Gas power-09

2
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.
ηth
net
in
W
Q
=

3
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 Description
1-2 Isothermal heat addition
2-3 Isentropic expansion
3-4 Isothermal heat rejection
4-1 Isentropic compression

4
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.

6
ηth Carnot
L
H
T
T
, = −1
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.

7
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).

8
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.

9
The compression ratio r of an engine is the ratio of the maximum volume to the
minimum volume formed in the cylinder.
r
V
V
V
V
BDC
TDC
= =
max
min
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.
MEP
W
V V
w
v v
net net
=
−
=
−max min max min

10
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?

11
The ideal air-standard Otto cycle is the ideal cycle that approximates the
spark-ignition combustion engine.
Process Description
2-3 Constant volume heat addition
4-1 Constant volume heat rejection
The P-v and T-s diagrams are shown below.

12
The actual and ideal air-standard Otto cycle
The P-v diagrams

13
Thermal Efficiency of the Otto cycle
ηth
net
in
net
in
in out
in
out
in
W
Q
Q
Q
Q Q
Q
Q
Q
= = =
−
= −1
Now to find Qin and Qout.

15
For constant specific heats,
For variable specific heats,
Apply first law closed system to process 2-3, (V = constant and neglecting
changes in kinetic and potential energies) yielding

16
Apply first law closed system to process 4-1, (V = constant and neglecting

17
Thermal Efficiency of the Otto cycle
or

18
Recall processes 1-2 and 3-4 are isentropic, so:
for variable specific heats:
for constant specific heats:

19
Since v3 = v2 and v4 = v1, we see that
T
T
T
T
or
T
T
T
T
2
1
3
4
4
1
3
2
=
=
and so becomes
For constant specific heats:

20
The Otto cycle efficiency becomes (for constant specific heats)
ηth Otto
T
T
, = −1 1
2
Since process 1-2 is isentropic,
For constant specific heats
where the compression ratio
is r = v1/v2 and then
ηth Otto k
r
, = − −
1
1
1
Is this the same
as the Carnot
cycle efficiency?

21
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.

22
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.

23
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,

24
The first law closed system for process 2-3 reduces
Q mC T Tin v= −( )3 2
Let qin = Qin / m and m = V1/v1
v
RT
P
kJ
kg K
K
kPa
m kPa
kJ
m
kg
1
1
1
3
3
0 287 290
95
0875
=
=
⋅
=
. ( )
.

25
q
Q
m
Q
v
V
kJ
m
kg
m
kJ
kg
in
in
in= =
=
⋅
=
−
1
1
3
3 3
7 5
0875
38 10
1727
.
.
.
then,
T T
q
C
K
kJ
kg
kJ
kg K
K
in
v
3 2
698 4
1727
0 718
31037
= +
= +
⋅
=
.
.
.

26
Using the combined gas law (V3 = V2) P P
T
T
MPa3 2
3
2
915= = .
Process 3-4 is isentropic; therefore,
1 1 1.4 1
3
4 3 3
4
1 1
(3103.7)
9
1288.8
k k
V
T T T K
V r
K
− − −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠
=
and

27
Process 4-1 is constant
volume. So the first law for
the closed system gives,
on a mass basis,
Q mC T T
q
Q
m
C T T
kJ
kg K
K
kJ
kg
out v
out
out
v
= −
= = −
=
⋅
−
=
( )
( )
. ( . )
.
4 1
4 1
0 718 12888 290
7171
The first law applied to the cycle gives (Recall Δucycle = 0)
w q q q
kJ
kg
kJ
kg
net net in out= = −
= −
=
( . )
.
1727 717 4
1009 6

28
The thermal efficiency is then:
The mean effective pressure is
max min max min
1 2 1 2 1 1
3
3
(1 / ) (1 1/ )
1009.6
1298
1
0.875 (1 )
9
net net
net net net
W w
MEP
V V v v
w w w
v v v v v v r
kJ
m kPakg
kPa
m kJ
kg
= =
− −
= = =
− − −
= =
−
As a check,
we note that:

29
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.

30
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.

31
Example 9-1b (Continued): Linear Interpolation

32
Now we still have
v
RT
P
kJ
kg K
K
kPa
m kPa
kJ
m
kg
1
1
1
3
3
0 287 290
95
0875
=
=
⋅
=
. ( )
.

33
Then, using Table A-17 and linear interpolation, we have,
We still 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.

34

35
Example 9-1b (Continued): Linear Extrapolation

36
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.

37
Example 9-1b (Continued): Linear Extrapolation

38

39

40
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.

41
The thermal efficiency
is then
compared to 0.585, assuming constant specific heats. The MEP is given by
compared to 1298 kPa assuming constant specific heats.
or

42
Summary Of
The Air-Standard Otto Cycle
The air-standard Diesel cycle is the ideal
cycle that approximates the Diesel
combustion engine
Process Description
2-3 Constant volume heat addition
The P-v and T-s diagrams are shown to
the right.

43
Air-Standard Diesel Cycle
The air-standard Diesel cycle is the ideal
cycle that approximates the Diesel
combustion engine
Process Description
2-3 Constant pressure heat addition
The P-v and T-s diagrams are shown to
the right.

