Scholars' Mine
Masters Theses Student Theses and Dissertations
1965
Optimization of flywheel design for internal combustion engines
Dady Jal Patel
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Recommended Citation Patel, Dady Jal, "Optimization of flywheel design for internal combustion engines" (1965). Masters Theses. 5715. https://scholarsmine.mst.edu/masters_theses/5715
This thesis is brought to you by Scholars' Mine, a service of the Missouri S&T Library and Learning Resources. This work is protected by U. S. Copyright Law. Unauthorized use including reproduction for redistribution requires the permission of the copyright holder. For more information, please contact [email protected]. OPTIMIZATION OF FLYWHEEL DESIGN FOR INTERNAL COMBUSTION ENGINES
BY DADiY JAL PATEL
A THESIS
submitted to the faculty of the UNIVERSITY OF MISSOURI AT ROLLA
in partial fulfillment of the requirements for the
Degree of MASTER OF SCIENCE IN MECHANICAL ENGINEERING
Rolla, Missouri
Approved by ~ f.~ (Advisor) ~ ii
ABSTRACT
A new approach to the problem of flywheel design has been developed, thus eliminating some of the problems of conventional design. The term flywheel design, in this thesis, refers to the the determination of the flywheel iner tia (mass moment of inertia) only. Two versatile computer programs were set up to obtain the flywheel inertia and the turning moment diagram for a large variety of internal
combustion engines, and use was made of these programs to
optimize the inertia of the flywheel with respect to speed.
As an illustration the data of a 1960 Co~vair engine were
selected, and the results obtained from the computer pro
grams agreed well with the ones used in practice. iii
ACKNOWLEDGMENTS
The author wishes to thank Professor Charles L. Ed wards for the idea that led to this work. He desires to express his ·gratitude to Professor Kenneth E. Spencer.
Dr. Harry J. Sauer . and Professor Lyman L. Francis for their guidance and criticism during the conduction of this research.
A vote of thanks goes to Professor Herbert R. Alcorn
for his assistance in programming. iv
TABLE OF CONTENTS
TITLE PAGE • • • • • • • • • • • • • • • • • • • i ABSTRACT • • • • • • • • • • • • • • • • • • • • 11 ACKNOWLEDGEMENTS • • • • • • • • • • • • • • • • iii TABLE OF CONTENTS • • • • • • • • • • • • • • • iv
LIST OF FIGURES • • • • • • • • • • • • • • • • vi LIST OF TABLES • • • • • • • • • • • • • • • • • vii I. INTRODUCTION • • • • • • • • • • • • • • • 1 II. REVIEW OF LITERATURE • • • • • • • • • • • 3 III. DISCUSSION • • • • • • • • • • • • • • • 9 A. Determination of the Torque Crank Angle Dia gram of the Engine and the Evaluation of the Flywheel Inertia •••••••••••••• 9
l. Logic Used in Developing the Computer Pro- gram for Inertia Evaluation · • • • • • • 9
a. Evaluation of the gas pressure torque • • • • • • • • • • • • lO b. Evaluation of the torque due to the inertia forces and the resulting torque due to gas pressure and iner- tia forces combined • • • • • • • 16
c. Determination ot the combined torque diagrams for the various cylinder combinations • • • • • • • • 18 d. Determination of the area under the torque crank angle diagram • • • • • 20 e. Determination of the mean torque line • • • • • • • • • • • • • • • • 22 f. Evaluation of the roots, that is, the intersection ot the torque curve with the mean torque line • • 22 v
g. Determination of the maximum and minimum speed points on the torque crank angle diagram • • ••• • 23 h. Evaluation of the largest positive net area above and below the mean torque line • • • • • • 25
i. Calculation of the rotating inertia 25
2. Programs and Their Handling • • • • 25 a.. Computer program for the evaluation of the flywheel inertia • • • • 25
(l} Notations for the Read State- ment of Program No. l • • 37 ( 2} The sequence in which the an- swers will be obtained • • • 38
(3} Column headings for the Punched Cards obtained in the output • • • • • • • • • 38
·b. Calcomp Plotter Program for obtain- ing the torque crank angle diagram of the engine • • • • • • 39 Notations used in the Program No. 2 • • • • • • • • • • 42
B. Optimization of Flywheel Inertia with Res- pect to Speed • • • • • • • • • • • • 42 c. Illustrative Problem • • • • • • • • • • 44
IV. CONCLUSIONS AND RECOMMENDATIONS • • • • • • 48
v. BIBLIOGRAPHY • • • • • • • • • • • • • • 49 VI. APPENDIX • • • • • • ... • • • • • • • • • • 50 56 VII. VITA • • • • • • • • • • • • • • • vi
LIST OF FIGURES
Figure Page
1 Arbitrary torque crank angle diagram of an internal combustion engine • • ••••• • • • 4 2 Torque crank angle diagram of a six cylinder engine • • • • • • • • • • • • • • • • • 6
3 Pressure volume diagram for Otto cycle • • • • 10 4 Pressure volume diagram for Diesel engine • • 13 5 Resolution of the net force in the engine mechanism • • • • • • • • 15 6 Torque crank angle diagram as a result of gas pressure alone in a four stroke engine • • • • • 16 7 Torque crank angle diagram as a result of pressure alone in a two stroke engine 16
8 Torque crank angle diagram showing the combina tion of the gas pressure torque with the iner- tia torque • • • • • • • • • • • • • 18
9 Combined torque crank angle diagram for a four cylinder, in line engine • •. • • • • ••• 19
10 Combined torque crank angle diagram for a Vee engine with two banks of one cylinder each • 20
11 Area bounded by a function: : y = f(x) • • 21 12 Magnified view of a curve between a and b, show- ing the presence of a root • • • • • 22
13 Arbitrary torque crank angle diagram showing the positive and negative areas, above and below the mean torque line • • • • • • • • • • • • • 23
14 Torque at various intervals of the crank • • • • 26 15 Horsepower versus speed relation for a six cylinder, opposed piston engine • • • • • • • • 43
16 Turning moment diagram of a 1960 Corvair engine. 55 vii
LIST OF TABLES
Table Page
1 Coefficient of speed fluctuation • • • • • • • • 8 2 Inertia variation with speed • • • • • • • • • • 47 l
I. INTRODUCTION
Flywheels have been known to man for ages, and have symbolized progress. One of their most outstanding appli cations has been in the field of internal combustion en- gines. Flywheels, although very simple by nature, have a very complicated design analysis. Each engine ~ requires an individual flywheel design and industries affiliated with the manufacture of internal combustion engines, find it a problem to make new designs for all their engines. The reason being the analysis is long, tedious, inaccurate and time consuming. Accuracy drops off, due to the choosing of large intervals in the torque crank angle analysis to ease repetitive calculation, slide rule application, and use of a planimeter or of graphical integration for area calculation. Besides "to err is human" and the designer is liable to commit mistakes at any stage in his calcula tions. Hence the usual procedure followed, is to take standard flywheels (specified by manufacturers) within the range of the approximate calculations of the designer, try them on the engine, and by trial and error come up with one which gives the smoothest performance of the engine.
It is evident, from above, that most engines are fitted with flywheels not having the best design. As nearly all the parts of a present day internal combustion engine have been optimized, the author feels it necessary to maintain this high degree of perfection in the flywheel, Thus, a 2
theoretical investigation was conducted and a method de veloped, which enables industries to have flywheels opti mized for their engines without appreciable la~or . or cost.
This investigation has been split into two parts, namely the determination of the inertia of the flywheel and the optimization of this inertia with respect to speed. Importance has been laid on the evaluation of the flywh el inertia, as it constitutes a major factor in fly wheel design nd it is this part of the analysis that of fers a blockade to moat designers. The author has made use of an IBM 1620 computer which offers a quick and reli able solution and has programming error detection facili ti s. Th program yields results tor either a two stroke or four stroke, diesel or gas, single cylinder or multi cylinder, radial or vee engine. Designers interested in obtaining the torque crank angle diagram for further appli cation can make uae of the Calcomp plotter program.
Th 1 cond part, which deals with the optimization of flyvh el inertia, invokes the knowledge that the horsepow er of an engin varies with speed and thus the inertia varie with speed. An inveatigation was made on an Cor vair engine vhoae apeed ranged from 1200 rpm to 3800 rpm, and the result eatabliahed that the optimum inertia oocura at idlins speed. 3
II. REVIEW OF LITERATURE
A flywheel as defined by Professor Joseph Shigley, is a mechanical filtering element in a circuit through which power is flowing. Energy is supplied by the engine at a variable rate and taken from the engine at a constant rate. This causes the shaft to vary in speed during one cycle. The flywheel helps to reduce this speed fluctua- tion and even out the ripples of the torque crank angle diagram by absorbing energy when it is delivered (by the engine) at a rate in excess of the load requirements and by releasing it when delivered by the engine at a rate less than load requirements. Thus, it acts as a smoothing or equalizing element in any mechanical power-transmission circuit which has a back and forth flow of energy.
