EGR 334
HW Set31_
Spring 2012
Problem 8: 2
Water is the working fluid in an ideal Rankin cycle. Superheated vapor enters the turbine at 10
MPa, 480 C and the condenser pressure is 6 kPa. Determine for the cycle
a) the rate of heat transfer to the working fluid passing through the steam generator, in kJ per kg
of steam flowing
b) the thermal efficiency.
c) the rate of heat transfer from the working fluid passing through the condenser to the cooling
water in kJ per kg of steam flowing.
---------------------------------------------------------------------------------------------------------------H20 with an Ideal Rankine cycle
State 1: p1 = 10 MPa = 100 bar T1 = 480 C
State 2 p2 = 6 kPa = 0.06 bar s2=s1
State 3; p3 = p2 = 0.06 bar Sat Liq.
State 4; s4 = s3 p4 = p1 = 100 bar.
Find other properties of states:
State 1
State 2
p [bar]
100
0.06
T [C]
480
x
super heated
76.92%
v [m3/kg]
h [kJ/kg]
3321.4
2009.8
s [kJ/kg-K]
6.5282
6.5282
Property values were found using the following steps:
1) Read h1 and s1 from Table A-4
2) Let s2 = s1, then calculate x2 using p2 and s2
3) using x2 at p2 = 0.06 bar, find h2
4) Find v3, h3, and s3 at 0.06 bar and sat. liquid
5) let s4 = s3 and find h4 using h4-h3 = v3(p4-p3)
State 3
State 4
0.06
100
sat. liq
0.0010064
151.53
0.5210
comp. liq
h4  h3  v3 ( p4  p3 )  151.53kJ / kg  0.0010064m3 / kg (10000  6)kPa
a) rate of heat transfer through the steam generator.
0  Qs.g .  m(h4  h1 )
Qs. g .
m
 (h1  h4 )  (3321.4  152.5)  3168.9kJ / kg
b) thermal efficiency:

Wturbine  Wpump
Qs. g .
where
Wturbine  m(h1  h2 )  m(3321.4  2009.8)  1311.6m
Wpump  m(h3  h4 )  m(151.53 160.59)  10.06m
161.59
0.5210
kN / m2 kJ
 161.59kJ / kg
kPa kN  m
then:

Wturbine  Wpump
Qs. g .

1311.5m  10.06m
 0.4107  41.07%
3168.9m
c) heat transfer through the condenser:
0  Qcond  m(h2  h3 )
Qcond
 (h3  h2 )  (151.53  2009.8)  1858.3kJ / kg
m
EGR 334
HW Set31_
Spring 2012
Problem 8: 7
Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 16
MPa and the condenser pressure is 8 kPa. The mass flow rate of steam entering the turbine is
120 kg/s. Determine
a) the net power developed in kW.
b) the rate of heat transfer to the steam passing through the boiler in kW.
c) the thermal efficiency.
d) the mass flow rate of condenser cooling water in kg/s,if the cooling water undergoes a
temperature increase of 18 C with negligible pressure change in passing through the condenser.
---------------------------------------------------------------------------------------------------------------H20 in Ideal Rankine Cycle: mdot = 120 kg/s
State 1: p1 = 16 MPa = 160 bar and sat. vapor.
State 2: p2 = 8 kPa= 0.08 bar and s2 = s1
State 3: p3 = p2 = 0.08 bar and sat. liq.
State 4: p4 = p1 = 16 MPa and s4 = s3
Find other properties of states:
State 1
State 2
State 3
p [bar]
160
0.08
0.08
T [C]
347.4
x
sat. vapor
0.6094
sat. liq
v [m3/kg]
0.0010084
h [kJ/kg]
2580.6
1638.2
173.88
s [kJ/kg-K]
5.2455
5.2455
0.5926
Property values were found using the following steps:
1) Read h1 and s1 from Table A-3 for p1 = 160 bar and sat. vapor.
2) Let s2 = s1, then calculate x2 using p2 and s2
3) using x2 at p2 = 0.08 bar, find h2
4) Find v3, h3, and s3 at 0.08 bar and sat. liquid
5) let s4 = s3 and find h4 using h4-h3 = v3(p4-p3)
h4  h3  v3 ( p4  p3 )  173.88kJ / kg  0.0010084m3 / kg (16000  8) kPa
State 4
160
comp. liq
190.01
0.5926
kN / m2 kJ
 190.01kJ / kg
kPa kN  m
-----------------------------------------------------------------------------------------------------------a) net power:
Wnet  Wturbine  Wpump
where
kW
 113088kW
kJ / s
kW
 m(h3  h4 )  (120kg / s)(173.88  190.01)kJ / kg
 1935.6kW
kJ / s
Wturbine  m(h1  h2 )  120kg / s (2580.6  1638.2)kJ / kg
Wpump
then:
Wnet  Wturbine  Wpump  113088 1936  111152kW
b) boiler heat transfer:
0  Qs.g .  m(h4  h1 )
Qs. g .  m(h1  h4 )  (120kg / s )(2580.6  190.01)kJ / kg
kW
 286871kW
kJ / s
c) thermal efficiency:

