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Masters Theses Student Theses and Dissertations

1965

Optimization of flywheel design for internal combustion engines Optimization of flywheel design for internal combustion engines

Dady Jal Patel

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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.

I.

II.

III.

TABLE OF CONTENTS

TITLE PAGE • • • • • • • • • • • • • • • • • • • ABSTRACT • • • • • • • • • • • • • • • • • • • •

ACKNOWLEDGEMENTS • • • • • • • • • • • • • • • • TABLE OF CONTENTS • • • • • • • • • • • • • • •

LIST OF FIGURES • • • • • • • • • • • • • • • •

LIST OF TABLES • • • • • • • • • • • • • • • • •

INTRODUCTION • • • • • • • • • • • • • • •

REVIEW OF LITERATURE • • • • • • • • • • •

DISCUSSION • • • • • • • • • • • • • • •

A. Determination of the Torque Crank Angle Dia­gram of the Engine and the Evaluation of the Flywheel Inertia ••••••••••••••

iv

i

11

iii

iv

vi

vii

1

3

9

9

l. Logic Used in Developing the Computer Pro-gram for Inertia Evaluation · • • • • • • 9

a. Evaluation of the gas pressure torque • • • • • • • • • • • •

b. Evaluation of the torque due to the inertia forces and the resulting torque due to gas pressure and iner-

lO

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

IV.

v.

VI.

VII.

g. Determination of the maximum and minimum speed points on the torque crank angle diagram • • •••

h. Evaluation of the largest positive net area above and below the mean

v

• 23

torque line • • • • • • 25

i. Calculation of the rotating inertia 25

2. Programs and Their Handling • • • •

a.. Computer program for the evaluation of the flywheel inertia • • • •

(l} Notations for the Read State-ment of Program No. l • •

( 2} The sequence in which the an-swers will be obtained • •

(3} Column headings for the Punched Cards obtained in the output • • • • • • • • •

•

·b. Calcomp Plotter Program for obtain-ing the torque crank angle diagram of the engine • • • • • •

Notations used in the Program No. 2 • • • • • • • • • •

B. Optimization of Flywheel Inertia with Res-pect to Speed • • • • • • • • • • • •

c. Illustrative Problem • • • • • • • • • •

CONCLUSIONS AND RECOMMENDATIONS • • • • • •

BIBLIOGRAPHY • • • • • • • • • • • • • •

APPENDIX • • • • • • ... • • • • • • • • • •

VITA • • • • • • • • • • • • • • •

25

25

37

38

38

39

42

42

44

48

49

50

56

Figure

1

2

3

4

LIST OF FIGURES

Arbitrary torque crank angle diagram of an internal combustion engine • • ••••• •

Torque crank angle diagram of a six cylinder engine • • • • • • • • • • • • • • •

Pressure volume diagram for Otto cycle • • •

Pressure volume diagram for Diesel engine

• •

• •

•

• •

5 Resolution of the net force in the engine

6

7

8

9

10

11

mechanism • • • • • • •

Torque crank angle diagram as a result of gas pressure alone in a four stroke engine • • •

Torque crank angle diagram as a result of pressure alone in a two stroke en gine

•

Torque crank angle diagram showing the combina­tion of the gas pressure torque with the iner-

•

•

tia torque • • • • • • • • • • • • •

Combined torque crank angle diagram for a four cylinder, in line engine • •. • • • • •••

Combined torque crank angle diagram for a Vee engine with two banks of one cylinder each •

Area bounded by a function: : y = f(x) • •

vi

Page

4

6

10

13

15

16

16

18

19

20

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

Table

1

2

LIST OF TABLES

Coefficient of speed fluctuation • • • •

Inertia variation with speed • • • • • •

vii

• • • •

• • • •

Page

8

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.

II. REVIEW OF LITERATURE

A flywheel as defined by Professor Joseph Shigley,

is a mechanical filtering element in a circuit through

3

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 num­bers you know something about it; but when you cannot express it in numbers, your know­ledge is of a meager and unsatisfactory kind: it may be the beginning or knowledge, but you have scaro~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

from first principles.

It has been known that I.e. engines deliver a very

"lumpy" t th orque, 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.

4

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.

