Class " ; : Book - (bpigM CfQPYRIGHT DEPOSIT COMPRESSED AIR g? McGraw-Hill Book Company Puj6fis/iers offioo^sfb?' Electrical World The Engineering and Mining Journal Engineering Record Engineering News Railway Age Gazette American Machinist Signal ELngini" = pxVz™ = pzV2 n ; observed law of gases 2 3 „ w P2W2 — PlVi . n 6. W = 7 h P2^2 — Pa^o 2 4 n — I 7. TF = -^-- (p 2 y 2 - PaVa ) 2 4 w — 1 7a. Tf = - 53.35 (i 2 — h) for one pound 2 4 n - 1 8. IF = —^ p„« (r w - l) 2 5 n - 1 8d. TF = T (^J-j) 53.35 (r B - l) 1 < = kt, for one pound 2 6 9. TP" = _ 1 1 + p2?>2 — p a vi ; incomplete expansion ... 3 8 10. m.e.p. = 2.3 p a logio r 4a 12 w - 1 10a. m.e.p. = — r p a ( r M — 1 1 4a 12 10b. m.e.p. = p 2 I — _ 1 ) — p a 4a 12 ( n) 11. E = 1 + c\l — r / ; volumetric efficiency 4b 13 /vi\ n ~ 1 "— ^ 12. i 2 = h .Ml - - iir » 7 18 7} 13. w = go og . ; dry air 8 19 00.00 t 13c. w> = —— — (m — 0.38 Hq) ; moist air 8a 20 14. 0*2 = —^i- and d 3 = — ; compounding 12 26 n-\ 15. W = — ^-r (p a Va) In ? — l) X 2 ; compound two-stage 13 27 n- 1 16. PT = — ^7 p a Wo m — 1] X 3 ; compound three-stage 13 27 xni 36 XIV INDEX TO FORMULAS Number Formula Art. Page 17. m = . tI^ — r — g , • exhaust pump 15 30 log (V + v) — log V r — 1 18. E = — -. ; efficiency without expansion 16 31 r log e r 19. p a = .4912[w - 0.0001 (F - 32)] ; pressure by barometer. 17 32 20. logio p a = 1.4687 — -199 a/™ , Anr,\ ! pressure at elevation 17 32 21. Q = c 0.1639 d 2 \l- rp a ) orifice measure 21 21a. w a = 1.321 -; weight by mercury column 21 37 22. di = -y ; orifice under pressure 25 44 I v a - 26. / = c -== — : friction in pipes 29 50 a 5 r I ch)a 2 \ y$ 27. d = I - ° ) ; friction in pipes 29 51 O0 , 0.102 5 Zt>.«. f ..... 00 K1 28. / = ir— : — : friction in pipes 29 51 I v a - 30. logio Pi = logio pi — c 2 -« -j — ; friction in pipes 30 58 32. E = . 2 ; efficiency of transmission 31 60 logri' loff r 33. E = — ^-r-; efficiency-hydraulic air compression. 32 63 o, v a ■ w(h + hi) . .... .„ on 34. ~ = ^r — t — — J air lift pump 40 80 Q E p a \og e r 36. gf = Jl\ + h ; air lift pump 40 80 Q 35 logio? 1 ' 37. H = ; fans — centrifugal pumps 49 96 „_ , „ w a lu~ + uV cos 0\ ,-"■. . CE in „ 37a. loff e i?i = — ( ) turbo-air compression .... 55 107 Pa\ g 1 38. R n = Ri n and log R n = n log Ei ; turbo-air compression . . 55 107 39. logx„ Ri = ( 2 . 3 x 5 3,3 5 x t x g ) ^ = ***i turbo-air com- pression 56 108 INTRODUCTION Compressed Air is a manufactured product of considerably greater value than the things used and consumed in its manu- facture. The things used in its production are power, machinery, and superintendence — chiefly power. Since the compressed air is then more costly than the power applied in its production it is but reasonable that we should give as much attention to its efficient use, or more, than we would to the use of steam, water power or electric energy. Yet in much of the practice in using compressed air this anticipation does not seem to be realized. This may be due to the fact that the exceeding convenience and safety of compressed air, and its labor-saving qualities in many applications, make efficiency as measured by foot-pounds, a secondary consideration; and from this a habit of wastefulness is formed. A further explanation may be found in its harmlessness and general good nature. A leaky air pipe, or excessive use at a motor, does not scald, suffocate, nor becloud the view as with steam; nor shock, burn nor start a fire as with electricity, nor flood and foul the premises as with water. Hence the user is apt to tolerate wastes of compressed air that should be checked to save the coal pile. In some lines of industry compressed air is supreme, in others electricity, and still in others steam, and water, each being specially adapted to do certain things better than any other, but in some lines the winner has not been decided, or even though decided in so far as present methods apply there may, any day, arise fresh competition by the invention of new devices or processes. COMPRESSED AIR CHAPTER I FORMULAS FOR WORK Art. 1. Temperature Constant or Isothermal Conditions. — From the laws of physics (Boyle's Law) we know that while the temperature remains unchanged the product pv remains constant for a fixed amount (weight) of air. Hence to determine the work done on, or by, air confined in a cylinder, when condi- tions are changed from piVi to p%v 2 we can write PlVl = p x v x = P2V2, sub x indicating variable intermediate conditions. 5 TF rfr 3 Fig. 1. Whence p x = and dW = p x Adl — p x dv x since Adl = dv; Vx dv A being the area of cylinder, therefore dW = piVi — -, and v x work of compression or expansion between points B and C (Fig. 1) is the integral of this, or 1 2 COMPRESSED AIR C V1 dv x W = PiVi I — - = P1V1 (l0g e Vi - log e V 2 ) = piVi log e -^ = pivi log e -* = pivi log e r = p 2 v 2 log e r. v 2 Pi Note that this analysis is only for the work against the front of the piston while passing from B to C. To get the work done during the entire stroke of piston from B to D we must note that throughout the stroke (in case of ordinary compression) air is entering behind the piston and following it up with pressure p\. Note also that after the piston reaches C (at which time valve / opens) the pressure in front is constant and = p 2 for the remainder of the stroke. Hence the complete expression for work done by, or against, the piston is PiVi log e r - piVi + p 2 v 2 ; but since piVi = p 2 v 2 , the whole work done is W = piVi log e r or p 2 v 2 log e r (1) Note that the operation may be reversed and the work be done by the air against the piston, as in a compressed-air engine, with- out in any way affecting the formula. Forestalling Art. 2, Eq. (4), we may substitute for pv in Eq. (1) its equivalent, 53.35 1, for 1 lb. of air and get for 1 lb. of dry air W = 53.35 1 Xlog e r (la) This may be adopted for common logs by multiplying by 2.3026. It then becomes W = (122.83 logio r) t (16) log 122.83 = 2.08930. Note that in solving by logs the log of log r must be taken. Values of the parenthesis in Eq. (16) are given in Table I, column 11. For the special temperature of 60°F. (16) becomes for 1 lb. of air W = 63,871 logio r (2) log 63,871 = 4.80536. Note that for moist air the coefficient in (la) is greater, being 53.87 for saturated air at 7Q°F. and under atmospheric pressure FORMULAS FOR WORK 3 = 14.7. For average conditions 53.5 would probably be about right. Example la. — What will be the work in foot-pounds per stroke done by an air compressor displacing 2 cu. ft. per stroke, com- pressing from p a = 14 lb. per square inch to a gage pressure = 701b.; compression isothermal, T = 60°F.? Solution (a). — Inserting the specified numerals in Eq. (1) it becomes W = 144 X 14 X 2 X log e 7 °^ 14 = 4,032 X 1.79 = 7,217. Solution (b). — By Tables I and II. By Table II the weight of a cubic foot of air at 14 lb. and 60° is 0.07277, and 0.07277 X 2 = 0.14554. The absolute t is 460 + 60 = 520, and r = 6.0. Then in Table I, column 11, opposite r = 6 we find 95.271, whence W = 95.271 X 520 X 0.14554 = 7,208. The difference in the two results is due to dropping off the fraction in temperature. Art. 2. Temperature Varying. — The conditions are said to be adiabatic when, during compression or expansion, no heat is al- lowed to enter in, or escape from, the air although the temperature in the body of confined air changes radically during the process. Physicists have proved that under adiabatic conditions the following relations hold: m = £ (3) P2V2 h and since for 1 lb. of air at 32°F. pv = 26,214 and t = 492, we get for 1 lb. of dry air at any pressure, volume and temperature, pv = 53.35f (4) While formulas (3) and (4) are very important, they do not apply to the actual conditions under which compressed air is worked, for in practice we get neither isothermal nor adiabatic conditions but something intermediate. Furthermore, moisture in the air will effect this coefficient. For such conditions physicists have discovered that the follow- ing holds nearly true : PiV! n = p x v x n = p 2 v 2 n (5) 4 COMPRESSED AIR sub x indicating any intermediate stage and the exponent n varying between 1 and 1.41 according to the effectiveness of the cooling in case of compression or the heating in case of expan- sion. From this basic formula (5) the formulas for work must be derived. dv As in Art. 1, dW = p x dv x = pivi n — - n = piVi n (v x ~ n ) dv x . v x Therefore W = pi«i" I v x ~ n dv x = p 1 v 1 n (^ 2 — J =piv 1 n { n _ ~ — j. Now since p%Vi n X v 2 l ~ n = p 2 v 2 n X v 2 l ~ n = p 2 v 2 and p\V\ n V]}-~ n = piVi the expression becomes w , = P2V2 - P1V1 n — 1 which represents the work done in compression or expansion be- tween B and C (Fig. 1). To this must be added the work of ex- pulsion, p 2 v 2 and from it must be subtracted the work done by air against the back side of the piston. In case of compression from free air this subtraction will be p a v a - Hence, the net work done in one stroke of volume v a is T jr P2V2 — PaVa , ,„. W = * _ h P2V2 - PaVa (6) This reduces to 71 W = ^ZZl (V*° 2 - VaVa) (7) By substituting from Eq. (4), Eq. (7) may be written, for work in compressing 1 lb. of air, Wi= — ^r 53.35 (h-t a ) (7a) n — 1 .' Whenn = 1.41, W x = 183 (t 2 - t a ) (76) When n = 1.25, Wi = 266 (t 2 - t a ) (7c) Equation (7) applies also to any cases of complete expansion, that is, when the air is expanded until the pressure within the cylinder equals that against which exhaust must escape. Equation (7) is in convenient form for numerical computations and may be used when the data are in pressures and volumes, but it is common to express the compression, or expansion, in terms of r. For such cases a more convenient form of equation is gotten as follows: FORMULAS FOR WORK 1 T) V V n From Eq. (5) by factoring out one v, p 2 v 2 = V2 n ~ L Al Vl V a n ,, - V a I Also r = — = — -. therefore - = r» p a v 2 n v 2 n—\ w — 1 n— 1 and — — t — r n , therefore p 2 v 2 = p a v a r n and Eq. (7) becomes w-l W = ^PaVa(r^- l) (8) In cases the higher pressure, p 2 , and the less volume, v 2 , are known, as may sometimes be the case in complete expansion en- gines, we would get by a similar process W = ^ T ^[l-( 1 r )^] (8a) Study of the derivation of Eqs. (7) and (8) shows that they are equally applicable to cases of complete expansion, that is, when the air within the cylinder is expanded until its pressure is equal to that of the air outside into which the exhaust takes place. In ordinary cases of expansion engines apply Eq. (9). In perfectly adiabatic conditions n = 1.41, but in practice the compressor cylinders are water-jacketed and thereby part of the heat of compression is conducted away, so that n is less than 1.41. For such cases Church assumes n = 1.33 and Unwin assumes n = 1.25. Undoubtedly the value varies with size and propor- tions of cylinders, details of water-jacketing, temperature of cooling water and speed of compressors. Hence precision in the value of n is not practicable. For 1 lb. of air at initial temperature of 60°F. Eq. (8) gives, in foot-pounds, Whenn = 1.41, W = 95,193 (r - 29 - 1) (86) When n = 1.25, W = 138, 405 (r - 2 - 1) (8c) Common log of 95,193 = 4.978606. Common log of 138,405 = 5.141141. Values of r - 2 and r - 29 are given in Table I, columns 5 and 6 respectively. . The above special values will be found convenient for approxi- mate computations. For compound compression see Art. 14. If in Eq. (8) we substitute for pv its value, 53.35 1, for 1 lb., we get for work on 1 lb. COMPRESSED AIR n-l where W = [(^l) 53 - 35 ( r n - 1 )] X ^ = Zi K =~yX 53.35 ( r ^ r - l). (8d) Note that Eq. (8cf) applies without change to cases of complete expansion provided that the temperature, h, of the exhaust be used and that r be determined to correspond, see Eqs. (12) and (12a). Table I gives values of K for n = 1.25 and n = 1.41 and for values of r up to 10, varying by one-tenth. The theoretic work *— sac= r (*)/ f e d c Fig. 2. in any case is K X Q X t, where Q is the number of pounds passed and t is the absolute initial temperature. Further explanation accompanies the table. The difference between isothermal and adiabatic compression (and expansion) can be very clearly shown graphically as in Fig. 2. In this illustration the terminal points are correctly placed for a ratio of 5 for both the compression and expansion curve. Note that in the compression diagram (a) , the area between the two curves aef represents the work lost in compression due to heating, and the area between the two curves aeghb in (6) repre- FORMULAS FOR WORK 7 sents the work lost by cooling during expansion. The isothermal curve, ae, will be the same in the two cases. Such illustrations can be readily adapted to show the effect of reheating before expansion, cooling before compression, heating during expansion, etc. For platting curves, see Art. 4a. Example 2a. — What horsepower will be required to compress 1,000 cu. ft. of free air per minute from p a = 14.5 to a gage pres- sure = 80, when n = 1.25 and initial temperature = 50°F.? Solution. — From Table II, interpolating between 40° and 60° the weight of 1 cu. ft. is 0.07686 and the weight of 1,000 is 76.86-. The r from above data is 6.5. Then in Table I opposite r = 6.5 in column 9 we find 0.3658. Then Horsepower = 0.3658 X 76.86 100 X510 = 143. The student should check this result by Eqs. (8) or (8d) and (106) without the aid of the table. Art. 3. Incomplete Expansion. — When compressed air is applied in an engine as a motive power its economical use requires that it _Y„ J Fig. 3. be used expansively in a manner similar to the use of steam. But it is never practicable to expand the air down to the free air pres- sure, for two reasons : first, the increase of volume in the cylinders would increase both cost and friction more than could be balanced by the increase in power; and second, unless some means of re- heating be provided, a high ratio of expansion of compressed air will cause a freezing of the moisture in and about the ports. The ideal indicator diagram for incomplete expansion is shown in Fig. 3. In such diagrams it is convenient and simplifies the demonstrations to let the horizontal length represent volumes. In any cylinder the volumes are proportional to the length. 8 COMPRESSED AIR Air at pressure P2 is admitted through that part of the stroke represented by v 2 — thence the air is expanded through the re- mainder of the stroke represented by v 1} the pressure dropping to pi. At this point the exhaust port opens and the pressure drops to that of the free air. The dotted portion would be added to the diagram if the expansion should be carried down to free air pressure. To write a formula for the work done by the air in such a case we will refer to Eq. (6) and its derivation. In the case of simple compression or complete expansion it is correctly written ttt V&2 — PaVa , W = ^ p- V&2 - PaVa, n — 1 which would give work in the case represented by Fig. 1 when there is a change of temperature, but in such a case as is repre- sented by Fig. 3 the equation must be modified thus : w P2V2 — P1V1 . , n s W = ^ h P2V2 - PaVl (9) it A. the reason being apparent on inspection. In numerical problems under Eq. (9) there will be known p2V2,n, and either pi or v\. The unknown must be computed from the relations from Eq. (5) : p^p^y or v 1 = v 2 (f) Table I, columns 1, 2, 3 and 4, is designed to reduce the labor of this computation. Example 3a. — A compressed-air motor takes air at a gage pressure = 100 lb. and works with a cut-off at J4 stroke. What work (foot-pounds) will be gotten per cubic foot of compressed air, assuming free air pressure = 14.5 lb. and n = 1.41? Solution. — Applying Eq. (9) and noting that all pressures are to be multiplied by 144 and that the pressure at end of stroke l\£\ 1.41 = p\ = 114.5 (y 1 ) = 16.3 and that vi = 4v 2 , we get / 114.5 X 1 - 16.3 X 4 \ 0.41 ^ 114.5 X 1 - 14.5 X 4) = 25,444. FORMULAS FOR WORK 9 Art. 4. Work as Shown by Indicator Cards. — Volumes have been written on indicators and indicator cards but more than the following brief notes would be out of place here : Let li = length in inches out to out, horizontally, of indicator card, = length in feet of piston stroke, = spring number = pounds per inch, = area in square inches of indicator diagram, = area in square inches of piston. Then work per stroke is (10) I s a A W h a As When a planimeter is available, it is the quickest and most reliable means of determining the area, a, provided the operator know the planimeter constant. This can easily be found by run- ning the planimeter several times round an accurately drawn figure of known area, as for instance a square of 2-in. sides, and averaging the readings. The repetition should be made without lifting the tracer from the paper. It is necessary only to read the vernier each time the tracer reaches a fixed point on the figure. Then subtract each reading from the next succeeding one. The several differences reveal whether or not the instrument, in the hands of the individual can be depended on to give reliable results. The sum of the differences divided by the number of Vernier Difference Error Per cent. Reading at start 794 184 578 964 362 753 390 394 386 398 391 1 392 3 392 5 392 7 392 392 Pass zero -0.25 After first round After second round -0.75 After third round -1.28 Pass zero After fourth round -1.80 After fifth round 0.00 Total for five rounds, 1,959 1 959 4 00 ' r = 392 = average and o^ = 1-03 = planimeter constant. 10 COMPRESSED AIR repetitions gives the average and the average divided into the known area gives the planimeter constant. Example. — The table, page 9, shows records and deductions for a planimeter tested on a rectangle of 4 sq. in.: To get the full information shown by an indicator diagram taken from an air compressor, there should be placed on it the clearance line and the isothermal curve. For direct determina- tion of clearance see Art. 46. A 10 Fig. 4. Referring to Fig. 4, the clearance line, OC, is placed at a dis- DO tance, DO, such that -ry. = percentage of clearance. If clear- ance has been determined by measurements in the machine, the line OC is set out by measuring a distance CB as determined above. If clearance has not been measured in the machine, the position of the line OC or point can be computed as follows : Scale one of the pressure ordinates where the curve is smooth. Represent this by p x ; the distance from its foot to the unknown point represent by x and the known distance from A to the foot of p x by k. FORMULAS FOR WORK 11 Then by Eq. (5) p a + k) n = p x x n or x + k = l—) n i 1 l X. \p Whence x =- j (P±)n- 1 \pj (a) This method is of course dependent on the assumed value of n. By the same principle, if can be correctly located independ- ently of n, the value of n can be computed thus: x is now known. So let x -f- k = I. rpi • I" Px j log Px - l Og Pa m Then — = — and n = ° 7 , (o) rc n p a log £ — log # In large well-designed air compressors the clearance should not exceed 1 per cent. Then OC would very nearly coincide with DB but would always be a little outside. Note that if x from Eq. (a) places inside of D, it is evidence that n has been assumed too small. In many books on steam engines and air compressors can be found instructions for locating the point graphically. Con- cerning these, the student is warned that they are all based on the assumption that the curve is the isothermal; and hence are apt to give very misleading results. For instance, one method is to construct a rectangle on the curve as developed by the indicator, as shown in mnqr, and the diagonal nr will pass through 0. This would be correct for the isothermal EF but evidently not so for the actual curve EG. Before passing, the student's attention should be directed to the fact that in steam engines the clearance is usually much greater than is allowed in air compressors. In steam engines it varies much and sometimes goes to 10 or even 15 per cent. ; and further, the curves of steam-engine indicator cards are much more erratic than for air compressors, due to condensation during expansion and compression without cooling, which may cause reevapora- tion. After all is said, if the investigator wants to know what the clearance is he should measure it in the machine. The curves, either isothermal or adiabatic, can most readily be set out by dividing the line OA into ten equal parts numbered as shown, then letting x = number of divisions from the pres- sure ordinate at the xth division will be, for the isothermal curve : 12 COMPRESSED AIR Px = and for the adiabatic curve p x = p a ( — ) x The following table applies where p a — 14.7: x = 10 9876543 2 Isothermal p x = 14.7 16.3 18.2 21.0 24.5 29.4 36.6 49.0 78.5 Adiabatic p x = 14.7 17.1 20.1 23.5 29.3 39.1 53.5 80.3 142.1 The isothermal curve is symmetrical about the middle line and the upper half can be set out from the other axis OB with the same ordinates used to plat the first half. Art. 4a. Mean Effect Pressures.— In much of the literature relating to work done in steam engines and air compressors, use is made of the term "mean effective pressure" — abbreviated m.e.p. A definition of the term is: A pressure which multiplied by the volume, or piston displacement, gives work. Then to find the m.e.p. in compression when the volume is v we have: In isothermal compression (or expansion) (m.e.p.)v = p a v log e r and m.e.p. = p a log e r = 2.3p a logio r. (10) In case of adiabatic compression (or complete expansion) (n — 1 . r n — lj (n — 1 . r n — lj (10a) n — 1 Values ofr - _ are given in Table I, column 5, for n = 1.25. n In case of incomplete expansion (Eq. 9) when the cutoff is at k per cent, of the stroke, or v 2 = kv. , v p 2 kv — piv 7 (m.e.p.) v = — f [- p 2 kv — p a v. lb J. From the condition that p\V n = p%vi n , we get whence the above equation reduces to p 2 (kn — k n ) (106) — --> — - - Pa n — 1 To reduce computations of m.e.p. in this case to simple arith- m.e.p. = - — ^j p a n — 1 FORMULAS FOR WORK 13 metic, the values of kn — k n are given below for n = 1.25 and for n — 1 a sufficient range in k to meet the demands of ordinary practice. These values will apply to any gas, including steam so long as there is no condensation of the steam. f Fraction {_ Decimal kn - k n 0.1250 0.3287 0.1875 0.4439 H 0.2500 0.5428 5ie 0.3125 0.6281 3 A 0.3750 0.7006 yi6 0.4375 0.7643 0.5000 0.8180 ?18 0.5625 0.8641 0.6250 0.9022 Art. 4b. Effect of Clearance in Compression. — It is not prac- ticable to discharge all of the air that is trapped in the cylinder. There are some pockets about the valves that the piston cannot enter, and the piston must not be allowed to strike the head of the cylinder. This clearance can usually be determined by measuring the water that can be let into the cylinder in front of the piston when at the end of its stroke; but the construction of each com- pressor must be studied before this can be undertaken intelli- gently, and it is not done with equal ease in all machines. To formulate the effect of this clearance in the operation of the machine, Let v = volume of piston displacement (= area of piston X length of stroke), Let cv = clearance, c being a percentage. Then v + cv is the volume compressed each stroke. But i the clearance volume cv will expand to a volume r n cv as the piston recedes, so that the fresh air taken in at each stroke will 'i be v + cv — r n cv, and the volumetric efficiency will be E v = v + cv i r n cv = 1 + c (1 - r n ) (ID Theoretically (as the word is usually used) clearance does not cause a loss of work, but practically it does, insomuch as it requires a larger machine, with its greater friction, to do a given amount of effective work. Example 46. — A compressor cylinder is 12-in. diameter by 16-in. stroke. The clearance is found to hold 134 pt- of water = -~ X 231 = 36 cu. in., therefore c = ^^ v 1fi = 0.02. 14 COMPRESSED AIR Then by Eq. (11) when r = 7 and n = 1.25. E = 1 '+ 0.02 (1 - 7 - 8 ) = 92 per cent. Such a condition is not abnormal in small compressors, and the volumetric efficiency is further reduced by the heating of air during admission as considered in Art. 6. Art. 5. Effect of Clearance and Compression in Expansion Engines. — Figure 5 is an ideal indicator diagram illustrating the effect of clearance and compression in an expansion engine. In this diagram the area E shows the effective work, D the effect of clearance, B the effect of back pressure of the atmosphere and C the effect of compression on the return stroke. The study of effect of clearance in an expansion engine differs from the study of that in compression, due to the fact that the b — bl— ^ Fig. 5. volume in the clearance space is exhausted into the atmosphere at the end of each stroke. If the engine takes full pressure throughout the stroke the air (or steam) in the clearance is entirely wasted; but when the air is allowed to expand as illustrated in the diagram some useful work is gotten out of the air in the clearance during the expansion. The loss due to clearance in such engine is modified by the amount of compression allowed in the back stroke. If the com- pression p c = p 2 , the loss of work due to clearance will be noth- ing, but the effective work of the engine will be considerably reduced, as will be apparent by a study of a diagram modified to conform to the assumption. While the formula for work that includes the effect of clear- ance and compression will not be often used in practice its deriva- tion is instructive and gives a clear insight into these effects. FORMULAS FOR WORK 15 The symbols are placed on the diagram and will not need fur- ther definition. The effective work E will be gotten by subtracting from the whole area the separate areas B, C and D. From Art. 2, after making the proper substitutions for the volumes, there results m x 1 ,rP2(c + fc) — pi(l+c) , / , 7 sl Total area = l\ — _ i h y% (c +.&) . Area 5 = Zp a , Area D = lp 2 c, Area C = l[ W ~ * <» + 3 _ pa5 ] Subtracting the last three from the first and reducing their results : Work 1 —j^ = ^ZT\ t c (Pa + Pa - 2? c - Pi) + n (p 2 & + pJ>-p a ) - (pi — p )] = mean effective pressure. The actual volume ratio before and after expansion is V2 _ c?i + kit _ c + k V\ eh + li c + 1 This is the ratio with which to enter Table I to get r and t and from r the unknown pressure pi. Similarly, for the compression curve the ratio of volumes is r> and p c can be found as indicated above. Art. 5a. Adjustment of Mechanically Operated Intake Valves. — While no attempt will be made to show details of valves, it is appropriate to call attention here to the fact that discharge valves to an air compressor are nearly always of the "poppet" type, that is they pop open automatically when the pressure in- side the compressor exceeds that in the receiver. Thus the time, or point of opening of the discharge valve, will adjust itself to any variation of pressure in the receiver, a condition evidently desir- able. But the intake valves may be mechanically operated, and are so operated in many of the larger machines. When so oper- ated, the machine works more efficiently, with less noise and there is less liability to breakdown. The correct adjustment of the point of opening of mechanically operated inlet valves depends on the clearance and the ratio of compression. 16 COMPRESSED AIR Figure 5a illustrates one class of inlet valve with its operating mechanism, the direction of motion being indicated by arrows. The piston d is at the left end of its stroke with compressed air in the clearance. If the port of valve A opens at this instant, the compressed air in the clearance will escape out into the inlet passage e. Evidently it is desirable to delay the opening of the port until the piston has receded enough to allow the air in the clearance to expand down to atmospheric pressure, thus letting the air in the clearance give back the work done in compressing it. Evidently, the opening should not be delayed longer, for there would result a suction (pressure below atmosphere) behind the piston which would cause a loss of work. When the adjust- ment is correct, there is no puffing or spitting at the inlet parts. When the adjustments are not correct, an experienced operator can detect the fact by the noise made by air puffing out or into the parts. SECTION .MN" SECTION QK Fig. 5a. If the clearance and the ratio of compression are known, the erector or operator can adjust the valves correctly. For example, assume the clearance as 1 per cent., r = 7 and n = 1.25. In Table I for r = 7, v 2 -=- Vi = 0.21 or say vi = 5v 2) that is the clear- ance should expand to five times its volume before the port opens. Otherwise stated, the piston should move back 4 per cent, of its stroke before the port opens. Thus, if the stroke be 18 in. the piston should be moved back 0.04 X 18 = 0.72 (or say Y± in.) and while the piston stands in that position bring the edge of valve and edge of inlet port to coincide by turning the rod 6 or a as the case may be. The manufacturers always put marks on the end of the valve and on the inclosing cylinder that will enable the operator to make this adjustment. In order that one adjustment may not interfere with another, it is necessary that the valve B be adjusted first by rotating FORMULAS FOR WORK 17 rod a; then adjust valve A by rotating rod 6. If the compressor be a compound tandem, adjust the valves in the order of their distance from the eccentric. Art. 6. Effect of Heating Air as it Enters Cylinders. — When a compressor is in operation all the metal exposed to the compressed air becomes hot even though the water-jacketing is of the best. The entering air comes into contact with the admission valves, cylinder head and walls and the piston head and piston rod, and is thereby heated to a very considerable degree. In being so heated the volume is increased in direct proportion to the absolute temperature (see Eq. (3)), since the pressure may be assumed constant and equal that of the atmosphere. Hence a volume of cool free air less than the cylinder volume will fill it when heated. This condition is expressed by the ratio V a "a "a - ■ = — or V a = Vc -> Vc Zc tc where v c and t c represent the cylinder volume and temperature. The volumetric efficiency as effected by the heating is V c t c Example 6. — Suppose in Ex. 4a the outside free air tempera- ture is 60°F. and in entering the temperature rises to 160°F., then t a 460 + 60 .. , ^ = 460 + 160 = 84perCent ' Then the final volumetric efficiency would be 92 X 84 = 77 per cent, nearly. The volumetric efficiency of a compressor may be further re- duced by leaky valves and piston. In Arts. 4& and 6 it is made evident that the volumetric efficiency of an air compressor is a matter that cannot be neglected in any case where an installation is to be intelligently proportioned. It should be noted that the volumetric efficiency varies with the various makes and sizes of compressors and that the catalog volume rating is always based on the piston displacement. These facts lead to the conclusion that much of the uncertainty of computations in compressed-air problems and the conflict- ing data recorded is due to the failure to determine the actual 18 COMPRESSED AIR amount of air involved either in terms of net volume and tempera- ture or in pounds. Methods of determining volumetric efficiency of air compres- sors are given in Chapter II. The loss of work due to the air heating as it enters the com- pressor cylinder is in direct proportion to the loss of volumetric efficiency due to this cause. In Ex. 6a only 84 per cent, of the work done on the air is effective. By the same law any cooling of the air before entering the compressor effects a saving of power. See Art. 10. Art. 7. Change of Temperature in Compression or Expan- sion. — Equation (4) may be written for any fixed weight of air PlVl = Ctl) 2?2^2 = cto and Eq. (5) may be factored thus, P\ViVi n ~ l = J>lV 2 V 2 n ~ l . Substituting we get CtiVi n ~ l = ct 2 v 2 n ~ 1 . Whence U = *i(^)"~ X (12) and . t 2 = *i(— )~"~ = tif»> (12a) i since from Eq. (5) — = ( — ) n - It is possible to compute n from Eq. (12) by controlling the V\ and v 2 and measuring t\ and t 2 . Table I, columns 5 and 6, is made up from Eq. (12a) and columns 3 and 4 from Eq. (5) as just written. Example 7. — What would be the temperature of air at the end of stroke when r = 7 and initial temperature = 70°F. ? Solution. — Referring to Table I in line with r = 7 note that 1.4758 when n = 1.25 :.h= (460 + 70) X 1.4758 - 460 = 322°F. 1.7585 when n = 1.41 :.t%= (460 4- 70) X 1.7585 - 460 = 472°F. From the same table the volume of 1 cu. ft. of free air when compressed and still hot would be respectively 0.21 and 0.25, FORMULAS FOR WORK 19 while when the compressed air is cooled back to 70° its volume would be 0.143. Art. 8. Density at Given Temperature and Pressure. — By Eq. (4) pv = 53.35 for 1 lb., and the weight of 1 cu. ft. = 1 lb. divided by the volume of 1 lb. Therefore w = ] = ^ (13) Note that p must be the absolute pressure in pounds per square foot, and t the absolute temperature. When gage pressures are used and ordinary Fahrenheit temperature the formula becomes w- 144 /^ + ?M 53.35 V460 + Fl Table III is made up from Eq. (13). Art. 8a. Weight of Moist Air. — In some cases where unusual refinement of calculations may be required it will be necessary to take cognizance of the fact that air containing water vapor is not equal in weight to pure air at the same pressure and temperature. In case of moving air at atmospheric pressure as in case of fans and blowers and where air is measured at atmospheric pressure by means of orifices, the error resulting from neglecting moisture may be as much as l}^ per cent. Since atmospheric pressure and water-vapor pressures are usually recorded in inches of mercury it will be convenient to re- tain in the formulas inches head of mercury instead of pounds per square inch. Let m = pressure (absolute) of mixture of air and water vapor in inches of mercury, q = pressure of saturated water vapor at given tempera- ture in inches of mercury (to be found in steam tables), H = percentage of humidity, K = ratio of weight of water vapor to dry air at given temperature, t = absolute temperature = 459.6 + F. To adopt Eq. (13), viz., W a = p .4- 53.35 £ to this case, note that 2.036 in. of mercury gives 1 lb. pressure and that Hq ( = 20 COMPRESSED AIR vapor pressure at H humidity) must be subtracted from p in order to get the true pressure of the air. _ 144 (m - Hq) 1.3253 (m - Hq) men w a 2m& x ^^ - f and weight of water vapor in a cubic foot is 1.3253 VTJ w w = — - — KHq. Then the combined weight is w a + w w = w = - 1 — — [m — Hq (1 — K)] (13&) t For values of K see Table Ilia, page 134, which is copied from Engineering News, June 18, 1908, or Compressed Air Magazine, vol. 13, p. 4967. These articles give also a very full treatment of the subject of moisture in air. The ratio K varies between 0.611 at zero degrees F. and 0.623 at 100°F. Some writers assume it constant. If we assume it constant and equal 0.62 (which is correct for temperature 74°), then the equation becomes w = ^^ (w - 0.38tfg) (13c) Example 13c. — Find the weight per cubic foot of air in a duct leading away from a fan when T = 70°F. Barometer reading in free air = 28.85-in.-water gage, (i), = 4 in. and humidity, (H), = 80 per cent. Solution. — 4 in. of water = 0.29 in. of mercury. Then m = 29.14. At 70° K = 0.6196 and 1 - K = 0.3804. At 70° q = 0.739 and Hq = 0.5912. 