Chapter 6
Example 6.3-33. ---------------------------------------------------------------------------------An insulated tank (V = 1.6628 L) is divided into two equal parts by a thin partition. On the
left is an ideal gas at 100 kP and 500 K; on the right is vacuum. The partition is ruptures and
the gas expands to fill the entire tank. Determine ∆suniv for the process.
Solution ------------------------------------------------------------------------------------------
Well insulated
V = 0.8416L
p = 100 kPa
T = 500 K
Vacuum
Process chosen
to calculate entropy
change for the
system
V = 0.8416L
p = 100 kPa
T = 500 K
Reversible
isothermal
expansion
Since s is a property of the system, it depends only on the final state and the initial state, not
on the process. To calculate ∆suniv, we choose a reversible path by which the system goes
from the same initial state to the same final state as the real system. Applying the energy
balance with negligible change in kinetic and potential energies to this reversible process, we
have:
∆U = Qrev + Wrev
Qrev = − Wrev =
V2
V1
pdV =
V2
V1
nRT
dV = n R T
V
V2
V1
dV
V
Solving with numerical values, we obtain
qrev =
Qrev
V
= R T ln 2
n
V1
qrev = (8.314 J/mol⋅K) (500 K) ln 2 = 2881 J/mol
The change in entropy for the system is given by
∆ssys =
3
δ qrev
T
=
qrev
2881 J/mol
=
= 5.76 J/mol⋅K
T
500 K
Koretsky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 119
6-9
Since there is no heat transfer to the surroundings in the real irreversible process
∆ssurr = 0
The change in entropy for the universe is then
∆suniv = ∆ssys + ∆ssurr = 5.76 J/mol⋅⋅K
6.4 Entropy Rate Balance for Open Systems
Entropy is an extensive property like mass and energy. Therefore streams of matter can
transfer it into or out of an open system or control volume. The unsteady state entropy
balance is give by
dS
=
dt
Qj
j
Tj
+
i
mi si −
me se + σ
e
(6.4-1)
In this equation
dS
= The rate of accumulation of entropy within the control volume
dt
Qj
= The rate of entropy transfer across the control volume boundary due
Tj
to heat tranfer
j
i
mi si −
e
me se = The rate of entropy transfer due to the mass flow rate into
and out of the control volume
σ = The rate of entropy production due to irreversibilities within the
control volume
For steady state, the rate of accumulation of entropy within the control volume is zero and
Eq. (6.4-1) becomes
Qj
0=
j
Tj
+
i
mi si −
e
me se + σ
(6.4-2)
For a control volume at steady state with one inlet and one exit we have
s2 − s1 =
1
m
Qj
j
Tj
+σ
(6.4-3)
6-10
When there is no heat transfer and no irreversibilities within the control volume, the unit
mass passes through the control volume isentropically (no change in entropy).
Example 6.4-14. ---------------------------------------------------------------------------------Steam enters a turbine with a pressure of 30 bar, a temperature of 400oC, and a velocity of
160 m/s. Saturated vapor at 100oC exits with a velocity of 100 m/s. At steady state, the
turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat
transfer between the turbine and its surroundings occurs at an average outer surface
temperature of 350 K. Determine the rate at which entropy is produced within the turbine per
kg of steam flowing, in kJ/kg · K. Neglect the change in potential energy between inlet and
exit.
Solution ------------------------------------------------------------------------------------------
For steady state, the rate of accumulation of entropy within the control volume is given by
Qj
0=
j
Tj
+
i
mi si −
e
me se + σ
For the turbine with heat transfer only at Tb = 350 K and m = mi = me , the entropy balance
becomes
0=
Q
+ m (s1 − s2 ) + σ
Tb
Solving for σ / m we have
σ
m
=−
Q/m
+ (s2 − s1)
Tb
The rate of heat transfer can be obtained from the energy balance
4
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 284
6-11
Q W
V 2 − V12
=
+ (h2 − h1) + 2
m
m
2
Table 6.4-1 Steam properties from CATT2
1
2
Temp
C
400
100
Specific
Pressure Volume
MPa
m3/kg
3
0.09936
0.1013
1.673
Internal
Energy
kJ/kg
2933
2506
Specific
Enthalpy
kJ/kg
3231
2676
Specific
Entropy
kJ/kg/K
6.921
7.355
Quality
Phase
1
Dense Fluid (T>TC)
Saturated Vapor
Using the value from Table 6.4-1 we have
Q
kJ
kJ
1002 − 1602
= 540
+ (2676 − 3231)
+
m
kg
kg
2
m2
1 kJ
2
3
s 10 kg ⋅ m 2 / s 2
Q
kJ
= − 22.8
m
kg
Solving with numerical values for the entropy production, we obtain
σ
m
σ
m
=−
Q/m
+ (s2 − s1)
Tb
=−
- 22.8 kJ/kg
kJ
kJ
+ (7.355 − 6.921)
= 0.4991
350 K
kg ⋅ K
kg ⋅ K
Example 6.4-25 ---------------------------------------------------------------------------------Calculate the horsepower for compressing 5,000 lbs/hr of ethylene from 100 psia, − 40oF to
200 psia. The adiabatic efficiency of the compressor is 75%. Include your calculations and
Mollier chart.
