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CANKAYA UNIVERSITY
FACULTY OF ENGINEERING AND ARCHITECTURE
MECHANICAL ENGINEERING DEPARTMENT
ME 212 THERMODYNAMICS II
CAHPTER 9
EXAMPLES SOLUTION
1) An air-standard Otto cycle has a compression ratio of 8.5. At the beginning of
compression, .K300TandkPa100p 11 The heat addition per unit mass of air is 1400
kJ/kg. Determine
a) The net work, in kJ per kg of air.
b) The thermal efficiency of the cycle.
c) The mean effective pressure, in kPa
d) The maximum temperature in the cycle, in K
To investigate the effects of varying compression ratio, plot each of the quantities
calculated in parts (a) through (d) for compression ratios ranging from 1 to 12.
Solution:
Known: An air-standard Otto cycle has a known compression ratio and a specified state
at the beginning of compression. The heat addition per unit mass of air is given.
Find: Determine (a) the net work per unit mass of air, (b) the thermal efficiency, (c) the
mean effective pressure, and (d) the maximum cycle temperature. Plot each of these
quantities versus compression ratio.
Schematic & Given Data:
Assumptions:
1) The air in the piston-cylinder assembly is the closed system.
2) The compression and expansion processes are adiabatic.
3) All processes are internally reversible
4) The air is modeled as an ideal gas
5) Kinetic and potential energy effects are negligible
Analysis: Begin by fixing each principle state of the cycle
State 1:
2.621v,kg/kJ07.214uK300T,kPa100p 1r111
State 2:
For isentropic compression
kg/kJ06.503u,K2.688T,Thus
082.735.8
2.621
V
Vvv
22
1
21r2r
State 3:
The specific internal energy 3u is found by using the energy balance for process 2-3
1
2
4
2
3
2
2
s = c
s = c
v
V1/V2 = 8.5
kg/kJ1400m
Q23
p1 = 100 kPa
T1 = 300 K
air
9192.1v,K3.2231T,Thus
kg
kJ06.1903
kg
kJ06.503
kg
kJ1400u
m
Qu
WQ)uu(m
3r3
2
23
3
0
232323
State 4:
For the isentropic expansion
kg/kJ91.892u,K3.1154T,Finally
3132.16)5.8)(9192.1(V
Vv
V
Vvv
44
2
13r
3
43r4r
(a) To find the net work, note that so,QW cyclecycle
kg
kJ16.721)07.21491.892(1400
)uu(m
Q
m
Q
m
Q
m
W14
234123cycle
(b) The thermal efficiency is
%)5.51(515.0)kg/kJ(1400
)kg/kJ(12.721
m/Q
m/W
23
cycle
(c) The displacement volume is ),vv(mVV 2121 so the mean effective pressure is
given by
)v/v1(v
m/W
vv
m/W
VV
Wmep
121
cycle
21
cycle
21
cycle
Evaluating 1v
kPa2.949
m/N10
kPa1
kJ1
m.N10
5.8
11
kg
m861.0
)kg/kJ12.721(mep
Thus
kg/m861.0kJ1
m.N10
m/N10
kPa1
)kPa100(
)K300(kgK
kJ
97.28
314.8
p
RTv
23
3
3
33
23
1
1
1
Plotting for compression ranging from 1 to 12,
3) The pressure and temperature at the beginning of compression of an air-standard
Diesel cycle are 95 kPa and 290 K, respectively. At the end of the heat addition, the
pressure is 6.5 MPa and the temperature is 2000 K.
Determine
a) The compression ratio.
b) The cut off ratio.
c) The thermal efficiency of the cycle.
d) The mean effective pressure, in kPa.
Solution:
Known: An air-standard Diesel cycle has a specified state at the beginning of
compression and a known pressure and temperature at the end of heat addition.
Find: Determine (a) the compression ratio, (b) the cut off ratio, (c) the thermal
efficiency, and (d) the mean effective pressure
Schematic and Given Data:
Assumptions:
1) The air in the piston-cylinder assembly is the closed system.
2) The compression and expansion processes are adiabatic.
3) All processes are internally reversible.
4) The air is modeled as an ideal gas.
5) Kinetic and potential energy effects are negligible.
Analysis: Begin by fixing each principal state in the cycle.
