Dynamics of Reciprocating

1

ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY

Dynamics of Reciprocating Engines

Dr. Sadettin KAPUCU

© 2007 Sadettin Kapucu

2Gaziantep University

Dynamics of Reciprocating Engine Dynamics of Reciprocating Engine This chapter studies the dynamics of a slider crank mechanisms in an analytical way. This is an example for the analytical approach of solution instead of the graphical accelerations and force analyses. The gas equations and models for combustion is not a concern of this chapter.

intake compression power exhaust

Crank angle

Pressure


Dynamics of Reciprocating EngineDynamics of Reciprocating Engine

Loop closure equation can be:

A1

4

B

2

3

P

lr

x

x

y

G3G2

I2,m2I3,m3

m4

sinsin lr

coscos lrx



A1

4

B

2

3

P

lr

x

x

y

G3G2

I2,m2I3,m3

m4

sinsin lr

coscos lrx

1cossin 22

2sin1cos

sinsinl

r

2

sin1cos

l

rlrx

2

sin1cos

l

r



2

sin1cos

l

rlrx

Dynamic analysis of reciprocating engines was done in late 1800’s and by that time extensive calculations had to be avoided. So there are many approximations in the analysis to simplify the arithmetic. In above equation square root term can be replaced by simplest expression. Taylor series expansion of square root term, first two term included is as follows:

22

22

sin2

1sin1l

r

l

r

Squaring is also an arithmetically difficult process:

2

2cos1sin 2



This equation defines the displacement of the slider. Velocity and acceleration expressions are by successive differentiation of this equation with respect to time. If we assume that the angular velocity of the crank is constant then velocity and acceleration of the slider become:

2

2cos1

21cos

2

2 l

rlrx t

t

l

rtr

l

rlx 2cos

4cos

4

2

t

l

rtrx 2sin

2sin

t

l

rtrx 2coscos

t

l

rtrx 2sin

2sin

t

l

rtrx 2coscos2



In dynamic force analysis, we put inertia and external forces on top of existing mechanism and then solve statically. Under the action of external and inertia forces, too many forces exist on the mechanism hence we use superposition

Gas force: Assume only gas force exists on the mechanism and calculate the torque on the crank by the gas force. Forces related with gas force will be denoted by a single prime.

A1

4

B

2

3

P

lr

x

x

y

m4


Gas Force ‘Gas Force ‘

A 4

B

C

2

3

P

B

CT'

x

y

+

F’23

F’43

F’34

F’14

F’32

F’12

P

F’34

F’14



A

4

B

2

3

P

lr

x

'F’12

P

F’34

F’14

F’14

14'' xF

t

l

rtr

l

rlx 2cos

4cos

4

2

P

F 14'tan tan'14 PF

cos

sin'14 PF

sinsinl

r

2

sin1cos

l

r

214

sin1

1sin'

l

rl

rPF

22

2

2sin

21

sin1

1

l

r

l

r

2

2

2

14 sin2

1sin'l

r

l

rPF



A

4

B

2

3

P

lr

x

'F’12

P

F’34

F’14

F’14

14'' xF

t

l

rtr

l

rlx 2cos

4cos

4

2

2

2

2

14 sin2

1sin'l

r

l

rPF

t

l

rt

l

rPt

l

rtr

l

rl 2

2

22

sin2

1sin2cos4

cos4

'

t

l

rt cos1sinPr'



t

l

rt cos1sinPr'


Inertia Forces’’Inertia Forces’’To obtain acceleration of third link in algebraic expression is a laborious task. After finding acceleration of the third link and putting the inertia force on center of mass of third link, doing force analysis is also laborious. To further simplify the problem, an equivalent mass approach can be used. In equivalent mass system problem, we generate a model which has two point masses rather than one.

A1

4

B

2

3

P

lr

x

x

y

G3G2

I2,m2I3,m3

m4


Inertia Forces’’Inertia Forces’’Equivalent massesEquivalent masses

In equivalent mass system problem, we generate a model which has two point masses rather than one. One of the masses will be at point C. The other is at P.

