EGR 334
HW Set32_
Spring 2012
Problem 8: 21
The figure provides steady state operating data for a vapor power plant using water as the
working fluid. The mass flow rate of water is 12 kg/s. The turbine and pump operate
adiabatically but not reversibly. Determine
a) the thermal efficiency
b) the rates of heat transfer Qin dot and Qoutdot each in kW.
----------------------------H20 system with mdot = 12 kg/s :
a) find Thermal Efficiency:

Wturbine  Wpump
Qs. g .
where
Wturbine  m(h1  h2 )
 (12kg / s)(3422.2  1633.3)kJ / kg
= 21466.8 kJ/s
Wpump  m(h3  h4 )
 (12kg / s)(191.83  199.4)kJ / kg
= -90.84 kJ/s
and
Qs. g .  m(h6  h5 )
 (12kg / s)(3545.3  167.57)kJ / kg
= 40532.8 kJ/kg
then:
W
 Wpump
  turbine
Qs. g .

21467  90.8
 0.5274  52.7%
40532.8
b) rate of heat transfer in and out.
Qs.gen.  m(h6  h5 )
 (12kg / s)(3545.3  167.57)kJ / kg
= 40532.8 kJ/kg
Qcondensor  m(h3  h2 )
 (12kg / s)(191.83  1633.3)kJ / kg
= -17297.6 kJ/kg
EGR 334
HW Set32
Spring 2012
Problem 8: 29
Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the
turbine at 10 MPa, 480 C and the condenser pressure is 6 kPa. Steam expands through the first
stage turbine to 0.7 MPa and then is reheated to 480 C. Determine for the cycle
a) the rate of heat addition in kJ per kg of steam entering the first stage turbine.
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 entering the first stage turbine.
------------------------------------------------------------------------------------------------------------------H20 as Ideal Rankine cycle with Reheat.
State 1: p1 = 20 MPa = 200 bar, T1 = 480 C
State 2: p2 = 0.7 MPa = 7 bar
and s2 = s1
State 3: p3 = p2 = 0.7 MPa = 7 bar and T3 = 480 C
State 4: p4 = 6 kPa = 0.06 bar and s4 = s3
State 5: p5= 6 kPa = 0.06 bar and sat. liquid.
State 6: p6 = 20 MPa
Find property values for the states:
State 1
State 2
State 3
State 4
p [bar]
100
7
7
0.06
T [C]
480
480
x
super heat
0.9619
super heat
0.9413
v [m3/kg]
h [kJ/kg]
3321.4
2684.8
3438.9
2425.6
s [kJ/kg-K] 6.5282
6.5282
7.8723
7.8723
Property values were found using the following steps:
1) Read h1 and s1 from Table A-3 for p1 = 200 bar and T1 = 480 C.
2) Let s2 = s1, then calculate x2 using p2 and s2
3) using x2 at p2 = 7 bar, find h2
4) Find h3, and s3 at 7 bar and T=480 C
5) let s4 = s3 and find x4
6) Using x4 and p4= 0.06 find h4
7) Find v5, h5, and s5 at 0.06 bar and sat. liquid
8) let s6 = s5 and find h6 using h6-h5 = v5(p6-p5)
h6  h5  v5 ( p6  p5 )  151.53kJ / kg  0.0010064m3 / kg (10000  6)kPa
State 5
State 6
0.06
sat liq
comp. liq
151.53
0.5210
161.59
kN / m2 kJ
 161.6kJ / kg
kPa kN  m
-----------------------------------------------------------------------------------------------------------a) rate of heat transfer through the steam generator.
0  Qs.g .  m(h6  h1 )  m(h2  h3 )
Qs. g .
m
 (h1  h6 )  (h3  h2 )  (3321.4  161.6)  (3438.9  2684.8)  3159.8  754.1  3913.9kJ / kg
p [bar]
T [C]
x
v [m3/kg]
h [kJ/kg]
s [kJ/kg-K]
State 1
State 2
State 3
State 4
State 5
100
480
super heat
7
7
480
super heat
0.06
0.06
0.9413
sat liq
comp. liq
3321.4
6.5282
2684.8
6.5282
3438.9
7.8723
2425.6
7.8723
151.53
0.5210
161.59
0.9619
b) thermal efficiency:

WHP _ turbine  WLP _ turbine  Wpump
Qs. g .
where
WhP _ turbine  m(h1  h2 )  m(3321.4  2684.8)  636.6m
WLP _ turbine  m(h3  h4 )  m(3438.9  2425.6)  1013.3m
Wpump  m(h5  h6 )  m(151.53 161.6)  10.1m
then:

WHP _ turbine  WLP _ turbine  Wpump
Qs. g .

