Question 5 of 8: Steam Cycle — 512 MW Reheat Plant Heat Balance

Question 5: Steam Cycle — 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. From the page-15 heat balance diagram (# = kg/s, h = kJ/kg, p = MPa, C = °C):

Find. The overall cycle efficiency on heat input and electrical output, and the shaft power developed separately by the HP and the IP cylinders.

[Figure not reproduced: The two cylinders redrawn as control volumes. The leak-offs taken at the HP inlet never expand in that cylinder and must come off its working flow; extraction A is drawn from the cold reheat pipe downstream of the exhaust flange, so it has already done its full HP work and stays in. The exhaust-end . See the official exam paper.]

Approach. Take the heat input as the boiler duty plus the reheater duty, both from the flows and enthalpies on the drawing; then take each cylinder as a control volume and evaluate the shaft power as the net enthalpy flux across it, taking care to count only the steam that actually expands.

1. Part (a) — boiler duty. The boiler takes the whole final feedwater flow from 1071 kJ/kg to main-steam conditions: $$\dot{Q}_{boiler} = \dot{m}_{ms}(h_{ms} - h_{fw}) = 430.11 \times (3396 - 1071) = 1\,000\,006\ \text{kW} = 1000.0\ \text{MW}$$
2. Part (a) — reheater duty. The reheater is fed by the 385.22 kg/s of cold reheat at 3046 kJ/kg plus the 0.50 kg/s front-gland leak-off, which the drawing returns into the reheat line at main-steam enthalpy. Taking energy out minus energy in, $$\dot{Q}_{reheat} = \dot{m}_{hrh}h_{hrh} - (\dot{m}_{crh}h_{crh} + 0.50\,h_{ms}) = 385.72(3536) - [385.22(3046) + 0.50(3396)] = 188.83\ \text{MW}$$ Treating the reheater as a single 385.72 kg/s stream through 3046 → 3536 kJ/kg would give 189.00 MW — a difference of 0.02 % in the final answer, so the distinction is a matter of bookkeeping rigour rather than of arithmetic consequence.
3. Part (a) — overall efficiency. Total heat added to the cycle is 1000.01 + 188.83 = 1188.83 MW against a measured electrical output of 512.008 MW, so $$\eta_{cycle} = \frac{\dot{W}_{elec}}{\dot{Q}_{boiler} + \dot{Q}_{reheat}} = \frac{512.008}{1188.83} = \boxed{43.07\ \%}$$ This is a cycle efficiency on heat into the steam, not a station efficiency: at a typical 88 % boiler efficiency the plant would burn about 1351 MW of fuel for a net heat rate near 9500 kJ/kWh, which is right for a 512 MW subcritical reheat unit.
4. Part (b) — decide which steam actually works in the HP cylinder. Three streams leave at inlet conditions and therefore develop no work in this cylinder: the 4.06 kg/s bypass that is admitted to the IP inlet, and the 0.50 and 0.11 kg/s inlet-end gland leak-offs. Everything else expands from 3396 to 3046 kJ/kg — including extraction A, which is drawn from the cold reheat pipe downstream of the exhaust flange, and the two exhaust-end glands. Hence $\dot{m}_{HP} = 430.11 - 4.06 - 0.50 - 0.11 = 425.44$ kg/s.
5. Part (b) — HP turbine power. With a uniform enthalpy drop of $3396 - 3046 = 350$ kJ/kg, $$\dot{W}_{HP} = \dot{m}_{HP}(h_{ms} - h_{crh}) = 425.44 \times 350 = 148\,904\ \text{kW} = \boxed{148.9\ \text{MW}}$$ Deducting extraction A as well — the commonest slip on this drawing — would give 135.5 MW, understating the cylinder by 13.5 MW or 9 %.
6. Part (c) — the IP inlet is a mixture. The IP cylinder receives the 385.72 kg/s of hot reheat at 3536 kJ/kg and the 4.06 kg/s HP bypass at 3396 kJ/kg, so $\dot{m}_{IP} = 389.78$ kg/s at a mixed enthalpy of $(385.72 \times 3536 + 4.06 \times 3396)/389.78 = 3534.5$ kJ/kg. That total is confirmed exactly by the four printed outlet flows, $11.49 + 24.54 + 13.36 + 340.39 = 389.78$ kg/s — the closure that proves the bypass belongs at the IP inlet and nowhere else on the drawing.
7. Part (c) — IP turbine power as an energy balance. Because the cylinder has one extraction at a different enthalpy from the exhaust, the power is the net enthalpy flux rather than a single mass times a single drop: $$\dot{W}_{IP} = \left[\dot{m}_{hrh}h_{hrh} + \dot{m}_{byp}h_{ms}\right] - \left[\dot{m}_B h_B + (\dot{m}_C + \dot{m}_{BFPT} + \dot{m}_{cross})h_{IPex}\right]$$ Substituting, energy in is $385.72(3536) + 4.06(3396) = 1\,377\,694$ kW and energy out is $11.49(3347) + 378.29(3228) = 1\,259\,577$ kW, so $$\dot{W}_{IP} = 1\,377\,694 - 1\,259\,577 = 118\,117\ \text{kW} = \boxed{118.1\ \text{MW}}$$
8. Check the two answers against the machine. HP and IP together develop 148.9 + 118.1 = 267.0 MW, which is 52.2 % of the 512.008 MW at the generator terminals. The balance is produced by the two LP cylinders on the same shaft, expanding 340.39 kg/s from 3228 kJ/kg down to the 0.0034 MPa condenser — a proportion entirely typical of a single-reheat subcritical unit, where the LP end normally contributes just under half the output.