# Example Problem with Complete Solution

9B-1 : The Ideal Rankine Power Cycle 12 pts
Water is the working fluid in a Rankine Cycle. Superheated vapor enters the turbine at 480°C. The net power output of the cycle is 100 MW. For this cycle :

a.) For a boiler pressure of 8 MPa, construct a plot of the mass flow rate of the working fluid, the rate of heat transfer into the working fluid in the boiler, the thermal efficiency of the cycle and the heat transfer rate out of the working fluid in the condenser as functions of condenser pressure over the range from 1 kPa and 100 kPa.  Assume the cycle is an ideal Rankine Cycle.

b.) For a condenser pressure of 8 kPa, construct a plot of the rate of heat transfer into the working fluid in the boiler, the thermal efficiency of the cycle and the heat transfer rate out of the working fluid in the condenser as functions of boiler pressure over the range from 1 MPa and 25 MPa.  Assume the cycle is an ideal Rankine Cycle.

c.) For a boiler pressure of 8 MPa and a condenser pressure of 8 kPa, determine the mass flow rate of the working fluid, the rate of heat transfer into the working fluid in the boiler, the thermal efficiency of the cycle and the heat transfer rate out of the working fluid in the condenser.  The isentropic efficiencies of the pump and turbine are 70% and 85%, respectively.

Read: The key to parts (a) and (b) 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 in parts (a) and (b) requires looking up a lot of data.  This can be done most efficiently using the Thermal/Fluids Toolbox plug-in for Excel.  This is the method I used for this solution.  The next best method would be to use the NIST Webbook to generate data tables that can be copied and pasted into Excel.
Given: T2 480 oC Wsh 100 MW
Part a.) Part b.)
P3 = P4 8 kPa P1 = P2 8000 kPa
Part c.)
P1 = P2 8 MPa ηs,pump 70%
P3 = P4 8 kPa ηs,turb 85%
Find: Part a.) Part b.)
QH ??? MW QH ??? MW
ηth ??? ηth ???
QC ??? MW QC ??? MW
P1 = P2 1 to 25 MPa P3 = P4 1 to 100 kPa
Part c.)
QH ??? MW
hth ???
QC ??? MW
Diagrams :
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 :
Part a.) Let's begin with a very detailed analysis of the problem for a single condenser pressure of 8 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.
State
T (oC)
P (kPa)
X
H (kJ/kg)
S
(kJ/kg-K)

1 Sub Liq
41.76
8000
N/A
181.46
0.5915
2 Super Vap
480
8000
N/A
3347.9
6.6576
3 VLE
41.51
8
0.7944
2082.4
6.6576
4 Sat Liquid
41.51
8
0
173.65
0.5915
Additional data that may be useful.
Sat Vap
41.51
8
1
2576.5
8.2278
Sat Liquid
41.51
8
0
173.6
0.5915
Sat Vap
295.06
8000
1
2757.5
5.7423
Sat Liquid
295.06
8000
0
1317.0
3.2072
The values in the table that are shown in bold with a shaded background are the values we will determine.
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 Ws 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 and we can look up all of the relevant properties.  I will begin at state 2.
Because the turbine is isentropic, I know that S3 = S2 :
S3 6.6576 kJ/kg-K
Now, we know the values of two intensive properties at state 3, so this state is completely fixed and we can look up all of its properties.  Because Ssat liq < S3 < Ssat vap , state 3 is a two-phase mixture, T3 = Tsat at 8 kPa and we must determine the quality so that we can determine the enthalpy.
Eqn 1
T3 41.51 oC
x3 0.7944 kg vap/kg
Eqn 2
H3 2082.4 kJ/kg
We can look up 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 8 kPa.
Next, we proceed to state 1 using the fact that the pump is also isentropic: S1 = S4.
S1 0.5915 kJ/kg
Now, we know the values of two intensive properties at state 4, so this state is completely fixed and we can look up all of its properties.  Because S4 < Ssat liq , state 4 is a subcooled liquid.  We can look up its properties in the subcooled liquid property table for water, the NIST Webbook or we can use the Thermal/Fluids Toolbox (TFT).  The data tables and NIST Webbook require interpolation, but the TFT does not !
At 8 MPa :
T (oC) H (kJ/kg) S(kJ/kg-K)
40 174.14 0.56846
T1 H1 0.59150 T1 41.76 oC
50 215.71 0.69915 H1 181.46 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 for each device.  This process is made easier by our assumptions that no shaft work crosses the boundary of the boiler and 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 :
Qboil 3166.4 kJ/kg Wturb 1265.4 kJ/kg
Qcond -1908.8 kJ/kg Wpump -7.818 kJ/kg
Wcycle 1257.6 kJ/kg
Now, we can determine the mass flow rate from the given value of the power output of the cycle and use the mass flow rate to evaluate the heat transfer rates in the condenser and boiler.
Eqn 9 Eqn 10
mdot 79.52 kg/s Qboil 251.8 MW
Qcond -151.8 MW
Finally, we can calculate the thermal efficiency of this cycle.
Eqn 11
ηth 39.72%

