peers re ae ret Sao eae Hine Ri Fe itera Tetra] i paling Mae Pau PaNS Honea esd Be ataats tinea its ij a 4 tt a ran man 4 ia + ea eS i seme Wire ra ne Parrett eee x + Stare 7 Sarees ne ae Energy and velocity diagrams of large ga ENERGY AND VELOCITY DIAGRAMS LARGE GAS ENGINES THEIR USE AND LAY-OUT By Paur Lb. Jostyn, M. E. CINCINNATI THE GAS ENGINE PUBLISHING CO. 1912 Copyright 1912 by The Gas Engine Publishing Co., Cincinnati, O. Contents Page Explanation of Technical Terms Used ...0ccceeseiaswensennng teas 1 Energy and Velocity Diagrams of Gas Engines...............0000- 5 Construction of the fadidatse GAAS! cceticseishenseec sergeant dalied aibettined ens ter ah 7 Inertia, (Diag ran tecss< cececeal arden ereuene neces sesame sel ees Gise e eas 20 Construction of Pressure Diagram ..............000005 ta pie ahane sth Construction of Tangential Effort Diagram...................000005 33 Diagrams for the Gas Blowing Engine ........ epee Sap gum woes Se aes - 50 Diagrams for the Main Bearing and Crosshead Guide Pressures..... 61 Diagrams for Single Acting Gas Engines ............6.0 00sec s ee eeees 65 Derivation of Formula for Flywheel and Safety Degree for Cycle Deviation for Engines Direct Connected to A, C. Generators. . .66 Formula for Safety Degree of Cycle Deviation.................... 68 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig, Fig. Fig. Fig. Illustrations Page 1. Diagram of Clearances in per cent of Cylinder volumes for IGOMIPTESSTOM: | cs cie sade sehen a ekae Sard ea oe Soe SRIOE RISE LE ta 9 2, 3 and 4. Construction of Compression Curves......... 10, 11 5, 6 and 7. Construction of Expansion Curves......... 13, 14, 15 8:9 and 10. Indicator Cards: .4:.... caves ree iaciaaian 17, TS, AS TOI PEWS. ad cscssuerecemcaieis odd denen €0 46 to 49. Main Bearing and fas Beata Guide Pressure Diagram 44 and 72x04. ENGine 4 co-5 vom icnle ore g hia rete ee anes 62, 64 50, 51 and 52. Piston Pressure, Tangential Crank Effort and Velocity Diagram 22x32 Single Acting Single Cylinder Engine 53, 54 and 55. Piston Pressure, Tangential Crank Effort and Velocity diagram 22x32 Single Acting Twin Cylinder Engine. 56 to 62. Cylinder Slide and Main Bearing Pressure Diagram 22x32 Single Cylinder, Single Acting Engine ......... Follow 64 EXPLANATION OF TECHNICAL TERMS USED. Absolute temperature is measured from the absolute zero which is obtained by assuming that it is possible to continue the cooling of a perfect gas until its volume is diminished to nothing, following hereby the law of expansion of heat. The absolute zero is considered to be equal to 491.2° F. below the melting point of ice on the air thermo- meter, thus corresponding to 32° F. of the freezing point of the latter. Absolute pressure is measured from a perfect zero of pressure that is 14.7 lbs. below the atmospheric pressure, or as is mostly used 14.2 lbs, of a metric atmosphere, which was taken for the construction of the diagrams. Adiabatic curve as expressed by the equation PX V"° = constant is obtained when the law of expansion of heat of a gas takes place with- out loss or addition of heat, i. e. at a constant temperature. The con- struction of the indicatcr diagram curves, as is shown, is merely a graphical representation of the equation PX V" = constant. Angular velocity is the angle through which any point of a body turns in a second, having a radius equal to unity 1, expressed in feet per second. Clearance space is the space left free, when the piston has reached its end position, between the piston and cylinder head in the cylinder proper and the passages and chambers of the inlet and exhaust valves. The clearance space is usually expressed in % of the total cylinder volume, or can be expressed of the piston displacement only. Circumferential velocity is the speed of a point revolving on a radius equal r and with n revolutions per minute measured on the : : 2*xRXr Xn. circumference in feet per second =v — 6). Crank radius is a term denoting the distance from the center line of the main shaft on the crank to the center of the crank pin. The diameter of the described circle is equal to the piston stroke. Crank angle is the inclination angle of the crank radius against the horizontal axis of the main shaft or the horizontal center line of the piston. Connecting rod angle is the inclination angle of the center line of connecting rod and horizontal axis of the piston at the crosshead end. Center of gravity of crank is the point about which, if the crank be suspended, it will be in perfect equilibrium, that is there will be no tendency of rotation. Center of gravity of connecting rod is the same point as defined for the crank. Division of gas engines is made first, according to the manner of work, four cycle or two cycle, single acting and double acting, and second, according to arrangements of the cylinders and the number of working cylinder ends: There can be (1) Single acting four cycle single cylinder engine, 6“ “cc “ce (2) he " twin or, (2a) ce ce oe ce tandem ce (3) Double *‘ “single cylinder ce “ce e “ce . ae (4) twin or, (5) ce ce “e ce tandem ce (6) . i" By “twin tandem * For two cycle engine the same division as for the four cycle is made. A four cycle engine is a term denoting that the period of work of an engine occurs every two revolutions or four strokes, i. e., the suction, compression, explosion and expansion and exhaust take place each during a full stroke. In a two cycle engine a period of work takes place during one revolution, or two strokes, working how- ever with the same principle as the four cycle, i. e. having the suction, compression, explosion and expansion and the exhaust. distinguishing mark between the two engines is the admission of the The main cylinder charge. Diagram is a term used to denote a graphical representation of a known variation of pressures or any quantities. Effective piston area is the net piston area exposed to the pressures occurring in the cylinder. Fluctuation or the degree ot regularity of the speed of an engine is the relation between the difference of the maximum and minimum speed Vmax — Vin, Vv Mechanical efficiency of an engine is a term denoting the relation of the actual effective work to the theoretical indicated, and is usually expressed as a coefficient given in % of the effective work of the en- to the average and is usually expressed gine. Moment of inertia in a general way is expressed through the equation [== (mr ), that is the moment of inertia of the weight or the mass of a body with respect to an axis is the algebraic sum of the products obtained by multiplying the weight or the mass of each ele- mentary particle by the square of the distances from its center of gravity to the axis. Mass of the body is obtained when its weight is divided by the coefficient of the acceleration due to gravity, equal to 32.1. Mean effective pressure, M. E. P., is the pressure which is obtained when the area of an effective indicator card (that is a card from which is deducted all the negative pressures) is transposed in a rectangular figure having the same length as the card and as well the same area, the height of such rectangle representing the pressures will be equal to the M. E. P Maximum circumferential speed, a term as used for engine speeds, is the highest circumferential speed obtained owing to the non uni- formity of the engine speed, the minimum being the lowest. The wi te ‘ .. Vmax + Vmin : average velocity is obtained from the expression —- ge and is usually termed the engine speed. Piston pressures as generally termed is an expression denoting all pressures acting on the piston. In a gas engine such pressures will be the following : (a) Suction pressures, occuring owing to the difference between the atmospheric pressure and the pressure inside of the cylinder, since during