Wnet 111152

 0.3874  38.7%
Qs. g . 286871
d) mass flow rate of cooling water.
0  msteam (h2  h3 )  mcooling (hA  hB )
0  msteam (h2  h3 )  mcooling ch 20 (TA  TB )
mcooling  msteam
(h2  h3 )
(1638.2  173.88)kJ / kg
 (120kg / s )
 2335kg / s
ch 20 (TA  TB )
(4.18kJ / kg o C )(18o C )
EGR 334
HW Set31_
Spring 2012
Problem 8: 13
The figure provides steady state operating data for a solar power plant that operates on a Rankine
cycle with R134A as its working fluid. The turbine and pump operate adiabatically. The rate of
energy input to the collectors from solar radiation is 0.3 kW per m2 of collector surface area with
60% of the solar input to the collectors absorbed by the refrigerant as it passes through the
collectors. Determine the solar collector surface area in m2 per kW of power developed by the
plant. Discuss possible operational improvements that could reduce the required collector
surface area.
R134A with a Rankine Cycle. Adiabatic Turbine and pump operation
Collectors bring in 0.3 kW/m2 and 60% is transmitted to the fluid.
b) thermal efficiency:

Wturbine  Wpump
Qs. g .
where
Wturbine  m(h1  h2 )  m(276.83  261.01)  15.82m
Wpump  m(h3  h4 )  m(86.78  87.93)  1.15m
Qcollector  m(h1  h4 )  m(276.83  87.93)  188.9m
and:
Wplant  Wturbine  Wpump  15.82 1.15  (14.67kJ / kg )m
and
Qs.g .  0.60Qin
Qin 
Qs. g .
0.60

188.9m
 (314.83kJ / kg )m
0.60
Area of collection
Qin  (0.3kW / m 2 ) Acollector
Qin
(314.83kJ / kg )m kW

 (1049.4m2  s / kg )m
2
2
0.3kW / m
(0.3kW / m ) kJ / s
Then the ration of Area to Net Power out is
Acollector 
Acollector (1049m2  s / kg )m kJ / s

 71.54m2 / kW
Wnet
(14.67kJ / kg )m kW
EGR 334
HW Set31_
Spring 2012
Problem 8: 17
Water is the working fluid in a Rankine cycle. Superheated vapor enters the turbine at 10 MPa,
480 C, and the condenser pressure is 6 kPa. The turbine and pump have isentropic efficiencies
of 80 and 70% respectively. Determine for the cycle.
a) the rate of heat transfer to the working fluid passing through the steam generator in kJ per kg
of steam flowing
b) the thermal efficiency
c) the rate of heat transfer from the working fluid passing through the condenser to the cooling
water in kJ per kg of steam flowing.
----------------------------------------------------------------------------------------------------------H20 as a Rankine cycle.
H20 in Ideal Rankine Cycle: mdot = 120 kg/s
State 1: p1 = 10 MPa = 100 bar and T1 = 480 C.
State 2: p2 = 6 kPa= 0.06 bar
ηturbine = 80%
State 3: p3 = p2 = 0.06 bar and sat. liq.
State 4: p4 = p1 = 100 bar
ηpump = 70%
Find other properties of states:
State 1
p [bar]
T [C]
x
v [m3/kg]
h [kJ/kg]
hs
s [kJ/kg-K]
State 2
State 3
State 4
100
480
super heated
0.06
0.06
100
x2s=76.92%
comp. liq
3321.4
2272.12
2009.8
s2s=6.5282
sat. liq
0.0010064
151.53
6.5282
0.5210
165.90
161.59
s4s=0.5210
Property values were found using the following steps:
1) Read h1 and s1 from Table A-3 for p1 = 100 bar and T1=480.
2) Use p2= 0.06 and s2s=s1 to find x2s
3) Use turbine efficiency to find h2:
h h

h2  h1  issen _ turbine (h1  h2 s )
issen _ turbine  1 2
h1  h2 s
h2  3321.4  0.8(3321.4  2009.8)  2272.12
4) Find v3, h3, and s3 at 0.06 bar and sat. liquid
5) let s4 = s3s and find h4s using h4s-h3 = v3(p4-p3)
6) Use pump efficiency to find the h4:
h h
1

issen _ pump  3 4 s
h4  h3 
h3  h4
issen _ pump
h4  151.53 
(h4 s  h3 )
1
(161.59  151.53)  165.90
0.7
------------------------------------------------------------------------------------------------------------
Find other properties of states:
State 1
p [bar]
T [C]
x
v [m3/kg]
h [kJ/kg]
hs
s [kJ/kg-K]
State 2
State 3
State 4
100
480
super heated
0.06
0.06
100
x2s=76.92%
comp. liq
3321.4
2272.12
2009.8
s2s=6.5282
sat. liq
0.0010064
151.53
6.5282
0.5210
a) rate of heat transfer through the steam generator.
0  Qs.g .  m(h4  h1 )
Qs. g .
m
 (h1  h4 )  (3321.4  165.90)  3155.5kJ / kg
b) thermal efficiency:

Wturbine  Wpump
Qs. g .
where
Wturbine  m(h1  h2 )  m(3321.4  2272.12)  1049.3m
Wpump  m(h3  h4 )  m(151.53 165.90)  14.37m
then:

Wturbine  Wpump
Qs. g .

1049.3m  14.37m
 0.3280  32.8%
3155.5m
c) heat transfer through the condenser:
0  Qcond  m(h2  h3 )
Qcond
 (h3  h2 )  (151.53  2272.12)  2120.6kJ / kg
m
165.90
161.59
s4s=0.5210