ES-7A Thermodynamics
Spring 2003
6-63
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 1 of 9
Entropy of Air
Given: Insulated piston/cylinder device with 300 L of air at 120 kPa and 17 °C. The air is heated by a
200 W resistance heater for 15 minutes at constant pressure.
Find: Entropy change during this process using a) constant specific heat, and b) variable specific heat.
Solution:
a) Using constant specific heat:
The properties of air are: R = 0.2870 kJ/kgK, cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK.
The mass of the air is found using ideal gas law:
m = P1V1/RT 1 = (120×0.3)/(0.2870×290) = 0.4325 kg.
The temperature at state 2 will be found using the 1st Law:
Q = m (u2 – u1) + W = m (h2 – h1) for constant pressure = m cp(T 2 – T 1).
Q = 200W × 15 min × 60 sec/min = 180000 W = 180 kW.
T 2 = Q/mcp + T 1 = 180/(0.4325×1.005) + 290 = 704.1 K.
The entropy change will be found by:

∆ S = m c

p
ln
T2
T1
− R ln
P2 

704 .1

 = 0 .4325 1.005 ln
− 0  = 0.3856 kJ/K.

P1 
290


b) Using variable specific heat:
The mass of the air is still found using the ideal gas law, so m = 0.4325 kg.
At state 1 (290 K), we look up h1 = 290.16 kJ/kg and s1° = 1.66802 kJ/kgK.
From the 1st Law, Q = m(h2 – h1), so h2 = Q/m + h1 = 180/0.4325 + 290.16 = 706.34 kJ/kg.
From the table, we see that this is between 690 K (h = 702.52 kJ/kg, s° = 2.55731 kJ/kgK) and 700
K (h = 713.27 kJ/kg, s° = 2.57277 kJ/kgK). We interpolate to find s2° :
s2° = (h2 – h@690)(s°@700 – s°@690)/(h@700 – h@690) + s°@690
= (706.34 – 702.52)(2.57277 – 2.55731)/(713.27 – 702.52) + 2.55731 = 2.5628 kJ/kgK.
∆S = m(s2° – s1°) = 0.4325(2.5628 – 1.66802) = 0.3870 kJ/K.
ES-7A Thermodynamics
Spring 2003
6-70
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 2 of 9
Reversible adiabatic process w/ air
Given: Air is compressed in a piston-cylinder device from 100 kPa and 17 °C to 800 kPa in a reversible,
adiabatic process.
Find: Final temperature and the work done using a) constant specific heat, and b) variable specific heat.
Solution:
a) Using constant specific heat:
The properties of air are: R = 0.2870 kJ/kgK, cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, k = 1.4.
Since this is a reversible and adiabatic process, we can use the isentropic ratio:
T2
T1
P
=  2
 P1




k −1
k
à T2
P
= T 1  2
 P1




k −1
k
0. 4
 800  1.4
= 290
= 525.32 K.

 100 
To find the work, we’ll use the 1st Law:
Q = 0 = m cv (T 2 – T 1) + W
à w = cv (T 1 – T 2) = 0.718(290 – 525.32) = -168.96 kJ/kg (work consumed).
b) Using variable specific heat:
At state 1, u1 = 206.91 kJ/kg and Pr1 = 1.2311.
Since this is an isentropic process, we’ll use the relative pressure ratio:
P2/P1 = Pr2/Pr1 à Pr2 = Pr1(P2/P1) = 1.2311(800/100) = 9.8488.
Interpolate between 520 K (Pr = 9.684, u = 374.36 kJ/kg) and 530 K (Pr = 10.37, u = 381.84 kJ/kg):
T 2 = (Pr2 – Pr@520)(530 – 520)/(Pr@530 – Pr@520) + 520
= (9.8488 – 9.684)(10)/(10.37 – 9.684) + 520 = 522.4 K.
u2 = (T 2 – 520)(u@530 – u@520)/(530 – 520) + u@520
= (522.4 – 520)(381.84 – 374.36)/10 + 374.36 = 376.16 kJ/kg.
From the 1st Law, m(u2 – u1) + W = 0 à w = (u1 – u2) = 206.91 – 376.16 = -169.25 kJ/kg.
ES-7A Thermodynamics
Spring 2003
6-95
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 3 of 9
Isentropic efficienfy of steam turbine
Given: Steam enters an adiabatic turbine at 8 MPa and 500 °C with a mass flow rate of 3 kg/s, and
leaves at 30 kPa. The turbine has an isentropic efficiency of 90 %.
Find: a) Temperature of the steam at the exit, and b) power output of the turbine.
Solution:
a) Inlet conditions of the steam: h1 = 3398.3 kJ/kg and s1 = 6.7240 kJ/kgK.
The “ideal turbine” has s2s = s1 = 6.7240 kJ/kgK and P2 = 30 kPa. This is a saturated mixture since
at P2, s2s falls between sf and sg.
sf = 0.9439 kJ/kgK, sfg = 6.8247 kJ/kgK, hf = 289.23 kJ/kg, hfg = 2336.1 kJ/kg.
x2s = (s2s – sf)/sfg = (6.7240 – 0.9439)/6.8247 = 0.8469.
h2s = hf + x2s hfg = 289.23 + 0.8469(2336.1) = 2267.78 kJ/kg.
For a turbine, the efficiency is defined as: η = (h2 – h1)/(h2s – h1)
à h2 = η(h2s – h1) + h1 = 0.9(2267.78 – 3398.3) + 3398.3 = 2380.8 kJ/kg.
This is a saturated mixture at 30 kPa, so T 2 = T sat = 69.1 °C.
b) The power output is found from the First Law:
Q& = 0 = m& (h2 − h1 ) + W& → W& = m& (h1 − h 2 ) = 3(3398.3 – 2380.8) = 3052.4 kW.
6-98
Isentropic efficienfy of gas turbine
Given: Argon gas enters an adiabatic turbine at 800 °C and 1.5 MPa at a rate of 80 kg/min and exits at
200 kPa. Power output of the turbine is 370 kW.
Find: Isentropic efficiency of the turbine.
Solution:
Using constant specific heat:
Properties of argon are: R = 0.2081 kJ/kgK, cp = 0.5203 kJ/kgK, cv = 0.3122 kJ/kgK, k = 1.667.
For the “ideal turbine,” use the isentropic ratio to find T 2s :
T 2s
T1
P
=  2
 P1