Work done by the engine torque during one cycle is

4>

/T(e)de =:"'<P, 0

vhere Tm is the mean torque required to satisfy load :req,ui:r ment,

<P = 2.11 radians in two stroke engines 1

¢ = 411 radians in four stroke engines.

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

5

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.. '

. .

Thus 6"

<.lb I/ de.;,. wdt w d.t ....

f(T(e)-T,...)de _ e ...

(.)"

Ij w c\ ~ ::: ~ I. ( w ~ - w~) w ...

I ( i.. .t ) - I w.- Wo. 2 •

The maximum work done by the accelerating torque occurs

between the minimum angular speed and maximum an gular

6

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

A,= j ( T c e)- T"') de = 1 I ( w ~ - wl.. . ) 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 d iagram 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

speed from Wmin to wmax is \ I _ I ( 2

w 'J) - 2. I tv IM.~ - ' ) <..,)"'.'"

W = i._ I ( w - w )(w + ) -:1) 2 ....... "'i'fl "' .. ,. w....... - I c.-f 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

I.- 'W'D c wa +

(1)*

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.

TABLE I*

Coefficient of Speed Fluctuation

Type of

Equipment

Crushing machiner~

Electrical machinery

Electrical M/c direct driven

Engines with belt transmission

Flour ~illing machinery

Gear wheel transmission

Machine tools

Pumping machinery

Spinning machinery

Textile machinery

Paper making machinery

*Taken from Reference 2.

Coeff i cient of

Fluctuation, Cf

0.200

0.003

0.002

0.030

0.020

0.020

0.030

0.03 - 0.05

O.Ol - 0.02

0.025

0.025

6

9

III. DISCUSSION

A. Determination of the Torque Crank An gle Di ag r a m of t he

En gine and the Evaluation of the Flywheel Inerti a.

l. Logic Used in Developing the Computer Pro g 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.

c.

d.

e.

f.

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.

Determination of the combined torque diagram for the various cylinder combinations.

Determination of the area under the torque crank

angle diagram.

Determination of the mean torque line.

Evaluation of the roots, that is, the intersec­tion of tbe torque curve with tbe mean torque line.

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.

10

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• Mechanical Efficiency = pb

•

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:

Indicated Mean Effective Pressure =

-o = r • I (K -l)(c.- i)

Work done

Length of the Diagram

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

Hence, v1 ( V1 - Ve) c = -(c - 1)

and, \fa=: v1 - S·a..

'

S · o.. ·c. (c.- 1)

Therefore, the clearance volume, Vc• can be evaluated as

v1 and v2 are known.

v = c

l PVK = constant Utilizing the general aw,

K

'P = ? (100 ;- Vc) (for- C!)(ra.t\s;ol'l .,.tro\(c.) :x:. e 1! X + Vc

where pxc =

Pxe =

pressure during compress ion stroke for piston position,

pressure during expansion stroke for any piston position,

ny

X = percentage of piston travel from the head end of the stroke.

13

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

Figure 4.

4

1

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:

?~ [ k (~o-1) - CL-\ ~: -1)]

(k-1)(c.-1) '

l4

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 •t ~ ll l. .t. 0 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

...

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

15

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 ~ .

t

I tl ~

Figure 6. Torque crank - angle diagram as a result of gas pressure alone in a four stroke engine.

(

rr~------~~~~L---~~------------~ 0

t-

Figure 7. Torque crank angle diagram as a result of gas pressure alone in a two stroke engine.

16

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.

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.

17

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.

t f

f

0

r-

For four stroke engine:

_, / \

I \ I \

I ' I \r~CI." rre,<»u.re tor,~e.

I \

Figure 8. Torque crank angle diagram showing the combination of the gas pressure torque with the inertia torque.

18

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.

..

( 5) Opposed Piston.

(6) Radial.

19

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 'th w~ two banks of one cy-

linder each, where the angle of th v e ee is 90°. The numb e 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 cy­linder 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

der the curve without actually drawing the curv

r un-

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 5 • h 6 , and 1 2 3 ..

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

12. Magnified view of a curve between a and Figure

of a root. b, showing the presence

In Figure 12. let y = f(x) be the torque curve and

the x-axis, be the mean torque line. the horizontal line,

signs in any interval, then the If the ordinate changes

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.

