Consider the ideal Rankine Cycle with superheat using water as the working fluid. 







Construct plots of the net
power output of the cycle and the thermodynamic efficiency as functions of the operating pressure
of the condenser.
Consider condenser pressures from 10 kPa to 200 kPa. 

Data: P_{1} = 10 MPa, T_{2} = 550^{o}C, m =
75 kg/s 














Read : 
The key is that the cycle is an ideal Rankine Cycle. This means that the pump and turbine operate isentropically and that the condenser effluent
is a saturated liquid. 













Constructing the plots
requires lookingup a lot of data. This can be done most
efficiently using the NIST
Webbook to generate data tables
that can be copied and pasted into Excel. 












Given: 
P_{1}
= P_{2} 
10 
MPa 



m 
75 
kg/s 


T_{2} 
550 
^{o}C 


















Find: 
Plot W_{cycle} and h_{th} as functions of P_{4}. 

















Diagram: 
























Assumptions: 
1  
The pump and the turbine are adiabatic and reversible and, therefore, isentropic. 



2  
Only flow work crosses the boundary of the boiler and condenser. 












Equations
/ Data / Solve: 




















Let's begin with a very detailed analysis of the problem for a single condenser pressure of 10 kPa. 













Then, we can present a
table of results for all of the other condenser pressures so we can construct the plot that is required. 













Let's organize the data that we need to collect into a table. This will make it easier to keep track of
the values we have looked up and the values we have calculated. 













Stream 
State 
T (^{o}C) 
P (kPa) 
X 
H (kJ/kg) 
S (kJ/kgK) 




1 
Sub Liq 
46.14 
10000 
N/A 
201.88 
0.64920 




2 
Super Vap 
550 
10000 
N/A 
3502.0 
6.7585 




3 
VLE 
45.81 
10 
0.8146 
2140.4 
6.7585 




4 
Sat Liquid 
45.81 
10 
0 
191.81 
0.64920 















Additional data that may be useful. 



















State 
T (^{o}C) 
P (kPa) 
X 
H (kJ/kg) 
S (kJ/kgK) 





Sat Vap 
45.81 
10 
1 
2583.9 
8.1488 





Sat Liquid 
45.81 
10 
0 
191.81 
0.64920 





Sat Vap 
311.06 
10000 
1 
2725.5 
5.6160 





Sat Liquid 
311.06 
10000 
0 
1408.1 
3.3606 















The values in the
table that are shown in bold
with a yellow background are the values we will determine in the following
solution. 













One approach to
solving cycle problems of
this nature is to work your way around the cycle
until you have evaluated all the properties to complete the table
shown above. Then, you can go back and apply the 1st Law to each process in the cycle to evaluate Q and W_{s} as need. That is the
approach I will take. 













In this problem, it
makes the most sense to begin at either state 2 or state 4
because these states
are completely fixed (this means we know the
values of two intensive properties and we can use
them to determine the values of any other intensive properties using the NIST Webbook. I will begin at state 2. 













H_{2} 
3502.0 
kJ/kg 



S_{2} 
6.7585 
kJ/kgK 













Because the turbine is isentropic, we know that S_{3} = S_{2} : 

S_{3} 
6.7585 
kJ/kgK 













Now, we know the
values of two intensive properties at state 3, so this state is completely fixed
and we can lookup all
of its properties. 













We can see from the additional data table that S_{sat
liq} < S_{3} < S_{sat vap} . Therefore, state 3 is a twophase mixture, T_{3} = T_{sat} at 10 kPa
and we must determine the quality so that we can determine the enthalpy. 




















Eqn 1 













T_{3} 
45.81 
^{o}C 



x_{3} 
0.8146 
kg vap/kg 


















Eqn 2 



















H_{3} 
2140.4 
kJ/kg 













We can lookup all the properties at state 4
because the state is completely fixed by the fact that the fluid is a saturated liquid leaving the condenser in an Ideal Rankine Cycle and because we
know the pressure is 10 kPa. 













Next, we proceed to state 1 using the fact that the pump is also isentropic: S_{1} = S_{4}. 



















S_{1} 
0.64920 
kJ/kg 













Now, we know the
values of two intensive properties at state 4, so this state is completely fixed
and we can lookup all
of its properties. 













