8.11 Plot each of the quantities calculated in Problem 8.

10 versus condenser pressure ranging

from 0.3 lbf/in.2 to 14.7 lbf/in.2 Maintain constant net power output. Discuss.
______________________________________________________________________________

IT Code
/*
Problem 8.10

KNOWN: Water is working fluid in an ideal Rankine cycle. Data are given in Problem 8.10.

FIND: Plot each of the quantities calculated in Problem 8.10 versus condenser pressure ranging from 0.3
to 14.7 lbf/in^2. Maintain constant net power output.

SCHEMATIC AND GIVEN DATA: See solution to Problem 8.10.

ENGINEERING MODEL:
1. Each component is analyzed as a control volume at steady state.
2. All processes of the working fluid are internally reversible.
3. The turbine and pump operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Condensate exits the condenser as saturated liquid.
6. There is no heat transfer between the outside of the condenser and the surroundings.

*/

p1 = 1400 // lbf/in^2
T1 = 1000 // deg F
p2 = 2 // lbf/in^2
p3 = p2
x3 = 0
p4 = p1
Wdotcycle = 1.0E09 // Btu/h
Twin = 60 // deg F
Twout = 76 // deg F

h1 = h_PT("Water/Steam", p1, T1) // Btu/lb

s1 = s_PT("Water/Steam", p1, T1) // Btu/(lb*deg R)
s2 = s1 // Btu/(lb * deg R)
h2 = h_Ps("Water/Steam", p2, s2) // Btu/lb
h3 = hsat_Px("Water/Steam", p3, x3) // Btu/lb
v3 = vsat_Px("Water/Steam", p3, x3) // ft^3/lb
h4 = h3 + v3 * (p4 - p3) * (144/778.17) // BTU/lb
pwin = Psat_T("Water/Steam", Twin) // lbf/in^2
hwin = hsat_Px("Water/Steam", pwin, 0) // Btu/lb
pwout = Psat_T("Water/Steam", Twout) // lbf/in^2
hwout = hsat_Px("Water/Steam", pwout, 0) // Btu/lb

mdot = Wdotcycle / ((h1 - h2) - (h4 - h3)) // lb/h

Wdott = mdot * (h1 - h2) // Btu/h
Wdotp = mdot * (h4 - h3) // Btu/h

Qdotin = mdot * (h1 - h4) // Btu/h

eta = Wdotcycle/Qdotin
mdotw = (mdot * (h2 - h3))/(hwout - hwin) // lb/h

1
Problem 8.11 (Continued) – Page 2

IT Results for p2 = 2 lbf/in.2

eta 0.3979
h1 1493
h2 934.1
h3 94.12
h4 98.32
hwin 27.63
hwout 43.67
mdot 1.80E+06
mdotw 9.43E+07
p3 2
p4 1400
pwin 0.2563
pwout 0.4446
Qdotin 2.51E+09
s1 1.609
s2 1.609
v3 0.01623
Wdotp 7.57E+06
Wdott 1.01E+09
p1 1400
p2 2
T1 1000
Twin 60
Twout 76
Wdotcycle 1.00E+09
x3 0

The results compare favorably with those of Problem 8.10.

2
 Problem 8.11 (Continued) – Page 3

Plots

Steam Mass Flow Rate versus Condenser Pressure Rate of Heat Transfer In versus Condenser Pressure
2,400,000
2,200,000 3,000,000,000

Heat Transfer Rate (Btu/h)

2,000,000
2,500,000,000
1,800,000
1,600,000
2,000,000,000
1,400,000
1,200,000
1,500,000,000
1,000,000
800,000 1,000,000,000
600,000
400,000 500,000,000
200,000
0 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 4 6 8 10 12 14
Condenser Pressure (lbf/in^2) Condenser Pressure (lbf/in^2)

Cycle Thermal Efficiency versus Condenser Pressure Cooling Water Mass Flow Rate versus Condenser Pressure
0.5
Cooling Water Mass Flow Rate (lb/h)

130,000,000
0.45 120,000,000

0.4 110,000,000
Thermal Efficiency

100,000,000
0.35
90,000,000
0.3 80,000,000
0.25 70,000,000
60,000,000
0.2
50,000,000
0.15 40,000,000
0.1 30,000,000
20,000,000
0.05
10,000,000
0 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 4 6 8 10 12 14
Condenser Pressure (lbf/in^2) Condenser Pressure (lbf/in^2)

Discussion
The plots indicate that as condenser pressure is increased while maintaining constant net power
output: steam mass flow rate increases, heat transfer rate into the cycle increases, and cycle
thermal efficiency decreases. As thermal efficiency decreases with increasing condenser
pressure, the rate energy is rejected by the condenser must increase, as evidenced by the
increasing cooling water flow rate.

From a practical standpoint, reducing condenser pressure while maintaining constant net power
output is desirable since steam mass flow rate decreases, heat transfer rate into the cycle
decreases, and cycle thermal efficiency increases. Since less energy is rejected in the condenser
at lower condenser pressure, required cooling water mass flow rate is reduced.