Figure l3. Arbitrary torque crank angle diagram show­ing the positive and negative areas, above and below the mean torque line.

• •

~ represent the roots, •• '='12

A represent the areas above and ~~ 2

l ·ne A2, Alt, AG' As' Alo' and Al2 below the mean torque 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 •

and if A2 + A3 + A4 ..(. A2 then w 3 ~ LJ 1 •

positive, then w > c.u and vice versa. 3 0

followed until W is obtained. ' 12

If A + A + A is 2 3 It

This procedure is

The sum of the areas, of all the loops, is zero, and

can be justified from the definition of the mean torque

line. Of the alaebraic sum , of all the

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 and Their Handlin&• • rogra.ms

a. tor the evaluation ot · the

Computer program

flywheel inertia. Vas written tor an IBM 1620

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.

Figure 14. i i ntervals of ' the crank. Torque at var ous

t P rogram is very general, and Although . this compu er

not limited to any particular engine, its application i~

t if used for a parti­still its solution becomes inaccura e

cular set of cylinder combinations. The problem that arises

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 c

I

I

c

\ ! i

\ ' . '

i

c

c

c I

c

STANDARDIZATION OF FLYWHEEL DESIGN FOR ANY VARIETY OF ENGINES 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 READ 2 NUMB NO

1 FORMAT (7F10.3) 2 FORMAT (118,118)

GAS PRESSURE TORQUE ANALYSIS FOR OTTO TWOSTROKE FOURSTROKE D=SPEED/E AREA=3.14159*(DIA**2)/~.0 BHP=HP /CYLIN · PB=t33000.*12.*BHP)/(D*AREA*STROK) PI=PB/EFF IF ·eNGIN 10 1 1 FOR OTTO CYCLE ONLY UPTO 333

10 XYZ=tCR-1.)*Pl/(((CR**EXA)-CR)*CARD) 333 4= EXA- * XYZ +PA

GO . TO 12 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 EVALUATING CLEARENCE VOLUME (OTTO AND DIESEL) UPTO 666

12 VOL=STROK*AREA Vl=VOL*CR/(CR-1.) V2=V1-VOL

666 CV=V2*100./VOL EVALUATING GAS PRESSURE TORQUE 1=1 PINQ=180.+PINT

I

I

!

I

1\)

(X)

r

I

· c !

: c l I

I

'

.. ' .

c

'

PIT=PINT 4 X_=p IT-PINT · ~~.-

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 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 .

FOR· DIESEL CYCLE UPTO 888 -------~-- - -. =--~- -· - -- ..:.: --: ~'--::···:::_---:--.:...·.--~: ~~_.:=."7'".:-:-~ - -= - -- _____ .._ ___ ··- -- _ __......__

22_PXE=P3*(((CUT*CV}/(Y+CV})**EXA)-PA RR8 PXf.=~*--LL1100 +f.Vl /(Y+f.Vl l**COMPl- PA · 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

. TXC(I)=PXC*R*RAD*AREA I=J_+_l '

PIT=PIT+PINT IF (PINQ-PIT)55,4,4

55 CONTINUE 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))

. .. .)

- .

·.-.-

I

1\) \()

..

c

:

- ·

1

. '

I

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 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) 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_

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 KAPO-KAPP+l

I

w 0

DO 27 J ZRA PO,N APP,l KP=NAPP+l-J

27 TXS(J)=-TXCCKP) GO TO 313

c COMBINATION OF GAS AND IN ERTIA TOR QUES FOR TWO STRO KE UPTO 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 SUM(JNAPP)=SUM(J) DO 143 J =JAPO,NAPP,1

143 TXE(J)-.000 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~ CYLINDER S

313 POT=O.O DO 9 L=1 7 NAPP,1 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) B=B+S

145

I

' ' :

: I :

I

I

I

w ....

c t

.

c

c

\

'

3 J U= J U-J A· 7_ZB_ PUMB=NUMB

TS UM(L ) =PUMB*(SUM {L)+B ) PRI NT 200 1 SP EED , PO T, TSUM {L),T( L), TXE (L), TXS( L) PUNCH 200~SP EED ,PO 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,Fl 4 .3,Fl2.3,Fl2.3,Fl2.3)

