Chapter 9
Gas Power Cycles
Our study of gas power cycles will involve the study of those heat engines in
which the working fluid remains in the gaseous state throughout the cycle.
We often study the ideal cycle in which internal irreversibilities and
complexities (the actual intake of air and fuel, the actual combustion process,
and the exhaust of products of combustion among others) are removed.
We will be concerned
with how the major
parameters of the cycle
affect the performance of
heat engines. The
performance is often
measured in terms of the
cycle efficiency.
Wnet
η th =
Qin
2
Carnot Cycle
The Carnot cycle was introduced in Chapter 5 as the most efficient heat engine that
can operate between two fixed temperatures TH and TL. The Carnot cycle is
described by the following four processes.
Process
1-2
2-3
3-4
4-1
Description
Isothermal heat addition
Isentropic expansion
Isothermal heat rejection
Isentropic compression
3
Note the processes on both the P-v and T-s diagrams. The areas under the process
curves on the P-v diagram represent the work done for closed systems. The net
cycle work done is the area enclosed by the cycle on the P-v diagram. The areas
under the process curves on the T-s diagram represent the heat transfer for the
processes. The net heat added to the cycle is the area that is enclosed by the cycle
on the T-s diagram. For a cycle we know Wnet = Qnet; therefore, the areas enclosed
on the P-v and T-s diagrams are equal.
4
Thus
or
since
5
We often use the Carnot efficiency as a means to think about
ways to improve the cycle efficiency of actual cycles. One of
the observations about the efficiency of both ideal and actual
cycles comes from the Carnot efficiency.
Thermal efficiency increases with an increase in the average
temperature at which heat is supplied to the system or with a
decrease in the average temperature at which heat is rejected
from the system.
η th , Carnot
TL
= 1−
TH
6
Air-Standard Assumptions
In our study of gas power cycles, we assume that the working fluid is air, and the air
undergoes a thermodynamic cycle even though the working fluid in the actual power
system does not undergo a cycle.
To simplify the analysis, we approximate the cycles with the following assumptions:
1.) The air continuously circulates in a closed loop and always behaves as an ideal
gas.
2.) All the processes that make up the cycle are internally reversible.
3.) The combustion process is replaced by a heat-addition process from an external
source.
4.) A heat rejection process that restores the working fluid to its initial state replaces
the exhaust process.
5.) The cold-air-standard assumptions apply when the working fluid is air and has
constant specific heat evaluated at room temperature (25oC or 77oF).
7
Terminology for Reciprocating Devices
The following is some terminology we need to understand for reciprocating
engines—typically piston-cylinder devices. Let’s look at the following figures
for the definitions of top dead center (TDC), bottom dead center (BDC),
stroke, bore, intake valve, exhaust valve, clearance volume and
displacement volume.
8
The compression ratio r of an engine is the ratio of the maximum volume to the
minimum volume formed in the cylinder.
V max VBDC
r=
=
V min VTDC
The mean effective pressure (MEP) is a
fictitious pressure that, if it operated on
the piston during the entire power stroke,
would produce the same amount of net
work as that produced during the actual
cycle.
Wnet
wnet
MEP =
=
Vmax − Vmin vmax − vmin
9
Otto Cycle: The Ideal Cycle for Spark-Ignition
Engines
Consider the automotive spark-ignition power cycle.
Processes includes:
Intake stroke
Compression stroke
Power (expansion) stroke
Exhaust stroke
Often the ignition and combustion process begins before the
completion of the compression stroke. The number of crank angle
degrees before the piston reaches TDC on the number one piston
at which the spark occurs is called the engine timing. What are
the compression ratio and timing of your engine in your car, truck,
or motorcycle?
10
The ideal air-standard Otto cycle is the ideal cycle that approximates the
spark-ignition combustion engine.
Process
Description
1-2
Isentropic compression
2-3
Constant volume heat addition
3-4
Isentropic expansion
4-1
Constant volume heat rejection
The P-v and T-s diagrams are shown below.
11
The actual and ideal air-standard Otto cycle
The P-v diagrams
12
Thermal Efficiency of the Otto cycle
Wnet Qnet Qin − Qout
Qout
η th =
= 1−
=
=
Qin
Qin
Qin
Qin
Now to find Qin and Qout.