We can see from the additional data table, above, that S_{4} < S_{sat liq} . Therefore, state 4 is a subcooled liquid. We can lookup its properties in the NIST Webbook. The NIST Webbook require interpolation or repeated temperature range selection to zeroin on the precise property
values for state 4. 













By zoomingin on a very narrow temperature range, I found : 
T_{1} 
46.14 
^{o}C 








H_{1} 
201.88 
kJ/kg 













Interpolation on a wider temperature range is shown below. The results are very similar. 













At 10
MPa : 
T (^{o}C) 
H (kJ/kg) 
S(kJ/kgK) 








45 
197.15 
0.63436 








T_{1} 
H_{1} 
0.64920 

T_{1} 
46.14 
^{o}C 




50 
217.94 
0.69920 

H_{1} 
201.91 
kJ/kg 













Now, we have all the information we need to apply the 1st Law to each
device in the cycle to determine Q and W_{S} for each device. This process is made
easier by our assumptions that no
shaft work crosses the
boundary of the boiler or condenser and that both the pump and the turbine are adiabatic. 













The four relevant
forms of the 1st Law are : 




















Eqn 4 













Boiler : 






Eqn 5 













Turbine : 






Eqn 6 













Condenser : 





Eqn 7 













Pump : 






Eqn 8 













Plugging values into Eqns 5  8 yields : 


















Q_{boil} 
3300.1 
kJ/kg 



W_{turb} 
1361.6 
kJ/kg 


Q_{cond} 
1948.6 
kJ/kg 



W_{pump} 
10.071 
kJ/kg 













Q_{boil} 
247.5 
MW 



W_{turb} 
102.12 
MW 


Q_{cond} 
146.1 
kW 



W_{pump} 
755.3 
kW 



















W_{cycle} 
101.36 
MW 













Finally, we can
calculate the thermal efficiency of this cycle. 























Eqn 9 



















h_{th} 
40.95% 














Repeating this
analysis for a variety of condenser pressures yield the following table of
results. 













P_{cond
}(kPa) 
H_{3
}(kJ/kg) 
H_{4
}(kJ/kg) 
S_{4
}(kJ/kgK) 
H_{1
}(kJ/kg) 
W_{cycle
}(MW) 
Q_{H
}(MW) 
h_{th} 



10 
2140.4 
191.81 
0.64920 
201.88 
101.4 
247.5 
41.0 



20 
2226.2 
251.42 
0.83202 
261.55 
94.9 
243.0 
39.1 



30 
2279.2 
289.27 
0.94407 
299.44 
90.9 
240.2 
37.9 



40 
2318.3 
317.62 
1.0261 
327.83 
88.0 
238.1 
37.0 



50 
2349.4 
340.54 
1.0912 
350.77 
85.7 
236.3 
36.3 



60 
2375.4 
359.91 
1.1454 
370.15 
83.7 
234.9 
35.6 



70 
2397.8 
376.75 
1.1921 
387.02 
82.0 
233.6 
35.1 



80 
2417.5 
391.71 
1.2330 
401.99 
80.6 
232.5 
34.7 



90 
2435.2 
405.20 
1.2696 
415.49 
79.2 
231.5 
34.2 



100 
2451.1 
417.50 
1.3028 
427.81 
78.0 
230.6 
33.8 



110 
2465.8 
428.84 
1.3330 
439.15 
76.9 
229.7 
33.5 



120 
2479.3 
439.36 
1.3609 
449.68 
75.9 
228.9 
33.2 



130 
2491.8 
449.19 
1.3868 
459.52 
75.0 
228.2 
32.9 



140 
2503.5 
458.42 
1.4110 
468.76 
74.1 
227.5 
32.6 



150 
2514.5 
467.13 
1.4337 
477.47 
73.3 
226.8 
32.3 



160 
2524.9 
475.38 
1.4551 
485.73 
72.5 
226.2 
32.1 



170 
2534.7 
483.22 
1.4753 
493.58 
71.8 
225.6 
31.8 



180 
2544.0 
490.70 
1.4945 
501.06 
71.1 
225.1 
31.6 



190 
2552.8 
497.85 
1.5127 
508.22 
70.4 
224.5 
31.4 



200 
2561.3 
504.70 
1.5302 
515.07 
69.8 
224.0 
31.1 













Verify: 
The assumptions made
in the solution of this problem cannot be verified with the given
information. 












Answers : 