State 1:
2311.1p,1.676v,kg/kJ91.206ukPa95p,K290T 1r1r111
State 2: For the isentropic compression
v
p2 = p3 = 6.5 MPa = 6500 kPa
T3 = 2000 K
p1 = 95 kPa
T1 = 290 K
1
4
2
Air
3 p
kg/kJ19.962h,K926T,59.31v233.8495
6500)2311.1(
p
ppp 222r
1
21r2r
State 3:
776.2v,kg/kJ1.2252hkPa6500P,K2000T 3r333
State 4:
For the isentropic expansion,
kg/kJ36.734u,K971T48.27vV
Vv
9.92000
926
59.31
1.676
T
T
v
v
T
T
V
V
V
V
V
V
V
V
443r
3
44r
3
2
2r
1r
3
2
2
1
3
2
2
1
3
4
(a) The compression ratio is
4.2159.31
1.676
v
v
V
Vr
2r
1r
2
1
b) The cut off ratio
16.2926
2000
T
T
V
Vr
2
3
2
3
c
c) The thermal efficiency is
%)1.59(591.091.1289
46.762
)19.9621.2252(
)91.20636.734()19.9621.2252(
hh
)uu()hh(
m/Q
m/W
23
1423
23
cycle
d) The mean effective pressure is given as
kPa913
m/N10
kPa1
kJ1
Nm10
4.21
11
kg
m8761.0
)kg/kJ46.762(mep
Thus
kg/m8761.0kJ1
Nm10
m/N10
kPa1
)kPa95(
)K290(K.kg
kJ
97.28
314.8
P
RTv
vevaluating
)v/v1(v
m/W
VV
Wmep
23
3
3
33
23
1
11
1
121
cycle
21
cycle
6) An air-standard dual cycle has a compression ratio of 9. At the beginning of
compression, K300TandkPa100p 11 . The heat addition per unit mass of air
is 1400 kJ/kg, with two thirds added at constant volume and the rest at constant
pressure.
Determine
a) The temperatures at the end of each heat addition process, in K.
b) The net work of the cycle per unit mass of air, in kJ/kg
c) The thermal efficiency.
d) The mean effective pressure, in kPa.
Solution:
Known: An air-standard dual cycle has a known compression ratio and a specified state
at the beginning of compression. The heat additions at constant volume and constant
pressure are also given.
Find: Determine (a) the temperatures at the end of each heat addition process, (b) the net
work per unit mass, (c) the thermal efficiency, and (d) the mean effective pressure.
Schematic and Given Data:
Assumptions:
1) The air in the piston-cylinder assembly is the closed system.
2) The compression and expansion processes are adiabatic.
3) All processes are internally reversible.
4) The air is modeled as an ideal gas.
5) Kinetic and potential energy effects are negligible.
Analysis: Begin by fixing principal state of the cycle
5
2
3 4
v
Qin/m = 1400 kJ/kg
Q23/m = 933.3 kJ/kg
Q34/m = 466.7 kJ/kg
V1/V2 = 9
p1 = 100 kPa
T1 = 300 K
1
Air
P
State 1:
2.621v,kg/kJ07.214uK300T 1r11
State 2:
For the isentropic compression, 022.69v)V/V(v 1r122r
Thus, kg/kJ4.514uandK7.702T 22
State 3: For the heat addition process from 2 to 3:
kg/kJ7.14474.5143.933um/Qu
or
WQ)uu(m
2233
0
232323
(a) Thus, kg/kJ1952handK5.1758T 33
State 4:
For the heat addition process from 3 to 4:
237.2vandK8.2132T,Thus
kg/kJ7.24187.4661952m/Qhh
or
hhm/Q
4r4
3434
3434
State 5: For the isentropic expansion
5997.16vT
T
V
Vv
V
V
V
Vv)V/V(v 4r
4
3
2
14r
4
3
2
14r455r
kg/kJ24.887uandK8.1147T,Thus 55
(b) For the cycle, Thus.QW cyclecycle
kg/kJ83.726)07.21424.887(1400
uum/Qm/Qm/Qm/Qm/W 15in513423cycle
c) The thermal efficiency is
%)9.51(519.01400
83.726
m/Q
m/W
in
cycle
d) The mean effective pressure is given by
kPa7.949
m/N10
kPa1
kJ1
Nm10
)9/11)(kg/m86096.0(
)kg/kJ83.726(mep
Thus
kg/m86096.0kJ1
Nm10
m/N10
kPa1
)kPa100(
)K300(kgK
kJ
97.28
314.8
P
RTv
vEvaluating
)v/v1(v
m/W
VV
Wmep
23
3
3
33
23
1
11
1
121
cycle
21
cycle
6) Air enters the compressor of an ideal air-standard Brayton cycle at 100 kPa, 300 K,
with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10. For turbine
inlet temperatures ranging from 1000 to 1600 K, plot
a) The thermal efficiency of the cycle.
b) The back work ratio.
c) The net power developed, in kW.
