The document describes four thermodynamic cycles: Otto, Diesel, Dual, and air-standard cycles. It provides equations for calculating work, heat transfer, and efficiency for each cycle. It explains that the Dual cycle generalizes the Otto and Diesel cycles by allowing both constant volume and constant pressure heat addition. It also notes that the Diesel cycle has lower efficiency than the Otto cycle at the same compression ratio but is used in combustion engines because it requires higher compression to ignite fuel.

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
12 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
−=−=
23 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.

9. Internal View of Diesel Engine

10. First Law Analysis of Diesel cycle
12 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 )()(
−
−+=−
AIR23 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

14. Air
TC
BC
Qin Qout
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
Dual
Cycle
Qin
Const volume
heat addition
Process
Thermodynamic Dual Cycle

15. Process 1  2 Isentropic compression
Process 2  2.5 Constant volume heat addition
Process 2.5  3 Constant pressure heat addition
Process 3  4 Isentropic expansion
Process 4  1 Constant volume heat rejection
Dual Cycle
Qin
Qin
Qout
1
1
2
2
2.5
2.5
3
3
4
4
)()()()( 5.2325.25.2325.2 TTcTTchhuu
m
Q
pv
in
−+−=−+−=

16. Thermal Efficiency
)()(
11
5.2325.2
14
hhuu
uu
mQ
mQ
in
out
cycle
Dual
−+−
−
−=−=η
( )





−+−
−
−= −
1)1(
11
1 1
c
k
c
k
cconst
Dual
rk
r
rv αα
α
η
1
1
1 −
−= kOtto
r
η
( )
( )





−
−
⋅−= −
1
111
1 1
c
k
c
k
const c
Diesel
r
r
krV
η
Note, the Otto cycle (rc=1) and the Diesel cycle (α=1) are special cases:
2
3
5.2
3 andwhere
P
P
v
v
rc == α

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