Chapter 7
Vapor and Gas Power Systems
In this chapter, we will study the basic components of common industrial power and
refrigeration systems. These systems are essentially thermodyanmic cycles in which a
working fluid is alternatively vaporized and condensed as it flows through a set of four
processes and returns to its initial state. We will use the first and second laws of
thermodynamics to analyze the performance of ower and refrigeration cycles.
7.1 Rankine Cycle
We can use a Rankine cycle to convert a fossil-fuel, nuclear, or solar power source into net
electrical power. Figure 7.1-1a shows the components of a Rankine cycle and Figure 7.1-1b
identifies the states of the Rankine cycle on a Ts diagram.
Wt
1
Fuel
air
Turbine
2
QH
Rankine cycle
Boiler
QC
Condenser
4
3
Wp
Pump
Figure 7.1-1a The ideal Rankine cycle
Figure 7.1-1b The path of an ideal Rankine cycle1
1
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 395
7-1
In the ideal Rankine cycle, the working fluid undergoes four reversible processes:
Process 1-2: Isentropic expansion of the working fluid from the turbine from saturated vapor
at state 1 or superheated vapor at state 1’ to the condenser pressure.
Process 2-3: Heat transfer from the working fluid as it flows at constant pressure through the
condenser with saturated liquid at state 3.
Process 3-4: Isentropic compression in the pump to state 4 in the compressed liquid region.
Process 4-1: Heat transfer to the working fluid as it flows at constant pressure through the
boiler to complete the cycle.
We will consider the Rankine cycle with water as the working fluid.
Process 1-2: Assuming that bulk kinetic and potential energy and heat transfer are negligible,
the power produced by the turbine is given by
Wt = m (h1 − h2)
(7.1-1)
In this equation m is the mass flow rate of the working fluid. If the steam enters as a
superheated vapor, it does not condense significantly in the turbine. If the steam is saturated
as it enters the turbine, a significant of liquid is formed which causes erosion and wear of the
turbine blades.
Process 2-3: The steam enters the condenser and exits in state 3 as saturated liquid water.
The change of phase occurs at constant pressure with the heat removed from the flowing
steam given by
Qc = m (h3 − h2)
(7.1-2)
Process 3-4: A pump raises the pressure of the liquid. High-pressure water exits the pump in
state 4 where the work delivered to the liquid is given by
W p = m (h4 − h3) ≈ m v3(p4 − p3)
(7.1-3)
In this equation the pump work is integrated from
Wp
m
=
4
3
vdp ≈ v3(p4 − p3)
The liquid volume is assumed to be constant. Since the specific volume of the liquid is
significantly less than that of the vapor, the work required by the pump is much less than that
produced by the turbine. Typically, a small fraction of the power produced by the turbine is
used to compress the liquid, and the remaining power is the net power obtained by the cycle.
7-2
Process 4-1: The high-pressure liquid is brought back to a saturated or superheated vapor
state in the boiler where the rate of heat transfer to the working fluid is given by
QH = m (h1 − h4)
(7.1-4)
The vapor exits the boiler in state 1 and the cycle is repeated.
Example 7.1-12. ---------------------------------------------------------------------------------Steam enters the turbine in a power plant at 600oC and 10 MPa and is condensed at a
pressure of 0.1 MPa. Assume the plant can be treated as an ideal Rankine cycle. Determine
the power produced per kg of steam and the efficiency of the cycle. Determine the efficiency
of a Carnot cycle operated between these two temperatures.
Solution -----------------------------------------------------------------------------------------Table E7.4-1 States of the ideal Rankine cycle
1
2
3
4
Temp
C
600
99.62
99.62
100.3
Specific
Pressure Volume
MPa
m3/kg
0.03837
10
1.566
0.1
0.001043
0.1
0.001039
10
Internal
Energy
kJ/kg
3242
2349
417.3
417.4
Specific
Enthalpy
kJ/kg
3625
2505
417.4
427.7
Specific
Entropy
kJ/kg/K
6.903
6.903
1.303
1.303
Quality
Phase
Dense Fluid (T>TC)
0.9246 Liquid Vapor Mixture
Saturated Liquid
0
Compressed Liquid
o
1 600 C, 10 MPa
T
4
2
3
0.1 MPa
s
Figure E7.1-1 The processes on Ts diagram.