G3B

CP

lB lC lB lClP

m3P G3 m3Cm3 I3

•The mass of the model and mass of the actual link should be equal.

Mass center of the model and mass center of the actual link should be at the same place.

Mass moment of inertia of the model and the actual link should be same.

CP mmm 333

CCPP lmlm 33

23

233 CCPP lmlmI


Inertia Forces’’Inertia Forces’’Equivalent massesEquivalent masses

G3B

CP

lB lC lB lClP

m3P G3 m3Cm3 I3

CP mmm 333

CCPP lmlm 33

23

233 CCPP lmlmI

P

CCP l

lmm 3

3 CP

CC ml

lmm 3

33

P

PC

P

CC

l

lm

l

lmm 33

3 P

PCC l

llmm

33

CP

PC ll

lmm

3

3

CP

CP ll

lmm

3

32323

3 CCP

PP

CP

C lll

lml

ll

lmI

CP

P

CP

CPC ll

l

ll

lllmI 33 PCllmI 33

CP

PC ll

lmm

3

3CP

CP ll

lmm

3

3C

P lm

Il

3

3

at that point second mass should be located. It is also known as centre of percussion. Centre of percussion is at point where there is no inertia moment. Only an inertia force exists. In a connecting rod, where mass centre is nearer to point B and distance between mass centre and point B is very little. P, the centre of percussion is somewhere in between centre of mass and B. So, P is nearly coinciding with point B.


Inertia Forces’’Inertia Forces’’Using equivalent mass concept slider crank mechanism can be converted into two mass system which are located at B and C.

A1

4

B

2

3

P

lr

x

x

y

G3G2

I2,m2I3,m3

m4

l

mlm C

B 3 l

mlm B

C 3

Bm3

Cm3


Inertia Forces’’Inertia Forces’’Second link mass amount assumed to concentrated can be found by:

A1

4

B

2

3

P

lr

x

x

y

G3G2

I2,m2I3,m3

m4

Bm3

Cm3

r

mrm G

B 2This equation satisfies the equality of mass and mass centre for the crank. Then total masses at B and C are;

Bm2

Then total masses at B and C are;

BBB mmm 32

43 mmm CC


Inertia Forces’’Inertia Forces’’Position vector defining point B;

A1

4

B

2

3

P

lr

x

x

y

G3G2

I2,m2I3,m3

m4

jtritrRB

sincos

jtritrVB

cossin

jtrtritrtraB sincoscossin 22

jtritraB sincos 22 it

l

rtrxaC

2coscos2


Inertia Forces’’Inertia Forces’’

jtritraB sincos 22 it

l

rtrxaC

2coscos2

A1

4

B

2

3

t

lry

xC

jtrmitrmam BBBB

sincos 22

itl

rtrmam CCC

2coscos2

BBam

CCam

immaterial from crank shaft torque point of view. Because this inertia force is directed radially and so does not produce any torque on the crank.



A 4

B

C

2

3

C

B

CT'

x

y

+

F’’23

F’’43

F’’34

F’’14

F’’32

F’’12

-mcac

F’’34

F’’14



A

4

B

2

3

C

lr

x

''F’’12

F’’34

F’’14

F’’14

14'''' xF

t

l

rtr

l

rlx 2cos

4cos

4

2

CCam

F 14'tan tan)('14 CCamF

cos

sin)('' 14 CCamF

sinsinl

r

2

sin1cos

l

r

CCam

CCam

t

l

rt

l

rt

l

rtrmt

l

rtr

l

rl C 2

2

22

2

sin2

1sin2coscos2cos4

cos4

''



A

4

B

2

3

C

lr

x

''F’’12

F’’34

F’’14

F’’14

CCam

CCam

t

l

rt

l

rt

l

rtrmt

l

rtr

l

rl C 2

2

22

2

sin2

1sin2coscos2cos4

cos4

''

t

l

rtt

l

rr

mC 3sin2

32sinsin

22'' 22


Example 1Example 1

AB=10 cm, AG3=BG3=5 cm, =60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2

In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec.