1 QQ'lS Then w = -~^ (29.14 - 0.5912 X 0.3804) = 0.07233. Pure air under the same pressure and temperature would have w a = 0.07287, a difference of less than 1 per cent. If the air were saturated the difference would be greater. Art. 9. Cooling Water Required. — In isothermal changes, since pv is constant, evidently there is no change in the mechanical energy in the body of air as measured by the absolute pressure and using the term " mechanical energy" to distinguish from heat energy. Hence evidently all the work delivered to the air from FORMULAS FOR WORK 21 outside must be abstracted from the air in some other form, and we find it in the heat absorbed by the cooling water. Therefore, pv log e r 778 = (B.t.u.'s) of work must be absorbed by the cooling water. If the water is to have a rise of temperature T° (T being small, else the assumption of isothermal changes will not hold), then Pv\oS e V 7R0 T = P oun< ^ s °f water required in same time. Example. 9 — How many cubic feet of water per minute will be required to cool 1,000 cu. ft. of free air per minute, air compressed from p a = 14.2 to p g = 90° gage, initial temperature of air = 50°F. and rise in temperature of cooling water = 25°? Solution. — / 90 -4- 14 2 \ 144 X 14.2 X 1,000 X log e ( ^ g ) 780 X 25 X 62.5 = 3 " 36 CU " ft per minute. It is practically possible to attain nearly isothermal conditions by spraying cool water into the cylinder during compression. In such a case this article would enable the designer to compute the quantity of water necessary and therefrom the sizes of pipes, pumps, valves, etc. Art. 10. Reheating and Cooling. — In any two cases of change of state of a given weight of air, assuming the ratio of change in pressure to be the same, the work done (in compression or expan- sion) will be directly proportional to the volume, as will be evident by examination of the formulas for work. Also, at any given pressure the volumes will be directly proportional to the absolute temperatures. Hence the work done either in compression or expansion (ratio of change in pressures being the same in each case) will be directly p: portional to the absolute initial temperatures. Thus if the temperature of the air in the intake end of one com- pressor is 150°F. and, in another 50°F., the work done on equal weights of air in the two cases will be in the proportion of 460 + 150 to 460 + 50, or 1.2 to 1; that is, the work in the first case is 20 per cent, more than that in the second case. This is equally true, of course, for expansion. 22 COMPRESSED AIR The facts above stated reveal a possible and quite practicable means of great economy of power in compressing air and in using compressed air. The opportunities for economy by cooling for compression are not as good as in heating before the application in a motor, but even in compression marked economy can be gotten at almost no cost by admitting air to the compressor from the coolest con- venient source, and by the most thorough water-jacketing with the coolest water that can be conveniently obtained. In all properly designed compressor installations the air is supplied to the machine through a pipe from outside the building to avoid the warm air of the engine room. In winter the differ- ence in temperature may exceed 100°, and this simple device would reduce the work of compression by about 20 per cent. Gas, Liquid or -y j Powdered Fuel — *" I — » L > A Hot Air Combustion JLJLfi_sx jUSLSLJSLju Fig. 6. For the effect of intercoolers and interheaters see Art. 11 on compounding. By reheating before admitting air to a compressed-air engine of any kind the possibilities of effecting economy of power are greater than in cooling for compression, the reason being that heating devices are simpler and less costly than any means of cooling other than those cited above. The compressed air passing to an engine can be heated to any desired temperature; the only limit is that temperature that will destroy the lubrication. Suppose the normal temperature of the air in the pipe system is 60°F. and that this is heated to 300°F. before entering the air engine, then the power is increased 46 per cent. Reheating has the further advantage that it makes possible a greater ratio of expansion without the temperature reaching freezing point. The devices for reheating are usually a coil or cluster of pipes through which the air passes while the pipe is exposed to the heat FORMULAS FOR WORK 23 of combustion from outside. Ordinary steam boilers may be used, the air taking the place of the steam and water. Unwin suggests reheating the air by burning the fuel in the compressed air as suggested in the cut. Even when the details are worked out such a device would be simple and inexpensive. The theoretic advantages of such a device are that all the heat would go into the air, the gases of com- bustion (if solid or liquid fuel be used) would increase the volume, and the combustion occurring in compressed air would be very complete. The author has no knowledge of any such devices having been used in practice. 1 The power efficiency of the fuel used in reheaters is very much greater than that of the fuel used in steam boilers. Unwin states that it is five or six times as much. The chief reason is that none of the heat is absorbed in evaporation as in a steam boiler. In many of the applications of compressed air reheating is im- practicable, and efficiency is secondary to convenience — but in large fixed installations, such as mine pumps, reheating should be applied. Art. 11. Compounding. — In steam-engine designs compound- ing is resorted to to economize power by saving steam, while in air compressors and compressed-air engines compounding is resorted to for the twofold purpose of economizing power and controlling temperature, both objects being accomplished by reducing the extreme change of temperature. The economic principles in- volved in compound steam engines and in compound air engines are quite different, the reasons underlying the latter being much more definite. The air is first compressed to a moderate ratio in the low- pressure cylinder, whence it is discharged into the "inter cooler," where most of the heat developed in the first stage is absorbed and thereby the volume materially reduced, so that in the second stage there will be less volume to compress and a less injurious temperature. The changes occurring and the manner in which economy is 1 Since the publication of the first edition a very promising device has appeared in which the current of compressed air automatically injects the fuel oil; thus, presumably, maintaining a constant proportion between the quantity of air and of oil, so that the temperature of the discharged air will be constant. 24 COMPRESSED AIR effected in compression may be most easily understood by refer- ence to Fig. 7, which represents ideal indicator diagrams from the two cylinders, superimposed one over the other, the scale being the same in each, the dividing line being kb. In this diagram, abk is the compression line in the first-stage or low-pressure cylinder, cds is the compression line in the second-stage or high-pressure cylinder, be is the reduction of volume in the intercooler, with pressure constant, e d f g Fig. 7. abf would be the pressure line if no intercooling occurred, The area cdfb is the work saved by the intercooler, ace would be the compression line for isothermal compression, aug would be the compression line for adiabatic compression. The diagram is correctly proportioned for r = 6. Figure 8 is a diagram drawn in a manner similar to that used in Fig. 7 and is to illustrate the changes and economy effected by compounding with heating when compressed air is applied in an engine. It is assumed that the air is "preheated," that is, heated once before entering the high-pressure cylinder and again heated between the two cylinders. FORMULAS FOR WORK 25 In this diagram, se is the volume of compressed air at normal temperature, sf is the volume of compressed air after preheating, fc is the expansion line in the high-pressure cylinder, cb is the increase of volume in the interheater, by is the expansion line in low-pressure cylinder, ezq would be the adiabatic expansion line without any heating, efcz is work gained by preheating, cbyx is work gained by interheating. In no case is it economical to expand down to atmospheric pressure. Hence the diagram is shown cut off with pressure still above that of free air. s e f \ * V \ V » V » V v V * \\ \ » * \ \ * p s ] > 1 Pa , 4 Pa "■"■- i y Fig. 8. The diagram, Fig. 8, is proportioned for preheating and reheat- ing 300°F. Art. 12. Proportions for Compounding. — It is desirable that equal work be done in each stage of compounding. If this con- dition be imposed, Eq. (8) indicates that the r must be the same in each stage, for on the assumption of complete intercooling the product pv will be the same at the beginning of each stage. If then ri be the ratio of compression in the first stage, the pressure at end of first stage will be rip = p h and the pressure at end of second stage = npi = ri 2 p a = p%, and similarly at end of third stage the pressure will be p 3 = ri z p a , or 26 COMPRESSED AIR In two T stage work r\ = (— Y = r 2 1/i# In three-stage work r x = ( — | ^ = r 3 * vp Let yi = free air intake per stroke in low-pressure cylinder or first stage, v 2 = piston displacement in second stage, v s = piston displacement in third stage, r.i = ratio of compression in each cylinder. Then, assuming complete intercooling, Vi , V 2 Vi v 2 = ~ and v s = — = — -j or IH 1 ,V 3 1 — = — and — = — 5* vi rx vi r x 2 The length of stroke will be the same in each cylinder; there- fore the volumes are in the ratio of the squares of diameters, or dl _ 1. A dl _ 1_ dji - ^ allCl di 2 ~ ri 2" Hence d 2 = -Aj and d z = — (14) ri /2 r x If the intention to make the work equal in the different cylin- ders be strictly carried out it will be necessary to make the first- stage cylinder enough larger to counteract the effect of volu- metric efficiency. Thus if volumetric efficiency be 75 per cent., the volume (or area) of the intake cylinder should be one-third larger. Note that the volumetric efficiency is chargeable entirely to the intake or low-pressure cylinder. Once the air is caught in that cylinder it must go on. Example 12. — Proportion the cylinders of a compound two- stage compressor to deliver 300 cu. ft. of free air per minute at a gage pressure = 150. Free air pressure = 14.0, r.p.m. = 100, stroke 18 in., piston rod 1% in. diameter, volumetric efficiency = 75 per cent. Solution. — From the above data the net intake must be 3 cu. ft. per revolution. Add to this the volume of one piston rod stroke ( = 0.025 cu. ft.), and divide by 2. This gives the volume of one piston stroke 1.512. The volume of 1 ft. of the cylinder will be FORMULAS FOR WORK 27 12 jg X 1.512 = 1.008 cu. ft. From Table X the nearest cylin- der is 14 in. in diameter, the total ratio of compression = 77 = 11.71, and the ratio in each stage is (11.71) I/§ = 3.7 = r h and by (14) v di 14 _ . _ = (nP = L92 = m '' say A lrlv for the high-pressure cylinder. Now we must increase the low-pressure cylinder by one-third to allow for volumetric efficiency. The volume per foot will then be 1.344, which will require a cylinder about 15% in. in diame- ter. Note that the diameter of the high-pressure cylinder will not be affected by the volumetric efficiency. Art. 13. Work in Compound Compression. — Assuming that the work is the same in each stage, Eq. (8) can be adapted to the case of multistage compression thus : In two-stage work ... n-l 1 = ^T- l VaVa(r^- l) X 2. (15a) In three-stage work n-l W = ~j p a v a (r^ - l) X 3 (16) W -^VaVain "-1 X2 (15) n — 1 n-l V n— i v a (r, *"" - l) X 3 (16a) Note that r 2 = — and r 3 = — and also that p a v a = ViVi = Pa Pa p 2 v 2 , etc., assuming complete intercooling. Laborious precision in computing the work done on or by com- pressed air is useless, for there are many uncertain and changing factors; n is always uncertain and changes with the amount and temperature of the jacket water, the volumetric efficiency, or actual amount of air compressed, is usually unknown, the value of p a varies with the altitude, and r is dependent on p a . Art. 14. Work under Variable Intake Pressure. — There are some cases where air compressors operate on air in which the in- 28 COMPRESSED AIR take pressure varies and the delivery pressure is constant. This is true in case of exhaust pumps taking air out of some closed vessels and delivering it into the atmosphere. It is also the con- dition in the "return-air" pumping system in which one charge of air is alternately forced into a tank to drive the water out and then exhausted from the tank to admit water. For full mathe- matical discussion of this pump see Trans. Am. Soc. C. E., vol. 54, p. 19. The formulas of Arts. 14 and 15 were first worked out to apply to that pumping system. In such cases it is necessary to determine the maximum rate of work in order to design the motive power. First assume the operation as being isothermal. Then in Eq. (1), viz., W = p x v\og e ~^> Px p x is variable, while v and py are constant. In this formula W becomes zero when p x is zero and again when p x = pi, since log 1 is zero. To find when the work is maximum, differentiate and equate to zero; thus differential of v (p x log e px - Px log e p x ) = v |^log e pidp x - (p x ^ + log, p x dp x .J J • Equate this to zero and get log e pi = 1 + log e p X) or log e — = 1, therefore ^ = e = 2.72. Px Px That is, when r = 2.72 the work is a maximum. When the temperature exponent n is to be considered the study must be made in Eq. (8), viz. W = p x v 71—1 (21)— -1 (8) ■Pi Differentiating this with respect to p x and equating to zero, w-l the condition for maximum work becomes ( -j j r = n. Insert this in (8) and the reduced formula becomes TXT VlV W = np x v. = — — FORMULAS FOR WORK 29 From the above expression for maximum the following results : When n = 1.41 the maximum occurs when r = 3.26. When n = 1.25 the maximum occurs when r = 3.05. When n = 1.00 the maximum occurs when r = 2.72. In practice r = 3 will be a safe and convenient rule. Exercise 14a. — Air is being exhausted out of a tank by an ex- haust pump with capacity = 1 cu. ft. per stroke. At the begin- ning the pressure in the tank is that of the atmosphere = 14.7 lb. per sq. in. Assume the pressure to drop by intervals of 1 lb. and plot the curve of work with p x as the horizontal ordinate and W as the vertical, using the formula W = p x vlog e —- Vx Exercise 146. — As in 14a plot the curve by Eq. (8) with n = 1.25. Art. 15. Exhaust Pumps. — In designing exhaust pumps the following problems may arise. Given a closed tank and pipe system of volume V under pres- sure po and an exhaust pump of stroke volume v, how many strokes will be necessary to bring the pressure down to p m ? The analytic solution is as follows, assuming isothermal condi- tions in the volume V. The initial product of pressure by volume is poV. After the first stroke of the exhaust pump this air has expanded into the cylinder of the pump and pressure has dropped to pi. Under the law that pressure by volume is constant; (V + v) pi = p V, or p! = ^r v at end of first stroke, PiV l V at end of second stroke, V ( V at end of third stroke, etc. Finally ViV I V \ 2 (V + v)p 2 = Vl V, or p 2 = y-^ = v*\Y+r v ) f second stroke, V I V \ (V + v) p z = p 2 V, or p 3 = p 2 y^±T v = P° \v^f~v) 1 Vr log r I V \ m , & p log (v + y 30 COMPRESSED AIR This is inconvenient for solution on account of the minus charac- teristics. Hence it is better to write it thus: _ log p m - log p . fro — 1.16136 log 101 = 2.00432 0.69897 log 100 = 2.00000 0.46239 0.00432 46239 432 = 107 = m. log V - log (V + v) Now change sign of both numerator and denominator and we get » = lo , s /°-, log f- (17) log ( V + v) — log V Example 15a. — A closed tank containing 100 cu. ft. of air at atmospheric pressure (14.7 lb.) is to be exhausted down to 5 lb. by a pump making 1 cu. ft. per stroke. How many strokes are required? Solution.— po = 14.7, p m ' = 5, V + v = 101 and V = 100. log 5 The results found under Arts. 14 and 15 serve well to illustrate the curious mathematical gymnastics that compressed air is sub- ject to, and should encourage the investigator who likes such work, and should put the designer on guard. Art. 16. Efficiency when Air is Used without Expansion. — In many applications of compressed air convenience and reliability are the prime requisites, so that power efficiency receives little attention at the place of application. This is so with such appara- tus as rock drills, pneumatic hammers air hoists and the like. The economy of such devices is so great in replacing human labor that the cost in power is little thought of. Further, the necessity of simplicity and portability in such apparatus would bar the complications needed to use the air expansively. There are other cases, however, notably in pumping engines and devices of vari- ous kinds, where the plant is fixed, the consumption of air con- siderable and the work continuous, where neglect to work the air expansively may not be justified. In any case the designer or purchaser of a compressed-air plant should know what is the sacrifice for simplicity or low first cost when the proposition is to use the air at full pressure throughout the stroke and then'exhaust the cylinder full of compressed air. FORMULAS FOR WORK 31 Let p be the absolute pressure on the driving side of the piston and p a be that of the atmosphere on the side next the exhaust. Then the effective pressure is p — p a and the effective work is (p ~ Pa) v, while the least possible work required to compress this air is pv log e r. Hence the efficiency is E = (P ~ Pa) V pv loge r Dividing numerator and denominator by p a v this reduces to T — 1 E = ~ — - (18) r loge r This is the absolute limit to the efficiency when air is used without expansion and without reheating. It cannot be reached in practice. Table VI represents this formula. Note that the efficiency de- creases as r increases. Hence it may be justifiable to use low- pressure air without expansion when it would not be if the air must be used at high pressure. Clearance in a machine of this kind is just that much com- pressed air wasted. If clearance be considered, Eq. (18) becomes E = . T ~ \ (18a) (1 + c) r loge r where c is the percentage of clearance. In some machines, if this loss were a visible leak, it would not be tolerated. Art. 17. Variation of Atmospheric Pressure with Altitude. — In most of the formulas relating to compressed-air operations the pressure p a , or weight w a , of free air is a factor. This factor varies slightly at any fixed place, as indicated by barometer readings, and it varies materially with varying elevations. To be precise in computations of work or of weights or volumes of air moved, the factors p a and w a should be determined for each experiment or test, but such precision is seldom warranted further than to get the value of p a for the particular locality for ordinary atmospheric conditions. This, however, should always be done. It is a simple matter and does not increase the labor of computa- tion. In many plants in the elevated region p a may be less than 14.0 lb. per square inch, and to assume it 14.7 would involve an error of more than 5 per cent. Direct reading of a barometer is the easiest and usual way of 32 COMPRESSED AIR getting atmospheric pressure, but barometers of the aneroid class should be used with caution. Some are found quite reliable, but others are not. In any case they should be checked by compari- son with a mercurial barometer as frequently as possible. If m be the barometer reading in inches of mercury and F be the temperature (Fahrenheit), the pressure in pounds per square inch is p a = 0.4912 m[l - 0.0001 (F - 32)] (19) Note. — One cubic inch of mercury at 32°F. weighs 0.4912 lb. The information in Table II will usually obviate the need of using Eq. (19). In case the elevation is known and no barometer available the problem can be solved as follows : Let p s = pressure of air at sea level, w s = weight of air at sea level, p x , Wx be like quantities for any other elevation. Then in any vertical prism of unit area and height dh we have dp x = w x dh. But — = — : therefore dp x = — p x dh, w, p s p 3 or dh = — -*-^, and therefrom h = — X log — > W s p x W s ° p a where p a is the pressure at elevation h above seal level. Sub- stitute for w s its equivalent w - = s£kt and we get 53§s* = Iog f; Whence l0g e p a = loge p s - 53.35 * Making p s = 14.745 and adopting to common logarithm and Fahrenheit temperatures, logio Pa = 1.16866 - J22^r+"460T (20) Table V is made up by formula (20). CHAPTER II MEASUREMENT OF AIR Art. 18. General Discussion. — Progress in the science of com- pressed-air production and application has evidently been hin- dered by a lack of accurate data as to the amount of compressed air produced and used. The custom has been almost universal of basing computations on, and of recording results as based on, catalog rating of compres- sor volumes — that is, on piston displacement. The evil would not be so great if all compressors had about the same volumetric efficiency, but it is a fact that the volumetric efficiency varies from 60 to 90 per cent., depending on the make, size, condition and speed of the machine, no wonder, then, that calculations often go wrong and results seem to be inconsistent. There are problems in compressed-air transmission and use for the solution of which accurate knowledge of the volume or weight of air passing is absolutely necessary. Chief among these are the determination of friction factors in air pipes and the efficiency of compressors, pumps, air lifts, fans, etc. Purchasers may be imposed upon, and no doubt often are, in the purchase of compressors with abnormally low volumetric efficiencies. Contracts for important air-compressor installation should set a minimum limit for the volumetric efficiency, and the ordinary mechanical engineer should have knowledge and means sufficient to test the plant when installed. There is little difficulty in the measurement of air. The calculations are a little more technical, but the apparatus is as simple and the work much less disagreeable than in measurements of water. At this date (1917) practice does not seem to have settled on a standard method of measuring quantities of air; but current literature shows that the subject is receiving what seems to be the long-delayed attention that it deserves. In any case where the air or gas to be measured will have a con- stant density and it is necessary only to get the rate of flow at any time, the apparatus and methods applicable would be as simple 3 33 34 COMPRESSED AIR as those applied in measuring water, but the problem is not so simple when it is necessary to record the total flow (weight) dur- ing a considerable time during which the pressure and density may vary between wide limits. Though there are some appara- tus that the makers claim will do this, the problem does not seem to have been solved in a satisfactory way. Art. 19. Apparatus for Measuring Air. — Several methods of measuring the rate of flow of air at the time of observation (or with pressure and temperature constant), that have been pro- posed and tried, will be briefly noted as follows: 1 (a) The Venturi Meter. — The principle is identical with that of the venturi water meter, but it is necessary to determine the coefficient over a range covering all pressures under which it may be used. This coefficient may not change with pressure, but if so the fact has not been ascertained. (b) The "Swinging Gate," Fig. 8a. — The air flowing in the direction of the arrow swings the gate open. The angle of opening depends on the weight of the gate, and on the density and velocity of the air. Every gate will have a special set of coeffi- cients and these would have to cover the whole field of velocities and densities. (c) The Thermal Method. — In this scheme the air is passed through an enlargement of the pipe in which there is placed an exposure of a great surface of wire, the wire being heated by a measured electric current. The temperature of the air is meas- ured before and after passing over the heated wire. The weight of air passing can be expressed in terms of the rise of temperature and the electric current absorbed. The objections are: Expen- sive apparatus, requiring great sensitiveness, and liability to error through various sources, among which is the humidity of the air. (d) Mechanical Meters. — This class includes common gas meters. They are satisfactory for commercial purposes and for such capacities as are covered by stock sizes. For large volumes they become expensive and the coefficient is always liable to variation, that is, the record may become inaccurate due to 1 See Compressed Air Magazine, vol. 16, p. 6255. Fig. 8a. MEASUREMENT OF AIR 35 corrosion or fouling of the mechanisms. Such meters show only the total volume that has passed between readings but unless the pressure and temperature are constant the record does not show the quantity or weight. As stated above, none of these methods will apply when it is necessary to determine the total weight passing during a pro- longed time in which the pressure varies. If in cases (a) and (&) the pressure is constant and the velocity only changes, a continuous recording apparatus could be attached to make a graph giving time and differential head in case (a) or time and swing of gate in case (b) from which cards the total volume could be integrated. If simultaneously another graph be taken show- ing time and pressure the two could be used to work out weights. If inventors could go this far, they could afford to neglect temperatures in commercial work. However, the cost of the apparatus and the labor of determining the proper coefficients seem to bar any of the above from general use. Art. 20. Measurement by Standard Orifices. — For reasons of economy, simplicity and accuracy, it seems that practice will settle on the standard orifice for determining the flow of air. For this reason the method and apparatus are described in detail. The standard orifice is the same as that specified for the meas- urement of water, that is, an orifice in a thin plate (or with sharp edges). In this article only circular orifices will be considered. These may be cut in any sheet metal up to ~% in. thick. The standard conditions shall be that the drop in pressure in passing through the orifice shall not exceed 6 in. head of water. With this restriction of conditions the change of temperature and of density of the air while passing the orifice may be neglected in commercial operations without appreciable error. This very much simplifies the formulas and reduces the chances of error. With these standards, experiments show coefficients for air more nearly constant than for water. Art. 21. Formula. Standard Orifice under Standard Condi- tions. — Let p = absolute pressure of air approaching the orifice = rp a , Q = weight of air passing per second, w = weight of a cubic foot of air at pressure p, d = diameter of orifice in inches, 36 COMPRESSED AIR i — pressure as read on water gage in inches, t = absolute temperature of air (F), c = experimental coefficient. When change of temperature and of density can be neglected, the theoretic velocity through an orifice is s = \2gh where h is the head of air of uniform density (w) that would produce the pressure head i. Hence i 62.5 ,, , L i 6.25 h = 77: > therefore s = \ 2g — - — • 12 w v 12 w But Q = w X a X s where a equals the area of orifice in d 2 square feet = t tttttt" Inserting these values and putting w under the radical, there results 6 " 4^144 V^ 62 ^ ^ but w = 53.35* therefore Q = 0.0136d 2 A I- rp a ' where p a ' is in pounds per square foot, = 0.1639d 2 A /- rp a where p a is in pounds per square inch. To this must be applied the experimental coefficient c so the formula becomes Q = cX 0.m9d 2 yjjrp a (21) For distilled water and dry air the equation would be Q = cX 0.1645d 2 ^ rp a . In very precise determinations the weight of air should be determined to accord with its humidity (see Art. 8a). This value of w would then go into Eq. (a) above. When working with an orifice set in a low-pressure drum, the MEASUREMENT OF AIR 37 product rp a can be most readily gotten by adding to p a the quantity 0.036* which is the pressure on a square inch due to a head i. Thus rp a = p a + 0.036*. If mercury be the liquid in the U-gage and barometer heights be inches of mercury, then li Q = c X 0.1U7d\-h where h = barometer height + i (i being inches of mercury). It will often be convenient to compute the weight of air when pressure is in inches of mercury. Then w a = 1.321 y (21a) The apparatus to be used in combination with this formula depends on whether the measured air is to be discharged directly into the free atmosphere or is to be retained in the pipe system under pressure. Art. 22. Apparatus for Measuring Air at Atmospheric Pres- sure. — This is the simpler of the two cases and is the one most easily applied in a single test of an air compressor. The essentials are indicated in Fig. 9. Fig. 9. A = compressed-air pipe, B = closed box or cylinder, T = throttle, b = baffle boards or screen, H = thermometer, C = cork, = orifice in thin metal plate (Standard), U = bent glass tube containing colored water, G = scale of inches. 38 COMPRESSED AIR The box B may be made of any light material, wood or metal. The pressure will be only a few ounces and the tendency to leak correspondingly slight. The purpose of the throttle T is to control the pressure against which the compressor works. The appropriate orifice can be determined by a preliminary compu- tation, assuming i at say 3 in., or use Plate I. Art. 23. Coefficients for Large Orifices. — Experiments were made at Missouri School of Mines in 1915 to determine the co- efficient, c, to apply in formula (21) in case of large orifices up to 30 in. in diameter and 30 by 30 in. square. The scheme being as follows: 1 Having a fan or blower of capacity and pressure sufficient for the purpose, direct the discharge into a conduit across which place one partition containing the appropriate number of small standard orifices for which the coefficient is known and in an- other partition place the large orifice. Then the same quantity of air passes through the group of small orifices and the single large orifice, and by observing the water gage at each partition the relation between the coefficients can be found thus : Let Ci be the unknown coefficient of the large orifice, c 2 be the known coefficient of the small orifices, n be the number of small orifices open, d be the diameter of the small orifices, D be the diameter of the large orifices. Then by formula (21) Q = a X 0.1639Z) 2 A /- 1 Vl = c 2 X 0.1639wd 2 J*- 2 p 2 Sub 1 and sub 2 indicating symbols at the large and small orifice partitions, respectively. Now it can be shown that where the drop in pressure is only a few inches (water gage) the factors may be taken as equal, especially so if the water gages be nearly equal at the two partitions. Hence we may express the relation of the two coefficients, thus r _ p (nd 2 fk 1 Missouri School of Mines Bulletin, vol. 2, No. 2, November, 1915. MEASUREMENT OF AIR 39 ZT----T-----V--" il\ :;: p iS :|::::::::::::|:- \ : 1 . : \ • ; - \ \-\ '• ■ i j ! • - :-:::■ j\ 4-LU ILL . - PLATE I " Appropriate Orifices E: for given- Volumes of Air:: Low Pressure Drum For High Pressure Drum :: Divide hv (r 1 Vi T^l 1 : + q * t* O O X _i_ co t- eo so . — ^_ — j^ — | -lili- — .-OHO |[ - 4- - . — saqoui u; saoutio jo ja;eureiQ 40 COMPRESSED AIR Similarly, when the large orifice is rectangular with area = a, nrd 2 to Ci — C2 1 \ 4a \ii For convenience let K represent the factor in parenthesis; then C\ = KC%. In the experiments referred to, the following results were obtained : Seventy-seven 3j^-in. orifices passing to one 30-in. round K = 1.01. Fifty 3%-in. orifices passing to one 24-in.>ound K = 1 .00. Twenty-six 3^-in. orifices passing to one 18-in. round K = 0.996. Fifty-eight 3JHri n - orifices passing to one 18 by 30-in. rectangle. K = 1 .005. Sixty 3^-i n - orifices passing to one 24 by 24-in. rectangle K = 1 . 014. Thirty-four 33^-in. orifices passing to one 18 by 18-in. rectangle . K = . 998. From the above it is evident that for commercial purposes the coefficients for these large orifices may be taken as equal that of a 33^-in. orifice (see Table VIII). Errors in reading water gages will probably exceed that made by such an assumption. Accepting the coefficients shown in Table VIII, those for large orifices are as shown in Table Villa. As a result of these experiments it is evident that large orifices, conforming to standard conditions, can be used with as much accuracy as in case of small ones. This being accepted, there is available for testing large fans and blowers the most reliable of all methods of measuring the flow of fluids, that is orifice measurement. Note that one 30-in. round orifice will pass about 25,000 cu. ft. per minute under 4-in. water pressure. Where very large fans are to be tested several orifices can be set in a conduit wall. For such cases accurately constructed wood orifices would probably be entirely reliable and could be put in at moderate cost. Art. 23a. Notes on Water Gages. — Experience with water gages, and in efforts to improve on the plain water gage, while doing this work may be of interest. In such a gage (any liquid) when oscillations (not gradual changes of pressure) interfere with the readings, a few bird shot (filling the tube about an inch) will prevent oscillations and yet permit sufficient sensitiveness under changing pressure. Any coloring matter is liable to cause error by changing the specific gravity of the water. MEASUREMENT OF AIR 41 Makers of some special gages recommend the use of gasoline of known specific gravity, instead of water, as it is lighter and therefore more sensitive. On trial it was found that if the two columns of the gage, above the liquid, are unequal in height, the presence of gasoline gas in the high column will unbalance the fluid columns and cause error. Often one arm of the gage is continued in a rubber tube. This will in effect be an extension of the column. In a gage in which the two columns have equal bore, or caliber, throughout, the sum of the two column readings will be constant as long as the volume of liquid in the gage does not change. In attempting to utilize this fact in a gage filled with gasoline it was found that the gasoline evaporated so fast as to render the scheme inapplicable. The same liability to inaccuracies exist in any of the combination gages in which both water and gasoline are used. Where much work is to be done while pressures are changing, the best scheme is to get a gage in which the sum of the readings is constant; use water or mercury; find the sum of the two column readings ^nd then read only one column. Let s = sum of column reading, h = reading of upper column of liquid, I = reading of lower column of liquid. Then i = 2 (h - y 2 s) or i = 2 (%s - I). Experience in this work in which thousands of readings of fluid pressure gages have been made under a variety of conditions and with a variety of gages, leads those who have done most of the work to the conclusion that most reliable results can be got- ten with pure water in a plain U-tube fastened vertically over a scale tacked to a plane board; the arms of the tube about 2-in. apart and the horizontal ruling of the scale extending under both arms of the gage. The readings to be taken with the assistance of a small draftsman's triangle held with the side resting against the vertical glass tube and edge against the scale, parallax being avoided by bringing the eye so that the upper edge of the tri- angle and the lines on the scale are projected parallel and both seen crossing the gage column as illustrated in the photograph. (Note that the eye of the camera was not in the correct position.) Art 24. Apparatus for Measuring Air Under Pressure with Standard Orifices. — In the ordinary case when it is desired to know the quantity of compressed air passing through a pipe with- 42 COMPRESSED AIR out sacrificing the pressure, the orifice drum must be made strong enough to withstand the high pressure and the U-gage de- scribed in the previous case must be replaced by a differential gage which must also be strong enough to withstand the pres- sure. The essentials are embodied in the illustration, Fig. 10, Fig. 9a. — Method of reading water gages. which also suggests a convenient scheme for attachment to an air main. The several essentials are: VxVzVz = valves for controlling the path of the air, U = unions for detaching apparatus, bbibi = baffles for steadying the current of air, MEASUREMENT OF AIR 43 = orifice, T = thermometer set through a gland, G = pressure gage, gg% = glass columns of the differential gage, C = cocks for convenience in manipulating the differ- ential gage. The manipulation of the apparatus Fig. 10, is as follows: To charge the differential gage close C\, C 4 and C 5 , open C 2 and C 3 and pour in the desired amount of liquid. Then close C 2 and C 3 and open d and C5. To pass the air through the measuring drum, open Vi and V3 and close V\. Note: Both legs of the gage should be tapt into the drum close beside the orifice. Fig. 10. Art. 25. Coefficients and Orifice Diameters for Measurements at High Pressures. — Unless evidence to the contrary is shown, it is reasonable to assume that the same coefficients would apply to the orifice in the high-pressure drum, Fig. 10, that have been determined for the low-pressure drum, Fig. 9. However, for the same Q, i, t and c the diameters, d, must differ according to the following : 44 COMPRESSED AIR Let di and p\ be the orifice diameter and air pressure respect- ively in the high-pressure drum, and note that the pressure in the low-pressure drum may be taken for this purpose as p a . Then . Qi = Q = C X 0.1QZ9d\ l - p a = cX 0.163W J.