13
The Otto Cycle
14
Apply first law closed system to process 2-3, (V = constant and neglecting
changes in kinetic and potential energies) yielding
For variable specific heats,
For constant specific heats,
15
Apply first law closed system to process 4-1, (V = constant and neglecting
changes in kinetic and potential energies) yielding
For variable specific heats,
For constant specific heats,
16
Thermal Efficiency of the Otto cycle
For variable specific heats,
For constant specific heats,
or
17
Recall processes 1-2 and 3-4 are isentropic, so:
for variable specific heats:
for constant specific heats:
18
For constant specific heats:
Since v3 = v2 and v4 = v1, we see that
T2 T3
=
T1 T4
or
T4 T3
=
T1 T2
and so
becomes
19
The Otto cycle efficiency becomes (for constant specific heats)
η th , Otto
T1
= 1−
T2
Is this the same
as the Carnot
cycle efficiency?
Since process 1-2 is isentropic,
For constant specific heats
where the compression ratio
is r = v1/v2 and then
η th , Otto = 1 −
1
r k −1
20
We see that increasing the compression ratio increases the thermal
efficiency. However, there is a limit on r depending upon the fuel. Fuels
under high temperature resulting from high compression ratios will
prematurely ignite, causing knock.
21
Example 9-1a (Constant Specific Heats)
An Otto cycle having a compression ratio of 9:1 uses air as the working fluid.
Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition
process, 7.5 kJ of heat are added. Determine all T's, P's, ηth, and the mean
effective pressure.
Process Diagrams: Review the P-v and T-s diagrams given above for the
Otto cycle.
22
Example 9-1a (Continued)
Assume constant specific heats with Cv = 0.718 kJ/kg ⋅K, k = 1.4. (Use the
300 K data from Table A-2)
Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9,
23
Example 9-1a (Continued)
The first law closed system for process 2-3 reduces
Qin = mCv ( T3 − T2 )
Let qin = Qin / m and m = V1/v1
RT1
v1 =
P1
kJ
(290 K ) 3
0.287
m kPa
kg ⋅ K
=
kJ
95 kPa
m3
= 0.875
kg
24
Example 9-1a (Continued)
then,
v1
Qin
qin =
= Qin
m
V1
m3
0.875
kg
= 7.5kJ
−3 3
38
. ⋅ 10 m
kJ
= 1727
kg
qin
T3 = T2 +
Cv
kJ
1727
kg
= 698.4 K +
kJ
0.718
kg ⋅ K
= 3103.7 K
25
Example 9-1a (Continued)
Using the combined gas law (V3 = V2)
T3
P3 = P2
= 9.15 MPa
T2
Process 3-4 is isentropic; therefore,
k −1
⎛ V3 ⎞
⎛1⎞
T4 = T3 ⎜ ⎟ = T3 ⎜ ⎟
⎝r⎠
⎝ V4 ⎠
= 1288.8 K
k −1
1.4 −1
⎛1⎞
= (3103.7) K ⎜ ⎟
⎝9⎠
and
26
Example 9-1a (Continued)
Process 4-1 is constant
volume. So the first law for
the closed system gives,
on a mass basis,
Qout = mCv ( T4 − T1 )
Q
qout = out = Cv ( T4 − T1 )
m
kJ
(1288.8 − 290) K
= 0.718
kg ⋅ K
kJ
= 717.1
kg
The first law applied to the cycle gives (Recall Δucycle = 0)
wnet = qnet = qin − qout
kJ
= (1727 − 717.4)
kg
kJ
= 1009.6
kg
27
Example 9-1a (Continued)
The thermal efficiency is then:
As a check,
we note that:
The mean effective pressure is
MEP =
=
Wnet
wnet
=
Vmax − Vmin vmax − vmin
wnet
wnet
wnet
=
=
v1 − v2 v1 (1 − v2 / v1 ) v1 (1 − 1/ r )
1009.6
kJ
kg
m3 kPa
=
= 1298 kPa
3
1
m
kJ
0.875
(1 − )
9
kg
28
Example 9-1b (Variable Specific Heats)
An Otto cycle having a compression ratio of 9:1 uses air as the working fluid.
Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition
process, 7.5 kJ of heat are added. Determine all T's, P's, ηth, and the mean
effective pressure.
Process Diagrams: Review the P-v and T-s diagrams given above for the
Otto cycle.
29
Example 9-1b (Continued)
Process 1-2 is isentropic. Since T1 = 17 + 273 = 290 K, we use Table A-17 to
get vr1 = 676.1, and u1 = 206.91 kJ/kg. Then r = v1/v2 = v4/v3 = 9, and
leads to vr2 = 676.1/9 = 75.122. Using Table A-17 and linear interpolation, we
find that
compared to T2 = 698.4 K, using constant specific heats.
30
Example 9-1b (Continued): Linear Interpolation
31
Example 9-1b (Continued)
Now we still have
RT1
v1 =
P1
kJ
(290 K ) 3
0.287
m kPa
kg ⋅ K
=
kJ
95 kPa
m3
= 0.875
kg
32
Example 9-1b (Continued)
We still have
Then, using Table A-17 and linear interpolation, we have,
Then, conservation of energy gives
which is off the charts in Table A-17. Linear Extrapolation gives
compared to T3 = 3103.7 K using constant specific heats.
33
Example 9-1b (Continued): Linear Interpolation
34
Example 9-1b (Continued): Linear Extrapolation
35
Example 9-1b (Continued)
We also use linear extrapolation to get
Then process 3-4 is isentropic so that
and yielding
vr4 = 9(0.9285) = 8.3565 which
leads to T4 = 1429.6 K compared to
T4 = 1288.8 K using constant
specific heats.
36
Example 9-1b (Continued): Linear Extrapolation
37
Example 9-1b (Continued): Linear Interpolation
38
Example 9-1b (Continued): Linear Interpolation
39
Example 9-1b (Continued)
Conservation of energy
gives
So that
qout = 1122.28 – 206.91 = 915.37 kJ/kg,
compared to 717.1 kJ/kg, for constant specific
heats. Then
wnet = qin – qout = 1727 - 915.37 = 811.63 kJ/kg,
compared to 1009.6 kJ/kg, assuming constant
specific heats.
40
Example 9-1b (Continued)
The thermal efficiency
is then
compared to 0.585, assuming constant specific heats. The MEP is given by
or
compared to 1298 kPa assuming constant specific heats.
41
Summary Of
The Air-Standard Otto Cycle
The air-standard Diesel cycle is the ideal
cycle that approximates the Diesel
combustion engine
Process
Description
1-2 Isentropic compression
2-3 Constant volume heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection
The P-v and T-s diagrams are shown to
the right.
42
Air-Standard Diesel Cycle
The air-standard Diesel cycle is the ideal
cycle that approximates the Diesel
combustion engine
Process
Description
1-2 Isentropic compression
2-3 Constant pressure heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection
The P-v and T-s diagrams are shown to
the right.
43
The Difference Between The Otto and Diesel Cycles
The Otto Cycle
The Diesel Cycle
44
Thermal Efficiency of the Diesel cycle
η th , Diesel
Wnet
Qout
=
= 1−
Qin
Qin
Now to find Qin and Qout.
45
Apply first law closed system to process 2-3, (P = constant and neglecting
changes in kinetic and potential energies) yielding
For variable specific heats,
For constant specific heats,
46
Apply first law closed system to process 4-1, (V = constant and
neglecting changes in kinetic and potential energies) yielding
For variable specific heats,
For constant specific heats,
47
Thermal Efficiency of the Diesel cycle
For variable specific heats,
For constant specific heats,
What are
T3/T2, T4/T1 and T1/T2?
48
What is T3/T2 ?
Treating the Air as an ideal gas leads to
PV
PV
3 3
= 2 2 where P3 = P2
T3
T2
T3 V3
=
= rc
T2 V2
where rc is called the cutoff ratio, defined as V3 /V2, and is a measure of the
duration of the heat addition at constant pressure. Since the fuel is injected
directly into the cylinder, the cutoff ratio can be related to the number of
degrees that the crank rotated during the fuel injection into the cylinder.