4

B

2A

3

G3

x

BABA VVV

?AV

smVB /1

ABtoVBA ?

BV

AV

BAV

smVA /5774.0

smVBA /1547.1


4

B

2A

3

G3

x

BABA VVV

BtoAfromsmAB

Va B

An

BA

22

2

/33.131.0

1547.1

?Aa

BABA aaa

0

BAA aa t

BA

n

BA aa

ABtoa tB

A?

Example 1Example 1In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec.


Example 1 contExample 1 cont

4

B

2A

3

G3

x

BtoAfromsmAB

Va B

An

BA

22

2

/33.131.0

1547.1

?Aa

BAA aa t

BA

n

BA aa

ABtoa tB

A?

2/698,7 sma tB

A

t

BAa

Aa

2/396.15 smaA n

BAa



4

B

2A

3

G3

x

t

BAa

Aa

2/396.15 smaA n

BAa

Equivalency of masses

BA mmm 333

Equivalency of mass center

BA mBGmAG 3333

AB

mAGm B

33

AB

mBGm A

33

kgm B 4.01.0

8.0*05.03 kgm A 4.0

1.0

8.0*05.03

kgmmm BB 9.04.05.043

kgmmm AA 9.04.05.023



4

B

2A

3

G3

x

t

BAa

Aa n

BAa

Ama

D’Alembert forces and moments

kgmmm AA 9.04.05.023

2/396.15 smaA

0908564.13396.15*9.0 NmaA



4

B

2A

3

G3

x

N8564.13

BF4

B

2A

3

G3

N8564.13

BF

12F

14F



4

B

2A

3G3

N8564.13

BF

12F

14F

1212 0;0 FFFFF BBx

NF

FFy

8564.13

08564.13;0

14

14

NF

FM B

80867.0

69282.0

060sin*1.0*60cos*1.0*8564.13;0

12

12

NFB 8

+

x

y



4

B

2A

3G3

N8564.13

BF

12F

14F

1212 0;0 FFFFF BBx

NF

FFy

8564.13

08564.13;0

14

14

NF

FM B

80867.0

69282.0

060sin*1.0*60cos*1.0*8564.13;0

12

12

NFB 8

+

x

y

4

B

2A

3G3

N698.7

BF

12F

14F

h

N1584.6

1212 0;0 FFFFF BBx

NF

FFy

86.13

01584.6698.7;0

14

14

NF

F

M B

11.150867.0

3087.1

060sin*1.0*60cos*1.0*698.7

60cos*)125.005.0(*1584.6;0

12

12

NFB 11.15

+

x

y


Example 1Example 1

AB=10 cm, AG3=BG3=5 cm, =60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2

In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force using algebraic approach required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec.

4B

2A

3

x

y

O

BABOAO

AO BA

BO

)180cos(*0 ABx cos*ABx )180sin(*0 ABy sin*ABy

sin* ABx cos* ABy

cos*sin* 2 ABABx sin*cos* 2 ABABy

sin*AB

x

cossin*

*AB

xABy

sin

cosxy

cos*sin*0 2 ABAB

sin

cos2

sin

sin*sin*cos

sin

cos* 2

2

ABABy

sin

* 2 ABy


Example 1Example 1

D’Alembert’s force

In the figure, an elliptic trammel mechanism is shown with appropriate dimensions, working in the horizontal plane. Put the coupler link into an equivalent form as two point masses concentrated at points A and B, on the basis of equivalency of mass and location of mass center. Then calculate the actuation force using algebraic approach required on the slider at B, parallel to the slide way if point B is moving rightward with constant velocity of 1 m/sec.

sin

* 2 ABy

4B

2A

3

x

O

xy

02

90sin

**

AB

mmaA

Ama

12F

14F

BF

1212 0;0 FFFFF BBx

sin

*0

sin

*;0

2

1414

2 ABmFF

ABmFy

cossin

*

0sin**cos**sin

*;0

2

2

12

12

2

ABmF

ABFABAB

mM B

Dynamics of Reciprocating

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