AREA UNDER _IQRQU E CRANK DIAGRAM " 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) EVALUAJE MEAN TORQUE . . TMEAN=SIMP/(RAT*3.14159) PRINT 160,TMEAN

160 FORMAT (Fl8.3) 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

NIPP-NAPP-1 ' w

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

\ JIVE - JACK-1 DO 28 I=l,JIVE,l

I MINT=MAD (I) MLAS- HAD (I+ 1) H=PINT* 3.141 59/180. BRAKl=O.O BRAK2=0 .0 MEAT=MINT+l .

- ----

... - - -

I i !

I

w w

c

c

I

:

-

ME A L = 1·1 L A S- 1 DO 23 J=I~EA T__t__UEA L 2

23 BRAK1=BRAKl+TAVG(J) H A S T = ~11 N T + 2 0 Q ___ 2_'!: __ J = 1·1 AS_ T , l:U:_~J__, 2

24 BRAK2=B RAK2+ TA VG (J) FO F E=TAVG(MINTl+TAVG(~LASl

2 8 S I M S ( I ) = (HI 3. ) ~' ( F 0 FE+ 4. * BR A K 1 + 2. ~'BRA K 2 ) FOR FIRST(SPLITlSMALL AREA MILL=MAD(1)-1 IF ( ~' I L L ) 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 SI~=(H/3.)*(FOF+4.*BRAK1+2.*BRAK2J

GO, TO 188 180 CONTINUE

FOR LAST SPLIT AREA 188 MOLT = MAD( lACK)

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

: : ,

I

w ~

c

:

l I

-

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 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

15 SAM=SIMS(J)+SAM :

SUN=O.O NV=WMAX I

DO 16 J-l.NV.l 16 SU~=SIMS(J)+SUN

SUMM=SUN+SAM --

...

..

'

w V1

36

.. I

:

-~ ....

.,~ ('()

g •

co

~ N

r-4

.. u_

z:7 I- ..

Wr<

'l o

-z

• t-4

_,(.

<l:CO

._u_ ..

.-~

<l:s

ClU

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w

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wrr\

-<l:::::

0..

• .....

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.-~

~ ~

OU

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r--

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<l:<

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0 II

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..... Z~-

ww

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0 0

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ZII ~0< z

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<-,...

Q.U

.L w

~ ,.... 0

.-4

,.... .-4

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-

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.

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.

38

(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-CRA NK 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

t 10

60

1-ol

3 60 0 0... ol

"' ~ 40 0

::r

30

~0 0

Figure 15.

" 'I\ 'PI"\

( hu..l\cir .. J.~)

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, Illustr a 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 rotati n

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

where I is the inertia in inch lb. sec 2. , w is in lbs, d is in

inches, g is 386 inches/sec 2 •

w g x~I06l< I· 141 = 2.0·15 I\, s = I~ X'~

the flywheel is t = 4W ::: 4x2.0·'6 :: .5{. ·,1\c.hn The thickness of -11.-

lid~ 'ii x I~JC 1'?1 )(•2.'ir '

where d is the diameter in inches, t is the thickness in

inches,~ is the density of the material in lb/in3 •

46

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 sec 2

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", McGraw­Hill, 1939.

5. Scarborough, J., "Numerical Mathematical Analysis", John Hopkins Press, 1962.

6. "Calcomp Plotter Notes", Computer Science Center, Uni­versity 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 . 05 5 o . ooo ----=11- -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 - 50.808 219 B. R50 n . Qo n 1200 . 000 30 . 000 2 219 . 956 - 55 . 888 2288 . 11 7 o . ooo 1200 . 000 35 . 000 2167.283 -58 . 641 2330 . 559 0 . 0 00 1200 , 000 40.000 208 8 .4 R6 - 59 . 006 2337 . 128 0 . 00 0 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 - . 489 174 8 . 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 . oo o

'' 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 . 86 0 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 . ooo • ·, .