Repeating this analysis for a variety of condenser pressures yield the following table of results.
Pcond
(kPa)
H3
(kJ/kg)
H4
(kJ/kg)
S4
(kJ/kg-K)
H1
(kJ/kg)
mdot
(kg/s)
QH
(MW)
ηth QC
(MW)
1 1864.6 28.58 0.10328 36.32 67.77 224.4 44.6% 124.4
2 1932.8 72.40 0.25684 80.12 71.06 232.2 43.1% 132.2
3 1974.6 100.06 0.35093 107.77 73.23 237.3 42.1% 137.3
6 2049.8 151.03 0.51903 158.75 77.50 247.1 40.5% 147.1
8 2082.4 173.65 0.59150 181.37 79.51 251.8 39.7% 151.8
10 2108.4 191.78 0.64874 199.52 81.19 255.6 39.1% 155.6
15 2157.0 226.24 0.75539 234.01 84.52 263.2 38.0% 163.2
20 2192.7 251.90 0.83309 259.70 87.16 269.2 37.2% 169.2
25 2221.2 272.54 0.89458 280.37 89.37 274.2 36.5% 174.2
30 2244.9 289.92 0.94564 297.77 91.31 278.5 35.9% 178.5
40 2283.2 318.31 1.02777 326.21 94.63 285.9 35.0% 185.9
50 2313.8 341.22 1.09287 349.15 97.46 292.2 34.2% 192.2
60 2339.4 360.55 1.14701 368.50 99.95 297.8 33.6% 197.8
65 2350.7 369.21 1.17106 377.18 101.10 300.3 33.3% 200.3
70 2361.4 377.34 1.19349 385.32 102.19 302.8 33.0% 202.8
75 2371.3 385.00 1.21451 392.99 103.25 305.1 32.8% 205.1
80 2380.7 392.24 1.23430 400.24 104.26 307.3 32.5% 207.3
85 2389.6 399.12 1.25300 407.12 105.24 309.5 32.3% 209.5
90 2398.1 405.67 1.27074 413.68 106.18 311.5 32.1% 211.5
95 2406.1 411.93 1.28760 419.94 107.09 313.6 31.9% 213.6
100 2413.7 417.92 1.30369 425.94 107.98 315.5 31.7% 215.5
Part b.) All of the analysis presented in part (a) applies just as well to part (b).
So, all we need to do here is build another table like the on ein part (a) so we can construct the plots.
The only difference is that the boiler pressure varies in this part of the problem.
Repeating this analysis for a variety of boiler pressures yield the following table of results.
Pboil
(MPa)
H2
(kJ/kg)
S2
(kJ/kg-K)
H3
(kJ/kg)
H1
(kJ/kg)
mdot QH
(MW)
ηth QC
(MW)
1 3434.7 7.7046 2411.9 174.32 97.83 319.0 31.4% 219.0
2 3423.0 7.3730 2307.5 175.33 89.78 291.6 34.3% 191.6
3 3411.0 7.1740 2244.9 176.34 85.95 278.0 36.0% 178.0
4 3398.9 7.0290 2199.3 177.35 83.62 269.4 37.1% 169.4
5 3386.5 6.9136 2163.0 178.35 82.05 263.2 38.0% 163.2
6 3373.8 6.8167 2132.5 179.36 80.93 258.5 38.7% 158.5
7 3361.0 6.7326 2106.0 180.36 80.11 254.8 39.2% 154.8
8 3347.9 6.6576 2082.4 181.37 79.51 251.8 39.7% 151.8
9 3334.5 6.5897 2061.1 182.37 79.07 249.2 40.1% 149.2
10 3320.9 6.5272 2041.4 183.38 78.75 247.1 40.5% 147.1
11 3307.1 6.4691 2023.1 184.38 78.54 245.3 40.8% 145.3
12 3293.0 6.4144 2005.9 185.38 78.41 243.7 41.0% 143.7
13 3278.6 6.3627 1989.6 186.39 78.35 242.3 41.3% 142.3
14 3264.0 6.3134 1974.1 187.39 78.36 241.1 41.5% 141.1
15 3249.1 6.2661 1959.2 188.39 78.42 240.0 41.7% 140.0