Solution:
Known: Air enters the compressor of an ideal Brayton cycle with known conditions. The
compressor pressure ratio is also known.
Find: Plot for various turbine inlet temperatures (a) the thermal efficiency, (b) the back
work ratio, (c) the net power developed.
Schematic and Given Data:
Assumptions:
1) Each component is analyzed as a control volume at steady state. The control
volumes are shown on the accompanying sketch by dashed lines.
2) The turbine and compressor processes are isentropic
3) There are no pressure drops for flow through the heat exchangers.
4) Kinetic and potential energy effects are negligible.
5) The working fluid is air modeled as an ideal gas.
Analysis: Sample calculations are given below for several turbine inlet temperatures.
First, fix each of the principal states.
P1 = 100 kPa
T1 = 300 K
(AV)1 = 5 m3/s
p2/p1 = 10 T3 = 1000 K
T3 = 1200 K
T3 = 1600 K
Compressor
T3’
T3
4
3
4’
3’
2
1
s
T
cycleW
4
3 2
1
outQ
Heat
Exchanger
Turbine
Heat
Exchanger
inQ
State 1: 3860.1pr,kg/kJ19.300hK300T 111
State 2: For the isentropic compression, 860.13prp/pp 1122r
Thus, kg/kJ86.579handK1.574T 22
State 3:
2.791pr,kg/kJ57.1757hK1600
0.238pr,kg/kJ79.1277hK1200
0.114pr,kg/kJ04.1046hK1000
T
33
33
33
3
State 4: For the isentropic expansion, 443344 h,Tprp/ppr
Thus
)K1600(kg/kJ65.945h,K3.911T,12.79
)K1200(kg/kJ84.675h,K665T,8.23
)K1000(kg/kJ45.548h,K9.543T,4.11
pr
44
44
44
4
(a) The thermal efficiency is
23
14
hh
hh1
(b) The back work ratio is
4343
12
hh
67.279
hh
hhbwr
c) The power is
67.279)hh(RT
p)AV(hhhhmW 43
1
111243net
The results are summarized in the plots, which give each of these quantities versus
turbine inlet temperatures ranging from 1000 to 1600K.
7) The compressor and turbine of a simple gas turbine each have isentropic efficiencies of
82%. The compressor pressure ratio is 12. The minimum and maximum temperatures are
290 K and 1400 K, respectively. On the basis of an air-standard analysis, compare the
values of
a) The net work per unit mass of air flowing, in kJ/kg.
b) The heat rejected per unit mass of air flowing, in kJ/kg, and
c) The thermal efficiency to the same quantities evaluated for an ideal cycle.
Solution:
Known: An air-standard gas turbine cycle has a known compressor pressure ratio and
specified minimum and maximum temperatures. The compressor and turbine each have
isentropic efficiencies of 82%.
Find: Determine (a) the net work per unit mass of air flow, (b) the heat rejected per unit
mass of air flow, and (c) the thermal efficiency and compare them to the same quantities
evaluated for an ideal cycle.
Schematic and Given Data:
Assumptions:
1) Each component is analyzed as a control volume at steady state.
2) The compressor and turbine are adiabatic.
3) There are no pressure drops for flow through the heat exchangers.
4) Kinetic and potential energy effects are negligible.
5) The working fluid is air modeled as an ideal gas.
Analysis: First, fix each of the principal states for each cycle using data from Tables.
