| 9B-2 : | Steam Power Plant Operating on the Rankine Cycle | 9 pts |
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| The steam power plant, shown below, operates on the Rankine Cycle. Steam leaves the boiler at 3.5 MPa and 400°C. The turbine exhausts to the condenser at 10 kPa. The turbine has an isentropic efficiency of 88% and the pump operates essentially isentropically. Both the pump and turbine are adiabatic. Consider liquid water to be an incompressible fluid. The boiler absorbs heat from the furnace which is at Tfurn = 500 K. The condenser rejects heat to cooling water which is at Tcw = 300 K. The temperature of the surroundings is Tsurr = 300 K. | |||||||||
| a.) | Construct a neat, fully labeled T-S diagram of the actual Rankine Cycle executed in this process. | ||||||||
| b.) | Calculate the specific work and specific heat transfer for each unit in the cycle. | ||||||||
| c.) | Calculate the thermal efficiency of the power cycle. | ||||||||
| d.) | Calculate the total entropy generation and the total lost work per kg for the cycle. | ||||||||
| e.) | Calculate the work that would be produced if this cycle were completely reversible and the state of all four streams remained the same as in the actual cycle. | ||||||||
| Read: | The TS Diagram is pretty standard. The pump is isentropic, but the turbine is not. In order to determine the Q's and WS's we will need to determine all the H's. H2 and H4 come straight from the Steam Tables. Then use entropy to determine H1 and H3. Plug the H's into the 1st Law for each unit to determine all the Q's and W's. Once we have all the Q's and W's in part (b), we can calculate the thermal efficiency from its definition. The key to part (d) is that, for a cycle, ΔS = 0. So, Sgen = ΔSuniv is just ΔSfurn + ΔScw. Because the furnace and cooling water behave as thermal reservoirs, we can evaluate the change in their entropy from the definition of entropy. Lost work is just Tsurr Sgen. The easiest way to evaluate WS,rev is to use the definition of WS,lost and the values of WS,act and WS,lost that were determined in parts (b) and (d), repectively. WS,rev = WS,act + WS,lost. | ||||||||
| Given: | P2 | 3500 | kPa | Q41 = Q23 = | 0 | kJ/kg | |||
| T2 | 400 | oC | |||||||
| ηs,turb | 88% | Tfurn | 500 | K | |||||
| P3 | 10 | kPa | Tcw | 273.15 | K | ||||
| x4 | 1 | kg vap / kg | Tsurr | 300 | K | ||||
| Find : | a.) | See TS-Diagram, below. | c.) | η | ? | % | |||
| b.) | Wstep | ? | kJ/kg | d.) | Sgen | ? | kJ/kg-K | ||
| Qstep | ? | kJ/kg | WS,lost | ? | kJ/kg | ||||
| For steps 1-2, 2-3, 3-4 & 4-1 | e.) | WS,rev | ? | kJ/kg | |||||
| Diagrams: |
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| Assumptions : | - Each component in the cycle is analyzed as an open system operating at steady-state. | ||||||||
| - All of the processes except the turbine are internally reversible. | |||||||||
| - The turbine is adiabatic and the pump is isentropic. | |||||||||
| - Condensate leaves the condenser as saturated liquid. | |||||||||
| - No shaft work in the boiler or condenser. | |||||||||
| - Changes in kinetic and potential energies are negligible. | |||||||||
| Equations / Data / Solve : | |||||||||
| Part b.) | In order to evaluate Q and WS for each process in the cycle, we will apply the 1st Law to each process. | ||||||||
| All the devices operate at steady-state and any changes in kinetic and potential energies are negligible. | |||||||||
| The boiler and condenser have no shaft work interaction and the pump and turbine are both adiabatic. | |||||||||
| Therefore, the relevant forms of the 1st Law for the four devices are: | |||||||||
| Boiler : |  |
Eqn 1 | |||||||
| Turbine : |  |
Eqn 2 | |||||||
| Condenser : |  |
Eqn 3 | |||||||
| Pump : |  |
Eqn 4 | |||||||
| In order to evaluate all the W's and Q's in Eqns 1 - 4, we must first determine H for all four streams. | |||||||||
| We can immediately evaluate H2 and H4 because we know both P2 and T2 and we know P4 and we know that it is a saturated liquid, x4 = 0. So, we can lookup H2 and H4 in the Steam Tables. | |||||||||
| H2 | 3223.2 | kJ/kg | H4 | 191.81 | kJ/kg | ||||
| In order to determine H1, we must make use of the fact that the pump is adiabatic and internally reversible and that changes in kinetic and potential energies are negligible. Under these conditions, the shaft work done at the pump can be determined from the Mechanical Energy Balance Equation: | |||||||||
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Eqn 5 | ||||||||
| Cancelling terms yields : |
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Eqn 6 | |||||||
| Because the liquid water flowing through the pump is incompressible, the specific volume is constant, and Eqn 6 simplifies to: | |||||||||
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Eqn 7 | ||||||||
| Now, we can lookup V4 and use it in Eqn 7 to evaluate WS,pump: |
V4 | 0.0010103 | m3/kg | ||||||
| WS,pump | -3.526 | kJ/kg | |||||||
| Now, we can use WS,pump and Eqn 4 to determine H1. |
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| H1 | 195.3 | kJ/kg | |||||||
| Next, we need to use the isentropic efficiency of the turbine to determine H3. |
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Eqn 8 | |||||||
| Solve Eqn 8 for H3 : |  |
Eqn 9 | |||||||
| Now, we must evaluate H3S before we can use Eqn 9 to determine H3. H3S is the enthalpy of the turbine effluent IF the turbine were isentropic. Therefore, S3S = S2. | |||||||||