49
What is T4/T1 ?
Treating the Air as an ideal gas leads to
PV
PV
4 4
= 1 1 where V4 = V1
T4
T1
T4 P4
=
T1 P1
Recall processes 1-2 and 3-4 are isentropic
So that (assuming constant specific heats)
k
k
k
k
PV
=
PV
and
PV
=
PV
1 1
2 2
4 4
3 3
Since V4 = V1 and P3 = P2, we divide the second equation by the first
equation and obtain
50
What is T1/T2 ?
Recall processes 1-2 is isentropic
So that (for constant specific heats)
51
Thermal Efficiency of the Diesel cycle
Therefore (for constant specific heats) we have,
η th , Diesel
1 T1 ( T4 / T1 − 1)
= 1−
k T2 ( T3 / T2 − 1)
1 T1 r − 1
= 1−
k T2 ( rc − 1)
k
c
= 1−
1
r
k −1
r −1
k ( rc − 1)
k
c
52
Comparison between the efficiencies of the
Otto and Diesel Cycles
(Assuming Constant Specific Heats)
η th , Otto = 1 −
1
r
k −1
When rc > 1 for a fixed r, we have
But, usually, one has
η th , Diesel < η th , Otto
rDiesel > rOtto
and it turns out that
η th , Diesel > η th , Otto
As rc goes to 1, the two become equal so that
53
Comparison between the efficiencies of the Otto and Diesel Cycles
54
Brayton Cycle
The Brayton cycle is the air-standard ideal cycle approximation for the gasturbine engine. This cycle differs from the Otto and Diesel cycles in that the
processes making the cycle occur in open systems or control volumes.
Therefore, an open system, steady-flow analysis is used to determine the
heat transfer and work for the cycle.
We assume the working fluid is air with the air-standard assumptions.
Process
1-2
2-3
3-4
4-1
Description
Isentropic compression (in a compressor)
Constant pressure heat addition
Isentropic expansion (in a turbine)
Constant pressure heat rejection
The T-s and P-v diagrams are the next slide
55
The Brayton Cycle
Process
1-2
2-3
3-4
4-1
Description
Isentropic compression (in a compressor)
Constant pressure heat addition
Isentropic expansion (in a turbine)
Constant pressure heat rejection
56
The Difference Between The Diesel and Brayton Cycles
The Brayton Cycle
The Diesel Cycle
57
The Brayton Cycle
Process
1-2
2-3
3-4
4-1
Description
Isentropic compression (in a compressor)
Constant pressure heat addition
Isentropic expansion (in a turbine)
Constant pressure heat rejection
58
Thermal Efficiency of the Brayton cycle
η th , Brayton
Wnet
Qout
=
= 1−
Qin
Qin
Now to find Qin and Qout.
59
Apply first law closed system to process 2-3, (P = constant and neglecting
changes in kinetic and potential energies) yielding
For variable specific heats,
For constant specific heats,
60
Apply first law closed system to process 4-1, (P = constant and
neglecting changes in kinetic and potential energies) yielding
For variable specific heats,
For constant specific heats,
61
Thermal Efficiency of the Brayton cycle
For variable specific heats,
For constant specific
heats:
or
η th , Brayton
( T4 − T1 )
= 1−
( T3 − T2 )
T1 ( T4 / T1 − 1)
= 1−
T2 ( T3 / T2 − 1)
What are
T3/T2, T4/T1 and T1/T2?
62
Recall processes 1-2 and 3-4 are isentropic, so
Since P3 = P2 and P4 = P1, we see that
T2 T3
T4 T3
=
or
=
T1 T4
T1 T2
The Brayton cycle efficiency then becomes
η th , Brayton
T1
= 1−
T2
Is this the same as the Carnot cycle
efficiency?
Since process
1-2 is isentropic,
where the
pressure ratio
is rp = P2/P1.
63
Thus we have (for constant specific heats)
η th , Brayton = 1 −
1
rp
( k −1)/ k
Extra Assignment
Evaluate the Brayton cycle efficiency by determining the net work directly from the
turbine work and the compressor work. Compare your result with the above
expression. Note that this approach does not require the closed cycle assumption.