Table E7.1-1 lists the states of steam in the ideal Rankine cycle with the bold values are the
two properties used to defined the states. The work produced by the turbine per kg of steam
is
Wt = (h1 − h2) = 3625 − 2505 = 1120 kJ/kg
2
Koretsky M.D., Engineering and Chemical Thermodynamics, Wiley, 2004, pg. 140
7-3
The work received by the pump is
Wp = (h4 − h3) = 427.7 − 417.4 = 10.3 kJ/kg
The net work produced by the plant is
Wcycle = Wt − Wp = 1120 − 10.3 = 1109.7 kJ/kg
The rate of heat transfer to the working fluid is given by
QH = (h1 − h4) = 3625 − 427.7 = 3197.3 kJ/kg
The efficiency of the cycle is
η=
Wcycle
QH
=
1109.7
= 0.347
3197.3
The efficiency of a Carnot cycle operated between these two temperatures is
TL
99.6 + 273.15
=1−
= 0.573
TH
600 + 273.15
--------------------------------------------------------------------------------------
ηCarnot = 1 −
Figure 7.1-2 Ts diagram showing irreversibilities in pump and turbine3.
The isentropic turbine and pump efficiencies are given by:
ηt =
h −h
h1 − h2
, and ηp = 4 s 3
h1 − h2 s
h4 − h3
The actual work obtained from the turbine is less than the isentropic work and the actual
work required for the pump is larger than the isentropic work.
3
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 402
7-4
Example 7.1-24. ---------------------------------------------------------------------------------Steam is the working fluid in a modified Rankine cycle where the turbine and pump each
have an isentropic efficiency of 85%. Saturated vapor enters the turbine at 8.0 MPa and
saturated liquid exits the condenser at a pressure of 0.008 MPa. The net power output of the
cycle is 100 MW. Determine for the modified cycle (a) the thermal efficiency, (b) the mass
flow rate of the steam, in kg/h, (c) the rate of heat transfer, QH , into the working fluid as it
passes through the boiler, in MW, (d) the rate of heat transfer, Qc , from the condensing
steam as it passes through the condenser, in MW, (e) the mass flow rate of the condenser
cooling water, in kg/h, if cooling water enters the condenser at 15°C and exits at 35°C.
Solution ------------------------------------------------------------------------------------------
Figure E7.1-2a The ideal Rankine cycle
Figure E7.1-2b The modified Rankine cycle5
Figure E7.1-2a shows the four states in the ideal Rankine cycle with the steam properties
listed in Table E7.1-2. Figure E7.1-2b shows the modified Rankine cycle on the Ts diagram
with the dash line represent the irreversibilities in the turbine and the pump. These
irreversibilities cause an increase in entropy across the turbine and the pump.
4
5
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 396
Moran, M. J. and Shapiro H. N., Fundamentals of Engineering Thermodynamics, Wiley, 2008, pg. 403
7-5
Table E7.1-2 Steam properties at various states in the Rankine cycle.