State 1:
2311.1p,kg/kJ16.290hK290T 1r11
State 2: First, determines2h . For isentropic compression
82.0c
T1 = 290 K
82.0t
T3 = 1400 K p2/p1 = 12
T1 = 290 K 1
2 2s
3
4 4s
T3 = 1400 K
s
T
Compressor
cycleW
4
3 2
1
outQ
Heat
Exchanger
Turbine
Heat
Exchanger
inQ
kg/kJ47.590h7732.14pp/pp s21r12s2r
Using the compressor efficiency; ;hh/hh 121s2c
kg/kJ39.656hh
hhc
1s2
12
State 3;
5.450p,kg/kJ42.1515hK1400T 3r33
State 4: First, determine s4h . For isentropic expansion
kg/kJ38.768h542.37Pp/pp s43r34s4r
Using the turbine efficiency; ;hh/hh s4343t
kg/kJ85.902h)hh(hh 4s43t34
(a) The net work per unit mass of air flowing is
kg/kJ3.246
kg
kJ)16.29039.656(
kg
kJ)85.9024.1515(hhhh
m
W1243
cycle
For the ideal cycle
kg/kJ7.446hhhhm
W1s2s43
ideal
cycle
Thus, irreversibilities reduce the net work by nearly one half
(b) The heat rejected per unit mass of air flowing is
kg/kJ7.612kg
kJ16.290
kg
kJ85.902hh
m
Q14
out
For the ideal cycle
kg/kJ2.478hhm
Q1s4
ideal
out
Thus, less heat is rejected for the ideal cycle.
c) The thermal efficiency is
%)7.28(287.039.6564.1515
7.6121
hh
hh1
23
14
For the ideal cycle
%)3.48(483.047.59042.1515
2.4781
hh
hh1
s23
1s4
Thus, irreversibilities cause a substantial decrease in thermal efficiency.
8) Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric
flow rate 5 m3/s. The compressor pressure ratio is 10 and its isentropic efficiency is 85%.
At the inlet to the turbine, the pressure is 950 kPa, and temperature is 1400 K. The
turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis
of an air-standard analysis, evaluate all availability inputs, destructions, and losses. Let T0
= 300 K, p0 = 100 kPa.
Solution:
Known: A simple gas turbine is analyzed on an air-standard basis from an availability
viewpoint. Data are known at various locations.
Find: Determine all availability inputs, destructions, and losses.
Schematic and given data:
p1 = 100 kPa
T1 = 300 K
s/m5AV 3
1
p4 = 100 kPa
p = 1000 kPa
p = 100 kPa
85.0c
88.0t
T3 = 1400 K
p3 = 950 kPa p2/p1 = 10
950 K
1
2 2s
3
4 4s
s
T
Compressor
cycleW
4
3 2
1
outQ
Heat
Exchanger
Turbine
Heat
Exchanger
inQ
Assumptions:
1) Each component is analyzed as a control volume at steady state.
2) The compressor and turbine are adiabatic.
3) Kinetic and potential energy effects are negligible.
4) The working fluid is air modeled as an ideal gas.
5) Let kPa100pandK300T 00
Analysis: Data are obtained for each principal state from Tables
State T(K) P(kPa) h(kJ/kg) 0s (kJ/kg.K)
1 300 100 300.19 1.70203
2 621.1 1000 629.21 2.44539
3 1400 950 1515.42 3.36200
4 873.9 100 903.72 2.81558
The increase in flow availability of the air passing through the heat exchanger is taken as
the net input of availability to the gas turbine.
Input: 23
0
2
0
30232f3f p/plnRss(T)hh(m)ee(m
Evaluating :m
s/kg807.5Nm10
kJ1
kPa1
m/N10
)K300(K.kg97.28
kJ314.8
)kPa100)(s/m5(
RT
p)AV(m
3
233
1
11
)Input(kW3524)ee(m
s/kJ1
kW1
kgK
kJ
1000
950ln
97.28
314.844539.236200.3)K300(
kg
kJ)21.62942.1515(
s
kg807.5)ee(m
2f3f
2f3f
Availability is destroyed due to irreversibilities in the compressor and turbine. Thus
Destructions:
kW6.173p/plnR)ss(mTI
kW8.143p/plnR)ss(mT)ss(mTI
34
0
3
0
40turb
12
0
1
0
20120comp
The net power developed by the cycle represents the output of availability from the cycle,
or
)Output(kW5.1641)hh()hh(mW 1243cycle
Finally, the air enters the gas turbine at00 Tandp . Thus, the change in flow availability
from inlet to exit represents the loss due to the hot air being discharged. Thus,
kW8.1564)p/plnRss(T)hh(m)ee(m:Loss 14
0
1
0
40141f4f
Summarizing
Input: 3524 kW
Disposition
Output: 1641.5 kW
Destroyed: 317.4 Kw
Loss: 1564.8 kW
3523.7 kW
9) A regenerative gas turbine power plant is shown in Figure. Air enters the compressor
at 1 bar, 270C and is compressed to 4 bars. The isentropic efficiency of the compressor is
80%, and the regenerator effectiveness is 90%. All the power developed by the higher-
pressure turbine is used to run the compressor and the lower-pressure turbine provides the
net power output of 97 kW. Each turbine has an isentropic efficiency of 87% and the
temperature at the inlet to the high-pressure turbine is 1200 K. Determine
a) The mass flow rate of air into the compressor, in kg/s.
b) The thermal efficiency
c) The temperature of the air at the exit of the regenerator, in K.
