a maximum. .'. 18^ b-b^ is to be a max. Differentiating and equating to zero. i82 - 3 62 = o = i82 = n/? = io"386 in. The depth being as above = iji8^ - i^ depth = \/324 - 108 = 1473 in. Noie. — From the above problem it is obvious that for any ■diameter (a) of circular log the breadth of the beam should be = V? Now if AP is drawn perpendicular to the diameter DB, the two triangles ADB and APB are similar triangles. .-. DB : AB : : AB : BP. 9^6 ' DEFLECTION OF BEAMS, substituting values : — = ^/ 3 "^ 3 -r a 3 "^3 3 3 This leads to a very simple geometrical construction for example 27. For if the diameter is divided into three equal parts and perpendiculars raised from the points of division to meet the circumscribing circle ; the corners of the strongest rectangular log are found. ^. Find the dimensions of the stittest beam which can be cut out of the 18 in. log of exercise 27. B,y making d^ b a. maximum, the dimensions work out as follows : — Depth 15-5 in.; breadth 9 in. — Ans. 29. Show that, if the diameter of the circle is divided into tour equal parts and perpendiculars raised, the corners of the stiffest log will be found by this construction. 12. — Stability of Ships. A good understanding of the elementary conditions influencingf the stability of a vessel may, perhaps, be obtained by considering a fevr simple experiments upon floating blocks of wood. Experiment 1. — Take a cylindrical piece of pine wood, some- what longer than its diameter, and float it in a vessel of water, figure 92. Pine having a specific gravity, about '5, the cylinder will float about half-immersed. We note that the block has no stability sideways, for the slightest touch makes it turn round and round in the water. Why is this ? We usually say the reason is that the centre of gravity of the dis- placed water, D, remains in all positions vertically under the centre of gravity, G, of the wood itself. Gravity pulls the wood down into the water, and the water pushes the wood up. Both forces act through these centres, which happen to be in the same vertical line in all positions. We notice particularly that the/orm of the cavity made by the cylinder in the water — the displacement — does not alter as the cylinder is turned round. Fig. 92. If a cylinder of hard wood is tried in the same way, the same conditions exist ; only the water line will be higher — say at HK instead of at AB. ' . Exp$riment 2. — Take the pine cylinder of experiment i and plane off a good third of its diameter, so as to produce a model like figiire 93. The thodfel will now be found to have stability, that is 248 STABILITY OF SHIPS. if pushed over it will roll back until -the flat is horizontal. Why is this ? The reason is that by removing- some portion of the wood of the cylinder, its centre of gravity is now below the geometrical centre of the curved surface, and, when the cylinder is pushed over, the centre of gravity, G, describes a circle about the geometrical centre C. But the centre of the displacement remains the same for all positions, viz. , at D ; hence G shifts to the right of the vertical through D, when the model is pushed over to the left and vice versa. We have now the downward and upward forces making a couple to bring the model back to the normal position as indicated by the It may also be noticed, with respect to the model in figure 93, that the lower th.& position of the e.g. on the line CV the greater the stability of the model, because the distance between the upward and down- ward forces — the arm of the couple [a) — is increased for the same inclination. And there is a limit to the height at which the centre of gravity may be, if the model is to be stable. The e.g. a must not be higher than the point C on ": the centre line. This is the only condi- tion necessary for stability. The e.g. may be above the centre of buoyancy, D, or it may be below D, but it must be Fig. 93. below C or the model will turn turtle. For the above reason the point C is the most important point in a floating body. It is called the Meta-centre and its distance above the e.g. of the body itself is known as the meta-centric height of the body (CG = meta-centric height). A few definitions of terms may now be given. Centre of Gravity of a floating body or ship is the point upon which the ship with her cargo and equipment (mast and rigging included) would balance in every position if supported in the air, or more correctly in a vacuum. STABILITY OF SHIPS. 249 Centre of Buoyancy is the centre of gravity of the cavity made •11 the water by the floating body. Or we may say it is the centre of gravity of the part of the ship below the water line considered as a homogeneous solid. In the case of a ship the centre of buoyancy is about f of her draught below the water line, or a little more, depending upon the ship's lines. Angle of Heel is the angle between the centre line of a ship, when she is inclined, and a vertical or plumb line. Righting Moment or Statical Stability is the measure of the effort to recover the upright position when a ship is inclined. It is equal to the weight of the ship (her displacement) multiplied by the distance between the two vertical forces, one through the e.g. of the ship and the other through the centre of buoyancy. In figure 93, if W is the weight of the model, W x « = righting moment. Meta-centre is the point where the vertical line through the centre of buoyancy meets the centre line of the ship. As already noticed, the meta-centre must lie above the ship's e.g., but the height of the meta-centre varies very much in different vessels, from 6 inches to 3 feet. A high meta-centre means a stiff ship, which may roll dangerously. A low meta-centre means a vessel with little stiffness and a slow period of rolling. Curve of Stability is a graphic representation of the righting ■moment of a vessel for different angles of heel. It is usual to set off the inclination or heel of the ship in degrees along the base line, and the ' 'arm" •of the righting moment [a in figure 92) in feet as vertical ordinates. Figure 94 is such a curve. To obtain the righting moment at an in- clination of 30 degrees, by means of the above diagram. 51 1^' 2 /^ / a \. 30 60 go ANGLES or HEEL Fig. 94. 250 STABILITY OF'SHr^S. we should measure the length of the ordinate, {a) upon the side scale of feet and multiply by the ship's diSplacerpent. The diagranj; also shows that the maximum stability is:at 50 degrees, and the; limit or range of stability for that particular vessel. There is noc stability left at 75 degrees on this diagram. , : Dynamic Stability is the measure of the work required to heel- the vessel to any particular angle (but excluding fluid friction).. Now apart from work done to overcome friction, the work required to change' the position of a body depends upon the amount which, its centre of gravity is raised. For instance in our ist experirrieiit^' as the cylinder rolled over and over in the water its centre of gravity rerriairied absolutely at the same level, for the e.g. was at the centre of motion; consequently no work was done, and the cylinder had! neither statical nor dynamical stability. As another instance, it may be noticed that before a stone column can be overturned, its centre of gravity rriust first be lifted; and until its e.g. reaches its highest point the column has stability. Also the inore the c.^. is raised' the more work is done, and so it happens that the dynamical stability of a column is greatest when the statical stability is on the- point of vanishing altogether. This is true also of a ship. If we have a "curve of stability',' drawn for a vessel, such as shown in figure 94, the dynamical stability is represented by the area enclosed by the curve. Thus the work done up to an inclina- tion of 30 degrees is the area ABC. And if the ship is inclined to- 75 degrees, the angle at which she is on the point of turning over. The work done is the area ABD. Experiment 3. — Take a block of pine whose section is rectangular and much wider than it is thick. When floating in water it will 1 be found to have considerable stability. Let it be pushed or heeled into the position of figure 95, the corners E and B just level With the surface. The e.g. of the model is at its geometrical centre; G,i and the first point to notice is that the e.g. does not change its position as the model heels ; for since the wood floats half immersed, the e.g. is just at the water-line in all positions. In this respect the case is similar to the ist experiment, and the STABILITY OF SHIPS., 2.51 questiioh : arises : — Why is there stability with this model and not with the ist? To explain this we must notice an essential difference between the two cases. In the ist and, 2nd the shape pr form of the 4ican now make a rough estimate of its dynamical stability. When 'Upright it is obvious that the centre of buoyancy is \ the depth of "the model below the water line. When inclined so as to displace a wedge of water AEB, the centre of buoyancy is on the middle line BR of the triangle, and at f BR from B. Now since the point R is \ the depth of the model below the surface of the water (approx.) it is seen that the new c.b., D, is f of ^ of the depth = \ depth rbelow the water line. And the difference between \ and \ (the ■distance when the model is upright) is the amount the centre of '■buoyancy is lowered. .". Dynamical stability = displacement X J^ depth. The reader will please notice that the values here given for the ■statical and dynamical stability of a block of wood — as well as being only approximately correct — have no value beyond illustrating the meaning of the terms themselves. It always leads to a clearer under- standing of a scientific term if we can bring it to an actual measured value. STABILITY OF SHIPS. ass- Summing' up conclusions : — We have learned from these experi- ments that the righting moment of a ship arises usually from twO' causes, viz. : — The movement of the ship's e.g. to windward and' the movement of the centre of buoyancy to leeward as the ship heels. And the actual value of the statical stability at any angle of- heel is the product of the ship's displacement multiplied by the horizontal distance between the vertical lines passing through these- two centres. Similarly the dynamical stability or work of heeling a ship is- usually due to two causes, viz. : — ^The lifting of the ship's e.g. higher with respect to the surface of the water, and the depression of the centre of buoyancy below its former depth from the surface. The actual value being found by multiplying the displacement by the sum of the two movements. The following formula is given here without proof, but it may be taken as approximately correct for angles of heel not exceeding, lo degrees or 15 degrees in most vessels. Dynamical stability = displacement X meta-centric height x versine angle of heel. Experiment 4. — Take a triangular prism of pine, the cross section, as in figure 96. Incidentally it may be noticed that when, placed in water it floats with one edge uppermost and has stability in that position. (A similar prism of hard wood floats with a flat side uppermost.) Now load the prism at one edge with a bit of, sheet lead so as to make it as near as possible of the same specific gravity as water. (A little salt put into the water is useful as a final adjustment.) We thus obtain a crude model of a sub- marine boat, and it is ound that the model has considerable stability although entirely im- mersed below the surface of the water. Fig. 90. ;254 -.lAtirUTY OF SHIPS. The special property possessed by a submarine body as regards stability is that no matter what its form is — circular, rectangTilar, •or ship shape — the centre of buoyancy does not change its position :in the body as the vessel heels. For since the body is entirely immersed its centre of buoyancy is the centre of its entire displace- iment, namely, a fixed point in all positions. Now the effect of load- ing an angle of the prism is to bring the centre of gravity of the 'body below its geometrical centre, which is the centre of buoyancy. ' Thus, in figure 96, C is the centre of buoyancy and G the centre of •-gravity. The righting moment is, therefore, W x a, as in the previous cases. But it should be noticed that for a submarine the 'Centre of buoyancy and the meta-centre are identical, and for stability it is essential that the e.g. of the vessel is below the centre ■ of buoyancy. As regards dynamical stability the case is similar to case 2, in which the dynamical stability is owing entirely to the lifting of the 'C.g. of the vessel. Hence — Dynamical stability = CG x versine angle of heel. More Advanced Principles. A principle which is sometimes very useful is the following: — The height of the meta-centre above the centre of iuoyancy can be I found by dividing the moment of inertia of the water line plane about . its centre line by the displacement of the vessel. In order to make accurate use of this principle we should ■■ require to have the ship's lines so as to calculate the I of the water line plane ; but for most steamers a near approximation is — I = -06 LBS • where L = length of water line and B = extreme breadth. Note. — It the virater line plane were a perfect rectaflgle the tmoment of inertia about its centre would be ^^ LB*, hence thfe - constant •oGshows that the actual moment is about y^^ of the rectangle;. ROLLING OF SHIPS, 255 Rolling of Ships. The rolling of a ship is a very complicated movement even in still water. There is, of course, some similarity between the rolling' of a vessel from side to side and the swing of a pendulum, but there is no fixed point in a ship round which she oscillates corresponding to the point of suspension of a pendulum. The meta-centre may or may not be such a point, but there is at every instant some point called the Instantaneous axis which remains fixed in space for a.n instant of time and round which the ship rolls. As an approximation it may be assumed that for a moderate degree of rolling the fixed point is the meta-centre of the ship. There is sufficient similarity between a rolling ship and a pendulum to use the principle applicable to the latter, when speaking of the period of oscillation of a ship's roll. In the first place, a " simple" pendulum means a length of fine thread with a small heavy bullet at the end. The time in which such a pendulum makes one vibration from side to side is given by the rule — = 'V^ Time " g Where /is the length of the thread in feet, and^=32'i6, n- = 3-i4i6. Any other body which swings to and fro under the action of g-ravity about a point of suspension is called a com-pound pendulum. Thus a batten of wood pivoted close to one end and allowed to swing is a compound pendulum, because the weight is distributed all along its length instead of being concentrated in a point at the lower extremity. Every particle in the batten really wants to swing G OF SHIPS. dividing the square of the radius of gyration by the radius to the centre of gravity. For example, a uniform bar has a radius of gyration = "577 of its length; and its centre of gravity is at \ its length from the same end. Let the bar be i foot long (a steel rule for instance). The length of the equivalent simple pendulum is : — (-577)^ - i = I- This means that a string, f of a foot long, with a bullet at the end will swing in the same time as the foot rule pivoted at one end. (The experiment is a simple one to try). Figure 97. ■- -0 Note. — The arc of vibration should not exceed a few degrees on each side of the vertical. I .©. y . eg:. -'S CO ^v 1 I • -is Now when these principles are j-'-. applied to the rolling of a ship (presumed to take place round her meta-centre as an axis), the radius of gyration means the radius from Fig. 97- this axis, say = R. Also the meta-centric height takes the place of the radius to the centre of gravity = m, The equivalent pendulum is now Length = R^ -i- ?w = /. And having found the length — Time of one swing = 3"i4i6 a/ — 32 13. — Exercises on Stability. 1. A ship's displacement is 6245 tons with her ballast tanks empty, the centre of gravity of the ship being 22 feet above the centre of the tanks. Find the new position of the centre of gravity after the tanks, which hold 772 tons, are filled. If we imagine the ship to be supported on her centre of gravity and, therefore, perfectly balanced, as in figure 98, upon the point A. And then add a weight at B representing the filling of the ballast tanks ; the effect is to destroy the balance about the point A, and to make the ship now balance on a point as C, which is the new e.g. of the whole. Taking moments about C, we have : — 6245 tons x CA CA. .'. 6245 X CA = 772 X (22 Fig. 98. 772 tons X CB, but CB = 22 CA) 6245 X CA = 772 X 22 - 772 X CA (6245 + 772) X CA = 772 X 22 CA = 772 X 22 7017 2 '42 feet. From this example we may obtain a rule for all such cases,. viz. : — The alteration in the centre of gravity of a ship due to any alteration of "weights is found by dividing the additional moment (weight X Aistaxics to e.g.') by the final weight of the ship. 258 STABILITY. 2. A steamer of 12500 tons displacement, on leaving port, burns 2500 tons of bunker coal, whose e.g. may be assumed at 8 ft. below the eg. of the ship with bunkers full. Find effect on e.g. of ship. The centre of gravity is lifted 2 feet. — Ans. 3. In a ship of 8000 tons displacement, if 20O tons of coal is taken out of port bunker, whose centre of gravity is 12 feet from the centre line of the ship, how much is the centre of gravity of the ship altered? •308 feet to starboard. — Ans. 4. Given dimensions of a steamer 3^7 ft. x 92 ft. x 25 ft. deep and draft 20 ft. (mean). Co-efficient of displace- ment -72. If 150 tons of coal is shifted 10 feet 8 inches lower down in this vessel, what effect will it have upon the meta-centric height and what effect upon the centre of buoyancy? 1st, increased by "2668 ft.; 2nd, no difference in centre of buoyancy. — Ans. ^. A ship, 420 feet long by M feet beam, has a draft of 20 ft. 6 in., with a co-efficient of displacement of -7. The centre of gravity of the ship is 1*5 feet above the plane of floiation. Required the alteration in the position of the e.g. of the ship, with respect to the water plane through filling the ballast tanks, which hold 350 tons and whose e.g. is 3*5 feet above the keel. Also state what effect this will have upon the meta-centric height. Note. — The area of the water plane may be taken as '72 of the circumscribing rectangle. The increased draft works out to "92 feet, which by itself would bring the e.g. of the ship that much nearer the water line. Also the filling of the tanks lowers the e.g. in the ship "817 feet, hente the e.g. is altogether -92 -|- -817 = 1-737 lower down with respect ■STABILiTY. '459 io the water line. The e.g. i^ now '237 ft. below the water line. The effect on the meta-centric height is to increase it by "817 feet, but it may be noticed 'that the ship ought to be heeled afresh with the water tanks full, to obtain a correct result. — Ans. 6. A Yessel is inclined for stability. Her displacement is 5000 tons. There is 17 tons of pig iron laid along the centre line of the deck and the ship is upright. This pig iron is now moved over to one side, a distance of 18 feet, and the vessel is found to incline 2 degrees. Find the height of the meta-centre above the centre of gravity Of the ship. In the figure 99, C is the centre of gravity of the ship when •upright, and A its new position as altered, owing to the shifting of the weight W to Wj. The ship is shown inclined (much , •exaggerated). Now the prin- » ■ciple of a stable floating body is that the centre of gravity ■must always be vertically above or below the centre of buoyancy. Ifwedrawaverti- *■" cal line through A we know that it will pass through the ■centre of buoyancy, although we may not know how far the latter is from A. But where Fig. 99. this vertical line cuts the ■centre line (or centre plane) of the ship is known as the meta-centre for that inclination. We have to find CM. First, as regards the distance CA, which is drawn parallel to the ship's deck, because the weights were moved parallel to the deck. CA, according to the previous examples, , ^ 17 X 18 ^ .^^^ f^^^_ 5000 JSJow the triangle CMA has a right angle at C, and angle M = 2 ■degrees. 26o STABILITY. Since CA tangent M CM CM = CA H- tang. M. This is the equation for finding the meta-centric height — CM = 'oeia -^ -035 = 175 feet. The position of the " meta-centre " changes with the inclination of a vessel, but when the term is used without qualification it is understood to mean the position for the least possible inclination. And it may be useful to know that for the first few degrees of inclination the height of the meta-centre is practically the same. 7. A ship of 2000 tons displacement and 1 foot meta-centric height has 50 tons of cargo shifted 10 feet to port. Find her permanent angle of heel. Also if the ship rolled 30 degs. each way before the cargo shifted, what will she roll to port and starboard now? C.G. moved "25 to port; tan. of heel = — ^; heel 14 degrees; roll 44 degrees to port, 16 degrees to star. — Ans. A steamer's displacement is 1800 tons, and 10 tons moved 7^ feet across the deck inclines the vessel 2 degrees- Find the height of meta-centre. Also if afterwards 200 tons of cargo is lifted 6 feet in the ship, find the righting moment at 6 degs of heel. The first meta-centre height works out to 1*19 feet. And the effect of lifting 200 tons 6 ft., is to raise the centre of gravity •666 ft. , thus reducing the meta-centre height by that amount, and making it only -524 ft. If we now refer to figure 100, representing a heel of 6 degs., and in which the same letters are used as before, it will be seen that whereas the weight of the ship is acting Fig. 100. STABILITY. 26t vertically downwards through the centre of gravity C, the buoyancy of the water acts vertically upwards through M. These equal forces form a couple with arm or leverage CR, and C R .— — = Sine of inclination CM .". CR = '524 X Sine 6 degs. = '0547 ft. The righting moment is 1800 x '0547 ,, ,, = 98'46 ton-feet. 9. GiYen displacement 2000 tons. Weight of 8 tons moved 20 ft. athwartships, inclines vessel If degs. Find meta- centric height and righting moment for 10 degs. heel. Height of meta-centre 2*62 ft. ; righting moment 909 "9 ton-ft. — Ans. 10. An excursion steamer of 1150 tons displacement is Inclined for stability by moving a weight of 8 tons 35 feet across the deck, when the heel is found to be 5 degrees. When put on the service she carries 300 passengers each weighing 110 lbs. The e.g. of the passengers may be considered as at 3 feet above the - -deck and the deck is 13 feet above the eg. of the ship lirithout passengers. Required the height of the meta- ■centre and the statical stability at 5 degs. heel, first without passengers; and secondly (approximately) with passengers on board. Tst, meta-centre 2*783 ft., stability 278*9 ton-ft.; and, meta- centre 2*783 - "256, meta-centre = 2*527 ft., stability = ii68'75 < "2202, stability = 257*3 ton-feet. — Ans. 11. A ship, iOO feet long and ^000 tons displacement, draws i ft. more aft than forward. Her longitudinal meta- centre height is 300 feet. How far must a weight of 100 tons be moved forward to bring the ship upon an even keel? Exactly the same principle applies to the inclination of a ship longitudinally as traversely, only owing to the centre of buoyancy shifting so much in a fore-and-aft direction with a small alteration of trim, the height of the meta-centre is very great. 262 STABILITY. In the example the inclinatioh Is 4 feet in 400 and whether we take the sine or tangent of the angle as ^^ makes practically no difference in the angle, which is o degs. 35 min. Now CM being known (figure 100), to be 300 ft. and angle M = 35 min., we can find CA = 3 feet. We may now suppose the inclination to have been caused by moving weights in a fore-and-aft direction, which has brought the centre of gravity from C to A. And to bring the ship on to an «ven keelwe mu^t shift the e.g. back to C again, a distance of 3 feet, hence 100 tons X X = 4000 X 3 ^ = 120 feet. The 100 tons must be moved 120 feet more forward. 12. A steamer $20 ft. x 45 ft. x 35 ft. floats at a draft of 19 ft. 6 in. forward and 21 ft. 6 in. aft. A cross bunker has 350 tons of coal put into it. Assuming that the longitudinal meta-centric height is equal to the length of the ship and that the cross bunker is iO feet from the centre of flotation of the ship towards the stern, find the new draft of the ship. The co-efScient of displacement is *7, and the water line co-efiScient '8> We first find the displacement of the ship from the data givea: T-v 420 X A? X 20'S D = S ^ ^ X 7 = 7749 tons. 35 Next the area of water line : Area = 420 x 45 x -8 = 15120 sq. ft. The extra draft due to the bunker coal is now 35° ^ 35 ^ .813 feet. 15120 If the bunker had been situated at the centre of flotation the new draft would, therefore, be "81 feet greater fore and aft, viz.: — 20*31 forward and 22"3i aft. It remains now to find the altered trim of the ship with a total displacement of 7749+350 = 8099 tons. STABILITY. '263. .Usi'ngf the same formulae as for the heelings of a vessel, viz. :— Weight X distance moved = displacement x movement of e.g. of ship. 350 X 40 = 8099 X X X = I '73 feet nearly. Then tan. of ansfle = m.-c. height But the tan. of angle is also equal to alteration of trim -^ half length of ship. . alteration of trim _ i "73 half length of ship m.-c. height alteration of trim _ i'73 210 • 420 alteration = '865 feet at each end. The final drafts are, therefore, i9'445 forward and 23"i7S aft, 13. A ship of 6000 tons displacement has a meta-centric height = 2 ft. 6 in. Find the loss of dynamical stability through rolling 5 degs. past the vertical instead of 10 degs. past the Yertical. Dynamical stability may be calculated by the rule : — Displacement x meta-centric height x versine angle of heel. For the ist case (10 degs. heel) Work = 6000 X 2*5 X '0152 And for the 2nd case (5 degs. heel) Work = 6000 X 2*5 X '0038 These work out to 228 and 57 foot-tons. The difference being 171 foot-tons, which, is the required answer. a64 Stability. 