Most books on theory of machines and dynamics of ma- chinery have dealt with the topic of flywheel inertia evaluation but very few have shown the mathematical ap- proach to this problem. Lord Kelvin has stated:
"I often say that when you can measure what you are speaking about and express it in num bers you know something about it; but when you cannot express it in numbers, your know ledge is of a meager and unsatisfactory kind: it may be the beginning or knowledge, but you have scar o~lx in your thoughts advanced to the stage of science, whatever the matter be."
The literature reviewed here has been of the kind
Lord Kelvin would greatly approve or, namely, evaluation 4
from first principles.
It has been known that I.e. engines deliver a very
"lumpy" t orque, th e reason being that there is only one power stroke for every cycle performed. Let Figure 1 represent an arbitrary torque crank angle diagram.
one c.ycl~
cro..nk o..n5\e.
Figure 1. Arbitrary torque crank angle diagram of an internal combustion engine.
The fundamental relation where the axis of rotation is fixed in inertial space is
(Euler's Equation) where T.Je) is the net accelerating torque applied to the shaft, I is the combined system inertia about its axis of rotation {which also is the principal axis), w is the angular acceleration about the axis of rotation. 5
Work done by the engine torque during one cycle is 4> /T(e)de =:"' Thus, the mean torque, Tm• vill be /~T (e) de T = o ______P\ ¢ ' ell vhere /T(e)Je physically represents the area under the 0 torque curve. A hypothetical engine delivering a steady torque Tm vould run at a constant angular speed ~. If the torque produced by the engine was larger than Tm• accelera- tion of the shaft vould result, and if less than Tm 1 re- tardation vould occur. Reconsidering Euler's Equation, the net accelerating torque is I~. T.._(e) ::. T(e)- T"" = I~. Work done by the accelerating torque in the limits 9~ to G~ro /(T(e) --r"')c!.e Q._ Since w = a.nd.. ' 6 <.lb (.)" . . de.;,. wdt I/ w c\ ~ ::: ~ I. ( w ~ - w~) w d.t Ij .... w... 6" Thus I ( i.. .t ) f(T(e)-T,...)de _ -2 I w.- Wo. • e... The maximum work done by the accelerating torque occurs between the minimum angular speed and maximum angular speed. This means that the greatest positive area about the mean torque line, in the torque crank angle diagram, occurs between Wmin and (..)max• In Figure 1, if the mini- mum speed occurs at S and maximum speed at 9 , then the 1 2 shaded area, A , represents the maximum work available , to 1 be absorbed by the rotating inertias, and is e qu a l to 1 I ( w ~ - wl.. . ) A,= j (T ce)- T"') de = 2 11"\0..)l f'f\tr\ • In case of a six cylinder engine the torque diagram resembles the one shown in Figure 2· Figure 2. Torque crank angle diagram of a six cylinder engine. 7 In this case maximum positive area about the mean torque line is between 9 and 8 and is equal to A + A + A + 6 ,13 6 7 8 A + A + A + A where A , A , and A are negative 9 10 11 12 7 9 11 areas. Maximum work absorbed by rotating inertias is The term rotating inertias covers the inertia of the fly- wheel, gears, pulleys, crankshaft, and all other rotating components of the load. Applying the general form, where in the work done by the accelerating torque in bringing \ I _ I ( 2 speed from Wmin to wmax is w 'J) - 2. I tv IM.~ - <..,)"'.'"' ) W = i._ I ( w - w )(w + ) - I c.-f w. , :1) 2 ...... "'i'fl "' .. ,. w...... - w where c+ • coefficient of spee·d fluctuation, (table 1) = Wmax - Wmin • w UJ = angular speed of a hypothetical engine deli vering a steady torque, Tm, Wmin. For low coefficient of speed fluctuations (1)* 'W'D I.- c wa + could also be call~d nominal speed of the engine and be represented by (550 x Horse Power/Mean Torque). *Numbers in parentheses are references listed in the Bibliography. 6 TABLE I* Coefficient of Speed Fluctuation Type Coeffi cient of of Equipment Fluctuation, Cf Crushing machiner~ 0.200 Electrical machinery 0.003 Electrical M/c direct driven 0.002 Engines with belt transmission 0.030 Flour ~illing machinery 0.020 Gear wheel transmission 0.020 Machine tools 0.030 Pumping machinery 0.03 - 0.05 Spinning machinery O.Ol - 0.02 Textile machinery 0.025 Paper making machinery 0.025 *Taken from Reference 2. 9 III. DISCUSSION A. Determination of the Torque Crank An gle Di agr a m of t he En gine and the Evaluation of the Flywheel Inertia. l. Logic Used in Developing the Computer Prog r am for Inertia Evaluation. The computer program has been set to give the magnitude of the rotating inertia from the follow- ing input data: Diesel cycle or Otto cycle, two stroke en- gine , or four stroke engine, horsepower, speed, diameter of cylinder, stroke, number of cylinders, mechanical efficien- cy, expansion index and comparison index, card factor, com- pression ratio, radius of crank, length of connecting rod, cut off ratio in diesel-cycle, weight of reciprocating parts, interval of crank chosen, and coefficient of flue- tuation for speed. The above data is partly fixed by the load specifications and the rest from design handbooks. Steps to be . followed in the building up of the pro- gram to determine the rotating inertia are: a. Evaluation of the gas pressure torque. Evaluation of torque due to the inertia forces and the resulting torque due to gas pressure and inertia forces combined. c. Determination of the combined torque diagram for the various cylinder combinations. Determination of the area under the torque crank d. angle diagram. e. Determination of the mean torque line. Evaluation of the roots, that is, the intersec f. tion of tbe torque curve with tbe mean torque line. 10 g. Determination of the maximum and minimum speed points on the torque crank angle diagram. h. Evaluation of the largest positive area about the mean torque line. i. Calculation of the rotating inertia. Torque on the shaft is due to ga.s pressure and inertia force. a. Evaluation of the gas pressure torque. The first step in the evaluation of gas pressure force, is the construction of the ~-V diagram. When the engine is in the design stage, it is necessary to estimate the indicator diagram from theoretical considerations. Consider the Otto Cycle, as shown in Figure 3: t 2 1 Volu.Me.. Figure 3. Pressure volume diagram for Otto Cycle. · d compression curve is PVK Gas law for both expans~on an = constant·. The relation between horsepower and dimension 11 of the engine is given by BHP- '3'3,000 )( 12. J where BHP = brake horsepower per cylinder, pb = brake mean effective pressure in psi, s = length of stroke in . inches, a = piston area in square inches, N :: =number of working strokes per minute. Knowing BHP, s, a, and N from the data, Pb can be calcu lated. If the indicated mean effective pressure is Pi' then• = pb Mechanical Efficiency • From tpis the indicated mean effective pressure can be calculated, knowing the mechanical efficiency. Lewitt (3) has obtained the results for tlie indicated mean effective pressure for Otto Cycle and is as follows: Work done Indicated Mean Effective Pressure = Length of the Diagram -or • = I (K -l)(c.