Solution:
Known: In a regenerative gas turbine power plant, a high pressure turbine runs the
compressor and the net power output is provided by a low pressure turbine. Data are
known at various locations.
Find: Determine (a) the mass flow of air into the compressor, (b) the thermal efficiency,
and (c) the temperature of the air at the exit of the regenerator.
Schematic and Given Data:
1 bar, 270C
inQ
T4 = 1200 K
netW Low pressure turbine
High pressure
turbine
Combustor
Compressor
P2 = 4 bar
1 bar
Regenerator
6
3
4
5 1
2
7
Assumptions:
1) Each component is analyzed as a control volume at steady state.
2) The compressors, turbines, and regenerator are adiabatic.
3) There are no pressure drops for flow through the heat exchangers.
4) Kinetic and potential energy effects are negligible.
5) The working fluid is air modeled as an ideal gas.
Analysis: First, fix each of the principal states.
kW97Wnet 87.0t
8.0c
9.0reg
1 bar, 270C
inQ
T4 = 1200 K
Low pressure turbine
High pressure
turbine
Combustor
Compressor
p2 = 4 bar
1 bar Regenerator
6
3
4
5 1
2
7
4 bars
T1 = 300 K
6s 6
5s 5
4
3 2
1
2s
s
T
7
T4 = 1200 K
State 1:
3860.1pr,kg/kJ19.300h,K300T 111
State 2:
For an isentropic compression, kg/kJ49.446h544.5pr)p/p(pr s2112s2
Using the compressor efficiency,
kg/kJ06.483hh
hhhh
hh
c
1s2
12
12
1s2
c
State 4:
238pr,kg/kJ79.1277hK1200T 444
State 5:
The work of the compressor and the high pressure turbine are equal. Thus,
kg/kJ92.1094hhhhh 55412
From the turbine efficiency; kg/kJ59.1067h)hh(hh s5s54t54
Thus, bars065.2pp/ppand86.122p 44rs5r5s5r
From 714.134pkg/kJ92.1094h 5r5
State 6: For isentropic expansion through the low pressure turbine
kg/kJ40.896h237.65)714.134)(065.2/1(pp/pp s65r56s6r
Using the turbine efficiency
kg/kJ21.922)hh(hh s65t56
State 3: Now using the regenerator effectiveness
kg/kJ3.878hhhhhh
hh226reg3
26
23
reg
a) the mass flow rate is the same for each component. Thus, for the low pressure
turbine
s/kg562.0kW1
s/kJ1
kg/kJ)21.92292.1094(
kW97
)hh(
Wm
or
)hh(mW
65
net
65net
b) The thermal efficiency is
%)2.43(432.0)3.87879.1277)(562.0(
97
)hh(m
W
Q
W
34
net
in
net
c) The specific enthalpy 7h is found from an energy balance for the regenerator,
kg/kJ97.526kg
kJ21.922
kg
kJ3.878
kg
kJ06.483hhhh
or
)hh(hh0
6327
7632
Thus, from Table; K2.523T7
10) Air enters the turbine of a gas turbine at 1200 kPa, 1200 K, and expands to 100 kPa
in two stages. Between the stages, the air is reheated at a constant pressure of 350 kPa to
1200 K. The expansion through each turbine stage is isentropic. Determine, in kJ per kg
of air flowing
a) The work developed by each stage
b) The heat transfer for the reheat process
c) The increase in net work as compared to a single stage of expansion with no
reheats.
Solution:
Known: Air expands in two stages through a turbine with reheat between the stages. The
states are specified at the inlet and exit of each component.
Find: Determine per unit mass of air flowing, (a) the work developed by each stage, (b)
the heat transfer for reheat, and (c) the increase in net work compared to a single stage of
expansion with no reheat.
Schematic and Given Data:
Assumptions:
1) Each component volume is at steady state.
2) The turbines operate isentropically.
3) Kinetic and potential energy effects are negligible.
4) The working fluid is air modeled as an ideal gas.
Analysis: First, fix each of the principal states.
