| 9B-1 : | Ideal Rankine Cycle Efficiency as a Function of Condenser Pressure | 8 pts |
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| Consider the ideal Rankine Cycle with superheat using water as the working fluid. | ||||||||||||||||||||
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| 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: | ||||||||||||||||||||
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| 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. | ||||||||||||||||||||
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Eqn 1 | |||||||||||||||||||
| T3 | 45.81 | oC | x3 | 0.8146 | kg vap/kg | |||||||||||||||
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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 : | ||||||||||||||||||||
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Eqn 4 | |||||||||||||||||||
| Boiler : | 
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Eqn 5 | ||||||||||||||||||
| Turbine : | 
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Eqn 6 | ||||||||||||||||||
| Condenser : | 
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Eqn 7 | ||||||||||||||||||
| Pump : | 
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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. | ||||||||||||||||||||
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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 : | 
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