- i) where pi = indicated mean effective pressure, p a pressure at the end of the expansion stroke, '+ p =atmospheric pressure (=14.7 psia), 1 K. = index of gas law, c = compression factor, 12 fc = card factor. Pressure at the end of the expansion stroke is calculated from this equation. Next to be determined is the clear- ance volume. In Figure 3 v volume a.t point l, 1 = v • volume at point 2. 2 s = stroke in inches, ( ~ a = area of cylinder. Therefore, piston displacement volume is , V - V ~ Sa. 1 2 The displacement volume can also be written as V1 (c-l) c ' where c • compression ratio ( V '. /V ) • 1 2 ( V1 - Ve) c S · o.. ·c. Hence, v1 = (c - 1) - (c.- 1) and, \fa=: v1 - S·a.. Therefore, the clearance volume, Vc• can be evaluated as v1 and v2 are known. vc = Utilizing the general l aw, PVK = constant K 'P = ? (100 ;- Vc) (for- C!)(ra.t\s;ol'l .,.tro\(c.) :x:. e 1! X + Vc 13 where pxc = pressure during compression stroke for piston position, ny Pxe = pressure during expansion stroke for any piston position, X = percentage of piston travel from the head end of the stroke. Thus pressure at every stage of piston travel has been de- termined for an Otto Cycle. A similar procedure is followed for the Diesel Cycle, shown in Figure 4. t 4 1 Figure 4. Pressure volume diagram for Piesel engine. In the case of the diesel cycle everything remains the same as in the Otto Cycle, except the equation for indicated mean effective pressure: l4 ?~ [ k (~o-1) - CL-\ ~: -1)] (k-1)(c.-1) ' where all the abbreviations have the same meaning as in the Otto Cycle (3) • ~o stands for cut-off ratio in Diesel Cycle and .K for the indeoc, I From the general law pyK = constant l.•t .t.~ 0 ll O'W'S that '· -(~:·f where X is the percentage of piston travel from the head end, and Yxe and Pxe are the volume and pressure at any stage of the expansion stroke· I~ ?. ( 100 + Vc ) 1 X + Vc ' where V and P are the volume and pressure at any stage XC XC of compression stroke. Thus, the pressures at the piston end are calculated for various percentages of the stroke as the piston moves. The next step in the case of Diesel and Otto Cycles is to convert this pressure into force and then transfer 15 it to the crank end of the connecting rod, as shown i.n Figure 5. q = force along the connecting rod axis Pa =pressure on the piston t 0 = tangential pressure at the crank end e a angle turned by the crank from ~op dead center position r = radius of crank 1 0 = length of the connecting rod Figure 5. Resolution of the net force in the engine mechanism. Lichty (4) gives the tangential pressure at the crank end as ' hence. the torque on the shaft due to gas pressure will be ... T ~ to· r ·a.. , where a is the area of the piston. It is interesting to note that this torque acts only over 180°, and has been shown in Figures 6 and ~ 7 ~ . 16 t Figure 6. Torque crank - angle diagram as a result of gas pressure alone in a four stroke engine. ( I tl ~ rr~------~~~~L---~~------~ 0 t- Figure 7. Torque crank angle diagram as a result of gas pressure alone in a two stroke engine. b. Evaluation of the torque due to the inertia forces and the resulting torque due to gas pressure and inertia forces combined. The reciprocating masses in an engine are chiefly responsible for setting · ~p these forces. 17 Lichty ( 4) has dealt· with this problem in great detail, and only the result is shown here: where F = force due to inertia of reciprocating parts at piston end, G = weight of reciprocating parts in lbs • n = speed in rpm, r = radius of crank, 1 0 = length of connecting rod, = angle turned by the crank from the top dead center position. This force is transferred to the · crank end of the connect- ing rod by the same· procedure followed in gas pressure tor- ces and yields, • Therefore, the torque on the crank due to inertia · torce will be: It should be noted that this torque, unlike the gas pres- sure torque, acts over the whole crank angle range of 720° in a four stroke engine and over 360° in a two stroke en gine. Its direction is as shown in Figure 8. 18 For four stroke engine: _, / \ I \ I \ I ' I \r~CI." rre,<»u.re tor,~e. t I \ f f r-0 Figure 8. Torque crank angle diagram showing the combination of the gas pressure torque with the inertia torque. The resultant torque on the shaft is obtained by combin- ing the gas pressure torque with the inertia torque for one cylinder. Nearly all engines have more than one cy- linder and the torque crank angle diagram is a combination of the torques of the different cylinders. c. Determination of the combined torque diagrams for the various cylinder combinations. Usual cylinder com- binations are: ( l) Single cylinder. ( 2) Multi cylinder in line. ( 3) v - type with two banks of one cylinder each. '4) v - type with two banks ot more than one cylin- der in each line. .. 19 ( 5) Opposed Piston. (6) Radial. In order to obtain the torque crank angle diagram for a multicylinder engine, all that is required is the torque crank angle diagram for one cylinder of that engine. Know ing the firing delay between cylinders, the torque crank angle diagram of a multicylinder engine can be obtained. This can be best understood with specific examples. (1) Consider a four cylinder, in-line engine: The way in which the cranks are arranged for perfect balancing is of no importance in this problem. What one should look for is the number of degrees of' crank angle between firing of two cylinders. Suppose the next cylinder fires after 180°, then the final torque crank angle diagram is obtained by superimposing the torque crank angle diagram after every 180° (Figure 9). Figure 9. Combined torque crank angle diagram for a tour cylinder, in-line engine. 20 ( 2 ) Consider a Vee-engine w~'th two banks of one cy- linder each, where the angle of th e vee is 90°. The numbe r of degrees between the firing of two cylinders is 450°, Hence, in this case the torque crank angle diagram for one cylinder repeats itself after· 450°, as shown in Figure 10. Figure 10, Combined torque crank angle diagram for a Vee-engine with two banks of one cy linder each. (3) Radial engine with five cylinders. The cylin- der next in the firing order will most probably fire af- ter 720/5 degrees of the crank. Therefore, the torque crank angle diagram for one cylinder is superimposed at every 144°, This diagram is a bit too complicated to draw. d. Determination of area under the torque crank angle diagram. From the above procedure, the torque for a small interval of crank a~gle is determined, and by the 21 use of Simpson's Rule it is possible to find th r un- der the curve without actually drawing the curv or know- ing its equation. s. t ~mpson a Rule is explained with th help ot Fi ure ll. t h, h, hs h, hs h, \,, . I I I I I I I ._H ~-H --H 4~ H ...,.._ H ...,._H -+ Figure ll. Area bounded by a function: y a f(x). If y f(x) is the curve and h • h • h • h • h • h , and = 1 2 3 .. 