State 1: 0.238p,kg/kJ79.1277hK1200T 1r11
State 1: kg/kJ11.912h417.69p)p/p(p 21r122r
State 3: 0.238pp,kg/kJ79.1277hhK1200T 1r3r133
State 4: kg/kJ85.906h68p)p/p(p 43r344r
a) The work developed by each stage is
4
a
2
3 1
s
T
1200 K
100 kPa
350 kPa
1200 kPa
inQ
4 p4 = 100 kPa
netW
p2 = p3 = 350 kPa
T3 = 1200 K
3 2 1
reheater
Turbine
2 Turbine
1
p1 = 1200 kPa
T1 = 1200 K
kg/kJ94.370hhm/W
kg/kJ68.365hhm/W
432t
211t
b) For the reheater
kg/kJ7.365)11.91279.1277(hhm
Q23
in
c) To determine the work for a single stage of expansion, determineah , as follows.
%2.151002.639
21.639)94.37068.365(increase%
and
kg/kJ21.639)hh(m/WThus
kg/kJ58.638h833.19PrP/PPr
a1
a11aa
11) A two-stage air compressor operates at steady state, compressing 10m3/min of air
from 100 kPa, 300 K, to 1200 kPa. An intercooler between the two stages cools the air to
300 K at a constant pressure of 350 kPa. The compression processes are isentropic.
Calculate the power required to run the compressor, in kW, and compare the result to the
power required for isentropic compression from the same inlet state to the same final
pressure.
Solution:
Known: Air is compressed into a two-stage compressor with intercooling between the
stages. Operating pressures and temperatures are given.
Find: Determine the power to run the compressor and compare this to the power required
for isentropic compression from the same inlet state to the same final pressure.
Schematic and Given Data:
Assumptions:
T1 = 300 K 1
c d
3 2 p
v
p1 = 100 kPa
pi = 350 kPa
p2 = 1200 kPa
cW
p2 = 1200 kPa
2
(AV)1 = 10 m3/min
T1 = 300 K
p1 = 100 kPa
Intercooler
outQ
Td = 300 K
d c 1
Compressor
stage 1
Compressor
stage 2
1) The compressor stages and intercooler are analyzed as control volumes at steady state.
The control volumes are shown on the accompanying sketch by dashed lines.
2) The compression processes are isentropic.
3) There is no pressure drop for flow through the intercooler. Kinetic and potential
energy effects are negligible. The air is modeled as an ideal gas.
Analysis: First, fix each of the principal states.
State 1: 3860.1p,kg/kJ19.300hK300T 1r11
State c: kg/kJ77.429h851.4p)p/p(p c1r1crc
State d: 3860.1p,kg/kJ19.300hK300TT rdd1d
State 2: kg/kJ21.427h752.4p)p/p(p 2rdd22r
The mass flow rate is
s/kg1936.0Nm10
kJ1
kPa1
m/N10
s60
min1
)K300(kgK
kJ
97.28
314.8
)kPa100min)(/m10(
RT
p)AV(m
3
233
1
11
The compressor power is
kW68.49W
)19.30021.427)(1936.0(s/kJ1
kW1
kg
kJ)19.30077.429(
s
kg1936.0W
)hh(m)hh(mWWW
c
c
d21c2c1cc
To find the power a single stage of compression, determine 3h as follows:
kg/kJ65.610h632.16pp/pp 31r123r
Thus, kW11.60)hh(mW 13
The decrease in power with two-stage, intercooled compression is
%35.1710011.60
68.4911.60decrease%
12) Air at 22 kPa, and 220 K, and 250 m/s enters a turbojet engine in flight at an altitude
of 10,000 m. The pressure ratio across the compressor is 12. The turbine inlet
temperature is 1400 K, and the pressure at the nozzle exit is 22 kPa. The diffuser and
nozzle processes are isentropic efficiencies of 85 and 88%, respectively, and there is no
pressure drop for flow through the combustor. On the basis of an air-standard analysis,
determine
a) The pressures and temperatures at each principal state, in kPa and K, respectively.
b) The velocity at the nozzle exit, in m/s
Neglect kinetic energy except at the diffuser inlet and the nozzle exit.
Solution:
Known: A turbojet engine is analyzed as an air standard basis. Data are known at various
locations.
Find: Determine (a) the pressures and temperatures at each principal state and (b) the
velocity at the nozzle exit.
Schematic and Given Data:
Assumptions:
1) Each component is analyzed as a control volume at steady state. The control
volumes are shown on the accompanying sketch by dashed lines.