(636.6  1013.3  10.1)m
 0.419  41.9%
3913.9m
c) heat transfer through the condenser:
0  Qcond  m(h5  h4 )
Qcond
 (h5  h4 )  (151.53  2425.6)  2274.1kJ / kg
m
State 6
EGR 334
HW Set32_
Spring 2012
Problem 8: 49
Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater
heater. Superheated vapor enters the turbine at 10 MPa, 480 C and the condenser pressure is 6
kPa. Steam expands through the first stage turbine where some is extracted and diverted to a
closed feedwater heater at 0.7 MPa. Condensate drains from the feedwater heater as saturated
liquid at 0.7 MPa and is trapped into the condenser. The feedwater leaves the heater at 10 MPa
and a temperature equal to the saturation temperature at 0.7 MPa. 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 entering the first stage turbine.
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 entering the first stage turbine.
----------------------------------------------------------------------------------------------------------------Regenerative Ideal Rankine Cycle:
State 1: super heated p1 = 10 MPa = 100 bar and T1 = 480 C
State 2: p2 = 0.7 MPa = 7 bar and s2 = s1
State 3: p3 = 6 kPa = 0.06 bar and s3 = s1
State 4: p4 = 6 kPa = 0.06 bar and sat. liq.
State 5: p5 = 10 MPa =100 bar and h5 = h4+v4(p5-p4)
h5  h4  v4 ( p5  p4 )  151.53kJ / kg  0.0010064m3 / kg (10000  6)kPa
kN / m2 kJ
 161.59kJ / kg
kPa kN  m
State 6: p6 = 10 MPa = 100 bar, T6 =Tsat(at p=7 bar) =165.0C, and h6 = hf(T7) = 997.22 kJ/kg
State 7: p7 = 7 bar and sat liq.
State 8: p8 = 7 bar and h8 = h7
p [bar]
T [C]
x
v [m3/kg]
h [kJ/kg]
s [kJ/kg-K]
State 1
State 2
State 3
State 4
State 5
State 6
State 7
State 8
100
480
super heat
7
0.06
0.06
100
100
7
0.06
0.9619
0.7692
3321.4
6.5282
2684.8
6.5282
2009.8
6.5282
sat liq.
0.0010064
151.53
0.5210
161.59
sat liq
0.001108
697.22
697.22
697.22
Mass balance:
m1  m2  m3
1st Law applied to the steam generator:
0  Qs.g .  m1 (h6  h1 )
1st Law applied to the turbine generator:
0  Wturbine  m2 (h1  h2 )  m3 (h1  h3 )
1st Law applied to the condensor:
0  Qcondensor  m2 (h8  h4 )  m3 (h3  h4 )
1st Law applied to the pump:
0  Wpump  m1 (h4  h5 )
1st Law applied to the closed heat exchanger:
0  m1 (h5  h6 )  m2 (h2  h7 )
6 equations with 7 unknowns: m1 , m2 , m3 ,Wturbine ,Wturbine , Qs.g. , Qcondensor
Let m1 = 1 kg/s and then solve for
the other unknowns using IT.
Therefore:
a)
Qs. g .
m1
 h1  h6
 3321.4  697.22  2624.2kW / kg
b)  
Wturbine  Wpump
Qs. g .

1130  10.06
 42.7%
2624
c) Qcondensor  m2 (h4  h8 )  m3 (h4  h3 )
Qcondensor m2
m
0.2695
0.7305