5 6 h 7 are the ordinates at equal interval H. then the area under the curve is as follows: The assumption used in Simpson's Rule is that the number of intervals is even. Thus the general case with N inter- vals would be: If the curve is plotted by a Calcomp plotter, then 22 one can find the area by using a · planimeter. This method is not recommended as the planimeter fails to give good accuracy due to slipping between the wheel and the paper. e. Determination of the mean torque. In case of a four stroke engine, divide the area obtained by 411 radians and in case of a two stroke engine, by 2 '\'\radians. The result obtained is the mean torque. f. Evaluation of the roots, that is, the inter- section points between the torque crank curve with the mean torque line. Scarborough (5) shows a way of deter mining approximately the values of the roots. t ,a.. f<~): X of a curve between a and Figure 12. Magnified view b, showing the presence of a root. the torque curve and In Figure 12. let y = f(x) be be the mean torque line. the horizontal line, the x-axis, any interval, then the If the ordinate changes signs in 23 root occurs in that interval, and in this mann r t h v r - ious roots are determined. g. Determination of the maximum and mi n imum sp ed points on the torque crank angle diagram. Th r e on f or fin~ing the exact position of the maximum ape d h b n dealt with already under the heading "Review of Liter - ture". What is now dealt with, is the method and the systematic procedure to be followed. A combined torque crank angle diagram, as shown in Figure 13, can now be constructed showing the mean torque line and the roots at the intersection of these two curves. Arbitrary torque crank angle diagram show Figure l3. ing the positive and negative areas, above and below the mean torque line. ~ represent the roots, •• '='12 A represent the areas above and • • ~~ 2 ·ne A2, Alt, AG' As' Alo' and Al2 below the mean torque l 1 • ~·· - 1 • 24 are termed positive areas, and A • A • A l 3 s• A7, Ag• nd Al l are termed negative areas. Simpson's Rule is made use of to calculate th r above and below the me a.n torque line. The:r tore, Al, A 2. • • • A12 are calculated. A systematic arithmetic method is utilized to deter mine the positions of the maximum and minimum speed. Let the speed at 9 1 represent a datum value (..)O. The speed at the end of the first loop will be c.l 1 • where c..\ > w 0 be- cause the area of the loop A2 . is positive. Physically this means torque produced by the engine is in excess of the load torque. At the end of the second loop the speed will be w2 where w2 <. w1 since area A 3 is negative. Whether w2 is greater or less than ~ is determined by the algebraic 0 sum of the areas. If A2 + A3 is positive, then w2 > t.)o~ Speed at the end of the third loop is 0 • where w > w2 as A is positive, but whether 3 3 It it is greater or less than w1 is determined by algebraic sum of the areas. If A2 + A 3 + A 4 > A2 then (.) 3 > w1 • If A + A + A is and if A2 + A3 + A4 ..(. A2 then w 3 ~ LJ 1 • 2 3 It This procedure is positive, then w > c.u and vice versa. 3 0 followed until W is obtained. ' 12 The sum of the areas, of all the loops, is zero, and can be justified from the definition of the mean torque Of the alaebraic sum , of all the line. The maximum value ·, o r eference point, gives the areas, from some datum or position of the maximum speed and the minimum v&lu 1v the minimum speed. h. Evaluation of the largest positiv above and below the mean torque line. The larg st po itiv net area between the minimum and maximum speed is the l • braic sum of the areas above and below the mean torqu line. i. Calculation of the rotating inertia. The e valuation of inertia has been dealt with in the "Review of Literature" section, Eq. l, and at thia atage only the introduction of units is illustrated. I - _e. C.f.W where I is the rotating inertia in lb inches sec2 , Wn is the work done in lb inches, cf is the coefficient of fluctuation, w is the nominal speed in rad/aec. Since - to ~ , the above equation becomes w i& equal 60 e 6o w:D !. a. - 41iCfN ' where N is the speed in rpm. 2 • P rogra.ms and Their Handlin&• tor the evaluation ot· the a. Computer program Vas written tor an IBM 1620 flywheel inertia. The program 26 digital computer with 60,000 positions of core storage. Be fore knowing the details of the "Read Statement" and the sequence of the answers, it would be best to know how to fix the interval "PINT". "PINT" is the interval of the crank angle at which the torques, over one complete cycle of 720° are calculated (Figure 14), It is common practice to make "PINT" as small as possible, so as to obtain an accurate analysis of the torque crank angle diagram. "PINT" should be chosen in a manner that makes 720 /PINT an integer. In most cases it is recommended to use 1°, , 0 , Very small intervals of 0,001° or 0,01° cannot be handled by this program as use is made of all the memory storage space set for subscripts in the computer. i i ntervals of ' the crank. Figure 14. Torque at var ous rogram is very general, and Although . this compu t er P not limited to any particular engine, its application i~ t if used for a parti still its solution becomes inaccura e The problem that arises cular set of cylinder combinations. 27 in these cases is that 720° divided by the number of cylin ders will not give an integer. This affects the logic used in combining the torque diagrams of the various cylinders and gives a slightly distorted torque crank angle curve. Luckily. the cylinder combinations which offer difficulty in obtaining an accurate solution with this computer pro gram also offer problems in balancing and thus engines with these combinations are never manufactured. c STANDARDIZATION OF FLYWHEEL DESIGN FOR ANY VARIETY OF ENGINES c FEED IN STATEMENTS UPTO 2 TSUM(500),TAVG(500),MAD(500),TROOT(500),SIMS(500) ADD(500) E TXE TR T UM T " M M READ 1, HP,SPEED,E,DIA,CYLIN,EFF,EXA,CR,CARD,PA,PINT,CONN,RAD READ l,CUT,COMP,WT,ANGLE,RAT,COEF,STROK,ENGIN,TYPE I READ 2 NUMB NO 1 FORMAT (7F10.3) I 2 FORMAT (118,118) c GAS PRESSURE TORQUE ANALYSIS FOR OTTO TWOSTROKE FOURSTROKE I D=SPEED/E I \ AREA=3.14159*(DIA**2)/~.0 ! ! i BHP=HP /CYLIN · PB=t33000.*12.