- Pl . Whence di = % (22) since p\/p a = r. By this relation the appropriate orifice can be determined from the curye, Plate I, by dividing the diameter ordinate by (r) Vi . The size drum necessary to measure a given volume of free air when under pressure is not as large as might be supposed before computations are made. For instance, with i = 3 in., T = 60°F. and c = 0.60, a 3-in. orifice will pass 570 cu. ft. of free air per minute when compressed to 100 lb. If this 3-in. orifice be placed in a drum 8 in. in diameter, the velocity of the compressed air within the drum will be 3.5 ft. per second, which is conservative. Example 25. — In a run with the apparatus shown in Fig. 9, the following were the records: d = 2.32 in., i = 4.6 in., T = 186°F. inside drum, T = 86°F. in free air, elevation == 1,200 ft. Find the weight and volume of air passing per minute. Solution. — From Table II interpolating for 86° in the line with 1,200 elevation we get w a = 0.0700 and p a — 14.1. Add to p a the pressure due to i ( = 0.036 X 4.6) and we get p a = 14.26. In Table VIII the coefficient for d = 2.32 and i = 4.6 is 0.599. These numbers inserted in Eq. (21) give Q = 0.599 X 0.1639 X (2.32) 2 ^/^ X 14.26 = 0.1684 lb. per second; and — — ~^ = 144.3 cu. ft. per minute of free air. Should there be doubt about the coefficients being the same for both high- and low-pressure drums, and we are willing to accept these now published for low-pressure drums, we can determine that of the high-pressure drum by placing the two drums in tandem, the same quantity of air passing through the high- and low-pressure drums in succession. Then letting sub 1 refer to the high-pressure drum we have the equation, Q = d X 0.1639di 2 J 1 - pi = c X 0.1639d 2 J~p a . MEASUREMENT OF AIR 45 Whence tf.j/tyi-hU (23) d\* i hpa Pi In extensive experiments at Missouri School of Mines in 1915, the coefficients proved to be equal so far as practical applications would be concerned though the high-pressure coefficients seemed to be slightly less. The experiments were not conclusive. See description of oil differential gage, Appendix D. In advocating the standard-orifice method of measuring air it should be noted that the coefficient of an orifice is not liable to change with time and that the necessary apparatus can be made up in any reasonably well-equipped shop of a compressed-air plant. The method as presented is adapted only to show the rate of flow at the time of observation. To determine the quantity passed during any prolonged period a continuous recording apparatus would have to be attached that would show both the value of i and of p. The factor t might be assumed constant in most cases in practice but even then the apparatus would be intricate, delicate and expensive. It may be stated then that there are no satisfactory means now available to measure the quantity of air passed during a definite time where pressure and velocities vary. However, the obstacles are not insurmountable. Art. 26. Discharge of Air through Orifice. Considerable Drop in Pressure. — Referring to Figs. 9 and 10, when the differ- ence in pressures pi and pi is considerable, we cannot neglect the change of density and of temperature. To analyze this case we must start from the equations of energy at sections 1 and 2, inside and outside the orifice, the energy in each case being part kinetic and part potential. Thus ~2g + PlVl = ~2g~ V%V2 ^ or Qsi 2 . n , Qs 2 2 . n , ~2f + Qdi = -^ + Qch where c = 53.35 for 1 lb. (see Art. 2). Whence Si 2 . ■ s 2 2 , 46 COMPRESSED AIR Now in any practical case the velocity of approach s x to the Si 2 orifice can be made so small that the numerical value of 77- is 2g so small as compared with ch 2 that it can be neglected, if desired, s 2 2 without appreciable error; but not so with the quantity -^" Hence we may write s 2 2 = 2gc(ti - t 2 ). Substituting for t 2 its value from Eq. (12a), viz., we get ft = VW, [l - {f) V] * - VWi (l - r-f) Yt where r x is the ratio — when the escape is into free air. The weight passing per second is Q = w a aS 2 where a is the area of orifice and w a = —r in which again substitute for t 2 its CI value as above. These substitutions give = api yJ^S. [r,« (l - r^) »*] (24) This is a max. when r x = B-+.1 When w = 1.41i Q is max. when r s = 0.526. When n = 1.25i Q is max. when r x = 0.555. Any such law as this could not have been suspected except by mathematical analysis, and seems contrary to what would other- wise have been supposed. Yet experiment seems to show that it is correct. Equation (24) is not recommended as a formula for practical application in measuring air. Art. 27. Air Measurement in Tanks. — The amount of air put into or taken out of a closed tank or system of tanks and MEASUREMENT OF AIR 47 pipes, of known volume, can be accurately determined by Eq. (3), viz., Pa" a "a P x"a V x - — = - or v a = - — y P xV x *x Pa *x The process would be as follows : Determine the volumes of all tanks, pipes, etc., to be included in the closed system, open all to free air and observe the free-air temperature; then switch the delivery from the compressor into the closed system; count the strokes of the compressor until the pressure is as high as desired; then shut off the closed tank and note pressure and tem- peratures of each separate part of the volume. Then the formula above will give the volume of free air which compressed and heated would occupy the tanks. From this subtract the volume of free air originally in the tanks ; the remainder will be what the compressor has delivered into the system. Note that the com- pressor should be running hot and at normal speed and pressure when the test is made for its volumetric efficiency. Usually the temperature changes will be considerable, but if the system is tight, time can be given for the temperature to come back to that of the atmosphere, thus saving the necessity of any temperature observations. Where a convenient closed-tank system is available, this method is recommended. This method— that is, Eq. (3) as stated above — was used to determine the quantity of air passing the orifices in the experi- ments by which the coefficients were determined as given in Art. 21, Table VIII. The varying volumetric efficiencies with changes of tempera- ture and pressures can be shown very impressively by starting with compressor cool and the air in tanks at atmospheric pres- sure. Then note the number of revolutions that bring the pres- sure up to say 20, 40, 60, 80 lb., and so get the data for volumetric efficiencies in each interval. In the first it may be found as high as 95 per cent, while in the last interval it may fall below 60 per cent, in small compressors. Of course, that in the last interval is that by which the compressor should be judged. Example 27. — A tank system consists of one receiver 3 ft. in diameter by 12 ft., one air pipe 6 in. by 40 ft., one 4 in. by 4,000 ft. and a second receiver at end of pipe 2 ft. in diameter by 8 ft. A compressor 12 by 18 in. with 1^-in. piston rod puts the air 48 COMPRESSED AIR from 1,250 revolutions into the system, after which the pressure is 80-gage and temperature in first receiver 200°, while in other parts of the tank system it is 60°. Temperature of outside air being 50°, p a = 14.5 per square inch. Find volumetric efficiency of the compressor. Solutions. — Volumes (from Table X) : First receiver 84.84 cu. ft. 6-in. pipe 7.841 4-in pipe 349.20 382.16 Second receiver. ..... 25. 12 J Total 467.00 in tank system. Piston displacement in one revolution = 2.338 cu. ft. (piston rod deducted). (T) t \ V — ^1 X — note that the quantity in paren- Pa ' lx thesis is constant and therefore a slide rule can be conveniently used, otherwise work by logarithms . _ . (80 + 14.5) (460 + 50) v , 84.84 , nf70 v a in first receiver = - ^-^ — X 46Q + 2QQ = 417.2 v a in 6-in. pipe, 4-in. pipe and second receiver with total volume 382.16 and t = 60° = 2,447. 1 Total 2,864.3 Original volume of free air , 467 . Volume of free air added 2,397.3 2,397.3 + 2.338 = 1,028. Therefore the volumetric- efficiency is E = 1,028 -T- 1,250 = 82 per cent. CHAPTER III FRICTION IN AIR PIPES Art. 28. — In the literature on compressed air many formulas can be found that are intended to give the friction in air pipes in some form. Some of these formulas are, by evidence on their face, unreliable, as for instance when no density factor appears; the origin of others cannot be traced and others are in incon- venient form. Tables claiming to give friction loss in air pipes are conflicting, and reliable experimental data relating to the subject are quite limited. In this chapter are presented the derivation of rational for- mulas for friction in air pipes with full exposition of the assump- tions on which they are based. The coefficients were gotten from the data collected in Appendix B. Art. 29. The Formula for Practice. — The first investigation will be based on the assumption that volume, density and tempera- ture remain constant throughout the pipe. Evidently these assumptions are never correct; for any de- crease in pressure is accompanied by a corresponding increase in volume even if temperature is constant. (The assumption of constant temperature is always permissible.) However, it is believed that the error involved in these assumptions will be less than other unavoidable inaccuracies involved in such computations. Let / = lost pressure in pounds per square inch, I = length of pipe in feet, d = diameter of pipe in inches, s = velocity of air in pipe in feet per second, r = ratio of compression in atmospheres, c = an empirical coefficient including all constants. Experiments have proved that fluid friction varies very nearly with the square of the velocity and directly with the density. Hence if k be the force in pounds necessary to force atmospheric air (r = 1) over 1 sq. ft. of surface at a velocity of 1 ft. per 4 49 50 COMPRESSED AIR second, then at any other velocity and ratio of compression the force will be F x = ks 2 r, and the force necessary to force the air over the whole interior of a pipe will be F = ^IX krs\ and the work done per second, being force multiplied by distance, is Work = — r X krs 3 . 12 Now if the pressure at entrance to the pipe is / lb. per square inch greater than at the other end, the work per second due to this difference (neglecting work of expansion in air) is Work = f-±- s. Equating these two expressions for work there results .ird 2 ivd 77 , / -j- s = ^2 lkrs ' or f-~A rsi (25) Now the volume of compressed air, v, passing through the pipe is, in cubic feet, 7T^ 2 V ~ 4 X 144 S and the volume of free air v a is rv. Therefore Va = 4X1U X rS and 2 = (4 X 144) V S " " 7r 2 d 4 r 2 Insert this value of s 2 in Eq. (25) and reduce and the results 4 _ /4 X 144\ 2 1 v a 2 '=12*( 7 & or I v a * / = c-h — (26) J d 5 r FRICTION IN AIR PIPES 51 where c is the experimental coefficient and includes all constants. From Eq. (26), /clv a 2 \ ** *-(-£■) (27) From the data recorded in the appendix the coefficients for formula (26) were worked out, first using the actual measured diameters, second using the nominal diameters. The average of the coefficients for each size pipe were then platted and the results tabulated as shown on Plate II. In studying this plate it should be borne in mind that the vertical scale is ten times that of the horizontal which exaggerates the irregularities of the coefficient. These studies reveal conclusively that c is practically independ- ent of r and of s (the velocity in pipes), and that it increases as the diameter decreases. If temperature has any effect, it could not be detected. Since the friction varies inversely as the fifth power of the diameter, it is very sensitive to any variation in the diameter. Hence, if the greatest possible accuracy is de- sired, the computations should be based on the measured diame- ter and the coefficient taken from the curve AB, Plate II. If the actual diameter is unknown and the computer must use nominal diameters, the coefficient should be taken from the line CD. In any case computations of friction loss in commercial pipes of less than 1 in. in diameter will be unreliable on account of the relative great effect caused by small obstructions and irregular diameters. Table IX is computed from Eq. (26) and is self-explanatory. It affords a direct and easy determination of friction losses in air pipes. A further study of the coefficients found by the curve AB, Plate II, shows that the logarithms of c and d plat to a straight line from which is obtained the relation = 0.1025 C d°- sl ' This inserted in Eq. (26) gives 10 rd 0.1025lv a > or , _ 0.1025 IvJ . J ~ 3,600 rd 5 - 31 {Zm) 52 COMPRESSED AIR c 6 o O a 5 - »TTTr „ _ . *, . Oj j a uo Suia't; OOOOoO OOOOoOO " j -i OOOOOOOO 1 01 O - ' as 9V eAJt»o no SuiX^j .110 .100 .089 .082 .072 .0G6 .032 .059 .056 .053 .050 .043 1 / / / pa^ndtaoQ sy rM 1— 1 O OOO OOO OO i f / pajtisBajv: 0.83 0.82 1.07 1.63 2.07 2.874 3.937 4.921 5.960 n / 1 / " JBECUOJJ ^ ro\ 1 " H ^' M^-^^sst-O'^^ 1 1 j / / , / 1 PLATE II „ 1 v Coefficients c for Formula f=c -p-jr O Indicates Average C Computed fron Measured Diameters A Indicates the C Adapted to Nomina Diameters -Computed from Values on Curve AB 1 / / / / / / / / / / 1 1 1 / / / 1 , j ' / / / / / , / I / 1 // ! / / / / / / / I / / / , / / , / / 1 > j s / / ' 1 ' _< d / (1 0„ }USPUP°D FRICTION IN AIR PIPES Chart for Solving Formula /= -1025 * _ 7-d 6 - 31 x3G00' /= Friction Loss in Pounds per Square Inch. I = Length, of Pipe in Feet. •y= Cubic Feet of Free Air per Minute. *"= Eatio of Compression d= Diameter of Pipe in Inches. The Dependent Factors {fr), v and d Lie in a Straight Lino. To get the Friction Loss in 1000 Feet; Divide the (fr) by r. Friction of Gasses will be Proportional the their Specific Gravities. -200 -190 -ISO -170 -ISO -150 -140 -130 -120 -no -100 or v"' = 35.13 (fr) d 80,000 7.-..HU0 70,000 65,000 60,000' 55,000 50,000 45,000 40,000 35,000 30,000 28,000 20,000 24,000 22 000 20,000 18,000 16,000 14,000 12,000 10,000 9000 8000 7000- 6000- 5000. 4500- 4000- 3500- 3000- 2500- 2000. 1800- 1600- 1400- 1200- 1000- 900- 800- 700- 600- 500- 450- 400- 350- 300- 250- 200- 53 12 1 10 6- 4V 2 - ZV 2 - 2V 2 ~ 2- 70- 60- 1 3 4- 1H- 1V 4 - t 1 Plate III. 1- 54 COMPRESSED AIR where v a is in cubic feet per minute. Log ' fif)f) = 5.4544. This equation gives results practically indentical with those from Eq. (26) when c is taken from the curve AB. It is almost as easy of solution and has the advantage that it is independent of a table of coefficients. Plate III is a logarithmic chart for solving Eq. (28a). Since such a chart can handle only three variables, the product fr is taken as a single variable and I as 1,000 ft. To solve the equation by this chart, lay a straight edge (or stretch a thread) over the chart. The three numbers under the line will satisfy Eq. (28a). Example 28a. — What pressure will be lost in a 4-in. pipe 5,000 ft. long when transmitting 1,200 cu. ft. of free air per minute compressed to 7 atmospheres (r = 7). A thread stretched over 4 in. and 1,200 cu. ft. crosses the fr line at 25, then 25 4- 7 = 3.6 and 3.6 X 5 = 18 lb. Since the process of designing such charts as Plate III has not appeared in any of the well-known text-books, the author has made it available in Appendix B. The following formula is that derived by Church for loss by friction in air pipes : 2 _ 4clQ 2 p 2 Vr - Vi gdA^Wz In this pi and pi are pressures at points on the pipe distance I apart, p\ being the less pressure, A is the area of the pipe and c some experimental coefficient. The other symbols are as used elsewhere in this article. Frank Richards recommends a simplification of Church's formula by assuming c constant and a temperature about 60°F. His formula is V a H V% - P\ L = 2,000d E In the experiments at the Missouri School of Mines in 1911 (described in Appendix C) effort was made to find the laws of resistance to flow of air through various pipe fittings. Facili- ties were not available for sizes above 2 in. in diameter and for the smaller sizes the results were erratic, doubtless due to the rela- tively greater effect of obstructions and variations in diameter FRICTION IN AIR PIPES 55 in the small pipes. The results are given below. Further re- search is needed along this line. Lengths op Pipe in Feet that GrvE Resistance Equal that op a Single Fitting Diameter of pipe, inches Elbows 90° Unreamed joints, 2 ends Reamed joints Return binds Globe valves y* 10.0 2-4 7 10.0 20 % 7.0 2-4 7 7.0 25 i 5.0 2-4 7 5.0 40 m 4.0 2-4 7 4.0 45 2 3.5 2-4 7 3.5 47 Tests on resistance in 50-ft. lengths of rubber-lined armored hose, with their end fittings such as is used to connect with com- pressed-air tools, were made with average result as follows : Diameter of hose, inches H 1 IK Resistance in 50-ft. length 20^ r 4.5 ~ r V a 2 2.6 — r Finally it is important to note that in cases where gases other than air are under consideration the friction losses will be di- rectly proportional to the specific gravity of the gas, for instance if the gas has a specific gravity of 0.8 the friction will be 0.8 of that for air under the same conditions. The rate of flow of air or gas through a long pipe of uniform diameter can be computed approximately by observing / for distance I; then in case of air, or cl Va = frd 5 c X 0.81 in case of gas of 0.8 specific gravity. This formula may be of value in determining the flow of natu- ral gas through long pipes. It may be well to note here that the deposit of solid matter 56 COMPRESSED AIR (paraffines and asphalts) out of natural gas may seriously obstruct the pipes and render such computations altogether inaccurate. Example. — 1,600 cu. ft. per minute of free air is supplied to a mine at a pressure of 7 atmospheres (r = 7) through a 4-in. main. At a distance of 2,840 ft. from the compressor is a 2-in. branch placed to take air to two 2}4 drills, requiring 100 cu. ft. each of free air per minute. The 2-in. pipe is 1,260 ft. long and has in that length two globe valves, four elbows, and eighteen unreamed (extra) joints. Each drill takes its air through 50 ft. of 1-in. hose. What will be the loss of pressure at the drills? Solution. — By formula (28a) : Loss in the 4-in. main: 5.4544 log 7 = 0.8457 log 2,840 = 3.4533 5.31 log 4 = 3.1972 2 log 1,600 = 6.4082 5.3159 4.0423 4.0423 log .A = 1.2736 .*. fi = 18.8 for 4-in. p Loss in 2-in. pipe. Note that the r is about 5.75 in the 2-in pipe : Effective length, straight 1,260 ft. Effective length, 2 glove valves @ 47 94 ft. Effective length, 4 elbows @ 3.5 14 ft. Effective length, 18 unreamed joints 3.0 54 ft. Total 1,422 ft. 5.4544 log 5.75 = 0.7597 log 1,422 = 3.1529 (5.31) log 2 = 1.5983 2 log 200 = 4.6020 2.3580 3.2093 2.3580 log/a = 0.8513 .'. / 2 - 7.10 for 2-in. pipe. v Loss in 50 ft. of 1-in. hose delivering 100 cu. ft., / = 4.5 — - • Note that the r in the hose is (after deducting the accumulated friction in the 4-in. and the 2-in. pipes) about 5.25. FRICTION IN AIR PIPES 57 log 4.5 = . 6532 log 5.25 = . 7202 2 log 100 = 4.0000 log 3,600 = 3.5563 4.6530 4.2763 4.2765 log/ = 0.3765 .*. 2.4 for the 50-ft. hose. Total loss of pressure = 18.8 + 7.1 + 2.4 = 28.3. Evidently such computations as this should not be accepted as giving precise results. Such matters as the varying r, varying density of air as effected by temperature and free air pressure, irregular qualities and changing conditions of the pipes, leaks, and irregular demands for air all more or less effect the resulting loss. Nevertheless such computations are the proper guides for the designer. Art. 30. Theoretically Correct Friction Formula. — The theo- retically correct formula for friction in air pipes must involve the work done in expansion while the pressure is dropping. Let pi and p 2 be the absolute pressures at entrance and dis- charge of the pipe respectively and let Q be the total weight of air passing per second. Then the total energy in the air at entrance is and at discharge the energy is 1 P2 , Qsi 2 p-^^ + w The difference in these two values must have been absorbed in friction in the pipe. Hence, using the expression for work done in friction that was derived in Art. 29, we get = VaVa (log ^ - log ^) - jr- (s 2 2 - Si 2 ). ^W- Numerical computations will show the last term, viz., is relatively so small that it can be neglected in any case in practice without appreciable error. Hence, by a simple reduc- tion we get , pi irk v , dlrs 3 , , ird 2 58 COMPRESSED AIR which when substituted gives . pi 4 X 144fc I 2 log e — = -7T X -j S 2 , p 2 I2p a d or considering p a as constant, I ' ■ Pi Z 2 P2 d or logio P2 = logio Pi -c x ^s 2 (29) In Eq. (29) C\ is the experimental coefficient and includes all constants, s is the velocity in the air pipe and varies slightly increasing as the pressure drops. All efforts so far have failed to get a formula in satisfactory shape that makes allowance for the variation in s. In working out Ci from experimental data s should be the mean between the s x and s 2 , and when using the formula the s may be taken as about 5 per cent, greater than Si. Note that in the solution of Eq. (29) common logarithms should be used for convenience, allowing the modulus, 2.3+, to go into the constant C\. The working formula may be put in a different and possibly a more convenient form, thus. In the expression i Pi t& . . dl l0g e — = rrr X rs 3 p 2 12 p a V a substitute for s its value = 4 X 144^ a ird 2 r and reduce and we get ha 2 p a d 5 r log p 2 = log pi - c 2 — j££ (30) Still another form is gotten thus. The whole weight of air passing is v a X w a = Q, and by Eq. (13) Q = v a rooci and therefore v a = 53.351Q 53.35f u p a Also r x — ^— and w a = p a 53.35^ Substitute these in (30) and it reduces to FRICTION IN AIR PIPES 59 log p 2 = log p'i - c 2 —^r b (—) (31) w a d b \pj ■ctice — done Eq. (31) becomes In ordinary parctice — may be taken as constant. If this be w a log p 2 = log Vi ~ c s j h [—) (31a) d 5 \p x If f a = 525 and w a = 0.075, then c 3 = 7,000 c 2 . In (31) and (31a) p s varies between pi and p 2 . Careful com- putations by sections of a long pipe show p x to vary as ordinates to a straight line. Modifying the formulas to allow for this variation renders them unmanageable. In working out the coefficient p x may be taken as a mean between pi and p 2 , and in using the formula p may be taken as p\ less half of the assumed loss of pressure. As before suggested, common logarithms should be used in all the equations of this article. A study of the data collected in Appendix B gives values for c 2 Eq. (31), that, for pipes 3 to 12 in. in diameter, conform closely to the expression. c 2 = 0.0124 - 0.0004d, which gives the following: d" = 3 4 5 6 8 10 12 C 2 = 0.0112 0.0108 0.0104 0.0100 0.0092 0.0084 0.0080 C 3 = 78.4 75.6 72.8 70.0 64.4 58.8 56.0 With these coefficients p x in Eqs. (31) and (31a) is to be taken in pounds per square inch. Equations (31) and (31a) are theoretically more correct than Eq. (26) and the coefficients of the former will not vary so much as those for the latter, but when the coefficients are correctly determined for Eq. (26) it is much easier to compute and can be adapted to tabulation, while Eq. (31) cannot be tabulated in any simple way. Finally it should be said that extreme refinement in computing friction- in air pipes is a waste of labor, for there are too many variables in practical conditions to warrant much effort at precision. Example 24a. — Apply formulas (26) and (31) to find the pres- sure lost in 1,000 ft. of 4-in. pipe when transmitting 1,200 cu. ft. 60 COMPRESSED AIR free air per minute compressed to 150 gage when atmospheric conditions are p a = 14.0, w a = 0.073 and t a = 540. Solution by Eq. (20).— r = 15 °^ U = 11.71. By Table IX divide 23.44 by 11.71 and the result, 2 lb., is the pressure lost per 1,000 ft. Solution of Eq. (31). — The coefficient for a 4-in. pipe is 0.0108, and log pi = log (150 -f- 14) = 2.214844. Then 1 nniAOAA nn-.no 54 w 1,000 /1,200 w 0.073\ * log p 2 = 2.214844 - 0.0108 ^X-^ (-% X w ) • The log of the last term is 3.791193 and its corresponding number is 0.006183. 2.214844 - 0.006183 = 2.208661 = log p 2 . Whence p 2 = 161.7 + and pi — p 2 = 2.3. Art. 31. Efficiency of Power Transmission by Compressed Air. — In the study of propositions to transmit power by piping compressed air, persons unfamiliar with the technicalities of compressed air are apt to make the error of assuming that the loss of power is proportional to the loss of pressure, as is the case in transmitting power by piping water. Following is the mathematical analysis of the problem : pi = absolute air pressure at entrance to transmission pipe, p 2 = absolute air pressure at end of transmission pipe, vi = volume of compressed air entering pipe at pressure pi, 02 = volume of compressed air discharged from pipe at pressure p2. Then crediting the air with all the energy it can develop in Pi isothermal expansion, the energy at entrance is piVi log — = Pa PiVi log ri, and at discharge the energy is p 2 V2 log — = p 2 02 log r 2 . Pa Hence efficiency E = W l °& r * = fe (32) P1V1 log e r x log ri Common logs may be used since the modulus cancels. The varying efficiencies are illustrated by the following tables: FRICTION IN AIR PIPES 61 p a = 14.5. pi = 87. n = 6. log n = 0.7781. 7>2 85 5.86 0.7679 0.987 80 5.52 0.7419 0.953 75 5.17 0.7135 0.917 70 4.83 0.6839 0.879 65 4.48 0.6513 0.837 60 7*2 4.14 log T2 0.6170 E 0.793 p a = 14.5. pi = 145. r x = 10. log r 1 = = 1.000. ■p 2 140 9.66 0.9850 0.98 135 9.31 0.9689 0.97 130 8.97 0.9528 0.95 125 8.62 0.9355 0.93 120 T2 8.28 log T% 0.9185 E 0.92 The above examples illustrate the advantage in transmitting at high pressure. Of course the work cannot be fully recovered in either case without expanding down to atmospheric pressure, and to do this in practice heating would be necessary. It should be understood also that by reheating this efficiency can be exceeded. It should be noted also that the above does not apply in cases where the air is applied without expansion. In such cases the efficiency of transmission alone would be E = (Pi — Pa) V2 = n 0*2 - 1) (Pi — Pa) V l r 2 0*1 — 1) Example 31a. — What diameter of pipe will transmit 5,000 cu. ft. of free air per minute compressed to 100 lb. gage, with a loss of 10 per cent, of its energy, in 2,500 ft. of pipe, assuming p a = 14.0? Solution. — r\ 114 14 = 8.15; then by Eq. (30) log r 2 log 8.15 90 100' Whence log r 2 = 0.8200; r 2 = 6.6, and 6.6 X 14 = 92.4. / = 114 — 92.4 = 21.6 = loss of pressure. By Eq. (27), log d=\ [log (0.06 X 2,500) X (^~) *- log (21.6X?^±M)] = 0.7602, whence d = 5.75 in. 21 6 Otherwise go into Table IX with loss for 1,000 ft. = -^j = 8.64, and 8.64 X r = 8.64 X 7.37 = 63 (7.37 being the mean r). 62 COMPRESSED AIR Then opposite 5,000 in the first column find nearest value to 63, which is 55 in the 6-in. column; showing the required pipe to be a little less than 6 in. Otherwise over Plate III stretch a thread passing over 63 on the fr line and 5,000 on the V a line. It will cut the d line at 5M. CHAPTER IV OTHER AIR COMPRESSORS Art. 32. Hydraulic Air Compressors. — Displacement Type. — Compressors of this type are of limited capacity and low effi- ciency, as will be shown. They are therefore of little practical importance. However, since they are frequently the subject of patents and special forms are on the market, their limitations are here shown for the benefit of those who may be interested. Omitting all reference to the special mechanisms by which the valves are operated, the scheme for such compressors is to admit water under pressure into a tank in which air has been trapped by the valve mechanisms. The entering water brings the air to a pressure equal to that of the water; after which the air is discharged to the receiver, or point of use. When the air is all out the tank is full of water, at which time the water discharge valves open, allowing the water to escape and free air to enter the tank again, after which the operation is repeated. Usually these operations are automatic. The efficiency of such compression is limited by the following conditions. Let P = pressure of water above atmosphere, or ordinary gage pressure, V — volume of the tank. Then P + p a = absolute pressure of air when compressed. The energy represented by one tank full of water is PV and by one tank full of free air when compressed to P + p a is PaV 10g e = p a V log e T. Pa Therefore the limit of the efficiency is PaV lOge r p a log e T E PV But P — pi — p a , where p± is the absolute pressure of the com- pressed air. Inserting this and dividing by p a the expression becomes E = log e r __ logio r X 2.3 r — 1 r — 1 - ' 63 64 COMPRESSED AIR Table VII is made up from formula (33) . The practical necessity of low velocities for the water entering and leaving the tanks renders the capacity of such compressors low in addition to their low efficiency. Should the problem arise of designing a large compressor of this class an interesting problem would involve the time of filling and emptying the tank under the varying pressure and head. Since it is not likely to arise space is not given it. It is possible to increase the efficiency of this style of com- pressor by carrying air into the tank with the water by induced current or Sprengle pump action — a well-known principle in hydraulics. At the begin- ning of the action water is entering the tank under full head with no resistance, and certainly additional air could be taken in with the water. Art. 33. Hydraulic Air Compressors. — Entanglement Type. — A few compressors of this type have been built compara- tively recently and have proven remark- ably successful as regards efficiency and economy of operation, but they are limited to localities where a waterfall is available, and the first cost of installa- tion is high. The principle involved is simply the reverse of the air-lift pump, and what theory can be applied will be found in Art. 39 on air-lift pumps. Figure 11 illustrates the elements of a hydraulic air compressor, h is the effective waterfall. H is the water head producing the pressure in the compressed air. t is a steel tube down which the water flows. S is a shaft in the rock to contain the tube t and allow the water to return. R is an air-tight hood or dome, either of metal or of natural rock, in which the air has time to separate from the water. A is the air pipe conducting the compressed air to point of use. OTHER AIR COMPRESSORS 65 6 is a number of small tubes open at top and terminating in a throat or contraction, in the tube t. By a well-known hydraulic principle, when water flows freely down the tube t there will occur suction in the contraction. This draws air in through the tubes 6, which air becomes en- tangled in the passing water in a myriad of small bubbles; these are swept down with the current and finally liberated under the dome R, whence the air pipe A conducts it away as compressed air. The variables involved practically defy algebraic manipula- tion, so that clear comprehension of the principles involved must be the guide to the proportions. Attention to the following facts is essential to an intelligent design of such a compressor. 1. Air must be admitted freely — all that the water can entangle. 2. The bubbles must be as small as possible. 3. The velocity of the descending water in the tube t should be eight or ten times as great as the relative ascending velocity of the bubble. The ascending velocity of the bubble relative to the water increases with the volume of the bubble, and therefore varies throughout the length of the tube, the volume of any one bubble being smaller at the bottom of the tube than at the top. For this reason it would be consistent to make the lower end of the tube t smaller than the top. Efficiencies as high as 80 per cent, are claimed for some of these compressors, which is a result hardly to have been expected. The great advantage of this method of air compression lies in its low cost of operation and in its continuity. Mechanical compressors operated by the water power could be built for less first cost and probably with as high efficiency, but cost of opera- tion would be much higher. Evidently there is a limit to the amount of air that can be taken down and compressed by this hydraulic air compressor. By the laws of conservation of energy we know that the energy in the compressed air as expressed by formula pv log e r cannot exceed that of the waterfall which is Wh where W is the weight of water passing, or in general . ■ Wh PaV a lOge T < Wfl Or V a < : * p a log e r The limitation can also be seen from the following considerations: 66 COMPRESSED AIR Let V represent the total volume of air in the whole length of the downcast pipe t and let A represent the area of that pipe. V Then when -r = h the downflow of water will cease, for the static pressure inside and outside the pipe will be equal — in this state- ment friction and velocity head in the pipe are neglected. A more correct statement would be that in order to be operative V S 2 where / is the head lost in friction and s the velocity in the downcast. Evidently in this, V is the dominant number and it can be controlled by opening or closing some of the inlet tubes at b. It is by such manipulation that the most efficient working can be secured. Art. 34. Centrifugal and Turbo Air Compressors. — With the development of the steam turbine it has become practicable to deliver air at several atmospheres pressure by means of centrif- ugal machines. The very high speed at which such machines are run (up to 4,000 r.p.m.) calls for the most perfect possible material and workmanship. Yet they are relatively simple, occupy small space, are of low first cost and are quite efficient, as compared with reciprocating machines to do equal service. These quali- ties assure this class of machine (which includes the "turbo air compressors") a popularity where large volumes of air are re- quired at a moderate and constant pressure. One very effective application of turbo air compressors is as a "booster" to large reciprocating machines, the scheme being to use the exhaust steam from the engines to run the steam tur- bines that actuate the turbo compressors. The air from the turbo compressors is delivered into the intake of the reciprocating machines. A relatively small increase in the intake pressure will materially influence the capacity and economy of opera- tion of the reciprocating machines. For example : Assume that the turbo machines deliver air at }?