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. o on 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 -2 8 .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 . o oo 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. oo o 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

120 0 . 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 1200 . 000 3'~0 . ooo 879 . 933 '~3 . 87 0 o. ooo o. ooo I I 1200 . 000 345 . 000 7 9'r • 3 76 3 ':- . 8 21 o. ooo . 0 . 000 I I 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 .00 0 18 2 3 . 433 4'~ . 053 o. ooo o. ooo

.. _ ... _- . --.-------.-J---... -. - - ~-·---··-- ·-- ---------.--------·--. -------------- - - - -- --·--

53

1200 . 000 5 00 . 00 0 205 8 . 534 'd . 120 o . o o o o . ooo ' 1200 . 000 50 5 . 000 21 7 ' h 9 7 5 · 3 7.'r2 8 o . o o o o . ooo I I 1200 . 000 510 . 000 2213 . 15~ 33. 109 00 (i() : 1200 . 000 5 15 . 000 219 8 . 116 2 8 . 2 87 o . o oo o . ooo

1200 . 000 5 20 . 00 0 214 6 . 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 1200 . 000 550 . 000 15 08 . 3lj-0 I - 11 . 646 o . o oo - 26 . 26 2 ·1 12oo . nno 555 . 000 llj-0 7.34 6 - 1 7 . 3 64 o . o n o - LeO . 5 ?. ?. .. 1200 . 000 560 . 000 1 301.3 8 5 - 22 . 87 8 0 . 000 - 55 . 6 55

1200 . 000 5 65 . 00 0 · 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 - 10 6 . 5 92 1200 . 000 580 . 0 00 88 0 . 256 . - LrO . 99 8 o . ooo - 125 . 495 1200 . 000 5 8 5 . 00 0 7 97 . 89 0 - 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 59 5 . 0 0 0 7 28 . 9 6'r · -Lr 7 o 06 9 0 . 000 - Hl8 . 053 i2oo . ooo 600 . 0 00 7 92 . 095 - ' :-6 . 95 0 o . oo o - 210 . 7 R3 120 0 . 00 0 605 . 000 660 . 333 - 45 . 566 0 . 000 - 234 . 368 1 200 . 00 0 610 . 000 1 LrO 1 . 34 8 - 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 . 00 0 62 0 . 000 2061 . Lr6 6 - 33 . 237 o . ooo - 309 . 2 9 2 1200 . 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 63 5 . 000 2202 . 9 3 8 - 9 . 532 o . oo o - 3 8 7 . 20 7

1200 . 00 0 6 Lr0 . 000 ' 2151 . 350 . 12 7 o . o oo - 4 1 2 . 80 0 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 . 00 0 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 . 58 5 · 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 . 00 0 6 75.0 0 0 142 7. 32'1· 5 6 .92 5 o . ooo - 5':-6 . 640

1200 . 00 0 6 8 0 . ooo ·. 1'322 . 032 5 8 . 975 o . ooo 55 1.7 0 1 . i

1200 . 00 0 6 8 5 . ooo 1213 . 8 '~- 2 5 8 . 6 96 0 . 000 -5 50 . 150 -I . 6 90 . 0 00 1105 . 3 7 2 56 . 030 o . ooo - 5 ' r0 . 13 2 ;_ 1200 . noo

I 1200 . 0 00 695 . 000 10 00 . 103 51 . 032 .0 . 000 - 51 9 . 05 9 ! I 1200 . 00 0 7 0 0 . 000 902 . 877 't-3 . 87 0 . o . ooo - 48 3. 171 I 1200 . 000 7 0 5.000 820 . 73 7 34 . 8 21 o. ooo - 426 .7 18 I

1200.000 7 10 . 0.00 · 7 64. 474 24 . 257 o . ooo - 34 0 . Lr 14

1200 . 0 00 715 . 00 0 • < 7 51 . 8 17 12 . 631 o . ooo - 20 8 . 21 7

1200 .000 720 .000 8 14 . 70 6 • 448 o . ooo o . oo o

1200 . 000 18 4 17. 3 11

· 1L~65 . 6 04

3 15

54

----------------~~~~------------------------------------------------40

1200.000

52 64 76 as

100 112 124

6 722.289

ooo

T o · R q u E

I LUU · U

Figure 16. Turning momen~ diagram of a 1960 Corvair engine.

'55

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.