the suction period a communication of the cylinder space and the outside air takes place. (b) Compression pressures, originated after the communication with the outside air is shut off, the admitted charge being com- pressed. (c) Explosion pressure obtained through ignition of the compressed charge. (d) Expansion pressures obtained through expansion of the exploded charge. (e) Inertia pressures originated through acceleration of the masses of the reciprocating parts. The action of the above pressures may be seen from the diagram. Reciprocating parts of an engine are such parts, through which the reciprocating motion of the piston is transformed into a revolving Eee motion; they are composed of the piston, piston rod, crosshead and the connecting rod. Tangential crank effort is the net revolving energy of the engine, obtained as a result of action of the net resultant piston pressures act- ing tangential on the crank circle. ah o ENERGY AND VELOCITY DIAGRAMS OF GAS ENGINES. THEIR USE AND LAY OUT. The so called piston pressure and tangential crank effort diagrams are a graphical representation of forces originated in the working cyl- inder of the engine and the manner in which they are transmitted through the reciprocating parts of the engine to the main shaft, during a period of time when one process or cycle takes place. From these diagrams it is possible in a simple manner to trace, step by step, the manner in which the variation of the forces occurs, how the masses, accelerated by the chosen speed of the engine, influence the above variation of the forces, and what results are obtained of these two actions at the point on the crankshaft where the energy is utilized for the purpose the engine is intended to serve. On the basis of similar investigations the right measures can then be easily taken, in order to assure the safety of the parts as to their stresses and to fulfill the requirements for steadiness of running of the engine as far as the regulation of the forces acting and the variation of speed are concerned, according to the demands of service for which the engine is to be used. For steam engines such measures were, on basis of similar diagrams, well worked out and adapted for any case. To a certain extent, there were even standards established, especially for such service of engines for which the requirements were of frequent occurrence and similar character. The gas engine, as is well known, is now successfully used for the same variety of service as the steam engine, and since the first is work- ing on the same reciprocating principle as the latter, in a way the same fulfillment of requirements for the individual case of service has to be observed, as is the case with the steam engine. Owing, however, to. the quite different principle of the utilization of the gas in the gas engine, the results obtained from diagrams will be different, and, therefore, the measures to be taken will vary from those required for the steam engine. = The object of this work is to show how to lay out the different diagrams for gas engines and to illustrate how and what measures were taken to fulfill certain requirements of some engines actually built. The gas engines to be considered are :— (1) —32’’x42” twin tandem, double acting, four cycle engine, running at 107 on blast furnace gas, with a rated maximum capacity of 1850 H. P. for each tandem engine, direct connected to a 1,000 K. W. A.C. generator of the three phase, 25 cycle type, in- tended for parallel operation. (2)—44” & 72’’x54’’ twin tandem, double acting, four cycle, gas blowing engine, running on blast furnace gas, with a rated maxi- mum capacity of 3750 H. P. for each tandem engine, and with a possible speed variation of from 20 up to 90 R. P. M. The piston pressures and tangential crank effort diagrams can be divided into :— (a)—Pressure diagram, obtained merely from the action of gas in the working cylinder, as represented by the indicator card. b)—Resultant pressure diagram, obtained by deducting the pres- sures acting against the expansion pressure in the remaining cylinder ends. (c)—lInertia diagrams, showing the pressures originated by the speed of the engine through the acceleration of the reciprocating parts. (d)—Net effective piston pressure diagram, obtained by deducting from the resultant pressure those due to inertia. (e)—Effort diagram, showing the net effective piston pressure act- ing tangential on the crank circle. (f)—Velocity diagram, showing the variation of the circumferential speed of crank pin. The sequence in laying out the diagrams conveniently has to be done in a certain order, since each one bears a certain relation to the other and, with the exception of the inertia pressure diagram, cannot be constructed without previously ascertaining the preceding diagram. The design of the gas engine is usually made in such a way, that the engine, by making a few minor changes, is able to run on either blast furnace, producer, or natural gases, only these three kinds of gases be- ing usually used for engines of larger sizes. In order to assure the re- quired degree of good service of the engine when running on one of the gases named, it is necessary to lay out diagrams for each of these cases. =6)S CONSTRUCTION OF THE INDICATOR CARD. The pressures as represented by the indicator diagram serve as a base from which the method has to commence in laying out of the dia- grams and investigations to be made. Usually a full load diagram is taken, and either a card taken from an engine, or, theoretically drawn, is used. In the latter case the dia- gram should, as near as possible, resemble the actual card. The four cycle diagram, as is well known, consists of the suction, compression, ignition and expansion, and exhaust periods, extending during two revolutions of the engine or four strokes. Suction. At the end of the exhaust period the suction will com- mence with a slight curve, which will quickly go over into the straight line. The suction pressure is on an average about 1.5 lbs. below the atmospheric pressure and remains constant up to the end of the stroke, at which point the compression begins ; it will therefore be represented by a straight line. Depending upon the kind of regulation of the en- gine, the suction pressure will vary slightly, though at full load of the engine it will be the same for any kind of gas. With light loads and especially with throttling regulation, the suction pressure changes greatly, and in some cases it may be found to be 8 to 10 lbs. below the atmospheric pressure. Compression. The compression for any kind of gas, is taken as following very near the adiabatic law, expressed by the equation P V"= Constant. Where, P= pressure, V =volume, and n= an exponent varying for different kinds of gas. The value of the quantity n, for gas engines depends upon the composition of the mixture, and the action of cylinder wall cooling. It varies from n=1.15 upton=—1.5. It is customary for the con- struction of the diagrams to take the average value for the exponent n=1.35, which will be but slightly in error. This value has often been found to give results approaching very near to the actual curvat- ure of compression as obtained from indicator cards. The maximum terminal compression pressure is influenced by the suction pressure, clearance space and the temperature of admitted and ies compressed charge, and slightly also by the exponent n. In a general way, it may be said that a larger exponent n and higher temperatures will give higher terminal compression pressures and vice versa. Since the composition of the charge varies greatly for different kinds of gases and the temperatures also correspondingly change, the terminal compression pressures, therefore will also have another value for each kind of gas. The latter value in