k −1
k
à T 2s
P
= T 1  2
 P1




k −1
k
0. 667
 200  1.667
= 1073
= 479.15 K.

 1500 
The power output of the ideal turbine is:
W& ideal = m& c p (T 1 − T 2 s ) = 80 ⋅
1 min
60 sec
⋅ 0.5203 (1073 − 479 .15 ) = 411.97 kW.
The isentropic efficiency is:
η=
W& actual
370
=
= 0.898, or 89.8 percent.
&
411 .97
W ideal
ES-7A Thermodynamics
Spring 2003
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 4 of 9
6-100 Compressor w/ R-134a
Given: R-143a enters an adiabatic compressor as saturated vapor at 120 kPa and exits at 1 MPa. The
flow rate at the inlet is 0.3 m3/min, and the isentropic efficiency of the compressor is 80 %.
Find: a) Temperature of the refrigerant at the exit, and b) power input in kW. Show the process on a Ts diagram with respect to the saturation lines.
Solution:
a) The properties at the inlet are: v1 = 0.1614 m3/kg, h1 = 233.86 kJ/kg, s1 = 0.9354 kJ/kgK.
For the ideal case, s2s = s1 = 0.9354 kJ/kgK.
At 1 MPa, this is a superheated vapor between 40 °C (s = 0.9066 kJ/kgK, h = 268.68 kJ/kg) and
50°C (s = 0.9428 kJ/kgK, h = 280.19 kJ/kg):
h2s = (s2s – s@40)(h@50 – h@40)/(s@50 – s@40) + h@40
= (0.9354 – 0.9066)(280.19 – 268.68)/(0.9428 – 0.9066) + 268.68 = 277.84 kJ/kg.
Using the definition for isentropic efficiency, we can find the actual h2:
η=
h 2s − h1
h 2 − h1
→ h2 =
h 2s − h1
η
+ h1 =
277 .84 − 233 .86
+ 233 .86 = 288.83 kJ/kg.
0.8
At 1 MPa, this is a superheated vapor between 50 °C (h = 280.19 kJ/kg) and 60 °C (h = 291.36
kJ/kg).
T 2 = (h2 – h@50)(60 – 50)/ (h@60 – h@50) + 50
= (288.83 – 280.19)(10)/(291.36 – 280.19) + 50 = 57.74 °C.
b) The power input is found from the 1st Law:
Q& = 0 = m& (h2 − h1 ) +W& → W& = m& (h1 − h2 )
The mass flow rate is:
1min
0.3 ⋅ 60
V&
sec
m& =
=
= 0.0309 kg/s.
v1
0.1614
W& = m& (h1 − h 2 ) = 0.0309(233.86 − 288.83) = -1.703 kW (work consumed).
o state 1 is saturated vapor.
T
P2
o ideal case is isentropic (vertical line).
o actual case slants to the right (entropy increases).
o P2 is same for actual and ideal cases.
P1
ideal
actual
The T-s diagram is shown on the right:
s
ES-7A Thermodynamics
Spring 2003
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 5 of 9
6-103 Air compressor
Given: Adiabatic air compressor with inlet conditions of 95 kPa and 27 °C and outlet conditions of 600
kPa and 277 °C.
Find: a) Isentropic efficiency and b) exit temperature for the ideal case, using variable specific heat.
Solution:
Using variable specific heat, the conditions at inlet (T 1 = 300 K) are: h1 = 300.19 kJ/kg, Pr1 = 1.3860.