64
Example 9-2a (Constant Specific Heats)
The ideal air-standard Brayton cycle operates with air entering the
compressor at 95 kPa, 22oC. The pressure ratio rp is 6:1 and the air leaves
the heat addition process at 1100 K. Determine the compressor work and
the turbine work per unit mass flow, and the cycle efficiency. Assume
constant specific heat properties.
65
Example 9-2a (Continued)
Apply the conservation of energy for steady-flow and neglect changes in
kinetic and potential energies to process 1-2 for the compressor. Note that
the compressor is isentropic.
or, per unit mass
66
Example 9-2a (Continued)
For constant specific heats, the compressor work per unit mass flow is
Since the compressor
(process 1-2) is isentropic
67
Example 9-2a (Continued)
The conservation of energy for
the turbine, process 3-4, yields
for constant specific heats
or
or (for constant specific heats),
68
Example 9-2a (Continued) Since process 3-4 is isentropic, we have
because P3 = P2 and P4 = P1. We thus
see that
Thus,
69
Example 9-2a (Continued)
Conservation of energy
applied to process 2-3 gives
70
Example 9-2a (Continued)
The net work done by the cycle is
wnet = wturb − wcomp
kJ
= (442.5 − 19815
. )
kg
kJ
= 244.3
kg
The cycle efficiency becomes
As a check, we note that
71
Example 9-2b (Variable Specific Heats)
The ideal air-standard Brayton cycle operates with air entering the
compressor at 95 kPa, 22oC. The pressure ratio rp is 6:1 and the air leaves
the heat addition process at 1100 K. Determine the compressor work and
the turbine work per unit mass flow, and the cycle efficiency. Assume
variable specific heat properties.
72
Example 9-2b (Continued)
Apply the conservation of energy for steady-flow and neglect changes in
kinetic and potential energies to process 1-2 for the compressor. Note that
the compressor is isentropic.
or, per unit mass
73
Example 9-2b (Continued)
For variable specific heats, the compressor work per unit mass flow is
We are given that P1 = 95 kPa and rp
= 6, so that P2 = 6(95) = 570 kPa. We
also have P2 = P3 = 570 kPa and P1 =
P4 = 95 kPa. We are also given that
T1 = 295 K (22oC) and T3 = 1100 K.
Using Table A-17 and T1 = 295 K, we
read h1 = 295.17 kJ/kg and
Pr1 = 1.3068.
Using Table A-17 and T3 = 1100 K,
we also read h3 = 1161.07 kJ/kg and
Pr3 = 167.1.
74
75
Example 9-2b (Continued)
Since the compressor (process 1-2) is
isentropic, we may write
Which leads to
Pr2 = 6(1.3068) = 7.8408.
Then using linear interpolation and
Table A-17, we find that h2 = 493.03
kJ/kg and T2 = 490.3 K (see next slide).
We may compare the temperature at 3
to T2 = 492.5 K, obtained using
constant specific heats.
76
77
Example 9-2b (Continued)
compared to 198.15 kJ/kg assuming
constant specific heats.
The conservation of energy for the
turbine, process 3-4, yields for
constant specific heats
or (per unit mass)
Leading to,
78
Example 9-2b (Continued) Since process 3-4 is isentropic, we have
Using: T3 = 1100 K, h3 = 1161.07 kJ/kg,
rp = 6, and Pr3 = 167.1, we get Pr4 = 167.1/6 =
27.85. Using Table A-17 and linear interpolation,
we then get h4 = 706.5 kJ/kg and T4 = 693.7 K,
(see next slide) compared to T4 = 659.1 K using
constant specific heats.
Thus,
compared to 442.5 kJ/kg, using constant specific heats.
79
80
Example 9-2b (Continued)
Conservation of energy
applied to process 2-3 gives
compared to 609.6 kJ/kg using
constant specific heats. We also
have
81
Example 9-2b (Continued)
The net work done by the cycle is
compared to 244.3 kJ/kg using constant specific heats. We can also get
this from qin – qout resulting in 668.04 – 411.33 = 256.71 kJ/kg.
The cycle efficiency then becomes
compared to 0.40 using constant specific heats.
82