question may be found from the following formula: C v\ - nol C \r TY pmol ye Ps Vv Ts Vv Ps Where: V~ total cylinder vol. (piston displacement + clearance space). C= terminal compression pressure absolute. Ps = suction pressure. v= volume of clearance space. n= exponent, as explained above. Tc= absolute temperature at end of compression. Ts = absolute temperature at end of suction. Good average values for the terminal compression pressures for the three different kinds of gas named previously, are given in Table 1, and also the clearance space necessary for each kind of gas. These values were verified from engines built. TABLE I. Kind of was Terminal compression pressure in Clearance space in percentage AE Oe ae lhs, above atmospheric pressure. of total cylinder volume. Blast furnace | 170 | 15. Producer 140 17. Natural | 100 | 21. Figure 1 shows a diagram by means of which by knowing either the terminal compression pressure, or the clearancé space in percentage of the total cylinder volume, the first or the second quantity can be conveniently read off. The curve is drgwn as an adiabatic with the exponent n ~1.35. Figures 2, 3 and 4 show the construction of compression curves on basis of values given above for blast furnace, producer and natural gases. The construction is made in a similar way in each case. Angle « can be taken of any size, however, the smaller it is chosen the more points will be obtained and the curve will be found more accurately. Lee By taking « = 15° good results are obtained, Angle 8 depends upon the exponent n and angle @ and may be found from: 1+tan. 8 =(1-+ tan.a)", Whena=15°, 8 will be found for the different values of the exponent n: n=1.15} 1.2 1.25 1.3 | 1.35 | 1.4 1.45 1.5 =17°30'118" 15° | 19° 5° (19° 55° | 26° 45° (91° 30'| 23° 35° 23° 107 The procedure of the construction of the curves is then the follow- ing: (see Fig. 2, blast furnace gas.) CLEARANCES 11% of TOTAL CXL VOLIME. The end compression pressure is for this gas 170 lbs. above the at- mospheric line; this pressure will correspond, when using the diagram Fig. 1, to a clearance volume of 15% of the total cylinder volume, i.e. piston displacement per stroke + the volume of the clearance space, Ove or 17.64% of the piston displacement only. The clearance space is to be first ascertained before commencing to draw the diagram. The scale for the stroke and pressures is then to be taken and it can be of any size. For the diagram Fig. 2 it was taken: for the stroke 5” and for pressures 1” =60 lbs. The lineal value of the clearance space is then to be found, for the stroke of 5”, fer this case it will be: 5 x 1764 = .8725’. In outlines the stroke and the clefrance distance is to be drawn and the zero line assumed. From the latter, in scale of the pressures, the atmospheric line is to be drawn, in this case at a distance of 2 — 2367", horizontally. The suction pressure as a hori- zontal line is to be drawn in, 1.5 lbs. = .025” below the atmospheric line. For the construction of the compression curve angle « is taken 15° and, with the average value of the exponent n , angle f will be BURST FURNACE GAS. D7 or retary, ey vou, Fig. 2. Construction of Compression Curves. Gas Indicator Cards. = 20° 45’. From the intersection point of the zero line and the clear- ance line, point ‘‘a’’, angle @ is laid off dgwnward and angle 8 upward, see Fig. 2. The intersection point ‘‘b’’ of the vertical end stroke line and the suction line will be the first point of the curve. The con- struction commences from the point ‘‘c’’ an intersection point of the zero line, with the end stroke line. From this point ‘‘c’’ at an in- clination of 45°, or any other angle (45° is more convenient to lay off) a line is drawn downward and prolonged until it bisects the side line of the = 10 PROBVCER GAS Uwe SVETION ae # ie Fig. 3. Construction of Compression Curves. Gas Indicator Cards. WAWRAL OAS DUCTION 1.5" ZAS OF TOTAL cY¥L.vor. Fig, 4. Construction of Compression Curves. Gas Indicator Cards. angle at a point ‘‘d’’, from which a vertical line is drawn upward. From the intersection point of the suction line and the clearance line, point ““e” another 45° inclined line is drawn upward and prolonged until it will bisect the vertical clearance line, point ‘‘f’’. Through this point “‘f”’ a horizontal line is drawn and prolonged until it will bisect the vertical line drawn from the point ‘‘d’’. The thus found intersection point ‘‘g”’ will be the second point of the compression curve. By repeating the foregoing proceeding and as shown in Fig. 2, a number of points of the curve will be found and the curve can be drawn in. The curve, if carefully constructed, will bisect the end stroke line at such a height =p which will represent the terminal compression pressure, in this case 170 Ibs., in the same scale as was taken for the pressures. The same method of construction is followed for compressions for the producer and natural gases, with the difference only that in each case the corresponding compression pressures and clearances were used. Ignition. The point at which the ignition takes place varies great- ly and cannot be found theoretically, it is more a matter of practical experience. Usually, provision is made for changing the point of igni- tion from about 37.7° before to 1° to 2° after the end of stroke. In general the ignition can be assumed as taking place at about 6 to 7% before end of stroke. Expansion. Depending upon a quicker or slower combustion of the mixture, the expansion will not commence from the end of the stroke, but more or less later. Indicator cards show that the expansion commences about 2 to 3% from the end of the stroke. The expansion curve also follows closely the adiabatic curve as ex- pressed by the equation PV" = Constant, and with the exponent n de- pending upon the same condition in the engine as for the compression, and for the same reason as given for the latter curve, the value of the exponent n may be taken as ==1.35 for the construction of the ex- pansion curves also. The same value should be used also for all three kinds of gases considered. The pressure at the beginning of expansion is influenced by the clearance space, compression pressure, temperature of the compressed and burning charge and by a quick or slow combustion; it cannot be theoretically computed with certainty. The following formula can serve to ascertain approximately the pressures at the beginning of ex- pansion : Po Tp P vy P= ———_ ; as, (tees > PexuV" ==Pvi?: Tec Pexu Vi Te. (< yo ( P ye Tes \vi Pex} — Where: P = pressure at beginning of expansion ; Pc = compression pressure ; Pexu == exhaust pressure at end of expansion; V = total cylinder volume ; vi — clearance space + space where expansion begins; Tp = absolute temperature at beginning of expansion ; Tc = absolute temperature at end of compression ; Texu ~= absolute temperature at beginning of exhaust ; n — exponent, as explained previously. 