£ atmosphere, gage pressure; that is r = 1)^. Then if the air be cooled to its original tem- perature before entering the reciprocating machine, the weight of air handled will be increased one-half. Now assume the re- ciprocating machine to have been designed to compress free air OTHER AIR COMPRESSORS 67 to a ratio r = 6 or about 75 lb. gage; then with the booster at- tached, and maintaining the same ratio (6) of compression within the compressor, the delivery ratio relative to atmosphere will be 9 or a gage pressure about 120 lb. This would be accom- plished without compounding and without development of any- more heat than without the booster. However, more work would be required of the reciprocating engines. Hence, in studying such an improvement the designer should determine whether the engines can meet the demand for increased power. The volume of air delivered by and the efficiency of centrifugal and turbo compressors, fans and blowers are matters understood by but few, seldom known, and often far from what is assumed or claimed. The theory underlying these subjects is somewhat difficult and is deferred to Chapters VIII and IX. CHAPTER V SPECIAL APPLICATIONS OF COMPRESSED AIR In this chapter attention is given only to those applications of compressed air that involve technicalities — with which the de- signer or user may not be familiar, or by the discussion of which misuse of compressed air may be discouraged and a proper use encouraged. Engine Fig. 12. Art. 35. The Return-air System. — In the effort to economize in the use of compressed air by working it expansively in a cylinder the designer meets two difficulties: first, the machine is much enlarged when proportioned for expansion; second, it is consider- ably more complicated; and third, unless reheating is applied the expansion is limited by danger of freezing. To avoid these difficulties it has been proposed to use the air at 68 SPECIAL APPLICATIONS OF COMPRESSED AIR 69 a high initial pressure, apply it in the engine without expansion, and exhaust it into a pipe, returning it to the intake of the com- pressor with say half of its initial pressure remaining. The diagram, Fig. 12, will assist in comprehending the system. To illustrate the operation and theoretic advantages of the system assume the compressor to discharge air at 200 lb. pressure and receive it back at 100 lb. Then the ratio of compression is only 2 and yet the effective pressure in the engine is 100 lb. Evidently then with a ratio of compression and expansion of only 2 the trouble and loss due to heating are practically removed ; and the efficiency in the engine even without a cutoff would be, by Eq. (18) 72 per cent. By the above discussion the advantages of the system are apparent, and where a compressor is to run a single machine, as for instance a pump, the advantage of this return-air system will surely outweigh the disadvantage of two pipes and the high pressure, but where one compressor installa- tion is to serve various purposes such as rock drills, pumps, ma- chine shops, etc., the system cannot be applied. There should be a receiver on each air pipe near the compressor. Art. 36. The Return-air Pumping System. — Following the preceding article it is appropriate to describe the return-air pumping system. The economic principle involved is different from that of the return-air system just explained. The scheme is illustrated in Fig. 13. It consists of two tanks near the source of water supply. Each tank is connected with the compressor by a single air pipe, but the air pipes pass through a switch whereby the connection with the discharge and intake of the compressor can be reversed, as is apparent on the diagram. In operation, the compressor discharges air into one tank, thereby forcing the water out while it is exhausting the air from the other tanks, thereby drawing the water in. The charge of air will adjust itself so that when one tank is emptied the other will be filled, at which time the switch will automatically reverse the operation. The economic advantage of the system lies in the fact that the expansive energy in the air is not lost as in the ordinary dis- placement pump (Art. 37). The compressor takes in air at vary- ing degrees of compression while it is exhausting the tank. The mathematical theory, and derivation of formulas for proportioning this style of pump are quite complicated but interesting. 70 COMPRESSED AIR Preliminary to a mathematical study for proportioning the installation it is well to follow a cycle in its operation: Referring to the two tanks, Fig. 13, as A and B, assume tank A to be full of air at a pressure sufficient to sustain the back pressure or head of the discharge water column and tank B to be full of water. The air compressor is running and taking the compressed air out of A and passing it over into B. At this stage (the beginning of a cycle) no work is demanded of the air compressor except w Water Supply Fig. 13. that necessary to overcome friction in the air and water pipes, but as the air is exhausted out of A the compressor must raise the pressure to that of the constant water head. This recom- pressed air goes into B and forces the water out. At a certain period in the cycle the air pressure in A will have dropped to a point when water will begin to flow in through the intake valves. After this point in the cycle we may assume that for every volume of air taken out of tank A an equal volume of water flows in, thus maintaining a constant air pressure in A until the tank is filled SPECIAL APPLICATIONS OF COMPRESSED AIR 71 with water. At this point the water will start up the air pipe and a sudden drop of pressure will occur in the intake pipe to the com- pressor. It is this sudden drop that is utilized to operate the reversing switch, which completes the cycle. From the foregoing it becomes evident that the mathematical analysis will involve the matter presented in Arts. 14 and 15, and there are two problems to solve for any installation: first, to determine the piston displacement of the compressor required to deliver a specified quantity of water per minute, say, second, to design the steam end of the compressor so as to meet the maximum demand for power which occurs once in each cycle. The first problem can be solved by Eq. (17), Art. 15, which may be modified thus: Let v a = the actual intake capacity of the compressor (usually about 70 per cent, of the piston displacement) ; this may be taken in cubic feet per minute. Let m = number of minutes required to bring the pressure down from p to p m . Then by Eq. (17) : log p - log p m m = log (7 + Va) - log 7 7 being the volume of one tank and the air pipe between tank and switch, p and p m being the highest and lowest pressures re- spectively occurring in a tank in one cycle. If a tank full of water (volume 7) is to be delivered in n min. the time n measures the length of a cycle, and is divided into two parts: first, that just noted as m; and second, that required to draw the tank full of water after water begins to flow in under pressure pw. This 7 latter is — Hence Va , 7 n = m H Va The solution must be made by trial. Thus assume v a and find m, then n by the equation next above. Repeat until a satis- factory n has been found. The second problem must be solved by the matter developed in Art. 14. There it is shown that, with sufficient accuracy for designing, the maximum rate of work occurs when r = 3. Hence, having determined v a , the maximum rate of work demanded of 72 COMPRESSED AIR the steam end can be gotten by Eq. (8) or Table I with r — 3, due allowance being made for efficiency, etc. Since the air pipes have an effect analogous to the clearance space in engines they should be made small, even at some sacrifice in friction. A velocity of 40 to 50 per second may be allowed in the air pipes. The tanks should have a volume not less than ten times the volume in the air pipes. Theoretically, the pump tanks may be placed above the water supply as shown in Fig. 13, but the cycle can be shortened by submerging the tanks and thus increasing the pressure p m . In any case where the tanks are to be submerged the valves can still be placed above water and the water siphoned over into the tanks as suggested in Fig. 13a. The most important application of this return-air pumping system has been in pumping slimes, sand and acids — such material as would soon ruin a reciprocating or centrifugal pump. A number of such pumps are in use pumping cement "slurry" which is a fine-ground alkaline mud or slime occurring in the process of manufacture of Portland cement. The agency or force usually utilized for automatically operating the switch is the suction or partial vacuum in the intake pipe which (as the tanks are usually installed) will suddenly increase when water starts up the air pipe as described above. Other means that could be used to move the switch are : The difference in pressures in the intake and discharge pipes. This difference gradually increases until the switch is thrown and would be util- ized when the tanks are deeply submerged. Another means would be to switch after an assigned number of revolutions of the compressor — the number being that necessary to run a cycle as described above. The switch is usually made in the form of a piston valve. Details would be inappropriate in this volume. Example 36. — Design a return-air pump to deliver 200 cu. ft. per minute under a head of 100 ft.; the tanks to be placed at water level (as in Fig. 13a) and air pipes to be 500 ft. long. Solution. — The ratio po -J- p n is the same as the ratio of the water heads (taking amtospheric pressure as a head of 33.3 ft.). Then this ratio is 133.3 -f- 33.5 = 4. Assume for first trial that the tanks have a volume of 400 cu. ft. each, and that effective SPECIAL APPLICATIONS OF COMPRESSED AIR 73 RETURN AIR PUMP Fig. 13a. 74 COMPRESSED AIR intake capacity of the compressor is 1.5 cu. ft. per stroke. Then the number of strokes required in a cycle is 400 log 4 n = T? + 1 „m : i Ana = 266 + 370 = 636 ' 1.5 log 401.5 - log 400 636 strokes = 318 revolutions to deliver 400 cu. ft., or 159 revolutions to deliver 200 cu. ft. per minute. This speed is excessive, but before we make another trial we will see what size air pipe will be necessary in order to prescribe the correct size of tanks. 159 X 2 X 1.5 = 477 cu. ft. per minute intake to the com- pressor. This is the maximum rate of passage of air through either air pipe and occurs once in a cycle, and that just at the end when the pressure in the air pipe is about that of the atmos- phere. It is at this time that friction in the air pipe is greatest and we may allow a drop (/) of 5 lb. in order to economize in size of both air pipes and tanks. Then by formula (27d) /o.60 X 500 X (^PfY or by Plate III we see that a 3^-in. pipe will give a drop of 4 lb. in 500 ft. The volume in a 3^-in. pipe 500 ft. long is 33.5 cu. ft. Since we may use tanks having ten times the volume of the air pipe, we will recalculate for tanks 335 cu. ft. and a compressor intake capacity of 2 cu. ft. per stroke. Then n = ^T + i — oo 7 l0g 1 oo. = 167 + 235 = 402, 2 log 337 — log 335 whence 402 strokes or 200 revolutions deliver 335 cu. ft. and 120 would deliver 200 — which is about the right speed for such a compressor. It remains to find the maximum rate of work required of the steam end of the compressor. The greatest ratio of compression occurring in a cycle is 4, but by Art. 14 we know that the great- est rate of work occurs when the ratio is about 3, and that this max. rate is w 144 poVa 144 X p V a _„ Tr W = —-3- = —j^sy- = 59poK, (ft) n - 1 V SPECIAL APPLICATIONS OF COMPRESSED AIR 75 where po is the constant delivery pressure in pounds per square inch, and V a is the effective intake capacity of the compressor in cubic feet per minute or seconds as desired. If we take V a in cubic feet per second and divide by 550 we get horsepower. Then approximately the max. horsepower rate is one-tenth poV a - This is a general rule when r goes up to 3. Note that po is in pounds per square inch and that n is not the same as in the next preceding equation. Applying this to the numerals above we get ,, , 133.3 X 0.434 X 120 X 2 X 2 ., Max. horsepower = - — nn w ... — - = 44. bO X 10 A description of the installation of a return-air pump and a full discussion of the theoretic design can be found in Trans. Am. Soc. C. E., vol. 54, page 1, Date, 1905. Art. 37. Simple Displacement Pump. — First Known as the Shone Ejector Pump. — In this style of pump the tank is sub- merged so that when the air escapes it will fill by gravity. The operation is simply to force in air and drive the water out. When the tank is emptied of water, a float mechanism closes the com- pressed-air inlet and opens to the atmosphere an outlet through which the air escapes, allowing the tank to refill. Various mechanisms are in use to control the air valve automatically. The chief troubles are the unreliable nature of float mechanisms and the liability to freezing caused by the expansion of the escap- ing air. Some of the late designs seem reliable. The limit of efficiency of this pump is given by formula (18) and Table VI. The pump is well adapted to many cases where pumping is necessary under low lifts. In case of drainage of shallow mines and quarries, lifting sewerage, and the like, one compressor can operate a number of pumps placed where con- venient; and each pump will automatically stop when the tank is uncovered and start again when the tank is again submerged. See page 120 for design of a system of such pumps. CHAPTER VI THE AIR-LIFT PUMP Art. 38.— The air-lift pump was introduced in a practical way about 1891, though it had been known previously, as revealed by records of the Patent Office. The first effort at mathematical analysis appeared in the Journal of the Franklin Institute in July, 1895, with some notes on patent claims. In 1891 the United States Patent Office twice rejected an application for a patent to cover the pump on the ground that it was contrary to the law of physics and therefore would not work. Altogether the discovery of the air-lift pump served to show that at that late date all the tricks of air and water had not been found out. The air lift is an important addition to the resources of the hydraulic engineer. By it a greater quantity of water can be gotten out of a small deep well than by any other known means, and it is free from the vexatious and expensive depreciation and breaks incident to other deep well pumps. While the efficiency of the air lift is low it is, when properly proportioned, probably better than would be gotten by any other pump doing the same service. The industrial importance of this pump; the difficulty sur- rounding its theoretic analysis; the diversity in practice and results; the scarcity of literature on the subject; and the fact that no patent covers the air lift in its best form, seem to justify the author in giving it relatively more discussion than is given on some better-understood applications of compressed air. Art. 39. Theory of the Air-lift Pump. — An attempt at rational analysis of this pump reveals so many variables, and some of them uncontrollable, that there seems little hope that a satis- factory rational formula will ever be worked out. However, a study of the theory will reveal tendencies and better enable the experimenter to interpret results. In Fig. 14, P is the water discharge or eduction pipe with area a, open at both ends and dipped into the water. A is the 76 THE AIR-LIFT PUMP 77 air pipe through which air is forced into the pipe, P, under pres- sure necessary to overcome the head D. 6 is a bubble liberated in the water and having a volume which increases as the bubble approaches the top of the pipe. The motive force operating the pump is the buoyancy of the bubble of air, but its buoyancy causes it to slip through the water with a relative velocity u. In one second of time a volume of water = au will have passed from above the bubble to below it and in so doing must have taken some absolute velocity s in passing the contracted section around the bubble. Equating the work done by the buoyancy of the bubble in ascending, to the kinetic energy given the water descending, we have wOu = wau ^— where w — weight of water, 2g or 2g (a) ■— -Qji' Fig. 14. ~— is the equivalent of the head h at top of the pipe which is necessary to produce s, therefore a Suppose the volume of air, 0, to be divided into an infinite number of small particles of air, then the volume of a particle divided by a would be zero and therefore s would be zero; but the sum of the volumes (=0) would reduce the specific gravity of the water, and to have a balance of pressure between the columns inside and outside the pipe the equation wO = wah must hold. Hence again h = -, showing that the head h depends only CI on the volume of air in the pipe and not on the manner of its subdivision. The slip, u, of the air relative to the water constitutes the chief loss of energy in the air lift. To find this apply the law of physics, that forces are proportional to the velocities they can produce in a given mass in a given time. The force of buoyancy wO' of the 78 COMPRESSED AIR bubble causes in 1 sec. a downward velocity s in a weight of water wan. Therefore wO s wau g Whence a s 2 u == - — But — = 7>~ as proved above. as a 2g Therefore 2 \o 2 This shows that the slip varies with the square root of the volume of the bubble. It is therefore desirable to reduce the size of the bubbles by any means found possible. If u = ~> then the bubble will occupy half the cross-section of the pipe. This conclusion is modified by the effect of surface tension, which tends to contract the bubble into a sphere. The law and effect of this surface tension cannot be formulated nor can the volume of the bubbles be entirely controlled. Unfor- tunately, since the larger bubbles slip through the water faster than the small ones, they tend to coalesce; and while the conclu- sions reached above may approximately exist about the lower end of an air lift, in the upper portion, where the air has about re- gained its free volume, no such decorous proceeding exists, but instead there is a succession of more or less violent rushes of air and foamy water. The losses of energy in the air lift are due chiefly to three causes: first, the slip of the bubbles, through the water; second, the friction of the water on the sides of the pipe; and third, 'the churning of the water as one bubble breaks into another. As one of the first two decreases the other increases, for by reducing the velocity of the water the bubble remains longer in the pipe and has more time to slip. The best proportion for an air lift is that which makes the sum of these two losses a minimum. Only experiment can deter- mine what this best proportion is. It will be affected somewhat by the average volume of the bubbles. As before said, any means of reducing this volume will improve the results. Art. 40. Design of Air-lift Pumps. — The variables involved in proportioning an air-lift pump are: THE AIR-LIFT PUMP 79 1 ff ti ff L r f 1 1 Q = volume of water to be lifted, per second, h = effective lift from free water surface outside the discharge pipe, I = D + h = total length of water pipe above air inlet, D = depth of submergence = depth at which air is liberated in water pipe measured from free water surface outside the discharge pipe, v a = volume per second of free air forced into well, a = area of water pipe, A = area of air pipe, = volume of the individual bubbles. The designer can control A, a, D + h and v a but he has little control over 0, and cannot foretell what D and Q will be until the pump is in and tested. When the pump is put into operation the free water surface in the well will always drop. What this drop will be depends first on the geology and second on the amount, Q, of water taken out. In very favorable conditions, as in cavernous lime stone, very porous sandstone or gravel, the drop may be only a few feet, but in other cases it may be so much as to prove the well worthless. In any case it can be determined by noting the drop in the air pressure when the water begins flowing. If this drop is p lb., the drop of water surface in the well is 2.3 X p ft. Unless other and similar wells in the locality have been tested, the designer should not expect to get the best proportion with the first set of pip- ing, and an inefficient set of piping should not be left in the well. The following suggestions for proportioning air lifts have proved safe in practice, but, of course, are subject to revision as further experimental data are obtained (see Figs. 16 and 17). Air Pipe. — Since in the usually very limited space high veloci- ties must be permitted we may allow a velocity of about 30 ft. per second or more in the air pipe. Submergence. — The ratio n _,_ ^ is defined as the submergence ratio. 1 Fig. 15. D + h Experience seems to indicate that this should be not less 80 COMPRESSED AIR ■> than one-half; and about 60 per cent, is a common rule of practice. Probably the efficiency will increase with the ratio of submergence, especially for shallow wells. The cost of the extra depth of well necessary to get this submergence is the most serious handicap to the air-lift pump. v Ratio jY-' — -When air is delivered into water under submergence D its ratio of compression will be D + 33.3 r = 33.3 33.3 ft. being a fair average for water head equivalent to one atmosphere. When air is mixed with water as in these pumps, it may be assumed to act under isothermal conditions. Then the energy in the air is p a v a log e r while the effective work realized by water delivered at top of discharge pipe is wQ i h+ B s being the velocity of discharge. Write ^— = Hi. Then if E be the efficiency of the pump (reckoned from energy in air deliv- ered) we have the equation wQ (h+hi) = E p a v a log e r. Whence Va w(h + hi) ..' Q ~Ep a log e r ^ In case of a pure-water air-lift pump w = 62.5, and we may take for average atmospheric conditions p a = 2,100. Then multiplying by 2.3 and using common logarithms the formula becomes th , v a h -\- Hi /0 _v For pure water ^ = 773ElogiQr (35) Complete data on several apparently well-designed air-lift pumps with ratio of submergence between 50 and 65 per cent, and total submergence between 350 and 500 ft. show E to have a value between 45 and 50 per cent, (see Art. 43). If we take E = 45 per cent., Eq. (35) becomes v a h + hi Q 35 logio r (36) THE AIR-LIFT PUMP 81 t — 1 -H- ■ — — ~T- : ■: J :■ :: : ■ jTT — 1 — 1 — 1 — 1 1 ;; :. : :.:: — " - vo --rtiTif" . co . .... j . : cr 1 :TTr> -■■; z - W - ||j7. I 1 ~ itti;:: \ \ —— ;.;■ - ; ::: ..:. — -* D r/5 -— — — — c, ^ ' , i o 1-1 --- o -H — i-i->-- o ".... "\ \ -T^V \ - -\ — rr :::: u " 1 1 L 1 1 ; 1 . u o — 1 h- o oo - \ \ :::.:: \ ■;, .:: :.. rtp — p-l= . : :; 3* V v ^?\ 73" fa\! o — ■ — + o to ^b :\iV : tA ' o . — . L. o <$■ V\ — — F" © \ \ v : ^__ _.'..:.. : ' .;•: . ,::; ' SE ~ ■ ~~~r~ ----]■ ._„_:_: IniL j-- \ \ \ V . : o — -— — - © : : : : n ■ o : --p-- : ; ; ■ - ... ■ :; J_ fun M^ rtTT + T: o 1— L o -irMlilllo t- cs q,o Plate IV. 82 COMPRESSED AIR Formula (36) is recommended -for the design of deep-well pumps. In this hi may be taken as about 6 ft. which is assum- ing a discharge velocity between 20 and 25 ft. For shallow wells hi may be taken as 1, which would correspond to a velocity of 8. ft. The curves on Plate IV represent Eq. (36) for ratios of sub- mergence as shown thereon. Note that in Plate IV an efficiency of 45 per cent, is assumed. When more data have been collected, some modification may be found desirable. In any case some excess air capacity should be provided; for should the free water surface in the well drop more than antici- pated, after prolonged pumping, more air will be needed to maintain the discharge. This is apparent on Plate IV. Note v that as submergence ratio S decreases ^ increases. Velocity in the Water Pipe. — This is the factor that most affects the efficiency, but unfortunately, owing to the usual small area in the well, the velocity cannot always be kept within the limits desired. The complicated action and varying conditions in a well make the designer entirely dependent on the results of ex- perience in fixing the allowable velocities in the discharge pipes. The velocity of the ascending column of mixed air and water should certainly be not less than twice the velocity at which the bubble would ascend in still water. This would probably put the most advantageous least velocity in any air lift at between 5 and 10 ft., and this would occur where the air enters the discharge pipe. The velocity at any section of the pipe will be Q + v s =-Z » a where Q and v are the volumes of water and air respectively and a the effective area of the water pipe, s increases from bottom to top probably very nearly according to the formula a \ ri I where K = increment of velocity, r — ratio of compression under running conditions, I = total length of discharge pipe above air inlet, x = distance up from bottom end of air pipe to section where velocity is wanted. THE AIR-LIFT PUMP 83 The formula is based on the assumption that the volume of air varies as the ordinate to a straight line while ascending the pipe through length I. As the volume of each bubble increases in ascending the pipe, the velocity of the mixture of water and air should also increase in order to keep the sum of losses due to slip of bubble and friction of water a minimum; but for deep wells with the resultant great expansion of air the velocity in the upper part of the pipe will be greater than desired, especially if the dis- charge pipe be of uniform diameter. Hence, it will be advanta- geous to increase the diameter of the discharge pipe as it ascends. The highest velocity (at top) probably should never exceed 20 ft. per second if good efficiency is the controlling object. Good results have been gotten in deep wells with velocities about 6 ft. at air inlet and about 20 ft. at top (see Art. 43). Figure 16 shows - the proportions and conditions in an air lift at Missouri School of Mines. The flaring or slotted inlet on the bottom should never be omitted. Well-informed students of hydraulics will see the reason for this, and the arguments will not be given here. The numerous small perforations in the lower joint of the air pipe liberate the bubbles in small subdivisions and some advan- tage is certainly gotten thereby. No simpler or cheaper layout can be designed, and it has proved as effective as any. It is the author's opinion that nothing better has been found where submergence greater than 50 per cent, can be had. Art. 41. The Air Lift as a Dredge Pump. — The possibilities in the application of the air lift as a dredge pump do not seem to have been fully appreciated. This may be due to its being free from patents and therefore no one being financially interested in advocating its use. With compressed air available a very effective dredge can be rigged up at relatively very little cost and one that can be adapted to a greater variety of conditions than those in common use, as the following will show. Suggestions. — Clamp the descending air pipe to the outside of the discharge pipe. Suspend the discharge pipe from a derrick and connect to the air supply with a flexible pipe (or hose). With such a rigging the lower end of the discharge pipe can be kept in contact with the material to be dredged by lowering from the derrick; the point of operation can be quickly changed within 84 COMPRESSED AIR the reach of the derrick, and the dredge can operate in very- limited space. In dredging operations the lift of the material above the water surface is usually small, hence a good submer- gence would be available. The depth from which dredging could be done is limited only by the weight of pipe that can be handled. In case of air-lift dredge pumps the ratio of submergence may be large and the weight per cubic foot lifted will be greater than 62.5 lb. In case heavy coarse material (such as gravel) is being lifted, the velocity should be high. Though the author has found no data by which the efficiency of such pumps can be determined, the following example is taken for illustration. v Example 35. — What is the ratio -^- for a dredge pump when submergence = 30 ft., net lift = 6 ft., velocity at discharge = 12 ft., percentage of silt = 33.3 per cent., weight of silt = 100 lb. per cubic foot, and efficiency = 0.333? Solution. — Referring to Eq. (34), w becomes 75, hi = 2, r = 1.9 (submergence 30 ft. in pure water). Whence we get Va_ = 75 X 8 + (100 - 62.4)^ X 30 Q ' 0.333 X 2,100 X 2.3 X 0.279 ' Note that the second term in the numerator is the work done in lifting the silt through the 30 ft. of water. Art. 42. Testing Wells with the Air Lift.— The air lift affords the most satisfactory means yet found for testing wells, even if it is not to be permanently installed. Such a test will reveal, in addition to the yield of water, the position of the free water surface in the well at every stage of the pumping, this being shown by the gage pressures. However, some precautions are neces- sary in order properly to correct the gage readings for friction loss in the air pipe. The length of air pipe in the well and any necessary correc- tions to gage readings must be known. The following order of proceeding is recommended. At the start run the compressor very slowly and note the pres- sure pi at which the gage comes to a stand. This will indicate the submergence before pumping commences, since there will be practically no air friction and no water flowing at the point where air is discharged. Now suddenly speed up the compressor to its prescribed rate and again note the gage pressure p 2 before THE AIR-LIFT PUMP 85 any discharge of water occurs. Then p 2 — Vi = P/ is the pres- sure lost in friction in the air pipe. When the well is in full flow the gage pressure p s indicates the submergence plus friction, or submergence pressure p s = p 3 — p/. The water head in feet may be taken as 2.3 X p s . Then, knowing the length of air pipe, the distance down to water can be computed for conditions when not pumping and also while pumping. ./£=> Fig. 16. Art. 43. Data on Operating Air Lifts. — In Figs. 16 and 17 are shown the controlling numerical data of two air lifts at Rolla, Mo. These pumps are perhaps unusual in the combination of high lift and good efficiency. The data may assist in designing other pumps under somewhat similar circumstances. The figures down the left side show the depth from surface. 86 COMPRESSED AIR The lower standing-water surface is maintained while the pump is in operation; the upper where it is not working. The broken line on the right shows, by its ordinate, the vary- ing velocities of mixed air and water as it ascends the pipe. The pump, Fig. 16, delivers 120 gal. per minute with a ratio j'l'pp O 1 T» : — = 6.0. The submergence is 58 per cent, and efficiency net energy in water lift = ; — = 50 per cent. pv log e r The pump, Fig. 17, delivers 290 gal. per minute with a ratio , = 5.2. Submergence = 64 per cent, and efficiency = net energv in water lift —, = 45 per cent. pv log e r The volumes of air used in the above data are the actual volumes delivered by the compressor. The volumetric effi- ciencies of the compressors by careful tests proved to be about 72 per cent. CHAPTER VII RECEIVERS AND STORAGE OF ENERGY BY COMPRESSED AIR Art. 44. Receivers for Suppressing Pulsations Only. — Every air compressor bf the reciprocating type has, or should have, an air receiver on the discharge pipe as close as possible to the com- pressor outlet. The chief duty of this receiver is to absorb or take out the pulsations caused by the intermittent discharge from the compressor in order that the flow of air through the discharge pipe (beyond the receiver) may be uniform, a condition evidently essential to efficient transmission. Incidently the receiver serves as a separator for some of the oil and water in the air and as a store of compressed air that may be drawn from when the demand is temporarily in excess of the compressor delivery. There is no standard rule, nor can there be one, for proportion- ing these receivers. However, the service demanded of the air will usually indicate whether or not a large receiver is desirable. The least size would apply in cases where the use of the air is continuous and the needed pressure constant — as in air-lift pumps. In such cases the requirement of the receiver is solely to suppress the pulsations. For such cases a rational formula v for the volume of the receiver would be V = c - in which V is r the volume of the receiver, v the piston displacement of one stroke (of low-pressure cylinder), r the total ratio of compression, and C an empyrical coefficient fixed by experiment or observation. Observe that v -f- r is the volume of compressed air per stroke, which being suddenly discharged causes the pulsations. The V ' question remains, what ratio of V to - will be necessary to suffi- ciently suppress the pulsations? The author finds no light on the subject in compressed air literature. Practice does not seem to distinguish between this case and the more usual one where some storage capacity is needed. The author is of the opinion that in such cases (the air lift for instance) a much smaller re- 87 88 COMPRESSED AIR ceiver could be used without detrimental effect, and thereby the cost reduced. Art. 45. Receivers. Some Storage Capacity Necessary. — In the majority of compressed-air installations the use of, or demand for, air will not be constant, as for instance in machine shops, quarries, mines, etc. In any such case the use of air may exceed the compressor capacity for a short time and then for a time may not be as great as the compressor capacity. In these (the more common) cases the receiver is intended to serve as a storage reservoir in addition to its several other duties. The problem of determining the necessary volume for the stor- age is simple, provided the maximum rate of use and its duration in time can be gotten ; but this is seldom possible as will be readily conceded when the complexity and irregularity of the service is considered. However, the problem may be better understood if expressed as a formula: Let V = volume of receiver, or storage reservoir, in cubic feet, v a — free-air capacity of compressor in cubic feet per min., Pi = highest pressure (absolute) permitted in the system, 7> 2 = lowest satisfactory working pressure permissible, R = maximum rate of usage, cubic feet free air per min., T = duration in minutes through which R is continued. To get a simple and sufficiently approximate relation assume isothermal changes and equate the pressure by volume products thus: PlV + TpaVa = P2V + TRp a . Whence, V = Pa T{R - v a ) Pi- Pi Example. — A shop has a compressor with v a = 300 and normal pressure p\ = 100 ft. A drop of 10 lb. (p 2 = 90) is permissible when the demand is double the average for 2 minutes. Then 1 f; V = ~ X 2 X (600 - 300) = 900. A calculation, such as above, applied to the more common in- stallations, will show the desirable receiver capacity much greater than is installed. The common practice seems to be to install a compressor RECEIVERS AND STORAGE OF ENERGY 89 capable of meeting the maximum demand without storage, and then let it run idle much of the time. Going to this extreme is profitable for the compressor makers, but expensive to the user in first cost of the compressor and still more so in the continual cost of extra fuel to run the larger compressor even though idle. Where a compressor has been installed with inconsiderable receiver (or storage) capacity and the business outgrows the capacity of the compressor as thus equipped, the addition of a considerable storage volume may defer the time for purchase of a larger compressor for several years, and at the same time get the needed additional air more economically than if a larger compressor were installed. This claim of economy is based on the fact that a small machine running more continually and with a nearly constant load is more economical than a larger machine running constantly but with intermittent load. The case is analogous to the use and duty of a distributing reservoir in a water-supply system. Art. 46. Hydrostatic Compressed-air Reservoirs. — In cases where it is desired to store large units of energy in the form of compressed air, and that energy is to be drawn out with but little drop in the air pressure, a computation of the volume necessary for tanks under conditions heretofore assumed is discouraging. Example. — Assume that storage is to be provided for air at 125 lb. (absolute) such that 100 hp. can be drawn from the storage for 1 hr. with a drop in pressure to 100 lb. What volume of storage is needed? For this example assume all changes to be isothermal. lot: inn Then 100 X 33,000 X 60 = p a V a log e ^| - p a V a log e ^ = 2,116 V a (2.140 - 1.917), whence V a = 4,200,000 and V = 50,000 approximately. Sup- pose now that all the compressed air in volume V at pressure 125 can be used without any reduction of pressure. What volume would give 100 hp. for 1 hr.? Then 100 X 33,000 X 60 = p a V a log 8.5, whence V a = 43,700 and V = 5,150 or about one-tenth of that required under the first assumption. This latter condition (making the entire volume of compressed air available without reduction of pressure) can be accomplished 90 COMPRESSED AIR simply and economically by the scheme illustrated by Fig. 18 which needs but little explanation. The water head against the air may be assumed constant. The dip down in the water pipe below the air reservoir is to prevent blowout through the water pipe should all the water be forced out of the air reservoir. A popoff, or an automatic, stop for the compressor, would be adjusted to act when the water line dropped down below the air tank as at C. Evidently the water pipe ABC need not be vertical nor in a vertical plane. The water reservoir can be economically placed on a hilltop in the neighbor- Fig. 18. hood, or the air reservoir can be placed in underground chambers (abandoned chambers in mines, for instance) and the water reservoir at the surface. This last suggestion naturally leads to that of using under- ground chambers naked, that is, without the steel tanks. This is quite feasible. To make the walls of such a chamber tight against escape of air into the rock the " cement-gun" is ideal. Note that there is no necessity for smooth or even surfaces. The cement-gun may be found more efficient if used while the chamber is under some pressure. The cement will thus be driven into every crevice and pore into which air may tend to escape. CHAPTER VIII FANS Art. 47. — The discussion in this article will apply to any centrifugal pump and to fans of the low-pressure type such as are applied in ventilation of mines and buildings, when change in density of the air may be neglected. Though the discussion is brief, the student entering the subject for the first time will find some difficulty in keeping in mind the several qualifying conditions such as relative velocities, velocity heads, pressures within and without the wheel, etc. He is warned not to jump at conclusions in this nor any other branch of fluid mechanics, but is urged to study and review each demonstration over and over again until familiar with it. We hope to encourage an interest in this subject by saying beforehand that there are several fallacious opinions that can be successfully contended with only by those who are well grounded in the following under- lying theory. We will first study the theory as revealed by the laws of pure mechanics and the conservation of energy, without considering the effects of friction, imperfection of design or improper opera- tion. Formulas thus obtained will not closely check with re- sults from a pump or fan in operation, but they tell what per- fection would be, and so show how far short we fall in practice; and they point to the best lines of improvement in design and in operation. In addition to the symbols shown on Fig. 19, the following will be used: w = weight of cubic unit of fluid, b = width of discharge at outer limit of vanes, 61= width of inlet at inner limit of vanes, Q = volume of fluid passing, cubic feet per second unless otherwise stated, W = total weight of fluid passing per second wQ, p = pressure head immediately after escape from revolving wheel, H = total head given to the fluid by the wheel. 