(h4  h8 )  3 (h4  h3 ) 
(151.5  697.22) 
(151.5  2010)
m1
m1
m1
1
1
 147.1  1357.6  1504.7kW / kg
EGR 334
HW Set32_
Spring 2012
Problem 8: 60
Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open
one as shown in the figure. Steam enters the first turbine stage at 12 MPa, 480 C and expands to
2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder
expands through the second stage turbine to 0.3 MPa where an additional amount is extracted
and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the
third stage turbine exits at the condenser pressure of 6 kPa.
Feedwater leaves the closed heater at 210 C, 12 MPa, and condensate exiting as saturated liquid
at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open
feedwater heater. Assume all pumps and turbine stages operate isentropically. 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 entering the first stage turbine.
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 entering the first stage turbine.
Summarize State information:
State
T [C]
p [kPa]
1
480
12000
2
2000
3
300
4
6
5
sat liq.
6
6
300
7
sat liq.
300
8
12000
9
210
12000
10
sat liq.
2000
11
300
v
x
h
s2=s1
s3=s1
s4=s1
0
s6=s5
0
s8=s7
0
For pump 1:
h6  h5  v5 ( p6  p5 )
For pump 2:
h8  h7  v7 ( p8  p7 )
For Trap:
h10  h11
Mass balance:
m1  m2  m3  m4
(1)
st
1 Law applied to the steam generator:
0  Qboiler  m1 (h9  h1 )
(2)
st
1 Law applied to the turbine generator:
0  Wturbine  m2 (h1  h2 )  m3 (h1  h3 )  m4 (h1  h4 )
(3)
1st Law applied to the condenser:
0  Qcondensor  m4 (h4  h5 )
(4)
st
1 Law applied to the pump:
0  Wpump _1  m4 (h5  h6 )
(5)
st
1 Law applied to the pump:
0  Wpump _ 2  m1 (h7  h8 )
(6)
1st Law applied to the open heat exchanger:
0  m4 (h6  h7 )  m3 (h3  h7 )  m2 (h11  h7 )
(7)
st
1 Law applied to the closed heat exchanger:
0  m1 (h8  h9 )  m2 (h2  h10 )
s
(8)
8 equations with 9 unknowns: m1 , m2 , m3 , m4 ,Wturbine ,Wpump1 ,Wpump2 , Qs. g. , Qcondensor
to calculate on a per mass flow basis, let m1 =1kg/s and then solve
Solve using IT:
// problem 8:60
//Set up state properties:
//state 1: superheated steam
T1=480 //C
p1=12000 //kPa
v1= v_PT("Water/Steam", p1, T1)
h1 = h_PT("Water/Steam", p1, T1)
s1 = s_PT("Water/Steam", p1, T1)
//State 2: superheated steam:
p2 =2000 //kPa
s2 = s1
s2 = s_PT("Water/Steam", p2, T2)
T2sat= Tsat_P("Water/Steam", p2)
v2 = v_PT("Water/Steam", p2, T2)
h2 = h_PT("Water/Steam", p2, T2)
//State 3: mixture
p3 =300 //kPa
s3 = s1
s3 = s_PT("Water/Steam", p3, T3)
T3sat= Tsat_P("Water/Steam", p3)
s3 = ssat_Px("Water/Steam", p3, x3)
h3 = hsat_Px("Water/Steam", p3, x3)
v3 = vsat_Px("Water/Steam", p3, x3)
// State 4: mixture
p4 =6 //kPa
s4 = s1
s4 = s_PT("Water/Steam", p4, T4)
T4sat= Tsat_P("Water/Steam", p4)
s4 = ssat_Px("Water/Steam", p4, x4)
h4 = hsat_Px("Water/Steam", p4, x4)
v4 = vsat_Px("Water/Steam", p4, x4)
// state 5: sat. liquid
p5=6
x5=0
T5 = Tsat_P("Water/Steam", p5)
s5 = ssat_Px("Water/Steam", p5, x5)
h5 = hsat_Px("Water/Steam", p5, x5)
v5 = vsat_Px("Water/Steam", p5, x5)
// State 6: compressed liq.
p6= 300 //kPa
h6=h5+v5*(p6-p5)
h6 = h_PT("Water/Steam", p6, T6)
v6 = v_Ph("Water/Steam", p6, h6)
s6 = s_Ph("Water/Steam", p6, h6)
// State 7: sat. liquid
p7=300
x7=0
T7 = Tsat_P("Water/Steam", p7)
s7 = ssat_Px("Water/Steam", p7, x7)
h7 = hsat_Px("Water/Steam", p7, x7)
v7 = vsat_Px("Water/Steam", p7, x7)
//State 8: compressed liquid.
p8= 2000 //kPa
h8=h7+v7*(p8-p7)
h8 = h_PT("Water/Steam", p8, T8)
v8 = v_Ph("Water/Steam", p8, h8)
s8 = s_Ph("Water/Steam", p8, h8)
//state 9: compressd liq.
p9=12000 // kPa
T9= 210 //C
T9sat= Tsat_P("Water/Steam", p9)
h9 = h_PT("Water/Steam", p9, T9)
v9 = v_PT("Water/Steam", p9, T9)
s9 = s_PT("Water/Steam", p9, T9)
//State 10: sat. liquid
p10=2000 //kPa
x10=0
T10 = Tsat_P("Water/Steam", p10)
s10 = ssat_Px("Water/Steam", p10, x10)
h10 = hsat_Px("Water/Steam", p10, x10)
v10 = vsat_Px("Water/Steam", p10, x10)
// State 11: mixture
p11 = 300 // kPa
h11=h10
// throttling process
x11 = x_hP("Water/Steam", h11, p11)
T11 = Tsat_P("Water/Steam", p11)
s11 = ssat_Px("Water/Steam", p11, x11)
v11 = vsat_Px("Water/Steam", p11, x11)
//Mass balance:
m1 = 1 // kg steam
m1=m2+m3+m4
// Energy balances
// Turbine
0 = -Wt+m2*(h1-h2)+m3*(h1-h3) +m4*(h1-h4)
// condenser
0 = Qcon + m4*(h4-h5)
// pump 1
0 = -Wp1+m4*(h5-h6)
// open heater
0 = m3*h3 + m4*h6 + m2*h11 - m1*h7
// pump 2
0 = -Wp2 + m1*(h7-h8)
// closed heater
0 = m1*(h8-h9)+m2*(h2-h100)
// boiler
0 = Qboiler + m1*(h9-h1)
// thermal Efficiency
n=(Wt+Wp1+Wp2)/Qboiler
Answers:
a) the rate of heat transfer to the working fluid passing through the steam generator, in kJ per kg
of steam entering the first stage turbine.
Qboiler = 2395 kJ/kg of steam
b) the thermal efficiency.:
η= 46.15%
c) the rate of heat transfer from the working fluid passing through the condenser to the cooling
water in kJ per kg of steam entering the first stage turbine: Qcondenser = -1290 kJ/kg of steam.