For the ideal case, we use the relative pressure ratio to find Pr2s:
Pr2s/Pr1 = P2/P1 à Pr2s = Pr1(P2/P1) = 1.3860(600/95) = 8.7537.
This value falls between 500 K (Pr = 8.411, h = 503.02 kJ/kg) and 510 K (Pr = 9.031, h = 513.52 kJ/kg).
T 2s = (Pr2s – Pr@500)(510 – 500)/(Pr@510 – Pr@500) + 500
= (8.7537 – 8.411)(10)/(9.031 – 8.411) + 500 = 505.53 K.
h2s = (T 2s – 500)(h@510 – h@500)/(510 – 500) + h@500
= (505.53 – 500)(513.52 – 503.02)/10 + 503.02 = 508.82 kJ/kg.
The actual h2 is found from T 2 = 550 K: h2 = 555.74 kJ/kg.
The isentropic efficiency is:
η=
h 2s − h 1
h2 − h1
=
508 .82 − 300 .19
= 0.8164, or 81.6 percent.
555 .74 − 300 .19
ES-7A Thermodynamics
Spring 2003
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 6 of 9
6-108 Nozzle with combustion gas – using variable specific heat.
Given: Hot combustion gas enters a nozzle at 260 kPa, 747 °C, and 80 m/s, and exits at 85 kPa. The
nozzle has an isentropic efficiency of 92 percent.
Find: a) exit velocity, and b) exit temperature.
Solution:
a) At the entrance, T 1 = 1020 K, so h1 = 1068.89 kJ/kg and Pr1 = 123.4.
For the ideal case, we will use the relative pressure ratio:
Pr2s = Pr1(P2/P1) = 123.4(85/260) = 40.34.
This is between 760 K (Pr = 39.27, h = 778.18 kJ/kg) and 780 K (Pr = 43.35, h = 800.03 kJ/kg).
h2s = (Pr2s – Pr@760)(h@780 – h@760)/(Pr@780 – Pr@760) + h@760
= (40.34 – 39.27)(800.03 – 778.18)/(43.35 – 39.27) + 778.18 = 783.92 kJ/kg.
To find V2s, we use the 1st Law:
q = 0 = (h 2 s − h 1 ) +
→ V 22s
V 22s −V 12
2000
− 2000 (h 2s − h 1 )
= 802 – 2000(783.92 – 1068.89) = 576340 (m/s)².
= V 12
The actual velocity is found from the isentropic efficiency:
η=
V 22
V 22s
(
→ V 2 = η ⋅V 22s
)
12
= (0 .92 ⋅ 576340 )
12
= 728.17 m/s.
b) From the 1st Law:
q = 0 = (h 2 − h1 ) +
V 22 − V 12
2000
→ h2 = h1 −
V 22 −V 12
2000
= 1068 .89 −
728 .17 2 − 80 2
= 806.97 kJ/kg.
2000
Interpolate between 780 K (h = 800.03 kJ/kg) and 800 K (h = 821.95 kJ/kg) to find T 2:
T 2 = (h2 – h@780)(800 – 780)/(h@800 – h@780) + 780
= (806.97 – 800.03)(20)/(821.95 – 800.03) + 780 = 786.34 K.
ES-7A Thermodynamics
Spring 2003
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 7 of 9
6-108 Nozzle with combustion gas – using constant specific heat.
Given: Hot combustion gas enters a nozzle at 260 kPa, 747 °C, and 80 m/s, and exits at 85 kPa. The
nozzle has an isentropic efficiency of 92 percent.
Find: a) exit velocity, and b) exit temperature.
Solution:
For constant specific heat, we’ll use cp = 1.005 kJ/kgK and R = 0.2870 kJ/kgK.
a) We can use the isentropic ratio to find T 2s :
T 2s
T1
P
=  2
 P1