243) = AGLT + AWOSS Bes = Ps = “pULeIBEIC] 10zeItpuy 8Ex) IOF e2ain) UoIsuedxy jo worpn44su0F Pee —~14- m~ % / By — WIOUSS “3 m4 INIT Wiy es y \ te ta reaonsd > 21 P— —— Warowrs Moisid- —— — f}—— —-}-— —}~--- 4 — |e sinnlawsna 4 (peat + SVS Baawoys a 2 x 9G 3 | Fe 3 8, m 4 < ‘ 2D m v 4 | | | I t ‘sULeIBeI(] IOEOIpU] sed) AOF sPAIN-y WOIeuedxY Jo worzoNIzsU07) wheaso pore = I > wS in x mH fi 2 Pe, x | “sy a oH Fe a a | 5%. fo [ ra TRY —e SRM LY a 5 bx: Se} Fe ee ees aA: ye ey a OS HOLSIG ——— | Lc b Sw AWN 5 § HY 3 L S43 Z \ we | 3 ct c 2 r 4 if ie & pope As good average values of pressures at beginning of expansion for the three kinds of gases mentioned, for the construction of diagrams, may be considered : for blast furnace gas: 375 lbs., for producer gas: 350 Ibs., and for natural gas: 310 Ibs. Figures 5, 6 and 7, show the construction of the expansion curves on basis of the above values for the three kinds of gas. From the diagrams is to be seen that the proceeding is a similar one to that of the compression curves, with the only difference that the construction commences from the point of the beginning of the expansion. The latter point is taken as being 3% from the end of the stroke for blast furnace gas and producer, and 24%% for natural gas. Angle « and £8 are taken the same as for the compression, since the same value for the exponent n is used, i. e. - 1.35. The construction may be easily followed from the diagrams. Exhaust. The exhaust begins in most cases at about 12% before end of stroke. At this point the exhaust pressures can be found from the diagrams, or approximately found from formula given for the high- est expansion pressures. Since the exhaust valve commences to open gradually, the exhaust pressure will also drop gradually from 40 to 50 Ibs. at the beginning to 1.5 lbs. at the end. Depending upon the speed allowed to pass through the valve and the manner of operation of the latter, the pressure of 1.5 Ibs. will be attained at a point about 10 to15% from the beginning of the next stroke, and it will remain constant to the end, representing a straight line. The curve showing the variation of exhaust pressure at its beginning until it remains con- stant, cannot be ascertained by computation, it is approximately shown In proportion freehand. Figures 8, 9 and 10 show indicator cards found by bringing together the different curves, drawn as previously explained, and rounding the sharp corners. The scales for pressures and stroke were retained the same for all constructions of the curves, and by means of a piece of tracing paper the different curves were conveniently brought into one diagram. The mean effective pressures whi@h each of the drawn indicator diagrams give are very near the average values which in practical de- sigr.ing of gas engines are taken as basis. For the gas blowing engine, previously mentioned, 44'’x54’’, the maximum rated capacity of each tandem engine was given as == 3750 H. P. By taking an average speed of piston <= 750 feet per minute, the mean effective pressure will be found to be, when the effective mee _—_---* —_—< 240 zo} \S —— 4 LE 4 ge\s SUCTION EsWAUST BLAST FEVANACE GAS Gas Indicator Diagram. ies ATH.UNE | + East suction | PRIDVLER G Ae Gas Indicator Diagram. T= piston be 85%, M. E. P. = area = 1416 sq. in. and assuming the mechanical efficiency to _ 3750 X 33000 __ = aoxiai66 OS The theoretically drawn card a M. E. P. = 67.8 lbs. per sq. in. The producer and natural cards are as near to those of pzactical cards as is proven for blast furnace gas. 240 imo—J eo 77.6% MEP =< ra GanAavst Suction ATR. Lint | WATY RAL GAD Gas Indicator Diagram. 24:= INERTIA DIAGRAM. The inertia pressures for any crank angle may be found from the approximate formula: w WV r P=—xX— | cos «@ +— cos 2c Ff, g r L Where P = inertia pressures W = weight of reciprocating parts (unbalanced), V =~ circumferential crank pin velocity, r==crank radius, a — crank angle, L = length of connecting rod. The positive sign is for inner, and the negative sign is for outer strokes. The weight of the reciprocating parts is composed of, for a tandem engine. Weight of two pistons, including cooling water; “3 “* piston rods; * crosshead; “* connecting rod; ‘“ piston rod coupling and internal guide head; ““ piston tail rod, rear end guide head; ““ parts attached to piston rod coupling, crosshead and rear guide head. The crank is usually provided with a counterweight for purposes of balancing, in most cases made in onegpiece with the crank. The counterweight, after reducing it to the crank radius, must be deducted from the weight of connecting rod, and the remaining portion added to the total weight of reciprocating parts. Table 2 gives figured weights of reciprocating parts for some sizes of engines built and in addition data necessary fcr ascertaining of the inertia pressure. The weights given are for tandem engines. _~ 20 - TABLE 2. t Size of engine......-. 6.05.0 0+ 464454" 32% 42" — 24x” 32” Weight of reciprocating parts with- out connecting rod....... ...39150 Ibs. 16330 Ibs. 8110 lbs. Weight of connecting rod........13250 Ibs. 5250 Ibs. 2350 lbs. Weight of crank and pin without counter-weight ..... haces 24400 Ibs. 10450 lbs. 4600 lbs. Center of gravity of crank........ 11.2” 6.2” 2s. Length of connecting rod........135” 105” 80" Effective piston area sq. in.......1416.6 740.6 413.7 R. P. M. of engine...... .....20-90 105 150 The values of the iacor( cos uw + R cos 2a ) in the formula L for the inertia pressures are given in table 3 for each advancing 10° of crank angle and R 1 cE 5 TABLE 3 FORWARD STROKE BACKWARD STROKE 1 1 0° cos uw + @ cos 2a =1.2 0° cos a — ae 2a == .800 10° 1. =1.1728 10° a = .7969 20° - =1.093 20° - = .7865 30° - = .966 30° - = .766 40° ae = .801 40° . = .7313 50° - = .608 50° i = .6775 60° 7 = .400 60° a = .600 70° a = .188 70° = 4952 80° e = 0143 80° a = .3616 90° < = .200 90° os = .200 100° es = .3616 100° . = .0143 110° os = .4952 110° - = .188 120° i = .600 120° es = .400 130" es = .6775 130° = .608 140° a = .7313 140° S = 801 150° e = .766 150° i: = .966 160° in = .7865 160° 2 =1.093 170° se = .7969 170° : =1.1728 180° = .800 180° =1.200 The values for the backward stroke are the same, but reversed as given in table 3. 21 - DIAGRAMS FOR THE 32x 42’’ Gas ENGINE. For this engine the inertia pressures will be found to be, when using data given in table 2: Weight of reciprocating parts without connecting rod = 16330 lbs. Weight of crank with counter weight reduced to crank radius will be with reference to Fig. 11 10440 _x 6.2 _ 3080 Ibs. 21 TG tt. —> Deducting this weight from that of the connecting rod, an unbalanced portion will remain: 5250— 3080 2170 lbs., which has to be added to the other weight of fhe reciprocating parts. The total weight will thus be: 16330 + 2170 = 18500 lbs. 18500 The above weight per sq. in. of effective piston area 740.6 7 25 Ibs. The circumferential crank pin velocity is an N = ft. per sec. V = 232M BXNT 19.6 ft. per. sec. Substituting the found values in the equation for the inertia pressure, we obtain: w Vv R 2 1 P-- —xX—f{ cose + — cos 24 ea 25 5 cosa + —cos2a hie i @ °° 1.75 5 1 By using values given in table 3 for ( 2 + COMPRESSION &- NET RESULTANT PISTON RABOSURES. F\q@ \o Piston Pressures Diagram, 32x42 . Tandem Gas Engine. expansion at this point, are not sufficient to overcome, and an outside source has to be provided, the latter being the rotary inertia of the flywheel. At the end of the stroke again the acceleration forces are still so high that after overcoming the resistance of compression and other pressures acting against them, they still retain comparatively high = 36S forces which have to be nullified again by the rotary inertia of the fly- wheel. The latter has also to help out at such points where the re- sultant pressures are negative. At such point of the stroke where the net resultant pressures cross the zero line going into positive or negative values, a change in direc- tion of forces acting in the connecting rod, 1. e., from tension to com- L LT | atl Lat a La] mN VL ><] SN A |i = Pe Th | Dated | | oH lee aoe " WL | | +x ; Th { a tf tprat aol WRAL GAS be a Ld hb LL LaeBbL ba Ww ws hod yp bb bal A- PRESS DUGTO ERPL & EAPANSION ‘S- PRESS DUE TO LOSS ow FRICTION ees * GAWAUST e- - + COMBINATION ef Ire a= e+ SUCTION Sw + INERTIA oF RECIPR PATS a 4 + COMPRESSION 1 HET 7: t SVAES. FIG \7 pression will take place, which will cause a shock in the connecting rod. This shock will be more or less severe depending upon the in- clination of the curve when crossing the zero line, and the clearance existing between the shells and pins at the crosshead and crank ends of the connecting rod. By