State
1
2s
3
4s
Specific
Temp Pressure Volume
C
MPa
m3/kg
8
0.02352
295.1
41.51
0.008
12.21
41.51
0.008 0.001008
41.76
8
0.001005
Internal
Energy
kJ/kg
2570
1697
173.8
173.9
Specific
Enthalpy
kJ/kg
2758
1795
173.9
181.9
Specific
Entropy
kJ/kg/K
5.743
5.743
0.5925
0.5925
Quality
Phase
1
Saturated Vapor
0.6745 Liquid Vapor Mixture
0
Saturated Liquid
Compressed Liquid
The isentropic work produced by the turbine per kg of steam is
(W / m )
t
s
= (h1 − h2s) = 2758 − 1795 = 963 kJ/kg
The specific enthalpy at the turbine exit, state 2, can be determined using the turbine
efficiency
Wt / m
h1 − h2
=
h1 − h2 s (Wt / m )
ηt =
s
h2 = h1 − ηt(h1 − h2s) = 2758 − 0.85(2758 − 1795) = 1939.5 kJ/kg
The specific enthalpy at the pump exit, state 4, can be determined using the turbine efficiency
ηp =
h4 s − h3
h4 − h3
h4 = h3 + (h4s − h3)/ηt
h4 = 173.9 + (181.9 − 173.9)/0.85 = 183.3 kJ/kg
The net power produced by the cycle is
Wcycle = Wt − W p = m [(h1 − h2) − (h4 − h3)]
The rate of heat transfer to the working fluid is given by
QH = m (h1 − h4)
(a) The thermal efficiency is
η=
η=
Wcycle
QH
=
( h1 − h2 ) − ( h4 − h3 )
h1 − h4
( 2758 − 1939.5 ) − (183.3 − 173.9 )
2758 − 183.3
7-6
= 0.314
(b) Determine the mass flow rate of the steam
Solving with numerical values, we obtain the net power produced by the cycle per kg/s of
steam
Wcycle / m = (h1 − h2) − (h4 − h3) = (2758 − 1939.5) − (183.3 − 173.9) = 809.1 kJ/kg
Since Wcycle = 100 MW, the mass flow rate of the steam is
m = Wcycle /809.1 kJ/kg =
(100 MW ) (1000 kW/MW)
809.1 kJ/kg
= 123.59 kg/s
m = (123.59 kg/s)(3600 s/h) = 4.449×
×105 kg/h
(c) Determine the rate of heat transfer to the steam, QH
QH = m (h1 − h4) = (123.59 kg/s) (2758 − 183.3) kJ/kg = 3.182×105 kW
QH = (3.182×105 kW)( 1 MW/103 kW) = 318.2 MW
(d) Determine the rate of heat transfer to the condensing water, Qc
Qc = m (h2 − h3) = (123.59 kg/s) (1939.5 − 173.9) kJ/kg = 2.182×105 kW
Qc = (2.182×105 kW)( 1 MW/103 kW) = 218.2 MW
(e) Determine the mass flow rate of the condenser cooling water, in kg/h, if cooling water
enters the condenser at 15°C and exits at 35°C.
1
2
Temp
C
15
35
Specific
Pressure Volume
MPa
m3/kg
0.001705 0.001001
0.005628 0.001006
Internal
Energy
kJ/kg
62.98
146.7
Specific
Enthalpy
kJ/kg
62.98
146.7
Specific
Entropy
kJ/kg/K
0.2245
0.5052
Quality
Phase
0
0
Saturated Liquid
Saturated Liquid
The mass flow rate of the condensing cooling water is given by
mcw =
Qc
2.182 ×105 kJ/s
=
= 2.61×103 kg/s
(146.7 - 62.98) kJ/kg
hcw,out − hcw,in
mcw = (2.61×103 kg/s)(3600 s/h) =9.39×106 kg/h
7-7
7.2 Refrigeration Cycle
The most common refrigeration cycle is the vapor compression cycle shown in Figure 7.2-1.
In step 4 → 1, heat is removed at the temperature TL from the system being refrigerated by
the evaporation of a liquid under the pressure PL. In step 1 → 2, saturated vapor at PL is
compress isentropically to PH where it becomes superheated vapor. In step 2 → 3, heat QH is
transferred to the surrounding by condensation at TH. In step 3 → 4, the cycle is closed by
throttling the liquid to the lower pressure PL.
T
PH
QH
3
Condenser
Throttle
valve
Evaporator
4
2
2
Compressor
Liquid
Isenthalp
1
W
QL
PL
3
4
a b
1
c
Figure 7.2-1 A vapor-compression refrigeration cycle and its Ts diagram.