2) There is no pressure drop for flow through the combustor.
3) The turbine work output is just sufficient to drive the compressor.
4) Except at the inlet and exit of the engine, kinetic energy effects can be ignored.
Potential energy effects are negligible throughout.
5) The working fluid is air modeled as an ideal gas.
Analysis: First, fix each of the principal states.
State a: 4690.0p,kg/kJ97.219hkPa22p,K220T raaaa
State 1: An energy balance for the diffuser gives; .h2
Vh0 1
2
a
a Thus,
7454.0pkg/kJ22.251Nm10
kJ1
s/kgm1
N1
2
s/m250
kg
kJ97.219
2
Vhh 1r32
2222
a
a1
Since the process is isentropic, kPa97.34pp/pp ara1r1
State 2: For isentropic compression, 9453.8pp/pp 1r122r
.kg/kJ9.511h s2 With the compressor efficiency
kPa64.419p,kPa12p,kg/kJ9.557hh
hh 12
c
1s2
12
State 3: 5.450p,kg/kJ42.1515hpp,K1400T 3r3233
State 4: For a turbojet, m/Wm/W tc . Thus
kg/kJ92.1166h)hh()hh( s4s43t12
Thus, kPa62.158)p/p(ppand28.170p 3rs4r34s4r
And
81.193pkg/kJ74.1208hhhh 4rs43t34
State 5:
kg/kJ48.699h;881.26p/ppp 5454r5r
b) An energy balance on the nozzle gives
s/m1009
N1
s/kgm1
kJ1
Nm10
kg
kJ)48.69974.1208(2)hh(2V
or
2
Vhh0
23
545
2
5
54
13) A combined gas turbine-vapor power plant has a net power output of 10 MW. Air
enters the compressor of the gas turbine at 100 kPa, 300 K, and is compressed to 1200
kPa. The isentropic efficiency of the compressor is 84%. The conditions at the inlet to the
turbine are 1200 kPa and 1400 K. Air expands through the turbine, which has an
isentropic efficiency of 88%, to a pressure of 100 kPa. The air then passes through the
interconnecting heat exchanger, and is finally discharged at 480 K. Steam enters the
turbine of the vapor power cycle at 8 MPa, 4000C, and expands to the condenser pressure
of 8 kPa. Water enters the pump as saturated liquid at 8 kPa. The turbine and pump have
isentropic efficiencies of 90 and 80%, respectively. Determine
a) The mass flow rates of air and water, each in kg/s.
b) The rate of heat transfer to the combined cycle, in MW.
c) The thermal efficiency of the combined cycle.
Solution:
Known: A combined gas turbine-vapor power plant has a known net power output. Data
are known at various locations in both cycles.
Find: Determine (a) The mass flow rates of water and air, (b) the rate of heat transfer to
the combined cycle, and (c) the overall thermal efficiency.
Schematic and Given Data:
Assumptions:
1) Each component is analyzed as a control volume at steady state.
2) The gas turbine is analyzed on an air-standard basis.
3) The turbines, compressor, pump, and interconnecting heat exchanger operate
adiabatically.
4) Kinetic and potential energy effects are negligible.
5) The working fluid for the gas turbine is air modeled as an ideal gas.
Analysis: First, fix each of the principal states. For the gas turbine cycle:
State 1: 3860.1p,kg/kJ19.300hK300T 1r11
State 2: For isentropic compression,
kg/kJ65.610h632.16pp/pp s21r122r
Using the compressor efficiency,
kg/kJ78.669hh
hhc
1s2
12
State 3: 5.450p,kg/kJ42.1515hK1400T 3r33
State 4: kg/kJ858h88.0,kg/kJ38.768h542.37pp/pp 4ts43r344r
State 5: kg/kJ49.482hK480T 55
For the vapor cycle;
State a: kgK/kJ3634.6s,kg/kJ3.3138hMPa8p,C400T aaa
0
a
State b: kg/kJ9.1989h,7557.0xss,kPa8p bsbsabsb
Thus, with the turbine efficiency, kg/kJ74.2104)hh(hh bsatab
State c: ,kPa8p c saturated liquid kg/kJ88.173hc
State d:
)pp(vhh cdccds
kg/kJ96.183hh
hh
kg/kJ94.181h
Nm10
kJ1
bar1
m/N10bars)08.080(
kg
m100084.1
kg
kJ88.173h
p
cds
cd
ds
3
2533
ds
a) To find the mass flow rates of water stm and air airm , begin with energy and mass
balances on the interconnecting heat exchanger
1217.0)kg/kJ(96.183)kg/kJ(3.3138
)kg/kJ(49.482)kg/kJ(858
hh
hh
m
m
or
)hh(m)hh(m0
da
54
air
st
adst54air
For the gas turbine cycle
)kg/kJ83.287(mhhhhmW air1243airgas
and for the vapor cycle
s/kg041.3m)1271.0(m
s/kg93.23
kg
kJ)5.1023)(1271.0(
kg
kJ)83.287(
s/kJ10000m
)5.1023(m
m)83.287(mW,Thus
)5.1023(mhhhhmW
airst
air
air
st
airnet
stcdbastvap
b) The rate of heat addition to the combined cycle is
MW24.20Q
s/kJ10
MW1
kg
kJ)78.66942.1515(
s
kg93.23hhmQ
in
323airin
c) The overall thermal efficiency is
%)4.49(494.024.20
10
Q
W
in
net
14) Air enters the compressor of an Ericsson cycle at 300 K, 1 bar, with a mass flow rate
of 5 kg/s. The pressure and temperature at the inlet to the turbine are 10 bar and 1400 K,
respectively. Determine
a) The net power developed, in kW.