choosing a too high engine speed a consid- erable hammering may take place, with the result that the shells are ee liable to wear rapidly and break, or the crank pinrun hot. Aside from this a heavier flywheel will also be required. Comparing the results found from the diagrams for the three dif- ferent gases, we see that the speed chosen is best for the engine for blast furnace gas, is not so good for producer gas and is still worse for natural gas. For the latter gas, it seems in consideration of the above, that a reduction of speed of the engine would be desirable. CONSTRUCTION OCF TANGENTIAL EFFORT DIAGRAM. It was previously explained that the tangential crank effort diagrams show the forces acting tangential on the crank circle and their variation. Since the net resultant piston pressures are transmitted to the crank shaft not directly, but through the piston rod to the connecting rod and from the latter to the crank, the full forces of piston pressures will not act as revolving efforts on the crank circle. Owing to the variable inclination of the connecting rod during a revolution, there will be a part of the forces lost, at first, at the crosshead point, which will act on the crosshead guide, and secondly, a part at the crank pin, where it will act through the crank on the main bearing. This division of the net resultant piston pressures can be found either graphically or by means of mathematical formulae. The first method has the advantage that for any crank anzle, aside from finding the values of the crank efforts, diagrams can be drawn for the main bearing and crosshead guide presssures, which enable us to obtain a clear idea about their variation in force and direction. The construc- tion of these diagrams will be shown later. Graphically the tangential crank efforts are found as follows: With the same scale for the stroke as was taken for drawing the piston pressure diagrams, draw in outlines the connecting rod dnd crank movements, Fig. 18. For any crank angle, as for instance, 50°, prolong the center line of connecting rod over the point 50° on the crank circle until it intersects the vertical axis of the latter, thus, find- ing point As connect As with O, and on this line carry up from O the ordinate representing the net resultant piston pressure in inner stroke of the diagram, at 50°; from the end point of the ordinate Ts draw a perpendicular on the horizontal axis of the crank circle, the distance from Ts to the latter axis, will be the tangential crank effort for the crank angle 50°. By repeating the foregoing proceeding for each division angle, the crank effort can be found for both strokes. In the above method, in finding the tangential crank efforts the loss of the net resultant piston pressures at the point of the crosshead was -33- lett out of consideration. The value of the latter loss is small and the error allowed is negligible for the consideration of the diagrams. The pressures on the crosshead guide are found from: Pe = P tan. B, where: (see Fig. 18). Pc = pressure on crosshead guide. P = net resultant piston pressure. 8 = connecting rod angle at crosshead end. tan. 8 will be for a connecting rod == 5x crank radius: u= O° tang. 8 = .000 a= 100° tang. 8 = .2008 “==10° “ = 084 = TO? = .1918 “== 20° ‘“* = _ .068 = 120° = 176 “= 30° “ = 099 “= 130° = 154 “== 40° “ = 129 “= 140° =“ = 129 re 50° 8° 1S Jib e507 = .099 “== 60° “ = 176 “= 160° ‘“ = .068 “= 79° “ = 1918 “S7170° = 034 “= 8g0° “ = _ .2008 “= 180° ‘ = .000 “== 90° “ = 203 The largest value is at about 90°, and since at this point the net resultant piston pressures are not the highest, especially for the outer stroke, where they are but a few pounds, the crosshead guide pressures in the construction of the tangential crank effort diagrams are usually left out of considcration. Mathematically the tangential crank @ffort can be found by means of the formula which was given when discussing the inertia pressures. T=p sin (a+8) cos the values for which for every 10° on crank circle are given in table 6. ey SHORT SUS VT WSINVL Tee WUBOWIT HUONG 180343 WI AWALNIONYL —— ————-—_—_ Was Tv -_ Nous Tay Mind \2 Ola ‘ge bia ‘ODA ~35- Figs. 20, 26 and 32 show tangential crank effort diagrams for a tandem engine, and Figs. 23, 29 and 35, those for a twin tandem engine, for engines running on blast furnace, producer and natural gases. The curves for the tangential crank effort for twin tandem engines are obtained by drawing in two efforts curves of a tandem engine in such a manner that the corresponding points of the latter will be so many degrees apart, as the crank of one engine leads the crank of the other, usually 90°, asitis taken fromthe diagrams. By algebraic com- bination of the ordinates at any crank angle a resultant curve will be found which will represent the combined tangential effort for a twin tandem engine. The length of the diagram is the developed crank circle 2xRx7, for which any scale may betaken. The ordinates repre- sent the pressures for which the same scale as for the piston pressure is retained. It was mentioned before that at such points of the stroke where the net resultant piston pressures are negative, or at the end of a stroke where they have a comparatively high force, the inertia of the flywheel must act as an outside source in order to keep the engine in a continuous motien in the first case, and help to nullify the bad effect in the latter. The flywheel will, therefore, act as a kind of regulator. This regulation will take effect in the speed of the engine, making it more or less steadier during a revolution, depending upon the power of the flywheel inertia. This power will, therefore, depend upon the variation of the net effective piston pressure acting on the crank circle and the desirable, or required degree of steadiness, or fluctuation, in speed of the engine. We see thus, that the net effective piston pres- sure acting on the crank circle which the tangential crank effort dia- grams represent, serves as a basis for determining the power of the fly- wheel inertia, and from the latter the weight of the flywheel. To attain a certain required degree of fluctuation in the speed of the engine, the regularity of the tangential crank effort, which varies greatly, depending upon the type of the engine, whether single acting, double acting, tandem or twin tandem, will, therefore, be the main factor in determining the weight of the flywheel. For purposes of ascertaining the weight of flywheels, the so called largest excess area in the tangential crank effort diagrams is used. This excess area is found in the following way: The area included by the curve, Fig. 19, and a line drawn from its lowest point of the length of the diagram, is transposed into a rectan- —36- _—_— Mois THY MN -_— BNOwIS DAMN ‘v2 ‘OL NV roid Ri. gular figure, having as a base the above line. The line representing the height of the rectangle will then intersect the curve in such a manner, that the sum of the areas included partially by the latter and the line will be the same above and below it. In Fig. 19 we have areas above = .24 + 1.12 + .52 + .11 = 1.99, and below = .21 + -71 + .70 + .37 = 1.99 sq. in. The last mentioned line is called the mean effective torsion line, and the area of the rectangle A—B—-C—D wwill represent the average effective work of the engine. The areas included by parts of the curve and the mean effective torsion line above or below the latter, will be the excess areas, representing the amount of work the flywheel has to give up by its inertia at the corresponding points. The largest excess area is taken into consideration for ascertaining the weight of the flywheel, which can be found by the use of the i : HPXFr~x 8 following formula: W R* +W; Ri? = 1,000,000 —— where: W = weight of flywheel; R ~= radius of gyration