Question 3 of 8: 512 MW reheat plant heat balance

Question 3: 512 MW reheat plant heat balance (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The page-11 heat balance diagram of a 512 MW reheat steam power plant. The readings that matter for the three parts asked are collected below.

Find. The overall cycle efficiency on heat in and electricity out, and the shaft power developed separately by the HP and the IP cylinders.

Approach. Heat input is a control volume drawn round the boiler and the reheater; each turbine cylinder is then a separate control volume, and its power is the enthalpy flowing in minus the enthalpy flowing out. The whole calculation is therefore three steady-flow energy balances, and each one is checked first by confirming that the corresponding mass balance closes on the printed flows.

1. Part (a) — the boiler duty. The boiler raises 430.11 kg/s from the final feedwater state to main steam: $$\dot Q_{\text{boiler}} = \dot m\,(h_{\text{ms}} - h_{\text{fw}}) = 430.11 \times (3396 - 1071)$$ $$\dot Q_{\text{boiler}} = 430.11 \times 2325 = 1\,000\,006\ \text{kW} = 1000.0\ \text{MW}$$ A round 1000 MW of boiler duty for a 512 MW machine is the first sign the readings are right.
2. Part (a) — the reheater duty. The reheater receives 385.22 kg/s of cold reheat at 3046 kJ/kg and, on the way to the IP turbine, the 0.50 kg/s HP front-gland leak-off at main-steam enthalpy joins it, so that 385.72 kg/s leaves at 3536 kJ/kg. Taking the whole reheat leg as one control volume, $$\dot Q_{\text{rh}} = 385.72 \times 3536 - (385.22 \times 3046 + 0.50 \times 3396)$$ $$\dot Q_{\text{rh}} = 1\,363\,906 - 1\,175\,078 = 188\,828\ \text{kW} = 188.8\ \text{MW}$$ Ignoring the small gland stream and writing $385.72 \times (3536 - 3046) = 189\,003$ kW changes the final efficiency by 0.01 percentage points, so either treatment is acceptable.
3. Part (a) — the overall cycle efficiency. Adding the two heat inputs and comparing with the generator terminals: $$\dot Q_{\text{in}} = 1\,000\,006 + 188\,828 = 1\,188\,834\ \text{kW}$$ $$\eta = \frac{P_{\text{elec}}}{\dot Q_{\text{in}}} = \frac{512\,008}{1\,188\,834} = \boxed{0.4307 = 43.07\ \%}$$ This is the cycle efficiency, not the station efficiency: it excludes the boiler loss up the stack, so a plant of this design would show an overall efficiency of roughly 40 % once a 93 % boiler is allowed for.
4. Part (b) — how much steam actually expands through the HP cylinder. Before any power is computed the HP mass balance is confirmed against the printed flows. Of the 430.11 kg/s admitted, 4.06 kg/s is a bypass leak-off taken at inlet conditions and admitted directly to the IP inlet, 0.50 kg/s and 0.11 kg/s are gland leak-offs taken at the inlet end, 0.96 kg/s and 0.83 kg/s leave the exhaust-end glands at exhaust enthalpy, and $385.22 + 38.43 = 423.65$ kg/s leaves through the exhaust: $$4.06 + 0.50 + 0.11 + 0.96 + 0.83 + 423.65 = 430.11\ \text{kg/s} \;\checkmark$$ The balance closes exactly, which validates every flow read off the drawing.
5. Part (b) — the HP turbine power. The three streams taken at inlet conditions do no work; everything else expands from 3396 kJ/kg to the cold reheat enthalpy of 3046 kJ/kg: $$\dot m_{\text{HP}} = 430.11 - 4.06 - 0.50 - 0.11 = 425.44\ \text{kg/s}$$ $$P_{\text{HP}} = \dot m_{\text{HP}}\,(h_{\text{ms}} - h_{\text{crh}}) = 425.44 \times (3396 - 3046)$$ $$P_{\text{HP}} = 425.44 \times 350 = \boxed{148\,904\ \text{kW} = 148.9\ \text{MW}}$$ Note that the 38.43 kg/s of extraction A is drawn off the cold reheat line downstream of the exhaust flange, so it has already done its full work in the cylinder and must not be deducted here.
6. Part (c) — the IP inlet stream. The IP cylinder receives the hot reheat plus the HP bypass leak-off, and the mass balance again closes exactly on the four printed outlet flows: $$\dot m_{\text{IP}} = 385.72 + 4.06 = 389.78\ \text{kg/s}$$ $$11.49 + 24.54 + 13.36 + 340.39 = 389.78\ \text{kg/s} \;\checkmark$$ The two inlet streams are at different enthalpies, so they must be carried separately (their mixed value is 3534.5 kJ/kg, slightly below the 3536 kJ/kg of the reheat alone).
7. Part (c) — the IP turbine power. Applying the steady-flow energy equation to the whole cylinder, with extraction B leaving at 3347 kJ/kg and the other three streams leaving together at the exhaust enthalpy of 3228 kJ/kg: $$\sum \dot m h\Big|_{\text{in}} = 385.72 \times 3536 + 4.06 \times 3396 = 1\,377\,694\ \text{kW}$$ $$\sum \dot m h\Big|_{\text{out}} = 11.49 \times 3347 + (24.54 + 13.36 + 340.39) \times 3228 = 1\,259\,577\ \text{kW}$$ $$P_{\text{IP}} = 1\,377\,694 - 1\,259\,577 = \boxed{118\,117\ \text{kW} = 118.1\ \text{MW}}$$
8. Cross-check the readings against the diagram's own arithmetic. Two mixing points on the drawing are printed with both their inlet and their outlet data, so they test the enthalpies the same way the mass balances tested the flows. Extraction C (24.54 kg/s at 3228 kJ/kg) is joined by the 0.96 kg/s exhaust-end gland drain at 3046 kJ/kg on the way to the deaerator, and the diagram prints the result as 25.50 kg/s at 3221 kJ/kg: $$\frac{24.54 \times 3228 + 0.96 \times 3046}{25.50} = 3221.1\ \text{kJ/kg} \;\checkmark$$ Likewise the last LP extraction (21.47 kg/s at 2585 kJ/kg) is joined by 0.76 kg/s of sealing steam at 3135 kJ/kg and the diagram prints 22.23 kg/s at 2604 kJ/kg, against a computed 2603.8. Every reading used above therefore reproduces a number printed elsewhere on the same drawing.
9. Put the two answers in context. Together the HP and IP cylinders make $$P_{\text{HP}} + P_{\text{IP}} = 148.9 + 118.1 = 267.0\ \text{MW}$$ which is 52.1 % of the 512.008 MW at the generator terminals. The balance comes from the two double-flow LP cylinders on the same shaft; the boiler feed pump turbine is driven by its own 13.36 kg/s of IP exhaust steam and does not contribute to the generator output at all, which is why its steam was counted as leaving the IP cylinder rather than as passing on to the LP cylinders.