*BHP)/(D*AREA*STROK) \ ' . ' PI=PB/EFF i IF ·eNGIN 10 1 1 c FOR OTTO CYCLE ONLY UPTO 333 10 XYZ=tCR-1.)*Pl/(((CR**EXA)-CR)*CARD) 333 4= EXA- * XYZ +PA I GO . TO 12 c FOR 'DIESEL CYCLE UP TO 999 1 EA T=l.-EXA XYA=EXA*(CUT-1.)-(CR**EAST)*((CUT**EXA)-1.) 999 P3~(EXA-l.)*(CR~l~)/XYA)PI c EVALUATING CLEARENCE VOLUME (OTTO AND DIESEL) UPTO 666 I 12 VOL=STROK*AREA Vl=VOL*CR/(CR-1.) V2=V1-VOL 666 CV=V2*100./VOL 1\) c EVALUATING GAS PRESSURE TORQUE (X) 1=1 PINQ=180.+PINT r PIT=PINT .) 4 X_=p IT-PINT · ~~.- . .. I Q=X*3.l4/180. . y = D I A;:~ ( ( t- c 0 sF (Q ) ) + ( ( D I A ;< s I N F ( Q) '*"*- 2 ) I ( 4 • * cONN ) ) ) I 2 0 Y=Y:#" 100./S TRO K . . -- - --· --· ------IF·(ENGIN)20,22,22 - . · c FOR OTTO CYC LE ONLY UPTO 777 : ! 20 XYP=((lOO.+CV)/(Y+CV)) PXE=P4.*(XYP**EXA)- PA .. 117 PXC=PA*(XYP**COMP)-PA . GO, TO 19 . : c FOR· DIESEL CYCLE UPTO 888 _____ .._ ··- -- ___...... _ _ ------~-- - -. =--~- -· - -- ..:.: --: ~'--::···:::_---:--.:...·.--~: ~~_.:=."7'".:-:- ~ - -= - -- ___ ·.-.- l I 22_PXE=P3*(((CUT*CV}/(Y+CV})**EXA)-PA RR8 PXf.=~*--LL1100 +f.Vl /(Y+f.Vl l**COMPl- PA I · 19 Q=X*3.14/180. ' S=SQRTF(((CONN/RAD)**2) - (SINF(Q)**2)) R=SINF lQ_}*J l.+COSF {Q) /S l TXE(I)=PXE*R*RAD*AREA I . TXC(I)=PXC*R*RAD*AREA I=J_+_l ' .. PIT=PIT+PINT IF (PINQ-PIT)55,4,4 ' . 55 CONTINUE c ESTIMATION OF INERTIA TORQU E FOR BOTH OTTO AND DIESEL . . K=l P_IN..G_=_360. +PINT l6=720./(2.*PINT) .. PIT=PINT 5 X=PIT-PINT Q=X*3.14/180. ' COSQ=COSF(Q) SINQ-SINF(Q) S=COSQ+(RAD*COSF(2.*Q)/CONN) F=-(.0000284*WT*(SPEED**2l*RAD*S) EGG-SQRTF ( .llC_ONN/RADl**2l-( SIN0**2)) 1\) \() R=SINQ*(l.+COSQ/EGG) l(K)=F*R*RAO KL6-K+L6 l(KL6)=T(K) .. K=K+l PIT=PIT+PINT . ·,; · IF(PlNG-PlT)66,5,5 66 CONTINUE IF(TYPE)311,312,312 c COMBINING THE GAS TORQUE WITH INERTIA TORQUE FOR FOUR STROKE ONLY 312 JAPP=(180./PINT)+l. KAPP=(540./PINT)+l. .. : NAPP=(720./PINT)+l. 00 6 J=l,JAPP,l SUM(·J)=T(J)+TXECJ) I JNAPP=J+NAPP 6 SUM(JNAPP)=SUM(J) JAPO=JAPP+l : 00 7 J=JAPO,KAPP,l SUM(J)=T(J) : JNA£1>_= ..J. +...N.A£..e. 7 SU~(JNAPP)=SUM(J) KAPO=KAPP+l . ~----::.·---- ·-- __ ....;·. -- -"··-=·-- - · ---- ....::....:.:..==-:::..::;._-.-.~~ ------00 8 J=KAPO,NAPP,l _KE_ =N A...e_2_ +~- j_ 1 SUM(J)=T(J)-TXC(KP) JNAPP=J+NAPP 8 SUM(JNAPP)-cSUM(Jl JAPO=JAPP+l 00 25 J=JAPO,NAPP,l 25 TXE(JJ=~ooo . 00 26 J=l,KAPP,l 26 TXS(J)=.OOO ' I KAPO-KAPP+l w 0 DO 27 J ZRA PO,N APP,l KP=NAPP+l-J I 27 TXS(J)=-TXCCKP) I GO TO 313 c COMBINATION OF GAS AND IN ERTIA TOR QUES FOR TWO STRO KE UPTO 145 311 JAPP=(180./PINT)+l. NAPP-(360./PINT)+l. DO 141 J=1 7 JAPP,1 SUM(J)=T(J)+TXE(J) JNAPP-J+NAPP 141 SUM(JNAPP)=SUM(J) JAPO=JAPP+1 DO 142 J-JAPO,NAPP,l KP=NAPP+1-J SUM(J)=T(J)-TXC(KP) JNAPP=J+NAPP 142 SU M(JNAPP)=SUM(J) DO 143 J =JAPO,NAPP,1 143 TXE(J)-.000 I ' DO 144 J=1,JAPP,1 ' ' 144 TXS(J)=.OOO : DO 14!> J-JAPO,NAPP 7 1 KP=NAPP+1-J 145 TXS(J)=-TXC(KP) c COMB INATION OF THt FINAL TORQUE~ OF VARIOU~ CYLIND ER S : I 313 POT=O.O : DO 9 L=1 7 NAPP,1 I B-0.0 JA=ANGL E/P INT JU=(NAPP-JA)-1 NUT=NO/NU MB IF (N0-2)728 7 729,729 729 DO 3 J=2,NUT,l LJU=L+JU S=SUM(LJU) w B=B+S .... 3 J U= J U-J A· 7_ZB_ PUMB=NUMB TS UM(L ) =P UMB*(SUM {L)+B ) PRI NT 200 1 SP EED , PO T, TSUM {L),T( L), TXE (L), TXS( L) PU NCH 200~ SP EED ,P O T,T S U M (L), T ( L ),TX E ( L )1T X SJ L ) 9 POT=POT+PINT 200 FORMAT (Fl0.3,Fl0.3,Fl4 .3,Fl2.3,Fl2.3,Fl2.3) c AREA UNDER _IQR QU E CRANK DIAGRAM " t H=PINT*3.14159/180. BRAKl=O .0 . _BRAK2=0__ .0 NIPP=NAPP-1 00 30 J=2,NIPP,2 30 BRAKl=BRAKl+~SUMC J) 00 40 J=3,NIPP,2 40 BRAK2=BRAK2+TSUM(J) FOFC=TSUMlli+TSUMCNAPP) SIMP=(H/3.)*(FOFC+4.*BRAK1+2.*BRAK2) ' PRINT 150,SPEED,SIMP ~ .. 150 FORMAT JF__l0_._3.Fl8.3) c EVALUAJE MEAN TORQUE . . TMEAN=SIMP/(RAT*3.14159) PRINT 160,TMEAN 160 FORMAT (Fl8.3) c DETERMINE ROOTS OF THE TORQUE CURVE WITH MEAN TORQUE LINE 00 77 J= l__t_NAPP.l 11 TAVG(JJ=TSUM(J)-TMEAN JUNK=2 \ JACK-O IF (TAVG(l))60,101,65 60 ISTAT=l GO TO 70 65 ISTAT=2 10 CONTINUE ' w NIPP-NAPP-1 I\) ' 00 75 I=JUNK,NIPP,l GO TO (80,85),ISTAT 80 IF (TAVG(!))75,90,90 90 JACK=JACK+l MAO(JACK)=l TROOT(JACK)-TAVG(!) I ISTAT=2 GO TO 75 85 IF (TAVG(I) )95,95,75 · 95 JA<;:K=JACK+l .. TROOT(JACKl=TAVG(I) ... 4 ...... ~:~=::..:-..:; --======--- - ··=--~~.:--~~~-= ~-=-r...-:- - - - , MAO(JACK)=I ISTAT=l 75 CONTINUE GO TO 100 101 JACK=l I 103 MAD(JACK)=JACK TROOTCJACK)=TAVGCJACK) IFf TAVG l .JACK+ 1)) 60.102.65 I 102 JACK=JACK+1 I JUNK=JACK+l ' I GO TO 103 100 CONTINUE 00 104 J=1,JACK,l .. 104 PRINT lOS.MADfJ) I 105 FORMAT (120) c TO EVALUATE AREA ABOVE AND BELOW MEAN TORQUE LINE i I JIVE- JACK-1 ! \ DO 28 I=l,JIVE,l I MINT=MAD (I) I MLAS - HAD (I+ 1) H=PINT * 3.14159/180. BRAKl=O.O BRAK2=0 .0 w MEAT=MINT+l . w - ---- ME AL = 1·1 L AS- 1 DO 23 J=I~EA T__t__UEA L 2 23 BRAK1=BRAKl+TAVG(J) HA S T = ~11 NT + 2 0 Q___ 2 _'!: __J = 1·1AS _T , l:U: _~J__, 2 24 BRAK2=B RAK2+ TA VG (J) FO F E=TAVG(MINTl+TAVG(~LASl 2 8 S I MS ( I ) = (HI 3. ) ~ ' ( F 0 FE+ 4. * BR AK 1 + 2. ~'BRA K2 ) c FOR FIRST(SPLITlSMALL AREA MILL=MAD(1)-1 IF ( ~' I LL ) 18 0 t 18 0__, 1 9 0 190 H=PINT*3.14159/l80. BRAK1=0.0 BRAK2=0.0 00 31 J=2,MILL,2 31 BRAK1=BRAK1+TAVG(J) ' 00 32 J=3,~1ILL,2 32 BRAK2=BRAK2+TAVG(J) : : , FOF=TAVG(l)+TAVG(MILL+ll I SI~=(H/3.)*(FOF+4.*BRAK1+2.*BRAK2J GO, TO 188 180 CONTINUE c FOR LAST SPLIT AREA 188 MOLT = MAD( lACK) I H=PINT*3.14159/l80. BRAKl=O.O _ BEM~ 2 =Q_, o MOLE=HOLT+l NIPP=NAPP-1 MlH F-MOl T+? : DO 51 J=MOLE 1 NIPP,2 51 BRAKl=BRAKl+TAVG(J) DO 52 J=I-1U LE ,N I PP t_2 w ~ - 52 BRAK2=8RAK2+TAVG(J) FORT=TAVG( MOL T)+TA VG ( NAPP ) SIMT=( H/3.) *(FORT+4,*8RAK 1+2, *BRAK2 ) I SIMS(JACK) =SI M+S I MT c TO FIND WMAX AND WMIN PO=O.O P=O.O A00(1)=0.0 DO 91 N=1,JACK,l ADD(N+ll=SIMS(N)+ADD(N) IF (P-ADO(N+1))35,35,76 76 P=ADD(N+l) WMIN=N-1 35 CONTINUED IF(PO-ADD(N+l) )92 1 91 1 91 92 PO=AOD(N+l) WMAX=N-1 91 CONTINUE IF (WMAX-WMIN)81,82 7 83 83 SUMM=O.O ' NQ=HMIN NV=WMAX .. DO 14 J=NO,NV,l 14 SUMM=SIMS(Jl+SUMM : GO TO 17 81 SAM=O.O NQN-WMIN NQV=JACK ... DO 15 J=NQN,NQV,l .. l 15 SAM=SIMS(J)+SAM : SUN=O.O I NV=WMAX I DO 16 J-l.NV.l 16 SU~=SIMS(J)+SUN SUMM=SUN+SAM w V1 - -- 36 .. I : - ~ .... .,~ ('() g co• Nr-4 ~ .. u_ I- .. z:7 Wr<'l z t-4o- _,(. • ._u_ - 37 (l) Notations for the Read Statement of Program No. l. It is recommended that the user of this program follow this page thoroughly and under no circumstances can any one of these notations be violated. The whole program has been set up on this basis: Floating Point Constants. HP = Horse Power in HP units. SPEED = Speed in rpm or average number of explosions per minute in a hit and miss governing system. E = Constant which is 2. for 4 stroke Constant which is l. for 2 stroke Constant which is l. for hit and miss governing. DIA = Diameter of the cylinder in inches. CYLIN = Number of cylinders of an engine. EFF = Mechanical Efficiency. EXP = Expansion Index. CR = Compression Ratio. CARD = Card Factor (usually 0.9 if not stated). PA -· Atmospheric pressure in psi a. PINT = Crank interval at which the torque is to be determined in the torque crank angle diagram. CONN = Length of connecting rod in inches. RAD = Radius of crank in inches. CUT = Cut off ratio in Diesel cycle. COMP = Compression index. WT = Weight of reciprocating parts in lbs. ANGLE = Angle turned by the crank after which the next cylinder (in the firing order) is fired. 