of flywheel; W, = weight of generator rotor, if direct connected to generator R = radius of gyration of rotor, if direct connected to generator HP = brake horsepower 5 = the fluctuation degree of speed, flywheel fluctuation, to be used as fraction. f : ’ Fr a 100 in which: f = largest excess area in tangential effort diagram F = total area representing the average effective work of engine. N = r.p.m. of engine. : 1 ; ir The fluctuation degree 3 depends upon the service the engine is xo ; 1 1 to be used for. For gas engines it varies fons =e a te 100" “275 and smaller. The more steady running of an engine is required, the smaller the fraction ; must be taken. A high degree of steadiness in the speed —38- FHoNssvs Ver saNs Ty Te “AMowls TEVA i] -_ Avowls Paving ——— 36. of the engine is required in engines intended for electrical service, and especially for engines direct connected to A.C. generators operating in parallel. As is well known, for the generating of a uniformly steady alter- nating current, the period of time and the length of each alternation must be theoretically exactly the same. The alternation itself varies in force, therefore, is a curve, representing actually the sine curve, and since during one revolution a certain number of these alternations combine, the number depending upon the type of the generator, giving a current of certain strength, the combination is found to be the most effective, when they curve in exactly the same manner. ‘This condi- tion is attained when a perfectly uniform motion is existing. In parallel operation the alternations must combine not only for one unit, but for a certain number of units, and therefore, the best eff- ciency for the strength of the resultant current will again be obtained, when the conditions for that purpose are fulfilled in the same manner. Practically, however, in a reciprocating engine a perfectly uniform motion cannot be obtained and a certain displacement of the alterna- tions will always be the result. The variation of the speed, will therefore, be the main factor which will cause the displacement to be larger or smaller and especially at such times when the load cf the engine changes much and with it the speed. Due to the latter conditions in the engine, and that the flywheel and the rotary part of the generator are placed on the crank- shaft at a distance from each other, the revolving inertia of the fly wheel and rotor will act in parallel planes at different radii, and with every pulsation in power or speed of the engine, the larger inertia of the flywheel will come out of step with the inertia of the generator rotor, and both by its shaking moment will cause a twisting of the crankshaft, and with it a displacement of the center lines of the fly wheel and rotor, with the result that a further increase in the devia- tion of the alternations will be added. In the design of the generators a certain displacement is taken into consideration, and its permissible limits are usually prescribed by the manufacturer of the generators. These limits depend upon the type of the generator, i. e. upon the number of poles or the number of cycles. Table 9 gives these limits in angular degrees for some A.C. gener- ators. The values are to be taken as being the allowance of deviation in either direction, positive or negative. -40- —_—_ NOU LA Levine -—————— o¢ HS 62 Hla eed sis TABLE Y. Permissible Permissible No. Displacement Per Cent. No, Displacement Per Cent. ae pire Bice ecee ites ee Girduimferenee 2 3.00 833 22 ad 075 4 1.50 .416 24 125 .064 6 1.00 277 26 23 .060 8 75 .208 28 221 0595 10 .600 .166 30 20 0555 12 500 .139 40 15 0415 14 -430 119 50 12 0335 16 375 .104 60 .10 028 18 333 092 70 086 024 20 .300 0835 80 075 9206 From the table is to be seen that as the number of poles increases the permissible displacement decreases. The limits for the displacements are often given in electrical de- grees, as for instance, a deviation of + 2% electrical degrees. For transposing the latter into angular degrees, it is necessary to know the number of cycles per revolution or, the number of poles. The generator direct connected to the 32” x 42” engine, before mentioned, was given as having 25 cycles per second, and allowing a deviation of -: 2% electrical degrees, the latter value in angular de- grees will be found as follows : The r.p.m. of engine = 107, hence, the time per one revolution 60 = 56 seconds, and the number of cycles per one revolution = ™~ 107 . . 360 5 7 25 & .56==14. One cycle is performed in 4 = 25.714°; the devia- tion in angular degrees is then = geet BE + .1787°. 2a. —_— BOWLS TAVALNO MA 20 Old \e Od oe The displacement is usually measured on the circumference of the rotary part of the generator or, onthe circumference of the flywheel. In order to assure that the prescribed limits for the displacement of the cycles are not overstepped not only at full loads of the engine, but also at lighter loads, a safety factor for the displacement is usually taken into account, however, so far only, as the variation in the speed is concerned. ‘The possible increase in the displacement due to the action of the inertia of the flywheel and rotor on the crank shaft is prevented to some extent by allowing but a slight deflection (about .0015” to .002’’) and arranging the flywheel, rotor and the main bear- ings as compactly as a convenient attendance will allow. The safety factor for the displacement of the cycles will depend upon the variation in the speed of the engine, the required degree of uniformity of which, as we have seen, is attained by the chosen fluctu- ation of the flywheel, i. e. its weight, and the diagrams representing the variation of the speed, according to the allowed uniformity, serve for ascertaining the safety factor in a similar way, as the tangential crank effort diagrams were used for determining the weight of the flywheel, i. e. by using the greatest excess area in the diagrams. Figures 20, 26 and 32, show velocity diagrams for tandem, and Figures 24, 30 and 36, those for twin tandem engines, supposed to run on blast furnace, producer and natural gases. The construction of the curves is in each case similar, and they are found on basis of the following considerations: It was previously explained that the speed of the engine as ob- tained on the crank circle is influenced by the energy of the flywheel, and that the latterin time, is influenced by the excess areas in the tangen- tial crank effort diagrams, which represent the work the flywheel has to develop. This work is a product of pressures, forces, and paths or, speeds per minute, and consequently, the speed at any point of the crank circle may be found by fede, where: V is the speed depend- ing upon the curve of the tangential efforts, therefore, also upon their ordinates and dt the time, which is to be taken on the length of the diagram, since the latter is the es crank circle. The solution of the integral cannot be made without knowing the nature of the curve, i. e. its equation expressing the law of its varia- tion, and as this cannot well be found, the solution is usually made graphically, and in an approximate way. The velocity curves were found in such a manner and as follows: (see Figure 20). -44- owe Tay Ma 7 — Taos twos | —— oe MA se Ola ve OU .