b) The thermal efficiency.
c) The back work ratio.
Solution:
Known: Air is the working fluid in an Ericsson cycle with data known at various
locations.
Find: Determine a) the net power developed, b) the thermal efficiency, c) the backwork
ratio
Schematic and Given Data:
Assumptions:
1) Each component is analyzed as a control volume at steady state.
2) All processes are internally reversible.
3) The compression and expansion processes are isothermal.
4) Kinetic and potential energy effects are negligible.
5) The air behaves as an ideal gas.
Analysis: a) The turbine power is evaluated using
kW4626s/kJ1
kW1
10
1ln)K1400(
kgK
kJ
97.28
314.8)s/kg5(W
)p/pln(RTmvdpmW
t
121
2
1t
and for the compressor
kW2.991s/kJ1
kW1
1
10ln)K300(
kgK
kJ
97.28
314.8)s/kg5(W
)p/pln(RTmvdpmW
c
343
4
3c
kW3635WWW ctcycle
b) The thermal efficiency is
%)6.78(786.01400
3001
T
T1
1
3
Alternatively, from an energy balance on the turbine;
tin WQ
Thus,
786.04626
3635
Q
W
in
cycle
c) The backwork ratio is
214.04626
2.991
W
Wbwr
t
c
15) Consider an ideal gas-turbine cycle with two stages of compression and two stages of
expansion. The pressure ratio across each stage of compressor and turbine is 3. The air
enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K.
Determine the back work ratio and the thermal efficiency of the cycle, assuming
a) no regenerator is used and
b) a regenerator with 75 percent effectiveness is used. Use constant specific heats at
room temperature
Solution:
Assumptions:
1) The air standard assumptions are applicable.
2) Air is an ideal gas with variable specific heats.
3) Kinetic and potential energy changes are negligible.
Analysis: a) The work inputs to each stage of compressor are identical, so are the work
outputs of each stage of the turbine since this is an ideal cycle. Then
kg/kJ26.411hh)386.1)(3(pp
pp
386.1p
kg/kJ19.300hK300T
421r
1
22r
1r
11
238p
kg/kJ79.1277hhK1200T
5r
755
1200 K
300 K
T
8
7
6
5
4
9
3
2
1
s
%8.36kg/kJ96.1197
kg/kJ72.440
q
W
kg/kJ72.44014.22286.662WWW
kg/kJ96.1197)36.94679.1277()26.41179.1277()hh()hh(q
%5.33kg/kJ86.662
kg/kJ14.222
W
Wr
,Thus
kg/kJ86.662)36.94679.1277(2)hh(2W
kg/kJ14.222)19.30026.411(2)hh(2W
kg/kJ36.946hh33.79)238(3
1p
p
pp
in
net
th
in,cToutnet
6745in
out,T
in,c
bw
65out,T
12in,c
865r
5
6
6r
b) When a regenerator is used, bwr remains the same. The thermal efficiency in this case
becomes
%3.55kg/kJ63.796
kg/kJ72.440
q
W
kg/kJ63.796kg
kJ33.401
kg
kJ96.1197qqq
kg/kJ33.401kg
kJ)26.41136.946)(75.0()hh(q
in
net
th
regenold,inin
49regregen