Solution -----------------------------------------------------------------------------------------The Mollier chart (Pressure-Enthalpy diagram) for ethylene can be obtained from the text by
Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 511.
5
Dr. Pang's Notes
6-12
1.6
4B
tu
/lb o
F
s=
100
o
T = - 40 F
Pressure, psia
200
1044 1064
Enthalpy, Btu/lb
Final stage
Initial state
Figure E1 Pressure-enthalpy diagram for ethylene6
The enthalpy of ethylene at 100 psia and − 40oF can be located from the diagram to give
h1 = 1044 Btu/lb
For isentropic compression, ∆s = 0, we follow the curve where s is a constant until it reaches
the line where p = 200 psia to locate a value for the enthalpy
h2 = 1064 Btu/lb
6
Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 511
6-13
Therefore
(h2 − h1)isentropic = 1064 − 1044 = 20 Btu/lb.
The actual increase in enthalpy of the gas is obtained from the adiabatic efficiency
(h2 − h1)actual =
1
(h2 − h1)isentropic = 26.7 Btu/lb
0.75
The horesepower supplied by the compressor to compress 5000 lb/hr of ethylene is
Power = 5000 × 26.7 = 133,500 Btu/hr
Power = (133,500 Btu/hr)(3.9301×10-4 hp/(Btu/hr)) = 52.5 hp
Example 6.4-3 ---------------------------------------------------------------------------------Hydrogen at a temperature of 20oC and an absolute pressure of 1380 kPa enters a compressor
where the absolute pressure is increased to 4140 kPa. If the mechanical efficiency of the
compressor is 55 percent on the basis of an isothermal and reversible operation, calculate the
pounds of hydrogen that can be handled per minutes when the power supplied to the pump is
224 kW. You can use ideal gas law and neglect kinetic energy effects.
Solution -----------------------------------------------------------------------------------------For isothermal compression, the work required for open system to compress the gas is
w = vdp
w=
R
M
p2
p1
p
TdP
RT
=
ln 2
p
M
p1
Using R = 1545 (lbf/ft2)ft3/lbmol⋅oR yields
w=
(1545)( 293 × 1.8) 4140
ln
= 444,040 ft⋅lbf/lbm
2.016
1380
The actual power received by the gas in English unit is given by
Power =
( 224)(0.55)(550)(60)
= 5,445,000 ft⋅lbf/min
0.746
Note: We have use the conversion 1 hp = 0.746 kW = 550 ft⋅lbf
Since Power = m w, the pounds of hydrogen that can be handled per minute is
m =
Power
5,445,000
=
= 12.2 lb/min
w
444,040
6-14
Example 6.4-4 ---------------------------------------------------------------------------------The following closed-loop steam cycle has been proposed to generate work from burning
fuel.
Q
Boiler
Pump
W
Turbine
Wnet
Condenser
Q
The temperature of the burning fuel is 1100oC, and cooling water is available at 15oC. The
steam leaving the boiler is at 20 bar and 700oC, and the condenser produces a saturated liquid
at 0.2 bar. The steam lines are well insulated, the turbine and pump operate reversibly and
adiabatically, and some of the mechanical work generated by the turbine is used to drive the
pump. The steam leaving the turbine is superheated vapor and the steam leaving the
condenser is saturated liquid.
Data:
T(oC)
p(bar)
h(kJ/kg)
s(kJ/kg.oK)
V(m3/kg)
700
20
3917.4
7.9487
60.1
20
254.3
0.8320
67
0.2
2623.2
7.9487
60.1
0.2
251.4
0.8320
0.00102
Determine:
a. The work supplied by the turbine per kg of steam generated in the boiler.