-45-- The excess area above and below the neutral line are considered as positive and negative, respectively, hence, also their ordinates at the corresponding crank angles. The division of the latter on the length of the diagram are taken to be dé, therefore, of the divided equal length. The integration, i. e. the addition, is made of the middle ordinates between two consecutive crank angles in the diagram and it can be commenced from any point. For the construction of the velocity curve, Figure 20, it was begun from 60°. The middle ordi- nate between 50° and 60°, measures — .02”, this is laid off downward from a line N-N drawn below the tangential crank effort diagram. The middle ordinate between 60° and 70°, is = + .07”, therefore, on ordinate 70° from the line N-N, + .05” is laid off upward. By proceeding in this manner the points of the velocity curve may be found at each advancing 10° of the crank angle. The ordinates are usually shown in a certain scale, since through the addition high values are obtained. In Figure 20 the ordinates are shown % of the found values. Care is essential, otherwise the end ordinates of the found curve will be of different heights, which will indicate an error made in addition, and it will be necessary to go over again. The highest and lowest points of the curve will correspond to the maximum and minimum velocities, and they will always be at such points in the diagram where the greatest excess area in the tangential crank efforts diagram will have its end points. The excess areas in the velocity diagrams are found in a similar manner as was previously explained for the tangential crank effort diagrams. The safety factor for the displacement of the cycles may be ap- proximately found by means of the following formula: = (W R?+W,Ri2) X NA H.P. x Fr x F, X 416 1000,000,000’ where: Y = number of cycles per revolution, W = weight of flywheel, W, = weight of rotor, R = radius of gyration of flywheel, Ri = radius of gyration of rotor, N —r.p.m. H.P. = brake horsepower matimum, (largest excess area in tangential crank effort a diagram) F (total area of average work) f (largest excess area in velocity diagram) F Fr = y= = (total area of average velocity) -46- This formula gives the number of cycles, which the uniformity of speed attained by the combined weights of the flywheel and generator rotor (if direct connected) allows, by considering the permissible dis- placement only, for which an average value in the derivation of the formula was taken. The coefficient indicating the relation between the number of cycles found by means of the formula and the actual ones, will be the safety factor for the cycles, therefore, also for the displacement. Table 10 gives weights of flywheels and rotors for some sizes of engines built: TABLE 10. Tandem Tandem Tandem Size of engine ................ecceceeee eee ees 42’'x54 32x42” 24x32” Weight of flywheel, in pounds........... 200000 93000 37599 Diameter of flywheel, in feet ............ 23 18 13 Radius of gyration of flywheel, in feet.. 8.58 6.62 4.166 Capacity of generator, K.W. ............ 2000 1000 500 Weight of rotor, in pounds............... 63000 30000 23500 Radius of gyration of rotor, in feet...... 6.25 4.166 1.92 Fer the 32” x 42” engine the coefficients of the greatest excess areas divided by the total areas were found to be in the diagrams for the crank efforts and velocities, as given in Table 11: TABLE 11. CRANK EFFORT DIAGRAM VELOCITY DIAGRAM Twin Twin Tandem Tandem Tandem Tandem Blast furnace gas...... .2155 1929 2448 .2276 Producer gas........... .2013 .1674 .2896 .2285 Natural gas ............ .2027 1776 .3145 .2171 Substituting the values for the 32° x 42” engine in the formula for the weight of the fywheel and the number of cycles, the fluctua- tion degree and the number of cycles will be obtained for the various cases as given in Table 12. FLUCTUATION NO. OF CYCLES SAFETY FACTOR Twin Twin Twin Tandem Tandem Tandem Tandem Tandem Tandem Blast furnace gas... 1-306 1-248 318 280 22.7 20 Producer gas........ 1-302 1-262 268 295 19.15 20 Natural Gas ........ 1-284 1-232 227 275 16.2 19.65 Ais The safety factors for the deviation of the cycles were found by dividing the number of the cycles as obtained by means of the formula by the actual number, as given for the generator, i. e. 14 cycles per revolution. The maximum and minimum crank pin velocities may be ascer- tained as follows: The flywheel fluctuation may be expressed 5 — Vmae.—— Vmina V= average crank pin velocity. The equation can also be written: : “ “ noo COMPRESSION >- “ > «© INERTIA OF RECS aso Ss os +s LOSS OF FRICTION WO ~ NET RESULTANT PISTON PP 0 Ss0 RPM 200 io vi + a) ‘ Yes to ei, — Ate ARD STROKE —————— ST ee Soa = == z abe ey eck EN a, 3 ae soe serene 26 a Ww os okt ae ee ok ee Se a i A eS ae Oe ae Se eecis es = 2 es tt fo eS he ee aed ee 8 ree paasee Seaview: eae 2 eee ces 2 N SAN catpetes recs = 2 os es, Ot ae a BAe) £ 7A $ we 7AR BM. FAQ 39 --57 jaa! Meo 130 140 owe fits 40) \— RESULTANT Piston PREts (CURIE 'B £1938) Zr PRESS Dot To INERTIA OF RECIPR PARTS D- NET RESULT PISTON Pass No sbo CUWART STROKE _—- = bp & © tb & & Wh yo Teo ly ub aT, bn OVIWARD STROKE, —————+#=> —- ———-_ ~rwt ARD OT ROKE, —- —_ a en \-RESULT. Piston PRESS. (CORTES FIG 58) Z- PRES. DUE TO INERTIA oF RECIFR. PARTS D-NET RESULT. Priston PRESS. eoREM ‘ FAGAN. -- 58 - Figs. NS / NY / Th / NN IN we PL! LAH LE VPN | TANG! CRE rrof we soft LJ 4 BT EhCe yds. Aw 4 _ 66 Ti beeen bas LAs =. p49 TN CEN Ny : / Pa TR AT N a |_-4 | RAN a =f) Ay ratte nen dd ole Ney NL Fig.4z. “h LJ | YH T 4 N 1 Vi) NY N eel aiecirdalelstecte ie be Naot ais hwilte aantis ¢ ankaplrange |” hdr RRONT TAN dar ren = oe Nap a hal ial ™~ S Undeuay wage p+ Ns RN Trethea ys) od - Phe my | | 74 Fg. 48. TNT : 7 AL Lae | 4 io ML wetad anda abalone] | > bh + wh PW bo to te we mie he ot to be te ty bo to be me be mo ho ee ke te od B-A-T eeu Ana ete ne, ie eee — SOBER TANGENTIALCR EFFORT. avian TAgRan Mani Aas a AME K Rag iuie AES 59 i sadgest eatesy’ rts — hiss go peal — uf f Fs] WA NZ pS SA i TA [e le LA Sr wel lal dhalacbele desde darts SS] Sh MAT WhO Tada Ch Ep roge NY A ‘ RPM a ha | “| passant ost pak ee is | |= A 2) AR i \ ARN AN HG.4S / si A port Pe 7 VA AT NEAT TE NEAT TT ENS) 1). | \ hl a Toth AREA = ba og |_| Vi ( At! 1 é | ‘ raga CR CEP Luh ro Ras Pier PRens lhssl TAR ' ; ; es | » |wepral ov gucda wdacs, Rojo | = Hex OVC TANG CW EPR OATy wa ee eee te a ee les eee la ee RA ee aaah a sr a Zz zp DT paw Ana stapne, [SSS EES ae EON — —— + _— _—- QR UANARNDLAL REET ONS AG as, amr DIAGRAMS FOR MAIN BEARING AND CROSSHEAD GUIDE PRESSURES. Figures 46 and 47 show diagrams for main bearing pressures, for inner and outer strokes. The radial lines indicate the direction of the pressures acting and the points of the curves show the variation, during one revolution. As it may be seen there is no pressure acting upward, indicating that no pressures will act on the bearing cap; they will be confined only as acting on the lower shell. The different points of the curves for each advancing 10° of crank circle were found in the following way: The pressures acting on the main bearing are composed of, first, the constant pressure due to weights of the crank shaft and fly-wheel, and second, the variable portion of the piston pressure. The weight of fly-wheel is = 200,000 lbs. Weight of shaft with cranks = 97300 Ibs. The total weights are then = 297300 lbs. This weight taken per square inch of effective piston area of the gas a = 210 lbs. or on each bearing 105 lbs. The piston pressure for the normal speed of the engine 75 revolu- tions, Fig. 42 was used. The outlines of the connecting rod and crank were then drawn. The main bearing pressure, for instance, at a crank angle 70° was found by constructing the pressure pazallelo- gram, Fig. 49. P. is the piston pressure at the point 70° in diagram Fig. 39, forward stroke, as acting in the direction of the connecting rod inclination; W is the constant weight pressure downward; and Px is the resultant pressure acting on the main bearing, giving at the same time the direction of action. The results were then reconstructed in a larger scale in Fig. 46. By repeating the foregoing proceeding the different values of the crank angle were fonnd. The maximum pressure is thus found to be 133 lbs. per sq. in., of gas piston area at 120° crank angle in forward stroke, and 153 lbs. at 20° in backward stroke. Fig. 48 shows the pressure on the crosshead guide for inner and —61-— cylinder = Zs tao m4 s70 OVIWART STROKE FAGQAb. 