16 3233.9 6.2206 1944.9 189.39 78.54 239.1 41.8% 139.1
17 3218.5 6.1765 1931.0 190.40 78.70 238.3 42.0% 138.3
18 3202.7 6.1336 1917.5 191.40 78.90 237.6 42.1% 137.6
19 3186.7 6.0918 1904.4 192.40 79.14 237.0 42.2% 137.0
21 3153.7 6.0108 1878.9 194.40 79.74 236.0 42.4% 136.0
22 3136.7 5.9714 1866.5 195.40 80.10 235.6 42.4% 135.6
23 3119.4 5.9325 1854.3 196.40 80.49 235.3 42.5% 135.3
24 3101.8 5.8941 1842.2 197.39 80.91 235.0 42.6% 135.0
25 3083.9 5.8560 1830.2 198.39 81.37 234.8 42.6% 134.8
Part c.) ηs,pump 70% ηs,turb 85%
The only difference between part (c) and the example case that we analyzed for part (a) is that the pump and turbine are NOT isentropic !  S3 > S2 and S4 > S3.
State T (oC) P (kPa) X H (kJ/kg) S (kJ/kg-K)
1 Sub Liq 42.57 8000 N/A 184.81 0.60203
1S Sub Liq 41.74 8000 N/A 181.37 0.59150
2 Super Vap 480 8000 N/A 3347.9 6.6576
3S VLE 41.51 8 0.7944 2082.4 6.6576
3 VLE 41.51 8 0.8734 2272.2 7.2609
4 Sat Liquid 41.51 8 0 173.65 0.59150
Let's consider the turbine first :
Eqn 12
The values we determined in part (a) apply for an isentropic turbine :
H3S 2082.4 kJ/kg
Eqn 13
Wturb,isen 1265.4 kJ/kg
Wturb,act 1075.6 kJ/kg
Eqn 14
H3 2272.2 kJ/kg
Now, we use H3 and P3 to determine S3 and x3.
Now, we know the values of two intensive properties at state 3, so this state is completely fixed and we can look up all of its properties.  Because Hsat liq < H3 < Hsat vap , state 3 is a two-phase mixture, T3 = Tsat at 8 kPa and we can determine both the quality and the entropy, but this is not really necessary.
Eqn 15
T3 41.51 oC x3 0.8734 kg vap/kg
Eqn 16
S3 7.2609 kJ/kg
Now, we can carry out a very similar analysis of the pump.
Eqn 17
The values we determined in part (a) apply for an isentropic pump :
H1S 181.46 kJ/kg Wpump,is -7.818 kJ/kg
Eqn 18
Eqn 19
Wpump,ac -11.169 kJ/kg H1 184.81 kJ/kg
T3 41.51 oC
Now, we can determine T1 and S1 by interpolation.  But again, it is not necessary.
At 8 MPa :
T (oC) H (kJ/kg) S(kJ/kg-K)
40 174.14 0.56846
T1 184.81 S1 T1 42.57 oC
50 215.71 0.69915 S1 0.60203 kJ/kg
Now, we can use Eqns 5 - 11 to complete this part of the problem.
Qboil 3163.1 kJ/kg Wturb 1075.6 kJ/kg
Qcond -2098.6 kJ/kg Wpump -11.169 kJ/kg
Wcycle 1064.5 kJ/kg
mdot 93.945 kg/s
Qboil 297.2 MW
Qcond -197.2 MW hth 33.65%
Compare these values to those from part (a).
You will notice that the WS,turb dropped by 15% while WS,pump increased by 42%.  This also caused an increase in both QH and QC.  But, the bottom line is that the thermal efficiency dropped by more than 15 % !!
Verify : None of the assumptions made in this problem solution can be verified.