Question 3 of 6: Power Plant Efficiency and Heat Discharge

Question 3: Power Plant Efficiency and Heat Discharge (15 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. Part I works from the page-9 heat balance diagram of a 600 MW subcritical-pressure reheat regenerative unit; Part II works from a short table of efficiencies for two 1000 MW stations.

Find. Part I: the steam cycle efficiency, the high-pressure turbine power and the boiler feed pump power, all from steam properties. Part II: for each of a CANDU and a coal-fired station of 1000 MW electrical output, the rate of heat discharged to the cooling water and the rate lost to the atmosphere.

Approach. Part I is three control-volume energy balances taken straight off the diagram, each validated against a figure the diagram itself prints. Part II is a routing problem: decide which loss goes to water and which to air, then work back from the electrical output through the generator, the cycle and the boiler or reactor.

1. Part I — first confirm the diagram's own mass balance. Before reading any enthalpy, check that the flows close, because a misread stream is far more likely than a misread number. Around the high-pressure casing:$$475.1+31.9+1.2+0.9+4.9+0.1=514.1\ \text{kg/s}$$which matches the primary steam flow exactly. The four small streams total 7.1 kg/s of gland leak-off and are easy to miss; counting them is what makes part (b) come out right.
2. Part I (a) — steam cycle efficiency. The thermal input is the heat added to the working fluid in the steam generator plus that added in the reheater:$$\dot{Q}_{sg}=514.1\,(3387-1107)=1\,172\,148\ \text{kW}$$$$\dot{Q}_{rh}=475.1\,(3530-3063)=221\,872\ \text{kW}$$$$\dot{Q}_{in}=1\,172\,148+221\,872=1\,394\,020\ \text{kW}$$The question defines the efficiency on electrical output, so the generator output printed on the diagram is the numerator:$$\boxed{\eta_{cycle}=\frac{608\,068}{1\,394\,020}=0.4362\ \text{or}\ 43.6\,\%}$$Equivalently the turbine heat rate is $3600\times 1\,394\,020/608\,068 = 8253$ kJ/kWh, a thoroughly typical figure for a 600 MW subcritical reheat unit and a good check that nothing has been dropped.
3. Part I (b) — high-pressure turbine power. Take the casing as a control volume. All 514.1 kg/s enters at 3387 kJ/kg. The cold reheat and the sixth-heater extraction both leave at the exhaust enthalpy of 3063 kJ/kg, while the 7.1 kg/s of gland leak-off leaves at the seal-header enthalpy of 3315 kJ/kg:$$P_{HP}=\dot{m}_{in}h_{in}-\sum \dot{m}_{out}h_{out}$$$$P_{HP}=514.1(3387)-(475.1+31.9)(3063)-7.1(3315)$$$$\boxed{P_{HP}=164\,779\ \text{kW}\approx 164.8\ \text{MW}}$$The cruder estimate that expands the whole flow to the exhaust enthalpy, $514.1(3387-3063)=166\,568$ kW, is 1.1 % high — precisely because it credits the leak-off streams with an expansion they never completed. The HP casing therefore produces about 27 % of the unit's output.
4. Part I (c) — boiler feed pump power. The pump is driven by its own steam turbine, so the cleanest route is the drive turbine's steam properties:$$P_{BFPT}=\dot{m}(h_{in}-h_{out})=16.0\,(3212-2480)=11\,712\ \text{kW}$$$$\boxed{P_{BFP}\approx 11\,712\ \text{kW}\ (11.7\ \text{MW})}$$The water side gives an independent confirmation: with the diagram's $\Delta h = 23$ kJ/kg at 19.99 MPa,$$P = 514.1\times 23 = 11\,824\ \text{kW}$$The two routes bracket the 11 682 kW printed on the diagram, one 0.3 % above it and the other 1.2 % above, which is as close as a heat balance diagram is drawn. The feed pump absorbs 1.9 % of the generator output — a reminder that at supercritical and near-supercritical pressures the feed pump is a major auxiliary in its own right, which is exactly why it is turbine driven.
5. Part II — establish the shaft power and the generator loss. Both plants deliver 1000 MW electrical at an electrical efficiency of 0.96, so each turbine shaft must produce$$W_{shaft}=\frac{P_e}{\eta_e}=\frac{1000}{0.96}=1041.7\ \text{MW}$$and each generator dissipates $1041.7-1000=41.7$ MW. The question directs that the electrical equipment is air cooled, so this 41.7 MW goes to the atmosphere in both cases.
6. Part II (a) — heat discharged to the cooling water. For each plant the heat entering the steam cycle follows from the cycle efficiency, and the condenser receives the remainder of it after the shaft work is taken out:$$\dot{Q}_{cycle}=\frac{W_{shaft}}{\eta_{cycle}},\qquad \dot{Q}_{cond}=\dot{Q}_{cycle}-W_{shaft}$$For the CANDU plant, $\dot{Q}_{cycle}=1041.7/0.33=3156.6$ MW and $\dot{Q}_{cond}=2114.9$ MW. The fission heat is $\dot{Q}_{f}=3156.6/0.99=3188.4$ MW, so the reactor itself loses 31.9 MW — and the question states the reactor is water cooled, so that loss joins the condenser duty:$$\boxed{\text{CANDU: }\dot{Q}_{water}=2114.9+31.9=2146.8\ \text{MW}}$$For the coal plant, $\dot{Q}_{cycle}=1041.7/0.41=2540.7$ MW and $\dot{Q}_{cond}=1499.0$ MW. Its fuel heat is $2540.7/0.94=2702.8$ MW, so 162.2 MW is lost in the boiler — but that leaves as flue gas, up the stack, not into the water:$$\boxed{\text{Coal: }\dot{Q}_{water}=1499.0\ \text{MW}}$$
7. Part II (b) — heat lost to the atmosphere. Collect the air-cooled losses for each plant:$$\boxed{\text{CANDU: }\dot{Q}_{air}=41.7\ \text{MW}\ \text{(generator only)}}$$$$\boxed{\text{Coal: }\dot{Q}_{air}=41.7+162.2=203.8\ \text{MW}\ \text{(generator + stack)}}$$Each plant's books must balance, and they do: for the CANDU station $1000+2146.8+41.7=3188.5$ MW of fission heat, and for the coal station $1000+1499.0+203.8=2702.8$ MW of fuel heat.
8. Part II — interpret the result, which is the point of the question. The nuclear station discharges 43.2 % more heat into the cooling water than the fossil station of identical electrical output. Two independent causes compound: its cycle efficiency is lower (0.33 against 0.41, because a CANDU delivers saturated rather than superheated steam at a much lower temperature), and its unit loss is water cooled rather than being carried away by flue gas. At a 10 °C permitted rise the two plants need 51.2 and 35.8 m³/s of cooling water respectively — which is why nuclear stations are so often sited on the coast or on a large lake, and why their thermal-discharge permits are the binding environmental constraint.