169 AD 140 20 159 Ot / es / 'b | Staves 30 =i of 180 wT’ 4 = fe See L 13 NWARD STROKE FAGAN, WAIN BEARING PRESSURES 62 outer strokes. The values of the curves at any crank angle were found by using the piston pressure diagram, Fig. 39, and constructing the parallelogram of pressures at the corresponding points of the piston stroke. From Fig. 48, the construction of the points of the curves may be easily understood. Mathematically the crosshead guide press- ure may be found by using the data given in table 7, and as previously explained. ‘oe OVA aus & Tana Jo HIM = Te BNWT NIWW NO 2 SSW" WASIA WS BULAN OL ENT TOU "NNOD NI“SSBYd= Zh ‘~ "By OV SARNSS Ada AMIN TWANSsoyd Bowls TAYMINd ° oe ———3r0"25 Tuyen —_ DIAGRAMS FOR SINGLE ACTING GAS ENGINES. Figures 50, 51, and 52 show piston pressure, tangential crank effort and velocity diagrams for a single acting, single cylinder gas engine, and Figs. 53, 54, and 55 show similar diagrams for a twin single acting engine. Figures 56 to 62 are diagrams of slide pressure on cylinder and pressures on main bearing. The number of revolutions per minute of these engines is 180, and the weight of the unbalanced portion of the reciprocating parts is =3.1 lbs. per sq. in. of effective piston area. Be DIAGRAMS FOR SINGLE ACTING GAS ENGINES. Figures 50, 51, and 52 show piston pressure, tangential crank effort and velocity diagrams for a single acting, single cylinder gas engine, and Figs. 53, 54, and 55 show similar diagrams for a twin single acting engine. Figures 56 to 62 are diagrams of slide pressure on cylinder and pressures on main bearing. The number of revolutions per minute of these engines is 180, and the weight of the unbalanced portion of the reciprocating parts is 3.1 lbs. per sq. in. of effective piston area. ee DERIVATION OF FORMULA FOR FLYWHEEL AND SAFETY DEGREE FOR CYCLE DEVIATION FOR ENGINES DIRECT CONNECTED TO A. C. GENERATORS. FLYWHEEL FORMULA. As it was previously explained the largest excess area in the tan- gential crank effort diagram represents the amount of work which the fly-wheel has to develop by its inertia or, by its kinetic energy. This 2 ax 2g W =the mass of the fly-wheel, supposed to be concentrated on its radius of gyration; force can be expressed by the equation: , where: V = circumferential velocity of fly-wheel on radius of gyration; g = 32.1 acceleration due to gravity. Since the kinetic energy as measured by the excess area, repre- sents a certain portion of the work of the engine, the following equa- tion may be written: WV? _ HP X 33000 De N X Fr where :...........0... (1) HP = brake horsepower. N = 1.p.m. of engine. f a ia = proportion between the excess area f and the total area F in the tangential crank effort diagram. The circumferential velocity V is not uniform during a revolution, therefore, the equation for the live force has to be expressed : ae. X (V max. # Vmin.)*, or, since (V max.— V min. )= 2g (V max. + V min.) X (V max. — V min.), we may write xX (V max. + V. min.) (V max. — V min.) V _ (Vv max. + V. min), or 2V = Vmax. + Vmin. (a) -6G- The average velocity V is= The fluctuation of a fly-wheel is expressed by the equation: 1 V max.— V. min., or 1 = ee —— *< V=Vmax.—Vmin. (b) The average circumferential velocity V HRN (c) Substituting the values found in the equations a, b, and c, in the equation 1, we obtain: WH sayy og a se eet a dy 2g 8 g 8 gi 8 2 XR? Xa? X N?_HP X 33000, 602 a N 2 Solving the above equation for WR’, we obtain: _ HPX8xFrx 33000 x 3600 x 32 2 = 8 WR NX4X3 14 96,780,000 y Eo The coefficient Fy = a> will always be a fraction, and using this value in per cent, i. e., multiplying it with 100, and making the numerical coefficient a round figure, the final form of tke formula will be obtained : HP X 8X Fr NE The fluctuation of the fly-wheel may be easily found from the WR? = 1,000,000 x above formula: 5— WR? < N3 1000,000 X HP X Fr When a generator is direct connected to the engine, the product WR? has to contain also the weight of the generator rotor, i. e. WR? = (WR? + W.R.?) 7 FORMULA FOR SAFETY DEGREE OF CYCLE DEVIATION. It was previously explained that the cycle deviation depends upon the variation of the crank pin velocity (hence also upon the fly-wheel fluctuation) and that the diagram for the deviation is found by inte- gration of the velocity curve. In the velocity diagram, (sce any F’g.), the length of the latter represents the time during which the pcricd of variation of the velccity is performed, i. e., one revolution. The ordi- nates of the curve (velocity) show its variation. We may therefore, write the relation: (V max.—V min.) t: fret. =F, : f, , where: V = velocity ; t = time for one revolution ; F, = total area in velocity diagram ; ce ae ae f; = excess arca : From the above proportion follows : fr= (V. max. — V. min.) t x seen fy For V max. — V min. we may write: V max. — V min =4V and for t = x, where: = fluctuation of flywheel ; N = R. P. M. of engine. Substituting the found values for V max. — V min., and 6 and v writing for V ee in the equation 1, we obtain; ae i te Fo PRR Hee. Svasv x te= 60 Mong og ee, hs Svae=2 x2 x Rx® XFL; or, since 2x RX = 360° 1 POPE cour osscreaeentontane (2) 8 fi From the velocity diagram we may approximately, use the pro- --68-- poron:( vie : f vat = 360° : 2* R 7, from which we obtain: Soi ie x 360 ee : (fu): =" Te ae = Jvatg ee Bete gens _ ( (i oS io 3 , ee 10 P, when P = number of poles, or, by using the number of alternations 10x N A W per min., the average displacement given as allowable, will be = A = 10 x N ; of ° 120 Xw’ which value may be expressed as equal ( wet) hence, ny 1 E pp kk equation for 6 = 1 360 Xe ccc cecceceee ceseeeeteteeteeees (3) By assuming that the average allowable displacement is /\ = ’ per minute A, A = , or, when using the number of cycles , from which follows when solving the When deriving the fly-wheel formula it was found: HP x 6 f 33000 « 3600 x 32 ' WR? = N * F x 43.18 from which follows: _ WRX 4X%3.142XN3_ F = TP x 33000 X 32 X 3600 °° f° and we may write the equation: 12¥ x 360 © f; _ WR?X 4X 3.14 « N3 x F N F, HP X 33000 x 32 x 3600 f Solving this equation for WR? , we obtain: 12 X 360 J fh y, HP _X 33000 x 3600 © 32 f Ni Fy 4x 3.14 F and by making a round ia of the numerical coefficient, we will obtain : WR* = NP ‘ Sa x ¥ X 416000000 000 --69-- WR? = From which equation the number of cycles can then be easily found : _. WR? X N4 Fr Yap ag KID When a generator is direct connected to the engine, the value WR” has to contain also the weight of the generator rotor, i. e., WR? == WR? + W.R? --70--, “HEADQUARTERS GAS ENGINE BOOKS” WE carry in stock or handle, all techni- cal books, and particularly those on the following subjects: Gas Engines Gasoline Engines Automobiles Motor Boats Aeronautics Ignition Batteries Etc., Etc. When you need any technical book we can fill your order at published price. Write for suggestions on any line you wish. THE GAS ENGINE PUBLISHING CO. CINCINNATI, 0. The Gas Engine MONTHLY Stationary, Automobile, Marine, Aeronautic, Farm 88 Pages Hlustrated Care and Operation of Engines. Fuels—Gas, Gasoline, Kerosene, Crude Oils. Ignition—Batteries, Plugs, Coils, Magnetos. Com- pression, Lubrication, Farm Tractors, Portables, Farm Light- ing Plants, Commercial Power Plants, Gas Producers. Answers to Inquiries of Readers. New Products of the Manufacturers. Marine, Aero- nautic and Automobile Motors. $1.00 Per Year Sample Free = THE GAS ENGINE PUBLISHING CO. CINCINNATI, 0. Missing Page Pinta a Seana seat eo n cere Dette SSR OSE obit Sa aeE at 2 it ue 3 iit pases rt Ht = Batt eae ES aot Hy ui Taessag ae ieee Beh ie