14. A steamer 420 x 45 x 32 feet. Loaded draft 23 feet. Displacement 6500 tons. The centre of gravity of the ship is 4 feet below the water line, and her centre of buoyancy is 8^ feet below the water line. Beqnired the initial meta-centre height, and the righting moment for 15 degrees of heel, if the meta-centre remains the same. Note. — The moment of inertia of the water line plane may be taken as -^-^ of the moment of inertia of a rectangle equal to the length and breadth of the ship. The above example introduces another principle of a floating body. The principle may be stated as follows : — In any floating body, if the displacement reckoned as a volume is ^multiplied by the height of the meta-centre above the centre of buoy- ancy, the result is equal to the m-omenc of inertia of the water line plane about the centre line of the ship as an axis. This principle enables us to locate the position of the centre of buoyancy, with respect to the meta-centre, in a vessel before she is inclined for stability. And inclining the vessel fixes the position of the centre of gravity with respect to the meta-centre. Thus the .relative positions of the three important centres can be found. In the example given we will first find the moment of inertia ■(I) of the water line plane, which if it were perfectly rectangular would be I = 420 X 45^ , but taking -^ of this 12 'S Tiy ' c3 ^ 4^ X i = 2,232,600 IQ Now the displacement in cubic feet is 6500 x 35 = 227,500. If the height of the meta-centre above the centre of buoyancy is called M, the principle gives us : — M X 227,500 = 2,232,600 M = 9-81 feet. STABILITY. 265 From figure loi it will be seen that the ■meta-centre is 5'3 feet above the e.g. of the ■x--A<' "«■*- ship. «--T-.^- 'fiVfe'r I ",1. * I- To get the righting moment at 1 5 degrees lieel we proceed as in previous examples : — ■ 5'3 X sine 15 degrees x 6500 tons = 89i6'3 ton-feet. Fig. loi. Ch 15. Given a vessel 347 ft. on the water line by i2 ft. beam and 20 ft. draft. Co-efiScient of displacement -72. If the centre of gravity of the vessel is 5 ft. above the centre of buoyancy, find the meta-centre height above the centre of gravity. Using the principle of the previous example, the moment of ■inertia of the water line plane works out to 1,499,670. And the •displacement is 209,870 cu. ft. The distance from the centre of buoyancy to the meta-centre is, therefore, 7 '14 ft., and the meta- -centric height 2"i4 feet. — Ans. d6. A rectangular tank or pontoon 300 ft. long, 30 ft. wide, and 15 ft. deep, made of uniform plate thickness and symmetri- cal construction, floats at a draft of 10 feet. It is closed on the top. State the position of the centres of gravity and buoyancy, and calculate the initial meta-centric height. C. G. is 2-5 ft. below water line; c.b. is 5 ft. below water line; ■meta-centric height 5 ft.^Ans. 17. Suppose that, in the previous question, the pontoon had been made of thinner plates and were made to float at the same draft by placing 300 cubic-foot blocks of iron (ea::h = 450 lbs.) down the centre line of the deck. What would be the position of the centre of gravity and how would the meta-centre be altered? C.G. is now "187 ft. higher and meta-centric height that amount Qess. — Ans. 26& STABILITY, 18. Salne pontoon as in laot question, but the 300 blocks are moved over ii ft. from the centre line. Find the^ angle of heel. * C.G. is •328 ft. out of centre line ; angle of heel 4 degrees nearly. — Ans. 19. A rectangular lighter, 120 feet long, iO ft. beam, and 12-" ft. deep, floats upon an even keel at a draft of 3 feet. There is a water-tight bulkhead 8 ft. from the forward end, and this compartment is holed and fills with water.; What do you estimate the approximate draft to becom& at each end? Figure 102 (which is not drawq to scale) represents roughly- what happens when the compartment is holed. AB is the former' water line. WL is the new water line. Fig. 102. We will first find what the draft would be if the lighter was- trimmed to still float upon an even keel. Original displacement = 120 X 40 X 3 = 14,400, dividing by the new area of water line^ J r-. 120 X 40 X -J „ , new draft = 2 9 = Q-ai feet. 1 1 2 X 40 We may now consider the lighter to be only 112 feet long, and in order to make her float upon an even keel we may conceive the damaged portion to be cut off' and brought into the middle of the new length. The weight of the damaged portion is jf^ of the total = jL of 411-4 = 27-4 tons, or we may find it from. ^ 40 X 8 X 3 the lost displacement = 35 27*4 tons. STABILITY, 26f Having got the lighter floating at adraft of 3*21 feet upon an even keel, the problem is simplified, for it is reduced to the problem of finding the alteration of draft, due to moving a weight of 27-4 tons from the middle of' the lighter to the forward end — a distance of 56 + 4 = 60 feet. ., , The movement of the centre of gravity is obtained from the usual equation 27"4 X 60 = 41 1 '4 X X X = 4 feet nearly. We shall also require the longitudinal meta-centric height, which is obtained from the moment of inertia of the water line plane T 40 X (112)* ^o I = t^ ^^ '- = 4,683,093 12 dividing by the displacement 14,400 cu. ft. heisrht of meta-centre = "^ — J'°93 = ,35 ft. ° 14,400 .'. Tangent of inclination = ^23- = -g-u nly. The inclination being i in 80, the extra draft at the forward end will be ^ of 64 ft. = '8, and the reduction at the after end -^ of 56 = 7. The final drafts are, therefore, forward 3.21 + -8 = 4-01 ft. after end 3-21 - 7 = 2 -51^ ft. 20. A ship 350 ft. long, iO ft. beam, and 20 ft. mean draft. A pendulum, 12 ft. long, swings 9'S in. to port when a weight of 25 tons is moved 36 ft. across the deck from starboard to port. The co-efScient of displacement being -68, find the meta-centric height. 2 "5 feet. — Ans. -268 STABILITY. ■Continued.— T)i& above ship has a ballast tank 35 ft. wide and 125 ft. long. The ship leaves port with this tank half full. What effect will this have upon the meta-centre and upon the stability of the vessel. Note. — The tank holds 500 tons when quite full. As a general statement it is obvious that a tank only half full ■of water must seriously reduce the stability of a vessel, because when the vessel heels the water will flow over to leeward, and have the same effect as any other weight (such as shipping a heavy sea) weighing the ship down on that side and perhaps preventing her ■from righting herself. To express the condition more mathematically, we may say that the movement of the water in the tank shifts the centre of gravity of the vessel some distance over to leeward and reduces the •'righting arm" (that is the distance between the vertical lines through the centre of gravity and the centre of buoyancy) and, therefore, the righting power. Following this out we may get a definite numerical answer to the problem. We note, in the first place, that the displacement of the ship is 190,400 cu. ft. and 5440 tons, and the ballast tank is 4 ft. deep. It will simplify the consideration of the changed stability of the ship if we take a definite heel to work from. Let us suppose the heel to be just sufficient to bring the water in the ballast tank to the top on the lee side, as shown in figure 103. The ship is now inclined to an angle whose tangent is^V = -1143. (The ■angle is 6\ degrees.) At this inclination •we can find the right- ing arm from the meta- centric height already known to be 2"5 feet. Fig. 103. STABILITY. 269. Righting- arm = 2 "5 x sine 6^ degrees = "283 feet. But this does not include the correction for the changed position. of the ship's centre of gravity due to the ballast water. In figure. 103, it will be noticed that the water now forms a triangular wedge,. whose e.g. is at B, which is ^ the breadth of the tank from the lee- side ; and the movement of the e.g. has been from A to B, a distance equal to ^ the breadth of the tank (approx.) say 6 feet. To find the corresponding movement of the e.g. of the ship, we have the total displacement already found, viz., 5440 tons and ballast. water 250 tons. . 2i;o X 6 ft. -. . . -^ = •275 ft- 5440 Now the original righting arm for the assumed inclination of 6J degrees being '283 ft., we take away "275 and there remains only •008 ft. at 6^ degrees heel. The righting moment is, therefore, only 5440 X 'ooS = 43*52 foot-tons, and would probably vanish altogether with a little greater roll one way or the other. A new meta-centric height can readily be found from the new righting arm, by dividing by sine of 6^ degrees. 14.— Heat. t. At what temperature will the degrees on Fahrenheit's scale be just three times the degrees Centigrade? Put C for the degrees Centigrade then §_ + 32 = degs. Fah. „'. to fulfil the required condition — + 32 = 3C 5 9C + I60 = 15C 6C = 160 C = 26f and Fah. degrees = 80 degrees.— ^wi. '2. If the temperature Centigrade is added to the temperature Reaumur and 32 degs. added to the sum, the result is the temperature Fahrenheit. Prove this. Putting X for the temperature Fah., the Centigrade temp, is ^ {x - 32) and Reaumur = ^ (x - 32). These added together and 32 as well, will equal x. 3. If the temperature of the sea reckoned on Fahrenheit's scale is equal to the temperature on Centigrade squared. What is the temperature? Fah. 43"93; Cent. 6-628. — Ans. -i. If the number of degrees Reaumur equals the degrees Fahrenheit squared minus 1000, what is the tempera- ture? 3i*6 Fah. — Ans. RADIATION OF HEAT AND LIGHT. 27 1 S, A red-hot shot at three feet from a screen imparts^ 126 units of heat to the screen per minute, Find the amounts of heat given to the screen when placed six feet and nine feet away. This is a question of radiation. Now heat radiates in all ■directions from a hot body in straight lines. In the figure 104, C is the hot body considered as a mere point. The screen AB inter- ■cepts all rays lying between the extreme ones CA and, CB, and in order that the screen CD, which is twice the distance away, may intercept as many rays from the same source C, it is obvious that it must have Fig. 104, twice the dimensions of the screen AB. That is to say, it must be four times the area of AB. It follows that if the small screen is moved from position AB to position CD it will only be able to intercept one-fourth of the rays. The law is : — The heat received by a screen varies inversely as the square of its distance from the source. Applying this law to the question : — 62 : 32 :: 126 : ^ = 31-5 9^ : 32 :: 126 : y = i4"o Note. — A practical application of the above law applied to light ■occurs in the method by which the illuminatin;^ power of two sources of light, such as two electric lamps or an electric lamp and a gas burner, are compared with each other. The method consists in hanging a small object, such as' a coin or a bullet, a few inches in front of a white screen, and throwing the shadow of the object upon the screen from each lamp. The lamps are moved about until the two. shadows come side by side and are of exactly equal intensity of darkness. When the equality is obtained, ,the distance of each lamp from the screen is measured and the distances squared. This gives the ratio of illuminating power in the two lamps. For example, if the two distances were 5 ft, and 7 . ft. j respectively, tha 272 TEMPERATURE OF STEAM. ratio would be as 25 to 49. In other words, the lamp at 7 feet: would have nearly twice the candle power as the one at 5 feet. It may seem strang-e that the illuminatingf power is measured by the- shadows ; the explanation is that each shadow is really the strength of illumination caused by the other lamp. This is apparent if th& lamps are obscured alternately. 6. Find the temperature of steam at 19S, 215, and 225 lbs, absolute pressure, from the formula: — p .222 ^^ = -00849. The formula may be written : — T = P -222 ^ -00849. Taking- the pressure = 195 and using logs. : — log. I 95 = 2-290034 -222 4580068 4580068 4580068 -. -00849 0-508387548 = 3-928908 subtract 379 •'1 = 2-579379 379-6 degs., 388 degs., 392 degs.—Ans. 7. A feed pump has a 3-in. ram and 16-in. stroke, single- acting, making 120 strokes per minute. The pump is half full each stroke. Find the amount of heat oiiits per hour necessary to raise the feed water, which is ad 110 degs. Fah., to the temperature of the boiler which is at 160 lbs. absolute pressure. Note.-r^k unit, of heat is the quantity of heat required to rais6 the temperature of i lb. of water one degree. j EVAPORATION. 273; The amount of feed water works out to 14726 lbs. per hour. Also the temperature of the boiler according to the formula in the previous question = 363"4 degs., hence the heat required = 1472& ^ (363*4 ~ no) = 3,731,568 thermal units. 8. One boiler is at a temperature of 222 degs. Fah. and the feed water is at 212 degs. Anotner boiler is at ai. temperature of 232 degs. and the feed water is at 222 degs. What is the ratio of the amounts of coal required to evaporate the same quantity of water in these two boilers? Using the formula 1115 + "3 T - ^ for the heat required to evaporate i lb. of water, the ratio comes out as 969 "6 : 962 "6 ; : or as I : "9928 nly. 9. A piece of lead weighing i lbs. at a temperature of ^00^ degs. Fah. is plunged into 3-2 pints of water at 72 degs. The specific heat of lead is '031. Find the resulting^ temperature. The principle to apply in all questions of mixtures, is that there will be as much heat reckoned in thermal units a/ier, as there was before the mixture took place. The specific heat of a substance is the quantity of heat necessary to raise the temperature of i pound of the substance I degree. A gallon of water weighs 10 lbs. and a pint i^ lbs., 3*2 pints = 4 lbs. ; Units of heat in water = 4x72 = 288 ;! ,, ,, lead =4X "031 x 400 = 49-6 i, The sum of these = 337'6 Now, if X is the resulting temperature of both lead and water after. immersion, s 274 TEMPERATURE OF FURNACE. The total heat is: — (4 X jc) + (4 X -031 X jc) = (4 + •124) X which must amount to the same as before, hence the equation is (4+ -124) X = 337-6 X = 81 '8 degrees. 10. An iron ball weighing 2 lbs. is heated in a furnace and then plunged into 3 gallons of water at 70 degs. Fah. The temperature of the water rises 8 degs. Find the temperature to which the ball was heated in the furnace. Sp. heat of iron MlG. li X is the temperature of the iron, the equation is: — water iron water iron (30 X 70) + (2 X -lie X jc) = (30 X 78) + (2 X "1 16 X 78) 1 1 12 degrees. — Ans. 11. A piece of iron weighing 5 lbs. is placed in a boiler uptake, and after acquiring the temperature of the gases, is plunged into 1^ lbs. of water at 72 degs. Fah. The water rises to 96 degs. Find the temperature of the uptake gases. Sp. heat of iron -116. 675-3 degrees. — Ans. 12. A lead bullet weighing | oz. falls 900 feet and retains the heat developed by the shock. It is then put into a pint of water at i degs. C. The original temperature of the bullet was 53 degs. F. Find the temperature of the water. Sp. heat of lead -031. The foot lbs. due to the fall is first expressed as units of heat according to Joule's equivalent (774 ft. lbs. = i unit). This heat, plus the heat originally in the lead plus the heat in the water, must all appear In the water and lead at the final temperature. 39-259 F.—Ans. LATENT HEATi, ^75 13. The steam fpom 10 lbs. of water is led into a vessel containing 120 lbs. of water at 32 degs. Fah, Find the lesulting temperature in degrees Centigrade. The principle to apply is that the final amount of heat in the tank is equal to the initial heat in both steam and water, including the latent heat of the steam. If x is the final temperature Fah., the equation is : — (id X ii78'6) + (i20 X 32) = 130 X X from which x = \20 degs. Fah. and = 49 degs. C. 14. Steam from 5 lbs. of water is passed into a vessel containing 60 lbs. of water at 30 degrees C. Find the resulting temperature Reaumur. 170*04 F.; 61 "35 R. — Ans. 15. If 30 lbs. of iron, at 1500 degs. temp., is plunged into 20 lbs. of water at 60 degs., how much of the water will go into steam? The specific heat of iron is -114. Put X for the amount of water evaporated ; leaving 20 — x for the rest of the water, which, although raised to the boiling point, is not evaporated. Using the formula for water evaporated at 2 1 2 degs. Heat in the steam = x y. (1115 + "3 x 212) Final heat in the water = (20 - x) x 212 Final heat in the iron = 30 x -114 x 212 The sum of these amounts is now equated to the heat in the water and iron at the commencement, viz. : — (20 X 60) + (30 X 1500) X -114 = 6330. The equation reduces to : — {x X 1178-6) + (4240 - 212 x) + 725 = 6330. X = I "412 lbs. 276. LATENT HEAT. 16. In an account of an eYaporation test of a certain boiler it was stated that 9*6 lbs. of water were evaporated from feed water at 85 degs., per lb. of coal. And this was equivalent to 11*3 lbs. under standard conditions, that is, from and at 212 degs. < What was the tem- perature of the boiler? Under standard conditions the heat required per lb. of water evaporated is 1115 +('3 x 212) — 212 = 966 '6, hence the heat supplied by i lb. coal is g66'6 x ii"3 (i). Similarly if T is the actual boiler temperature, heat supplied by 1 lb of coal is (1115 + '3 T - 85) x 9-6 (2). Equating (i) and (2) gives us the value of T = 359*2 Fah. 17. How many gallons of water at 15-5 degrees C. must b& added to 1 lb. of ice at 32 degs. Fah., so as to leava all water at i degs. Reaumur? The additional fact necessary to know is the amount of heat which disappears when ice is melted, viz., 144 British thermal units for each lb. of ice melted. This number is called the latent heat of water or ice. If the temperatures are all expressed in Fah., and x^ — pounds of water required, the equation is: — (x X 60 degs.) + (1 X 32) - 144 = (;c + i) X 41 from which x = 8-052 lbs. and in gallons water required is '8052 galls. 18. A tank contains 300 lbs. of water at 60 degs. Fah. Steam at 37^ degs. F. is blown into the tank until the weight increases to 310 lbs. when the temperature has risen to 90 degs. F. Find the percentage of water in the steam and the "dryness fraction." . Let there be x pounds of water in the steam; then as the total weight of water and steam which inters the tank; is 10 lbs. (310 - 300), the weight of the steam alone is 10 - x. WET STEAM. 277 To find the total heat of the steam we may use the rule : — Total heat = 1115 + "3 T per lb. whicli works out to 1115 + "3 x 374 = 1227 -2. In lo - x lbs. there is heat = {10 - jc) x i227"2 and in x lbs. of hot water there is heat = ^ x 374. Stating the equation of equal quantities of heat before and after mixture : — iio — x) x 1227 + {x y. 374) + (300 X 60 degs.) = 310 x 90 12270 - 1227 X + 374 X + 18000 = 27900 — 853 X = 27900 - 12270 - 18000 - 853 :r = - 2370 X ■= 277 There is, therefore, 2*77 lbs. of water in 10 lbs. of nominal steam. The percentage is hence 27*7. Also the dryness fraction = 72*3 per cent. 19. If one cwt. of steam at a temperature of 370 degs. F. flows from a boiler into a tank containing 1 ton of water at 64 degs., what will be the final temperature. The steam contains i per cent, water. ii3'4 degrees. — Ans. 20. A steel bloom is increased ririnrth part of its length by the application of heat. The co-efficient of expansion is ■0012 for 180 degs. Fah. and the specific heat of the metal is '116. Express the work-equivalent of the heat giyen to the bar. The bloom weighs 200 lbs. It will be observed that the only purpose of the co-efficient of expansion is to enable us to find the increase of temperature. The units of heat can then be calculated as usual. •0012 : "ooi :: 180 degs. : increase of temperature Increase =150 degs. Fah. Heat added = 200 x 150 x •116 ,, = 3480 thermal units or = 3480 X 774 foot-lbs. 2,693,520 — -Ans. ■ i 278 EXPANSION OF METALS. 21. The co-eflFicient of expansion of brass being -000019 for- each degree Centigrade. What amount in inches will a brass rod expand when heated from 58 degs. to 180 degs. Fah.? The length of the rod being 1-65 metres. Working on the metric system, we have a rise of temperature of 122 deg-s. Fah. which equals 6'j"j'j degrees Centigrade. The expansion is i'65 x -000019 x 67-77 = "0021248 or 2-I248 m.m. which = 2'i248 X '03937 inches = '08365 inches. — Ans. 22. The apparent co-efScient of expansion of mercury in a. glass tube is ■^^. The absolute expansion is -g-g-^, both reckoned for 1 degree. Find the expansion of th& glass. The difference of the two fractions is what is wanted. ssTm ^°^ sach degree. — Ans. 23. In a U tube a column of mercury, at 212 degs. Fah., is. balanced by a column of mercury at 32 degs F. If the hot column is 16-35 in. high what is the height of the cold column? Co-efiBcient of expansion (cubical) ■00018 per degree Centigrade. Two columns of liquid will balance in a U tube when their heights are inversely proportional to their specific gravities. Now as there is 100 degrees difference of temperature Centigrade between the two columns, the density of the cooler mercury will be ■00018 X 100 = 'oiS greater than the hotter mercury, whose density may be considered as unity, therefore if x is the height of the cool column X X I'oiS = i6'34 X I X = 16-05 inches. 24. A brass sphere 14 ins. diameter is at the temperature of 2 degs. C. What will be its volume at the temperatura of 7S degs. C? The co-eflBcient of linear expansion i» ■000019 per degree Centigrade. An obvious way to work this out would be to find the increased diameter from the co-efficient given and then the volume EFFICIENCY OF ENGINE. 279 by the rule for volume of a sphere. But a better way is to make use of the principle that the co-efficient of cubical expansion is three times the linear (approx.); (proved in part III). Thus the co-efficient to use is "000057, which for 73 degs. rise of temperature = •004161. Now find volume of 14 in. sphere and multiply by •004161, The result is the increase of volume due to expansion. Increase 5*978 ; volume i442"66 — Ans. Find the eflSciency of a compound engine if 18-38 pounds of feed water are used per I.H.P. per hour. The pressure in the boiler is 69 lbs. (temperature 315 degs.), and the feed water is at 1^1 degs. The efficiency required in this question is the comparison between the thermal units put into the feed water in the boiler and the work given out by the engine (horse-power) also reckoned in thermal units. It should be noticed in all such questions that the horse-power is reckoned per minute, but the feed water is g-enerally reckoned per hour, and in comparing the two they must both be brought to the same period of time. Taking the time as one hour — ■ Units of heat put into feed water = (1115 H- -3 X 315 - 141) X 18-38 = 19639 Also in one hour the engine performs units of work = 33000 X 60 = igSoooo or in thermal units = 1980000 -r- 'j'jS ,. .. = 2545 The efficiency is, therefore, = 2545 -r- 19639 = -1296 £8o EFFICIENCY OF BOILERS. 26. One boiler evaporates 9 lbs. of water per lb. of coal from 160 degs. feed temperature at 370 degs. boiler tempera- ture. Another boiler evaporates 8*5 lbs. of water per lb. coal the feed being at 120 degs. and the boiler at 331 degs. Compare the efficiencies of the boilers. The calorific value of the coal is 13500 thermal units per lb. In the I St boiler, heat put into 9 lbs. of feed water = 9 x (1115 + -3 X 370 - 160). This amount divided by 13500 = '71 The 2nd boiler efficiency = '688 The ratio is i : "969 Con^nued.— It tliQ 1st boiler and engine requires I'B lbs. of coal per I.H.P. per hour, and the 2nd, 2 lbs. per I.H.P per hour, find the combined efficiency of boiler and engine in each case. Taking the first case, t'6 lbs. of coal gives out 13500 x i'6 thermal units. This is the heat set free per hour in the boiler furnaces. The work done by the engine in the same time is 33000 X 60 units of work. If this is expressed in thermal units (divide by 774), and compared with 13500 x i '6, the combined efficiency will come out "iiS. And for the 2nd case, efficiency = "095. 27. Find the evaporative power of a sample of coal contain- ing -83 carbon, '0^ hydrogen, -025 oxygen, and the rest ash in 1 lb. coal. First we get the free hydrogen, which is -04 - -^^ = "0369. 8 Now a lb. of hydrogen gives out 62500 units of heat and a lb. ■of carbon 14500. The total heat is, therefore, — (•85 X 14500) -I- (-0369 X 62500) which comes out to 14631 units; dividing by 966-6, the latent heat of steam, we get the evaporative power 15' 136 lbs. of water, from and at 212 degrees. i COMBUSTION. '281 28. Find the weight of air to burn 1 lb. coal of composition: 81 per cent, carbon, 3 par cent, hydrogen, 5 per cent, oxygen, and 10 per cent. ash. Now when hydrogen burns it forms HjO, and since the symbol H represents i part of hydrogen and the symbol O represents 16 :parts of oxygen (atomic weights as they are called), the formula expresses the fact that 2 parts of hydrogen combine with 16 parts of oxygen. The ratio is i to 8. It follows that the free hydrogen in the coal is only 3 — f = af per cent. The oxygen required to burn the free hydrogen is, therefore, ^1 X 8 = ig. Similarly with the carbon which combines with oxygen to form COg (C = 12). The ratio here is 12 : 32 or i : 2|, •hence the oxygen required to burn the carbon in the coal is 81 x 2§ = 216. Total oxygen required = 216 + 19 = 235. Lastly, air contains by weight 23 parts oxygen to 77 nitrogen in every 100. We must, therefore, supply air amounting to 235 x ■3^^ = io2f7 per cent., this, of course, is io'2i7 lbs. per lb. coal. 29. Coal burnt, 52 tons per day. Fannel, 9 feet diam. Temperature ia funnel 650 degs. F., and outside 60 degs. F. Estimate the velocity of gases passing up funnel and weight per day. No analysis of the coal being given we may take the quantity ■of air required for combustion as that found in the previous example, viz., io"2 lbs. per lb. coal. And this number must be doubled in practice to allow for the air which enters the furnace and is not required for combustion. So 20.4 lbs. of air may be expected to pass up the funnel per lb. coal. Now at 32 degs. F. a cubic foot of air weighs "08 lbs., hence 50*4 lbs. equals 255 cu. ft. But this air when it reaches the funnel is at 650 degs. Fah., and, therefore, expanded in the ratio of the absolute temperatures. ' '•' (32 + 460) : (650+460) :: 255 : V ■'' : V = 575 cu. ft. of funnel gas per lb. coaU ' 282 VELOCITY OF GASES IN FUNNEL. It now remains to find the total volume of gases passing uj> the funnel per second. Lbs. coal per second used = — 5 . tl._. 