# Example Problem with Complete Solution

9B-1 : Ideal Rankine Cycle Efficiency as a Function of Condenser Pressure 8 pts
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: P1 = 10 MPa, T2 = 550oC, 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 looking-up 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: P1 = P2 10 MPa m 75 kg/s
T2 550 oC
Find: Plot Wcycle and hth as functions of P4.
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 (oC) P (kPa) X H (kJ/kg) S (kJ/kg-K)
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 (oC) P (kPa) X H (kJ/kg) S (kJ/kg-K)
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 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 (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.
H2 3502.0 kJ/kg S2 6.7585 kJ/kg-K
Because the turbine is isentropic, we know that S3 = S2 : S3 6.7585 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.
We can see from the additional data table that Ssat liq < S3 < Ssat vap . Therefore, state 3 is a two-phase mixture, T3 = Tsat at 10 kPa and we must determine the quality so that we can determine the enthalpy. Eqn 1
T3 45.81 oC x3 0.8146 kg vap/kg Eqn 2
H3 2140.4 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 10 kPa.
Next, we proceed to state 1 using the fact that the pump is also isentropic: S1 = S4.
S1 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 look-up all of its properties.
We can see from the additional data table, above, that S4 < Ssat liq . Therefore, state 4 is a subcooled liquid.  We can look-up its properties in the NIST Webbook.  The NIST Webbook require interpolation or repeated temperature range selection to zero-in on the precise property values for state 4.
By zooming-in on a very narrow temperature range, I found : T1 46.14 oC
H1 201.88 kJ/kg
Interpolation on a wider temperature range is shown below. The results are very similar.
At 10 MPa : T (oC) H (kJ/kg) S(kJ/kg-K)
45 197.15 0.63436
T1 H1 0.64920 T1 46.14 oC
50 217.94 0.69920 H1 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 WS 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 :
Qboil 3300.1 kJ/kg Wturb 1361.6 kJ/kg
Qcond -1948.6 kJ/kg Wpump -10.071 kJ/kg
Qboil 247.5 MW Wturb 102.12 MW
Qcond -146.1 kW Wpump -755.3 kW
Wcycle 101.36 MW
Finally, we can calculate the thermal efficiency of this cycle. Eqn 9
hth 40.95%
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)
Wcycle
(MW)
QH
(MW)
hth
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 :  