Question 2 of 6: Power Plant Efficiency and Heat Discharge

Question 2: 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.

Part I — Power plant efficiency (9 marks)

Given (read from the page-8 heat balance diagram, Fig. 7, a 600 MW subcritical-pressure reheat regenerative unit; mass flows in kg/s, enthalpies in kJ/kg).

Find. (a) the steam cycle efficiency defined as electrical output over thermal input, (b) the HP turbine power from an energy balance on steam properties, and (c) the power absorbed by the boiler feedwater pump, again from steam properties.

Approach. Take the working fluid as the accounting boundary: heat in is what the steam generator and the reheater add to the fluid, and each machine's power comes from a steady-flow energy balance \(\dot W = \sum \dot m h|_{in} - \sum \dot m h|_{out}\) written on the streams the diagram labels.

1. Part (a) — confirm the mass balance before using any enthalpy. The HP turbine takes 514.1 kg/s and the diagram labels six streams leaving it: \[ 475.1 + 31.9 + 1.2 + 0.9 + 4.9 + 0.1 = 514.1\ \text{kg/s} \] The balance closes exactly, which validates the readings and fixes the total gland leak-off at 7.1 kg/s. Doing this first is worthwhile because the diagram is the only source of data and an arithmetic closure is the cheapest possible check on it.
2. Add the heat supplied in the steam generator. Feedwater enters the economizer at 1107 kJ/kg and leaves as primary steam at 3387 kJ/kg, all 514.1 kg/s of it: \[ \dot Q_{sg} = \dot m\,(h_{ms} - h_{fw}) = 514.1\,(3387 - 1107) = 1\,172\,148\ \text{kW} \]
3. Add the heat supplied in the reheater. Only the cold reheat stream is reheated, from 3063 to 3530 kJ/kg: \[ \dot Q_{rh} = 475.1\,(3530 - 3063) = 221\,872\ \text{kW} \] Summing the two, the total thermal input to the working fluid is \[ \dot Q_{in} = 1\,172\,148 + 221\,872 = 1\,394\,020\ \text{kW} = 1394.0\ \text{MW} \] Note that this is heat into the steam, not fuel heat — the question defines the efficiency on thermal input, so boiler losses lie outside the boundary.
4. Form the steam cycle efficiency. The electrical output is the labelled generator output of 608 068 kW: \[ \eta_{cycle} = \frac{P_{gen}}{\dot Q_{in}} = \frac{608\,068}{1\,394\,020} \] \[ \boxed{\eta_{cycle} = 0.4362 \;\;\text{(43.62\%)}} \] Equivalently a turbine heat rate of \(3600 \times 1\,394\,020/608\,068 = 8253\) kJ/kWh, which is squarely in the expected band for a 17 MPa single-reheat subcritical unit and so corroborates the readings. For reference the shaft power is \(608\,068 + 1914 + 8182 = 618\,164\) kW, the fixed and generator losses being the two loss arrows on the diagram.
5. Part (b) — energy balance on the HP turbine. With the control volume drawn in the figure above, and noting that the 31.9 kg/s extraction leaves at the same 3063 kJ/kg as the exhaust while the 7.1 kg/s of gland steam leaves on the seal header at about 3315 kJ/kg: \[ \dot W_{HP} = \dot m_{in}h_{in} - \big[(\dot m_{rh}+\dot m_{ext})h_{cr} + \dot m_{leak}h_{seal}\big] \] Substituting, \[ \dot W_{HP} = 514.1(3387) - 507.0(3063) - 7.1(3315) \] \[ \dot W_{HP} = 1\,741\,257 - 1\,552\,941 - 23\,537 \] \[ \boxed{\dot W_{HP} = 164\,779\ \text{kW} \approx 164.8\ \text{MW}} \]
6. Sanity-check the HP result two ways. Treating the whole 514.1 kg/s as expanding to the exhaust enthalpy gives \(514.1(3387-3063) = 166\,568\) kW, only 1.1% high — the difference is exactly the gland steam that does not expand the full way, \(7.1(3315-3063) = 1789\) kW. The HP section therefore takes about 27% of the 618 MW shaft total, which is the normal share for the high-pressure cylinder of a single-reheat machine.
7. Part (c) — feed pump power from the steam side of its drive turbine. The feed pump on this unit is turbine-driven, so its power is most directly obtained from the steam expanding through the drive turbine, 16.0 kg/s from 3212 to 2480 kJ/kg: \[ \dot W_{bfpt} = \dot m\,(h_{in}-h_{out}) = 16.0\,(3212-2480) \] \[ \boxed{\dot W_{bfp} = 11\,712\ \text{kW} \approx 11.7\ \text{MW}} \]
8. Cross-check on the water side. The diagram also gives the pump's own enthalpy rise, Δh = 23 kJ/kg on 514.1 kg/s at 19.99 MPa, so \[ \dot W_{bfp} = 514.1 \times 23 = 11\,824\ \text{kW} \] The two independent routes give 11 712 kW and 11 824 kW, and the diagram itself prints 11 682 kW against the pump. All three agree within 1.2%, the residual being drive-coupling loss and the round-off in a Δh quoted to the nearest whole kJ/kg. Taking the printed value as the reference, the pump absorbs a little under 2% of the plant's gross output — the largest single auxiliary on the unit.