38 RAT = 2. for 2 stroke engines, 4. for 4 stroke engines. COEF = Coefficient of fluctuation. STROK = Stroke of the cylinder in inches. EN GIN = -3. for Otto cycle, +3. for Diesel cycle. TYPE = +3. for four stroke, -3. for two stroke. Fixed Point Constants. NUMB = Number of cylinders fired together. NO = Number of cylinders. (2) The sequence in which the answers will be obtained: l) The first series of answers will appear according to Format 200, which appears in the programs. SPEED, CRANK ANGLE, FINAL TORQUE• INERTIA TORQUE, GAS TORQUE DURING EX~ PANSION, GAS TORQUE DURING COMPRESSION. 2) Second type of answers will be according to Format 150. SPEED, AREA UNDER TORQUE CRANK DIAGRAM. 3) Third type ~ of answers will be according to Format 160. MEAN , TORQUE. 4) Fou~th ~· series of answers according to Format 104. CRANK ANGLES AT WHICH ROOTS OCCUR. 5) Fifth type of answers, according to Format 170. SPEED, WORK DONE BY ACCELERATED TORQUE, INERTIA. (3) Column headings for the Punched Cards ob- tained in the output. These punched cards have application 39 in Program 2, set up for plotting the torque crank angle dia gram on a Plotter. They constitute the input data to Pro gram No. 2 and appear according the Format 200, SPEED, CRANK ANGLE, FINAL TORQUE, INERTIA TORQUE, GAS TORQUE DURING EXPANSION, GAS TORQUE DURING COMPRESSION. b. Calcomp Plotter Program for obtaining the tor que crank angle diagram of the engine. The torque crank an gle diagram can be plotted by means of a Calcomp 566 Plotter in conjunction with a 1625 Plotter control unit and 1620 Central Processing unit, The input data for this program comes in the form of punch cards from Program No. 1. The FORMAT used to read these cards is given by Statement Num ber 100 and it enables the computer to read SPEED, CRANK ANGLE , and FINAL TORQUE only. A better understanding of this program is acheived by refering to the Calcomp Plotter Notes (6) and by viewing the p~otter unit itself. An important factor is the fixing of the overall size of the torque crank angle diagram. The maximum plotting width in the y- direction is 11 inches and the maximum plot ting length in the x-direction is 499.99 inches, as limited by the size of the machine. It would be absurd if the whole length was made use of, because instead of obtaining a smooth curve, a series of straight lines joining points would appear, Besides one might want to measure its area by means of a planimeter and the existence of such a big planimeter, is seriously doubted. The recommended length is 7.2 inches 40 in the x-direction and 10 inches in the y-direction, the reason being it gives convenient scaling factors. C PROGRAM TO PLOT TOR QUE-CRANK DIAGRAM PROGRAM NO 2 C NAME OF PROGRAMMER DIME NSION POT(NAPP) ,TSUM(NAPP) 1 READ 100 I SPEED I (POT (I) t 'l'SUM( I L I =l,NAJ'P) CALL PLOT (l,XMI N,XMAX,XL,XD,YMIN,YMAX,YL,YD) DO 10 I=1,NAPP 10 CALL PLOT. (90.POT(I CALL PLOT l99) CALL PLOT l90~~A~P 1 YMAX) CALL CHAR _(t •. 4 I 0 I SPEED) 102 FORMAT CfHSPI!!l(D__ =-.-F7.1) CALL PLO'll rfl ------______100 FORMAT (2Fl0.3,Fl4.3/(iOX~F~0.3,Fl4.3)) 20 CALL EXIT END ::- 1-' 42 Notations used in the Program No. 2. These values are to be substituted directly into the "CALL PLOT" State ment and other following statements: XHIN = 0.0. XMAX = 720. for four stroke, 360. for two stroke. XL = 14.4. XD = Increment of crank angle "PINT". YMIN = 0.0. YMAX = Maximum torque rounded off (explained a bit later). YL = 20. YD = Increment of torque in the y-direction, over a total length of 10 inches. Example, if YMAX was 20,000 one may choose YD as 500 so as to get 40 divisions over the total length of 10 inches. NAPP = (720/PINT) + l for four stroke, (360/PINT) ~ l for two stroke. YMAX is usually fixed by finding the maximum Torque un- der the FINAL TORQUE column in the "Read Statement''. If this torque is 18,800 then the rounded off YMAX must be taken as 20,000 as it helps scaling. B. Optimization of Flywheel Inertia with Respect to Speed. Most engines run at a variable speed and since the horse power varies with speed, it is hard to say at what particu- lar speed the optimum inertia of the flywheel will prevail. This could be further explained by reconsidering the equa- tion: As speed increases, horsepower increases and this evidently makes "A'! increase. At what rate "A" increases is not known and thus, a special investigation was carried out to evaluate the speed at which optimum inertia might occur. The enGine chosen for this evaluation was a. 6 cylinder, opposed piston, Corvair engine. Specifications (7): 3.375 inches bore, 2.6 inches stroke, 1.462 lb total weight of reciprocating parts, 8: l compression ratio, 1.3 as expansion and compres- sion index. Its HP versus Speed curve is shown in Figure 15. 'iSO 10 t 60 1- ol 3 60 0 0... ol "'~ 40 ::r0 30 ~0 0 'I\ 'PI"\ " ( hu..l\cir .. J.~) Figure 15. Horsepower versus speed relation for a. six cylinder, opposed piston engine. Torque crank angle diagrams were computed, for speeds ranging between 1,200 rpm and 3~800 rpm (at intervals of 200 rpm), by the logic formulated before. From this the rotating inertias were calculated at the various speeds 44 with the help of Computer Program No, 1, The results ob tained are shown in Table No, 2, From the results it may be concluded that the parameter " N" is the deciding factor and the parameter "A" h a s very little control over the results obtained, Thus, the idling speed in automobile engines, is the prime consideration for flywheel design, C, Illustra tive Problem It is desired to find the inertia and the turning mo ment diagram of a 1960 Corvair opposed piston engine, whose specifications are as £o11ows: 6 cylinder, horizontally opposed with overhead valves; air cooled; 3,375 in. bore; 2.90 in. stroke; 140 in. 3 piston displacement; compression ratio, 8:1; 80 bhp (max) at 4,400 rpm; torque, 125 lbft (max) at 2,400 rpm; numbering system (front of vehicle to rear) is 6, ~ 4, 2 for left bank, 5, 3, 1 for right bank; firing order 1, 4, 5, 2, 3, 6, weight of piston assembly 1.266 lb; con necting-rod length, 7.2 in.; weight of reciprocating parts per cylinder, 1,462 lb; six throw crankshaft with the throws arranged in pairs; members of each pair are 180° apart, and the pairs are 120° apart; idling speed, 1,200 rpm; horsepower at idling speed, 23 bhp; mechanical efficiency, .77; expan