24 X 60 X 60 ,. „ = i'348- Volume of gases = i "348 x 575. IT 1 -i. Volume 1*348 X ijy? o r<. Velocity per sec. = = —^2 -21^ = 12-18 ft. area Now, the heat put into each lb. of feed water in the boiler is 1 1 15 + ('3 X 370) - 130 = 1096 .". Heat turned into work by a perfect engine = 1096 x "29 = 3I7-8 ;per pound of feed water. But to produce one horse-power for one hour requires ~ thermal units = 21:1x8 774 ^^ Feed water required for perfect engine = 2558 -j- 317-8 = 8-05 nearly. Comparing this with the result of (i), The efficiency of the actual engine is only °5 _ .-g- of the •' 13-76 ^ ^ ■^maximum (iii). — Ans. EFFICIENCY AND HORSE-POWER, 287 2. Triple-expansion engine with cylinders 6|, 11, and 21 in. diam. and strolie 16 in. Revolutions 65. Boiler pressure 165 lbs. Back pressure in condenser 2-5 lbs. Assume the cut-off in the h.p. cylinder at ■? of the stroke and find I.H.P. One estimate of the I.H.P. which could be made here would be to draw the probable indicator diagrams and measure the mean pressure therefrom. Another way is to use some rule of thumb for the mean effective pressure referred to the low pressure cylinder such as: — ^ Referred mean pressure = square root of six times boiler pressure. A third method is to calculate the mean pressure for the total ratio of expansion as if it occurred all in the low-pressure cylinder, using Boyle's law, and then to multiply by a co- efficient obtained from practice. The result would be the referred mean pressure approximately. We will adopt this method. Total ratio of expansion is from -^ the capacity of the h.p. cylinder to the whole capacity of the l.p. This amounts to (2i)2 -^ (6A)2 X 7 = i6-i = r. Now, the most satisfactory method of finding mean pressure by Boyle's law is by using hyperbolic logs. m.p. = (i + log. e ?-) X ■? (The hyperbolic log. can always be found, even if no table is available, by multiplying the common log. by 2 "3026). m.p. = (i -f 27726) X i-^^ = 42*44 lbs. Effective pressure = 42*44 — 2*5 = 40 nly. The next thing is to assume a co-efficient for obtaining the reduced mean pressure owing to loss of temperature as the steam passes through the engine. The engine being of small dimensions we will use '6. Actual mean pressure = 40 x "6 = 24 lbs., which is the pressure to use in the l.p. cylinder to obtain the total horse-power of the whole engine. T Tj r> 21^ X 78,4 X 24 X 2# X 615 -■-- - I.H.P, = t_J2 5 s o _ 4,.66 — Ans. 33000 , ' 288 EFFICIENCY AND HORSE-POWER. 3. A triple-expansion engine has cylinders 80, 40, and 30 in; diameter with 58 in. stroke. The boiler pressure is 150' lbs. and the cut-off is at ^Sy stroke. Find the I.H.P. when running at 80 rcYolutions per minute. Also find diameter of shaft if working under a stress of 7500 lbs. per tq. inch with a ratio of maximum to mean twist of 1*3. Using- co-efficient "66 for mean effective pressure, H.P. = 3600; shaft i3'58 inches. — Ans. Continued.— \i the cylinders are % 7, and 10^ in. diameter,^ stroke 8 in. Revolutions 260 per minute. Boiler pressure 160 lbs. Cat-off at ^^. Find H.P. and diameter of shaft^ using same constants as in previous example. ( H.P. 30 (m.p. 33 lbs.); shaft i'95 in. — Ans. 4. The boiler pressure is 250 ibs. The engine is triple-^ expansion and uses 30 cu. ft. of boiler steam every second. Estimate the indicated horse-power developed. We can assume a reasonable number of expansions (say 18) and obtain a mean effective pressure as in the previous examples, using- a co-efficient of say "SS and back pressure 3 lbs. The next step is to invent a cylinder large enough at some number of strokes per minute to take away the amount of steam which is given in the question and expand it. As the same horse-power will h6 obtained by the calculation whatever proportions and speed are invented it is as well to choose simple numbers. Thus, a cylinder of 30 sq. ft. area of piston, cutting off at i foot and expandintj to a full stroke of 18 feet, will just take away 30 cu. ft. and expand it the required number of times. Also, if the piston of this engine makes one stroke per second it will take away 30 cu. ft. of boiler stearn per second which is what is required to be done. The speed of piston will then be 18 x 60 = 1080 feet per minute. Working upon these lines we obtain, a mean effective pressure of 34'8 lbs. EFFICIENCY AND HORSE-POWER. 289' The area of piston is 30 x 144 = 4320 sq. inches hence H.P. = 432o x 34-8 x 1080 33000 ,, = 4902 — Ans. S. An engine uses 120,000 cabib feet of steam per hour at 175 lbs. gross pressure. The vaoaum is 26 inches and the steam is expanded 12 times. Estimate the I.H.P. developed. m.p, 31 "5 (using co-efficient "66); I. H.P. 3299 — Ans. i6. — Ship's Perfortaance. i. The speed of a ship on trial is taken three times with the current and three times against the current. Tho results are: — 1, with 16-3 knots against 15-0 2, „ 15-6 „ „ 147 3, „ 162 „ „ 15-1 Find the correct rate of steaming. If we place the runs in a column, the 2nd column gives the means taken in pairs, and the 3rd column gives the means of the -2nd column, and the 4th the means of the 3rd, etc. J 1st 2nd 3rd 4th Last pee s. Means Means Means Means Mean 16-3 15-65 15-0 1 5 '30 15-475 15-35 iS-6 15-225 15-306 15-15 15-262 15-32 •47 1 5 '45 15-300 15-425 15-343 i6-2 1 5 '65 15-550 15-1 2. A steamer measures 420 ft. x ii ft., and draws 20 ft. 6 in. of water, with a co-efBcient of displacement of ■?. She is propelled at 12| knots. Estimate the I.H.P. necessary to drive her at this speed, and assuming that at a speed of 10 knots every 100 square feet of wetted surface requires 5 horse power, find the area of wetted surface. The Admiralty formula may be used, viz. : — I.H.P. = S" X D^ where S = C speed in knots, D = displacement, and C is a constant (say 250) ship's performance. 2gi The displacement = 4^0 x 44 x 20-.s ^ 35 )> = 7577 tons nearly. The formula may now be written : — I.H.P = (12-5)" X (7577)s 250 And working this out by logarithms : — S = i2'5log. - = 1-0969 3 7577 log. = 3-8795 Log. of S" - - =3-2907 r Log. of D^ - - =2-5863 ^ J ^'^^^° 2 Log. of numeration = 5*8770 C = 250 log. - = 2-3979 Log. D^ = 2-5863 Log. of LH.P. - = 3-4791 = 3014 LH.P. The next step is to find the LH.P. for 10 knots in the same ■way. It will be found to work out to 1543. Then as 5 H.P. corresponds to 100 sq. ft. of wetted surface. 5 : 1543 : : 100 : x (surface) which gives wetted surface = 30860 sq. ft. 3. Find the speed of a vessel whose length is 385 feet; breadth 45 feet; mean draught 23 feet; co-efficient of displacement '65; and I.H.P. 2095. Use Admiralty constant 200. The formula to use is : — I.H.P. = S^ X D^ . ^here S is 200 -speed in knots, and D is displacement in tons. aga ship's performance. The displacement is 385 x 45 x 23 ^ .g 35 ,, = 7400"2 tons. Putting this for D in the formula and transposing, we get: — ^v 2095 X 200 (7400)* This is best worked out by logarithms. Speed = 10-33 knots. — Ans. i. Given a vessel whose length is 300 ft., breadth 38 ft., draught 13-1 ft., and displacement 2506 tons; find hep co-efflcient of fineness; and if 4300 I.H.P. drives her 18 knots, what is her Admiralty constant? Co-efficient '587 ; constant 250. — Ans. 5. A vessel's displacement is 3000 tons and the I.H.P. of her- engines is 1200. What may her speed be expected to be? And if she discharges 1000 tons of cargo and the- I.H.P. remains the same, find the probable new speed. Take C = 260. ist speed 11 "4; 2nd speed 12 -52. — Ans. 6. If I.H.P. varies as ¥» D^ and it takes 8000 I.H.P. to- drive a certain vessel 19^ knots when the displacement- is 3500 tons, what I.H.P. will be necessary to drive the same vessel 21 knots when the displacement is reduced, to 2500 tons. A straight-forward way to work this example is to first find the- Admiralty constant C from the formula I.H.P. = V^ D^ C ■which transposed is : — C = V^ D^ I.H.P. V, D, and I.H.P. being known. SHIP S PERFORMANCE. 293 The new horse power can then be found for the altered speed and displacement, as the constant will not change for the same vessel. Constant 21 3 "66; I. H. P. 7984. — Ans. 7. Two vessels are built on similar lines. The first on a displacement of 9200 tons, steams 18 knots with 10,000 I.H.P. The second on a displacement of 10,000 tons, steams 21 knots with 18,000 I.H.P. Compare the «£9ciency of the engines and propellers of the two vessels. Since the vessels have similar lines, the constant should be the same for each, hence if the constant works out greater in one ship than in the other, it is probably due to a greater efficiency of pro- peller or of machinery. C = 256 and 238'5 ; ist vessel more efficient in ratio of i "07 : I. — Ans. S. A ship of 25,000 sq. feet wetted surface has a resistance known to be '3 lbs. per sq. foot, at a speed of 600 feet per minute. What do you estimate the total resistance to be at 18 knots? Also what horse power will be required? The wave resistance is neglected. From Froude's experiments we know that t-he skin friction varies as the i'825 power of the speed (see Part III.). Now 18 knots corresponds to 1824 ft. per minute, hence 1-825 "'825 (600) : (1824) : : -3 : x IBy logarithms x = 2*265 lbs. = resistance per sq. ft. Total resistance = 25000 x 2*265 ,, = 56617 lbs. Horse power at 18 knots = 3129*3. 9. What horse power is exerted in towing a vessel when the pull upon the tow rope is 15^ tons and the speed ii knots? H.P. 1153*85.— ^«f. 294 ship's performance. 10. Given speed of steamer 20 knots. Diam. of ppopeller 19 ft. Diam. of boss i ft. Slip 10 per cent. Find thrust on ship and horse power when making 90 revolutions per minute. ' Using Rankine's formula; : — Area of stream from propeller = , (D2 - B^) X (-7854 - ^) = A ; and Thrust = A x 5-67 x K x k , . The first thing is to find the pitch of propeller P. - ! ' T.T P X QO X 60 , J, „ Now -2_ = speed of screw = K 6080 '^ : ^ . P X 00 X 60 qo , . . -~: X -2— = 20 knots 6080 100 from which we obtain pitch = 25 feet.' ' The ist formula works out to ii3"5. For the 2nd formula we ave K = 22-5, k = 2-1 Thrust = 113-5 X 5'67 x 22-2 x 2-S , , „ = 31780 lbs. , . :'M Ik, The effective horse power is now calculated from the thrust, and the speed of ship in feet per minute. HD 31780 X 2027 - .P. = ^~i L = 1951-6 33000 11. The speed of a screw steamer is 14 knots, slip 12 per cent.y diam. of propeller 16 ft.« diam. of boss 3-5 feet. Find thrust exerted upon ship. Pitch = 20 feet. \ ' " '■ ' ■' '.' * ' 22472 lbs.— 4wJ. 17. — The Valve Diagram. The. eccentric being- merely a crank whose length is equal to the throw of the eccentric, the movement of a slide valve having- a vertical motion will be represented by the figure 105, where o, .1,. 2, 3, etc., are successive positions of the eccentric centre corres- ponding- to successive equal divisions of the stroke of valve (in this case id). The obliquity of eccentric rod being neglected. 'Now, if the slide has no lap when at mid-travel, that is position 5, it will begin to open the port exactly at this ppsition of the eccentric, and tlie port will remain open (top port) as the eccentric pras:^es through positions 6, 7, 8, etc., until it' is full open at lo,^ after this .it- will begin to close the port and finally shut at position 15, having been open exactly half; a, revolution of the eccentric, and, therefore, of ' Fig. 105, the enginp. From 15 too, and thence to 5, the top port iwill remain closed and the bottpmi port ;wilt be open idstead. \^ f ; Again, let the slide have some lap, say as much as two-tenths of its stroke. It is evident now that the port (top port) cannot be o^eai -until the slide has moved down below its half-travel position;. aniStaOunb eqttal , to the. lap given. The top port will open: at 7 instead of at 5, because 7 is two-tenths below 5. For Ihe Sanses 2g6 THE VALVE DIAGRAM. reason the port must close at position 13, the same two-tenths below the mid position on the up stroke. Hence from 7 to 13 the port is open and from 13 over the top to 7 it is closed. The above argument is the foundation of the valve diagram for it amounts to this : — Let a circle be drawn with the radius of the eccentric throw, figure 106, and from the same centre another circle of radius equal to the lap of the slide. Then any straight Kne as AB or CD or XY drawn to touch the lap circle will cut the travel circle in two points which divide the circumference into two unequal parts, the lesser arc representing the period during which the port is open, and the greater aire the period when it is close'd. It is-easily seen also that with negative lap the arcs are reversed,, the larger arc representing the open period and the smaller arc the closed period. So far in the explanation of the valve diagram we have not considered the position •of the crank or piston at all, but merely the action of the slide in opening and closing a port. But it is necessary to kefep in mind the fact that no matter what position the crank may be in, or what position the eccentric may occupy withf respect to the crank, both crank and eccentric revolve through the same number of ■degrees in the same period of time. Hence, if we find that the eccentric moves through 120 degs., for instance, between the opening arid cloising of a port, then also the crank moves thi/ough 120 degs. between the opening and closing of the port. , Thesimple diagram in figure 106, therefore, fixes the arc described by the crank' from the. admission of steam to the cut-off, bat it does not tell us at what part. of the revolution either event occurs. i, ■,) ui THE VALVE DIAGRAM. 297 Now, if the steam port opened just on a dead centre, the diagram for the cut-ofF would be as in figure 106, where the line AB is drawn from the position of one dead centre to touch the lap •circle. The arc AB gives the angle described from the centre A when the port opens to the point B when it closes. But this allows for no lead. If the valve has lead some modification is necessary. Now, referring back to figure 105, we have seen that the port i^ •open in every case for an arc of revolution represented by 7 to 13 •or AB in the subsequent figures. It is also true that the amount •of port opening is given by the same figure, for as the port begins to open at 7 it must be just one-tenth of the travel open at 8, two- tenths at 9, and three-tenths at 10. In other words, the port ■opening at any position, subsequent to the opening, is measured toy the perpendicular distance from the line AB to the travel circle. In figure 106, for example, when the crank or eccentric has Hioved the arc Ai the port has opened la, and the lines 2b, 3c, are the port openings for the positions 2, 3, etc., until the port closes at B. The modification to be made for lead is now easily seen, for if the port is tQ be open on the dead centre, the, dead centre canoot be at A, but must be some distance along the aire from A to B; at which the amount of lead is obtained — say at O, so that OC equals the lead on that centre — the arc of revolution performed from the dead centre until the cut-off is thus OB in- stead of AB. But in constructing a valve diagram it is con- venient to first fix the dead ■centre position, as in figure 107, at the top of the diagram, then draw the lead and lap •circles, and touching these •draw the steam line AB, rwbich gives the arc described front the dead c^entre. O to the cut-off at B. agS THE VALVE DIAGRAM. .There is no fresh principle involved for the position of the release and compression epochs. As these depend upon the exhaust lap, it I is only necessary to describe a circle round, C with radius equal to the exhaust lap, and then draw a line DC touching- this circle and parallel to the steam line AB. It is in the treatment of the exhaust lap that Zeuner's diagram is practically defective, for it requires a line to be drawn through two points so close together generally that the direction is very- uncertain. The above construction is free from this defect. It should be noted here that the exhaust lap in figure is positive. ' '' Now the student should not have any difficulty in constructing- a valve diagram from any data that may be given. For instance : — ^ 1. Travel, lead, and angle of crank at cut-ofF, given. Having described the travel and lead circles, mark the position of the cuti- ofF, and draw the steam line from the point ot cut-ofF to touch the lead circle. 2. Given travel, lead, and advance of eccentric. The angle between crank and eccentric fixes the position of the axis CR ii» figure 107, hence the steam line can be drawn at right angles t6 CR and tpuching the lead circle, 3. G'we.n lap, 7naxifnuni opening, and. lead. Lap + maximun* opening = \ travel, hence travel Is known also. 4. Given maximum opening, lead, and angle of crank at cut-off. I This is a difficult case, but it is also a very practical problemf. 1^1 fact it is the draughtsman's problem ; for the maximum opening i^ fixed by the speed of piston and area of piston, the lead is fixed from experience of the type of engine, etc., and the cut-ofF is fixeii by the number of expansions required. ' What is waited is a slide valve which will fulfil the purposes The following cohstruction has not, to the writer's knowledge, beeft published before except in a letter, by the writer, to the Engineer'^ some years ago. > , f ,THE V4I.VE- DIAGRAM. 299 a-n^Ce at cut- First draw any horizontal line AB, figure 108, and at any point in this line, set ofF by protractoi: one-half the angle which the crank has made from the dead centre at the cut-off. In, the figure, the angle of the crank is taken at 1 10 degrees and the angle BCX has. been made = 55 degs. = )^. Next dra,w DY parallel to CX, and distant from it an amount equal to the lead (^ in. in figure). Also draw EZ parallel to CX, and distant from it an amount equal to- the maximum open- |ing(ijin. in figure). > Find the middle point between C and D and draw mno^ at right angles to AB. In fhw find a centre' so that a circle will pass through C and D and touch the line EZ. (There is a simple construction for doing this, but it 'is quicker to find the' centre by trial). This circle is the travel circle and D ;:i ;• is the dead centre. The cut-oif is at P,,and CP is the steam Hnej, which fulfils the conditions required. The travel is found to be 4^ in. and the lap i in. Fig. 108. It may be noticed that, in this constructipn, the dead centre D' does not come exactly at the -top of the trayeli circle. If this is an objection it is easy to start afresh, as soon, as the trave) and lap- have been found and draw the figure upon a true vertical centre line. The above construction is perfectly sound geometricajly and can CEtisily be demonstrated. Jn figure 108, all the given dataware figuredi on the diagram. The lead is excessive to,tnake cohstruction plain,. Now, in drawing valve ^sdiagrams, since the lead and lap'are- not, as a rule,, the Same at each end, it is necessary to dr^w iai ^oo THE VALVE DIAGRAM. -separate diagram for each end of the cylinder. They are often both represented upon the same figure, but the plan is not to be 'recommended. Piston Slides. — The diagram for a piston slide is the same as 5for an ordinary D valve ; for even if the steam is admitted between the pistons and the steam lap is inside instead of outside there is no 'need to draw the diagram in a different way. Of course, the position of the eccentric requires to be moved back i8o degrees, but the angle of advance remains the same. Double-Ported Slides. — If a single-ported slide is first designed "to give the required port opening, and cut-off, etc. , etc. Then the •double-ported slide requires to have just one-half the travel and one- half the lap (steam and exhaust) and one-half the lead at each face. The advance of the eccentric remaining the same as for the single- ported slide. But if the action of an existing valve is to be shown on a diagram, it is simpler to use the actual travel and the actual laps and lead of one face of the valve, keeping in mind the fact that the maximum opening for steam and exhaust, as well as the lead, ■are virtually twice what the diagram gives. Li ti king up. — An approximate diagram for a link motion with the sliding block at any inter- mediate position may be drawn as follows : — In 'figure 109, A and B are the centres of the head and stern eccentrics when the •crank is at the dead centre ■C. Also ab is the link, ■a being the position of the block when in line with one eccentric rod, and b the position when in line Fig. 109. THE VALVE DIAGRAM. SOI* with the other rod. Now divide the link centre line db and also the- straight line between eccentric centres AB, into the same number of equal parts (6 in figure 109). Let the gear be linked up until the block comes into position d. Take the corresponding position D- between the eccentric centres and join OD. Next construct a valve diagram for an eccentric, whose throw is OD, and angle of advance FOD; using the same lap and lead as in full gear. This diagram. will give an approximate cut-ofF, etc., for the linked-up position. The assumption made in the above construction is that the lead' remains the same when linked-up, which does not hold good with. Stephenson's link. As is well known, the lead increases by linking- up when the centre lines of the rods cross each other with the- engine on the top centre, but the error is not serious with long'; eccentric rods. i8.— The Piston and Crank Diagram. In valve diagrams it is not usual to take into consideration the •obliquity of the eccentric rods, because the rods are so long in pro- portion to the throw of the eccentric (20 to 30 times) that the correction would be inappreciable. When, however, we come to make a diagram of the relative positions of the crank and piston in which the ratio of connecting rod to crank is often only 4 to i, and inot generally more than 5 to i, in marine work, the obliquity of the • rod cannot be disregarded, as its effect is to bring the crosshead . and piston several inches lower for a great part of the stroke than with a rod infinitely long. Perhaps the most useful "piston diagram" is Zeuner's (figure ii 10) which is constructed as follows : — Describe the large circle with a radius equal to the length of ■•the connecting rod plus the crank ; and from the same centre O describe a smaller circle with a radius equal to the connecting rod : minus the crank. The radial distance between the two circles is thus equal to the stroke of the piston or twice the length of the crank. The third circle is described with a radius equal to the -connecting rod itself, and its centre A is taken so that it touches both the others — inside the large one and outside the small one. "The diameter joining the two centres is the vertical centre line of the engine, and the point A is the position of the crank when on ^the top centre, O being the centre of the shaft. The diagram is used as follows : — If it is required to know the -position of the piston in the cylinder for any angle of the crank — say ■60 degrees from the top centre — draw a radius OT from O making' THE PISTON AND GRANK DIAGRAM. 303 thegivqn angle 60 degrees with the vertical, and cutting the three circles at B, P, and T respectively. Then PT is the distance of the; piston from the top, and PB its distance from the bottom of its stroke. We may look upon the outer circle as the cylinder cover itself, and the inner circle as the cylinder bottom, then as the' crank swings round from the top centre A, we can watch the intersection of its radial line with the oblique circle for a whole revolution. We can see the point P (which represents the piston itself) leave the <;over at the top centre C and gradually approach the bottom of the cylinder at V, after : , half a revolution of the crank. And so on for the up-stroke. When the crank is pointing in direction O/for instance, which is 90 degs... from the bottom centre, the piston is distant pb from the bottom and pt from the cover. The difterence be- tween/^ and ^J shows the maximum disturb- ance due to the obli- •quity of the connect- ing rod, for at 90, if it were not for obliquity, the piston would be exactly at half- stroke, whereas the diagram shows that the distance pt from the top of the stroke is considerably greater than the distance pb from the bottom. In order to show the practical utility of the valve and piston diagrams, we now propose to get out the particulars of a slide valve for given conditions of cut-off, etc. , in a cylinder, as follows : — Fig.. 1 10. Given. — Stroke of piston 2 ft. 9 in. ; length of connecting rod 6 ft. ; maximum opening of top port i in. ; cut-off 9.1 ^ on bailt, 304 A SLIDE VALVE PROBLEM. Strokes; compression to commence at -^ on the up- stroke and at 8i ID on the down-stroke ; lead at top -j^ in. To find. — I, travel of slide; 2, lap at each end (steam and exhaust) ; 3, lead at bottom and maximum port opening ; 4, advance of eccentric; 5, position of piston at release on each stroke. We commence with the piston diagram, figure iii. The radius for the outer circle is 6 ft. + (i ft. 4^ in.) = 7 ft. 4^ in. The radius Fig. III. for the inner circle is 6 ft. - (i ft. 4^ in.) =4 ft. 7^ in., and for the oblique circle 6 feet. Next divide the stroke into tenths, and from the centre O sweep the six-tenth division round to cut the oblique circle at P. The radius OPT is the position of the crank at the cut-oif on the down stroke. Similarly, sweep the eight-and- a-half tenth division round to cut the oblique circle at Q. The radius OQM is the position of the crank at compression on the down stroke. A SLIDE VALVE PROBLEM. 