k −1
k
à T 2s
P
= T 1  2
 P1




k −1
k
 85 
= 1020 

 260 
1. 4 −1
1. 4
= 741.09 K
To find V2s, we use the 1st Law:
q = 0 = (h 2 s − h 1 ) +
V 22s −V 12
2000
→ V 22s = V 12 − 2000c p (T 2s − T 1 )
= 802 – 2000×1.005(741.09 – 1020) = 567190 (m/s)².
The actual velocity is found from the isentropic efficiency:
η=
V 22
V 22s
(
→ V 2 = η ⋅V 22s
)
12
= (0 .92 ⋅ 567190 )
12
= 722.4 m/s.
b) From the 1st Law:
q = 0 = c p (T 2 − T 1 ) +
V 22 − V12
2000
→ T 2 = T1 −
V 22 −V 12
2000 ⋅ c p
= 1020 −
722 .4 2 − 80 2
= 763.5 K.
200 ⋅ 1 .005
ES-7A Thermodynamics
Spring 2003
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 8 of 9
6-152 Air compressor
Given: Air is compressed by a compressor from 100 kPa and 17 °C to 700 kPa at a rate of 5 kg/min.
Find: Minimum power required if the process is a) adiabatic, and b) isothermal. Use variable specific
heats.
Solution:
Using variable specific heat:
a) The initial conditions are: h1 = 290.16 kJ/kg, Pr1 = 1.2311.
For an adiabatic process, minimum power is achieved when the process is isentropic.
Using relative pressure ratios, find Pr2:
Pr2 = Pr1(P2/P1) = 1.2311(700/100) = 8.6177.
Interpolate between 500 K (Pr = 8.411, h = 503.02 kJ/kg) and 510 K (Pr = 9.031, h = 513.32 kJ/kg):
h2 = (Pr2 – Pr@500)(h@510 – h@500)/(Pr@510 – Pr@500) + h@500
= (8.6177 – 8.411)(513.32 – 503.02)/(9.031 – 8.411) + 503.02 = 506.45 kJ/kg.
From the 1st Law:
1 min
Q& = 0 = m& (h 2 − h 1 ) + W& → W& = m& (h1 − h 2 ) = 5 ⋅ 60
(290 .16 − 506 .45 )
sec
= -18.02 kW (consumed).
b) For isothermal process, there is no change in entropy (∆h = 0), so the 1st Law becomes Q& = W& .
The work is minimized when the process is reversible. For a reversible, isothermal process:
Q& = m& T (s 2 − s 1 ) = m& T
 o
P
 s 2 − s 1o − R ln 2

P1



700 
 = 5 ⋅ 1 min ⋅ 290  − 0.2870 ln
 = -13.4965 kW

60 sec
100 


Q& = W& , so the work is -13.47 kW (consumed).
ES-7A Thermodynamics
Spring 2003
HW 4: 6-63, 70, 95, 98, 100, 103, 108, 152, 157
Page 9 of 9
6-157 Compressor with R-134a
Given: A 0.5 kW adiabatic compressor with R-134a. Inlet conditions are 140 kPa and -10 °C, exit
conditions are 700 kPa and 60 °C.
Find: a) isentropic efficiency of the compressor, b) volumetric flow rate at the inlet in L/min, c) maximum
volumetric flow rate at the inlet that this compressor can have without violating the Second Law.
Solution:
a) At the inlet, we have superheated vapor with h1 = 243.40 kJ/kg, v1 = 0.14549 m3/kg, and s1 =
0.9606 kJ/kgK.
At the exit, we have superheated vapor with h2 = 296.69 kJ/kg.
For an ideal compressor, s2s = s1 = 0.9606 kJ/kgK, and at 700 kPa this is superheated vapor between
40 °C (s = 0.9539 kJ/kgK, h = 275.93 kJ/kg) and 50 °C (s = 0.9867 kJ/kgK, h = 286.35 kJ/kg).
Interpolate to find h2s:
h2s = (s2s – s@40)(h@50 – h@40)/(s@50 – s@40) + h@40
= (0.9606 – 0.9539)(286.35 – 275.93)/(0.9867 – 0.9539) + 275.93 = 278.06 kJ/kg.
The isentropic efficiency is:
η=
h 2s − h 1
h2 − h1
=
278 .06 − 243 .40
= 0.650, or 65 percent.
296 .69 − 243 .40
b) The flow rate is round using the 1st Law:
V&
− v 1W&
− W&
− 0 .14549 (− 0.5 )
Q& = 0 = m& (h 2 − h 1 ) + W& → m& = 1 =
→ V&1 =
=
v1
h 2 − h1
h2 − h1
296 .69 − 243 .40
= 0.001365 m3/s
0.001365
c)
m3
sec
sec 1000L
⋅ 60
⋅
= 81.9 L/min.
3
1 min
1m
The maximum volumetric flow rate is achieved when the process is reversible and adiabatic. We
repeat the calculation in part (b) using h2s instead of h2:
− v 1W&
− 0 .14549 (− 0 .5)
V&1 =
=
= 0.002099 m3/s
h 2s − h1
278 .06 − 243 .40
0.002099
m3
sec
sec 1000L
⋅ 60
⋅
= 125.9 L/min.
3
1 min
1m