Vapor
s
The heat transfer between the system and the surroundings can be obtained from the Ts
diagram. Since Q = Tds , the heat effect is the area under the curve representing the path. In
Figure 7.2-1 the heat QH transferred from the refrigerator to the high temperature
environment is the area 2-3-a-c, which is negative. The heat QL removed from the low
temperature system is the area 4-1-c-b and is positive. For the cyclic process
∆U = QL + W − QH = 0
W = QH − QL
(7.2-1)
The efficiency of the refrigeration cycle, called the coefficient of performance COP, is given
by
COP =
QL
W
(7.2-2)
The work required for the refrigeration cycle can be obtained from the Ts diagram
W = QH − QL = (Area 2-3-a-c) − (Area 4-1-c-b)
W = QH − QL = (Area 1-2-3-4) + (Area 3-4-b-c)
The refrigeration cycle shown in Figure 7.2-1 is a semi-reversible cycle since all steps except
throttling are reversible.
7-8
Example 7.2-15---------------------------------------------------------------------------------A vapor compression refrigeration process using NH3 as the working fluid is to operate
between 20 and 80oF. Determine the coefficient of performance for the semi-reversible
operation.
Solution -----------------------------------------------------------------------------------------T
PH
2
Liquid
h4
4
o
h1
Evaporator
QL
h2
1
W
80 F
PL
3
Isenthalp
o
20 F
4
a b
We will first locate the four states of the refrigeration cycle:
State 3: Saturated liquid at 80oF, h3 = h4 = 131.7 Btu/lb.
State 1: Saturated vapor at 20oF, h1 = 616.8 Btu/lb.
State 4: Liquid and vapor mixture at h4 = 131.7 Btu/lb and 20oC
State 2: Superheated vapor at s2 = s1 = 1.295 Btu/lb⋅oR, PH = P2 = 153.1 psia
Making energy balance around the evaporator yields
0 = h4 − h1 + QL
QL = h1 − h4 = 616.8 − 131.7 = 485.1 Btu/lb
Making energy balance around the compressor yields
0 = h1 − h2 + W
W = h2 − h1 = 686.6 − 616.8 = 69.8 Btu/lb
The coefficient of performance is then
COP =
5
QL
485.1
=
= 6.95
W
69.8
Kyle, B.G., Chemical and Process Thermodynamics, Prentice Hall, 1999, pg. 656
7-9
1
c
Vapor
s
Example 7.2-2---------------------------------------------------------------------------------A vapor-compression refrigeration process using ammonia as the working fluid is to operate
between 0oC and 30oC. In step 4 → 1 heat is supplied to the fluid at 0oC under the pressure
p1. The saturated vapor at P1 is then compressed isentropically to p2, where it becomes
superheated vapor, state 2. Removal of heat from this vapor leads to cooling at constant
pressure followed by condensation at 30oC, step 2 → 3. Throttling the saturated liquid at
state 3 to the lower pressure at state 4 closes the cycle. The four states are given (not in any
particular order) as follow
T(oC)
p(MPa)
h(kJ/kg)
s(kJ/kg⋅K)
0
.4293
1442.32
5.3313
0
1.2315
30
1.1668
322.52
1.2005
69.8
1581.49
Determine (a) The heat transfer (kJ/kg) in step 4 → 1, (b) the heat transfer (kJ/kg) in step 2
→ 3, and (c) the work supplied (kJ/kg) by the compressor.
Solution -----------------------------------------------------------------------------------------T
PH
QH
3
Condenser
Throttle
valve
Evaporator
4
Compressor
Liquid
PL
3
Isenthalp
1
W
QL
State
1
2
3
4
2
2
4
a b
Table E7.2-2 The four states in proper order
p(MPa)
h(kJ/kg)
0
0.4293
1442.32
69.8
1.1668
1581.49
30
1.1668
322.52
0
0.4293
322.52
T(oC)
(a) The heat transfer (kJ/kg) in step 4 → 1
QL = 1442.32 − 322.52 = 1119.8 kJ/kg
(b) The heat transfer (kJ/kg) in step 2 → 3
QH = 1581.49 − 322.52 = 1259.0 kJ/kg
(c) The work supplied (kJ/kg) by the compressor.\
Wc = 1581.49 − 1442.32 = 139.17 kJ/kg
7-10
1
c
Vapor
s
s(kJ/kg⋅K)
5.3313
5.3313
1.2005
1.2315