305 ■ ' For the up stroke the sixth and ninth divisions (counting from the inner circle) are swept up to cut the oblique circle' at P and Q, thus fixing- the corresponding crank angles for cut-ofF and compres- sion on the up stroke. It may be noticed, in passing, that the crank angles for the cut-off on the two strokes are not the same, although the piston positions are both at yg-. We have now to construct a valve diagram for the top valve face working over the top port. We have given the lead ^ in., and maximutp opening i in., and from the piston diagram we have tt(e angle of the crank at the cut-off, which measures 95 degrees, and the angle when the top port closes to the exhaust = 35 degrees. i ' ' In figure irz, the angle BCX is made equal to half the angle ■of the crank at the ciit-off = 47-|' degrees. Also the parallel lines •DY and EZ are drawn, giving the maximurii opening and lead; ri'h'e perpendicular frbm the middle point C and D is dropped, and tHfe travel circle 'described to touch EZ. The travel will be found ^ti'tn^stsure 5I in. ' Renlembering that the top centre is the point D in this diagram, we draw DW through the centre of "the travel o 306 A SLIDE VALVE PROBLEM. •circle. The steam lap OL is found to measure lif . To find the exhaust lap, lay off the crank angle at the top compression obtained from the piston diagram (DOM = 35 degs.). From M draw M/R a line parallel to CX. This is the exhaust line and Ol is the exhaust lap required for the top port = f in., which is negative in this case. The point R is the crank position for release at the top port. If the angle WOR is measured and laid off upon the piston diagram, figure III, it would be found to coincide with the lineOQM, which is accidental, but in any case the piston diagram would give us the position of the piston when release occurs at the top port. In the present case the piston has made "85 of its stroke from the top at release. We have next to draw the valve diagram for the bottom port. The data to work with are : — I St. The travel of the slide and the position of the eccentric, "which are the same as in the top diagram. 2nd. The angle of the crank at the cut-off, which was found toy the piston diagram to be 108 degs. from the bottom centre. 3rd. The angle of the crank when compression begins, also found on the piston diagram, viz., 50 degs. from the bottom centre. In figure 113, the travel circle is first drawn (diam. sf in.). Next the axis of the diagram making the same angle with the dead centre as in figure 112 (angle WOF is the same in both figures, being angle the between crank and eccentric). The cut-off is now marked at 108 degs. from the bottom centre, and the steam line PC drawn at right angles to the axis OF from the point P. The lap and lead can now be measured off at the bottom port. For the exhaust lap we must mark the compression point M, at 50 degs. from the bottom centre and draw the exhaust line MR parallel to the steam line, which enables the exhaust lap to be measured {^ in. positive). Also the exhaust line shows the release at R, which, when trans- ferred to the piston diagram, figure 11 1, gives the piston position at xis stroke from the bottom. A SLIDE VALVE PROBLEM. 307 Note. — Some may prefer to draw figure 113 the other way up, so as to correspond better with figures 1 1 1 and 112, BOTTOM 19> — Indicator Diagrams. 1. To make a diagram of "effective pressures" for the np and down strokes of the engine, from the indicator diagrams given in figure Hi. We have to consider that in order to obtain the effective- pressure npon a piston, it is necessary to subtract the back pressure- acting upon the opposite side from the forward steam pressure,. which is driving the piston. Taking the up stroke for example, the- the bottom diagram gives us the steam pressure below the piston, which at -^ the stroke is represented by the line ab, reckoning from, atmospheric pressure. Now at the same part of the up stroke, the- top diagram gives us the pressure above the piston, viz., cb. Subtracting cb from ab leaves ac, which is the effective pressure at ■^ of the up stroke. The rule, therefore, to follow in measuring effective pressures- is to take ihe distance from the steam line of one diagram to the back pressure line of the other diagram.. It may be noticed in passing that the common measurements of" a diagram taken in finding I.H.P are not effective pressures, either for the up or down strokes. To follow the system out and make a diagram of effective pressures, draw a base line CCC divided into lo parts, the same as the indicator card. At every tenth of the up stroke, measure with dividers the distances a^c", aV, aV, etc., and lay them off from the straight common base line CCC. Thus A^Cq = a^c^, AjCj = a c , etc. When the crossing of the lines is reached at about 8^ tenths- ■EFFECTIVE PRESSURES. 309 •of the stroke, the effective pressure is nil, and at 9 tenths the resist- ance to the piston is a^c^, which is set off as A9C9 beloia the base line, the same with ffjo'^io — -^lo^io' which shows the cushion resist- ance at the top dead centre. A fair curve is now drawn through the points Ao, Aj, Aj, etc., which is a curve of effective pressures for the up stroke. A similar construction is made for the down stroke. ytea-n effetti(;e bresA 330WYI iiro\t,-.%b'% "Uff Atrelvc - j5-» Fig. ti4. The use of these curves is that a correct estimate can then be made of the work done during each stroke of the piston. In the present case, for instance, the area enclosed between the curve and the base line (when the part below the base line is deducted) is very much less for the up stroke than for the dovs^n stroke, which is a defect in the engine not immediately apparent oh inspection of thei 3IO EFFECT OF THROTTLING. two indicator diagrams, in-as-much as they have very nearly the same mean pressure when measured in the ordinary way. The ordinary mean pressure of the two diagrams, of figure 1 14, gives 20'6 for one and 20-8 for the other. Whereas, the mean effective pressures are 26'2 and i5'i pounds. The average pressure for a whole revolution is 207 lbs. very nearly, whether we measure in one way or the other. 2. To show the effect of a set of indicator diagrams of throttling the steam to the H.P. cylinder. When the valve gears are not altered in any way, but only the quantity of steam to the H. P. valve box crippled by closing the stop valve or throttle valve nearly down, it is clear that the points of admission, cut-off, release, etc. , shown on the diagrams will remain the same in all the cylinders as before. Hence the amount of expansion of the steam at various points in the cycle will also be unaffected by the throttling. This being the case we should expect to find the new pressures shown on the diagrams from the throttled engine a certain fixed fraction of the full load pressures in each diagram. For instance, if the engine is throttled so that the pressure in the H.P. cylinder at cut-off (when expansion begins) happens to be \ the pressure at cut-off when full open, then all the subsequent pressures should be also \ the fuh load pressures until we get to the back pressure in the L. P. cylinder, which would be practically unaltered as it depends upon the vacuum in the condenser. This theory does not take into consideration the fact that the effect of throttling is to dry the steam and thus alter its law of expansion. It also ignores other differences, such as the altered temperature of the cylinder walls and their effect on cylinder con- densation ; but the above simple theory is a help in estimating the new pressures resulting from a less quantity of steam entering the H.P. cylinder when throttling. EFFECT OF THROTTLINGi 311 The full line diagrams, in figure 115, are from a triple-expansior* engine using steam of about 200 lbs. in the boilers. The dotted* line diagrams have been drawn to show the effect of throttling. In order to get a definite case it is supposed that the effect of throttling is to reduce the pressure in the H.P. cylinder at the cut-off to just one-half its former pressure (absolute), that is from 170 lbs. absolute * _v . _ . ^ „ f>erjecli i/aeuunv lint- Fig. 115- to 85 lbs. absolute. Now, according to what has been said at the outset, we may expect the pressures to be reduced to one-half all through the engine, so it will be observed that the back pressure in the H.P., as shown by the dotted line, has been made just one-half the former amount. Also the M.P. card, which showed an initial pressure of 66 lbs. when full open is now reduced to 33 lbs. Its back pressure- now falling below the atmospheric line. In the L.P, diagram the initial pressure is reduced from 20 lbs. absolute to 10 312 EFFECT OF LINKING UP. lbs', absolute ; ' but no alteration is made in the back pressure aS the vacuum might be assumed to be the same as before. ;i Looking at the respective areas of diagrams, shown by the full and dotted lines respectively, it may be noticed that the L.P. dialgrani suflfers the greatest proportional loss of effective pressurfe; amount- ing approximately to 75 per cent. The M.P. suffers a loss of about 50 per cent., and the H.P. perhaps 40 per cent. 3. To show the diiference in the diagrams obtained, 1st, when the' power is reduced by using the reversing wheel, and 2nd, when closing the stop or throttle valve. Figure 116 {i) is an _^^ - -^j) ^ actual h. p. diagram taken '•'***"'-^... ''' / when the revolutions were "•,. --;■:•=-„-- ,.,--•::'_' reduced from 62 (at full power) to 33 by closing- -"^ ....^ , ^; . , the stop valve. The boiler ".'v-, , ' • ,---.-• pressure was kept at 175 \ "•, / / lbs., and the m.p. gauge '\ '■-. V '^ / .'',/' came down to"" 15' lbs.', iand ''., '""--,_ ,,--''',,-'' the l.p. gauge fell to - 5 _-_^ --"f.'im^-^-V.-.-.-,-,--- lbs. The vacuum being 26 inches. The other dia- - ^.^..^,. -. grams are missing, but Fig. 116. they could easily be drawn from the pressures given, and would bear out the conclusions arrived at in No. 2. '"■■■■'''"' ".''': .-r'j ■;'''■ r,1 Figure 116 (2) is the actual h.p. diagram, taken when, th^ stop. valve was kept full open, and the revolutions reduced to 33 (as before!) by the reversing wheel instead by the stop valye., Boiler gauge as before at 175 lb?; The m.p., gauge now showed 58, .and the, l.p.' gauge 4-5. yac. 26 inches. ;Since all the links wppld be affected similarly by the reversing wheel, we may conclude that the othefl diagrams would bej similar to. the h.p., diagram. 1 . ;, ,'1 ,!.;:» EFFECT OF LINKING UP. 31S - It would be interfesting to know under which set of conditions the consvimption of steam would be, the least. The total I.H.E. is probably very nearly thesame, as the revolutions are the same viijder the two conditions. 4. To show the effect npon the indicator diagrams of a triple-expansion engine of linking up the medium engine's va,lYe gear. r, i 1, In figure 117, the full lines show the: diagrams under n&rmal conditions, and the dotted tines show the effect of linking ifp the medium gear. It will be noticed that _the..exp.ansiQQ . „ curve of the medium card is not altered except that -it is carried back to an earlier cut-off. The reasorji being that inras-mueh as there is the same quantity of steam passing' fhrbugh'the whole engtee^a^ before the altera- . . tion, the pt^^urejwhen the steam ts lillirig. the. i^ediiim cylinder at the end of the - stroke cannot be any different. A higher pres- ' ■ 1 sure would mean a greater amount of steam by isseight, used by.tti^ engine each stroke, and a lower pressure would mean a less quantity of steam by weight, used by the engine ; both of which are ippppsistent with tb^.fonditions of the problem. It is, for ^ similar reason that the \.-g. :^ia.gr&vri remains unaltered. , The increased bgpk pi;essiire,-;in the h.p. diagram jfequires no explanation; ,: , ; ,,(| Q'. To -coUibine' t^e diagrams of a multiple-expansion engine into one simple diagram. ' - The 'diagtariliSv figure 118, are from a five-crank engine." with cylinders,, 1 7,ini,,.:24cin.'; 34 iow, -and two of 42 in. diametert- ' Strtik^ Fig. 117. 3H COMBINED DIAGRAMS. 3 ft. 6 in. The boiler pressure is 26b lbs., and the engine makes 78 revolutions per minute. The cylinder clearance volumes may be taken as -^ of the working volumes in each case. In combining diagrams, three things have to be done : — 1st. AH the diagrams must be drawn afresh to a common scale of pressures. It is generally convenient to choose the same Hi^V. Preisure dijlr i^ iTitcT'Yvicd.la.U lo>v Pressure fiyl - ■m-fs.ij-b Fig. 118. scale as the intermediate cylinder diagram, but the only thing that limits the scale is the size of the paper which is used tor. the combined diagram. and. In re-drawing the diagrams, the lengths of the several diagrams, instead of being nearly equal as taken off the indicator barrel, must be of various lengths : — These lengths must be in the same proportion to each other as working volumes ot the cylinders 3rd. The diagrams with the cylinder clearances set off are plotted to a common base line representing "no pressure," and so COMBINED DIAGRAMS. 315 that the ordinates representing "no volume" are in the same vertical line. In the case before us the engine is quadruple expansion, the two low-pressure cylinders dividing the steam between them ; and since the stroke is the same for all the ratio of volumes is : — 17^ : 24^ : 34^ : 2 x 42^ or I : 2 : 4' : i2'2 If we now increase each volume by ^ for clearance we get : — i'o83 : 2'i65 : 4:33 : i3"22 In figure 119, OX is the line of no pressure (vac. line) and OY the line of no volume. The distance aa is made = i '083 to any scale, and represents the volume of the high-pressure cylinder including clearance. Similarly bb is made = 2"i66 representing' the volume of the ist intermediate and its clearance. Also cc = 4*33 = 2nd intermediate, and OX = total volume of both low-pressure eylinders = 13 '22. What we have now to do is to draw the indicator diagrams in the spaces thus set out for them. Beginning with .the high-pressure diagram the scale now used is 40 lbs. to the inch. A T is the atmo'spheric line assumed as 15 lbs. above the vacuum line OX. To save time the diagrams may be divided into five spaces only. At each ordinate the pressure above the atmosphere is measured upon its proper scale (120 lbs. = i inch) and set off on figure 119 at the rate of 40 lbs. = i inch. (As it happens the ordinates will be just three times their original length iii the case of the H.P. diagiranji). This is done for both steam and back-pressure lines and the points connected by fair curves. In plotting the diagrams upon the combined diagram we must be careful- to leave the clearance spaces for each cylinder untouched; and only divide the actual working volumes by the 10 ordinates. -4- ♦ ,^ 1 ■ — r-j — « — . . . - a :^ ri COMBINED DIAGRAMS. 317 When all the diagrams have been transferred to the combined figure, the general outline will suggest an indicator diagram with a very early cut-ofF, in the present case at about -jy of the stroke, which should agree with the number of expansions of the whole engine. Thus the cut-ofF in the H.P. cylinder being at "] and the ratio of H.P. to both L.P. cylinders being i : 12 '2, the total number of expansions is 12 "2 -r 7 = 17*4. Indeed, the combined diagram is constructed with the object of showing graphically how the pressure of the steam falls as the expansion proceeds through the successive cylinders, the actual volume at every instant being apparent to the eye. Having obtained a combined diagram such as figure 118, it is- usual to draw upon it an expansion curve showing some theoretical law^ of expansion for comparison with the indicator diagram. The dotted line in the figure is an isothermal curve (or Boyle's law curve) commencing at the pressure and volume of the steam at cut-off in the H.P. cylinder and continued for a ratio of volumes =17. The curve is very easily set out as follows : — The volume of steam in the H.P. cylinder at the point (i) is represented by the distance O i which includes the clearance space. We set off this volnme from i to 2 and 2 to 3, etc., up to 17 volumes, and at (2) the pressure by Boyle's law should be one-half the pressure at (i) ; at (3) the pressure should be one-third ; at (4) it should be one-fourth ; at (5) one-fifth, etc. Since the pressure at (i) is 232 lbs. (reckoning from the vacuum line OX), if we set up 116 lbs. at (2), 74 lbs. at (3), and 58 lbs. at (4), etc., we readily obtain a succession of points on the expansion curve, and a fair curve can be drawn through them. It will be noticed that the expansion in the H.P. cylinder as shown by the indicator diagram follows pretty closely the isothermal curve, but by the time the steam reaches the ist intermediate the diagram pressure is much below the isothermal, due, of course, to the lower temperature of the expanded steam. Later on in the expansion we notice that the pressure is reduced to atmospheric pressure after 10 expansions as proved by the L.P. diagram, whereas by Boyle's law atmospheric pressure is reached only after 16 expansions. 3l8 INDICATOR DIAGRAMS. 6. Draw a perfect high-pressure indicator diagram. Also draw the usual diagram showing the difference. Boiler pressure 160 Ihs.; cut-off at -6 stroke. Explain fully why the two do not coincide. The valve may he assumed to he in good order. 7. What travel would you give the slide valve for the above engine, and what angle of advance to cut off at -6 stroke? The lap is 2^ in. and the lead ^ in. S. The h.p. valve gear is linked up 2^ in, towards the centre of the link which is 2i inches long. The m.p. gear is full over and the l.p. gear is linked up the same as the h.p. If under these conditions each engine develops 1000 H.P. you are required to state the probable effect of opening the gears full out. Draw the indicator diagrams you would expect under the two sets of conditions and estimate the horse-power of the second set. FART III. PROOF OP RULES AND FOR/AUL/E. 1.— Comiixon Logarithms. If we take the number loand write down several powers of lo as under : — r.0 = lO" I ID (a) IO'= = lOO lO^ = lOOO ID* = 10,000 we have a set of numbers and their logarithms. When the power is I the number obtained is 10 ; when the power is 2 the nunlber is 100; when the power is 3 the number is 1000, etc. If we agree to refer everything to the resulting numbers then the indices i, 2, 3, 4, etc., are called the common logarithms of their respective numbers,. 10, 100, 1000, 10,000. Common logarithms are said to have a "base" of 10, because that is the number which is raised to the various powers ; and every common logarithm is the index of some power of 10. Compare the statement (a) with the following state- ment (5). Both mean the same thing : — log- 10 I log. i„ 10 log. 10 100 log. 10 1000 log. 10 10,000 = o = I = 2 = 4 1 (b) It will be observed that the "ba^e" (10) is written immediately, after and below the word log. ; but in the case of common logarithms it is not usual to .Write the base at all. It is understood to be there. Althpugb the : calculation , of thfi 'l;9gari;thm of any given number isjijot an fi^^ m,atter, it is easy to find ^wwe logarithms, between the numbers in («) and (5). I,;,;; ..j!. , j V 322 COMMON LOGARITHMS. Beginning again with the base lo, we can readily obtain some fractional powers such as lo*, for \/io = 3'i62. .". io'5 = 3M62, and log. 3"i62 = •^. Again ioi'= Vio = ijyi62 = I'yjS. .'. lo'^^ = 1778, and log. 1778 = •25. Also 10^ = Vio^ = Viooo = 5'623. ,-. 10 •'^s = 5-623, and log. 5-623 = -75. " •v ^ ^ -at^ ^, «9 5-6 / ¥ . K / / X ■ / K^ / C . 2 1 ^ 1 1 \ ^ : [ \ 1 ■ ' " i A S 6 N U M B £. R & Fig. 120. la We have now sufficient data to construct a logarithmic curve {as in figure 120) as far as the number 10. The numbers are set off along the bottom and their logarithms at the side. The ringed spots have been set off from the logs., and numbers just calculated, and after drawing a fair curve through these points, the logarithms of the whole numbers from i to 10 have been read off to the second decimal place, thus : — COMMON LOGARITHMS. 323 log:. I = O" log-. 2 = •30 log-. 3 = •475 log. 4 = •60 log-. 5 = •69s log. 6 = •775 log. 7 = •845 log. 8 = ■91 log. 9 = •995 log. 10 = I -co If the reader has obtained from the above arithmetical con- siderations an idea of -what a table of logarithms means, it is all that has been aimed at. Of course, logarithms are tabulated to an extreme degree of accuracy — sometimes to 6i decimal places — -which enables calculations to be made also of great accuracy. The use of logarithms in actual work depends upon a fe-w principles of algebra. In order to multiply two different powers of any number we add their indices, thus — fl '" X a" = «"■ + " 10^ X 10* = 10^ But since the indices (tw) and («) in the one case and 3 and 4 in the other are the logarithms of the quantities themselves, it follows that to multiply the quantities we must add their logarithms. The sum being then the logarithm of the product. For a similar reason to divide one quantity by another we subtract their logs. 1. Find by logarithms the value of ^'"""^^^^g^^ ^ ^ "" By writing the decimal number as a vulgar fraction and putting- the number 2"* in the denominator the formula becomes f 8754 A * ^ I V,1 0000000/ ■ (6567* X 2* 324 COMMON, LOGARITHMS. We now find the square root of both numerator and denominator of the first fraction. 8754 log- 2J 3-942206 6567 log. 4J 3-817369 I 0000000 log. 2J 1-971103 7 -oooooo 3-500000 0-954341 8 log. 0-903090- 3'5 5"35743i bring down 1-971103 subtract 5 '357431 - 4-613672 number = -0004108 2. Find the value of P from the formula p_ 806300 T^-'^ L X D where T = ^V. L = 90, D = 39. The above can only be worked out by logarithms. In order to. avoid the log. of ^s = -4375, write the formula as under p ^ 80600 X (7)^-1" 90 X 39 X (i6)2-i" 37"56-— ^w-J. 3. In the formula P = ( 71 J . P is the absolute pressure of steam, and t is the boiling point of the water (Fahrenheit). Find the temperature corresponding to. a pressure of 60 lbs. above the atmosphere. First take the 5th root of each side of the equation. pi = ^ + 4<^ '47 Now multiply each side by 147 and change sides V+ 40 = 147 P* COMMON LOGARITHMS. 325 We have now to work out by logs, the value of 147 x 75^ and subtract 40 from the result. / = 308-6 degs. — Ans. i. Work oat the above formula when P = atmospherio pressure, and also wh^n P = 200 lbs. above the atmosphere. ^ 2i2'6 degs. and 421 degs. — Ans, ■'■■:■■ i . ;:. ■ ; '■ ■ ,'; ril. 5. Find the absolute pressure of steam in a cq>nd6nter when the temperature is 128 Fah. ' fN 1 ,i ;■'-' '■> a. — Hyperbolic Logarithms. In hyperbolic logarithms the base is the number 2*7183, etc., so that if we raise 2'7i83 to various powers and note the numbers obtained, the indices of the powers are the hyperbolic logarithms of the numbers obtained. As a single example, since (27183)2 = 7-39 (nly.) hyp. log. 7-39 = 2 But by writing the Greek letter e (epsilon) for the "base" the word "hyperbolic" may be omitted, thus log. £ 7-39 = 2 Tables of "hyperbolic" logarithms (called also "natural" or "Naperian," are not so convenient for ordinary calculations as the common logarithms, because we cannot so readily obtain the logarithm of a number with the decimal point shifted as we can in common logarithms. For example, we have above log. e 7*39 = 2, but it does not follow that log. « 73*9 = 3 as it would in common logarithms (by adding i to the indpx). To obtain the log. of 73*9 from the logarithm of 7*39 we must add the log. of 10, and it is only in the common system that the log. of 10 is I. In order to obtain the hyperbolic logarithm of a number from its common logarithm it is necessary to multiply the common logarithm by 2*302 which is the hyperbolic log. of 10. The name of "hyperbolic" logarithms has arisen from a property of these logarithms of much use in calculating the area enclosed by a hyperbolic curve. HYPERBOLIC LOGARITHMS. 327 The rectangular hyperbola, figure 121, is a curve symmetrical with respect to two lines at right angles, called " assumptotes, "^ OX and OY. The curve, when produced, continually approaches the assumptotes but never meets them. A property of the curve is that if any point P is taken in tl^e curve, the product of its distance from OX and its distance from OY is a constant quantity, no matter where the point may be situated in the curve, thus Pa x P& = constant = V^a^ x P^^i. Fig. 121. It is owing to this property, of course, that the expansion curve of an indicator diagram is an approximate hyperbola. Now, where hyperbolic logarithms come in is in calculating- the area of a portion of the figure such as P Pj a-^ a. Let the ratio of 0«, to Oa be r or y^ = r. Oa Then it can be proved that area P Pj Cj a = Pa x Pd x log. e r.. .(I). , 328 «YPERBOLIC LOGARITHMS. Another way of putting it is that the ,hyperbolici k>g;b.rithm ■ of (r) area M' i r. area N , By means of a planimeter it would be easy to make up a table of hyperbolic logarithms in, this way. Thus- having set^ out a hyperbolic curve as accurately as possible, take r=_i, 2, 3, ,4,,, 5, etc., and with each position of P, measure the areas M and N (N would be the same every time). THen by dividing M ■ by N the correspondiiig logarithm would be found. 3. — The Trigonometrical Ratios and Angular Measures. Whenever two lines meet together or intersect each other an angle is formed. Two planes meeting or intersecting also form an angle, and three planes under similar circumstances form a solid angle. The corner of a cube is an example of a solid angle. But the angle wliich the trigonometrical ratios apply to is an angle formed by two sfraight^lJnes. Now every angular point may be considered as the centre of a circle of any radius we please; which at once gives us the universal method of measuring angles. The circumference of th^ circle /being divided into 360 jequal parts called degrees, we ha\je only to count the number of dejgrees inter- cepted between the two lines, which form, the angle, in order to obtain the measurement of the angle. And it, is evident that, whether we choose a small or a large circle, the resulting' number of degrees will be the same. Protractors and theodolites as well as sextants andazimuth compasses measttfe angles in this manner. In figure 122, we have a quadrant of a circle, which, of course, fcoqtains 9adegrfe.es. The two lines CASndiC'B foi ""^^^^^ \i^ JS Z \ .^^ % li X >3^ T 5 ^0- Scne ^-rq \ ^ <& e" y «9 ^r \ rs y^^ \ l< \ s \ \ Q \ •<• X '^ \ ^ <5a versln _ CL s ■21 A, Fig. 122. Again, the angle BCD in figure 122 being the amount which ACB is less than 90 degs. is called the complement of ACB, and in the figure it is equal to 52 degs. Now, by drawing BE perpen- dicular to CD we get the sine of the angle BCD, that is the sine of the complement of A CB the original angle. This expression is contracted to cosine of A CB, A similar explanation gives DN the co-tangent and CN the co-secant. Their approximate values are marked on the diagram. It should be noticed that although we speak of the line BS as the sine of the angle, etc., what is really meant is not the absolute THE TRIGONOMETRICAL RATIOS AND ANGULAR MEASURES. 