91 92 COMPRESSED AIR The reason for using heads instead of pressures is that the formulas are thereby simplified. Note that head must be in C cdx B 1 mm^m Fig. 19. feet of the fluid passing. In case of air the air head is to be of constant density. FANS 93 Art. 48. Purely Centrifugal Effects. — Consider a prism of fluid of unit area and extending between the limits r\ and r, in a revolving wheel without outlet, as CB, Fig. 19. The centrifugal force of an elementary disk across this prism, of thickness, dx, and distance x, from the axis of revolution, is by well-known laws of mechanics 7/ . wu x 2 dx df = gx where u x is the velocity of revolution at the distance x from the center. Thus 3/ u x = - u r and therefore wu df = — r xdx gr 2 and the total centrifugal force/, which is effective at the outer end of this prism, is the integral between the limits x = r and x = rj. Hence wu 2 ,-„ „ N w 2 gr 2 2 g since Mi = — U. r This is the pressure on a unit area at the circumference of the wheel, and, evidently, it is independent of the form or cross- section of the arm CB. Now, pressure divided by weight gives head. Hence, the pressure head against the walls of the wheel at the circumference is 7 . u 2 — Mi 2 •v h 1 = 2g Note that if ri = then h = „ — 2g Note that this h does not include velocity of rotation. Now, if an orifice be opened at the circumference, in any di- rection whatever, and the pressure outside be the same as at the entrance, the velocity of the discharge, relative to the revolving walls of the wheel, will be V = V2~gh or V 2 = u 2 - u x \ Note that if r = 0, V =-u. Note also that the absolute velocity v of discharge is made up 94 COMPRESSED AIR of the two components V and u and in amount v 2 = u 2 + V 2 + 2 uV cos /3 = 2 u 2 + 2 wF cos when ri = 0. v 2 Total head, i/, in the departing fluid = ~— or H _ u 2 + uVgos fl , _. When there is a discharge, there must be an initial velocity, Vi, at the entrance, and this must be considered in the final head within the wheel. Thus, the total relative head at B will now be 7 Vi 2 , u 2 — Ui 2 h = ^ h 2flf ' 2g and the velocity of the discharge, relative to the revolving parts, will have the relation v 2 = y, 2 + u 2 - wi 2 (i) Suppose, now, that CB is a radial frictionless tube, open at both ends, and that a particle of matter starts from a state, Vi, relative to the tube, and moves out, without change of pressure, from radius r x to r, in obedience to the laws of centrifugal force (or acceleration). Its radial acceleration, when distant x from the center, is, by well-known laws of mechanics : Acceleration u x 2 _ dV x x dt Also Vx = dt' Therefore, by eliminating dt, we get u x 2 dx but, as before, V x dV x = x X U x = — U r (sub x indicating the conditions at the distance x from the cen- ter) . Therefore, V x dV x = —r xdx. FANS 95 Integrating between the limits V and Vi, on one side, and r and ri, on the other, we get as before 7 2 -Fi 2 = U 2 _ Mi s (I) Art. 49. Impulsive or Dynamic Effects. — We have now to study the effect of picking up the fluid at entrance to the moving parts of the wheel. This will be studied by a method somewhat different from that preceding: Assuming the fluid to be at rest until influenced by the wheel, we see that during each second there is a weight, W, brought to a velocity, v, Fig. 20. Now the reaction against the wheel due to Wv the creation of the velocity, v, is F = — and the component of this velocity opposite in direction to the rotation, u, is v cos and this equals u — V cos (180° — 8) = u + V cos /3 and since work is force multiplied by distance, the work done in overcoming the reaction is W W u(u -f- V cos 13) — = (u 2 + uV cos 8) — g g If H be the total head given the fluid up to the point considered, 96 COMPRESSED AIR then work done = WH, since all the head has been imparted by the wheel. Hence, H = W<1 + uV cos A (37) 9 Note that it is preferable to use the angle /3 rather than a for /3 is fixed and is an element in the design of the machine, while a varies with u and V. The demonstration above evidently applies, however short the radial depth of the vanes (r — r{). So we may say that it applies at the entrance where r — r x = 0. Here, then, we find Eq. (37) applies in case of purely impulsive, or dynamic action with neither centrifugal force nor centrifugal acceleration. It is now apparent that no matter at what distance from the center of rotation the fluid is engaged by the wheel, it will have imparted to it a head the same as if it had been under influence of the wheel from the center out. Art. 50. Discharging against Back Pressure. — Note that so far we have assumed that the pressure against the discharge is the same as at intake. Under this condition the relative velocity of escape will be V = u, no matter in what direction V may be, relative to the wheel. We have now to establish a general formula for H when pres- sure head against outlet is p. and at inlet p\. Note that pi may be and generally is negative in centrifugal pumps, but in fans it is usually zero. We have established the fact that the pressure head developed within the wheel, when no discharge is allowed, is h = ^— ■ Now i & J 2 g if an orifice be opened through the periphery of the wheel into the discharge duct where pressure head = p, the velocity of escape (relative) will be V 2 = 2g(~+p i -p)=u* + 2g(p 1 -p) (II) Whatever the absolute velocity of escape v may be, the total absolute head added to the fluid by the machine will be H=£g + V-Vl (Hi) Note that when pi is negative (or suction), it becomes a plus addition in (III). FANS 97 From pure trigonometry we have in any case v 2 = u 2 + V 2 + 2 uV cos j(3. Now in (III) replace v 2 from the last expression, then replace V 2 with its value from (II) and we get as before H = u 2 + uV cos ft , g We have now proven Eq. (37) to be correct for both purely centrifugal and impulsive action, and to be independent of entrance and discharge pressures. Equations (II), (III) and (37) are the theoretic relations when effects of friction and imperfections of design are neglected. Results in practice may be, and often are, quite different from what this theory would indicate, due to imperfections of design, some of which cannot be overcome entirely. v 2 Note that if p\ = P, then H = ^— and V = u; and if pi = p in 2 and = 90° then v 2 = 2u 2 . When = 90°, H = — irrespective of u 2 pressures. Also, when (3 is less than 90°, H is greater than — u 2 and when /3 is greater than 90°, H is less than — • The pressure that a pump or fan can produce depends on H. P = wH when the whole energy is transformed into pressure head. Otherwise, in general at or near outlet, friction being neglected. The work required of the machine (neglecting friction, etc.) is WH = wQH where W and Q are total weight and total volume passed, respectively, and this will in actual performance nearly equal the theoretic, regardless of friction and other losses. qi 2 Note that V may be zero and still H = — . In this case the 9 u 2 fluid revolving in the wheel has a pressure head = ^— , and a ve- u 2 locity head = „— , the total head being the sum. In this case the work is zero since W = 0. If the pressure head, p, in the discharge duct be ^— , there will be no discharge, the pressure ^g 7 98 COMPRESSED AIR inside and outside the wheel balancing. As p decreases V in- creases (see Eq. (II)) and therefore also Q increases. In case of pumps when /3 is less than 90° the pump cannot start a discharge under full back pressure, but if /3 be less than 90° the pump may start under its full head. Art. 51. Designing. — The dimensions of any pump or fan must conform to the following formulas, which hold in all cases: Q = 2 wrbV sin ,3 = 2 vrb sin p^u 2 - 2 g(p x - p) (V a ) Also Q = 2 TTibiVi sin (V b ) Note that V sin (5 and Vi sin 8 are the radial components of velocity of discharge at outlet and inlet respectively. In designing a fan or pump, the chief factors are H and Q. By equation (37) these factors are seen to be interdependent (except where j8 = 90°), since for any completed machine Q is directly proportional to V. Unfortunately there seems no rational formula for V. The formula V 2 = u 2 — 2 g(pi — p) is theoretically correct, but there is no satisfactory way to determine or fix p in this formula pre- liminary to the design. This fact necessitated dependence on the cut-and-try process by the pioneers in this field; though now data have been gotten for so many pumps and fans of various styles, showing the relation between head and discharge, that designers can proceed with tolerable confidence except where some radical departure in design is to be tried (see Art. 53) . Assuming that the designer has the data showing relation between H and Q (or V) for the style of machine he is going to copy, he has equations V a and Vb to conform to. The angle /3 should be selected with due regard to the service required of a machine and method of propulsion. Notice that, assuming u constant, when : Angle j8 is less than 90°, H increases as Q increases. Angle fi is equal to 90°, H is independent of Q. Angle j(3 is greater than 90°, H decreases as Q increases. It is common to assume the fluid as approaching the inner limits of the fan blades in a radial direction (see Fig. 19) even when no guide vanes are provided, though in that case the as- sumption may be far from the truth. Note also that with H fixed, u must increase as /3 increases. This fact is taken into consideration when a machine is to be FANS 99 driven by a high-speed electric motor or a steam turbine. In such cases the embarrassing condition in the design is to apply the high rotative speed without getting excessive head; hence, in such cases the angle /3 is made greater than 90°. Another advantage in turning the vanes backward (0 greater than 90°) in case of electric-driven machines, is that the machine will not be overloaded when the head or resistance is suddenly thrown off, with the resulting great increase in discharge (which increases V) (see Fig. 21). In cases where a machine is to be run by a reciprocating steam engine direct-connected and the designer has trouble in *9 07 24 f? <90° SI LUt-C ffH sad ^^ ■^.^ ft' = 90° ^ '^90 > "^> N 's N \ s 123456789 10 Cubic Feet per Minute -Thousands Fig. 21. — Characteristic curves. Constant speed. Varying discharge. getting the desired head with the limited speed, he will find it advantageous to turn the vanes forward. Where a constant head (or pressure) is desired with varying quantity as in sewage pumps and ventilating fans for buildings, the most rational design would be to provide an adjustable dis- charge with radial vanes. Another condition that should receive consideration in de- signing, or selecting, a machine is where a pump is to force water through a long pipe and where a fan is to force air through a mine. In such cases the greater portion of the resistance to be over- come is friction head, and it is well known that this varies with the square of the velocity, and, therefore, any increase in the 100 COMPRESSED AIR quantity will be accompanied with a relatively greater resisting head. This case would be best met by setting the vanes radial (at discharge). Then, theoretically, H = — and quantity varies directly with u, when running under most favorable con- ditions. Now, as stated before, friction will vary as quantity squared. Hence, H will vary directly with the friction. This very nearly meets the most desired condition in mine ventilation where practically all the resistance is due to friction. To illustrate numerically: Suppose it is decided to double the quantity of air passing through a mine. If we double the speed of the fan we get double the flow of air, four times the pressure, four times the friction and eight times the power will be neces- sary to run the fan. Probably the engine or motor in the above example would be incapable of developing eight times its normal power. How then can the problem be solved? Will it be effective to install a duplicate of the first fan and discharge both into the mine? No, for we would be trying to put through double the quantity without increasing the pressure. The result would be a some- what greater pressure and quantity, but both fans would now be working inefficiently, if they were properly adopted to the first condition. Would the problem be solved by placing another fan in series (or tandem) with the first? No, for now we would be proposing twice the head with no increase of volume. There would be some increase in volume, but not double, and again the fans would not be working under best conditions. By this simple example it is evident that if there is a radically different quantity to be sent through a long conduit (pipe, flume or mine), the only scientific solution is to install a new machine adapted to work efficiently under the new conditions. Art. 52. Testing. — The following discussion refers to fans or blowers. Manufacturers of recognized standing have facilities for testing their machines and should know, with sufficient accuracy for commercial purposes, what their machines will do and the condi- tion under which they will work most efficiently, and the pur- chaser of a machine for any important serivce should demand a guaranteed performance chart for the machine, this chart to give information equivalent to that shown on Fig. 22. Then, in the FANS 101 acceptance test the engineer for the purchaser might be content with checking a few points on the performance chart under con- ditions approximating those under which the machine is to work. In the purchaser's test, the data wanted are quantity, head and efficiency. Too often the purchaser is content with determining the first two (or even with no test at all if the machine runs and does some work). A test will require the service of a technical man, but under competent direction should not be difficult nor expensive. The head will be measured in inches of water in a U-tube (see precautions, Art. 23a). The quantity may be determined by measuring velocity and area. Where very great quantities are passing, the annamometer is the most convenient instrument for measuring velocity, but it should not be depended on in unskilled hands. It will need careful rating and should be applied in all parts of the cross-section of the conduit, and the total quantity found by summing the products of small areas by their respective velocities. In doing such work the operator's confidence in the method is apt to be shaken by the discovery that the velocity will vary considerably over the area and will also vary with time at a fixed point. In any case, the section at which the velocity is taken should be well away from the machine, else the irregular currents will render the observations altogether unreliable. The author is partial to orifice measurement, even for testing the largest fans. Orifice coefficients are now available up to 30 in. diameter or 30 in. square (see Art. 23). It is the author's opinion that a coefficient of 0.60 will result in errors well within those made in reading water gages, and quite certainly with errors less than are apt to enter any annamometer or petot-tube measurements. Note that the orifice coefficient is constant, while that of a petot tube or a revolving annamometer must be found for each instrument and may change with the slighest injury or misuse of the instrument, and note further that with reasonable care that cross-currents are not allowed in front of the orifice, its discharge is not effected by unequal velocities in the cross-section of the conduit. Omitting any discussion of apparatus for measuring velocities, quantities and pressures, with their calibration and defects, the engineer will need to determine : 102 COMPRESSED AIR v = the velocity of air passing the section of area, a (feet per second), W = weight of air passing (pounds per second), P = pressure of air at section = 5.21 i where i is pressure in inches of water (pounds per square feet), N — revolutions per minute, u = 2rrN, Q = volume passing = av (cubic feet per second). Ei = power put into the fan (foot-pounds per second), E 2 = the useful work done by the fan. Then, and efficiency E% E~i Wv 2 He should also have all dimensions and angles of the fan in order better to interpret results. D P< I at 12Q0 R.P .M. ' $ &/ i i — p -atJ nnri 04 < i i i 6^. A rt §s oo i i i i Pi it 30 u V \ \ U l\ ^ x P a t COC // \J ii \ — .. '/ / ^ "^ A V. ^ *^> ^^ p at 4U0 1 1 3 4 5 G 7 Cubic Feet per Minute in Thousands Fig. 22. The variables are N, v, and P. In a thorough test to get the performance chart of a fan, the preferable method is to maintain a constant N throughout a series of runs in which P is varied at will by the operator, v measured and E computed. Then, with another N another series is run as before and so covering the desired range for the fan. From these notes the performance curves can be drawn. The most important of these are those for efficiency and for quantity (see Fig. 22) . FANS 103 The measures of v and P should be taken in a section of smooth straight conduit some distance from the fan. The pressure is controlled by placing some kind of shutter in the conduit beyond the section at which v and P are measured. A performance chart of the class shown in Fig. 22 shows in remarkable completeness what can, and should, be accomplished by the machine and under what conditions it will work most efficiently. On this chart the pressure curves and efficiency curves are plotted in the usual way as suggested above, then the efficiency contours are located thus. To locate the 55 per cent, contour, find the two points where the 1,000-r.p.m. efficiency line crosses the 55 per cent, line (at about the 5.2 and 7.7 ver- ticals). Shift these points vertically to the 1,000-r.p.m. pressure line and mark 55. Similarly find where the 800-r.p.m. efficiency line cuts the 55 per cent, efficiency line (at 3.6 and 6.). Shift these up (or down) to the 800-r.p.m. pressure line and mark 55 as before, etc. Connect the points of equal efficiency by a curve. Similarly the 60 per cent, contour can be drawn. Then evidently the best combination for operating the machine is within the area surrounded by the 60 per cent, efficiency con- tour. For instance, if we want 7,000 cu. ft. per minute, the machine should be speeded to about 1,000 r.p.m. and at these rates the pressure would be about 7 in. Of if we want 5,000 cu. ft. per minute the speed should be about 800 and the pressure about 5 in. Art. 53. Suggestions. — The following suggestions seem to be the rational conclusions pointed to by theory, the difficulties in controlling operation, and complications in analyzing the results of tests. Observing the rules as to smooth curves, polished surfaces, and correct angles, design a high-speed fan with characteristics as revealed in Fig. 23. DBE is an adjusting tongue hinged at D. By this area AB can be varied at will. The area of the sections gh, etc., are so proportioned as to maintain the velocity u in the volute until the throat, or switchpoint, A, is passed, after which the velocity is gradually reduced and pressure increased (by the well-known laws of fluid dynamics) in the trumpet-shaped outlet. In operation the intent would be to keep a constant pressure in the volute up to AB, this pressure head being approximately u 2 sr- * Then the whole theory of this machine would be 104 COMPRESSED AIR _ _. u z , ' wau 6 Q = au, H = — and E 2 = • 9 g A factory test of such a machine would reveal the most favor- able relation between u and p. Then a size would be selected that would give the desired Q. In operation there would be a Adjuster Sliding Contact - Fig. 23. water gage tapped into the section AB to show p at any time. When the operator notes that p is low, he will open up the area at AB and vice versa. Note particularly that in such a design Q can be controlled independently of H. CHAPTER IX CENTRIFUGAL OR TURBO AIR COMPRESSORS Art. 54. Centrifugal Compression of an Elastic Fluid. — The demonstrations given in the preceding chapter apply to any case where there is no change of density of the fluid while passing through the machine, and this includes the case of centrifugal acceleration without change of pressure, and purely impulsive action. We have now to study the case where compression due to centrifugal force within the wheel, or runner, is so great that it must be considered in the formulas. We will assume isothermal conditions, since the ratio of com- pression in each stage is low and intercooling can be applied between each stage. The formulas thus gotten are simpler than can be gotten otherwise, and are as accurate as is justified by other considerations. In Fig. 19 assume the cylinder CB filled with a compressible fluid, as air. The weight of a unit volume will not be constant, but will depend on the distance x from the center and on the velocity of rotation. Let w x be the weight of a unit volume at distance x from the center. Then the centrifugal force due to a disk of unit area and radial thickness dx will be Also 1 iV x^x 7 UJ %W 7 " **S ap x = dx = — r- xdx since u x = - u. gx gr z r Wi pi where p x is absolute pressure in the air at distance x from the center, and Wi and p x are the weight and pressure respectively of the air at entrance. Substituting and dividing by p x there results dj) x _ WiU 2 I Pi Px ViQT 2 Jri - L — = 5 xdx, then I - J — = 1 xdx. p x Pigr 2 Jpi p x Pigr- Whence 105 106 COMPRESSED AIR . p Wiu 2 /„ ,\ Wi/u 2 —Ui 2 \ . r log c — = — - — -lr 2 — ri 2 ) = — ( — = ) since n = -Mi. pi pi'2gr 2 \ I p x \ 2g I u If ri = and we consider a single-stage machine taking in free air we will have where R' is the ratio of compression at the periphery but within the revolving wheel. Assume that the machine is in 1 second putting a volume of free air = v a into the state of pressure and motion indicated above. Then the work done per second will be ■1/2 9/ 2 PaV a l0g c R' + W a V a 7T~ = W a^a ~ (&) ^9 9 when the value of log c R' from (a) is inserted. Note that this is the same as would result if the machine were working on an inelastic fluid of weight w a (see Art. 50). Note also that the work done in compression is equal to that done in giving velocity. Art. 55. A More Direct Derivation of Equation (37). — Appli- cable also to Centrifugal Air Compressors: After a study of Arts. 49, 50 and 54 the student will be pre- pared for the following more general and more direct demonstra- tion of Eq. (37) and its application in case of considerable com- pression. Referring to Figs. 24 and 20. The static pressure in the fluid changes as it passes out of the rotating part into the fixed out- let passage. It is this drop in pressure that induces the relative discharge velocity, V. This difference in pressure offers no resistance to the rotation of the wheel; as will be readily seen if we imagine the perifery of the wheel closed while rotating in a frictionless fluid. The pressure in the frictionless fluid must be normal to the perifery and therefore does not resist its rotation. Then in all cases (regardless of change of pressure at outlet) the resistance to rotation is due solely to the reaction of the departing jet. This reaction is in direction opposite to that of the absolute velocity of discharge, v, (Fig. 19) and in amount v is W-. But the component opposed to rotation (that is in direc- v tion opposite to u) is W~ cos 6 and as is apparent on the dia- CENTRIFUGAL OR TURBO AIR COMPRESSORS 107 grams v cos 6 = u + V cos 8. Therefore the force opposed to W rotation is — (u -\- V cos 8) and since work done by the wheel equals force multiplied by distance. Then W Work = — (u 2 + uV cos 8). g Evidently this is independent of the radial depth (r — r x ) of the vanes. Then the radial depth of vanes is a matter of convenience or expediency. In case of a fluid of uniform density (in low pressure fans we may neglect change of density) if the machine imparts a head, H, to the fluid, then work = WH: Whence W WH = — (u 2 + uV cos 8) (J and = u 2 + uV cos 8 g In case of a compressible fluid (as air) and we are to consider the work done in compression. Let Ri be the final ratio of compression when the air has been brought to rest after one stage. Then work = p a v a log c R: where v a is the volume of free air compressed. Then and y a Va log c #i = — — (u 2 + uV cos 8) w a (u 2 + uV cos 8\ log e Bi = —(- -— -) (37a) This is the formula for ratio of compression produced by one stage of a centrifugal air compressor. All the discussion in Art. 51 concerning the effects of the angle 8 applies also to equation (37a). The student should read that article as a part of this study. If there are n stages in a machine, each giving an additional ratio, R h and the final ratio from free air be R n , then R n = R\ n and log c R n = n log c R\ (38) Art. 56. Working Formula. — The very great centrifugal force developed in these machines prompts manufacturers generally to prefer to set the outer tips of the propeller blades radial {8 = 90°) 108 COMPRESSED AIR to avoid cross bending. This is good practice for other reasons (see Art. 51) one being that formulas for designing and analysis are much simplified, as the following will show: In Eq. (37a) assume p = 90°. Then . W a U 2 l0&Ri = y«7 7) W Note that w = ,. Q Q - . and, if t be assumed constant, — - will be constant. To adopt the formula to common logarithms (which will be more convenient) divided by 2.3026. In such a machine perfect cooling cannot be accomplished. We will assume for this study an average temperature of 580 (120°F.). Then the formula becomes l0gl ° Rl = (2.3026 X 53.35X 580 X 3T2) U * = fcu ' (39) log k = 7.6398. In the following examples (taken from practice) j8 = 90°. Example 1. — A three-stage machine, 54 in. in diameter, r.p.m. 2,500, in operation, gives 15 lb. gage pressure and delivers 35,000 cu. ft. per minute of free air, the power necessary being 2,700 hp. Determine the efficiency of the machine as to pressure and power. Here u ^^ X 3.14 X ~ and logw 2 = 5.5368 u 590 60 12 log k = 7.6398 log of log #x = T.1766 logfli = 0.1500 #1 1.40 3 log R n = 0.4500 R n 2.83 Assuming p a = 14.4 where the machine is in operation, then p = 2.83 X 14.4 = 40.75 and gage pressure = 40.75 — 14.4 = 26.3. 15 Then efficiency as to gage pressure ^-5 = 57 per cent, and theoretic efficiency as to work would be, by Eq. 32, log R log 2.04 0.3096 log R n log 2.83 0.4518 68 per cent. The report of the test of the machine gave the "shaft" effi- ciency as 71 per cent., the meaning not being further defined. CENTRIFUGAL OR TURBO AIR COMPRESSORS 109 Example 2. — A single-stage machine, 34 in. in diameter with 3,450 r.p.m. gave 3.25 gage pressure and the horsepower was 350 for 18,000 cu. ft. per minute. What efficiency as to pressure and power did the machine show? q 4 f^O 34 lL = ^p X 3.14 X ^2 and logu 2 = 5.4048 log k = 7.6395 log of log Ri = T.0443 logfli = 0.1107 Ri = 1.29 Assuming 14.5, then P = (1.29 - 1) 14.5 = 4.2 nearly. 14.5 + 3.25 The ratio efficiency would be — ... - J — = (1.22) divided by 3.25 1.29 = 95 per cent, and gage pressure efficiency = -j^ = 77 per cent. Horsepower necessary to compress 18,000 cu. ft. per second to R = 1.22 is 218. Therefore, the efficiency as to power = 218 ^~. = 63 per cent. Example 3. ■ — A six-stage machine 27 in. in diameter, r.p.m. 3,450, gives 15 lb. gage and 340 hp., capacity 4,500 cu. ft. per minute. 