1 me 302 dynamics of machinery dynamics of reciprocating engines dr. sadettin kapucu © 2007...
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1
ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY
Dynamics of Reciprocating Engines
Dr. Sadettin KAPUCU
© 2007 Sadettin Kapucu
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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
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3Gaziantep University
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
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4Gaziantep University
Dynamics of Reciprocating EngineDynamics of Reciprocating Engine
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
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5Gaziantep University
Dynamics of Reciprocating EngineDynamics of Reciprocating Engine
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
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6Gaziantep University
Dynamics of Reciprocating EngineDynamics of Reciprocating Engine
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
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7Gaziantep University
Dynamics of Reciprocating EngineDynamics of Reciprocating Engine
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
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8Gaziantep University
Dynamics of Reciprocating EngineDynamics of Reciprocating Engine
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
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9Gaziantep University
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
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10Gaziantep University
Gas Force ‘Gas Force ‘
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
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11Gaziantep University
Gas Force ‘Gas Force ‘
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'
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12Gaziantep University
Gas Force ‘Gas Force ‘
t
l
rt cos1sinPr'
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13Gaziantep University
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
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14Gaziantep University
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
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15Gaziantep University
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.
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16Gaziantep University
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
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17Gaziantep University
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
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18Gaziantep University
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
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19Gaziantep University
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.
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20Gaziantep University
Inertia Forces’’Inertia Forces’’
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
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21Gaziantep University
Inertia Forces’’Inertia Forces’’
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
''
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22Gaziantep University
Inertia Forces’’Inertia Forces’’
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
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23Gaziantep University
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
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24Gaziantep University
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.
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25Gaziantep University
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
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26Gaziantep University
Example 1 contExample 1 cont
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
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27Gaziantep University
Example 1 contExample 1 cont
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
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28Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
x
N8564.13
BF4
B
2A
3
G3
N8564.13
BF
12F
14F
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29Gaziantep University
Example 1 contExample 1 cont
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
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30Gaziantep University
Example 1 contExample 1 cont
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
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31Gaziantep University
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
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32Gaziantep University
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