sion and compression index, 1.3; card factor, .9; coefficient of fluctuation, ,02, The symbols in the "Read Statement" of the Computer pro g ram 1 are assigned numerical numbers as follows: 45 HP = 23., SPEED= 1,200., E = 2.0, DIA = 3.375, CYLIN = 6 .0, EFF = .77, EXP = 1.3, CR = 8.0, CARD= .9, PA = 14.7, PINT= 5.0, CONN = 7.2, RAD = 1.3, CUT = 0.0, CO MP = 1.3, WT = 1.462, ANGLE = 120.0, RAT = 4., COEF = .02, STROK = 2.6, ENGIN = -3.0, TYPE = 3.0, NUMB = l, No = 6. The following are the numerical values that have been used for the Calcomp Plotter Program 2: XMIN = 0.0, XMAX = 720.0, XL= 14.4, XD = 5.0, YMIN = 0.0, YMAX = 4000.0, YL = ~ 20.0, YD = 200.0, NAPP = 145. The results obtained from the Computer Program, along with the turning moment diagram, have been grouped in the Appendix. From the results it can be seen that the total rotatin inertia is 2.282 inch lb.sec2. Assuming about half of it to be external inertia, the remaining 1.141 inch lb.sec 2 is the flywheel inertia. For a disc type flywheel of 13 inches diameter, the in- ertia is I 2. where I is the inertia in inch lb. sec , w is in lbs, d is in inches, g is 386 inches/sec2 • g I· 141 w = x~I06l< = 2.0·15 I\, s I~ X'~ 4W ::: 4x2.0·'6 :: .5{. ·,1\c.hn The thickness of the flywheel is t = -11.- lid~ 'ii x I~JC 1'?1 )(•2.'ir ' 46 where d is the diameter in inches, t is the thickness in inches,~ is the density of the material in lb/in3 • The reason this engine has a small flywheel is because there are six firing strokes during one cycle and the combined torque at any instant does not deviate very far from the mean torque line as can be seen in the T~rning Moment dia gram obtained from the Calcomp Plotter. TABLE II Inertia Variation With Speed Speed in Net Are a in Inertia in rpm 1b inches 1b inch sec2 1200.00 722.289 2.282 1400.00 717.649 1.665 1600.00 716.066 1.272 1800.00 635.182 0.892 2000.00 593.374 0.674 2200.00 582.455 0.547 2400.00 662.095 0.523 2600.00 663.213 0.446 2800.00 550.295 0.319 3000.00 514.789 0.260 3200.00 558.171 0.248 3400.00 469.160 0.184 3600.00 399.431 0.140 3800.00 439.896 0.138 48 IV. CONCLUSIONS AND RECOMMENDATIONS It has been shown that the inertia analysis• which was initially so long and inaccurate. has been made very simple just by the use of one versatile computer program. Also. the problem or optimizing the inertia was made simple by the fact that the optimum inertia occurred at the idling speed of the engine. It is suggested that this problem be run on a computer which has a larger core storage and is faster than the I.B.M. 1620. This enables smaller intervals to be chosen in the torque crank angle analysis, thus giving a more accurate solution and even permitting the use ot cylinder combinations in which 720 divided by PINT is not an integer. V. BIBLIOGRAPHY 1. Mischke, c. R., "Elements of Mechanical Analysis", Addison-Wessley, 1963. 2. Kent., "Mechanical Engineers' Handbook", (Twelfth Edition). 3. Lewitt, E. H., "Thermodynamics Applied to Heat Engines", Pitman and Sons, 1958. 4. Lichty, L. c., "Internal Combustion Engines", McGraw Hill, 1939. 5. Scarborough, J., "Numerical Mathematical Analysis", John Hopkins Press, 1962. 6. "Calcomp Plotter Notes", Computer Science Center, Uni versity of Missouri at Rolla. 7. Shigley, J., "Dynamic Analysis of Machines", McGraw-Hill, 1961. 50 VI. APPENDIX 51 1Zoo.-o-oo -----o-. o -oo~ · 68 2 . 534 ---:--~o-. -o-o~o----0 -._o_o_o ____-o . _0_0_0 __ 1200 . 000 5 . 000 1422. 977 -12 . 192 882 . 055 o . ooo ----=1- -2~-0Q . nno · 1o . ooo 1846 . 374 Z3 . 8't9 146.2 . 072 o . noo oo . ooo 15 . 000 2081 . 525 - ~4 ~ 459 1807 . 6 73 o . ooo 1200 . 000 20 . 000. 2197 . 775 -43. 571 2046 . 818 o .ooo ..---._... :::·err~ . -=:--:=:-::--:=-----::--:-::--::-~------120Q . ()()0 25 . 000 2235 . 555 - 50.808 219B. R50 n . Qo n 1200 . 000 30 . 000 2 219 . 956 - 55 . 888 2288 . 117 o . ooo 1200 . 000 35 . 000 2167.283 -58 . 641 2330 . 559 0 . 000 1200 , 000 40.000 2088 .4R6 - 59 . 006 2337 . 128 0 . 000 1200 . 000 45 . 000 1991 . 076 - 57. 040 2315 . 689 o . ooo 1200 . 000 50 . 000 1880 . 279 ~52 . 900 2272 . 114 o . ooo l~no.ooo 55.000 1759.772 - 46.838 2210.950 n. nnn 1200 . 000 60 . 000 1632 . 195 - 39. 181 2135 . 826 o . ooo 1200 . 000 65 . 000 1535. 921 -30. 307 2049 . 722 o . ooo -,---'1 2 0 0 • 0 0 0 7 0 • 0 0 0 1 4 3 3 • 3 9 5 - 2 0 • 6 2 't l 9 5 5 • J 3 3 0 • 0 0 0 1200 . 000 75 . 000 1325. 043 - 10. 551 1854.186 o . ooo 1200 . 000 80 . 000 1212. 570 - . 489 1748 . 711 o . ooo 120o . nno n5 . ooo 1098 . 019 9.190 1640. 294 o.noo 1%00 . 000 90 . 000 984 . 029 18. 166 1530. 312 o . ooo 1200,000 95 . 000 874. 151 26 . 173 1419. 952 o . ooo 1200 . 000 100 . 000 773 . 334 33 . 011 1310.232 o . ooo '' 1200 . 000 105 . 000 688 . 734 ' 38 . 548 1202 . 016 o.ooo '' 1~00 . 000 110 . 000 631 . 246 42. 711 1096.022 o . ooo 1200 . 000 115.000 618.671 45.490 992. 837 o. ooo 1200 . 000 120 . 000 682 . 983 46.923 . 892 . 925 o . ooo ·1200,000 125 ~ 000 553 . 115 47 . ~86 796 . 637 o . ooo .---'1200 . 000 130 . 000 1298 . 0l't 46.087 704 227 o . ooo 1200 . 000 135 . 000 1726. 534 44 . 053 615 . 860 o . ooo 1200 . 000 140 . 000 1967. 742 ' 41 . 120 531 . 62 5 o . ooo 1200 . 000 145 . 000 2090 . 760 37. 428 451 .547 o . ooo 1200 . 000 150 . 000 2135 . 769 33. 109 375.597 o . ooo 1200 . 000 155 . 000 212 7~ 594 28 . 287 303 . 704 ' o . ooo ~ ---~1~2~o~o~.~o~o~o~. ~1~6~0~-~o~o~o~---~2~o~Bu2~·u2~P~J2~--~2~3~-~0~7~4~-~2~3~5~·~7~6~8~----~o~·~o~o~,o~---l ·, . 12oo . ooo 165 . oao 2010·. 549 17. 569 171. 664 o . ooo ~ 1200 . 000 170 . 000 1919 . 714 11~ 858 111. 254 o . o oo • 12no . ooo 175 . 000 1814. 8 75 6.015 54 . 396 n . noo 1200.000 180 . 000 1699.661 . ,108 . 945 o . ooo 1200 . 000 18 5 . 000 1613. 151 - 5 . 800 o . ooo o . ooo 12oo . ooo 19o . ooo 151A . 302 - 11 . 646 o . ooo o. oon 1200 . 000 195~000 1415. 917 ' - 17. 364 o . ooo o.ooo 1200 . 000 200 . 000 · 1307 . 983 -22 . 878 o.ooo o . o oo 1200.000 205 . 000 1196.756 -28 .103 o .ooo o . ooo 12oo . ooo 21o . ooo 1085 . 035 -32 . 941 · · O.ooo ·o . ooo 1200 . 000 . 215 . 000 976 . 504 -37. 280 o.ooo o . ooo 1200 . 000 220 . 000 876 . 215 -40.998 o.ooo o . ooo 1200 , 000 225 , 000 '791 , Lt1Lt "':'ll-3 , 961 0 , 000 0 , 000 1.200.000 230 . 000 733 . 076 -46. 030 0.000 o . ooo ' 1200 . 000 235 . 000 719 . 077 -47. 069 o . ooo o . oo o 1200.000 240 . 000 . 781 . 463 -46. 950 o. ooo o . ooo 1200.000 245 . 000 649 . 239 -45.566 o.ooo o . ooo 1200.000 250.000 1391.422 -42 . R37 0,000 0. 000 1200 . 000 255.000 1816. 93:::> - 38.724 o.ooo . o . ooo 1200 . 000 260.000 ' 2054 . 909 . -33. 237 o.ooo o.ooo l .-~-~- --:_, -~- ""-;:"~-~-- --:--~--...:~~--~-- ,------·------·.--~------~-~-- -·------·-.... - ... '• -I I. 52 1200 . 000 265 . 000 '21 7 '~ . 5 31 - 26 . 444 o. ooo o. ooo 1200 . 000 270 . 