3311 length of the line BS but its length compared with the radius of the circle ; thus, whatever radius we take for the circle, BS is '62 of that radius. The sine of an angle is, therefore, correctly said to be a "ratio," for it is ratio of BS to BC. , And if we merely look at the triangle BSC and call BS the "perpendicular," CS the. "base," and CB the "hypothenuse," as in figure 123, the sine of the angle is the ratio of the perpendicular to the hypothenuse, or as it is written : — per. sine = f — hyp. ,(i). The other ratios may be expressed in a similar manner. Thus the tangent of the angle BCS is the ratio of AT to the radius. AT that is -^— and this becomes CA tangent per base .(2). CT Also the secant is the ratio of CT to the radius, that is — ^ or secant = base' •(3)- Modern works on Trigo- nometry commence by giving equations (i), (2), (3) as the definitions of the sine, tangent, and secant of an angle with- out any reference to figure 122. Starting in this way may have advantages, but figure 122 enables many relations between the ratios to be seen at a glance which are not so easily seen without it. For instance: — CS being equal to EB is also the cosine of ACB, and SA, which is called the versed sine, = CA - CS, that is (CA being unity) ,_332 THE TRIGONOMETRICAL RATIOS AND ANGULAR MEASURES. Versine = i - cosine (4). Similarly co-versine = i - sine (5). Again, B S^ + C S^ = C B^ = i^ which is ^ine^ + C0S.2 = I (6). or sine^ = i - cos.^ (7). and C0S.2 = I — sine^ (8). Also note in the triangle CAT CT2 = CA2 + AT2 which is secant^ = 1^ -j- tan.^ (9). or sec. ^ — tan.2 = i (lo)- Then again by similar triangles ' ■'■-' ' '■ -BS AT u-'u u ' '' ' -=— = -— which becomes CS CA ji^ , ; J; . sine tan. . sine , n or tan. = (!•)• cosine ...I ,,, , , I cos. Such relations are very useful for changing one ratio into its equivalent to-aaother,^ and- wjth, figure (ip 2 [b^fpc^ one |tljey/can be picked out generally when wanted without committing so many of thepi- to. memory» --•..' ' ■ '''• 1. The circular measure of an angle is 1'5§, lioi|ir many degrees does it contain? ' X,^ , ' V J.; ■ I. Ill The unti-qi circular measure is an arc of a circle the same length as the radiuSr. An angle equal to 2 in circular measure has an arc twice the length of the radius, and so on. It is easy to •change from one measure to the other if it is remembered that, the owhole length of the circumference of a circle js 6'2832 times the radius, \Khich shows that an angle or arc of 36P degs. is th^ same as an arc of 6'2832 in circular measure. In other words 360 idegs. -corresponds to ztt, and, therefore, by proportion 27r : I '54 : : 360 degs. : x I •1:4 X 360 00 •* = %. o "^ — = 88-235 6-2832 •'^ , ■ ; r : .. ■■>::<] . ■■ ; The angle is 88 degs. 14 min. — Aits" CIRCULAR MEASURE. 33J: 2. Express the following angles in degrees and minutes:— ^ Circular measure 2, ^, tr, 1. D ii4degs. 35^ min., 30 degs., 180 degs., 57 degs. 17! min. — Ahs. 3. A circle of 20 in radius. Find the actual length of the sine and cosine of 15 degrees. Also find the circular- measure of the angle. First for the circular measure, since 360 degs. = 27r = 6*2832 cir. m. TT l> 30 ,, = 6 = "5236 15 » ^ — = '2818 To find the lengths of these lines we must refer to a table of natural sines and cosines in some pocket book or other work. "From Chambers' tables we find Sine 15 degs. = "2588 Cosine 15 degs. = "965. These numbers are the lengths of the sine and cosine when the radius of the circle is unity, hence for a radius of 20 in., Sine 15 degs. = "2588 x 20 in. = 5'i76 in. Cosine 15 = '966 x 20 in. = i9'32 in. 4. If sine A = '5, find the value of the tangent. Referring to figure 122, BS is now -5 when the radius CA or CB = I, it follows that CS = \/i^ - (•5)^ which works out to •866 (the cosine). Now the two triangles, CPB and CAT are similar triangles and we can state a proportion as follows : — CS : BS :: CA : AT or -866 : -5 : : i : AT from this AT (the tangent) works out to "5773. 334 TRIGONOMETRICAL RATIOS. 5. Show by a diagram that the tangent of an angle is to 1 as 1 - co-versine is to 1 - yersine. All the radii of the circle being equal to unity, i - co-versine = CE which = BS or the sine, also i - versine = CS or the ■cosine. The proportion in the question, therefore, becomes AT : CA : : BS : CS "which is obviously true. 6. ProYe that sine^ + cosine^ = 1. This is obvious from the right-angled triangle CBS whose hypothenuse is a radius and, therefore, unity. 7. If the co-versine of an angle is ^, what is the value of the secant? -% = I-347-— ^w*- 8. If the versine of an angle is §, what is the value of the tangent ? ^8 = 2-828.— A?is. 9. Given sine A = -66. Cosine & = '78. Find tangent A and angle A. Tangent A = •8461 ; angle A = 40 degs. 14 min. — Ans. No/e.— Tangent A = p^^^A. Cosme A 4« — Acceleration. To prove that with a constant acceleration (/) the final Yelocity (Y) is related to the initial velocity (u) by the equation Y2 = o2 + 2/s In this equation (s) represents the space or distance passed over by the body while the velocity increases from u to Y. Velocity is the raie at which a body is moving. A body may have a constant or steady velocity of so many feet per second or so many miles per hour, as for example, a ship at full speed. Also a body may have a velocity which is being accelerated every second, as for example, a railway train just starting from a station. Thirdly, a body may be moving in such a way that its velocity is being accelerated by exactly an equal amount every second of time. This is the condition known as "constant acceleration of velocity." It is an experimental fact that a falling stone (no matter what its weight), increases its velocity by 32 feet per second after every second that elapses from the time it is let fall. This is the most familiar instance of constant acceleration. But all bodies acquire a constant acceleration if acted upon by an unbalanced force which does not change in amount. It is almost certain that a railway train and a ship have a constant acceleration for a few seconds after starting from rest. The steady rate of acceleration does not continue for many seconds because the resistances to motioa increase with the speed gained. The excess of the propelling force over the resistance thus diminishes continually, with the result that the rate of acceleration gets less and less until the various resistances to motion just equal the propelling force. When this equality is obtained the velocity of the train or ship remains constant until such time as the equality is again disturbed. 33^ ACCELERATION. Questions on acceleration can g-enerally be worlsed out arith- metically from first principles as follows : — Required the distance moved by a steamer from rest in 12 seconds if the rate of acceleration is '5 ft. per second every second. The acceleration being supposed constant, the velocity after 12 seconds has elapsed is "5 x 12 = 6 feet. The average velocity = -^t — = 3 ft. per sec. .*. Multiplying by the time we find distance moved = 3 x 12 = 36 feet. Putting the same argument into an algebraic form, let/be the acceleration each second, i the time, V the velocity at the expira- tion of the time, and j the space moved over. Final velocity V = _/ x ^. Average velocity = -LJL-E 2 Distance moved ^ = <^- x if = <-— The equation' is true for any rate of acceleration whatever, but ■ if we write g = ,32 for/ we obtain the rule for a falling stone, viz. : — distance fallen in time, i = ^— = 3^^" =15^2 ?■ 2 ■ ■ ].. With respect to the formula at the head of this article, we first note that since the acceleration of velocity is constant, we know that the velocity for the whole time is the arithmetical mean of the initial aijd final velocities^ hence : — , , '■; Average velocity = —+_!? and if (i) is the time occupied ' ' ' (S) space! mOVed = '^ ^ x /...'; t^^ '.'" ACCELERATION. 337 In order next to get rid of (i), we know that the gain of velocity == f y. t. .', Y - v = ft and V - 1) . ,j substituting this value cS, t in equation (i). Space = — X — - — = S, multiplying out .. xl-., -■-■"'-■' ^-'J / Y!_1J:! = S; W - v^ = 2/S, or V2 = w2 + 2/S. ('''Hxaitnple.^ — A isteamer going full speed, 16 knots, reverses her engine's, and thus checks her speed at the rate of i knot every 5. s^lGoads. Required the distance gone before her speed is reduced to 4 knots. J'' 'The retardation is at the rate of i knot in 5 seconds, or la kttots a minute, or 720 knots an hour,' which is (/"). i62 = 42 + (2 X 720 X S) .'. S (in knots) = — ^ = Tr of a mile. ' 2 X 720 Note I. — ^The formula under consideration is true, no matter what units are used. But we must be consistent throughout. In the example worked out, by taking, the velocities in knots per hour, we were compelled to take the retardation of velocity also in knots p^r hour. Note 2. — Strictly speaking, the proper formula for a retardation of velocity is V2 = 1/2 - 2/S, but the original formula \srith the plus sigh will answer all cases if (V) represents the greater speed and (v) the lesser speed. 5* — Kinetic Energy. To prove the formula for accumulated work or kinetic energy, ¥iz., Bfork = ^^ = ^' (nearly). In the first place it is evident that the kinetic energy of a body must be the same whenever its velocity is the same, no matter how i, the velocity has been acquired (whether by firing out of a guo or by falling from a height for instance.) Let a body of weight (W) fall from a height (H) under the action of gravity. The work accumulated in the body due to .gravity is W x H. Also let (i) be the time taken in falling and (V) the velocity acquired. Then V = p- ^ or / = — . g- But the height fallen is equal to the average velocity multiplied by the time, that is 2 2ff The work, therefore, W x H becomes W V2 Work = 2^- Noie I. — If a body has its velocity increased from (v) to V, the work put into it W V^ _ W«2 _ w (V2 - gg) 2^" 2^ J^ ENGINEERING ESSAYS. 339 and the same amount of work will be given out when its velocity falls from (V) to (v). Note 2. — In using- the above formula it must be remembered that the acceleration due to gravity {g) is generally reckoned in feet per second {^g = 32), and if this number is used it is necessary to reckon the velocities also in feet per" second. r:i b.— Cubical Expansion. To prove that the co-efficient of cubical expansion is approxi- mately three times the co-efficient of linear expansion. Let the length of the side of a cube be (a) and its linear expansion (A) so that the length of the side becomes (a + A)^ The volume of the cube is now {a + Kf {a + hf = a^ + 3*% + j,ah? + h^ Now take away the original volume of the cube, viz. : — cfi remainder = 3<2^A + 3^^^ + h^ which is the amount of cubical expansion, and the co-efficient of cubical expansion will be found by dividing by the original volume e^ co-efficient = =5 L^ 1 = 3- + 3 + Now this result is true for any amount of expansion great or small, but as the expansion of solids due to heat is always- comparatively a small amount the terms in the above expression after the 1st, namely 3-5 and -5 may be neglected without appreci- a a, y able error, hence co-efficient of cubical expansion = 3- a The linear co-effieient being - it is proved that the cubical is three a times the linear. Note I. — To show what error is involved in the above approximation we may take an expansion due to a rise of tempera- ture of 100 degs. in iron which is about 0007 of the linear dimensions.- ENGINEERING ESSAYS. 341 Co-efE. cubical expansion = 3 x '0007 + 3 x ("0007)^ + ("0007)* ,, ,, = "002 1 + '00000147 + etc. Here the second term is only a little over one millionth. Note 2. — The calculus enables us to find very readily the limit towards which the ratio of the two co-efficients tends when the «xpansion becomes indefinitely small, as follows: — In the language of the calculus the linear co-efficient of expansion is r differential of (a) ■'< • \ « and the co-efficient of cubical expansion is ; differential of cfi _ 2,0^ x diif. (a) _ 3 x differential of a 1 a\ a' - ' •: « ] .1 which is absolutely three times the linear co-efficient. 1 7. — Centre of Gravity of a Triangle. To prove that the centre of gravity of a triangle is at one- third the height of the triangle from the base. ist proof — geometrical — ABC is a triangle, figure 1 24. Take the middle point of AC, and join Bx. It is evident that the triangle would balance upon the line Bx, for if we suppose the triangle built up of strips of material, all parallel to AC (as shown), then since each strip has its centre on Bx, each will balance separately, and, therefore, the whole will balance. Now, in figure 125, take the middle point of AB and join CZ. The triangle will also balance upon CZ for the same reason as before. The centre of gravity being, therefore, in Bx, and also in CZ, must be at their intersection, viz., at O. We have to prove that OZ = JCZ. JoinOA. The proof depends upon the fact that if two triangles have the same length of base and the same height, they have equal areas. Fig. 125. ENGINEERING ESSAYS. 343 For this reason, triangle ABx = triangle CBat also \Ox = COx, subtracting the latter from the former triangle ABO = CBO but ABO = twice ZBO .-. CBO = twice ZBO ; Now thelse two, triangles have each the same vertex B, therefore, the base CO must be twice thef base ZO, and OZ = J CZ. In the same way it may be proved that xO = J Bx. 8. — Centre of Gravity of a Semi-Circle. To prove that the centre of gravity of a semi-circsle is at ■424 of its radius , from the base. ' ' It is a principle in mensuration that "if a plane figureis reVolved about any side as an axis, the solid of revolution thus described has a volume, which can be calculated by multiplying the area of the plane figure by the distance, which its centre of gravity describes in one revolution." Now, if the semi-circle ACB, figure 126, is made to revolve about its diameter AB the figure described will be a sphere. Hence, by the above principle, the volume of the sphere will be found by multiply- ing the area of the semi-circle by the circumference of the circle described by the centre of gravity G, Fig. 126. Let the radius of the semi-circle = R, and the distance of its centre of gravity from the base = x. Area of semi-circle = R^ tt -^ 2. Circum. of circle described by centre of gravity = 2 xv. 1D2 ^ volume of sphere = x 2 X'lr = R^ n^^ x. 2 But the volume of a sphere is known to be f of the cylinder of same dimensions. ENGINEERING ESSAYS. 345 .'. Volume of sphere = R''7rx2Rxf = 5 ^. 3 Equating these two expressions for volume. R^ T^ X = 4 , cancelling, we get, ' ' 4R . ' 4 R ■T X = 3 — and X = 2 — » « 3 3'^ ),■" !■>■'--■■ -^ X = — 2i—- X R =. -434 R;;! ,...':[ .^.■. !'■ l'.'^''.:' ■ , ■ 9'4.248 ,. .. A'ofe. — ^The same argun;ient applies ;to the centre of, gravity of a quadrant of a cfi-cle, which, is at •424: the, radius from either side. {Shown by dotted circle in figure 126.) ■'. ,■ . r,,,.;,;^ 9* — Centre of Gravity of a Pyramid. To prove that the centre of gravity of a pyramid or cone is at one-fourth the height of the body above the base. ABC, figure 127,' is a section through a square pyramid, side of base = b, h = height. b y. X mn — — = — h and area of layer at mn = [mnY Solidity I ^ / Z» X x Y^ ^ dx of layer J fb X x\^ Moment about A = ( — - — j d x x x A2 Integrating this d X Total moment / 62 X? , 62 X^ d X = C-'t- 4^2 Fig. 127. when X becomes = h, moment = A* 62 A2 4^2 .(I). 62 h Now solidity of pyramid = and if R = distance from A 3 to the centre of gravity Moment = x R, equating this to (i) 62 A R 62^2 and R 62^2 X -J- 3 4 4 62A . R = f A The centre of gravity is, therefore, \ the height above the base. lo.— Moment of Inertia of a Rectangle. To prove that the moment of inertia (I) of a rectangle about an axis coinciding with one edge is = ^. 1st proof — approximate — ABCD is the rectangle, figure 128, XY the axis which coin- cides with the side AD. AD = the breadth {b) and AB = the length (/), no matter which happens to be the longer of the two. / 2 1 3 1 4- s Bl^ rC cr Fig. 128. Definition. — "The geometrical moment of inertia of any plane figure is the sum. of all the products, obtained by multiplying every element of the area by the squaie of its distance from the axis." Working upon this definition, we proceed to divide the rect- angle into a number of (say 5) equal parts, by lines parallel to the axis. The greater the number of parts taken, the more nearly will the result come out correct. But five parts will give a' fairly close approximation. Taking No. i, its area is 6 x - and its centre of gravity is — distant from the axis, therefore, the product of its area by the 5 \io/ 500 S Vio/ SCO square of its distance is - « In the same way No. 2 gives ^48 ENGINEERING ESSAYS. No. 3 gives - - - 5 Vu>/ 500 No. 4 gives - - - 5 Vio/ 500 No. 5 g-ives - - - bl ^ hiy^ _8r6/* 5 \io/ 500 Summing up we get 1656/3 500 A/3 ■which reduces to which is only one per, dent.: too great in 3 '03 the denominator. 2nd proof-^by the calculus ; — figure 129, in this proof we take an indefinitely narrow strip ■or element of the area at {x) •distance from the axis, its width being (dx), and we ■express its moment of inertia according to the meaning already g'iven. Fig. 129. Its area being b x dx and its distance x, the product cjf area ■by square of distance is b X dx X x^ == bx^ X dx. We call this "the differential of the total moment of inertia of the rectangle," because it is the exact addition to total moment, which results from increasing the length x by an indefinitely small amount dx. Now using the well-known rule for integration / bx^ X dx = — . 3 which gives the expression for the morn^e^nt of inertia of a rectangle of length x; but as x may be any length, let x become = /, and the moment of inertia = — for the complete rectangle. ENGINEERING ESSAYS. 349- Note J. — If the axis is taken throug-h the middle of the rect- angle instead of coinciding with one edge, the rule becomes — . Figure 130." To prove this we have only to write - for I in the i st rule and we 2 get the moment of inertia of each half of the whole rectangle, viz. , 3 ^3 ^ IS and for the whole rectangle x 2 = . 24 12 1 \ t Fig. 130. 11. — ^Radius of Gyration. To prove that the radius of gyration of a plane rectangle revolving about one edge as an axis is = -—. — -571 U V 3 Definition. — "The radius of gyration of a revolving body is the radius at which, if all the mass were concentrated, the total kinetic energy would be the same at the same speed of i-evolutian." The above definition is the most useful one that could be g-iven, because it indicates why the radius of gyration is wanted for engineering problems. It is not, however, necessary to bring in the idea of energy in finding the radius of gyration of a body. The moment of inertia is sufficient. If a radius is found at which the moment of inertia remains the same when all the mass is con- centrated there, the energy will be the same too, for energy depends on the mass and on the square of the velocity, and velocity depends on radius when the speed of revolution does not alter, so the moment of inertia which depends on the mass and the square of the radius is always proportional to the kinetic energy. Referring to No. lo for the moment of inertia of a rectangle bl^ we write I = — 3 Then if R is the radius of gyration and we suppose all the area to be concentrated at that radius. I also = 6/ X R2 :. bi X R^ =1L 3 R2 = *Z! ^ 6/ = {! 3 3 ENGINEERING ESSAYS. 351 Note. — Since a rectangle may be supposed to be built up of a great number of lines each = /, the radius of gyration of a line of leng'th (/) revolving- about one extremity is also •577 / and approxi- mately the radius of gyration of a uniform rod revolving about one end is "577 cf the length. 12. — Moment of Inertia of a Cirele. To prove that the moment of inertia of a circle about its centre is exactly twice the moment of inertia about its diameter. We must first demonstrate a general property of the moment of inertia of a particle or element of an area, viz. : "If the moment of inertia of a particle is taken about two- separate axes at right angles to each other, and in the same plane, then the sum of these two moments will be equal to the moment taken about an axis at right angles to the plane, and passing; through the intersection of the first two axes." In figure 131, OX and OY are two axes in the plane of the paper and P is a particle or element of area of magnitude The I of P about OX is ^ x a^, where (a) is the perpendicular distance of the particle from the axis OX. Also the I of P about OY is ^ x b^, where (6) is the distance from the axis OY. Fig. 131. Their sum \s {p -x. a^) -\- {p y. l)^) =^ p y. {a^ + W) (i). Now imagine a third axis passing through O, but at right angles to the paper. The distance of P from this third axis is = C in the figure, hence the I = /> X c^. But c* = a2 + ^2 .-. I =/ X {a2+*2) (2). The identity of (i) and (2) proves the proposition. MOMENT OF INERTIA OF A CIRCLE. 353 In the case of a plane figure, such as a circle, figure 132, what is true for any single particle of area, is true for every other par- ticle, hence the total moment of the whole figure about the axis XY, added to the total moment about the axis WZ, will be equal to the total moment about an axis through the intersection O, which in this case is the centre of the circle. Now, since XY and WZ are both diameters of the circle, the moment about XY must be equal to the moment about WZ ; and their sum is equal to twice either one singly, but their sum being equal to the moment about the central axis, the latter is equal to twice the moment about a diameter. Note. — An important application of the above principle occurs in- the relative moment of resistance of a bar to bending and, twisting strains. It is proved in No. 15 that the moment of resistance can always be obtained from the moment of inertia of a section by multiplying by (which is a constant for every part of the section). Now, the axis for twisting strains is the centre litie of the bar, whilst the axis for bending strains is a diarmter of .the; bar, so it follows that the moment of resistance for twisting straitiiS is just twice the moment for bending strains. In other words, a bar; i§;twice as strong to resist twisting as it is to resist bending. 13> — Moment of Resistance of Rectangular Beams. To prove the rule for the moment of resistance of a rect- angular beam, viz.: — BD^S M.R. Where 6 B = breadth. D = depth. S = stress at outer surface. I St proof — in the annexed dia- gram, figure 133, is represented in a crude manner the effecting of bending a rectangular beam. The vertical cross section at (a b) becomes inclined and takes the posi- tion (oi bi) (the movement being, of course, very much exaggerated in the diagram). Hence aa-^ represents the extension or stretch of the top surface of the beam, and bbj^ the compression of the bottom surface, due to the behding action. Also, since there is no external longitudinal force acting upon the beam when loaded, the sum of the tensile forces from the upper surface to the neutral axis must exactly balance the sum of the compression forces below the neutral axis. F'g. 133. RECTANGULAR BEAMS. 3SS But by Hooke's law "stress is proportional to strain," there- fore the sum of the tensile strains will also equal the sum of the compressive strains. In the diagram this amounts to a proof that the ^ aoai is equal to the A bob^^, hence a«j = bbj^, and the neutral axis is in the centre of the depth of the beam. Now to make an approximate estimate of the value of these tensile and compressive forces, which resist the tendency of the load ^W) to bend the beam, let us suppose the beam marked off, as in .diag-ram, into lo equal layers — 5 above the neutral plane and 5 below — -and further suppose the intensity of stress to remain constant for the depth of each layer, its value being that due to the centre position in the layer BD Sectional area of each layer = ID Stress at centre of top layer = -^ S. 10 n 11 .1 BD qS Pull upon top layer = x < — . 10 10 -Similarly — r. ,1 J 1 BD 7 S Pull upon and layer = x t — . ■^ ' 10 10 Pull upon 3rd layer = x 5 — T, 1, .LI BD 3 S Pull upon 4th layer = — x ^ — BD S Pull upon sth layer = x — . This brings us to the neutral axis below which the stresses are repeated but with a change of direction. Dealing with the upper half of the beam first, we have now to convert the above forces into moments acting about the neutral plane. 3^6 RECJANGULAR BEAMS. ment of top layer j> and' it »> 3rd )» «> 4th M BD io X 10 9.D 20 81 BD^S 2000 BD 10 X 7S, 10 Xd 20 _ 49 BD2 S 2000 BD 10 X SSx 10 20 _ 25 BD2 S 2000 BD 10 X 10 ^D 20 _ 9 BD2 S 2000 BD 10 X S x 10 ±D = 20 BD^S 2000 Sth „ Adding these moments together gives us the moment of resistance of the upper half of the beam, its amount is 165 BD2 S 2000 and as the lower half has an equal effect in resisting bending. . -r ^ , <- 330 BD2 S BD2 S . . Total moment = ^^ = -: — -rzr- 2000 0"0DD The above approximation is not very close, but with a greater number of layers the denominator would approach nearer to the proper value, viz. : — 6. V////A ^m ■w Fig- 134. 2nd proof — using the calculus. Having established the- principle that the neutral axis is in the centre of the depth, and that the stress increases as the distance from the neutral axis, both- in tension and compression, we proceed as follows: — See figure 134 Take an indefinitely thin layer of the beam at (x) distance: from the neutral axis, the thickness being flf;v. RECTANGULAR BEAMS. 357 Sectional area of layer is B x dx, and the stress at this position X 2 ^ X 2 S X being _- X S = — =- — the pull on this layer = B x dx x — =j— The moment about the central axis T, j' aSji; 2 "R S x^ , = B X dx X y. X = — X dx I This has to be integrated to obtain the expression: for the sam of all such moments. ■ i' v , " / 2BS^2 2BS^« dx = D 3D Now, when x becomes equal to - " , 2BSD3 BD^S Moment = — =p = ^ 8 X 3 D 12 and the moment of resistance for the whole beam BD S 14. — Moment of Resistance of a Shaft.^ >7o prove the rule for the moment of resistance of a solid shaft of circular section, viz., .R. = or more accurately M.R. = where D is diameter of shaft, and S is stress at surface of shaft. 1st proof — approximate — figure 135. Suppose the section of the shaft to be divided into five concentric rings as in diagram. Then if R is radius of shaft, the sectional area of outer ring (a) may be calculated from its mean radius, which is j^ of outside radius. 51 D^ S X TT 16 Fig- 135- Area of outer ring = ARx2irx-x llj_^^ 5 50 Area of 2nd ring = ^ R x 2 tt x - = '^ ^ ^\ 5 50 Area of 3rd ring = T^Rx27rx?