3 450 27 u = -^- X 3.14 X j2 lo S w2 = 5 - 1976 log k = 7.6395 log of log R x = 2.8371 logfli = 0.0687 Ri = 1.17 6 log R n = 0.4122 R n = 2.58, P n = 37.5 P g = 23 lb. The ratio accomplished by the machine is 2 very nearly; there- 2 fore, the ratio efficiency ^-^ = 77 per cent. The work necessary to compress 4,500 cu. ft. per second to R = 2 is 198. Therefore, the work efficiency is 58 per cent. When pressure is low, as in Ex. 2, the estimated efficiencies will be materially effected by the atmospheric pressure and the pres- 110 COMPRESSED AIR sure developed should be determined by a water or mercury column; otherwise only rough approximations will be obtained. Art. 57. Suggestions. — The following considerations point to the conclusion that best results will be gotten from centrifugal air compressors when the air is held in the machine until every par- ticle is under full centrifugal pressure regardless of its position relative to the propellers. Then it will escape through the outlet passages with uniform velocity and pressure, a condition evi- dently essential to high efficiency. Otherwise, if the air is still under the impulsive pressure of the vanes as it escapes from the machine, those particles next the propeller as at d, Fig. 24, must be under greater pressure than those at c, and the velocity of Fig. 24. escape relative to the revolving machine will be greater at d than at c. In these machines the velocity of rotation, u, is always very high and any moderate relative velocity of discharge (say within 100 ft. per second) will leave the absolute path of the escaping air nearly on a tangent to the perimeter, as at ab. This being the case, a flaring fixed receiving passage about as shown at (6), Fig. 24, would cause the velocity to be gradually checked. It is not apparent that any advantage will be gotten by putting tongues ef in this outlet passage. They would increase friction without apparent compensating benefit. Note that a section of the flaring outlet on a horizontal plane through ab will show a much longer path than the radial section (see (c), Fig. 24). The very great centrifugal stress in these machines lead manu- CENTRIFUGAL OR TURBO AIR COMPRESSORS 111 facturers generally to prefer to set the outer end of the propellers radial, and this is good practice for other reasons, one being the simplified formula for designing. Art. 58. Proportioning. — Let v a = volume of free air to be compressed, cubic feet per second, r = radius to outlet of propeller, ri = radius to inlet of propeller, u = velocity of rotation at outlet, Uy = velocity of rotation at inlet, S = radial component of the velocity at outlet, Si = radial component of velocity of air entering at radius ri, 4> = angle between forward direction of U\, and tangent to vane at inlet, b = width of outlet, 61 = width of inlet. All linear units in feet. R' = ratio of compression of air within the wheel at the outlet but before escaping, Ri = ratio of compression when brought to rest at end of first stage. Then tan d> == — Mi and V a = 2 wribiSi. Usually 61 is to be determined by this relation, all other factors being known. Note that this equation holds only when Si is the radial com- ponent of outward movement into the vanes. When there are no guide vanes at entrance, Si becomes uncertain and erratic. When it becomes the practice to put guide vanes at entrance, much of the uncertainties in the design of such machines will be removed. If there are to be n stages and a final ratio of compression 1 R n , then Ri = R n n and R' = Ri*. u will be fixed by the re- lation from (38) 112 COMPRESSED AIR When u is determined, r and r 1 can be assigned between limits found advisable by experience and the necessity of having pas- sages of sufficient area. At the outlet the width b is fixed by the relation ^- = 2irrbS. The greatest difficulty in the theoretic design of this class of machines is in correctly predicting the factor S (or relative ve- locity of discharge) . It is, of course, quite sensitive to changes of pressure in the discharge ducts. It is this doubtful factor chiefly that forces the designer to depend on results of tests. Fortunately, the width can be varied without affecting any other factors except V a . Hence, after test of a design of machine the desired capacity can be gotten by varying b and &i. The discharge of a centrifugal blower can be made adjust- able without varying the pressure by the simple device shown in Fig. 23. Finally, the student should be reminded that the above mathematical formulas do not include losses due to friction nor imperfections of design. Their chief value is to show what would be realized in a perfect machine and so reveal the short- comings of a machine and guide the designer in modification for improvements. CHAPTER X ROTARY BLOWERS Art. 59. — In certain lines of manufacture, there is necessary a supply of air in great volume, and at pressures not found prac- ticable for fans and yet so low, that to build reciprocating com- pressors to meet the demand would seem extravagant, when the cost is compared to the power demanded. Fig. 25. This demand has, in the past, been most economically met by the class of machines known as rotary blowers. These vary in details and there are several patterns on the market. Perhaps the simplest and best known is illustrated in Fig. 25. The two propellers revolve as shown by the arrows, and as is apparent by inspection the pockets of fluid (air or water) are forced upward. The flow is continuous, but not uniform ; neither is the tort on the shafts constant. The irregular discharge and tort tend to cause 8 113 114 COMPRESSED AIR vibrations but this is met by making the machines heavy and rigid (and in case of pumps by putting air chambers near both in- let and outlet). There are no valves in the machine, and the makers do not design them to rub or move in contact at any sur- face within the casing, but depend on accurate workmanship to make the clearance between the surfaces so small as to render the leakage so small as to be tolerable. Then, having no valves nor rubbing surfaces, the machines handling air should be quite durable, but this cannot be said of them, when used to pump water containing any grit. It is a good practice to supply a liberal quantity of thick oil to a blower, not for lubrication, but to reduce clearance between the surfaces. It is necessary to note one important difference in the working of this class of machine as a blower, and as a pump. This is due to the reexpansion of the compressed air out of chamber B into A, as soon as communication is opened between the two chambers. This is lost work and would limit the pressure at which the ma- chine could operate economically, even if slippage did not increase with pressure. For these reasons the useful range of pressures on such machines seems to be between }4 and 5 lb. For pres- sures below }4 lb. fans are usually selected, on account of the less cost. For pressures above 5 lb. reciprocating compressors are usually selected, on account of the better efficiency. Apropos to this phase of the subject, read Chapter IX on "Centrifugal Air Compressors." CHAPTER XI EXAMPLES AND EXERCISES Art. 60. — The following combined example includes a solution of many of the types of problems that arise in designing com- pressed-air plants. The student will find it well worth while to become familiar with every step and detail of the solutions, which are given more fully than would be necessary except for a first exercise. Example 60. — An air-compressor plant is to be installed to operate a mine pump under the following specifications : 1. Volume of water = 1,500 gal. per minute. 2. Net water lift = 430 ft. 3. Length of water pipe = 1,280 ft. 4. Diameter of water pipe = 10 in. 5. Length of air pipe = 1,160 ft. 6. Atmospheric pressure = 14 lb. per square inch. 7. Atmospheric temperature 50°F. 8. Loss in transmission through air line = 8 per cent, of the pv log e r at compressor. 9. Mechanical efficiency of the pump = 90 per cent, as re- vealed by the indicators on the air end and the known work delivered to the water. 10. Average piston speed of pump = 200 ft. per minute. 11. Mechanical efficiency of the air compressor = 85 per cent, as revealed by the indicator cards. 12. Revolutions per minute of air compressor = 90 and vol- umetric efficiency = 82 per cent. 13. In compression and expansion n = 1.25. Preliminary to the study of the problems involving the air we must determine: (a) Total pressure head against which the pump must work. By the methods taught in hydraulics the friction head in a pipe 10 in. in diameter, 1,280 ft. long, delivering 1,500 gal. per minute, is about 20 ft. Therefore, the total head = 450 ft. 115 116 COMPRESSED AIR (b) Total work (Wi) delivered to the water in 1 min. Wi = 1,500 X SM X 450 = 5,625,000 ft.-lb. (c) Total work (W) required in air end of pump. By specifica- tion 9, W = ^ = 6,250,000 ft.-lb. = 190 hp. For the purpose of comparison, two air plants will be designed; the first, designated (d) as follows: (d) Compression single-stage to 80 lb. gage. No reheating. No expansion in air end of pump. Pump direct-acting without flywheels. Determine the following: (dl) Air pressure at pump and pressure lost in air pipe. By specification 8 and Eq. (32), 100 " — 80 + 14' ° r l0g 14 = 0!)2 l0g 6 ' 72 - log ™T4— P 2 Whence, using common logs, log y^ = 0.76118 and p 2 = 80.78. Then lost pressure = p x - p 2 = 94 - 80.78 = 13.22 = f, and gage pressure at pump = 80 — 13.22 = 66.78. (d2) Ratio between areas of air and water cylinders in pump. The pressure due to 450 ft. head = 450 X 0.434 = 194.3, say 195 lb., per square inch; and since pressure by area must be ■ , . area air end 195 _ . equal on the two ends, r -j — nn no = 3 nearly. M area water end 66.78 (d3) Volume of compressed air used in the pump. Cubic feet per minute. Evidently from solution (d2) the volume of com- pressed, air used in the pump will be three times that of the water pumped, or v = I t c X 3 = 601.6 cu. ft. per minute. 7.48 (d4) Diameters of air cylinder and of water cylinder. Since the piston speed is limited to 200 ft. per minute (specification 10) and the volume is 1,500 gal., we have, when all is reduced to inch units and letting a = area of water cylinder, a X 200 X 12 = 1,500 X 231. Whence a = 144 sq. in. which requires a diameter of about 13% in. EXAMPLES AND EXERCISES 117 The area of air cylinder is by {62) three times that of the water cylinder, which gives a diameter 23)^ in. for the air cylinder. {65) Volume of free air. From {61) r at the pump = 5.76. Therefore v a = 601.6 X 5.76 = 3,465 cu. ft. per minute. {6Q) Diameter of %ir pipe. The mean r in the air pipe is - — -^— = 6.24. Using this in Eq. (27) with c = 0.06, we get 6 = 5 in. 13 22 Or using Plate III with r X 13.22 -f- 1.160 or r X yJqq on the fr line and 3,465 on the V a line, the intersection falls near the 5-in. point on the d line. {67) Horsepower require6 in steam en6 of compressor. By Table II the weight per foot of free air is 0.07422 lb. per cubic foot. Total weight of air compressed = Q. Q = 0.07422 X 3,465 = 257 lb. per minute. In Table I opposite r = 6.72 in column 9 find by interpolation 0.3736. Then Horsepower = 2.57 X 0.3736 X (460 + 50) = 489.6 in air end, and tt^V - = 576 in steam end. 0.85 The second plant will be designated by the letter (e) and will be two-stage compression to 200 lb. gage at air compressor, will be reheated to 300° at the pump and used expansively in the pump; the expansion to be such that the temperature will be 32° at end of stroke. (el) Air pressure at pump. Apply Eq. (32) as in {61). In this case r\ (at the compressor) == 15.3 and r 2 (at the pump) = 12.3. Therefore pressure at the pump = 12.3 X 14 = 172.3 and the lost pressure = 214 - 172.3 = 41.7 = f. (e2) Point of cutoff in air en6 of pump = fraction of stroke 6uring which air is a6mitte6. By Eq. (12), viz., — = (— ) , t\ \Vy 492 /vA ° -25 Vi putting in numbers we get =ttx = ( — ) whence — = 0.176, which ^ ° 760 \vil v 2 is the point of cutoff, and v 2 = 5.68 V\. Or go into Table I in column 5, find the ratio ^qo = 1-545, and in same horizontal line in column 3 find 0.176. 118 COMPRESSED AIR (e3) Volume of compressed hot air admitted to air end of pump. Apply Eq. (9), viz., work = — 1 f- piVi - p a v 2 . In this we have work = 6,250,00, v 2 = 5.68 Vi, p\ = 214, n — 1 = 0.25, p a = 14, and p 2 must be found by Eq. (12a), or it may be gotten from Table I by noting that for a tempera- ture ratio of 1.545 the pressure ratio is 8.8 and - = 0.1136. Therefore p 2 = 0.1136 X 172.3 = 19.57. This would give gage pressure = 5.57. Inserting these numerals in Eq. (9) we get 6,250,000 = 144^( 172 - 3 ~^ X 19 ' 57 + 172.3 - 14 X 5.68) . Whence v\ = 128.6 cu. ft. per minute. (e4) Diameter of air cylinder of pump when air and water pistons are direct-connected. Since expansion ratio is 5.68 (see (e2)) and the volume before cutoff is 128.6, the total piston dis- placement is 128.6 X 5.68 = 730.8 cu. ft. per minute. When the air and water pistons are direct-connected they must travel through equal distances, therefore, the air piston travels through 200 ft. per minute (specification 10). Then if a = area of piston in square feet we have 200 a = 730.8 and a = 3.654 sq.ft. By Table X the diameter is 26 in. nearly. (e5) Volume of cool compressed air used by pump, cubic feet per minute. By (e3) the volume of hot compressed air is 128.6, and since under constant pressure volumes are proportional to absolute temperatures, we have v 510 Whence v = 86.3 cu. ft. per minute. 128.6 760" (e6) Volume of free air used. From (el) the ratio of compres- sion at the pump is 12.3 and from (e5) the volume of cool com- pressed air is 86.3, therefore, the volume of free air is 86.3 X 12.3 = 1,061.6. (e7) Diameter of air pipe. The r for Eq. (27) is 12 - 3 + 153 = 13.8. EXAMPLES AND EXERCISES 119 Applying Eq. (21) with coefficient c = 0.07 we have 1,061. 6\V* 0.07 X 1,160 x (i5"±) d = \~ 41.7 X 13.8 / = 2 - 13in - (e8) Horsepower required in steam end of compressor. By (d7) the weight per cubic foot of free air is 0.07422 and by (e6) the volume of free air compressed is 1,061.6 Therefore, the total weight compressed is 0.07422 X 1,061.6 = 78.8 lb. per minute, and the initial absolute temperature is 510. In the two-stage compression r 2 = 15.3, and assuming equal work in the two stages the r\ = \/l5.3 = 3.91 nearly (see Art. 13). Then going into Table I with r = 3.91 in column 9 find 0.2525. Hence horsepower = 0.2525 X 78.8 X 510 = 101.5 for one stage, and for the two stages 101.5 X 2 = 203, 203 and (specification 11) w~Qk = 238.8 hp. in steam end. (e9) Diameter of air compressor cylinders, assuming 3-ft. strokes and 2%-in. piston rods, equal work in the two cylinders and allowing for volumetric efficiency. By (e6) the free air volume is 1,061.6 and (specification 12) the volumetric efficiency = 82 per cent. Therefore, the piston displacement = ' ' = 1,294.6 cu. ft. per minute. By specification 12 the r.p.m. = 90. Therefore, the displace- ment per revolution = 14.7 nearly, for the low-pressure cylinder. Add to this the volume of one piston rod length of 3 ft. which is 3 X 0.341 = 0.1023. Whence the volume per revolution must 7.4 be 14.8 or, for one stroke, 7.4. Whence the area = -tt = 2.466 sq. ft. By Table X the diameter is 213^ in. nearly for low- pressure cylinders. The high-pressure cylinder must take in the net volume of air compressed to r = 3.91 (see (e8)). Therefore, the net volume per revolution = Q J ^ Q1 = 3.02. Add one piston rod volume and get 3.12 per revolution or 1.56 per stroke and an area of 0.53 sq. ft. By Table X this requires a diameter of 10 in. nearly. (elO) Temperature of air at end of each compression stroke. In Table I the ratio of temperatures for r = 3.91 is 1.313. Hence the higher temperature = 510 X 1.313 = 669 absolute = 209°F. 120 COMPRESSED AIR DESIGN OF A SYSTEM OF DISPLACEMENT PUMPS The water in a mine is to be collected by a system of displace- ment pumps, one each at B, C, D and E, delivering into a sump at A. The data are shown on the sketch (Fig. 26) and include: lengths of pipes (I) ; elevations (El) and quantities (Q) of water in cubic feet per minute. The lengths of water pipes and air pipes will be assumed equal. The pipes may change diameter at junctions. Assume one-third of time consumed in filling the tanks with water. Then the maximum rate of discharge must be three-halves of the average. The problem is to specify the free air volume (V a ) for the com- pressor and the gage pressure (P) of delivery. Also, the diame- ters of all pipes both for water and for air. 1=800- ^ -1=800- -1 = 900- El=60 ■1=1400 El =45 Q = 12 t>U Qj El=40 B Q= io El=35 El =30 EH Q =20 e Fig. 26. Solution. — In order to avoid putting in reducer valves and for economy in piping we will as nearly as practicable, design the system so that static head + friction head in air pipe + fric- tion head in water pipe shall be the same for each unit. Evi- dently this sum will be fixed by conditions at E, since it has greatest lift and greatest length of pipe. We will, therefore, first fix diameters for lines EH and HA, giving them liberal dimensions in order to keep down the pressure at A for we will find that some of the pressure at A must be wasted when working the pumps at B, C and D. The following computations were made with the aid of slide rule and logarithmic friction charts (Plate III), such method being sufficiently accurate for the purpose. EXAMPLES AND EXERCISES 121 Water line EH : 6 in. diameter Water line HA : 6 in. diameter Air pipe EH: 2 in. diameter Air pipe HA : 2 in. diameter [ Length 1,400ft., Q = ^ X 20 =30 Pressure loss in 1,400 ft. (pounds per square inch = 4.0 Length 800, Q =42 Pressure loss in 800 ft. = 3.8 Water friction E to A = 7.8 Static pressure E to A = 13.0 Pressure on water at E = 20.8 For 20.8 lb. gage r = 2.44, but at A the air pressure must be somewhat greater. Hence, we may assume r = 2.5 for estimating friction in pipes leading from A. Length 1,400 ft. volume of compressed air V c = 30 V a = r X 30 = 75, air friction E to H,f = 2.3 Length = 800 ft. V c = 42, V a = 105, / = 2^4 Air friction E to A = 4.7 Air pressure at A = 20.8 + 4.7 = 25.5 r at A = 2.78 Air pipe A to Compressor 2 in. diameter ;i2 + Length 500 ft., V c = 10 + 8 + 20) = 75 V a = r X 75 = 210, /= 4.8 At compressor P = 30.3 The assumption that all pumps will discharge simultaneously is extreme. Hence a compressor of 200 cu. ft. per minute and gage pressure = 30 lb. will be ample. Now with air pressure = 25.5 at junc- tion A , we have to design the air and water pipes to pumps B, C and D so as to about use up this pressure. Air pipe GA: ) Length 800 ft., V c = 33, V a = 2.5 V c , 1% in. diameter J V a = 82, / = 3.2 Air pipe BG: \ L th 200 ft y = 15 y = 37 j = 2A 1 m. diameter J Air pipe CG: 1 th = = 45, / = 5.4 1>4 in. diameter J From the above we find air pressure in tank B = 25.5 - (3.2 + 2.4) = 18.9 tank C = 25.5 - (3.2 + 5.4) = 16.9 122 COMPRESSED AIR Water pipe AG : J Length = 800 ft., Q = 33, friction loss 5 in. diameter J (pounds) = 6.1 Static pressure at B (20 ft.) = 8.7 and 18.9 - (8.7 + 6.1) = 4.1 for water friction BG Water pipes BG: Length = 200 ft., Q = 15, loss of pres- Z}i in. diameter sure = 2.2 Leaving a margin of 1.9 lb. Water pipe CG : Length 800 ft., Q = 18, static pressure 4 in. diameter (15 ft.) = 6.5 and 16.9 - 6.5 = 10.4 that can be lost in friction in the water pipe. A 4-in. pipe will take up 6.2 leaving a margin of about 4 lb. This is the nearest commercial size that can be used. Air pipe DH : Length = 100 ft., V c = 12, V a = 2.5 X Y± in. diameter 12 = 30, / = 3.6 Air pressure in D = 25.5 — air friction in AH and HD = 25.5 - (2.4 + 3.6) = 19.5 Static water pressure at D (25 ft.) = 10.9 Available for water friction = 8.6 Water pipe DH: Length 100 ft., Q = 12, loss in 2^-in. 2}i in. diameter pipe = 5.9 Nearest commercial size, Margin 2.7 EXERCISES In the following exercises, where not otherwise specified, atmospheric con- ditions may be taken as T = 60°F. and p a = 14.7. The article of the text on which the solution chiefly depends is indicated thus ( ) and the answer thus [ ]. 1. (a) Assuming isothermal conditions, how many revolutions of a com- pressor 16-in. stroke, 14-in. diameter, double-acting, would bring the pressure up to 100 lb. gage in a tank 4 ft. diameter by 12 ft. length, atmospheric pres- sure = 14.5 per square inch? (1) [361]. (b) What would be the horsepower of such a compressor running at 100 r.p.m.? (1) [37.3]. (c) What would be the horsepower if the compression were adiabatic? (2) [51.0]. (d) What weight of air would be passed per minute when r.p.m. = 100 and T = 60°F.? (8) [21.4]. 2. The air end of a pump (operated by compressed air) is 20 in. in diameter by 30-in. stroke, r.p.m. = 50, cutoff at Y± stroke, free air pressure = 14.0, T a = 60°, compressed air delivered at 75 lb. gage, T = 60° and n = 1.41. (a) Find work done in horsepower. (3) [70]. EXAMPLES AND EXERCISES 123 (6) Find weight handled per minute. (8) [56]. (c) Find temperature of exhaust (degrees F.). (7) [ — 165]. 3. With atmospheric pressure, p a = 14.7, and T a = 50°, under perfect adiabatic compression, what would be the pressure (gage) and temperature (F.) when air is compressed to: (a) % its original volume? (7) [210]. (6) % its original volume? -(7) [435]. (c) }i its original volume? (7) [603]. (d) H its original volume? (7) [737]. (e) Ko its original volume? (7) [852]. 4. With Pa = 14.1 and T a = 60° what will be the pressure of a pound of air when its volume = 3 cu. ft.? (8) [51.4]. 5. What would be the theoretic horsepower to compress 10 lb. of air per minute from p a = 14.3 and T a = 60° to 90 lb. gage? (a) Compression isothermal. (1) [16.7]. (b) Compression adiabatic. (2) [22.7]. 6. Find the point of cutoff when air is admitted to a motor at 250°F. and expanded adiabatically until the temperature falls to 32°F. (7) [0.41]. 7. What is the weight of 1 cu. ft. of air when p„= 14.0 and T a = - 10°? (8) [0.84]. 8. A compressor cylinder is 20 in. in diameter by 26-in. stroke double- acting. Clearance = 0.8 per cent., piston rod = 2 in., r.p.m. = 100, atmospheric pressure, p a = 14.3, atmospheric temperature = T a = 60°F., and gage pressure = 98 lb. Determine the following: (a) Compression isothermal. la. Volume of free air compressed, cubic feet per minute. (46) [891]. 2a. Volume of compressed air, cubic feet per minute. (1) [1,144]. 3a. Work of compression, foot-pounds per minute. (1) [3,757,000]. 4a. Pounds of cooling water, Ti = 50°, T 2 = 75°. (9) [193]. (b) n = 1.25 and air heated to 100° while entering. lb. Volume of free air compressed per minute. (46) [830]. 2b. Volume of cool compressed air per minute. (1) [106.5]. 36. Work done in compression. (1) [4,658,000]. 46. Temperature of air at discharge. (7) [385°F.]. 9. The cylinder of a compressed-air motor is 18 by 24 in., the r.p.m. = 90, air pressure 100 lb. gage. In the motor the air is expanded to four times its original volume (cutoff at %), with n = 1.25. (a) Determine the horsepower and final temperature when initial T = 60°F. (3 and 7) [hp. = 132, T = -90]. (6) Determine the horsepower and final temperature when initial T = 212°F. (3 and 7) [hp. = 132, T = +17]. 10. Observations on an air" compressor show the intake temperature to be 60°F., the r = 7 and the discharge temperature = 300°F. What is the n during compression? Hint.— Use Eq. (11a) with n unknown. (7) [1.25]. 11. In a compressed-air motor what percentage of power will be gained by heating the air before admission from 60° to 300°F.? (2) [46 per cent.]. 124 COMPRESSED AIR 12. If air is delivered into a motor at 60°F. and the exhaust temperature is not to fall below 32 °F., what ratio of expansion can be allowed? What could be allowed if initial temperature were 300°? n = 1.25. (2 and 7) [1.31, 8.8]. 13. A compressed-air locomotive system is estimated to require 4,000 cu. ft. per minute of free air compressed to 500 lb. gage in three stages with complete cooling between stages. Assume n = 1.25, p a = 14.5, T a = 60°, vol. eff. = 80 per cent., mech. eff. = 85 per cent, and r.p.m. = 60. Compute the volume of piston stroke in each of the three cylinders and the total horsepower required of the steam end. (13 and 14) [41.5, 12.7, 3.87, 1,220]. 14. A compressor is guaranteed to deliver 4 cu. ft. of free air per revolu- tion at a pressure of 116 (absolute). To test this the compressor is caused to deliver into a closed system consisting of a receiver, a pipe line and a tank. Observed conditions are as follows: Receiver Pipe Tank Pressure at start (ab.) . . . Temperatures at start (F. Pressures at end (ab.) . . . Temperatures at end (F.) Volumes (cubic feet) .... 14.5 60.0 116.0 150.0 50.0 14.5 60.0 116.0 90.0 10.0 14.5 60.0 116.0 60.0 100.0 How many revolutions of the compressor should produce this effect? (27) [264]. 15. Find the discharge in pounds per minute through a standard orifice when d = 2 in., i = 5 in., t = 600° and p a = 14.0. (21) [8.03]. 16. What diameter of orifice should be supplied to test the delivery of a compressor that is guaranteed to deliver 1,000 cu. ft. per minute of free air? (21) [6.5]. 17. What is the efficiency of transmission when air pressure drops from 100 to 90 lb. (gage) in passing through a pipe system? (31) [95.5]. 18. A compressor must deliver 100 cu. ft. per minute of compressed air at a pressure = 90 lb. gage, at the terminus of a pipe 3,000 ft. long and 3 in. in diameter. p a = 14.4, T a = 60°F. (a) Assuming a vol. eff. = 75 per cent., what must be the piston displace- ment of the compressor? [967]. (b) What pressure is lost in transmission? (29) [17]. (c) What horsepower is necessary in steam end of compressor if n = 1.25 and the mech. eff. = 85 per cent.? (29 and 2) [141]. (d) What would be the efficiency of the whole system if air is applied in the motor without expansion, the efficiency to be reckoned from steam engine to work done in motor? (6) [27 per cent.]. 19. It is proposed to convey compressed air into a mine a distance of 5,000 ft. The question arises: Which is better, a 3-in. or a 4-in. pipe? Compare the propositions financially, using the following data: Nominal EXAMPLES AND EXERCISES 125 capacity of the plant = 1,000 cu. ft. free air per minute, vol. eff. of com- pressor = 80 per cent., n = 1.25 gage pressure at compressor = 100, weight of free air w a = 0.074, p a = 14.36, weight of 3-in. pipe = 7.5 and of 4-in. pipe = 10.7 lb. per foot. Cost of pipe in place = 4 cts. per pound. Cost of 1 hp. in form of pv log r for 10 hr. per day for 1 year = $150. Plant runs 24 hr. per day. Rate of interest = 6 per cent. (29) [Economy of 4-in. pipe capitalized = $86,260]. 20. Air enters a 4-in. pipe with 60 ft. velocity and 80 lb. gage pressure; the air pipe is 1,500 ft. long. (a) Find the efficiency of transmission. (31) [91 per cent.]. (b) Find horsepower delivered at end of pipe in form pv log r. (31) [224]. (c) Find horsepower delivered at end of pipe in form P g X v. (31) [73.5]. 21. An air pipe is to be 2,000 ft. long and must deliver 50 hp. at the end with a loss of 5 per cent, of the pv log r as measured at compressor. The pressure at compressor is 75 lb. gage. p a = 14.7. Find diameter of pipe. (29) [2%]. _ 22. Modify 21 to read: 50 hp. . . with loss of 5 per cent, of the energy in form P g X v, where P g is gage pressure, and find diameter of air pipe. (29) [SHI 23. In case 21 let pressure at compressor be 250 lb. gage and find diameter of air pipe. (29) [1.4]. 24. The air cylinder of a compressed-air pump is 20 in. in diameter by 30- in. stroke. The machine is double-acting and makes 50 r.p.m. The cutoff is to be so adjusted that the temperature of exhaust shall be 30°. p a = 14.5 and the r at pump =8. n = 1.25. (a) Find cutoff when initial temperature is 60°F. [0.78]. (b) Find cutoff when initial temperature is 250°F. [0.226]. (c) Find horsepower in case (a). [223]. (d) Find horsepower in case (b). [112]. (e) In case (a) find efficiency in applying the pv log r of cool air. [55 per cent.]. if) In case (b) find efficiency in applying the pv log r of cool air. [85 per cent.]. (g) Find the volumes of free air used in cases (a) and (b). [3,400 and 732]. 25. A compound mine pump is to receive air at 150 lb. gage; this is to be reheated from 60° to 250°F., let into the H.P. cylinder of the pump and ex- panded until the temperature is 32°, then exhausted into an interheater where the temperature is again brought to 250°. It then goes into the L.P. cylinder and is expanded down to atmospheric pressure = 14.5 (ab.). (a) Find point of cutoff in each cylinder, n — 1.25. [0.23 and 0.61]. (b) If the air is compressed in two stages with n = 1.25, what will be the efficiency of the system, neglecting friction losses? [1.06]. (c) How much free air will be required to operate the pump if it is to deliver 250 hp., assuming the efficiency of the pump to be 80 per cent, reckoned from the work in the air end? [1,686]. (d) If the pump strokes be 60 per minute and 60 in. long, fix diameters of air cylinders in case (c). [23 in. and 35 in.] 26. Compute the horsepower of a motor passing 1 lb. of air per minute 126 COMPRESSED AIR admitted at 200°F. and 116 lb. (ab.) r = 8, the air to be expanded until pressure drops to 29 lb. (ab.), r = 2. n = 1.25. (3 and 7) [1.727]. 27. A pump to be operated by compressed air must deliver 1,000 gal. of water per minute against a net head of 200 ft. through 800 ft. of 10-in. pipe. The pump is double-acting, 30-in. stroke, 50 strokes per minute. The air is reheated to 275°F. before entering the pump. The cutoff is so adjusted that with n = 1.25 the temperature at exhaust = 36°F. . Mec. eff. of pump = 80 per cent. Air pressure at compressor = 80 lb. gage, p a = 14.4. Length of air pipe = 2,000 ft. Permissible loss in transmission = 7 per cent, of the pv log r at compressor. Mec. eff. of compressor = 85 per cent. Vol. eff. = 80 per cent. (a) Proportion the cylinders of the pump. [Water 14 in., air 26 in.]. (b) Determine the volume of free air used. [444]. (c) Determine the diameter of air pipe. [3^]. 28. Compare the volume displacement of two air compressors, one at sea level and the other at 12,000 ft. elevation; the compressors to handle the same weight of air. [9.45 -5- 14.7]. 29. (a) An exhaust pump has an effective displacement of 3 cu. ft. per revolution. How many revolutions will reduce the pressure in a gas tank from 30 to 5 lb. absolute, volume of tank = 400 cu. ft.? (15) [239]. (b) If the pump is delivering the gas under a constant pressure of 30 lb. (ab.) what is the maximum rate of work done by the pump — foot-pounds per revolution? n = 1.25. (15) [5,433], 30. An air-lift pump is to be designed to elevate gravel from a submerged bed. Specifications as follows: Depth of submergence = 50 ft. ; lift above water surface = 10 ft.; volume lifted to be y± gravel and % sea water; specific gravity of gravel = 3; weight of sea water = 65 lb. per cubic foot; volume of gravel = 1 cu. yd. per minute. (a) Determine the ratio ~q>Q = volume of mixed water and gravel. (b) Determine the ratio of compression and horsepower of compressor. (c) Recommend diameters for water pipe and for air pipe. (41). TABLES Notes on Table i The table is designed to reduce the labor of solution of formulas 12, 120, Sd and 1a. When the weight of air passed and its initial temperature are known, the table covers all conditions such as elevation above sea level, reheating and com- pounding, but it does not include the effect of friction and clearance. In compound compression the same weight goes through each cylinder. Then knowing the initial t and the r for each cylinder, find from the table the work done in each cylinder and add. Usually the r and / are assumed the same in each cylinder. In that case take out the work for one stage and multiply by the number of stages. The columns headed "Work Factor" are applicable in cases of expansion, only when the expansion is complete, that is, when final pressure in the cylinder is equal that outside. (In free air or in a receiver.) Example. — Air is received at such a pressure that r = 8. What should be the cutoff in order that the temperature drop from 60° to 32°F. when expansion is adiabatic? The ratio of absolute temperatures is 1.057 which by linea interpolation corre- sponds to a volume ratio 0.871 or cutoff is at %. What would be the pressure at exhaust? The two ratios above are in the horizontal line with - = .825 therefore the r final pressure = .825 X initial pressure. To find the foot-pounds per pound of air, multiply the number opposite r in columns 7, 8 or 11 as the case may be by the absolute lower temperature. To find the weight compressed, go into Table II with known atmospheric con- ditions and cubic feet capacity of the machine. To find the horse-power per 100 of air per minute multiply the number oppo- site r in columns 9, 10 or 12, as the case may be, by the absolute lower temperature. 128 Table L- -General Table Relating to Air Compression and Expansion Ratio of Ratio of Work Factor. Work Factor for Isothermal (3 $"0 K Less to Greater Volume — Greater to Less Tem- perature — Air Heated by Compression Compression u 2 OO Tempera- Tempera- K n H.P. Fac- 2 » p.' is o.fci tures tures Ab- tor per 100 K a Pi Pi \ 1 pound 330 *5 n = n = n = n — n = n = K 1.25 1. 41 I. 25 1. 41 n = 1-25 n = 1 .41 1-25 r.14 330 r 1 V2 Vl V2 Vl h h Ft. -Lbs. Ft.-Lbs. H.P. 9 H.P. Ft.-Lbs. H.P. i 2 3 4 5 6 7 8 10 11 12 i 1 . 0000 1. 000 I. OOO 1 . 000 1 .000 O 0.0 .0 .0 0.0 .0 1. 1 .9091 .927 935 1. 019 1.028 5 131 5.140 •OI55 •OI55 5.068 •oi53 1.2 • & 333 .862 877 1.037 1.054 9 863 9-932 .0298 .0301 9.694 .0293 i-3 .7692 .812 830 1.054 1.079 14 329 14-45° •0434 •0437 I3-950 .0422 1.4 •7i43 .764 787 1.070 1-103 18 5°3 18.766 .0560 .0568 17.890 •0542 i-5 .6667 •7 2 3 75° 1.085 1-125 22 465 22.827 .0680 .0691 21-559 •0653 i.6 .6250 .687 717 1. 100 1. 146 26 186 26.704 •0793 .0809 24.991 •o757 i-7 .5882 •654 686 1. 112 1. 166 29 775 3 -4i7 .0902 .0921 28.214 •0855 i.8 •5555 .625 659 1-125 1. 186 33 178 33-985 .1005 .1029 31-252 •0947 1.9 •5 26 3 •598 634 1 -137 1.205 36 421 37.422 .1104 •I 134 34.127 .1034 2.0 .5000 •574 612 1. 149 1.223 39 53° 40.733 .1198 •1235 36.855 .1117 2. 1 .4762 •552 59° 1. 160 I.240 42 536 43-897 .1289 •1330 39-450 .1196 2.2 •4545 •532 57i 1. 171 1.259 45 407 46.988 .1376 .1424 41.912 .1270 2 -3 .4348 .514 553 1. 181 i ; 273 48 199 49.970 .1461 •1514 44-287 •1342 2.4 .4166 .496 537 1. 191 1.289 5° 884 52.878 .1542 .1602 46.548 .1411 2 -5 .4000 .480 522 1.202 1.304 53 462 55-676 . 1620 .1687 48.720 .1476 2.6 .3846 .466 508 1. 211 I-3I9 55 988 58.402 .1697 .1769 50.805 •1539 2.7 •37°4 • 452 493 1.220 1-334 58 434 61.054 .1771 .1850 52.811 . 1600 2.8 •3571 •439 481 1.229 1.348 60 800 63-651 .1843 .1929 54-745 .1659 2.9 .3448 •427 469 1.237 1.362 63 086 66.175 .1912 .2006 56.612 •1715 3-° •3333 .415 458 1.246 i-375 65 3J9 68.626 .1979 .2080 58.414 .1770 3-i .3226 .405 448 1.254 1.388 67 499 71.158 .2045 ■2156 60.157 .1823 3-2 •3125 •394 438 1 .262 1. 401 69 626 73 • 4oo .2110 .2224 61.845 .1874 3-3 • 3°3° .385 428 1.270 i-4i4 7i 700 75,686 .2173 .2294 63.481 .1924 3-4 .2941 •376 419 1.277 1.426 73 720 77-936 •2234 .2362 65.087 .1972 3-5 •2857 ■ -3 6 7 411 1.285 1.438 75 688 80.131 .2294 .2428 66.610 .2019 3-6 .2778 •359 403 1.292 1.450 77 628 82.307 •2352 .2494 68.108 .2064 3-7 .2703 •35i 395 1.299 1. 461 79 5i6 84.411 .2410 •2557 69.564 .2108 3-8 .2632 •343 388 1.306 i-473 81 35° 86.496 .2465 .2621 70.982 .2151 3-9 .2564 •337 381 I-3I3 1.484 83 158 88.544 .2520 .2683 72.364 .2193 4.0 .2500 ■33° 374 i-3i9 1-495 84 939 90.510 •2574 •2 743 73-7io .2234 4.1 .2439 •323 3 6 7 1.326 1.506 86 694 92.472 .2627 .2802 75-023 .2274 4.2 .2381 •3i7 361 1 -33 2 1. 516 88 395 94-434 .2678 .2862 76.304 .2312 4-3 .2326 •311 355 i-339 1.526 90 043 96.346 .2729 .2919 77-555 •2350 4-4 .2273 .306 349 1-345 1-537 9i 691 98.202 •2779 .2976 78.776 .2387 4-5 .2222 .300 344 i-35i i-547 93 312 100.012 .2828 •303 1 79.972 .2424 4.6 .2174 •295 338 i-357 i-557 94 882 101.823 •2875 •3085 81. 141 •2459 4-7 .2128 .290 333 *-3 6 3 1.566 96 424 103.616 .2922 .3140 82.284 •• 2494 4.8 .2083 • 285 328 1.368 1-576 97 966 io5-37i .2969 •3193 83.404 .2528 129 130 COMPRESSED AIR Table I. — (Continued) I 2 3 4 5 6 7 8 9 10 11 » 4-9 5-° 5-i .2041 .2000 .1961 .280 .276 .272 •324 •319 •315 i-374 1.380 i-385 1.586 i-595 1.604 99.481 100.943 102.405 107. 109 108. 811 110.493 •3015 •3059 •3103 .3246 •3297 •3348 84.50c 85.574 86.627 .2561 •2593 •2625 5-2 5-3 5-4 .1923 .1887 .1852 .267 .263 •259 .310 .306 .302 1 -391 1.396 1. 401 1-613 1.622 1-631 103.841 105.260 106.673 112. 157 113.830 115.440 •3147 .3180 •3 2 32 •3398 •3449 •3498 87.660 88.673 89.666 .2657 .2687 .2717 5-5 5-6 5-7 .1818 .1786 •i754 .256 .252 .248 .298 •294 .291 1.406 1. 411 1. 416 1.640 1.648 1-657 108.013 io9-353 110.683 117. 010 118.570 120. 114 •3273 •33i4 •3354 •3546 •3593 •3640 90.642 9 1 . 600 92.541 .2747 .2776 .2805 5-8 5-9 6.o .1722 .1695 .1667 •245 .242 .238 .287 .284 .280 1. 42 1 1.426 i-43i 1.665 1.673 1. 681 112.003 113-305 114-581 121.632 123.150 124.640 •3394 •3433 •3472 .3686 •3732 •3777 93.466 94-375 95.27I •2833 .2860 .2887 6.i 6.2 6-3 .1639 .1613 •1587 •235 .232 .229 .277 • 274 .271 1.436 1.440 i-445 1.689 1.697 1-705 115-831 117.080 118.303 126. 113 127.576 129.030 •35io •3548 •3585 .3822 .3866 .3910 96.147 97.012 97.863 .2914 .2940 .2966 6. 4 6-5 6.6 .1562 •1538 •1515 .226 .223 .221 .268 • 265 .262 1.449 1-454 1.458 *-7*3 1. 72 1 1.728 H9-573 120.723 121.920 130.466 131.880 133-30° .3622 •3658 •3694 •3953 •3997 •4039 98.700 99-5 2 4 100.336 .2991 .3016 .3040 6.7 6.8 6.9 .1492 .1471 .1449 .219 .216 .213 •259 .256 •254 1.464 1.467 1. 471 1.736 1.744 1 -75i 123.063 124.205 125.348 134.710 136.090 I37-450 •3729 ■3764 •3799 .4082 .4124 .4165 101.134 101.920 102.700 •3065 .3088 .3112 7.0 7-i 7.2 .1428 .1408 .1389 .211 .208 .206 .251 .249 .246 1.476 1.480 1.484 i-758 1.766 i-773 126.492 127.608 128.708 138.800 140. 120 141.430 •3833 .3867 .3900 .4206 .4246 .4286 103.465 104.219 104.963 •3135 •3158 .3181 7-3 7-4 7-5 •i37o •i35i -.1333 .204 .202 .199 • 244 .241 •239 1.488 1.492 1.496 1.780 1.787 1.794 129.789 130.878 i3i-94i 142.710 !43-979 145-239 •3933 .3966 •399 8 •4327 ■4363 .4401 105.696 106.420 107.133 •3203 •3225 .3246 7.6 7-7 7.8 .1316 .1299 .1282 .197 •195 •193 •237 •235 •233 1.500 1.504 1.508 1. 801 1.807 1. 814 I32-995 134-043 135-063 146.489 147.732 148.976 ■ 4030 .4062 •4093 •4439 •4477 •45 I 4 107.837 108.539 109.219 .3268 •3289 •33io 7-9 8.0 8.1 . 1266 • 1250 .1236 .191 .189 .188 .231 .228 .226 1. 512 1. 516 i-5i9 1. 821 1.828 1.834 136.091 137. no 138. in 150.217 151.427 152.633 .4124 •4i55 .4185 •4552 •4589 .4625 109.896 110.565 in. 225 •333° •335° •3370 8.2 8-3 8.4 . 1220 .1205 . 1 190 .186 .184 .182 .224 .223 .221 I-523 1-527 i.-53i 1. 841 1.847 1.854 I39-093 140.076 141.060 153-823 155.010 156.178 •4215 • 4245 •4275 .4661 .4698 •4733 in. 875 112.522 113-158 •3390 .3410 •3429 8-5 8.6 8.7 .1176 .1163 .1149 .180 .179 .177 .219 .217 .215 i-534 i-538 1 -541 1. 861 1.867 i-873 142.017 142.974 I43.93I I57-348 158.508 159.658 •4304 •4333 •4362 .4768 .4804 •4838 113.788 114. 410 115.023 •3448 •3465 •3487 8.8 8. 9 9.0 .1136 . 1124 . 1111 .176 .174 .172 .214 .212 .210 !-545 1.548 i-55 2 1.879 1.885 1. 891 144.862 145.780 146.700 1 60 . 800 161.927 163.041 •4390 .4418 • 4446 •4873 .4906 .4941 II5-633 116.233 116.827 •3504 •3522 •3540 9.1 9.2 9-3 .1099 .1087 . 1072 . 171 .170 .168 .208 .207 •205 i-555 i-559 1.562 1.897 1.903 1.909 147.627 148.557 149-554 164.147 165.236 166.334 •4474 .4502 •4532 •4974 •5007 .5041 ii7-4i5 117.996 118. 571 •3558 •3576 •3593 9.4 9-5 9.6 . 1064 .1058 . 1042 .167 .165 .164 .204 .202 .201 1-565 1.569 i-572 i-9i5 1. 