b. Heat discarded in the condenser per kg of steam generated in the boiler
c. The work used by the pump per kg of steam generated in the boiler
d. Heat absorbed in the boiler per kg of steam generated
e. The efficiency of a Carnot cycle operated at this cycle temperature range
Solution -----------------------------------------------------------------------------------------a. The work supplied by the turbine per kg of steam generated in the boiler is
Wturbine = 3917.4 − 2623.2 = 1294.2 kJ
b. Heat discarded in the condenser per kg of steam generated in the boiler is
Qcondenser = 2623.2 − 251.4 = 2371.8 kJ
c. The work used by the pump per kg of steam generated in the boiler is
Wpump= 254.3 − 251.4 = 2.9 kJ
6-15
d. Heat absorbed in the boiler per kg of steam generated is
Qboiler = 3917.4− 254.3 = 3663.1 kJ
e. The efficiency of a Carnot cycle operated at this cycle temperature range is
η=1−
TL
15 + 273.15
=1−
= 0.79015
TH
1100 + 273.15
Example 6.4-57. ---------------------------------------------------------------------------------Components of a heat pump for supplying heated air to a dwelling are shown in the
schematic below. At steady state, Refrigerant 22 enters the compressor at −5°C, 3.5 bar and
is compressed adiabatically to 75°C, 14 bar. From the compressor, the refrigerant passes
through the condenser, where it condenses to liquid at 28°C, 14 bar. The refrigerant then
expands through a throttling valve to 3.5 bar. The states of the refrigerant are shown on the
accompanying T–s diagram. Return air from the dwelling enters the condenser at 20°C, 1 bar
with a volumetric flow rate of 0.42m3/s and exits at 50°C with a negligible change in
pressure. Using the ideal gas model for the air and neglecting kinetic and potential energy
effects, (a) determine the rates of entropy production, in kW/K, for control volumes
enclosing the condenser, compressor, and expansion valve, respectively. (b) Discuss the
sources of irreversibility in the components considered in part (a).
Solution ------------------------------------------------------------------------------------------
Table E6.4-5 Properties of Refrigerant 22 from CATT2
Type
1
2
3
4
7
R-22
R-22
R-22
R-22
Specific Internal Specific Specific
Temp Pressure Volume Energy Enthalpy Entropy Quality
Phase
C
MPa
m3/kg
kJ/kg
kJ/kg
kJ/kg/K
-5
0.35
0.06791
225.9
249.7
0.9566
Superheated Vapor
75
1.4
0.02078
265
294
0.9815
Superheated Vapor
28
1.131 0.000846 77.74
78.7
0.2921
0
Saturated Liquid
-10.34
0.35
0.01487
73.5
78.7
0.3062 0.216 Liquid Vapor Mixture
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 288
6-16
(a) Determine the rates of entropy production, in kW/K, for control volumes enclosing
the condenser, compressor, and expansion valve, respectively.
Condenser: Entropy balance around the condenser gives
0 = mref (s2 − s3) + mair (s5 − s6) + σ cond
The air mass flow rate is given by
mair =
( AV )5
= (AV)5
v5
mair = (0.42 m3/s)
p5
RT5
1 kJ
105 N/m 2
= 0.5 kg/s
3
10 N ⋅ m
8.314 kJ
(293 K)
28.97 kg ⋅ K
The refrigerant mass flow rate is determined from the energy balance around the condenser
o
20 C
5
Air
3
6
o
50 C
2 75oC
Condenser
mref = mair
h6 − h5
h2 − h3
The change in air enthalpy can be evaluated from (h6 − h5) = cp(T6 − T5) with cp = 1.005
kJ/kg⋅K. The properties of compressed liquid refrigerant 22 at 28oC and 14 bar are
approximated by the properties of saturated liquid refrigerant 22 at 28oC (11.31 bar saturation
pressure).
kJ
(50 - 20) C
kg ⋅ K
= 0.070 kg/s
(294 - 78.7) kJ/kg
1.005
mref = (0.5 kg/s)
The change in specific entropy of the air is
s6 − s5 = cp ln
T6
p
− R ln 6
T5
p5
6-17
s6 − s5 = 1.005
kJ
323
1
ln
− R ln = 0.098 kJ/kg⋅K
kg ⋅ K
293
1
Solving with numerical values for the entropy production in the condenser, we obtain
σ cond = mref (s3 − s2) + mair (s6 − s5)
σ cond = (0.07 kg/s)(0.2921 − 0.9815) kJ/kg⋅K + (0.5 kg/s)(0.098 kJ/kg⋅K)
σ cond = 7.42×10-4 kW/K
Compressor: Entropy balance around the compressor gives
0 = mref (s1 − s2) + σ comp
Solving with numerical values for the entropy production in the compressor, we obtain
σ comp = mref (s2 − s1) = (0.07 kg/s)(0.9815 − 0.9566) kJ/kg⋅K = 17×10-4 kW/K
Valve: Entropy balance around the valve gives
0 = mref (s3 − s4) + σ valve
Solving with numerical values for the entropy production in the valve, we obtain
σ valve = mref (s4 − s3) = (0.07 kg/s)(0.3062 − 0.2921) kJ/kg⋅K = 9.87×10-4 kW/K
(b) Discuss the sources of irreversibility in the components considered in part (a).
Components
Compressor
kW/K
17×10-4
Valve
Condenser
9.87×10-4
7.42×10-4
Source of irreversibility
Fluid friction, mechanical friction of the moving parts, and
internal heat transfer.
Fluid friction due to the expansion across the valve
The temperature difference between the air and refrigerant
streams.
6-18