= I£J!LA\ S 50 Area of 4th ring = ^»^Rx2jrx^ = ^ '^ ^\ 5 50 Area of 5th ring =j^Rx27rx?- = ^ "" ^.^ 5 50 STRENGTH OF SHAFTS. '359 The next step is to multiply the area of each ring by the mean stress, and then by the radius or leverage with which it acts ii) resisting torsion. For the ist ring the mean stress is j^ S and the leverage -^ R. Moment of ist ring = l^^H^ x A S x A R = ^58'^^^' SO 5000 50 '■" 5000 „ 3rd,, =^.gJLR!xASx AR = ^5o^SR3 50 " 5000 „ 4th „ =l^^xASxAR = 54zAR! 50 5000 „ sth „ =i^xASxAR = ^-^^^' 50 5000 \. f The sum of all these moments is the moment of resistance of the shaft to torsion, the result is — H, r, 2450 a- S R3 TT S R3 , M. R. = —32 = very nearly. 5000 2 By writing — for R, we get M.R. = — = — = nearly, 2 X 8 16 5-1 ^ ' NoU. — For a hollow shaft of diam. (D) outside and diam. (dy inside, we have to subtract the strength of the portion removed from the strength of the outside diameter. Now it might appear at first as if the strength of the portion removed (of diam. = d) would be S d^ represented by the formula — ; — and hence the strength of the hollow shaft = SD! - S^^ = SPi,^^), 5'i 5-1 5"i But upon looking closer, we notice that the same symbol (S) cannot represent the stress in both formulas correctly, for the stress^ in the formula is the stress af the surface of the shaft considered, and the value of this stress must be less for the portion renioved 360 STRENGTH OF SHAFTS. than for the outer diameter. In order to make this correction we must write — S for the stress in the formula involving- the inn^V d - S d' diameter. Then strength of portion removed = D The strength , of the hollow shaft is'now , , ^,f STD^ - Sd* SB^ _ ^S d^ S"i 5"i D X 5-1 ^ S (D^ - d*) S-i X D ' which is the correct rule for a hollow shaft. 2nd proof — by the calculus: — figure 136. Consider an indefinitely narrow ring, as in diagram, whose radius is x. The breadth of this ring- is necessarily = dx and Its area = zxtt, x dx, multiplying by the stress at this position which is — of stress at surface, and again by the leverage of the ring- to resist torsion, we get !Fig. 136. (': moment = (2^:77 x dx) x - S x x = ^'^Sy^^^ ^ ' R R' We have now to integrate this expression. f 2'7r S X^ , 2 IT S X^ X ax = R 4R which when .«; = R becomes — — — , — and writing- — for R ' 4 2 moment of resistance = 2 7rSD» IT D» S 4x8 16 as before. ( > Noie.^By integrating between the limits x = r and « = R an expression is obtained for the strength of a hollow shaft. / X = R 2«-Sa;8 ^^ ^ 2 7rSR* _ 2-irSi^ _ jtS (R* - r^ x=r R 4R 4R 2 R 15* — Moment of Resistance of any Beam. To prove that the moment of resistance of a beam of any section, whatever, can be found from the moment of inertia of the siaction about its centre of gravity, by 'multiplying by the stress at any point and dividing by . , the distance of that ppint from the centre of gravity ^r; neutral plane. M = moment of resistance. In symbols M = I x -i! ■! I = moment of inertia. ' S = stress at (y) dist. from axis. In the first place, since the intensity of stress in a beam varies directly as the distance from the neutral axis, it is clear that the fraction — is constant for every point in the section of a beam, no matter at what distance from the neutral axis the value of (s) may be taken. We will assume then that s is the stress at the outer surface and consequently y = distance of neutral axis from surface. Now, if we look into the meaning of the "geometrical moment of inertia" we shall find that it is the sum of all products obtained by multiplying every element of the area by the square of its distance from the axis. Thus, if (a) is such an element of the area and (r) its distance from the axis, we want the sum of all such terms as a X r^ (i). Again, if we look into the meaning of "moment of resistance" we shall find that it is the sum of all products obtained by multi- plying every element of the sectional area by, ist, the stress at that point and, 2nd, by the distance from the axis. Thus (a) being an element of the area as before, we want the sum of all such terms 362 BEAMS. as a X stress x r, but the stress at this point = - x S, where y S is the stress at the surface, and j/ the distance of the surface from the axis. Moment of resistance, therefore, is a X - S X ;- = ar^ - (2). y y s Comparing (i) and (2) we see that the fraction — is a constant multiplier for every element of the area of the section if we want to convert moment of inertia into moment of resistance, and what is s true of every part must be true of the whole, hence MR = I x -. y i6. — Bending Moments in Beam Problems.. To prove the bending moments (M) used in beam problems, (page 226). 1. Cantilever, load at end, M = WL. This requires no proof. 2. Cantilever, load distributed, M = — because the eg-, of 2 the load is at a distance only — from the support. 2 3. Beam with supports at the ends and load in centre. In> W this case the reaction at each support = — and, therefore, the- 2 u ^- \ ..u . W L WL bending' moment at the centre = — ^ x — = . i 224 i. Beam with supports at the end and load distributed.- The reaction at each support = — , but as w^ll as the moment — ■ ■ ■ 2 2 X — there is a moment in the opposite direction due to the- 2 1 distributed load on one half the beam = — x —. The difference 2 4 WL WL WL u- u • ^u u A- . ■ ^u <= — -—— = — -— which IS the bending moment in the centre- 488 of the span. S. Beam fixed at the ends and loaded in the centre. Figure- 137 shows' the form of deflection (much exaggerated) which the lieam assumes in case 5. What we have to notice is that the curvature of the beam at the supports A and B is exactly the same (but reversed) as the curvature at the centre C. Both curvatures- -364 BENDING MOMENTS. •are due to similar causes. This fact is better seen if we extend the beam beyond the supports as shown by dotted line to Cj, so as to repeat another half-span. By turning the diagram upside down the reaction at A becomes a load, and C and Cj become supports, under identical conditions as the part AB. It follows from this that at points X and Y, midway between the centre and the two ends respectively, the beam will be bent neither in one direction •nor in the other. In other words there is no bending moment at .X and Y. The beam is, therefore, merely supported 2X X and Y -{by pure shearing forces). These forces or their reactions are each W . , W L WL = — and the bendmg moment at the centre = — • x - = — — 2 ^ 248' which is just one-half as much as in case 3. The positions X and Y are also known as places of "contrary flexure" or merely as ■"virtual joints." C' Fig. 137- 6. Beam fixed at the ends and load distributed. This case •is the most difficult to investigate. Of course, as in case 5, there -must be a point somewhere between the centre and the end of the Jbeam where there is only a shearing force ; but the point is not •half-way unless the "fixing" at the ends should happen to "give" -a little. It would make the beam all the stronger if the fixing did give a little, so as to equalise the strain at the ends and centre / might be the bending moment in that case). However, by ■means of the calculus the problem can be solved with really fixed •ends. The positions of X and Y are known to be "577 (or accurately J^ of the half span from the centre, each way ; that is •nearly six-tenths of the distance from the centre towards each end. BENDING MOMENTS. 365: For the proof of this fact readers are referred to Cotterill's or- Perry's work on Applied Mechanics. We proceed to find the bending- moment at the supports A or- B when the position of the point of contrary flexure is ^u °^ ^^^' half-span fromithe centre. Or if / = half-span, CX = '6 /, and AX = '4 /. The portion of the beam AX is now a simple- cantilever with two loads, ist an end load at X equal to the load upon CX, transferred to the cantilever by pure shearing force at X; and 2nd, by a distributed load upon its own length AX. Let the- distributed load upon the whole beam be w lbs. per foot of length. Then the end load at X is ?« x CX = w x "6 / and its moment about A is OT x "6 / x "47 = "24 w P. Also the distributed load upon AX = w X % I and its moment about A is w x "4 / x '2 /' = 'oS w l^. Adding the moments together total moment = '32 "W l^. If we now write — for / so as to express the bending moment- 2 in terms of the full-span, we ^t.i , J. ^ "\Z TO L^ .... TO L^ bendmg moment = -^ ' which is Also, since w L = total load = W. Bending moment = • (approx.) 12-5 Note I. — It is a good mathematical, exercise to work out the- above moment using the exact value ^ instead of *6. The result - <.i WL IS exactly 12 Note 2. — ^To find the bending moment in the centre of the- span, we notice that the portions of the beam CX and CY are- cantilevers turned upside down, so that the end reactions at X andi. Y act upwards and against the distributed loads on CX and CY. Total moment at C due to loads on cantilever CX is (w x '6 / X '6 /) - (to X -6 / X "3 ^ /i) - = -iS TO/2 = ,2 -18 WL WL (The exapt denominator is ,24)^ 17- — Radius of Curvature. To prove the equation connecting the stress in a bent elastic bar or beam with tne radius of curvature, viz.: — stress _ ^ £ R where E is the modulus of elasticity, R is radius of curvature, y is distance from neutral axis at which the stress is taken. We must assume at the outset that the bar takes the form ■shown in sketch in all respects. The curve must be circular and ■ every cross-sectional plane must after bending be still at right -angles with the axis as Illustrated by the flat ends AC and BD. 'The dotted line represents the neutral axis or plane (figure 138). Consider the under surface of the bar — C V^ D. We see that -after bending it is longer than the neutral axis plane. If the bar ■ made a complete circle (which it would do if long enough) the length of the neutral axis would be = R x 2 tt, and the length df ENGINEERING ESSAYS. 367 the under surface would be (R + _j/) x 2 tt = (R x 2 ir) + (y x 2 ir), hence the amount of stretch is J/ x 2 tt. We can thus arrive at the amount oi stress in the material at the under surface by proportion. Remembering that the modulus of elasticity E is the stress which would stretch the material an amount equal to its original length Rx27r : y y. 2 TT :: E : stress . D i TT stress y . . R X stress = _j/ x E, or — — — = ^ E R The same argument applies to the upper surface in compression. It will be noticed that the form of the section of the bar does not enter into the investigation. It may be round, square, or triangular, or any form of flanged girder. An example may be given. Suppose a flanged girder is loaded so as to bend to a radius of 1000 ft. Required the stress at 10 in. from the neutral axis, the modulus being 30,000,000. i stress _ 10 30,000,000 1000 X 12 •^00,000,000 ,, stress = - — — - = 25000 lbs. 1 2000 i8.— Bending Moment and Curvature. To prove the equation connecting the bending moment of a beam with the radius of curvature, viz. : — M ^ E I R where M is the bending moment, I is the moment of inertia of the section, E is the modulus of elasticity R is the radius of curvature. The mechanical couple tending to bend a beam must be resisted by the strength and distribution of the material in the beam itself in such a manner that bending moment = moment of resistance. Now, in a previous investigation (No. 15) it has been proved that moment of resistance = I x y hence M (the bending moment) = I x " y where y is the distance from the neutral axis to the part of the beam where the stress is acting. But from the previous investigation, No. 17, we know that stress _ E y ~ R substituting, therefore, we have AT T E ME M = Ix— or _= — R I R As an example of the use of this equation let it be required to find' the bending moment which will deflect a rectangular bar of steel I in. deep and 5 ins. broad to a curvature whose radius is 50 ft. I (the moment of inertia of the section) = 5 12 BENDING MOMENT AND CURVATURE. 369 E is 30,cx30,ooo and R is 600 ins. I X E the equation may be written M R . -^ ^ 5. ^ 30,«>o,ooo = 20,800 nly. 1 2 600 The result is pound inches and would represent a bending' action in which 2080 lbs. acted at a leverage of 10 ins. . It must not be supposed that the results or equations of the last two articles are restricted to cases of beams which happen to be deflected to a circular arc for a considerable portion of their length. Such cases are rare in practice. But in all beams, no .matter what may happen to be the nature of the curve assumed under their load, any small portion of the curve will coincide sufficiently near with a circular arc of some radius to make the «quations practically true for that portion of the beam. The stress and bending moment given by the formulse will, therefore, be true for that portion of the beam and that portion only. And as the radius of curvature changes so will the corresponding stresses and bending moments change. 19* — ^Deflediion of Beiams. To prove that the defection at the &ee end of a cantilevep of uniform section is where W is load at end. 3I£' / is length, I is moment of inertia of the siBctioh, E is Tonng's modulus of elasticity. ' lUsing' the calculus — figure' 1 39 — ^Take a position along the canti- lever at {x) distance from the fixed end and let (r) be the radius of curvature at that position. ^1 If a tangent is drawn at this position to the neutral surface of the beam and continued to the point (d), the distance {ab) will measure the deflection of the beam below the horizontal, due to the bending of the first portion of the beam x in length ; the remainder of the beam being supposed unbent. Also if another point be taken DEFLECTION OF BEAMS. 371;" further, along the , beam, and ^ tangent drawn as before ~ and continued to (c), the distance {be) will measure the additJonaJi;7 deflection due to the bendingin that extra portion of |the beam. Now let the two positions be taken so near together that the- distance between them is only ■(^), the distance {be)' becomes thei* the differential of the deflection. 1 . Keeping in mind how much the figure is exaggerated, we see,, that Afl, Bb, Ce, Bd, are all equal in length — practically; for they only differ by an ipfinitely small quantity, namely, dx. The lerngth of these lines = I — x. Also, owing to the same cause, the two- , radii, r and f^, are practically equal. We proceed to find a relation between dx and be. Since each tangent is perpendicular to its own radius, it follows- that the angle between the two tangents is equal to the angle- between the two radii and the distance apart of the tangents at their extremities which = be, can be obtained from the similar tri- angles OBC and B&. r : dx : : {a - x) : be dx {I - x) ,. > •'^= be (')■ "' Again it has been proved, in the last article, that — = — wh^ M is the beiidmg' moment. juh I. .i, ,.., ■ i; i ,. ' '. • ,' :. 't . ' ' -v Our bending moment at B or C, = W x (I - x), hence , , , W (/ - X) ^ E _ _ ^^^_ - Take the value of r from equation (i), viz. : — i ; ;: ' (I ^ x) dx r = 5^ -1 be and substituting. in (a) gives us W (/ - x) _ R X &c .7 -.(I I (I - x) dx ■ he - w ki - xY dx 372 DEFLECTION OF BEAMS. ' Now be is the differential of the deflection of the beam, and we proceed to integrate ; J , = =-Y ( ^^^ ~ 1" ~ ) ^'^'^ when x = I which Was to be proved. Note I. — For a cantilever uniformly loaded with {w) per unit of length,' the bending moment at B or C is A/r fi \ I - X IB (I - xY M = w (I - x) X = — i '— 2 2 If this, is used and the proof followed out on the previous lines, the •deflection will come out deflection = 8E I 8EI Noie 2. — A beam, supported at each end and loaded in the centre, can be treated as two cantilevers upside down, when the ist rule can be applied ■ W/8 deflection = 48 E I" Note 3. — A beam, supported at each end and loaded uniformly, is a combination of the 1st and 2nd, thus: — Upward deflection due to — at each end 3EI Downward deflection due to uniform load on half beam = 2 \2/ = —5-^ 8E I 128 E I 3 The difference of (i) and (2) is the true deflection = 5 .. 384 E I 20.— Centrifugal Force. To prpve the rale for centrifagal force, namely, F = W X Y» g y^ r ' ' 'A body at A ia supposed to be revolving in a circular patli, round ■thp centre G, with a' linear velocity of V feet per second, figure 140. Now if it were not for some controlling force directed towards the centre, the body, instead of following the cir- cular path AD, would follow the straight path AB. Let the body arrive at B in, the time (/), when uninflu- enced by the centripetal force, so that distance AB = V/. We have to find what centri- petal force will be necessary to pull the body from B to D in the same time {i). . If, then, this fo,rpe,is acting from the instant the body leaves A, it will result that the body will not leave the circular path at all. It wili be understood, of course, that the distance AB is supposed to be infinitely small and is equal in length to the arc AD. Fig. 140. We must first get an expression for the distance BD. radiu^.of the circle beiag. (?). ;, >, The 374 CENTRIFUGAL FORCE. In the triangle ABC, CB^ = CA" + AB^ {r + BD)2 = r^ + (AB)^ r^ + 2r X BD + (BD)2 = r^ + (AB)2 .-. 2?" X BD + BD2 = AB2 Now the term BD^ is sosmall compared with 2r x BD that it may be neglected. then 2r x (BD) = (AB)2 ■ ^ndBD = (J^^m... (I). ^^^ 2r zr {"-•■ Ne'xt to find the force required'to draw the body firobi' S' to D in the time (/). The distance moved by a body in any time fitii^ rest is : — f t^ distance = •'■ — (see article 4) ~ • > - , 2 , where /"is the acceleration per second. Substituting ,this value, of distance for BD in equation (i) ". _ ^ ' 2 2r .^. f ^- _ . ,,;,.•, ,^,. ,,;j which cancels outto f= — (2). , r. This gives the acceleration towards the centre. " Np^y if* a force F acts upon a body of weight W, the acceleration is always (g) multiplied by the fraction that F is of W; tliat is, ' , F ' ■' '.' ' '"' ' "\' /=^x..- So substituting' again, we get fi'om equation (2) L£ = 2_ • F = 21-1., fi) JVoie. — Another rule for centrifugal force is W;-N2 2935 where N is revolutions ^Jer minute. CENTRIFUGAL FORCE. '375 This rule Is obtained from the former, as follows : — ^ The diameter of the circle being 2r, the circumference is 2r v, and the velocity per minute is zr ir N, also velocity per second = V _ 2r JT N 6o~" , Substituting this in {3). F = ^ x (^I^Y. : ; ct ^ X 3600 2935 31. — Fly Wheels. To prove the rnle for the force required to hold the two halves of a fly-wheel together due to centrifugal force, viz.: — Total force = Also stress = A Y^ 5-226 Y2 for cast iron 10-452 g = acceleration due to gravity = 32-16 ft., A = area (in sq. ins.) of section of rim, R = mean radius (in ft.) of wheel, ¥ = linear velocity (in ft. per second) of centre of gravity of rim section. i ty \ 7 i Fig. 141. Consider a portion of the rim only 1 in. circumferentially, illustrated in sketch from atob (figure 141). The cubic inches in this portion = A x i in. = A and its weisrht, if the material is cast iron = — 3*9 FLY WHEELS. 377 The centrifugal force of this mass = — x — — - = 3'9 ^-R s'g^-R Now we have here a similar thing to the steam pressure in a cylindrical boiler, for every inch of the circumference is urged out- wards radially with an equal force, viz., =-, just the same as if 3'9^R it had a steam pressure upon it. The resolved force acting at right angles to any plane, such as XY, through the centre of the wheel is, therefore, found by multiplying by the number of inches in the diameter, which is 12 R x 2 (see No. 22). Total force to burst wheel = = x -^ — = _ 24 A V^ 3 9;?' A V^ This works out to and proves that the two halves of the 5-226 " ; ,i wheel tend to separate with a force which depends solely upon the sectional area of the rim and its linear velocity. The diameter of the wheel only influencing the bursting tendency by increasing the linear velocity (when the number of revolutions is constant). Secondly, for the stress in the material of the rim. ., To obtain this we have only to divide the total force by the sectional area at two opposite sides, viz., 2 A. A V^ ■ V* Stress per sq. mch = -r t- 2 A. Stress = •226 io'452 Note I. — ^The above results are only true for cast-iron rims, and take no account of the effect of the arms or spokes, which may increase or even decrease the strength of the wheel. Note 2. — The constant, io-452, is easily changed to suit different materials. For instance, hard wood being about 10 times lighter than cast iron, will have a constant in the formula 10 times. the above or 104*5, thus proving the greater safety of wooden fly wheels. Timber is much stronger in proportion to its weJglit than cast iron. 22.—T strength of a Cylindrical Pipe or Boiler. To prove the i-ole for the strength of a cylindrical pipe or boiler, Yiz.: — p _ 2TS where P is pressure, T is thickness, D is diameter* S is stress per square inch. Take a hoop of the pipe or boiler just i in. wide, then as the •effect of the ends of the boiler ig not considered in this rule, the strength of any such hoop to resist the,, pressure upon its inner -surface will be the same as the strength of thfe pipfe or boiler. ' ' ' Fig. 142. Now, in figure 142, the pressure aicting radially, or at right angles to the surface, upon any small length pf the circumfelrencei ^isabisp X ^d,wher^(^) is the pressure per sq. inch.,, and if/ thb pressure is resolved vertically, it becomes 1 /(il p y. ab X sine BaG (i.). STRENGTH OF CYLINDRICAL PIPE. 379 Also in the small triangle abc, which is similar to BCa or BG«, and consequently has angle abc — angle BaG. ab X sine abc = or substituting this in equation (i). - ' -Vertical force =/ x ac ■= p x xy. • ' --' This proof merely amounts to showing that the pressure upon an. oblique surface, when resolved in a direction at - riglit' angled to a plane, is the same as the pt^essure calculated for the projected area upon that plane. The pre^ssure, upon the surface ab when resolved 'Vertically, that is at fi^ht angles to fee plane MN, becomes the same as the pressure upon xy. ' ^ "A vpry tonvincing proof of the truth of this principle is pbtainied 'without matherhatics, by considering what would happen if a, box were made of the form abxy and subjfected to internal fluid pressure. It would be absurd to suppose that the box would have any tendency to move of itself in any direction whatever, hence the pressures on opposite ends must balance each other, as demonstrate,d above. Now s^nce the same thing is true for every portion of the, l:^oop MVN, it is clear that the total vertical effect of the internal pressure upon this half of the hoop is equal to the pressure upon the diameter MN. Vertical force = p y. diameter. We have now only to equate this force to the resisting force of the material of the hoepj Which cornsists in the tensile strength of the material at M and N. Since the hoop is i in. wide^area tofrJicture at M and N = 2 T x i and resistan;ce = 2.T x Stress. Our equation becomes — [• . -, r, '. • p-x. di^m. ;=, 2T X Stress. ' from which ^ = iX§, ;:i 23* — I^ongitudinal Strength of a Boiler. ITo, prove that the longitndinal strength of a boiler shell i&: where p = pressure, D = diameter, T = thickness of plate, S = stfess. To avoid any question concerning the buckling of the end plates it is as well to consider the boiler to have hemispherical ends as in diagram, figure 143. Then, no matter what the form of the ends may be the resolved longitudinal force due to the pressure upon each 6nd is:-T 0* X 7854 X p '= force longitudinally. Fig- H3- This force is resisted by the tensile strength of a transverse ;5ectior^ of the boiler , which = D X 3"i4i6 x T x S Equating these •7854 D» p = 3-1416 D T S and ^-4DTS ^ 4TS ^ D» D iVbfe I. — By supposing the two' hemispherical ends joined together without a cylindrical part between; it is seen that the STRENGTH OF BOILER SHELL. 381 strength of a sperical vessel of thin material is the same as just found, namely : — Note 2. — The formula for the transverse strength of a cylinder 2 T S being p = — =- — (see No. 22) we have now proved that a cylindrical boiler is just twice as strong longitudinally as it is transversely. 24* — Compressive Stress on Boiler Tube r^iihir^, Plates. I To pFoira the foFmuIa for compression stress on tube plates, viz.: — Working pressure = ^° " ^ '^/p ^^°°° Except in cases where the combustion chamber top is stayed from the crown of the boiler shell, or from the back plate of the boiler, it is clear that the total steam pressure upon it must be carried by the back tube plate and c. c. back, and be divided equally between these two. The side plates of the c.c. would also take their share of the load were it not for the c.c. girder stays which transfer the load to the tube plate and back of chamber alone. This is true, more particularly of the central portion of the c.c. tops. In figure 1 44 (6) consider a portion of the top of a combustion chamber equal to the pitch of the tubes, one way (D) and the width of the chamber (W) the other way. The area of this portion = W X D and if p is the working pressure, the load carried is W X D X p (i.) Now, one-half this load is carried by the tube plate and the other half by the c.c. back plate. Fig. 144. The tube plate although gene- rally thicker than the c.c. plate is more severely crushed by its load because the metal left between the tube holes is so small TUBE PLATES. 383. From figure 144 (a) it is seen that the amount of metal to support the load we are considering- is {D — d) (where d is the inside- diameter of the tubes). Multiply by the thickness of tube plate T and we have (D — ^) x T = sectional area of metal. If the- stress allowed is denoted by S we can equate the crushing resist- ance of the plate to the load it has to carry (equation i). (D -^ X T X S = W ^ P ^-^ From this equation * = ^ -_- <=r W X D 0, . b' . . -' ■:. Since the B.T. rule puts 28000 in place of 2 S it proves that the crushing stress allowed on steel tube plates is 14000 lbs. perr sq. in. 25.