921 1.927 150.312 151. 188 152.066 167.431 168.520 169.589 •4555 .4582 .4609 •5074 •5107 •5139 119-138 119.702 120.259 .3610 •3627 •3644 9-7 9.8 9.9 ■ 1031 . 1020 . IOIO . 162 .161 .160 .199 .198 .196 i-575i-933 i-578i-939 1. 5821. 944 152.944 153-794 154-645 170.650 171.700 I72.754 •4635 4661 4686 •5i7i •5213 5235 120.810 121-355 121.895 .3661 •3677 •3693 10. . IOOO •iS9 • 195 n. 5851. 950 C55-495 E 73. 789 47 r2 5266 122.429 .3710 TABLES 131 Notes on Table II The purpose of this table is to determine the weight of air compressed by a machine of known cubic feet capacity. It is to be used in connection with Table I for determining power or work. The barometric readings and elevations are made out for a uniform tempera- ture of 6o°F. and are subject to slight errors but not enough to materially affect results. Table V gives more accurately the relation between elevation tem- perature and pressure. Table II.- -Weights of Free Air Under Various Conditions Approximate Baro- metric Reading. T=6o 4) O w £ 1 II 1 6 p. "^ < -20 oo° 20° 40° 6o° 8o° IOO° I 2 3 4 5 6 7 8 9 10 30-5 2 3°-3 2 30.12 15.O 14.9 14-8 .09211 .09150 .09089 .08811 •08753 .08694 . 08444 .08388 •08331 .08108 .08054 . 08000 .07796 .07744 .07693 .07508 .07458 .07408 .07240 .07192 .07144 —600 —400 — 200 29.91 29.71 29.50 14.7 14.6 14-5 .09027 .08965 .08903 .08635 .08576 •08517 •08275 .08219 .08163 •07945 .07895 .07837 .07640 .07589 •07536 •073S8 .07308 .07258 •07095 •07047 . 06999 00 200 400 29.30 29.10 28.90 14.4 14-3 14.2 .08842 .08781 .08719 .08458 .08400 .08341 .08107 .0805c •o7994 .07783 .07729 .07675 .07484 •07432 .07380 .07208 .07158 .07108 .06950 .06902 .06854 600 800 1000 28.69 28:49 28.28 14. 1 14.0 13-9 .08659 .08597 •08535 .08282 .08224 .08165 .07938 .07882 .07825 .07621 •o75 6 7 •07513 .07329 .07277 .07225 •07058 .07008 .06957 . 06806 •06758 .06709 1200 1400 1600 28.08 27.88 27.67 13.8 13-7 13.6 .08474 .08412 .08351 .08106 .08048 .07989 .07769 •07713 ■07656 •07459 .07405 ■07350 •07173 .07120 .07068 .06907 .06857 .06807 .06661 .06612 .06564 1800 2000 2100 27.47 27.27 27.06 i3-5 13-4 13-3 .08289 .08228 .08167 .07930 .07871 .07813 .07600 •07544 •07487 .07296 .07242 .07189 .07016 .06965 .06913 .06757 .06707 .06657 .06516 .06468 .06420 2300 2500 2700 26.86 26.66 26.45 13.2 i3-i 13.0 .08106 . 08044 .07983 •07754 .07695 .07637 •07431 •o7375 .07319 •o7i35 .07080 .07026 .06861 . 06809 .06757 .06607 •o6557 •06507 .06371 .06323 .06274 2900 3100 33°° 26.25 26.05 25.84 12.9 12.8 12.7 .07921 .07860 .07798 .07578 ■07518 .07460 .07262 .07206 .07150 .06972 .06918 .06862 •06705 .06652 . 06600 •06457 .06407 •06357 .06226 .06178 .06130 35°° 3700 4000 25.64 25-44 25-23 12.6 12.5 12.4 •07737 .07676 .07615 .07401 ■07343 .07284 .07094 .07038 .06981 .06810 .06756 .06702 .06549 .06497 .06445 .06307 •06257 .06207 .06082 •06033 •05985 4200 4400 4600 132 COMPRESSED AIR Table II- -{Continued) I 2 3 4 5 6 7 8 9 10 25-°3 24.83 24.62 12.3 12.2 12. 1 •07553 .07492 .07430 •07225 .07166 .07108 .06925 .06868 .06812 . 06648 .06594 .06540 • 06393 .06341 .06289 .06157 .06107 .06057 •05937 .05889 • 05840 4800 5000 5200 24.42 24.22 24.01 12.0 11. 9 11. 8 .07369 .07307 .07246 .07049 . 06990 .06932 .06756 . 06699 . 06643 .06486 .06432 .06378 .06237 .06185 .06133 .06007 •05957 .05907 .05792 •05744 .05696 5400 5600 5800 23.81 23.60 23.40 11. 7 11. 6 ii-S .07184 .07123 .07061 •06873 .06812 • o6 755 .06587 .06530 .06474 .06324 .06270 .06216 .06081 .06029 •05977 •05857 .05807 •05757 •05647 ■05599 •o555i 6100 6300 6500 23.20 22.99 22.79 11. 4 "•3 11. 2 .07000 .06938 .06877 . 06693 .06638 •06579 .06418 .06362 .06305 .06161 .06108 .06054 •05925 •05873 .05821 .05707 .05656 .05606 .05502 •05454 -05406 6800 7100 7300 22.59 22.38 22.18 11. 1 11. 10.9 .06816 .06754 .06692 .06520 .06462 .06403 .06249 •06193 .06136 . 06000 •o5945 .05891 ■05769 •05717 .05665 •05556 •05506 •05456 ■0535 8 •05310 .05261 7600 7900 8100 21.98 21.77 21-57 10.8 10.7 10.6 .06632 •06571 .06510 •06344 .06285 .06226 . 06080 .06024 .05968 •05837 •05783 .05729 •05613 •05561 •05509 .05406 •05356 .05306 •05213 •05164 .05116 8400 8600 8900 21.37 21.16 20.96 10.5 10.4 10.3 . 06448 .06386 .06325 .06168 .06109 .06050 .05911 •05855 •o5799 •05675 .05621 •05567 •o5457 •05405 •o5353 •05256 .05206 •05156 .05068 .05020 .04972 9100 9400 9600 20.76 20.55 20.35 10.2 10. 1 10. .06263 .06202 .06141 .05991 •05933 .05874 •o5743 .05686 .05630 •05513 •05459 •05405 •05301 •05249 •05198 .05106 .05056 .05006 .04923 .04875 .04827 9900 IOIOO 10400 20. 15 19-94 19.74 9.9 9.8 9-7 .06079 .06017 ■05956 .05816 •o5757 .05698 •05572 •05517 .05461 ■05351 .05297 •05243 .05146 .05094 .05041 •04956 . 04906 .04856 .04779 ■04730 .04682 10700 I IOOO 1 1200 19-53 ^ 19-33 I9-I3 9.6 9-5 9-4 .05894 •05833 •05772 •05639 .05580 •05522 .05404 •05348 .05292 .05188 •05134 .05081 . 04990 •04937 .04886 .04806 •04756 .04706 • 04633 •04585 •0453 8 1 1 500 1 1 800 I2IOO 18.93 18.72 18.52 9-3 9.2 9.1 .05711 ■05649 •05587 •05463 •05404 •05345 .06236 •05179 •05123 •05027 .04972 .04918 • 04834 .04782 •04730 •04655 .04605 •04555 . 04489 . 04440 .04392 I24OO I270O 13OOO 1S.31 9.0 ■05526 .05286 .05067 .04864 .04678 •04505 • 04344 I34OO Note on Table III The table is designed to compute readily weights of compressed air by formula 12, Art. 8, viz., w becomes w 53-17 * 144 X p 53-^7 X t If p is given in pounds per square inch the formula Table III. — Weights of Compressed Air Pounds per Cubic Foot P The Ratio - is for absolute pressure in pounds per square inch and absolute temperature Fahrenheit. (See Note at foot of previous page.) i t w P t w t t w t W .000 .005 .010 0000 0135 0271 •255 .260 .265 .6906 .7041 .7177 5io 515 520 1-3813 1-3947 1 . 4083 765 77o 775 2.0718 2.0853 2.0988 .015 .020 .025 0406 0542 0677 .270 •275 .280 •7312 •7447 •7583 525 53° 535 1 .4219 1-4355 1 . 4490 780 785 79o 2-1125 2. 1260 2-1395 .030 •035 .040 0813 0948 1083 .285 .290 •295 .7719 •7852 .7989 54o 545 55o 1.4625 1.4760 1.4895 795 800 805 2-153° 2. 1665 2.1798 ■045 .050 •055 1218 1354 1489 .300 •3°5 .310 .8125 .8260 •8395 555 560 565 1 ■ 5°3° 1. 5166 i-53i2 810 815 820 2.1950 2.2071 2 . 2207 .060 .065 .070 1625 1760 1896 •315 .320 •325 •8531 .8666 .8801 57° 575 580 1-5437 I-557 2 i-57°7 825 830 835 2-2343 2 . 2480 2.2615 •075 .080 .085 2031 2166 2302 ■33° •335 •34o •8937 .9072 .9208 585 59° 595 1 ■ 5843 1.5980 1-6115 840 845 850 2.2750 2.2885 2 .3020 .090 '•095 . 100 2437 2573 2708 •345 •35o •355 •9343 .9478 .9613 600 605 610 1.6250 1.6385 1.6520 855 860 865 2-3155 2.3290 2-3425 .105 . no ■115 2843 2979 3"4 .360 ■3°5 •37° •9749 .9S84 1.0020 615 620 625 1 . 6654 1.6792 1.6927 870 875 880 2-356i 2.3698 2-3 8 33 . 120 ■ 125 .130 3250 3385 352° •375 .380 •385 i-oi55 1 .0290 1.0425 630 635 640 1 . 7062 1. 7198 1-7333 885 890 895 2.3970 2.4105 2.4240 • 135 .140 • 145 3656 3792 3927 •39° ■395 .400 1. 0561 1.0697 1-0833 645 650 655 1 . 7468 1 . 7603 1-7739 900 905 910 2-4375 2.4510 2 ■ 4645 .150 • 155 . 160 4062 4197 4333 •405 .410 •415 1 .0968 1. 1 103 r. 1240 660 665 670 I-7875 1 .8010 1-8145 915 920 925 2.4780 2.4917 2.505 2 .165 .170 • 175 4468 4603 4739 .420 •425 •43° 1 • 1375 1. 1510 1-1645 675 680 685 1.8280 1-8415 1-8550 930 935 940 2.5187 2-5323 2-5459 .180 .185 .190 4875 5010 5145 •435 .440 •445 1 . 1 780 1 . 1917 1.2052 690 695 700 1 . 8680 1.8822 1.8959 945 95o 955 2 5594 2-573° 2.5865 • 195 .200 .205 5281 54i6 555i •45° •455 .460 1 .2177 1.2323 1-2457 705 710 7i5 1 . 9094 1.9229 1-9365 960 965 970 2 . 6000 2-6135 2 . 6270 . 210 • 215 .220 5687 5822 •5958 •465 .470 •475 1-2594 1.2730 1.2865 720 7- 7 5 73° 1.9500 ;-9G35 1.9770 975 980 985 2 ■ 6405 2.6541 2 6670 .225 .230 • 235 .6094 6229 .6364 .480 .485 .490 1 . 3000 i-3i35 1.3270 735 740 745 1 9905 2 . 0042 2.0177 1 990 995 000 2.6813 2 . 6949 2 7084 .240 • 245 .250 • 6499 6635 6771 •495 .500 ■ 505 1. 3416 1-3542 1-3677 75o 755 760 2.0312 2.0448 2.0582 133 Table Ilia— Giving The Values of "K" and "H" Corresponding to Each 2 «. 3 M 'S u so m O to CU 3 > 3 CO CO 4-> CU <-< h t" £ BO'S a CO cu 3 > O 3 > CO ^ +3 3 to co p. i) b to co > &3 O CD <-■ *3 3 to'S 2 "c 2 &>£ SQ-S cu cL O a) "(4 > S > u f! 3 co'S ■P u S cu ho &3 3 "3 > -30 .6082 .0099 17 .6132 .0941 64 .6188 .5962 III .6251 2.654 158 ■6323 9.177 -29 .6083 .0105 18 .6133 .0983 65 .6189 .6175 112 .6253 2.731 159 .6325 9.400 -28 .6084 .0111 19 .6134 .1028 66 .6190 ■ 6393 113 • 6255 2. 811 160 .6326 9.628 -27 .6085 .0117 20 • 6135 .1074 67 .6192 .6617 114 .6256 2.892 161 .6328 9.860 -26 .6086 .0123 21 .6136 .1122 68 .6193 .6848 115 .6257 2.976 162 .6330 10.10 -25 .6087 .0130 22 ■ 6137 .1172 69 .6194 .7086 Il6 .6258 3.061 163 .6331 10.34 -24 .6088 .0137 23 ■ 6139 .1224 70 .6196 ■ 7332 117 .626O 3.149 164 .6333 10.59 -23 .6089 .0144 24 .6140 .1279 7i .6197 .7585 118 .6261 3.239 165 .6335 10.84 — 22 .6090 .0152 25 .6141 .1336 72 .6198 .7846 119 .6263 3-331 166 .6336 n. 10 — 21 .6091 .0160 26 .6142 .1396 73 .6199 .8114 120 .6264 3.42s 167 .6338 n .36 — 20 .6092 .0168 27 .6143 .1458 74 .6201 .8391 121 .6266 3.522 168 .6340 11.63 -19 .6093 .0177 28 .6144 .1523 75 .6202 .8676 122 .6267 3.621 169 ■ 6341 11 .90 -18 .6094 .0186 29 .6146 • 1590 76 .6203 .8969 123 .6269 3.722 170 ■ 6343 12.18 -1.7 .6095 .0196 30 .6147 .1661 77 .6205 .9271 124 .6270 3.826 171 • 6345 12.46 -16 .6096 .0206 31 .6148 ■1734 78 .6206 .9585 125 .6272 3.933 172 ■ 6346 12. 75^ -15 .6097 .0216 32 .6149 .l8ll 79 .6207 .9906 126 .6273 4.042 173 .6349 13.04 -14 .6098 .0227 33 .6150 .1884 80 .6209 I .024 127 .6275 4-153 174 .6350 13.34 -13 .6099 .0238 34 .6151 .i960 81 .6210 1. 057 128 .6276 4.267 17s .6352 13.65 — 12 .6100 .0250 35 .6153 .2039 82 .6211 1.092 129 .6278 4.384 176 .6353 13.96 — II .6101 .0262 36 .6154 .2120 83 .6213 1. 128 130 .6279 4-503 177 ■ 6355 14.28 — 10 .6102 .0275 37 .6155 .2205 84 .6214 1. 165 131 .6281 4-625 178 • 6357 14.60 - 9 .6103 .0289 38 .6156 .2292 85 .6215 1.203 132 .6282 4-750 179 .6359 14.92 - 8 .6104 .0303 39 ■ 6157 .2382 86 .6217 1.242 133 .6284 4.877 180 .6360 15.27 - 7 .6105 .0317 40 .6158 .2476 87 .6218 1.282 134 .6285 5.008 181 .6362 15.62 - 6 .6107 .0332 41 .6160 .2572 88 .6219 1.324 135 .6287 5.142 182 ■ 6364 15.97 - 5 .6108 .0348 42 .6161 .2673 89 .6221 1.366 136 .6288 5.280 183 .6365 16.32 - 4 .6109 .0365 43 .6162 .2776 90 .6222 1 .410 137 .6290 5. 420 184 .6367 16.68 - 3 .6110 .0382 44 .6163 .2883 9i .6223 1.455 138 .6291 5.563 185 .6369 17.05 — 2 .6111 .0400 45 .6164 .2994 92 .6225 1. 501 139 .6293 5.709 186 • 6371 17-43 — I .6112 .0419 46 .6166 .3109 93 .6226 I.S48 140 .6294 5-859 187 .6373 17.81 .6113 0439 47 .6167 .3227 94 .6227 1.597 141 .6296 6. on 188 .6374 18.20 + I .6114 .0459 48 .6168 • 3350 95 .6229 1.647 142 .6298 6.167 189 .6376 18.59 2 .6115 .0481 49 .6169 ■ 3477 96 .6230 1.698 143 .6299 6.327 190 • 6377 19.00 3 .6116 0503 50 .6170 .3608 97 .6232 r:75i 144 .63OI 6.490 191 .6380 19.41 4 .6117 .0526 5i .6172 ■ 3743 98 ■6233 1.805 145 .6302 6.656 192 .6381 19.83 5 .6118 .0551 52 .6173 .3883 99 .6234 1. 861 I46 .6304 6.827 193 .6383 20.25 6 .6120 .0576 53 .6174 .4027 100 .6236 1. 918 147 ■6305 7 .001 194 .6385 20.69 7 .6121 .0603 54 ■ 6175 .4176 101 .6237 1.976 I48 .6307 7.178 195 .6387 21.13 8 .6122 .0630 55 .6177 ■ 4331 102 .6238 2.036 149 .6309 7-359 196 .6389 21.58 9 .6123 0659 56 .6178 .4490 103 .6240 2 .098 150 .63IO 7-545 197 .6391 22 .04 10 .6124 .0690 57 .6179 .4655 104 .6241 2 .161 151 .6312 7-736 198 .6393 22.50 ii .6125 .0722 58 .6180 .4824 105 .6243 2.226 152 ■6313 7.929 199 ■ 6394 22.97 12 .6126 0754 59 .6182 • 4999 106 .6244 2 .294 153 .6315 8.127 200 .6396 23.46 13 .6127 0789 60 .6183 .5180 107 .6246 2.362 154 .6317 8.328 201 .6397 23-94 14 .6128 .0824 61 .6184 ■ 5367 108 .6247 2.432 155 .6318 8.534 202 .6400 24.44 15 .6130 .0862 62 .6185 ■ 5559 109 .6248 2.504 156 .6320 8.744 203 .6402 24-95 16 .6131 .0900 63 .6187 ■ 5758 no .6250 2.578 157 .6322 8.958 204 .6404 25-47 134 Fahrenheit Degree of Temperature From 30 Below to 434 Above Zero ■^ &3 en > CD CU > > > > > > > > 205 1 .640525.99 251 .6499,61 .89 297 .6607 130.8 343 .6736 250.9 389 .6890 444-4 206 .640726.53 252 .6501,62.97 298 .6610 132.8 344 .6739 254.2 390 ■ 6893 449.6 207 .640927.07 253 .650364.08 299 .6612 134-8 345 .6741 257.6 391 .6897 454-9 208 .6411 27 .62 254 .650565.21 300 .6615 136.8 346 .6745 261 .0 392 .6901 460.2 209 .641328.18 255 .6508 66.34 301 .6617 138.9 347 .6749 264.5 393 .6905 465.6 210 .6415 28.75 256 .6510 67.49 302 .6620 141 .0 348 ■ 6751 268.0 394 .6908 47O.9 211 .6417 29.33 257 .651268.66 303 .6623 143. 1 349 ■ 6754 271.5 395 .69H 476.4 212 .6419 29.92 258 .651469.85 304 .6625 145.3 350 .6757 275.0 396 .6915 48I.9 213 .6421 30.53 259 .6516 71. OS 305 .6628 147.4 351 .6760 278.6 397. .6919 487.4 214 .6423 31.14 260 .6518 72 .26 306 .6631 149.6 352 .6763 282.2 398 .6923 493-0 215 .6424 31.76 261 .6521 73.50 307 .6633 151. 8 353 .6767 285.9 399 .6927 498.7 216 .6426 32.38 262 .6523 74-75 308 .6636 I54-I 354 .6770 289.6 400 .6931 504.4 217 .6428 33-02 263 .6525 76.02 309 .6639 156.3 355 .6773 293-3 401 .6935 510. 1 218 .6430 33.67 264 .6528 77-30 310 .6641 158.7 356 .6776 297.1 402. .6939 515.9 219 .6432 34-33 265 .6530 78.61 311 .6644 161. 357 .6780 300.9 403 •6943 521. 7 220 .6434 35-01 266 .6532 79-93 312 .6647 i63-3 358 .6783 304.8 404 .6947 527.6 221 .6436135-69 267 .6534 81.27 3i3 .6650 165.7 359 .6786 308.7 40s .6951 533-5 222 .643836.38 268 .6537 82.62 314 .6652 168. 1 360 .6789 312.6 406 .6955 539-5 223 .6440,37.08 269 -6539 84.00 3i5 .6655 170.5 36l .6792 316.6 407 .6958 545.6 224 .6442 37.80 270 .6541 85.39 316 .6658 173-0 362 .6795 320.6 408 .6962 551.6 225 .6444 38.53 271 • 6543 86.83 3i7 .6661 175-5 363 ■ 6799 324.6 409 .6966 557.8 226 .6446 ! 39.27 272 .6546 88.26 318 .6663 178.0 364 .6803 328.7 410 .6970 564.0 227 .644840.02 273 .654889.71 319 .6666 180.6 365 .6806 332.8 411 .6975 570.2 228 .645140.78 274 .6551 91.18 320 .6669 183. 1 366 .6809 337-0 412 .6979 576.5 229 .6453 41.56 275 .6553 92.67 321 .6671 185.7 367 .6813 341-2 413 .6983 582.8 230 .645542.34 276 .6555 94-18 322 .6674 188.3 368 .6816 354-4 414 .6987 589.3 231 .6457 43.14 277 ■655895.71 323 .6677 191. 369 .6820 349-7 415 .6991 595-7 232 .6458 43-95 278 .656097 .26 324 .6680 193.7 370 .6822 354-0 416 .6995 602 .2 233 .6460 44-77 279 .656398.83 325 .6683 196.5 371 .6825 358.4 417 .6999 608.8 234 .6463 4S-6i 280 .6565 100.4 326 .6686 199-2 372 .682g 362.8 418 ■7003 615.4 235 .6465 46.46 281 .6568 102.0 327 .6689 202.0 373 .6832 367.3 419 .7007 622.1 236 .6467J47.32 282 ■ 6570 103.7 328 .6691 204.8 374 .6836 371.8 420 .7012 628.8 237 .6469148.19 283 ■ 6572 105-3 329 .6694 207.7 375 .6839 376.3 421 .7016 635.6 238 .647149.08 284 • 6575 107.0 330 .6697 210.5 376 .6843 380.9 422 .7021 642.5 239 .6473 49.98 285 ■ 6577 108.7 33i .6700 213-5 •377 .6847 385.5 423 .7025 649.4 240 .6475 50.89 286 .6580 no. 4 332 .6703 216.4 378 .685c 390.2 424 .7029 656.3 241 .6477 51.83 287 .6582 112. 1 333 .6707 219.4 379 .6853 394-9 425 .7033 663.3 242 .6479 52.77 288 .6584 II3-9 334 .6709 222.4 380 .6857 399-6 426 .7037 670.4 243 .6481 53-72 289 .6587 II5- 8 335 .6712 225.4 381 .6861 404-3 427 .7042 677.5 244 .6484 54-69 290 .6590 II7-5 336 .6715 228.5 382 .6865 409.3 428 .704C 684.7 245 .6486 55-68 291 .6592 119- 3 337 .6717 231.6 383 .6868 414.2 429 .7051 691.9 246 .6488 56.67 292 .6594 121 .2 338 .6721 234-7 384 .6871 419. 1 430 .7055 699-2 247 .6490 57.69 293 ■ 6597 123. 1 339 .6724 237-9 385 .6875 424.1 431 • 705C 706.5 248 .6492 58.71 294 .6600 125.0 340 .6727 241 .1 386 .687? 429 .1 432 .7064 713-9 249 .6494 59.76 295 .6602 126.9 341 .6730 244-3 387 .6882 434-2 433 .7068 721.4 250 .649660.81 296 .6604 128.8 342 • 6733 247-6 388 .688C 439-3 434 .7073728.9 135 136 COMPRESSED AIR Table IV.* — Special Table Relating to Stage Compression From Free Air at 14.7 Pounds Pressure and 62 Temperature Compression adiabatic but cooled between stages H 3 1 *w 0) a ° 3 4J b •2 3 ft t) 60 n! O 3 .9 •5 i, ° 3 .5? En -2 4) 1 8 a -a * a.* w S 11 0) 0) «> Is • 2 a +> .H S W 4, H •« 1 B W g £ B 1- 0) 0) fe a 3 s <8 v u bo ft cS O fa ~ O ^ 3 o u fa s "3 4J +3 °fa* fe C h O < fa 1/1 10 7 565 920 *272 621 968 312 653 992 327 66l 992 32O 646 969 29O 609 925 "239 551 817 *242 662 +078 490 898 =302 7°3 *IOO 493 883 *269 652 *Q32 ♦151 5i8 882 *243 600 955 *3°7 656 *oo3 346 687 *025 361 694 *024 352 678 *OOI 322 640 95 6 ^270 *i68 473 776 *o77 376 673 967 260 55i Pp. Pts. 5°3 806 *io7 406 702 997 289 580 41 4-1 8.2 12.3 16.4 20.5 24.6 28.7 32-8 36.9 38 3-8 7.6 11. 4 15-2 19.0 22.8 26.6 3°-4 34-2 42 4-2 8.4 12.6 16.8 21.0 25-2 29.4 33-6 38.7 37.8 40 4.0 8.0 12.0 16.0 20.0 24.0 28.0 3?.o 36.0 37 3-7 7-4 11. 1 14.8 18. 5 22.2 25-9 29.6 33-3 35 34 3-5 7.0 3 6 4 8 10.5 10 2 14.0 13 6 17-5 17 21.0 20 4 24-5 28.0 23 27 4i *o7o *o99 *I27 *i 5 6 152 18 184 213 241 270 298 327 355 384 412 441 29 28 153 469 498 526 554 583 611 6 39 667 696 724 1 2 2.9 5 8 2.8 5-6 8.4 154 752 780 808 837 865 893 921 949 977 *oc>5 3 8.7 iS5 19 o33 061 089 117 145 173 201 229 257 285 4 11. 6 11. 2 156 312 340 368 396 424 451 479 5°7 535 562 5 6 14-5 17.4 16^8 157 59o 618 645 673 700 728 756 783 8n 838 7 20.3 19.6 158 866 893 921 948 976 *oo3 *o3o *o58 *o85 *II2 8 9 23.2 26. 1 22.4 25. 2 159 20 140 167 194 222 249 276 3°3 33° 358 385 160 412 439 466 493 520 548 575 602 629 656 161 683 710 737 7 6 3 790 817 844 871 898 925 162 95 2 978 *oo5 *C>32 *°59 *o8 S *H2 *i39 *i6 5 *I92 1 27 2.7 5-4 26 2.6 163 21 219 245 272 299 325 352 378 405 43i 458 2 5-2 164 484 5" 537 564 59° 617 6 43 669 696 722 3 8.1 7.8 165 748 775 801 827 854 880 906 932 958 985 4 5 10.8 13-5 10.4 13-0 166 22 on °37 063 089 ii5 141 167 194 220 246 6 16.2 15.6 167 272 298 324 35° 376 401 427 453 479 505 7 8 18.9 21.6 18.2 20.8 168 531 557 583 608 634 660 686 712 737 7 6 3 9 24-3 23.4 169 789 814 840 866 891 917 943 963 994 *oi9 170 23 °45 070 096 121 147 172 198 223 249 274 171 300 325 35° 37 6 401 426 452 477 5° 2 528 id 172 553 578 603 629 654 679 704 729 754 779 I 2.5 173 805 830 855 880 9°5 93° 955 980 *oo5 ^030 2 s 174 24 o55 080 105 130 i55 180 204 229 254 279 3 4 7 10 5 175 3°4 329 353 378 403 428 . 452 477 502 527 5 12 S 176 55i 576 601 625 650 674 699 724 748 773 6 7 IS 17 5 177 797 822 846 871 895 920 944 969 993 *oi8 8 20 178 25 042 066 091 115 139 164 188 212 237 261 9 22.5 179 285 310 334 358 382 406 43i 455 479 5°3 180 527 55i 575 600 624 648 672 696 720 744 181 768 792 816 840 864 888 912 935 959 983 24 23 182 26 007 031 °55 079 102 126 i5° i74 198 221 1 2.4 2.3 183 245 269 293 316 34o 3 6 4 387 411 435 458 2 3 4.8 7. 2 4.6 6.9 184 482 505 529 553 576 600 623 647 670 694 4 9.6 9.2 185 717 74i 764 788 811 834 858 881 905 928 5 6 7 12.0 11. 5 T? 9. 186 95i 975 998 *02I *°45 *o68 *09i *H4 * I3 8 *i6i 14.4 16.8 13.0 16. 1 187 27 184 207 231 254 277 300 323 346 37° 393 8 19. 2 18.4 188 416 439 462 485 5°8 53i 554 577 600 623 9 21.0 20. 7 189 646 669 692 715 738 761 784 807 830 852 190 875 898 921 944 967 989 *OI2 *°35 *o58 *o8i 191 28 103 126 149 171 194 ' 217 24O 262 285 3°7 22 21 192 33° 353 375 398 421 443 466 488 5" 533 1 2. 2 4-4 6.6 2. I 193 556 578 601 623 646 668 691 7i3 735 758 3 i'-l 194 780 803 825 847 870 892 914 937 959 981 4 5 6 8.8 8.4 195 29 003 026 048 070 092 ii5 137 159 181 203 13-2 10. 5 12.6 196 226 248 270 292 3i4 33 6 358 380 403 425 7 iS-4 14.7 197 447 469 491 513 535 557 579 601 623 645 8 9 17.6 10.8 16.8 18.0 198 667 688 710 732 754 776 798 820 842 863 199 885 907 929 95i 973 994 *oi6 *o38 *o6o *o8i 30 TABLES Table XIV. Continued. — Logarithms of Numbers 153 No. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 200 30 103 125 146 168 190 211 233 255 276 298 20 1 320 34i 3 6 3 384 406 428 449 471 492 5i4 202 535 557 578 600 621 643 664 685 707 728 22 21 203 75° 771 792 814 835 856 878 899 920 942 1 2. 2 4. 3 6. 2 2.1 * 4-2 5 6.3 204 963 984 *oo6 *02 7 *ch8 *o69 *o9i *II2 *i33 *i54 205 31 175 197 218 239 260 281 302 323 345 366 4 8. i 8.4 206 387 408 429 45° 471 492 513 534 555 576 5 ii- 6 13. 3 10. s 212.6 207 597 618 639 660 681 702 7 2 3 744 7 6 5 785 7 is.- 4 14-7 208 806 827 848 869 890 911 93i 952 973 994 8 17. 9 19- 6 16. S S 18.0 209 3 2 OI 5 °35 056 077 098 118 139 160 181 201 2IO 222 243 263 284 3°5 325 346 366 387 408 211 428 449 469 490 5io 53i 55'2 572 593 613 212 634 654 675 695 7i5 736 756 777 797 818 20 213 838 858 879 899 919 940 960 980 *OOI *02I 2 4.0 214 33 041 062 082 102 122 143 163 183 203 224 3 6.0 215 244 264 284 3°4 325 345 365 385 405 425 i 8. 10. 2l6 445 465 486 506 526 546 566 586 606 626 6 12.0 217 646 666 686 706 726 746 766 786 806 826 7 g 14.0 16.0 2l8 846 866 885 9°5 925 945 965 985 *oo5 *025 9 18.0 219 34 044 064 084 104 124 143 163 183 203 223 220 242 262 282 301 321 34i 361 380 400 420 221 439 459 479 498 5i8 537 557 577 596 616 19 1.9 222 635 655 674 694 7i3 733 753 772 792 811 1 223 830 850 869 889 908 928 947 967 986 *oo5 2 3-8 224 35 ° 2 5 044 064 083 102 122 141 160 180 199 3 4 5-7 7.6 225 218 238 257 276 295 3i5 334 353 372 392 5 9-5 226 411 43° 449 468 488 5°7 526 545 5 6 4 583 6 7 11. 4 13.3 227 603 622 641 660 679 698 717 736 755 774 8 15-2 228 793 813 832 851 870 889 908 927 946 965 9 17. 1 229 984 *oo3 *02I *040 *°59 =1=078 *o97 *n6 *i35 *i54 23O 36 173 192 211 229 248 267 286 3°5 324 342 231 361 380 399 418 43 6 455 474 493 5ii 53° 18 232 549 568 586 605 624 642 661 680 698 717 1 1.8 233 736 754 773 791 810 829 847 866 884 9°3 2 3 3-6 5-4 234 922 940 959 977 996 *oi4 *o 33 *°5i *o"jo *o88 4 7-2 235 37 107 125 144 162 181 199 218 236 254 273 5 6 9.0 10.8 236 291 310 328 346 3 6 5 383 401 420 438 457 7 12.6 237 475 493 5ii 53° 548 566 585 603 621 639 8 14.4 16. 2 238 658 676 694 712 73i 749 767 785 803 822 9 239 840 858 876 894 912 93i 949 967 985 *oo3 24O 38 021 °39 o57 o75 °93 112 130 148 166 184 241 202 220 238 256 2 74 292 310 328 346 3 6 4 17 242 382 399 417 435 453 47i 489 5°7 525 543 1 2 i-7 3.4 243 56i 578 596 614 632 650 668 686 7°3 721 3 5-i 244 739 757 775 792 810 828 846 863 881 899 4 5 6 6.8 8-5 10. 2 245 917 934 952 970 987 *oo5 *023 *04i *o58 *c>76 246 39 °94 in 129 146 164 182 199 217 235 252 7 8 9 11. 9 13.6 15-3 247 270 287 305 322 34o 358 375 393 410 428 248 445 463 480 498 5i5 533 55° 568 585 602 249 620 6 37 655 672 690 707 724 742 759 777 154 COMPRESSED AIR Table XIV. Continued. — Logarithms of Numbers "No. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 250 39 794 811 829 846 863 881 898 9i5 933 95° 251 967 985 *002 *oi9 *o 3 7 *o54 *o7i *o88 *io6 *I2 3 252 40 140 157 175 192 209 226 243 261 278 295 18 253 312 329 346 3 6 4 381 398 4i5 432 449 466 I 1.8 3-6 5-4 254 483 500 5l8 535 552 5 6 9 586 603 620 637 3 255 654 671 688 7°5 722 739 756 773 790 807 4 7.2 256 824 841 858 875 892 909 926 943 960 976 5 6 9.0 10.8 257 993 *OIO *02 7 *o44 *o6i #078 *o 9 5 *iii *I28 *I45 7 12.6 258 41 162 179 196 212 229 246 263 280 296 3*3 8 Q 14.4 259 33° 347 3 6 3 380 397 414 43° 447 464 481 260 497 5i4 53i 547 5 6 4 581 597 614 631 647 261 664 681 697 714 73i 747 764 780 797 814 262 830 847 863 880 896 913 929 946 963 979 17 263 996 *OI2 *029 *Q45 *o62 =1=078 *°95 *iii *I27 *I44 2 1-7 3-4 264 42 160 177 193 210 226 243 259 275 292 308 3 5-1 265 3 2 5 341 357 374 39° 406 423 439 455 472 4 5 6.8 8-5_ 266 488 5°4 521 . 537 553 57° 586 602 619 635 6 10. 2 267 6Si 667 684 700 716 732 749 765 781 797 7 8 11. 9 268 813 830 846 862 878 894 911 927 943 959 9 IS- 3 269 975 991 *ooS *024 *04o *o56 +072 *o88 *io4 *I20 270 43 i3 6 152 169 185 201 217 233 249 265 28l 271 297 313 329 345 361 377 393 409 425 441 16 1.6 272 457 473 489 5°5 52i 537 553 5 6 9 584 60O 1 273 616 632 648 664 680 696 712 727 743 759 2 3 2 274 775 791 807 823 838 854 870 886 902 917 3 4 4 6 8 4 275 933 949 965 981 996 *OI2 *028 *o44 *°59 *°75 5 8 Q 276 44 091 107 122 138 154 170 185 201 217 232 6 7 8 9 6 277 248 264 2 79 295 3" 326 342 358 373 389 12 8 278 404 420 43 6 45i 467 483 498 514 529 545 9 14 4 279 560 576 592 607 623 638 654 669 685 700 280 716 73i 747 762 778 793 809 824 840 855 281 871 886 902 917 932 948 963 979 994 *OIO 15 282 45 ° 2 5 040 056 071 086 102 117 t-33 148 163 1 1-5 283 179 194 209 225 240 255 271 286 301 317 2 3 3 4 5 284 332 347 362 378 393 408 423 439 454 469 4 6 285 484 500 5i5 53° 545 56i 576 59i 606 621 5 6 7 7 S 286 637 652 667 682 697 712 728 743 758 773 9 10 s 287 788 803 818 834 849 864 879 894 909 924 8 12 288 939 954 969 984 *ooo *oi5 *o3o *
3 417 62 014 024 034 °45 °55 066 076 0S6 097 107 418 118 128 138 149 i59 170 180 190 201 211 419 221 232 242 252 263 273 284 294 3°4 315 420 3 2 5 335 346 356 366 377 387 397 408 418 421 428 439 449 459 469 480 490 500 5" 521 10 422 53i 542 552 562 572 583 593 603 613 624 I 1.0 423 634 644 655 665 675 685 696 706 716 726 2 2.0 424 737 747 757 767 778 788 798 808 818 829 3 4 3-° 4.0 425 839 849 859 870 880 890 900 910 921 93i 5 5-o 426 941 951 961 972 982 992 *002 *OI2 *022 *o33 6 7 6.0 7.0 427 63 043 •053 063 o73 083 094 IO4 114 124 134 8 8.0 428 144 155 165 175 185 i95 205 215 225 236 9 9.0 429 246 256 266 276 286 296 306 317 327 337 430 347 357 3 6 7 377 387 397 407 417 428 438 431 448 458 468 478 488 498 508 518 528 538 432 548 558 568 579 589 599 699 619 629 639 433 649 659 669 679 689 699 709 719 729 739 434 749 759 769 779 789 799 809 819 829 839 435 849 859 869 879 889 899 909 919 929 939 9 436 949 959 969 979 988 998 *oo8 *oi8 *028 #038 1 0.9 437 64 048 058 068 078 088 098 108 118 128 137 2 1.8 438 147 157 167 177 187 197 207 217 227 237 3 4 2 . 7 3-6 439 246 256 266 276 286 296 306 316 326 335 5 g 4-5 5-4 6.3 440 345 355 365 375 385 395 404 414 424 434 7 441 444 454 464 473 483 493 5°3 5i3 523 532 8 7.2 8.1 442 542 552 562 572 582 59i 601 611 621 631 9 443 640 650 660 670 680 689 699 709 719 729 444 738 748 758 768 777 787 797 807 816 826 445 836 846 856 865 875 885 895 904 914 924 446 933 943 953 9 6 3 972 982 992 *002 *OII *02I 447 65 031 040 050 060 070 079 089 O99 108 Il8 448 128 137 147 157 167 176 186 I96 205 21S 449 225 234 244 254 263 2 73 283 292 302 312 158 COMPRESSED AIR Table XIV. Continued. — Logarithms of Numbers No. 1 2 3 4 5 6 7 8 9 Pp.Pts. 450 65 321 33i 34i 35° 360 3 6 9 379 389 398 408 451 418 427 437 447 45 6 466 475 485 495 5°4 452 514 523 533 543 552 562 57i 581 59i 600 453 610 619 629 639 648 658 667 677 686 696 454 706 7i5 725 734 744 753 763 772 782 792 455 801 811 820 830 839 849 858 868 877 887 456 896 906 916 925 935 944 954 963 973 982 457 992 *OOI *OII *020 #030 !=o 39 *049 =1=058 *o68 *o77 458 66 087 096 106 115 124 134 143 153 162 172 1 10 1 . 459 181 191 200 2IO 219 229 238 247 257 266 2 2.0 460 276 285 295 3°4 3i4 323 332 342 35i 361 3 3-° 461 37° 380 389 398 408 417 427 43 6 445 455 4 5 4.0 5-0 462 464 474 483 492 502 5" 52i 53° 539 549 6 6.0 463 558 5 6 7 577 586 596 605 614 624 633 642 7 8 7.0 8.0 464 652 661 671 680 689 699 708 717 727 736 9 9.0 465 745 755 764 773 783 792 801 811 820 829 466 839 848 857 867 876 885 894 904 9*3 922 467 93 2 941 95° 960 969 978 987 997 *oo6 *°i5 468 67 025 034 043 052 062 071 080 089 099 108 469 117 127 136 145 154 164 173 182 191 201 470 210 219 228 237 247 256 265 274 284 293 47i 302 311 321 33° 339 348 357 367 376 385 472 394 403 413 422 43i 440 449 459 468 477 1 0.9 473 486 495 5°4 5i4 523 532 54i 55° 560 5 6 9 2 1.8 474 578 587 596 605 614 624 633 642 651 660 3 4 2.7 3-6 475 669 679 688 697 706 7i5 724 733 742 752 5 4-5 476 761 770 779 788 797 806 815 825 834 843 6 7 8 5-4 6.3 477 852 861 870 879 888 897 906 916 925 934 7.2 478 943 95 2 961 970 979 988 997 *oo6 *oi5 *024 9 8.1 479 68 034 043 052 061 070 079 088 097 106 115 480 124 i33 142 151 160 169 178 187 196 205 481 215 224 233 242 251 260 269 278 287 296 482 3°5 3i4 323 332 34i 35° 359 368 377 386 483 395 404 413 422 43i 440 449 458 467 476 484 485 494 502 5ii 520 529 538 547 556 56s 485 574 583 592 601 610 619 628 637 646 655 g 486 664 673 681 690 699 708 717 726 735 744 I 0.8 487 753 762 771 780 789 797 806 8i5 824 833 2 1.6 488 842 851 860 869 878 886 895 904 9i3 922 3 4 2.4 3-2 489 93i 940 949 958 966 975 984 993 *002 *OII 5 6 7 4.0 4.8 5-6 490 69 020 028 °37 046 °55 064 °73 082 090 099 491 108 117 126 135 144 i5 2 161 170 179 188 8 6.4 492 197 205 214 223 232 241 249 258 267 276 9 7.2 493 285 294 302 3ii 320 329 338 346 355 3 6 4 494 373 381 39° 399 408 4i7 425 434 443 45 2 . 495 461 469 478 487 496 5°4 513 522 53i 539 496 548 557 566 574 583 592 601 609 618 627 497 636 644 653 662 671 679 688 697 7°5 7i4 498 723 732 740 749 758 767 775 784 793 801 499 810 819 827 836 845 854 862 871 880 888 TABLES Table XIV. Continued. — Logarithms of Numbers 159 No. 1 2 3 4 5 6 7 8 9 Pp. Pt». 500 69 897 906 914 923 93 2 940 949 958 966 975 50I 984 992 <=OOI l=OIO : 1=018 " 1=027 : ^03 6 " 1=044 • "053 : 1=062 502 70 070 079 088 096 i°5 114 122 131 140 148 503 157 165 174 183 191 200 209 217 226 234 504 243 252 260 269 278 286 295 3°3 312 321 505 3 2 9 338 346 355 3 6 4 372 381 389 398 406 506 415 424 432 441 449 458 467 475 484 49 2 507 5 01 5°9 5i8 526 535 544 552 561 5 6 9 578 508 586 595 603 612 621 629 638 646 655 663 I 9 d. 9 509 672 680 6S9 697 706 714 723 73i 740 749 2 1.8 510 757 766 774 783 791 800 808 817 825 834 3 2.7 7 fi 5" 842 851 859 868 876 885 893 902 910 919 4 5 3-° 45 512 927 935 944 952 961 969 978 986 995 +003 6 7 8 5-4 6.3 7.2 5i3 71 012 020 029 °37 046 054 063 071 079 088 5i4 096 105 ii3 122 130 139 147 155 164 172 9 8.1 5i5 181 189 198 206 214 223 231 240 248 257 5i6 265 273 282 290 299 3°7 3i5 324 332 34i 517 349 357 366 374 383 39i 399 408 416 425 5i8 433 441 45° 458 466 475 483 492 500 5°8 519 5i7 525 533 542 55° 559 5 6 7 575 584 59 2 520 600 609 617 625 634 642 650 659 667 675 521 684 692 700 709 717 725 734 742 75° 759 g 522 767 775 784 792 800 809 817 825 834 842 1 0.8 523 850 858 867 875 883 892 900 908 917 9 2 5 2 1.6 524 933 941 95° 958 966 975 983 991 999 *oo8 3 4 2.4 3- 2 525 72 016 024 032 041 049 °57 066 074 082 090 5 4.0 526 099 107 ii5 123 132 140 148 156 165 173 6 7 4.8 5.6 527 181 189 198 206 214 222 230 239 247 255 8 6.4 528 263 272 280 288 296 3°4 3*3 321 329 337 9 7-2 S29 346 354 362 37° 378 387 395 403 411 419 530 428 43 6 444 452 460 469 477 485 493 5oi 53i 5°9 5i8 526 534 542 55° 558 567 575 583 532 S9i 599 607 616 624 632 640 648 656 665 533 673 681 689 697 7°5 7i3 722 73° 738 746 534 754 76a 770 779 787 795 803 811 819 827 535 835 843 852 860 868 876 884 892 900 908 7 0.7 536 916 925 933 941 949 957 965 973 981 989 1 537 997 *oo6 *oi4 *02 2 +030 *o38 ^046 *054 *o62 *o7o 2 1.4 538 73 °7 8 086 094 I02 in 119 127 135 143 151 3 4 2 !8 539 159 167 175 183 191 199 207 215 223 231 5 6 7 3-5 540 239 247 255 263 272 280 288 296 3°4 312 4. 2 4.9 541 320 328 33 6 344 352 360 368 37 6 384 392 8 5-6 542 400 408 416 424 43 2 440 448 45 6 464 472 9 6.3 543 480 488 496 5°4 512 520 528 536 544 552 544 560 568 57 6 584 592 600 608 616 624 632 545 640 648 656 664 672 679 687 695 7°3 711 546 719 727 735 743 75i 759 767 775 783 791 547 799 807 815 823 830 838 846 854 862 870 548 878 886 894 902 910 918 926 933 941 949 549 957 965 973 981 989 997 *oo5 *°i3 *020 *028 74 160 COMPRESSED AIR Table XIV. Continued. — Logarithms of Numbers No. 1 2 3 4 5 6 7 8 9 Pp.Pts. 550 74 °3 6 044 052 060 068 076 084 092 099 107 551 US 123 131 139 147 155 162 170 178 186 552 194 202 210 218 225 233 241 249 257 265 553 273 280 288 296 3°4 312 320 3 2 7 335 343 554 35i 359 367 374 382 39° 398 406 414 421 555 429 437 445 453 461 468 476 484 492 500 556 5°7 5i5 5 2 3 53i 539 547 554 562 57° 578 557 586 593 601 609 617 624 632 640 648 656 558 663 671 679 687 695 702 710 718 726 733 559 741 749 757 764 772 780 788 796 803 811 560 819 827 834 842 850 858 865 873 881 889 56i 896 904 912 920 927 935 943 95° 958 966 562 974 981 989 997 *oo5 *OI2 *020 *028 *°35 *043 8 8 563 75 051 °59 066 074 082 089 097 i°5 113 120 I 564 128 136 i43 151 159 l66 174 182 189 197 2 1 6 565 205 213 220 228 236 2 43 2 5i 2 59 266 274 3 4 2 3 4 2 566 282 289 297 3°5 312 320 328 335 343 35i 5 4 567 358 366 374 381 389 397 404 412 420 427 6 7 8 4 s 6 8 6 568 435 442 45° 458 465 473 481 488 496 5°4 4 569 5ii 5i9 526 534 54 2 549 .557 565 572 580 9 7 2 570 587 595 603 610 618 626 633 641 648 656 57i 664 671 679 686 694 702 709 7i7 724 732 572 740 747 755 762 770 778 785 793 800 808 573 8i5 823 831 838 846 853 861 868 876 884 574 891 899 906 914 921 929 937 944 95 2 959 575 967 974 982 989 997 *oo5 *OI2 *020 *027 *°35 576 76 042 050 °57 065 072 080 087 °95 103 no 577 118 125 i33 140 148 155 163 170 178 185 578 193 200 208 215 223 230 238 2 45 253 260 579 268 275 283 290 298 3°5 313 320 328 335 580 343 35° 358 365 373 380 388 395 403 410 58i 418 425 433 440 448 455 462 470 477 485 1 7 0.7 582 492 500 5°7 5i5 522 53° 537 545 55 2 559 2 1 4 583 567 574 582 589 597 604 612 619 626 634 3 4 2 2 1 8 584 641 649 656 664 671 678 686 693 701 708 5 3 5 585 716 7 2 3 73° 738 745 753 760 768 775 782 6 7 8 4 2 586 790 797 805 812 819 827 834 842 849 856 4 5 9 6 587 864 871 879 886 893 901 908 916 9 2 3 93° 9 3 588 938 945 953 960 967 975 982 989 997 *oo4 589 77 012 019 026 °34 041 048 056 063 070 078 590 085 093 100 107 ii5 122 129 137 144 .^ 59i 159 166 i73 181 188 195 203 210 217 225 592 232 240 247 254 262 269 276 283 291 298 593 3°5 313 320 3 2 7 335 34 2 349 357 3 6 4 37i 594 379 386 393 401 408 415 422 43° 437 444 595 45 2 459 466 474 481 488 495 5°3 5i° 517 596 525 532 539 546 554 56i 568 576 583 59° 597 597 605 612 619 627 634 641 648 656 663 598 670 677 685 692 699 706 7i4 721 728 735 599 743 75° 757 764 772 779 786 793 801 808 TABLES Table XIV. Continued. — Logarithms of Numbers 161 No. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 600 77 8l 5 822 830 837 844 851 859 866 873 880 601 887 895 902 909 916 924 93i 938 945 952 602 960 967 974 981 988 996 *oo3 *OIO *oi7 *025 603 78 032 °39 046 °53 061 068 °75 082 089 O97 604 104 in 118 125 132 140 147 154 161 168 605 176 183 190 197 204 211 219 226 233 24O 606 247 254 262 269 276 283 290 297 3°5 312 607 319 326 333 34° 347 355 362 369 376 383 608 39° 398 405 412 419 426 433 440 447 455 8 s 609 462 469 476 483 490 497 5°4 512 5i9 526 2 1 6 6lO 533 54o 547 554 561 569 576 583 59° 597 3 2 4 6ll 604 611 618 625 6 33 640 647 654 661 668 4 5 3 4 2 6l2 675 682 689 696 704 711 718 725 732 739 6 4 8 613 746 753 760 767 774 781 789 796 803 810 7 8 S 6 6 4 614 817 824 831 838 845 852 859 866 873 880 9 7 2 615 888 895 902 909 916 923 93° 937 944 95i 616 958 965 972 979 986 993 *ooo *oo7 *oi4 *02I 617 79 029 c 3 6 043 050 °57 064 071 078 085 O92 618 099 106 "3 120 127 134 141 148 155 l62 619 169 176 183 190 197 204 211 218 225 232 62O 239 246 253 260 ' 267 274 281 288 295 302 621 3°9 316 323 33° 337 344 351 358 365 372 622 379 386 393 400 407 414 421 428 435 442 1 / O *7 623 449 45 6 463 470 477 484 49 1 498 5°5 5" 2 I 4 624 5i8 525 532 539 546 553 560 567 574 58l 3 4 2 2 1 8 625 588 595 602 609 616 623 630 637 644 650 5 3 S 626 6 57 664 671 678 685 692 699 706 713 720 6 4 2 627 727 734 741 748 754 761 768 775 782 789 7 8 4 5 9 6 628 796 803 810 817 824 831 837 844 851 858 9 6 3 629 865 872 879 886 893 900 906 9 J 3 920 927 630 934 941 948 955 962 969 975 982 989 996 631 80 003 010 017 024 030 °37 044 051 058 065 632 072 079 085 092 099 106 "3 120 127 134 633 140 147 i54 161 168 i75 182 188 i95 202 634 209 216 223 229 236 243 250 257 264 271 635 277 284 291 298 3°5 312 318 3 2 5 33 2 339 5 636 346 353 359 366 373 380 387 393 400 407 I 0.6 637 414 421 428 434 441 448 455 462 468 475 2 1 2 638 482 489 49 6 502 5°9 5i6 523 53° 536 543 3 4 1 2 8 4 639 55° 557 564 57° 577 584 59i 598 604 611 5 3 640 618 625 632 63S 645 652 659 665 672 679 6 7 8 3 6 641 686 693 699 706 7i3 720 726 733 740 747 4 8 642 754 760 767 774 781 787 794 801 808 814 9 5-4 643 821 828 835 841 848 855 862 868 875 882 644 889 895 902 909 916 922 929 93 6 943 949 645 95 6 963 969 976 983 990 996 *oo3 *OIO *oi7 646 81 023 030 °37 043 050 o57 064 070 077 084 647 090 097 104 in 117 124 131 137 144 151 648 158 164 171 178 184 191 198 204 211 218 649 224 231 238 245 251. 