| S2 | 6.8427 | kJ/kg-K | S3S | 6.8427 | kJ/kg-K | ||||
| Now, we know the values of two intensive properties at state 3S: S3S and P3S, so we can fix this state and determine H3S by interpolation in the Steam Tables. First we must determine the phase of state 3S. | |||||||||
| At 10 kPa : | Tsat | 45.806 | oC | ||||||
| H (kJ/kg) | S (kJ/kg-K) | Since Ssat liq < S3S < Ssat vap, state 3S is a saturated mixture. |
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| Sat Liq: | 191.81 | 0.64920 | |||||||
| Sat Vap : | 2583.9 | 8.1488 | T3S | 45.806 | oC | ||||
| Determine x3S from the specific entropy, using: |
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Eqn 10 | |||||||
| x3S | 0.8258 | kg vap/kg | |||||||
| Then, we can use the quality to determine H3S, using: |
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Eqn 11 | |||||||
| H3S | 2167.30 | kJ/kg | |||||||
| Now, plug H3S back into Eqn 9 to determine H3 : |
H3 | 2294.01 | kJ/kg | ||||||
| Now that we know the specific enthalpy of all four streams, we can use Eqns 1 - 3 to evaluate the WS,turb and the two remaining Q's. | |||||||||
| Q12 | 3027.86 | kJ/kg | |||||||
| WS,turb | 929.19 | kJ/kg | Q34 | -2102.20 | kJ/kg | ||||
| Part c.) | The thermal efficiency of a power cycle is defined by: |
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Eqn 12 | ||||||
| Where : |  |
Eqn 13 | |||||||
| Now, we can plug values into Eqns 12 & 13 to complete this part of the problem. |
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| Wcycle | 925.66 | kJ/kg | |||||||
| η | 30.57% | ||||||||
| Part d.) | The total entropy generation for the cycle is equal to the entropy change of the universe caused by the cycle. | ||||||||
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Eqn 14 | ||||||||
| Because the furnace and the cooling water are isothermal (they behave as thermal reservoirs) we can evaluate the change in their entropy directly from the definition of entropy. | |||||||||
 |
Eqn 15 |  |
Eqn 16 | ||||||
| Now, we can put values into Eqns 15, 16 & 14 : | ΔSfurn | -6.0557 | kJ/kg-K | ||||||
| ΔScw | 7.6961 | kJ/kg-K | |||||||
| Sgen | 1.6404 | kJ/kg-K | |||||||
| We can calculate lost work using : |  |
Eqn 17 | |||||||
| WS,lost | 492.12 | kJ/kg | |||||||
| Alternatively, we can compute the lost work using : |
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Eqn 18 | |||||||
| WS,lost | 492.12 | kJ/kg | |||||||
| Part e.) | The reversible work can be determined from the definition of last work : |
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Eqn 19 | ||||||
| Solving for WS,rev yields : |  |
Eqn 20 | |||||||
| WS,rev | 1417.79 | kJ/kg | |||||||
| Verify : | The assumptions made in the solution of this problem cannot be verified with the given information. | ||||||||
| Answers : | a.) | See the diagram at the beginning of this solution. | |||||||
| b.) | c.) | η | 30.6% | ||||||
| Device | Step | Q (kJ/kg) | WS (kJ/kg) | ||||||
| Boiler | 1 - 2 | 3030 | 0 | d.) | Sgen | 1.640 | kJ/kg-K | ||
| Turbine | 2 - 3 | 0 | 929 | WS,lost | 492 | kJ/kg | |||
| Condenser | 3 - 4 | -2100 | 0 | ||||||
| Pump | 4 - 1 | 0 | -3.53 | e.) | WS,rev | 1418 | kJ/kg | ||
| HEY ! | Why is it that: |
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??? | WS,rev / QH | 46.8% | ||||
| 1-TC/TH | 45.4% | ||||||||
| Because the reversible cycle also exchanges heat at the turbine ! | |||||||||
| In fact, the reversible turbine absorbs heat reversibly from a third thermal reservoir. This reservoir is a very special thermal reservoir because it is always at the same temperature as the fluid in the turbine. This is pretty tricky because the temperature of the fluid changes as it passes through the turbine ! | |||||||||
| Consider the definition of entropy generation: |
 |
Eqn 21 | |||||||
| But, Tsys is NOT constant ! |  |
Eqn 22 | |||||||
| The value of Q23 is equal to the area under the path for Step 1-2 on the TS Diagram, but we cannot evaluate it because we do not know the equation of the path, T = fxn(S). | |||||||||
| We could evaluate Q23 by applying the 1st Law to the entire reversible cycle, because Q12, Q34 and Q41 are the same for reversible cycle as they were for the actual cycle. So, we can use the values calculated in part (c). | |||||||||
| 1st Law, Reversible Cycle : |  |
Eqn 23 | |||||||
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Eqn 24 | ||||||||
| Q23 | 492.12 | kJ/kg | |||||||
| Let's double check that this cycle is indeed reversible when it GAINS heat from the surroundings at the turbine. | |||||||||
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Eqn 25 | ||||||||
| We already calculated ΔSfurn and ΔScw, above. So, now we need to calculate ΔSsurr. | |||||||||
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Eqn 26 | Plug in values: | ΔSsurr | -1.6404 | kJ/kg-K | ||||
| Now, plug values into Eqn 25 : | ΔSuniv | 0.0000 | kJ/kg-K | ||||||
| There is one impossible aspect about our imaginary reversible cycle. The surroundings (at 300 K) must reversiblby supply heat to the steam inside the turbine (which is always at a higher temperature than 300 K). That would require a heat pump and that would be a very complicated turbine, indeed ! This is ok because this is just a hypothetical reversible turbine anyway. It just seems strange. | |||||||||