— stress allowed in B.T. Formula for Girders. To find what stress is allowed in the B.T. formula for o.c girders, viz.: — Working pressure = ^y^^Jp^Q j^ W = Width of combustion box in inches. P = Pitch of supporting bolts in inches. D = Distance between girders in inches. L = Length of girder in feet, of = Depth of girder in inches. T = Thickness of girder in inches. N = Number of supporting bolts. C = ^J^ ^^°° when the number of bolts is odd. M + 1 fN -I- 1^ 1200 " = ' M I o when the number of bolts is even. The value of C is readily found to be 6oo when N is i 900 ,, N is 2 or 3 1000 ,, N is 4. The next point to notice, in considering c.c. girders, is that the g'irders do not carry the entire steam pressure upon the c.c. tops. It is the stays which put load upon the girders, and it will be seen that the more stays there are in each girder the greater the load which the girder carries. In figure 145, if there were only one stay bolt in each girder, the girder would only carry one-half the total steam pressure upon the portion of the c.c. top, from one girder to the next. The area actually supported is (W - P) x D, what- ■ever the number of stays may be, P being the distance from the COMBUSTION CHAMBER GlRDERSt 3^§ extreme bolt to the lap joint on tube plate or c.c. back. P is | ffTl fTTi fffil •■ ' practically the pitch of the stay bolts .when there are more than one. r iBi iS) i^j ^ The load upon each girder can 1^ f>_>Lj_p_>!,_/>_>V(^/> now be written | •* vV ■ Load = (W - P) D j> where p is working pressure. Fig. 145. Now, taking the case of one stay bolt, the bending moment of a beam loaded in the centre is — — ^ — . 4 In the B.T. formula the length (L) is in feet. . „ .. ^ (W - P) D/ X 12 L . . Bendmg moment = ^ '- £- ■? 4 T d^ S The moment of resistance is ' — - — . Equating these moments O ' (W - P) D/ X 12 L ^ T^" S 4 6 , " ^ 18 (W - P) D L ' Comparing this formula with the B.T. formula we see that 600 S . or C in the B.T. rule corresponds to — m the other; we may, s therefore, write — = 600 and S = 10800 lbs., which is the stress 18 allowed. We will next take the case of 3 stay bolts in each girder. The value of the B.T. constant C. is now 3_iL-L^2? = 900 3+1 and the working pressure by the B.T. rule is: — _ 900 X ^'^ X T '^'^' ~ (W - P) D X L ^86 COMBUSTION CHAMBER GIRDERS. Proceeding to find the working pressure from first principles : — As already explained, the load carried by the girder is (W — P) x E)i X p, and since there are 3 stay bolts each bolt carries one-third of this amount. Now the greatest bending moment will be at the centre of the girder, and is that due to the reaction of the upward pressure at one end minus that due to the downward pull of the adjacent bolt or '^ load L _ load L _ load x L 2234 6 ' substituting the value of the load and expressing the length L in inches. ... ^ rW - P) D * X 12 L bending moment = -^ '- — ^-^ Equating the bending moment and the momeht of resistance we have : — (W - P) D ^ X 12 L _ d^ T S :-P = 6 6 rf'^ T S (W - P) D L X 12, Comparing this with the B.T. rule it is seen that - corresponds to the constant 900. .'. S = 10800 lbs. as before. The proof of the constant when there are 4 or more stays is similar to the above. 26.— Safety Valve Springs. To prove that in the Board of Trade rule for safety valve springs of round steel, the stress allowed is 20000 lbs. per sq. Id. approximately. The rule is: — Load = »00Q ^ SB a where S = diam. of wire, d = mean diam. of coiL j The first thing to notice about a safety valve spring is that ■when loaded and compressed there is a twisting action put upon' the wire. The same also if the spring is extended. If we take a length 'of' stout lead wire atid cdiil it up like a spring; figure 146; and draw a chalk mark along the wire from end to end at the outer surface of the coil, and then pull the coil out into a straight wire again by taking hold of the two ends, it ^'S' ''*^' will be found that the wire has now just as many turns or twists round its own axis as there were •coils at the commencement. -WW- The second point to notice is that the leverage to twist the ■wire at the very commencement of the pulling out is just the radius •of the coil. The subsequent behaviour has, of course, no relation to a real spring. Now, these facts show — if they do not actually prove mathe- ■matically — that the twisting moment put upon the wire of which a spring is made, is equal to the load multiplied by the radius of the •coil. We, therefore, use the shaft formula in treating of a helical spring. 388 ENGINEERING ESSAYS. Twisting moment = Load x — ., i r ■ >. S^ X stress Moment of resistance = 5-1 Equating' we get T J S' X stress •< 2 , » Load = = 11). 5-1 X rf Bringing forward the B.T. rule for load and equating to (i), S^ X stress x 2 _ 8000 x S^ 5"i -K d d Cancelling common factors : — stress X 2 and stress = 204000 lbs. 8000 Note. — ^The constant for square steel is iiooo instead of 8000 because the moment of resistance to twisting is greater for square steel. 37* — Angle of Twist in a Shaft. To prove the rule for the angle of twist in a shaft when working under a given twisting moment, viz.: — angle of twist e (in degrees) = lU^ where M is the twisting moment in lb. feet, L is the length of the shaft in feet, and d is the diameter in inches. , Referring- to figure 147 and to pages 220-223 for meaning of ^o-efHcient of rigidity (symbol K), we first notice that the twisting ktrain is C c -i- DC = Cc , if C c is taken in inches and L in feet. ,'. Since 12 L stress stram stress = co-efiicient K Cc K X strain = K x — ~ (i) 12 L ' 20 JeeL-- Fig. 147. But Q, c = a 5 (in the figure) and we have the proportion : 360 : 6 :: circumference : ab ot C c IT d y. B :.Cc = 360 substituting this in (i). K ird y. e stress = — - X 12 L 360 .(2). 39° ANGLE OF TWIST. Again, from the formula for the strength of a shaft we knon that if moment of resistance reckoned in foot pounds is M, stress = ~ ^""^ (3)' Equating the two expressions for stress in (2) and (3) K T d d 12 M X i:'i 5^ ^ *j 12 L 360 d^ From this ; equation we have to find 6, - ■ a _ 12 L X 360 X 12 (M) X 5-1 K IT d^ . and using the value of K given on page 221, viz., 12,000,000 g _ L X M ~ 142 "5 d* The constaint thus obtained -is somewhat greater than 140111 the usual formula ; but as a matter of fact the constant 140 was fixed by Mr Denny, .the eminent , shipbuilder,- as agreeing with numerous experiments on actual shafts. 2S. — Compression of a Safety Valve Spring. To prove the rale for the compression of a safety valve spring: viz.: — Compression = — =rr — ^ where W = load on Spring, d = mean diameter of ooils, N = number of coils, S = diameter of wire in sixteenths,^ a = & constant = 22 for round steel and 28 for square steel. If the experiment with the lead wire of No. 26 is tried with a single coil of wire it will be obvious that to produce one complete turn or twist in the wire we must pull out a distance equal' to the circumference of the coil We will assume (although this fact ought to be proved) that the amount of twist produced in the wire is proportional to the amount the coil is extended or compressed. That is to say, since an extension oi it d produces a twist of 360 degs. , we will assume this relation between extension (or compression) and angle of twi^ to hold good for angles of twist however small. Professor Perry has a very pretty experiment bearing on this point in his book on Mechanics. Now, we have already proved the rule for the angle of twist of a shaft (see No. 27). Twisting moments in foot-pounds = -3 la where 6 = angle of twist in degrees, D = diameter of shaft or wire in inches, L = length in feet. 392 COMPRESSION OF SPRINGS. In the case of a loaded spring the twisting moment is W x - ■( x Rj : p x R^ )> i> : : Kj : Kg Also, it is known that horse-power varies as the cube of the speed of the vessel, .-. istH.P. : 2nd H.P. :: Sj^ : S/ It follows that :- — Rj : R, :: ^ ■= S,* This gives the relation between revolutions of engines and speed of ship under the condition of constant mean pressure. Again, the number of revolutions may be obtained by dividing the speed per minute of the ship by the pitch of the screw if slip is neglected.' - , . Sj . ^2 . . c 3 . c 3 cancelling and multiplying mean and extreme terms we ha,ve the equation S' 2 G 2 2 = ''l or P S 2 = P S 2 — - — — or f-j Oj — ^^2 •'a This proves that P S^ is constant. 31* — Speed against a Current. To prove that the most economical speed against a oarrent is one-and-a-half times the speed of the current. To express the above principle more fully, let us suppose that a steamer is proceeding up a river against a strong ebb tide of C knots per hour. Then the steamer will reach her port with the least expenditure of coal by proceeding at a speed of i^ >^ C knots ,per hour through the water. ,1 To investigate the principle of the rule. Let the distance the •vessel has to steam to reach her port = D. Speed of current = C, arid the required speed of vessel through the water = S. The distance made good over the ground per hour is now = S - C ; ■and, therefore, the time taken to reach port = = ^ in hours. O — K^ Now the amount of coal burnt to reach port will vary directly -as the time. But it will also vary as the cube of the speed through ■the water (the same as horse power). Expressing this is symbols : Consumption varies as ^ -=, x S^. And whatever value of S makes this expression have the least ■value is the speed required ; and by trying different values for S, such as ij times C, twice C, three times C, etc., it may easily be proved tentatively that the least value occurs when S = ij C. However, the proper method is to use the calculus. DifFeren- D S^ tiating the fraction = ^, we obtain : — D X 3 SMS - C) - SS X I (s - c)2 ■which has to be equated to o. SPEED AGAINST A CURRENT. 397" .-. 3 SMS - C) - S3 = o 3 (S - C) - S = o 2 S = 3 C S = i^C Note. — It may be pointed out that in any vessel the consumption. only varies as the cube of the speed for a limited range of speeds. Hence the above rule would only hold good if the normal speed of the vessel happened to be not far different from once-and-a-half the: current. The rule has not much application in practice. 3a. — Horse-Power from Wetted Surface. One of the methods of estimating' the indicated hofse-power •necessary to obtain the required speed in a proposed vessel is based upon the amount of surface exposed to the water when the ship is at her normal draft. The amount of this surface is called the "wetted surface." It is found that for every loo sq. ft. of wetted surface about 5 I.H.P. is usually required to drive the ■vessel 10 knots an hour. Thus, if a vessel has 25,000 sq. ft. wetted surface she will require about 5 per cent, of this number for 'her indicated horse-power, which equals 1250 to go 10 knots. For ■higher or lower speeds the rule of the cubes of the speeds can be ■used. For example, at 1 1 knots the power would be 10' : 11^ : : 1250 : x 1000 : 1331 : : 1250 : 1663 I.H.P. In order to apply this method it is, however, first necessary to -obtain the wetted surface, and to find accurately the wetted surface •of a ship is a tedious process requiring the lines of the ship to be first drawn so that the mean girth to the load water-line can be ■measured. The mean girth multiplied by the length on the water- .line then gives the square feet of wetted surface. But various short methods may be employed to find an -approximate wetted surface sufficiently accurate for the purpose. "The simplest is Mumford's method, which has the formula Wetted surface = L (17 D + B x C) In words the rule reads as follows : — Multiply the draft by 1 "j and ■multiply the greatest breadth by the co-efficient of fineness of the ship, add the two results together for the mean girth, and then ■•multiply by the length of the ship between perpendiculars. HORSE-POWER FROM WETTED SURFACE, 399 For example: — To find the wetted surface of a vessel 460 feet •ong, 52 ^eet broad, and 267 feet draft, whose co-efficient of fine- ness is '72. j 267 X 17 = 45-39 52 X 72 = 37-44 mean girth = 82 -83 82-83 ^ 460 = 38,101 sq. ft. wetted surface. If we now take 5 per cent, of this for the l.H.P. at 10 knots we get 1905 l.H.P. And at 12 knots we get 10* : 12^ : : 1905 : 3292 l.H.P. nly. (In. Mr S^atpn's Manual, this vessel is quoted as steaming 12-05 knots with 33^2 l.H.P. , which agrees pretty well.) Second Method — Kirk's Analysis. Suppose a block model of the ship to be constructed in the Iprm of figure 148, consisting of a middle body and pointed fends, .The depth or thickness of the model is, uniform from end to end and equal to the mean draft of water of the ship, i 1 The middle body is rectangular and has a cross sectional area equal to the midship section ofl! the : ship. Thirdly, the total volume of the block model is equal to the displacement of the ship in cubic feet and the total ■* ci . !a_ 2 32 6 1 r -/lS-7— *< 1 ^-l* R E' 1 F i- ;'»'--iiii///|l^^ 111 II lllllllll lit lllllll M P Fig. 1.48, 400 KIRKS ANALYSIS. length equals the length of the ship. All dimensions of the model having been calculated the surface area of the bottom and sides of the model is taken as the wetted surface of the ship. ' To apply Kirk's analysis to any proposed ship, it is necessary to know her length on the water line, her draft of water, the area of her midship section, and her displacement. Of course, the displacement can be obtained from the three dimensions and the co-efficient of fineness. We will take the same example as before, viz., a vessel 460 teet long, and 26 "7 feet draft, whose displacement is 13,080 tons, and area of midship section 1322 sq. ft. First to find the displacement-volume 13080 X 35 = 457800 cu. ft. Referring to figure 148, it will be seen that the breadth of the model can be found by dividing the area of the cross section by the thickness, and as the cross section must be equal to the midship section of the ship. Breadth = 1322 -j- 26*7 = 49*51 feet. We proceed to find the lengths CA, AB, and BD in figure 148. Since the area of a triangle is one-half the area of the enclosing rectangle, and CA = BD, we may consider the total volume of the model to be equal to the rectangular solid C B P N. or displacement = area of cross section >« CB .-. CB = 457800 ^ ^^g.^ f^^^ 1322 .'. BD = 460 - 346"3 = ii3*7 feet also AC = ii3"7 feet and AB = 346'3 - 1137 = 232"6feet. Next we find the length of the oblique lines VA = VE = BX = FX VA2 = VR2 + RA2 = (1137)2 + (24755)2 VA = 116-36 feet. The dimensions are now all found and it only remains to calculate the areas for the wetted surface. kirk's analysis. 401 The bottom area is made up of a rectangle and two triangles, but we have already seen that it is equal to a rectangle of length CB and breadth BF. Area of bottom = 346*3 x 49"5i = 17145 sq. ft. The side area is of uniform depth, so we may use the total girth, which amounts to (232-6 X 2) + (116-36 X 4) = 930-64. Side area = 930*64 x 26-7 = 24848 sq. ft. Total wetted surface 17 145 + 24848 „ „ „ = 41993 sq. feet. It will be observed that in this particular case Kirk's method gives a greater wetted area than Mumfords by 10 per cent. Mr Seaton, in his Manual, states that Kirk's method may give as much as 8 per cent, greater area than the actual wetted surface of the ship. So the real amount may possibly be betweeen the two results obtained above. As another example taken from the same source, we will find the horse-power required for the vessel whose particulars are : — Length 311 -5 ft. = L; Extreme breadth 36 ft. = B; Mean draft 11-5 ft. = D; Displacement 1780 tons; Area, midship section, 392 sq. ft.; Speed 18-7 knots. Working by Mumford's rule we first require the co-efficient of fineness. Block measurement = 311*5 x 36 x 11-5 ,, = 128971 cub. ft. Actual displacement = 1780 x 35 ,, = 62300 cub. ft. Co-efficient = -^^ = -483 = C I 28971 Now, using the formula Wetted surface = L (1*7 D + B x C) M = 3ii"S(i9'5S + i7*388) ,, = 11500 sq. ft. nly. AI 402 HORSE-POWER FROM WETTED SURFACE. I.H.P. at lo knots = 11500 x. -j^ = 575 io« : i87» :: 575 : 3760 I.H.P. The same vessel worked out by Kirk's method has the block model somewhat like figure 149, with hardly any middle body. The wetted surface comes out = 12628 sq. ft., and the I.H.Pj = 4128. The actual I.H.P. is given as 4000, which is not far from the average of the two methods. Fig. 149. 33* — Tank Kxperiments for predicting the speed and horse-po'vsrer of vessels. The late Mr Froude by means of experiments carried out in a tank with models of ships arrived at certain laws which connect the performance of the model with the performance of the ship. In the case of both ship and model the total resistance is made up of two parts, viz., "frictional resistance" and "wave-making resistance," the latter being sometimes called "head resistance." The frictional resistance depends upon the kind of surface exposed to the water. Also, with certain exceptions, it is propor- tional to the area of the wetted surface, and thirdly, it is propor- tional to a certain power of the speed. The kind of surface that gives the least frictional resistance is one of tinfoil, next to this is paraffin wax, then varnish, and clean painted iron, and so on to sandy surfaces, which have about double the friction of varnish or paraffin. The amount of frictional resist- ance per square feet of wetted surface is much less in the case of a large surface than with a small one. Froude experimented with thin boards ^ in. thick, 19 in. wide, and varying from 2 ft. long to 50 ft. long. The boards were thinned off to a knife edge at each «nd, and were towed through the tank water at various speeds, and their resistances accurately measured. The following is an example of the kind of result obtained with varnished boards towed at 10 ft. per second (about 6 knots) the frictional resistance with a board 2 ft. long was •41 lbs. per sq. ft. )) ° )> >i '3^5 " ,, 20 M >i '278 ,, I) 5*^ >> '» '25 M 404 SPEED AND HORSE-POWER OF VESSELS. It is easy to explain the smaller resistance per square ft. in the case of the long- boards, for it is obvious that the friction of the first few feet from the cut water must set up a certain amount of current in the water next the surfaces, which, moving in the same direction as the board will reduce the friction of the rear portion. But it seems as if for lengths over 50 ft. there is not much further change to be expected, and the frictional resistance of a ship's skin may be assumed as about "25 lbs. per square foot at a speed of 6 knots. As regards the law of frictional resistance with change of speed, it has been found to be proportional to the square of the' speed for lengfths of 2 or 3 feet, but for 50-feet lengths the index is reduced to i"83, and for actual vessels the index is taken as i'825. We are. now in a position to calculate the frictional resistance of any vessel if we know her wetted surface and speed. Let the wetted surface be = S and the speed = V in knots per hour. At 6 knots the friction is "25 lbs. per sq. ft. .'. total friction = "25 S and changing the speed 61-826 . ylB25 . . .3^ s . R 26-3 : Vi-825 : : -25 S : R VI '825 V -oc S R_ V x^sa _ -oogSVisss (i). 26-3 ^ ' If we take a ship with a wetted surface of 30,000 sq. ft, and a speed of 18 knots, this Works out to a friction resistance of 52750 lbs. Next, as regards the wave resistance. It is here that the ' special use of an experimental tank in connection with a ship- building establishment comes in. Froude found that the wave resistance of a ship can be predicted if the -wave resistance of an exact' model of the ship is first ascertained. The model must be driven at a particular speed in the first place. This speed is related to the ship's speed as follows : — Let SPEED AND HORSE-POWER OF VESSELS. 4OS (L) and ( /) be the lengths of ship and model respectively, and (V) .and (v) their respective speeds, then JL ■■ sjl :: \ : V For instance, let the same ship be 400 ft. long and steam 18 knots. Let the model be made 9 feet long, then the speed of the model in the experimental tank is found by the proportion V400 : s/g : : 18 : » 20 : 3 : : 18 : V V = ^ = 2"j knots. 20 Thus 18 knots for the ship and 27 knots for the model are called corresponding speeds, for at these speeds the waves and eddies produced by both ship and model will be alike. The next step is to make a calculation of the skin friction of the model. For this we use the formula (i) with a slightly larger constant, viz., .012 r = "012 S V ^'^^ We still require the wetted surface of the model, but this can be deduced easily from the wetted surface of the ship by comparing the squares of the lengths thus : — 400^ : 9^ : 300000 : S „ 81 X 30000 ,. f. S = ~ = IS 2 sq. ft. 160000 The skin friction r is now = '012 x i5"2 x 27 ^'^^^ r = '012 X 15*2 X 6T3 = i"ii8 lbs. We cannot go any further without the tank. The total resist- ance of the model must be measured by actually drawing it at the proper speed in a tank. Let us suppose that this has been done and that the mean value of a number of runs gives a total resistance of 1-668 lbs. in the case of the g-foot model towed at 27 knots. Bringing forward the skin friction of the model (i*ii8) and subtracting it from the total resistance, we find the residual wave resistance — wave resistance (w) = i-668 - i-ii8 = 0-55. 4o6 SPEED AND HORSE-POWER OF VESSELS. Now the law of wave resistance discovered by Froude, is that, at corresponding speeds of ship and model, the wave resistances ate proportional to the cubes of the lengths. Hence if we put (W) for the wave resistance of the ship and (w) for the wave resistance of the model. /3 : L3 :: TO : W. Applying this rule to our case 9^ : 4008 :: -55 : W W = 4°°^ ^ -55 = 48280 lbs. 93 It only remains now to add the friction and wave resistances together; Total resistance = 52750 + 48280 = 101030 lbs. Next, to find the horse-power exerted by the propeller in driving the ship at the rate of 18 knots or 1804 feet per minute. HTj loio-^o X 1804 .P. = ^ -2. = 5523. 33000 And allowing for an efficiency of the propeller of 60 per cent Indicated horse-power = 5523 x -^^ = 9205. 34* — Cut-off by Slide "Valve. To prove that the fraction of the stroke performed by thfi piston after the cut-off is given by the rule, as follows: — / twice cap + lead y \ travel / * In figure 150, let the centre of the eccentric be at E,rwhen the; crank is on the dead centre A. The point £ is known to be aa amount equal to the lap + lead below CH, as shown by the two horizontal lines. Now since the valve cuts off when the eccentric arrives at D, the arc ED measures the angle described by the eccentric (and, therefore, by the crank) from the dead centre to the cut-ofF. There- fore AB being made equal to ED gives us CB as the position of the crank at the cut-ofF. Draw the horizontal line BP. Then neglecting the obliquity of the connect- P TV las: rod, -— is the fraction of the stroke performed after the cut-oiF. * AN Next draw the dotted lines cd and ce to contain the same angle as CD and CE, in which case de will lie very nearly midway between the lap and lead lines, that is cp = lap + ^ lead. Now, since the angle ACB is equal to the angle dee, the perpendidular C Q is equal to cp and, therefore, also equal to lap + ^ lead. 408 CUT-OFF BY SLIDE VALVE. We have now to find P N in terms of C Q and the travel. C P j^-= = cosine < A C B CO . r./-T3 .ACB _-^ = cosine < Q C B = cosine C B 2 It is a simple trigonometrical theorem that i + cos. A 2 A 9 COS.'* — CB + CP j /c_Qy C B Vc B/ sincfe CB = CN C N Vc B/ • L^ = {9-QY = /lap + jlead y ••AN Vc B/ V throw / or P N _ /twice lap + leadN AN ~ V travel / 2 35» — Difference in Cut-off. To prove that with equal leads on top and bottom centres the cut-off on the down stroke is later than the cut-off on the up stroke by an amount equal to : B where K = distance moved before the cut-off, B = distance moved after the cut-off, C == distance &om crosshead to shaft at given cut-off. The piston diagram of Zeuner is convenient for this proof* (See page 303 for description.) ^'ig. iSi- In figure 151, let oa be the top dead centre and ob the position of the crank at the cut-off from top. The leads being the same, the position of the crank when cutting off on the up stroke will be diametrically opposite, namely, at ob^, so that the crank describes 4IO DIFFERENCE IN CUT-OFF. the same angle from the bottom centre as it does from the top centre ; any difference in the piston movements being due to the obliquity of the connecting- rod. Now, the eccentric circle representing the position of the crosshead tells us that when the crank is at ob the distance of the crosshead from the shaft is oc which = C in the rule. Also cb = A in the rule and cA = B. , ^ There is a proposition in the third book of Euclid which is often useful when circles are involved. It is there proved that when two straight lines intersect each other in a circle, the product of the segments of one line is equal to the product of the segments of the other line. Looking at the cirblte ac^^ we see that oc •><. og = oa x of. (i). But oc = C in the i-iile and oa = ob = oc -V cb = C -\- h. of = oh = oc - ch = C - B og = ok + X = oh + X = C — B + a; substituting these equivalents in ( i ) C X (C - B + ^) = (C + A) X (C - B) .'