000 2216 . 039 - 18. '~- 7 7 o. ooo o. ooo ___1_7 ()!) • 0 () 0 275 . 000 2204. 308 - 2.522 Q.QOQ Q.QQO lf:OO . OOO 280 . 000 215 5 . ' :- 31 . 127 o. ooo o. ooo 1200 . 000 285.000 2080 . 158 10 . 181 o. ooo o.ooo 1200.000 290 . 000 1905 . 835 20 . 261 0200Q n.nno 1200 . 000 295 . 000 1877 . 577 29 . 965 o. ooo o .. ooo 1200 . 000 300 . 000 17 59 . 019 38 . 876 o. ooo o. ooo 1200 . 000 305 . 000 1620 . 004 ':-6 . 585 o.oQQ Q.QQO 1200 . 000 310 . 000 1 5 2 ':· • l'd3 52 . 711 0 . 000 0 . 000 1200 . 000 315 . 000 l'r2l. 63 5 56 . 925 o. ooo o. ooo 1200.000 320.000 13 J. 3 • L:-9 7 58.975 O.()OQ Q1 nOQ 1200 . 000 325 . 000 1201 . 981 58 . 6 96 0 . 000 0 . 000 ; 1200 . 000 330 . 000 108 9 . 8 7 '~ 56 . 030 o.ooo o. ooo I . 1200 . 000 335 . 000 980 . 844 51. 032 0.000 OtQOO I I 1200 . 000 3'~0 . ooo 879 . 933 '~3 . 87 0 o. ooo o. ooo I I 1200 . 000 345 . 000 I 7 9'r• 3 76 3 ':- . 8 21 o. ooo . 0 . 000 I J,200.00Q ;250.000 1:22 a 1'~5 2'ta 2 ') 1 Q.QQQ Oai!OQ 1200 . 000 355 . 000 720 . 116 12 . 631 o. ooo o. ooo 1200 . 000 ' 360 . 000 7 8 1. 3'r5 • ~-4 8 o. ooo o. ooo I 1?00 . 000 365.000 6~!'1.825 - 12.122 Q. QQQ Q. QQQ i 1200 . 000 370 . 000 138 8 . 69 3 - 23 . 849' o. ooo 0 . 000 l 1200 . 000 375 . 000 . 1812 . 822 - 34 . ~ · 59 0 . 000 0 . 000 i' 1200 . 000 380.000 2049 2 4?.2 - ~:2~211 QaOQQ Q. QQQ 1200 . 000 385 . 000 2167.738 - 50 . 808 o. ooo o. ooo 1200 . 000 390-. 000 2208 . 07 1 - 55 : 888 o. ooo o. ooo 1200 . 000 395.000 219 5·. 3 64 :.... 58.641 0.0()0 o.ooo 1200 . 000 L:-00 . 000 2145 .772 - 5'9 . 006 o. ooo o. ooo '. 1200 . 000 405 . 000 2070 . 104 - 57 . 040 0 . 000 0 . 000 1200 . 000 Lr10 . 000 1975.755 - 52 . 900 o.ooo o.noo 1200 . 000 415 . 000 1867 ~ 873 -Lr6 . 838 0 . 000 o. ooo 1200 . 000 420 . 000 17 50 . 108 - 39 . 181 o. ooo o. n.oo 1200 . 000 425 . 000 1612 . 296 - 30 . 307 o.ooo Q.QQQ 1200 . 000 ':·30 .ooo 1518 . 023 - 20 . 624 o. ooo o. ooo 1200 . 000 ~-35 . ooo ·l'r1 7. 420 - 1'0 . 551 o. ooo o. ooo 12 00· . 000 '~-':-0 .000 . 13 11 • L:-L;. 7 - . 4119 0.000 o,ooo 1200 . 000 445 . 000 1202 . 260 9 . 190 o. ooo o. ooo 1200 . 000 '1·50 . ooo 1092 . 539 18. 166 0 . 000 o. ooo 1200 . 000 ,1 ~ 55 . 000 985 . 8'~1 26 . 173 o.ooo 0~000 1200 . 000 ~·60 . 000 887 . 095 33 . 011 o. ooo o. ooo 1200 . 000 Lr65 ."000 8 0 3 . '•2 6 3 8 . 5';- 8- o. ooo o. ooo 1200.000 L:-70.000 7 Lr5 . 709 42 . 711 o.ooo 02 000 1200.000 ,, 75 . 0.00 7 3 1. 743 45 • L;-90 o. ooo o. ooo 1200.000 480 . 000 7 93 . 527 46 . 923 o.ooo 0 . 000 1200.000 485 . 000 660 . 496 L:-7 • 086 o.ooo o.ooo 1200.000 490 . 000 1400 . 349 46.087 o. ooo o. ooo 1200 . ooo . 495 .000 18 2 3 . 433 4'~ . 053 o. ooo o. ooo .. _... _- . --.------.-J ---... - . - - ~- ·---··-- · ------.------· --. ------·-- 53 1200 . 000 5 00 . 000 2058 . 534 'd . 120 o . o o o o . ooo 1200 . 000 50 5 . 000 I' 21 7'h 9 7 5 · 3 7.'r2 8 o . o o o o . ooo I 1200 . 000 510 . 000 2213 . 15~ 33. 109 00 (i() : 1200 . 000 5 15 . 000 2198 . 116 2 8 . 2 87 o . o oo o . ooo 1200 . 000 5 20 . 000 2146 . 13 7 2 3 . 071.~ o. ooo n . ooo 1200 . 000 525 . 001) 2068 . 13 7 J. 7 . 5n9 0. 00 1) o . oon 1200 . 00 0 530 . 0 00 19 7 1 . 6 15 11. 85 8 0 . 0 0 0 o . ooo 1~00 . 000 535 . 000 18 61. 8 12 6 . 015 o . o o o o . ooo 1200 . 000 5Lr0 . 0 00 17 Lc2 . 4 5 1 .1 08 o . noo o . ooo 1200 . 000 545 . 000 160 3 . ' r21 - 5 . 8 00 o . o o o - 12 . 8 40 I I 1200 . 000 550 . 000 1508 . 3lj-0 - 11 . 646 o . o oo - 26 . 26 2 ·1 12oo . nno 555 . 000 llj-0 7.346 - 1 7 . 3 64 o . o no - LeO . 5 ?. ?. .. 1200 . 000 560 . 000 1 301.38 5 - 22 . 87 8 0 . 000 - 55 . 6 55 1200 . 000 5 65 . 000 · 1192 . 5 80 -2 8 . 103 0 . 000 - 71 . 6 92 __),__?- 0 0 • 0 () 0 5 7 0 . 000 10R3 . 'i 6 'i - 3?.gtd 0 . 00 0 - RA . (, f-, ':) 1200 . 000 5 7 5 . 0 00 9 7 7. [l 3 L1- - 37. 2 80 o . o oo - 106 . 5 92 1200 . 000 580 . 0 00 88 0 . 256 . - LrO . 99 8 o . ooo - 125 . 495 1200 . 000 5 8 5 . 000 7 97 . 890 - Lc3 . 96 1 o . ooo - J L:-5 • 37 9 1200 . 00 0 590 . 0 0 0 7'1-1 . 5Lr5 - 46 . 030 0 ~ 0 0 0 - 16 6 . 239 ~200 . 00 0 595 . 0 0 0 7 28 . 9 6'r · -Lr 7 o 06 9 0 . 000 - Hl8 . 0 53 i2oo . ooo 600 . 0 00 7 92 . 095 - ' :-6 . 95 0 o . oo o - 210 . 7 R3 120 0 . 000 605 . 000 660 . 333 - 45 . 566 0 . 000 - 234 . 368 1 200 . 00 0 610 . 000 1 LrO 1 . 348 - Lr2 . 83 7 o . ooo - 25 8 . 726 1200 . 000 6 15. 000 18 2 5 • Lr 6 7 - 3 8 . 72Lr o . ooo - 2P.3 . 7 ".- 7 1200 . 000 620 . 000 2061 . Lr6 6 - 33. 237 o . ooo - 309 . 2 9 2 12 00 . 0 00 625 . 00 0 2 1 78 . 66 7 - 26 . ' t- 'r4 0 . 000 - 3 35 . 193 1200 . 00 0 6 30 . 0 0 0 221 7. ' :-71 - 1 8 . 4 7 7 o . o oo - 3 6 1 . 24L!- 1200 . 00 0 635 . 000 2202 . 9 3 8 - 9 . 532 o . ooo - 3 8 7 . 20 7 1200 . 00 0 6 Lr0 . 000 ' 2151 . 350 . 12 7 o . o oo - 4 1 2 . 800 1200 . 000 6'r5 . 000 20 73 • 6L:- 2 1 0 .1 8 1 o . ooo -Lr3 7. 698 1200 . 000 650 . 0 0 0 1977·. 326 20 . 261 o . oo o ' :-6 1 . 528 1 200 . 000 65 5 . 000 '18 67. 654 29 . 9 65 o . o o o - Lr8 3 . 8 5 6 1200 . 00 0 660 . 0 00 17 Lr8 . 35 8 3 8 . 87 6 o. ooo - 5 O' r . 18 2 ' 1200 . 0 0 0 6 6 5 . 000 1622 . 170 46 . 585 ·o . oo o 5 2 1. 9 15 1 200 . 00 0 . 670 . 000 1527. 609 52 . 711 . 0 . 000 - 5 36 . 35 Lr 1200 . 000 6 75.00 0 142 7. 32'1· 5 6 .925 o . ooo - 5':-6 . 640 1200 . 00 0 6 8 0 . ooo ·. 1'322 . 032 5 8 . 975 o . ooo 55 1.70 1 . i 1213 . 8 ' ~- 2 5 8 . 6 96 0 . 000 -550 . 150 I . 1200 . 000 6 8 5 . ooo - ;_ 1200 . noo 6 90 . 0 00 1105 . 3 7 2 56 . 030 o . ooo - 5'r0 . 13 2 I 1200 . 0 00 695 . 000 1000 . 103 51 . 032 .0 . 000 - 519 . 05 9 ! 't-3 . 87 0 . I 1200 . 00 0 7 0 0 . 000 902 . 877 o . ooo - 48 3. 171 I 820 . 73 7 34. 8 21 o. ooo - 426 .7 18 I 1200 . 000 7 0 5.000 24. 257 o . ooo - 340 . Lr 14 1200.000 7 10 . 0.00 · 7 64. 474 • < 12 . 631 o . ooo - 208 . 21 7 1200 . 0 00 715 . 00 0 7 51 . 8 17 • 448 o . ooo o . oo o 1200 .000 720 .000 8 14 . 70 6 1200 . 000 18 4 17. 3 11 · 1L~65 . 6 04 54 3 15 ------~~~~------40 ------52 64 76 as 100 112 124 6 1200.000 722.289 ooo I LUU · U '55 Figure 16. Turning momen~ diagram of a 1960 Corvair engine. T o · R q u E VII. VITA The author, Dady Jal Patel, was born on December 5, 1943 in Bombay, India. He received his primary and second ary education from St. Mary's High School, Bombay, India. After completing his high school studies, he joined St. Xavier's College of Arts and Sciences, Bombay, for a period of two years. He did his undergraduate work in mechanical and electrical engineering at the Walchand College of Engineering, Sangli, India, where he received his Bachelor's degree in electrical (June, 1963) and mechanical engineering (June, 1964). He has been enrolled in the Graduate School of the Uni versity of Missouri at Rolla, since September, _1964, and has held the Foreign Education Scholarship and Parsi Public School Scholarship for the period of September, 1964, to September, 1965.