Cheap Rate Call girls Malviya Nagar 9205541914 shot 1500 night
Ankit
1. Process 1 2 Isentropic compression
Process 2 3 Constant volume heat addition
Process 3 4 Isentropic expansion
Process 4 1 Constant volume heat rejection
v2
TC
TC
v1
BC
BC
Qout
Qin
Air-Standard Otto cycle
3
4
2
1
v
v
v
v
r ==
Compression ratio:
2. First Law Analysis of Otto Cycle
12 Isentropic Compression
)()( 12
m
W
m
Q
uu in
−−=−
2
1
1
2
1
2
v
v
T
T
P
P
⋅=
AIR
)()( 1212 TTcuu
m
W
v
in
−=−=
23 Constant Volume Heat Addition
m
W
m
Q
uu in
−+=− )()( 23
)()( 2323 TTcuu
m
Q
v
in
−=−=
2
3
2
3
T
T
P
P
=
AIR Qin
TC
1
1
2
1
1
2 −
−
=
= k
k
r
v
v
T
T
3. 3 4 Isentropic Expansion
AIR)()( 34
m
W
m
Q
uu out
+−=−
)()( 4343 TTcuu
m
W
v
out
−=−=
4
3
3
4
3
4
v
v
T
T
P
P
⋅=
4 1 Constant Volume Heat Removal
AIR Qoutm
W
m
Q
uu out
−−=− )()( 41
)()( 1414 TTcuu
m
Q
v
out
−=−=
1
1
4
4
T
P
T
P
=
BC
1
1
4
3
3
4 1
−
−
=
= k
k
rv
v
T
T
4. ( ) ( )
( )23
1243
uu
uuuu
Q
W
in
cycle
th
−
−−−
==η
( ) ( )
23
14
23
1423
1
uu
uu
uu
uuuu
−
−
−=
−
−−−
=
Cycle thermal efficiency:
th
in
th
incycle
r
r
u
mQ
kr
r
VP
Q
P
imep
VV
W
imep ηη
−
−
=
−
=→
−
=
1
/
1
1
1 111121
Indicated mean effective pressure is:
Net cycle work:
( ) ( )1243 uumuumWWW inoutcycle −−−=−=
First Law Analysis Parameters
1
2
1
23
14 1
11
)(
)(
1 −
−=−=
−
−
−= k
v
v
rT
T
TTc
TTc
5. Ideal Diesel Cycle
Air
BC
Qin Qout
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
6. Process 1 2 Isentropic compression
Process 2 3 Constant pressure heat addition
Process 3 4 Isentropic expansion
Process 4 1 Constant volume heat rejection
Air-Standard Diesel cycle
Qin
Qout
2
3
v
v
rc =
Cut-off ratio:
v2
TC
v1
BC
TC
BC
7. 23
1411
hh
uu
mQ
mQ
in
out
cycle
Diesel
−
−
−=−=η
( )
( )
−
−
⋅−= −
1
111
1 1
c
k
c
k
const c
Diesel
r
r
krV
η
For cold air-standard the above reduces to:
Thermal Efficiency
1
1
1 −
−= kOtto
r
ηrecall,
Note the term in the square bracket is always larger than one so for the
same compression ratio, r, the Diesel cycle has a lower thermal efficiency
than the Otto cycle
Note: CI needs higher r compared to SI to ignite fuel
8. How it Works
Comprises of 4 Stages:
• Intake: Inlet valve opens, exhaust closed.
• Compression: Both valves closed. Piston
compresses air upwards. Fuel injected.
• Power: Fuel ignites. Gas forces piston
downwards.
• Exhaust: Inlet valve closed. Exhaust valve
opens. Piston travels upward.
10. First Law Analysis of Diesel cycle
12 Isentropic Compression
)()( 12
m
W
m
Q
uu in
−−=−
2
1
1
2
1
2
v
v
T
T
P
P
⋅=
AIR
)()( 1212 TTcuu
m
W
v
in
−=−=
1
1
2
1
1
2 −
−
=
= k
k
r
v
v
T
T
11. ( )
m
VVP
m
Q
uu in 232
23 )()(
−
−+=−
AIR23 Constant Pressure Heat Addition
)()( 222333 vPuvPu
m
Qin
+−+=
)( 23 hh
m
Qin
−= cr
v
v
T
T
v
RT
v
RT
P ==→==
2
3
2
3
3
3
2
2
Qin
12. )()( 34
m
W
m
Q
uu out
+−=−
AIR
3 4 Isentropic Expansion
)( 43 uu
m
Wout
−=
3
4
3
4
v
v
v
v
r
r
= note v4
=v1
so
cr
r
v
v
v
v
v
v
v
v
v
v
=⋅=⋅=
3
2
2
1
3
2
2
4
3
4
cr
r
r
r
v
v
v
v
==
3
4
3
4
cr
r
T
T
P
P
T
vP
T
vP
⋅=→=
3
4
3
4
3
33
4
44
13. 4 1 Constant Volume Heat Removal
AIR Qout
m
W
m
Q
uu out
−−=− )()( 41
)()( 1414 TTcuu
m
Q
v
out
−=−=
1
1
4
4
T
P
T
P
=
BC
17. The use of the Dual cycle requires information about either:
i) the fractions of constant volume and constant pressure heat addition
(common assumption is to equally split the heat addition), or
ii) maximum pressure P3.
Transformation of rc and α into more natural variables yields
−
−
−
−
−= − 1
111
1 1
11 krVP
Q
k
k
r k
in
c
α
α
1
31
P
P
rk
=α
For the same inlet conditions P1, V1 and the same compression ratio:
DieselDualOtto ηηη >>
For the same inlet conditions P1, V1 and the same peak pressure P3
(actual design limitation in engines):
ottoDualDiesel ηηη >>
18. For the same inlet conditions P1, V1
and the same compression ratio P2/P1:
For the same inlet conditions P1, V1
and the same peak pressure P3:
∫
∫
−=
−=
3
2
1
41
1
Tds
Tds
Q
Q
in
out
thη
Diesel
Dual
Otto
Diesel
DualO
tto
“x” →“2.5”
Pmax
Tmax
Po
Po
Pressure,P
Pressure,P
Temperature,T
Temperature,T
Specific Volume
Specific Volume
Entropy Entropy