. multiplying out and omitting x signs. C2 -CB + Qv; = C2 + CA - CB -AB cancelling, etc. , ' C^ = CA - AB A AB X = h. — — — C This proves that the cut-off from the bottom [x] is less than the cut-off from the top (A) by — — — Note. — If A, B, and C apply to th& cut-off on the up stroke, it can be proved similarly that the cut-off on the down stroke '^ =A + ^L^ 36. — Formula connecting the Pressure, Temperature, and Volume of Air. To proYe the formula connecting the pressure, temperature^ and volume of air, viz.:— PV ^ = S3-2 vhere P = gross pressure per square foot, T = absolute temperature Fah., Y = volume in cu. ft. of one pound of air. P V We assume that -— - is constant for all temperatures, pres- sures, and volumes, and find the value of the constant. We must know some fact about the volume and weight of air obtained from actual weighing- or we cannot proceed. The specific gravity of air at a given temperature and pressure would be enough, Or better still the density per cubic foot which is '0807 lbs. at 32 degs. and i4'7 lbs. pressure. Using this number the volume of one lb. of air is -rz; — = I2'39 cu. ft. = V. •0807 '"^ , Also the pressure per sq. ft. = 21 16 lbs. = P, and the absolute temp, is 461 + 32 = 493 = T. Substituting these values in the formula we obtain : — PV 21 16 X I2"'JQ , . T 493 " 37* — Steam Consumption. IRule for weight of steam accounted for by the low pressure diagrams, viz.: — Tons in 24 hours = i^i+^^AJi where Pi and p^ are the absol'ute pressures at ^ the stroke and L is length of stroke in feet. Taking one diagram at a time, representing one stroke of the •engine. — X (^^-x(Lx 12)) = capacity of cylinder in cu. inches 10 \ 4 / ■up to ^ stroke. This is the volume of steam used at pressure J>^ each down stroke of the piston. In cu. ft. per day this amounts to 8«/2irLxi2xRx6ox24 ,, j „ . = 2 a^ TT L, K, 4 X 10 X 1728 Now the weight of one cubic foot of steam at atmospheric (pressure = '0387 lbs., and for a few pounds less or greater, the weight varies with the absolute pressure (see table in appendix). ,•. at pressure p^ (absolute) i cu. ft. weighs — 2_Z £1. Weight -of steam = 2 ^^ ;r L R x '2^M. - ^ 2 X -0387 IT hot air = m:E The difference in the weight of the two columns of air outside and inside the funnel, is : — H X lip- H X 195^5- (1.) which is the pressure per square foot causing draft. In order to convert this pressure into aft equivalent velocity we must find the equivalent head of hot air; that is, the height of a column if hot air which would of itself balance the difference of pressure. We, therefore, now divide by the weight of a cubic foot of hot air which is a2i — Head = H x I - H = H ( i) = — ^^ ■ (everything else cancels). If there was no obstruction to the draught its velocity in the funnel would be : — Velocity = J2g k head = k/ 2 g H x — (2.) But there is always a very great obstruction to draught in the furnaces and tubes of a boiler, making the above formula quite useless by itself. According to Peclet the "head" to be used in calculating the velocity is less than ^ of the real head just stated ; hence the actual velocity is more nearly _ / H T -/ , . BI 4i8 FUNNEL DRAUGHT. This formula works out to a velocity of i6 ft. per sec. for a funnel £2 ft, high. However, the formula shows that the velocity in the funnel varies as the square root of (T - /) ; (all the other terms being constant). We have already seen that the weight of gases discharged also varies as the density, hence the maximum effective draught must vary as the product of velocity and density, that is as T — i squaring numerator and denominator we get to be a maxi- mum. Differentiating this fraction (T being the variable) and equating to o. I X T2 - 2 T (T - /) ■ ~ ^ o •pi .-. T2 = 2 T (T - /) T = 2 T - 2 ^ T = 2 if Proving that the most effective draught is obtained when the absolute temperature of the funnel is twice the temperature of the external air. JVbie. — It is usually said that the ratio of temperatures should be as 12 to 25 (see Rankine's steam, page 285). This result is obtained by taking into consideration the greater specific gravity of the funnel gases compared with atmospheric air. The correction is very trifling. 40. — ^Work done during Expansion. To prove the formula for the work done daring the expansion of any elastic fluid, viz.: — Work = ^1^1 ~ ^i^i n - 1 where P^ and Y^ are the pressure and Tolnme respeet- ively at the commencement', and Pj and T^ are the pressure and volume at the end of the expansion respectively; n being the index in the law of expansion. Let the law of expansion be PV" = a constant, that is PiVj " = PgVj", the particular value of the constant not being wanted at present. In figure 152 (which repre- 1*"^' — >« sents the pressure and volume as the expansion proceeds) take a position on the expan- sion curve at a, where ab repre- -sents the pressure and x the -volume of the expanding gas. Now the pressure ab may be isupposed constant for an in- finitesimal increase of the volume, represented by dx ■(using the language of the cal- ■culus). Hence the increase in the area of the diagram = ab -x. dx. But Pj X V," = {ab) X :i;" by the law of expansion. Fig. 152- .-. ab = PiVi» -r X". The increment of area may now be written -PiV X dx. The integral of this expression taken between the limits :« = Vj and X = Vj, will give us the area of the diagram under the expan- 420 ENGINEERING ESSAYS. sion curve, and, therefore, the work done during the expansion portion of the stroke. I ?lZl2 X dx = I (PiVi ") (x-") dx — n -\- 1 When^ = V,, Work = ^ ^i ^ ' ^^ > ^ 1 — n When X = V„, Work = (^i"^! ") (^2' ") Subtracting, we have — Work (during expansion) = -^ — ^ — ^^ — ^ * — * ^ I - ra In order to simplify this expression we notice that P^Vj^" = PgVj"; substituting this in the first term of the expression and multiplying both terms out, by adding the indices {n + i — n= i). Work = P^V^ - Pi^i = P^Vi - P^V^ (,.) I — K TO — I To find the total work of one stroke of an engine, we must add the work done before expansion commences, which is P^Vp P V _ P V Total Work = RV, + i-LlJ £ilJ (2). , 11-1 In applying formula (i) or (2) to any actual case, it must be noted that Pj is the initial pressure per square foot and Vj the volume of fluid used in cubic feet. Also the index in the law of expansion must be known (w), not only for putting the correct value of n - i in the denominator, but also for calculating the terminal pressure or volume (Pj) or (V2) whichever is unknown. The following are the values of n usually quoted : — For adiabatic expansion of air (or other \ •,-'■• r \ > w = 1-4 permanent gas) ... ... ... J ^ For adiabatic expansion of steam gaS w = i -3 For adiabatic expansion of common steam ra = i-i For steam when there is no condensation re = i -06 The formulas (i) and (2) cannot be used for isothermal expan- ■ sion of air or steam, because in Boyle's Law n = i (PV constant). The formula of the following article is used instead. 41. — Mean Pressure of Steam. To proYe the rale for mean pressure of steam in a cylinder when expanding according to Boyle's Law (pv constant). P Mean pressure = - (1 + log. e r) In figure 152 we have the ideal indicator diagram for expand- ing steam. The first step is to find an expression for the area of the diagram, and with this object we may conveniently take it in two portions. The area up to the point of cut-off = Pj Vj. The area after the cut-off is bounded by the hyperbolic curve of expansion, and in article No. 2 of part III. it is mentioned that the ratio between these two areas is known as the hyperbolic logarithm of the ratio of V^ to V-^ in the figure, which is the ratio between the full stroke and the cut-off, which is the ratio of expansion = r. We have, therefore area after cutting off _ . area before cutting off or area after cutting off = P^ Vi x log. e r The total area = Pi Vj -|- Pj Vj x log. e r „ = Pi Vi (I -I- log. e r) Now to obtain the mean height of the diagram (which is the mean pressure) we must divide the area by the length Vg or, which is the same thing, by Vj x r When V^ cancels out and leaves p Mean pressure = —^(14- log. e r) r 422 MEAN PRESSURE. If we use Pj for the initial pressure per square foot we obtain a convenient rule for the work done in one stroke of an eng^ine, for Let A = area of piston in square feet Then work = mean pressure x length of stroke M M = ^ (i + log. e r) X A X Vj T P, X A X Vo / , , = — 1 ? (i + log- ^ r but A X Vg = volume of cylinder and ? = volume of steant admitted each stroke = Vj Work in one stroke = Pj Vj (i + log. c r) where P^ is the pressure per square foot and Vj is cubic feet of steam used at pressure Pj. 42* — ^Work done during Isothermal Expansion. To proYe the role for the work done during isothermal expansion, yIz.: — ^Work = PY x log. c r (using the calculus). In this formula P is the initial pressure per square foot, and V is the volume of the steam at the initial pressure which is used ia one stroke. If there is no expansion PV is the work done. For total pressure = P x area of piston, and Work = P X area x stroke but area x stroke = volume = V .-. Work = P x V. Also if the steam is expanded, PV still equals the work done during the admission part of the stroke, and the object of the present investigation is to find a formula for the work done during the expansion portion of the stroke. In figure 152, representing the expansion diagram of the steam» let the volume of the steam at some epoch of the expansion be x. The pressure at this point = ab, and the work done during the next infinitely Small portion of the stroke is ah x dx. Now from Boyle's law P V P.V, = ab y. }c. :. ab = i-m. '■ ^ X P V The differential of the work now = -i — 1 y. d x and we have X to integrate this expression between the limits jc = Vj and x = Y^ The integral of - is the hyperbolic logarithm of x. X 424 ISOTHERMAL EXPANSION. /: 'x = V2 Pi Vi ^^ = p^ Vi X log. € Vg - P V X log. e V^ = Pi Vj (log. € V^ - log. . Vi) = P V log. c Yg If we call =^ the ratio of expansion = r. Work during expansion = Pi V^ x log. e r. For the work done during the whole stroke we must add the work done during admission which = Pj Vj ! ' Total work = Pi V^ + Pj Vj log. € r „ = Pi Vi (i + log. 6 r) 43* — I^inear Velocity of Escaping Steam. Given the experimental fact that the weight of steam which escapes in 70 seconds, through an orifice one square inch in area, is just equal to the gross pressure of the steam in pounds. It is required to find the linear velocity of the issuing steam. The statement above should be qualified by the further statement: — "When the pressure on the discharge side of the orifice does not exceed 58 per cent, of the steam pressure ; " for if the back pressure exceeds this amount, less weight of steam will escape, and as the difference between the pressures on opposite sides of the orifice is made less and less, so will the linear velocity of the issuing steam debrease, until when the pressures are exactly alike no flow can occur. But the present problem has nothing to do with this condition of things, but only with the case where the back pressure is less than 58 per cent, of the initial pressure. In the first place, some common-sense explanation may be offered of the experimental fact first mentioned. It is as follows : — As the steam passes through the orifice to the discharge side it expands considerably in volume, and the more it expands the greater volume there is to pass away. In this way we can see the proba- bility of the discharged steam blocking the way, as it were, for the steam to follow. The result appears to be that, no matter how low the pressure may be at the exhaust side of the orifice, the pressure immediately in front of the orifice is never less than 58 per cent, of the initial pressure. At this back pressure the maximum discharge is obtained by weight. Let P = gross pressure of steam. p = gross pressure of steam as it escapes = "58 P. Let TO = weight of a cubic foot of escaping steam. 426 VELOCITY OF STEAM. Now from experiment, weight of steam which escapes per p second = — when the orifice is i sq. inch, and the volume of the 70 p escaping' steam = r w (cu. ft.)- If this volume is divided by the sectional area of the orifice in sq. ft. we shall have an expression for the velocity in feet per second. P . I _ 144 P velocity = 70 TW 144 70 W To take a particular case, such as steam from a boiler at 185 lbs. pressure by gauge escaping into the atmosphere Here P = 200 absolute p = 200 X "58 = 116 absolute and referring to a table of steam in the appendix, w is found to be 'aSs lbs. ■.T , .. 144 X 200 r^ Velocity — !3 = i-ra ft. per sec. ■^ 70 X -265 ^^ '^ Next, we will take a very low pressure, say steam of atmospheric pressure escaping into a condenser (where the pressure is sure to be less than 58% of the atmosphere). Here P = 15 lbs. ;> = IS X -58 = 87 lbs. from the table w = "023. Velocity = '*^ — 15 = 1341 ft. per sec. 70 X "023 These two results show that there is not very much difference in the velocity of escaping steam whether it issues from a high- pressure boiler into the atmosphere or from a low-pressure valve box into the condenser. 44«— The Action of Giffard's Injector. There are many types of injector, all modifications of thfr original " GifFard. " An illustration will be found in any elementary work on the steam engine. The action may be briefly described as follows : — ^A small jet of live steam is allowed to impinge upon comparatively cool water entering at a side orifice somewhat behind the nozzle of the jet. The steam is all condensed and the mixed steam and water is driven by the force of the original Impact straight forward through the injector, the passage- way being made with a gradually- increasing sectional area. By this means the velocity of the steam decreases and by a principle in hydraulics its pressure incresLses until it equals the boiler pressure and passes through a check valve as feed water. To show that there is nothing contrary to dynamical principles in the action of an injector we may offer the following investigation. I St. — In order that the steam may be condensed it must mix with a certain number of times its own Fig- 153- 428 THE INJECTOR. weight of water. A pound of steam requires at least about 7 lbs. of water at 60 degs. to condense it. To prove this, let the steam be at 200 lbs. gross pressure. Its temperature (see table) is 382 degs. and its total heat is : — 1115 + "3 X 382 — - 1229 units per lb. The heat in 7 lbs. of water at 60 degs. is 7 X 60 = 420 units Total heat in the mixed jet = (i X 1229) + (7 X 60) = 1649 dividing by 8 lbs. (the total quantity) Temperature = —~ = 206 degrees. o It is, therefore, clear that unless there is about 7 lbs. of water to I lb. of steam the temperature will not be reduced below boiling point. 2nd. — Assuming the above proportion of water to steam we ■have now a case of momentum. From the previous article we know that steam of 200 lbs. ■gross pressure escapes with a velocity of 1500 feet per second. The momentum of i pound of it is, therefore, = ^52? And since all the momentum passes into the mixed jet 8 X V^ _ I X 1500 "Where V = velocity of mixed jet from this equation, V = 187 feet per sec. 3rd. — The question is now, what pressure will a jet of water ■moving at 187 feet per second overcome? If all the energy is utilized in over-coming pressure, available energy of each lb. = — ' ' = 546 ft. lbs. 64 Now, a pound of water will fill a tube i sq. inch section and -27'7 inches long (= 2*3 feet long). THE INJECTOR, 429- If p = pressure against water,/ x 2'3 = work to discharge I lb. of water into boiler. .'. p X 2 "3 = 546 work available p = 237 lbs. According to the above principles the injector would pump against a greater pressure than its own boiler if the feed water was at 60 degs. Fah. This is not a theory of the injector, but merely a proof that there is sufficient momentum in the steam to do the- work. APPENDIX. TABLES AND EXA/v\l NATION PAPERS. 432 Chemical Symbols and Atomic Weights of Common Elements. Gases — Hydrogen Nitrogen Oxygen Chlorine Non-Metallic — Carbon Silicon Sulphur Phosphorus Metallic — Magnesium Aluminium Calcium Manganese Iron Nickel Copper Zinc Tin - Platinum - Mercury - Lead H. = I N. = 14 CI. -- 16 35*5 c. ^ 12 Si. = 28 S. = 32 p. = 31 Mg. . 24 Al. = 27-5 Ca. = 40 Mn. = 55 Fe. = 56 Ni. = 58-5 Cu. = 63-5 Zn. = 65 Sn. = 118 Pt. = 197 Hg. = 200 Pb. = 207 specific Gravity and Weight of Solids an I/iquids. Sp.gr. Water = 1. Founds Cubic Pound Nakb. per cu. inch. inches per -lb.' per cu. foo Platinum - - 21-53 •77 ' 1-29 1342 Mercury (fluid) - 13-6 •49 2*04 84c Lead (sheet) - II-4 •41 2-44 7IJ Copper (sheet) - - 8-8 ■33 3-II SSc Nickel (cast) '■ - 8.28 •3 3-14 5'e Brass (cast) - 8-1 ;-29 3*42 . so; Steel (average) -' - 7-85 •283 3-53 48c Wro't Iron (average) 77 -277 3-6 a 48; Cast Iron (average) - 7-2 •26 3."9 ; - : '. 45c Tin - - - - 7-2 ; -26 3-9 4SC Zinc (cast)i - ' 6-86 — 4'o <42f Aluminium (caSt) - 2-56 •096 io"4 i6c Linfestone - 2-84 — — 17^ Coal - '- - 1-37 — — — Bees Wax - !' -96 •03s 28-8 6c Yellow Pine , - T, '47 -,, — — ,. . 2C Cw-k - , - - •24 '!' — , "5 ;,'-r H Liquids — Specific Gravity and Weight of Various^ Gases. Lbs, per Specific cu. ft. at 32* Cubic feet ,; Name. Symbol. Gravity and. to the fl • ' air = I. atmospheric press. pound. Hydrogen - H •069 -0056 i>8-8 Coal Gas - — •438 •03S4 28-2(7 Marsh Gas - - CH^ •552 •0446 22*41 Ammonia - NH3 •589 •0476; 21-01 Acetylene CjHj •900, •0727 1375 Carbonic Oxide - - CO •967 . -0781 1 2 -So Nitrogen N •973 -0786 12-72 Air - - - ^ — !• -0807 I2-38 Oxygen I'lOS •0893 1 1 -2 Carbonic Acid - COj 1-529 •1234 8-» Chlorine - ■ - CI. 2-440 •1972 507 43S Specific Heat of Substances (-vsrater at 39 degrees being = 1). ;_■, I Specific Heat ',■■■. , 1 Specific Heat Substance. at constant at constant pressure. volume. Gases — Hydrogen - - 3 •405 2 ■416 Atmospheric Air - •238 •169 Nitrogen - •244 •■173 Steam Gas , - - - ^48 •37' Carbonic Acid •217 :nx Liquids — ■ ;■'■ A Water - - - ' " I. , 1 -, — ' ; Alcohol - , - - . "659 1. "— J Pet;-,oleum , - - - » "434 .-.' : — i" Oliye Oil - - ' '31- ..-jK '-^,'' Merpury •,.-.. - •033 (L.J;l .-rr'l Solids — Ice - •504 — Aluminium - - ■234 — Coke - - •203 — Cast Iron - •13 — Wrought Iron - •113 — Copper - - •095 — Lead - • •031 — 436 Melting Point of Metals, etc. ^UBSTANCB. 'i '-:■'.).<.■ 'Melting tomt Fah. Ice - - - 32 degrees. Tallow - .., - 92 M Paraffin Wax .;; - 114.: ., . >> Bees Wax - ' '- ■ ■' '- - 150 ^y ■ Tin - - 2 ■ ^^'"^'^ Lead, - 3 - Metal 1 . I. Bismuth 5 - 199 ; Tin - ;- - 449 »> ■'^■' Lead - - 620 yy Zinc - ■■- ' - - 786 >^ Aluminium - - - 1214 >)■ Brass - - - 1650 yy- Cast Iron . .: . - 1920 to 2012 )^ ' Copper - - 1943 jy Tool Steel - -, - 2370 :-■■■ >» Mild Steel - - ' - - 2550 ' jV Pure Iron - ¥ • - - 2737 ■'.• >>• 437 Temperature and Weight of Steam. Absolute pressure. I lb. Temperature Fah. 102 degrees Pounds per en. •003 ft. 2 , 126 , •006 3 > 142 •008 4 158 , , •on S 162 . "014 6 , > '' 170 , , -oie 8 - 183 , , •021 10 , 193 , -026 IS . 212 , > "039 20 , , 228 , •051 25 , 240 , . -063 30 , 250 , . -074 35 , 259, , •086 40 , 267 , •097 4S 274.- , > '109 SO , 281 , J r ''^° SS , 287 , •131 60 , 293 , > ■143 «S , 298 , •154 70 303 I 1-165 75 . 307 . •176 80 312 •187 85 , ;: 316. , . . '198 90 . 320 , . "209 95 ■ 324 , -220 100 , 327. . . - '230, loS 331 •241 no , 334 . •252 "S 338 , •263 120 , 341 ' -274 I2S , 344 •28S 130 . 347 •295 13s 350 , •306 140 , 353 •316 I4S . : 356 . •327 150 358 , •337 »55 . 361 , ■348 t6o 363 : "359 165 , 366 . •369 m , '- 370 . ■390 18S , 375 . 1 '412 195 , 379 . t -433 205 . 384 . •453 21S , 388 , •474 22s 392 •495 235 395 •51S 24s . 399 . •536 2SS . 403 , •556 ^65 . 406 , •577 438 Breaking Strength of Materials. Tension Compression Shear Material in tons per sq. in. in tons - per sq. in. in tons per sq. iiw Nickel Steel (soft) - - - 33 — — Steel Plates (i% c.) - 28 — 23 Aluminium Bronze ( :95% copper) 27 — — Phosphor Bronze (cast) - 25 — — Wrought Iron Bars - 23 21 iS Iron Boiler Plates - 21 21 16 Muntz Metal - - 20 — Copper (sheet) - - - - 15 — — Gun Metal - . •' - 13 — — Zinc (sheet) - - - 13 — — Brass (cast) - - 9 4 — Cast Iron - - - - 7 40 5 Ash (timber) - - - 7 — — Oak „ - - 6 4 I Aluminium (cast) - - - 6 — — Pitch Pine - - - 5- ^ 3 — Fir - - - S — — Hemp Rope - - - 4i — — Tin - - - - - 2 — — Lead - - -,. - li" — — 439 Blastic Strength of Materials in Tension. Steel Plates (4% carbon) - 19 tons per sq. im Nickel Steel (soft) 16 „ J M ■ - Wrought Iron - - - 13 kOm J;, J ) I. .' Phosphor Bronze (cast) 9 ».» "■/.? Cast Iron - - - 5 ,, '" ) Gun Metal - 3 )> . A Copper (sheet) - - - 4 ,/ J> J r -;, ; Faaiors of Safefy^ (from Unwin). Material. A dead load.. A live load. Stress of one kind only. " • A live load. Stresses of opposite ' kinds. A live load. Subject to Shpcks. Wrought Iron and Steel tons per sq. in. 3 tons per sq. in. 5 tons per ' ; sq. in. 8 tons per sq. in. 12 Cast Iron 4 6 . 10 15 Timber - 7 10 '5 20 440 Young's Modulus of Elasticity (E)« Cast Steel (tempered) Steel Plates Wrought Iron :-" Cast Iron . '. . Copper (rolled) ,- Phosphor Bronzef, Brass Wire {'- Lead - - - Wood 36 million lbs. 30 1- 29 » 17 1 15 % 14 > 14 > V r 2h > 'i >> > Modulus of Rigidity (K). Cast Steel (tempered) - - - 14 million lbs, Steel Plates - - - - . t-? ': Wrought Iron ., ^ t - - 1 1 , , Cast Iron --/:-.-. 5 ,, Phosphor Bronze - - - - si ,> ■ Copper (rolled) - : - - - ci „ Questions set at the B. T. Examination oi : held in January 190Q, for Extra- First-Class Engineers. t ' : I. What is energy? Distinguish between potential and tsirietic einergy. What is meant by the phrase "Conservation of Energy"? : ! , , 2. Find the amount of energy in a fly-wheel 3 tons weight. JE^adiua of gyratipn rtiay be ta|£pn as 7 feet. . The wheel makes 65 revolutions per minute. , 3. Find the length of a hoUowi shaft (ste^l), to have the same ^pergy as the aboye wheel when making the same, number of revolutions per minute. Diameter pf shaft- 20 in., diameter pf hole 8 in. Radius of gyration may be taken »/ -^—X — . The weight .of a cubic foot of steel = 496 lbs. ' ' 4. A hollow spherical shell has an internal diameter "j of the ^xternal, and the weight , is 98 lbs. Find the inside and outside diameters, if a cu. inch — '26 lbs. . ' < ■ - 5. Find the weight of a .triangular steel plate, whose sides measure 12, 11, ant^y feet respectively.! Thickness of plate ijin., and th^weight of a cubic foot of steel 490 IbS. , . ' f -'''''' ' ' ' .' - ' * ,/ .'6.'i IRectgcngular iron' beam is, 2 in. broad aUd 6 in. deep, find maximum stress when a weight of 2 tons is placed in the centre,' if the distance between the supports is 10 ft. 442 B.T. EXAMINATION gUESTlONS. Also find the stress if the beam is replaced by one having' same sectional area, but with flanges 4-J- in. wide, joined by a web f in. thick, the over-all depth being as before. Sketch the section of the beam, taking care sectional area is same as at first. 7. A dead weight of 10 tons in the centre of a beam causes a deflection of i| inches. From what height should a weight of 180 lbs. fall on to the centre of the beam to have the same eff^ect. 8. A combustion chamber is 3 ft. wide and the length of the g-irder may be taken as the same length.. There are three stays to each girder pitched at 9 in. apart and 9 in. from back and front of chamber. Find the bending moment at each stay when working pressure is 175 lbs. per sq. inch, and girders are 8 in. apart, centre to centre. ■ , Sketch the girder and work the problem, both by calculation and graphically. ' '9. A rule given to second-class candidates is 2^ d'^' \fp = cu. ft. Of water per min. discharged from a hole, where d is diam. in inches and p is head in pounds per sq. inch.' Investigate this from the rule V = ijz'g S, ^ being taken as 32 and usual co-efficient allowed for contraction of the orifice. Find how nearly above rule approximates to this. 10. You arrive at a port too early to enter and are ordered to steam dead slow. Draw the probable indicator diagrams, ist, throttling the steam by nearly closing the stop valve or throttle valve ; and, by linking up the engines and keeping stop valve full open. Mark estimated mean pressure, I.H.P., and revolutions ori each card, and state clearly the causes of difference, if any, in the form of the diagrams. Give boiler pressure, diameters of cylinders, assuming usual ratios, and engines triple-expansion, and I.H.P., and revolutions when ;steaming full speed. B.T. EXAMINATION QUESTIONS. 443. 11. Draw also diagframs from above eng-ine when steaming' full speed, and estimate and mark on the .cards the H.P. of eacT* engine, and mean pressure. You need not work up the cards, but can make your meaning- clear in the statements. ' 12. Steam for a feed heater is taken from the L.P. casing, •where temperature is 240 degs. F. The feed enters the heater at 120 degs., and enters the boiler at 200 degs. F. What percentage of steam is required for the heater. The steam is supposed to be dry. 13. Suppose the steam contains 20 per cent, of Water. What will be the percentage required now. 14. Two plates are joined by a single butt strap, having 21 rivets on each side of the butt, and rivets are pitched 4 iri. apart. The total width of plate is 2 feet. Find percentage of strength of plate (the butt strap is not to be considered).' ■J 15. The rule for the amount of compression on a safety valve- spring is ^ . load X (f^ X N Compression = • , ^ C X S* Assume the modulus of rigidity = 17,000,000 lbs., and find the- value of the constant. 16. A ship's engines develop 45,000 H.P. using 14 lbs. steam per H.P. per hour. The value of the coal is 12,500 T.U. per pound, and the efficiency of the boiler is "j. Find the coal used per H.P. per hour. 17. In the above case, compare the work done with the thermal value of the coal burnt. 18. Given H.P. 4500. Revolutions 72 per minute. Ratio of maximum to mean twist on shaft if. Shaft hollow, hole '5 outside diam. Stress allowed 8000 lbs. Find diameter of shaft. 444 B-T. EXAMINATION QUESTIONS. 19. Given displacement of ship, also hSr leftgth, breadth, and depth, and mean draft. Also centre of gravity 4 ft. below water line and centre of buoyancy 8*5 ft. below water Has. Moment of inertia of water line plane "6 of rectang-le. Find meta-centric height and statical stability for 7 degrees of heel. 20. A weight of 30 tons, placed upon a ship's deck 18 ft. from the centre line, makes a pendulum 12 feet long to s\ving 4^ in. Displacement 5000 tons. Find rn.c. height. > 21. A steamer leaves a salt water port, and burns 150 tons of coal before arriving at a fresh water port. She has exactly the same draft on the two occasions. Find the original displacement. The two specific gravities are 1024 and 1005. 22. The connecting rod is 14 ft. long, the engine stroke is 6 ft. Find the angle between rod and crank at half stroke. Also the position of the piston when the crank is at right angles to the piston rod. 23. Sketch an accumulator plant and give full description. Or— Draw a reducing valve to 'Scale. 445. Books Recommended for Study and Reference. Algebra. As far as quadratics, indices, and logarithms. By Hall and Knight. Price 4/6. Trigonometry. (For solution of triangles.) By Lock. An Introduction to the Calculus. By W. J. Millar, C.E.. (Blackie & Sons.) 1/6. Mechanics for Engineering Students. (For simple exercises- in the mechanical powers, beams, shafts, etc.) By A. N.. SOMERSCALES. (R. C. AnNANDALE, HuLL.) 1/6. Formulae, Rules, and Questions in Steam. (For exercises; in steam generally.) By A. N. Somerscales. (R. C. Annandale, Hull.) 1/6. Theoretical Mechanics, Solids and Fluids. By Taylor.. (Longmans.) 2/6 each. Simple Problems in Marine Engineering Design, By Sothern. (Munro & Co.) 2/6. Indicator Diagrams for Marine Engineers. By W. G M 'Gibbon. (Munro & Co.) 7/6. Mathematical Tables. Chambers. 4/6. Any Mechanical Engineers' Pocket Book.