258 265 271 278 28^ 11 162 COMPRESSED AIR Table XIV. Continued. — Logarithms or Numbers No. 1 2 3 4 5 6 7 8 9 Pp. Pts. 650 81 291 298 3°5 3ii 318 3 2 5 33i 338 345 35i 651 358 365 37i 378 385 39i 398 405 411 418 652 425 43i 43 8 445 45i 458 465 47i 478 485 653 491 498 5o5 5" 5i8 525 53i 538 544 55i 654 558 564 57i 578 584 59i 59 8 604 611 617 655 624 631 637 644 651 657 664 671 677 684 656 690 697 704 710 717 723 73° 737 743 75° 657 757 763 770 776 783 790 796 803 809 816 658 823 829 836 842 849 856 862 869 875 882 659 889 895 902 908 9i5 921 928 935 941 948 660 954 961 968 974 981 987 994 +000 *oo7 *oi4 66l 82 020 027 °33 040 046 °53 060 066 °73 079 662 086 092 099 i°5 112 119 125 132 138 145 663 151 158 164 171 178 184 191 197 204 210 1 7 1 664 217 223 230 236 243 249 256 263 269 276 2 1 4 665 282 289 295 302 308 3i5 321 328 334 341 3 4 2 2 1 8 666 347 354 360 3 6 7 373 380 387 393 400 406 5 3 5 667 413 419 426 432 439 445 452 458 465 471 6 4 2 668 478 484 49 1 497 5°4 5i° 5i7 523 53° 536 8 5 6 669 543 549 556 562 5 6 9 575 582 588 595 601 9 6 3 670 607 614 620 627 6 33 640 646 6 53 659 666 671 672 679 685 692 .698 7o5 711 718 724 73° 672 737 743 75° 756 763 769 776 782 789 795 673 802 808 814 821 827 834 840 847 853 860 674 866 872 879 88 5 892 898 9°5 911 918 924 675 93° 937 943 95° 95 6 963 969 975 982 988 676 995 *OOI *oo8 *oi4 *020 *02 7 *°33 +040 =•=046 *052 677 83 o59 065 072 078 08 5 091 097 104 no 117 678 123 129 136 142 149 155 161 168 174 181 679 187 193 200 206 213 219 225 232 238 245 680 251 257 264 270 276 283 289 296 302 308 e. 681 3i5 321 327 334 340 347 353 359 366 372 1 0.6 682 378 385 39i 398 404 410 417 423 429 43 6 2 1 2 683 442 448 455 461 467 474 480 487 493 499 3 4 1 2 8 4 684 506 512 5i8 525 531 537 544 55° 556 5 6 3 5 3 685 5 6 9 575 582 588 594 601 607 613 620 626 6 7 8 3 6 68'6 632 639 645 651 658 664 670 677 683 689 4 4 8 687 696 702 708 715 721 727 734 740 746 753 9 5 • 4 688 759 765 771 778 784 790 797 803 809 816 689 822 828 835 841 847 853 860 866 872 879 690 88 S 891 897 904 910 916 923 929 935 942 691 948 954 960 967 973 979 985 992 998 *oo4 692 84 on 017 023 029 036 042 048 o55 061 067 693 °73 080 086 092 098 i°5 in 117 123 130 694 136 142 148 155 161 167 i73 180 186 192 695 198 205 211 217 223 230 236 242 248 255 696 261 267 273 280 286 292 298 3°5 311 317 697 323 33° 33 6 342 348 354 361 3 6 7 373 379 698 386 392 398 404 410 417 423 429 435 442 699 448 454 460 466 473 479 485 491 497 5°4 TABLES Table XIV. Continued. — Logarithms of Numbers 163 No. 1 2 3 4 5 6 7 8 9 Pp. Pt8. 700 84 510 516 522 528 535 54i 547 553 559 566 701 572 578 584 59° 597 603 609 615 621 628 702 . 634 640 646 652 658 665 671 677 683 689 703 696 702 708 714 720 726 733 739 745 75i 704 i 757 763 770 776 782 788 794 800 807 813 705 j 819 825 831 837 844 850 856 862 868 874 706 880 887 893 899 9°5 911 917 924 93° 936 707 942 948 954 960 967 973 979 985 991 997 708 85 003 009 Ol6 022 028 034 040 046 052 058 I 7 709 065 071 077 083 089 o95 IOI 107 114 120 2 1 4 710 126 132 138 144 i5° 156 163 169 175 181 3 2 1 8 S 711 187 193 199 205 211 217 224 230 236 242 4 5 3 712 248 254 260 266 272 278 285 291 297 3°3 6 4 2 713 3°9 315 321 327 333 339 345 352 358 3 6 4 7 8 4 5 9 6 714 37° 376 382 388 394 400 406 412 418 425 9 6 3 715 43i 437 443 449 455 461 467 473 479 485 716 491 497 5°3 5°9 5i6 522 528 534 54o 546 717 552 558 564 57° 576 582 588 594 600 606 718 612 618 625 631 637 643 649 655 661 667 719 673 679 685 691 697 7°3 709 7i5 721 727 720 733 739 745 75i 757 763 769 775 781 788 721 794 800 806 812 818 824 830 836 842 848 722 854 860 866 872 878 884 890 896 902 908 1 fi 723 914 920 926 932 938 944 95° 95 6 962 968 2 1 2 724 974 980 986 992 998 *oo4 *OIO *oi6 *022 *028 3 4 5 1 8 4 725 86 034 040 046 052 058 064 070 076 082 088 3 726 094 100 106 112 118 124 130 136 141 147 6 7 8 3 6 727 153 iS9 165 171 177 183 189 195 201 207 4 4 8 728 213 219 225 231 237 243 249 255 261 267 9 5 4 729 273 279 285 291 297 3°3 308 3i4 320 326 730 332 338 344 35° 356 362 368 374 380 386 731 392 398 404 410 4i5 421 427 433 439 445 732 45i 457 463 469 475 481 487 493 499 5°4 733 Sio 5i6 522 528 534 54o 546 552 558 564 734 57° 576 58i 587 593 599 605 611 617 623 735 629 635 641 646 652 658 664 670 676 682 736 688 694 700 7°5 711 717 723 729 735 741 1 5 0.5 737 747 753 759 764 770 776 782 788 794 800 2 1 738 806 812 817 823 829 835 841 847 853 859 3 4 1 2 5 739 864 870 876 882 888 894 900 906 911 917 5 2 S 740 923 929 935 941 947 953 958 964 970 976 6 3 5 74i ; 982 988 994 999 *oc>5 *OII *oi7 *023 *029 *°35 8 4 742 87 040 046 052 058 064 070 °75 081 087 °93 9 4-5 743 099 i°5 in 116 122 128 134 140 146 151 744 157 163 169 175 181 186 192 198 204 210 745 216 221 227 •233 239 245 251 256 262 268 746 274 280 286 291 297 3°3 3°9 315 320 326 747 332 338 344 349 355 361 3 6 7 373 379 384 748' 39° 396 402 408 4i3 419 425 43i 437 442 749 448 454 460 466 47i 477 483 489 495 500 164 COMPRESSED AIR Table XIV. Continued. — Logarithms of Numbers No. 1 2 3 4 5 6 7 8 9 Pp.Pts. 750 87 506 512 5i8 523 529 535 54i 547 552 558 751 564 57° 576 58i 587 593 599 604 610 616 752 622 628 6 33 639 645 651 656 662 668 674 753 679 685 691 697 7°3 708 714 720 726 73i 754 737 743 749 754 760 766 772 777 783 789 755 795 800 806 812 818 823 829 835 841 846 756 852 858 864 869 875 881 887 892 898 904 757 910 9i5 921 927 933 938 944 95° 955 961 758 967 973 978 984 990 996 *OOI *oo7 *oi3 *oi8 759 88 024 030 036 041 047 °53 058 064 070 076 760 081 087 °93 098 104 no 116 121 127 133 761 138 144 150 156 161 167 173 178 184 190 762 195 201 207 213 218 224 230 235 241 247 6 0.6 763 252 258 264 270 275 281 287 292 298 3°4 1 764 3°9 3i5 321 326 332 338 343 349 355 360 2 1 2 765 366 372 377 383 389 395 400 406 412 4i7 3 4 1 2 8 4 766 423 429 434 440 446 45i 457 463 468 474 5 3 767 480 485 49 1 497 502 508 5i3 519 525 53° 6 3 6 768 536 542 547 553 559 564 57° 576 581 587 7 8 4 4 8 769 593 598 604 610 615 621 627 632 638 643 9 5 4 770 649 655 660 666 672 677 683 689 694 700 771 705 711 717 722 728 734 739 745 75o 756 772 762 767 773 779 784 790 795 801 807 812 773 818 824 829 835 840 846 852 857 863 868 774 874 880 885 891 897 902 908 913 919 925 775 93o 93 6 941 947 953 958 964 969 975 981 776 986 992 997 *oo3 *oo9 *oi4 *020 *025 *03i *Q37 777 89 042 048 °53 o59 064 070 076 081 087 092 778 098 104 109 ii5 120 126 131 137 143 148 779 154 159 165 170 176 182 187 193 198 204 780 209 215 221 226 232 237 243 248 254 260 781 265 271 276 282 287 293 298 3°4 310 3i5 I 5 0. <: 782 321 326 332 337 343 348 354 360 365 37i 2 1 783 376 382 387 393 398 404 409 415 421 426 3 4 1 2 5 784 43 2 437 443 448 454 459 465 470 476 481 5 2 5 785 487 492 498 5°4 5°9 515 520 526 53i 537 6 3 786 542 548 553 559 5 6 4 57° 575 S81 586 592 7 8 3 4 S 787 597 603 609 614 620 625 631 636 642 647 9 4 5 788 653 658 664 669 6 75 680 686 691 697 702 789 708 7i3 •719 724 73° 735 741 746 752 757 790 7 6 3 768 774 779 785 790 796 801 807 812 791 818 823 829 834 840 845 851 856 862 867 792 873 878 883 889 894 900 9°5 911 916 922 793 927 933 938 944 949 955 960 966 971 977 794 982 988 993 998 *oo4 *oc>9 =1=015 *020 *026 *o 3 i 795 90 037 042 048 o53 059 064 069 °75 080 086 796 091 097 102 108 "3 119 124 I29 135 140 797 146 151 157 162 168 173 179 184 189 195 798 200 206 211 217 222 227 233 238 244 249 799 255 260 266 271 276 282 287 293 298 3°4 TABLES Table XIV. Continued. — Logarithms of Numbers 165 No. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 800 90 309 314 320 325 33* 336 342 347 352 358 801 3 6 3 369 374 380 385 39° 396 401 407 412 802 417 423 428 434 439 445 45° 455 461 466 803 472 477 482 488 493 499 5°4 5°9 5i5 520 804 526 53i 536 542 547 553 558 5 6 3 5 6 9 574 805 580 585 59° 596 601 607 612 617 623 628 806 634 639 644 650 655 660 666 671 677 682 807 687 693 698 7°3 709 714 720 725 73° 736 808 741 747 752 757 7 6 3 768 773 779 784 789 809 795 800 806 811 816 822 827 832 838 843 8lO 849 854 859 865 870 875 881 886 891 897 8ll 902 907 913 918 924 929 934 940 945 95° 8l2 95 6 961 966 972 977 982 988 993 998 *oo4 5 813 91 009 014 020 025 030 036 041 046 052 °57 1 0.6 814 062 068 °73 078 084 089 094 100 i°5 no 2 1 2 815 116 121 126 132 t-37 142 148 153 158 164 3 4 1 2 8 4 8l6 169 i74 180 185 190 196 201 206 212 217 5 3 817 222 228 2 33 238 243 249 254 259 265 270 6 7 3 6 818 275 281 286 291 297 302 3°7 312 318 323 8 4 8 819 328 334 339 344 35° 355 360 365 37i 376 9 5 4 82O 381 387 392 397 403 408 413 418 424 429 821 434 440 445 45° 455 461 466 47i 477 482 822 487 492 498 5°3 508 5i4 5i9 524 529 535 823 54o 545 55i 556 56i 566 572 577 582 587 824 593 598 603 609 614 619 624 630 635 640 825 645 651 656 661 666 672 677 682 687 6 93 826 698 7°3 709 7i4 719 724 73° 735 740 745 827 75i 756 761 766 772 777 782 787 793 798 828 803 808 814 819 824 829 834 840 845 850 829 855 861 866 871 876 882 887 892 897 9°3 83O 908 913 918 924 929 934 939 944 95° 955 831 960 965 971 976 981 986 991 997 *002 *oo7 1 5 <; 832 92 012 018 023 028 o33 038 044 049 054 o59 2 1 833 065 070 °75 080 085 091 096 IOI 106 in 3 4 1 2 S 834 117 122 127 132 i37 143 148 153 158 163 5 2 s 835 169 i74 179 184 189 195 200 205 2IO 215 6 7 8 3 836 221 226 231 236 241 247 252 257 262 267 3 4 5 837 273 278 283 288 293 298 3°4 3°9 314 3 J 9 9 4 5 838 324 33° 335 34o 345 35° 355 361 366 37i 839 376 381 387 392 397 402 407 412 418 423 84O 428 433 438 443 449 454 459 464 469 474 841 480 485 490 495 500 5°5 5ii 5i6 521 526 842 53i 536 542 547 552 557 562 567 572 578 843 583 588 593 598 603 609 614 619 624 629 844 634 639 645 650 655 660 665 670 675 681 845 686 691 696 701 706 711 716 722 727 732 846 737 742 747 75 2 758 7 6 3 768 773 778 783 847 788 793 799 804 809 814 819 824 829 834 848 840 845 850 855 860 865 870 875 88l 886 849 891 896 901 906 911 916 921 927 932 937 166 COMPRESSED AIR Table XIV. Continued. — Logarithms of Numbers No. 1 2 3 4 5 6 7 8 9 Pp. Pts. 850 92 942 947 952 957 962 967 973 978 983 988 851 993 998 *oo3 *oo8 *oi3 *oi8 *024 *029 *°34 *039 852 93 °44 049 o54 °59 064 069 °75 080 085 090 853 °95 100 i°5 no ii5 120 125 131 136 141 854 146 151 156 161 166 171 176 181 186 192 855 197 202 207 212 217 222 227 232 237 242 856 247 252 258 263 268 273 278 283 288 293 857 298 3°3 308 3*3 318 323 328 334 339 344 858 349 354 359 364 3 6 9 374 379 384 389 394 1 6 i 859 399 404 409 414 420 425 43° 435 440 445 2 1 2 860 45° 455 460 465 470 475 480 485 490 495 3 1 8 86l 500 5°5 5i° 515 520 526 53i 53 6 541 546 4 5 2 3 4 862 551 556 561 566 57i 576 58i 586 59i 596 6 3 6 863 601 606 611 616 621 626 631 636 641 646 7 8 4 2 8 864 6Si 656 661 666 671 676 682 687 692 697 9 5 4 865 702 707 712 717 722 727 732 737 742 747 866 752 757 762 767 772 777 782 787 792 797 867 802 807 812 817 822 827 832 837 842 847 868 852 857 862 867 872 877 882 887 892 897 869 902 907 912 917 922 927 932 937 942 947 870 95 2 957 962 967 972 977 982 987 992 997 871 94 002 007 012 017 022 027 032 °37 042 047 872 052 °57 062 067 072 077 082 086 091 096 5 873 IOI 106 in 116 121 126 131 136 141 146 2 1 874 151 1S6 161 166 171 176 181 186 191 196 3 4 5 1 5 875 201 206 211 216 221 226 231 236 240 245 2 5 876 250 255 260 265 270 275 280 285 290 295 6 7 8 3 877 300 3°5 310 3i5 320 325 33° 335 34o 345 3- 4 5 878 349 354 359 3 6 4 369 374 379 384 389 394 9 4 5 879 399 404 409 414 419 424 429 433 438 443 880 448 453 458 463 468 473 478 483 488 493 881 498 5°3 5°7 512 5i7 522 527 532 537 542 882 547 552 557 562 5 6 7 57i 576 58i 586 591 883 596 601 606 611 616 621 626 630 635 640 884 64S 650 655 660 665 670 675 680 685 689 885 694 699 704 709 714 719 724 729 734 738 886 743 748 753 758 763 768 773 778 783 787 1 4 0.4 887 792 797 802 807 812 817 822 827 832 836 2 8 888 841 846 851 856 861 866 871 876 880 885 3 4 1 1 2 6 889 890 895 900 9°5 910 9i5 919 924 929 934 5 2 890 939 944 949 954 959 963 968 973 978 983 6 7 8 2 4 8 891 988 993 998 *002 *oo7 *OI2 *oi7 *02 2 *02 7 ^032 3 2 892 95 36 041 046 05I 056 061 066 071 075 080 9 3-6 893 085 090 °95 IOO i°5 IO9 114 119 124 129 894 i34 139 143 -I48 153 158 163 168 173 177 895 182 187 192 197 202 207 211 2l6 221 226 896 231 236 240 245 250 255 260 265 270 274 897 279 284 289 294 299 3°3 308 313 318 323 898 328 33 2 337 342 347 352 357 361 366 37i 899 376 381 386 39° 395 400 405 4IO 415 419 TABLES Table XIV. Continued. — Logarithms op Numbers 167 No. 1 2 3 4 5 6 7 8 9 r Pp. Pts. 900 95 424 429 434 439 444 448 453 458 463 468 901 472 477 482 487 492 497 5°i 506 5ii 5i6 902 5 2 i 525 53° 535 540 545 55° 554 559 5 6 4 903 S69 574 578 583 588 593 598 602 607 612 904 617 622 626 631 636 641 646 650 655 660 905 665 670 674 679 684 689 694 698 7°3 708 906 713 718 722 727 73 2 737 742 746 75i 756 907 761 766 770 775 780 785 789 794 799 804 908 809 813 818 823 828 832 837 842 847 852 909 856 861 866 871 875 880 885 890 895 899 910 904 909 914 918 923 928 933 938 943 947 911 952 957 961 966 971 976 980 985 990 995 912 999 *oo4 *oo9 *oi4 *oi9 *023 *028 *°33 *o 3 8 #042 5 0.5 913 96 047 052 057 061 066 071 076 080 085 090 1 914 095 099 104 109 114 118 123 128 133 i37 2 1 915 142 147 152 156 161 166 171 175 180 185 3 4 1 2 5 916 190 194 199 204 209 213 218 223 227 232 5 2 5 917 237 242 246 2 5 J 256 261 265 270 275 280 6 7 3 5 918 284 289 294 298 3°3 308 3*3 3i7 322 327 8 4 919 332 33 6 34i 346 35° 355 360 3 6 5 369 374 9 4-5 920 379 384 388 393 398 402 407 412 4i7 421 921 426 43i 435 440 445 45° 454 459 464 468 922 473 478 483 487 49 2 497 5°i 506 5ii 5i5 923 520 525 53° 534 539 544 548 553 558 562 924 567 572 577 581 586 59i 595 600 605 609 925 614 619 624 628 £>33 638 642 647 652 656 926 661 666 670 675 680 685 689 694 699 7°3 927 708 7i3 717 722 727 73i 736 741 745 75° 928 755 759 764 769 774 778 783 788 792 797 929 802 806 811 816 820 825 830 834 839 844 930 848 853 858 862 867 872 876 881 886 890 4 4 931 895 900 904 909 914 918 923 928 932 937 1 932 942 946 95i 956 960 965 970 974 979 984 2 8 933 988 993 997 *002 *oc>7 *OII *oi6 *02I *025 *o3o 3 4 1 1 2 6 934 97 °35 °39 044 O49 °53 058 063 067 072 077 5 2 935 081 086 090 °95 100 104 109 114 118 123 6 7 8 2 4 8 936 128 132 137 142 146 151 155 l6o 165 169 3 2 937 174 179 183 l88 192 197 202 206 211 216 9 3 6 938 220 225 230 234 239 243 248 253 257 262 939 267 271 276 280 285 290 294 299 3°4 308 940 3*3 3i7 322 327 33i 336 34o 345 35° 354 941 359 3 6 4 368 373 377 382 387 39i 396 400 942 405 410 414 419 424 428 433 437 442 447 943 45i 45 6 460 465 470 474 479 483 488 493 944 497 5°2 506 5" 5i6 520 525 529 534 539 945 543 548 552 557 562 566 57i 575 580 585 946 589 594 598 603 607 612 617 621 626 630 947 635 640 644 649 653 658 663 667 672 676 948 681 685 690 695 699 704 708 7i3 717 722 949 727 73i 736 740 745 749 754 759 7 6 3 768 168 COMPRESSED AIR Table XIV. Continued. — Logarithms of Numbers Ko. I 2 3 4 5 6 7 8 9 Pp. Pts. 950 97 772 777 782 786 791 795 800 804 809 8i3 9Si 818 823 827 832 836 841 845 850 855 859 952 864 868 873 877 882 886 891 896 900 90S 953 909 914 918 923 928 932 937 941 946 95° 954 955 959 964 968 973 978 982 987 991 996 955 98 000 005 009 014 019 023 028 032 °37 041 956 046 050 °55 °59 064 068 073 078 082 087 957 091 096 100 i°5 109 114 118 123 127 132 958 137 141 146 150 155 159 164 168 173 177 959 182 186 191 i95 200 204 209 214 218 223 960 227 232 236 241 245 250 254 259 263 268 961 272 277 281 286 290 295 299 3°4 308 3i3 962 3i8 322 327 33i 336 34o 345 349 354 358 963 3 6 3 367 372 376 381 385 39° 394 399 403 1 5 O- c 964 408 412 417 421 426 43° 435 439 444 448 2 I 965 453 457 462 466 47i 475 480 484 489 493 3 4 I 2 5 966 498 502 507 5ii 5i6 520 525 529 534 538 5 2 5 967 543 547 55 2 556 56i 565 57° 574 579 583 6 7 8 3 968 588 592 597 601 605 610 614 619 623- 628 3 4 5 969 632 637 641 646 650 655 659 664 668 673 9 4 5 970 677 682 686 691 695 700 704 709 7i3 717 971 722 726 73i 735 740 744 749 753 758 762 972 767 771 776 780 784 789 793 798 802 807 973 811 816 820 825 829 834 838 843 847 851 974 856 860 865 869 874 878 883 887 892 896 975 900 9°5 909 914 918 923 927 93 2 936 941 976 945 949 954 958 9 6 3 967 972 976 981 985 977 989 994 998 *oo3 *oo7 *OI2 *oi6 *02I *025 *029 978 99 °34 038 043 047 052 056 061 065 069 074 979 078 083 087 092 096 IOO i°5 IO9 114 118 980 123 127 131 136 140 145 149 154 158 162 981 167 171 176 180 185 189 193 I98 202 207 1 4 O A 982 211 216 220 224 229 233 238 242 247 251 2 O 8 983 255 260 264 269 273 277 282 286 291 295 3 4 I I 2 6 984 300 3°4 308 313 3i7 322 326 33° 335 339 5 2 985 344 348 352 357 361 366 37° 374 379 383 6 2 4 8 2 986 388 392 396 401 405 4IO 414 419 423 427 7 8 3 987 43 2 43 6 441 445 449 454 458 463 467 47i 9 3 6 988 476 480 484 489 493 498 502 506 5ii 5i5 989 520 524 528 533 537 542 546 55° 555 559 990 5 6 4 568 572 577 58i 585 59° 594 599 603 991 607 612 616 621 625 629 634 638 642 647 992 651 656 660 664 669 673 677 682 686 691 993 6 95 699 704 708 712 717 721 726 739 734 994 739 743 747 752 756 760 765 769 774 778 995 782 787 791 795 800 804 808 813 817 822 996 826 830 835 839 843 848 852 856 861 86 S 997 870 874 878 883 887 891 896 900 904 909 998 9i3 917 922 926 93° 935 939 944 948 952 999 957 961 96S 970 974 978 983 987 991 996 APPENDIX A The following notes and tables relating to drill capacities are taken from the Ingersoll-Rand catalog. DRILL CAPACITY TABLES The following tables are to determine the amount of free air required to operate rock drills at various altitudes with air at given pressures. The tables have been compiled from a review of a wide expe- rience and from tests run on drills of various sizes. They are in- tended for fair conditions in ordinary hard rock, but owing to varying conditions it is impossible to make any guarantee with- out a full knowledge of existing conditions. In soft material where the actual time of drilling is short, more drills can be run with a given sized compressor than when working in hard material, when the drills would be working continuously for a longer period, thereby increasing the chance of all the drills operating at the same time. In tunnel work, where the rock is hard, it has been the expe- rience that more rapid progress has been made when the drills were operated under a high air pressure, and that it has been found profitable to provide compressor capacity in excess of the requirements by about 25 per cent. There is also a distinct ad- vantage in having a compressor of large capacity, in that it saves the trouble and expense of moving the compressor as the work progresses, and will not interfere with the progress of the work by crowding the tunnel. No allowance has been made in the tables for loss due to leaky pipes, or for transmission loss due to friction, but the capacities given are merely the displacement required, so that when select- ing a compressor for the work required these matters must be taken into account. Table I gives cubic feet of free air required to operate one drill of a given size and under a given pressure. Table II gives multiplication factors for altitudes and number 169 170 COMPRESSED AIR of drills by which the air consumption of one drill must be multi- plied in order to give the total amount of air. Table 1.- — Cubic Feet of Free Air Required to Run One Drill of the Size and at the Pressure Stated Below u * § cS O 60 70 80 90 100 SIZE AND CYLINDER DIAMETER OF DRILL A35 2" 50 56 63 70 77 A32 2\" 60 68 76 84 92 B 68 77 86 95 104 C 2f" 82 93 104 115 126 D 3" 90 102 114 126 138 D ol " 95 108 120 133 146 D o 3 II °16 97 110 123 136 149 E si" F 3|" 108 124 131 152 166 F a s 113 129 143 159 174 G 4|" 130 147 164 182 199 H 5" 150 170 190 210 240 H9 6*"" 164 181 207 230 252 100 113 127 141 154 GLOBE VALVES, TEES AND ELBOWS The reduction of pressure produced by globe valves is the same as that caused by the following additional lengths of straight pipe, as calculated by the formula: Additional length of pipe = 114 X diameter of pipe 1 -f- (36 -T- diameter) Diameter of pipe 1 1 1^ 2 2^ 3 3M 4 5 6 inches Additional length J 2 4 7 10 13 16 20 28 36 feet 7 8 10 12 15 18 20 22 24 inches 44 53 70 88 115 143 162 181 200 feet The reduction of pressure produced by elbows and tees is equal to two-thirds of that caused by globe valves. The following are the additional lengths of straight pipe to be taken into account for elbows and tees. For globe valves multiply by ^. Diameter of pipe \ Additional length J 1 W 2 2 2V 2 3 3K 2 3 5 7 9 11 7 8 10 12 15 18 30 35 47 59 77 96 4 5 6 inches 13 19 24 feet 20 22 24 inches 108 120 134 feet These additional lengths of pipe for globe valves, elbows and tees must be added in each case to the actual lengths of straight pipe. Thus, a 6-inch pipe, 500 ft. long, with 1 glove valve, 2 elbows and 3 tees, would be equivalent to a straight pipe 500 + 36 + (2 X 24) + (3 X 24) = 656 feet long. APPENDIX A 171 o rt o t^ o H i-H § o « N H H <1 W H Ch o o iJ H H Q > W P < C7 H H V PI M M H o £ W) H « H ft « a <1 o Ph o S o U rfl !h K H a O P < H ft H J u -1 H H fc # ■*CO-*OOlO-^lTto«oco'ONi^coooa lOcNr^-tfiOi^OJ-'tfcs'+icoco NOlOOOratOOCONOMlON i— l(M(M(MCOCO-*l-*-^lOlOCOCO 00 f^ lO CO CM OO t~ CO -># CM 00 » O) o p oh h rtin eq eq «ico ■*NHCJOON(OHK5ffl 00 iHCOtOO)(N'l l t>01IN'*l®HiO oocooocooiffloiaoooHH O ■— iffir- i CM CO lO CO t^ CO irt Hn»ooowQNo>Hnt»rH NNNNOOCOOOCCOOffiOiOO OllOlOr-l r-l 05 05 00 00 intoair-i-*(Dcc®'-im"naico 00 CM TtTcM ooto^Ncqcc Ort^lOWONCOlONOlMiO COCD!OCOCONNNNNN»W co oo ■* io cq oo ■* co co cm ■*ifJNOJ'-lcNNCO OOHHrt(NCNCNCNCOC0 1< ooooooooooooo oooooooooooo oooooooooooo HINCCiOCJCONOOOiOCNlO w a si CO •** H o &, H ft „, wo O 3 • o O 1 p° CO ■" ' !>> M o CD £ CO Ml O (H CM CD >> td -° M . CD PI d ^ ,-, * o -+^ CD J3 i-H O b0£ ~ 2^ 00 «-" o o d o3 O -2 O p| ^ Q Pi "S o o J5 'Hi f^h P3 w o J3 T3 co CO ^ O -- 1 "- -d CO O • ^ a CD a -I d _7 *=> "d -2 CD ^3 T3 bi) a a> +j 53 3 d s ^ CD HH ro pi - cd •<-' I o J2 CD h£h 03 co - ® d * £ a 2 "- -d ti rd d ^ CD o o « ?> ^ !>. CO 00 d CO ^3 .5 O APPENDIX B DESIGN OF LOGARITHMIC COMPUTING CHARTS Problem. — Design a chart for determining values of x, y and z in equations of the form: x n = ay m z r or whence 1 m r^ x = a n y n z n i TYl V logo; = - log a + - logy + - log z (I) IV IV IV As a preliminary and introductory study assume n — m + r and construct a chart as follows (See Fig. 27) : Tabulate values of x, y and z to be covered or included in the chart. Take out the logarithms of these numbers. Plat these logarithms, to some convenient scale, on the vertical lines marked x, y and z setting the zero of scale at A and B for the y and z lines respectively, but for the x line set the zero of scale at F making FG = - log a. On the lines x, y and z mark the num- bers whose logs have been scaled. Then evidently wherever the line CD may be placed across the chart the proportions thereon written will hold and Eq. (I) is completely satisfied' — that is — given any two of x, y and z the third will be found on the line CD laid over the two given. Note that the line AE is unnecessary — it being placed in the figure for demonstration only. Note also that the line FG is not to appear on the chart and that the factor C effects only the width of the chart and may be taken to suit convenience. Evidently this solution applies only to the special case when n = m + r. It has the further objection that if the corre- sponding numerical values of x, y and z are very different then the lengths of the x, y and z lines will be different, though not in the same proprtion. The chart will have a better appearance if the three lines are nearly equal. The general solution is as follows (See Fig. 28) : 172 APPENDIX B Let I = desired length of chart in inches, k = desired width of chart in inches, Xi = greatest value of x to appear on the chart, f x be the necessary multiplier for logs x, to give desired length to the x line. 173 log z Then /*(log xi log a) = I Whence f x Let Z\ — be the value of z corresponding with X\. (II) 174 COMPRESSED AIR Let f z be the nearest whole number to I (HI) log Z\ that being the most convenient multiplier for logs of z to make the z line nearly equal to the x line. f z lo e z F-l Fig. 28. Let f y be the necessary multiplier of logs y. We have yet to find p, q and f y . Imposing the condition that Eq. (I) must be satisfied and remembering that all values of log x are to be multiplied by f x . Then must .APPENDIX B 175 ^f, log z = V f x log z. Whence p = -fy-k (IV) /c ft w/ z also | /.log y = | /, log y. Whence fv=^fh (V) Evidently q = k — p (VI) Example.- — -Design a chart to solve the formula for friction in air pipes, viz., 0.1025 vH rd 5 - n X 3,600 in which / = loss of pressure in pounds per square inch, I = length of pipe in feet, v = cubic feet of free air per minute, r = ratio of compression = number of atmospheres, d = diameter of air pipe in inches. Here we find five variables while our chart can provide for three only. We will, therefore, take I = 1,000 feet and replace the product fr with a single variable and represent it by h. The equa- tion will now become Whence fr = h = 35J3 dP" ° r V ' = 35 ' 13 M5 ' 31 log v = ^ log 35.13 + \ log h + -^ log d (VII) which is in the same form as Eq. I. We will design the chart to be about 12 in. long (I = 12) and 8 in. wide (k = 8) and will provide for a Max. v = 50,000 (=vi) log 50,000 = 4.6990 and log 35.13 = 1.5456 whence by II, 1.5456\ /. (4.699 -^P)=12, whence /» = 3 (nearest whole number) . The value of d corresponding to v = 50,000 is about 12 in. 12 log 12 = 1.0792. Whence by III, f d = 1 Q?92 = 12 (nearest whole number). Then by IV, P = 5 f X8X^ = 5.31 and q = 8 - 531. = 2.69 176 COMPRESSED AIR and by V } - = 2 X 2^9 X 3 " 4Mh See Plate III, page 53, for the completed chart. To lay out such a chart, make a table such as is indicated below. Then measure out, on the respective lines, with a scale of inches and decimals (engineers scale) the quantities in columns headed f v log v, fh log h and f d log d remembering that log 1 = 0.0. Hence the bottom of each line A, F and B will be marked 1. As each multiplied log is marked on its line, write there the corre- sponding number in columns headed v, h and d respectively. On a chart thus laid out a thread stretched as at CD will lie over the three quantities that will satisfy the equation, hence any two being known the third can be found. Table for Charting Equation, v 2 = 35.13 hd 6 - 31 Note Y% log 35.13 = 0.7728 and 3 X 0.7728 = 2.318 v line h line d line V log V fv log V h log h fh log h d log d fd log d The following notes may help the student when designing charts for other equations of the form given in Eq. I. (a) If the constant (a, Eq. I) becomes a proper fraction its log. is minus and the point F must be set above G instead of below; the zero of scale to be set at F when measuring out the X logs. (6) If the equation takes the form x n _ y m Z r then TYl T log x = - log a log y log z n n n APPENDIX B 177 This can be satisfied by reversing the direction of the measure- ments on the x line and placing the zero (or F) point above the line AB a distance - log af x as indicated in Fig. 29. it (c) When values of any one of the variables must be fractional the log is minus and must be measured in the opposite direction from A, B or F as the case may be. For instance if in the ex- Fig. 29. ample above d = % in. then log d = T.8751 = —0.1249 and we must set Y± at 0.1249/,* below B. (d) If circumstances are such that in the solution of such prob- lems as occur in practice; the figures on the lower portion of one of the lines (say the y line) will never be needed; the chart may be set out as suggested in Fig. 30 and only that portion above BR retained on the finished chart. Thus the scale may be enlarged and accuracy increased thereby. 12 178 COMPRESSED AIR Evidently the essential proportions of Fig. 27 are not changed by putting the chart in the shape of Fig. 30. i i i 1 |-F Fig. 30. (e) It will be found convenient to let x, in the above discus- sion, represent the largest factor in the equation. APPENDIX C During 1910 and 1911, an extensive series of experiments were made at Missouri School of Mines to determine the laws of fric- tion of air in pipes under three inches in diameter; the chief object being to determine the coefficient c" in the formula / = c -r 5 (See Art. 29.) The general scheme is illustrated in Fig. 31, in which the parts are lettered as follows: a, is the compressed-air supply pipe. b, a receiver of about 25 cu. ft. capacity. c, a thermometer set in receiver. d and d, points of attachment of differential gage. Fig. 31. — Diagram illustrating assembled apparatus. / and /, lengths of straight pipe going to and from the group of fittings. e, the pressure gage. g, the group of fittings — varied in different experiments. h, the throttle valve to control pressure. I, the orifice drum for measuring air, with the attachments as in Fig. 9. Experiments at Missouri School of Mines — 1911 On each set of fittings there were made ten or twelve runs with varying pressures and quantities of air in order to show the rela- V tion of / to — over as wide a field as possible. 179 180 COMPRESSED AIR £ e8 II ft ft w pi ft CO ±t T— 1 « O H pQ H o S 50i-Ht^O«ooseooO'^e<3t>-oseo»oeo i-HCO'tf'Oi— I(Mt-HO>-<©Ot-HOO©00>-h cc iOLOtDt^00C5-*C5(N(N00TtH00»O'-i00O500 CO «D «>■ i-H OOffiiOH«iiOOONO i-H^ttiOO'-H(M'-HO-i.-i^H £' OO^OSOSOO-^COCOCOCSOSCSCOCDO^^-* lONNffliOfflCOCCCOOOOMCOOOtOOCO CO C* 00_i-lt> iH CO OS co CO 00 H eo to Tjt rHCO^ l Oi-H(MT-iO(NOO'-IOO'-HOO'-l 1 '3 N 5. o g lO O t- CO "5 t^ CO lO CO 00 OS CO 1> 00 CO 00 ■* OS ONOlHCOiOINH^OHCCOHClOHiN 6 .-HCOOCiO>-HO'*O-i(NOONOO(N CDOOWNOON(N(NHNiOMO(N(MON l(NC0^iO^it>00O5O^H*ii0CDt^00 APPENDIX C 183 y Pq O 00 II , , r/l Tl PL, Pi .H ^H Tl £ H o s 1 o sd £ 01 !K w s= s co u 3 PL, o Ph m fa g O ci cd w H to W 9 3 a ^ 2 2 .5 ■b ^ -a Q «, .a i -^ I -^ >-, NOTfhOOWNOHNiONiOOONffl CC01'OtH1003HCO!00(N'*OIMMOhC^ i— 1 CQ NHNOMiHIMOiOMHCO^OONOiOO ^NOH^OHN'JIHNNHHNHHH 00-^iOiOT-HCN)fO00-*^COCO'>*t^t^"#Ol«3 '-h^OOO(MiOOt-hCOO^htDC0iOI>03OlN-*00 H^i>>.0 , *«N'*NH'^SHrl ■ i\ - • -J-l-1 -.%\j±r.v. 1 ' 1 14 L -Uj-j-" j- wi": \i i i i r \ | i i ! 1 1 1 \ \ |'|| \ _ ; _ _j__ Ji i i w \ \ mi > ..V 1 ' 1 ? "n P \ \\ 1 \\ rt - mi M 1 1 1 1 1 --- --H s S - — di = MM 3 a rj -rj ; ; -_ ~ 1 1 1 1 ! V f |- " -s 3 i - "" N I H o3 02 ^ 5 JS i 1 I 1 I + I ISO if,-M « - ,» 1 1 \t 4 ditt - i-i »3 «-i s a ■< s -^ _ jjjjj : - J * " .5 ^ £ -a .5 i 1 :: — ± _ " I I ::: ±:: SLL R7 \ S « u § a 3 1 1 I 1 i :::::-±:: 1T " ;:::x _ ~ M 7! =J P. - £ "1 llli fr 1 1 . *i 1 1 f=! - R 8 II <•) LI O | J II 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 II 1 1 Values of "f"' From this plate we get the following three equations : 80.0 K + 2e + 5u+4:g= 18.3, 80.0 K + 10 e + 9u = 11.8, 80.0 K + 2 e + 13 m = 6.8, in which APPENDIX C 185 v K— = resistance due to one foot of pipe, v e — = resistance due to one elbow, r v m— = resistance due to one extra ferrule or joint r with ends reamed, v u— = resistance due to one extra ferrule or joint r g — = resistance due to one globe valve. with ends unreamed, 2 i r So by attaching other lengths or fittings we get other equations and by simple algebra can find the numerical value of each symbol. Then v j i y 2 Kl — = c -= — — or c = d 5 K. r d b r Also the length of pipe giving friction equal to that of one elbow g is t, and so with other fittings. These experiments covered standard galvanized pipes of 2, 1}^, 1, %, and 3^-inch diameter. With each size pipe, runs were made to find friction loss in ordinary elbows, 45° elbows, globe valves, return bends, unreamed joints, and reamed joints. For each combination, data were taken for platting twelve to eighteen points, altogether about eight hundred. The results as a whole are satisfactory for the 2-, IjHj-, and 1-inch pipes. For the %- and 3^-inch pipes, especially the 3^-in pip e ; the results were so irregular, erratic, and conflicting that the results finally recorded cannot be accepted as final. In the light of these results, it is not probable that a satisfactory coefficient will ever be gotten for pipes under 1 inch; the reason being that in pipes of such small diameter, irregularities have relatively much greater effect than in larger pipes, and the probability of obstructions lodging in such pipes is relatively greater. In the 3^-inch pipe and fitting, unreamed joints were found at which four-tenths of the area was obstructed, and this with a knife edge. No doubt consistent results could have been gotten by using only pipes that had been "plugged and reamed," and selected fittings, but these results would not have been a safe guide for practice unless such preparation of the pipe be specified. 186 COMPRESSED AIR The results of these researches are embodied in Plate II. They show the averages of such data as seem worthy of consideration. The data for pipes exceeding 2-in. diameter are taken from various published data. Verification of these by the use of the sensitive differential gage is desirable. In the series of experiments referred to, the results worked out for the resistance of fittings were more erratic than those for straight pipes. Hence no claim is made for precision or finality in the results here presented. However, two important conclu- sions are reached. One is that the resistance of globe valves has heretofore been underestimated, and the importance of reaming small pipe has not been appreciated. . Table op Lengths of Pipe in Feet That Give Resistance Equal That op Various Fittings Diameter of Pipe 90° Elbows Un reamed Joints, Two Ends Reamed Joints, Two Ends Return Bends Globe Valves 1 2 3 4 1 11 2 10.0 7.0 5.0 4.0 3.5 2 to 4 it it tt tt 1.0 1.0 1.0 1.0 1.0 10.0 7.0 5.0 4.0 3.5 20.0 25.0 40.0 45.0 47.0 A series of runs was made on 50-foot lengths of rubber-lined armored hose such as is used to connect with compressed-air tools. The scheme was the same as that described for pipes and fittings ; v ^ and the range of — was the same. The average results are here given. This includes the resistance in a 50-foot length with the metallic end couplings. In these end connections a considerable contraction occurs. For the half-inch hose the end couplings are quarter-inch. The excessive resistance in the half-inch hose may have been due to these end contractions or to some other ob- struction. It is a further illustration of the fact that reliable coefficients cannot be gotten for pipes of half-inch diameter and less. Diameter of hose in inches Resistance in 50-ft. lengths i 2 950.0^ r 3 4 20.0 — r 1 4.5^ r 14 2.6^ r APPENDIX D THE OIL DIFFERENTIAL GAGE Examination of Eq. (21) shows that the greatest liability to inaccuracy lies in determining i since it is relatively small as com- pared with t and p and the conditions are such that the scale can- not be read with precision. To better determine i the oil differ- ential gage may be used. A special design suitable to this purpose is illustrated in Fig. 32. l The special fittings are inserted in place of the two plain glass tubes of Fig. 32. The cocks being lettered similarly in the two figures. The special features of the gage are the two reservoirs G\ and Gi the capacity of each being controlled by the movable piston. The manipulation is as follows : With water in the low pressure side and oil in the high pressure side and with C 2 and Cz open, the specific gravity of the oil is -~- = s. Now with C 2 and C 3 closed and d and C5 open the higher pressure will depress the surface, Ai, of the oil and raise A 2, that of the water. Now, by manipu- lating the piston G\ oil can be forced in or withdrawn from the gage tube until A v and A\ are in the same horizontal plane, or on the same scale line. While this coincidence holds i = Z (1 — s). Proof.— Let w equal pressure due to 1 in. of water head and equal that due to 1 in. of oil head. Then since the two pressures become equal on the line BB, we have p -f- OZ = p% + wZo and pi — p 2 = Z Q (w — 0) but i = Vl ~ V2 - Therefore, i = Z (1 - s). w .- '. . With kerosene oil in the gage i equals one-fifth of Z very nearly. The length of oil and water in the gage tubes can be further con- trolled by the drain cocks, on the reservoirs. The length of oil should be about five times as much as the anticipated i and this (i) must be kept within the limits specified by the standards of 1 This gage was designed and is used in the laboratories of the Missouri School of Mines. 187 188 APPENDIX D Cs S ? ( i } " C