44
The Difference Between The Otto and Diesel Cycles
The Diesel CycleThe Otto Cycle

45
Thermal Efficiency of the Diesel cycle
ηth Diesel
net
in
out
in
W
Q
Q
Q
, = = −1

46
Apply first law closed system to process 2-3, (P = constant and neglecting

47
Apply first law closed system to process 4-1, (V = constant and
neglecting changes in kinetic and potential energies) yielding

48
What are
T3/T2, T4/T1 and T1/T2?

49
What is T3/T2 ?
PV
T
PV
T
P P
T
T
V
V
rc
3 3
3
2 2
2
3 2
3
2
3
2
= =
= =
where
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.
Treating the Air as an ideal gas leads to

50
What is T4/T1 ?
PV
T
PV
T
V V
T
T
P
P
4 4
4
1 1
1
4 1
4
1
4
1
= =
=
where
Recall processes 1-2 and 3-4 are isentropic
So that (assuming constant specific heats)
PV PV PV PVk k k k
1 1 2 2 4 4 3 3= =and
Since V4 = V1 and P3 = P2, we divide the second equation by the first
equation and obtain
Treating the Air as an ideal gas leads to

51
What is T1/T2 ?
Recall processes 1-2 is isentropic
So that (for constant specific heats)

52
Therefore (for constant specific heats) we have,
ηth Diesel
c
k
c
k
c
k
c
k
T T T
T T T
k
T
T
r
r
r
r
k r
,
( / )
( / )
( )
( )
= −
−
−
= −
−
−
= −
−
−−
1
1 1
1
1
1 1
1
1
1 1
1
1 4 1
2 3 2
1
2
1

53
As rc goes to 1, the two become equal so that
When rc > 1 for a fixed r, we have
ηth Otto k
r
, = − −
1
1
1
Comparison between the efficiencies of the
Otto and Diesel Cycles
(Assuming Constant Specific Heats)
η ηth Diesel th Otto, ,<
But, usually, one has r rDiesel Otto> and it turns out that
η ηth Diesel th Otto, ,>

54
Comparison between the efficiencies of the Otto and Diesel Cycles

55
Brayton Cycle
The Brayton cycle is the air-standard ideal cycle approximation for the gas-
turbine 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 Description
1-2 Isentropic compression (in a compressor)
3-4 Isentropic expansion (in a turbine)
4-1 Constant pressure heat rejection
The T-s and P-v diagrams are the next slide

56
The Brayton Cycle
Process Description

57
The Difference Between The Diesel and Brayton Cycles
The Diesel CycleThe Brayton Cycle

58
The Brayton Cycle
Process Description

59
Thermal Efficiency of the Brayton cycle
ηth Brayton
net
in
out
in
W
Q
Q
Q
, = = −1

60
Apply first law closed system to process 2-3, (P = constant and neglecting

61
Apply first law closed system to process 4-1, (P = constant and
neglecting changes in kinetic and potential energies) yielding

62
For constant specific
heats:
Thermal Efficiency of the Brayton cycle
What are
T3/T2, T4/T1 and T1/T2?
ηth Brayton
T T
T T
T T T
T T T
,
( )
( )
( / )
( / )
= −
−
−
= −
−
−
1
1
1
1
4 1
3 2
1 4 1
2 3 2
or

63
Recall processes 1-2 and 3-4 are isentropic, so
Since P3 = P2 and P4 = P1, we see that
3 32 4
1 4 1 2
or
T TT T
T T T T
= =
The Brayton cycle efficiency then becomes
ηth Brayton
T
T
, = −1 1
2
Is this the same as the Carnot cycle
efficiency?
Since process
1-2 is isentropic, where the
pressure ratio
is rp = P2/P1.

64
Thus we have (for constant specific heats)
ηth Brayton
p
k k
r
, ( )/
= − −
1
1
1
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.

65
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.

66
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

67
For constant specific heats, the compressor work per unit mass flow is
Since the compressor
(process 1-2) is isentropic

68
The conservation of energy for
the turbine, process 3-4, yields
for constant specific heats
or
or (for constant specific heats),

69
because P3 = P2 and P4 = P1. We thus
see that
Example 9-2a (Continued) Since process 3-4 is isentropic, we have
Thus,

70
applied to process 2-3 gives

71
The net work done by the cycle is w w w
kJ
kg
kJ
kg
net turb comp= −
= −
=
( . . )
.
442 5 19815
244 3
The cycle efficiency becomes
As a check, we note that

72
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.

73
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

74
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.

76
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.

78
The conservation of energy for the
turbine, process 3-4, yields for
constant specific heats
or (per unit mass)
Leading to,
compared to 198.15 kJ/kg assuming

79
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
Example 9-2b (Continued) Since process 3-4 is isentropic, we have
Thus,
compared to 442.5 kJ/kg, using constant specific heats.

81
applied to process 2-3 gives
compared to 609.6 kJ/kg using
constant specific heats. We also
have

82
The net work done by the cycle is
The cycle efficiency then becomes
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.
compared to 0.40 using constant specific heats.

Gas power-09

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Gas power-09

Similar to Gas power-09 (20)

More from UthsoNandy

More from UthsoNandy (20)

Recently uploaded

Recently uploaded (20)

Gas power-09