Part II — Heat discharge (6 marks)

Given.

Find. For each plant, the total rate of heat discharged into the cooling water and the total rate of heat lost to the atmosphere.

Approach. Work backwards from the fixed 1000 MW of electricity through each efficiency in turn to build a complete energy account, then post each loss to the cooling water or to the atmosphere according to how the question says that component is cooled.

1. Part (a) — step back from electricity to shaft power. The generator's 0.96 electrical efficiency applies to both plants, so \[ \dot W_{shaft} = \frac{P_e}{\eta_{elec}} = \frac{1000}{0.96} = 1041.67\ \text{MW} \] and the generator loss is \(1041.67 - 1000 = 41.67\) MW for each plant. Because the electrical equipment is air cooled, this 41.67 MW goes to the atmosphere, not to the cooling water.
2. Step back from shaft power to the heat taken in by the steam cycle. The steam cycle efficiency is work out over heat into the cycle, so \[ \dot Q_{cycle} = \frac{\dot W_{shaft}}{\eta_{cycle}} \] For the CANDU plant, \(1041.67/0.33 = 3156.57\) MW; for the coal plant, \(1041.67/0.41 = 2540.65\) MW.
3. Get the condenser duty from the cycle balance. The condenser rejects whatever the cycle took in and did not turn into shaft work: \[ \dot Q_{cond} = \dot Q_{cycle} - \dot W_{shaft} \] CANDU: \(3156.57 - 1041.67 = 2114.90\) MW. Coal: \(2540.65 - 1041.67 = 1498.98\) MW. This is the single most important line in the question — the condenser is credited with the cycle loss, never with the fuel-side loss.
4. Step back once more to the fuel or fission heat. Reactor and boiler thermal efficiency is defined in the question as heat output via steam or coolant over heat input from fuel, so \[ \dot Q_{fuel} = \frac{\dot Q_{cycle}}{\eta_{unit}} \] CANDU: \(3156.57/0.99 = 3188.45\) MW, leaving a reactor loss of 31.88 MW. Coal: \(2540.65/0.94 = 2702.82\) MW, leaving a boiler loss of 162.17 MW.
5. Post each loss to its cooling medium and total the cooling water. The reactor is water cooled, so the CANDU plant's 31.88 MW of reactor loss joins the condenser duty in the cooling water; the coal plant's 162.17 MW of boiler loss leaves up the stack with the flue gas, so it does not. \[ \boxed{\dot Q_{water,CANDU} = 2114.90 + 31.88 = 2146.8\ \text{MW}} \] \[ \boxed{\dot Q_{water,coal} = 1499.0\ \text{MW}} \]
6. Part (b) — total the heat lost to the atmosphere. For the CANDU plant only the air-cooled generator loss qualifies; for the coal plant the stack loss adds to it. \[ \boxed{\dot Q_{air,CANDU} = 41.7\ \text{MW}} \] \[ \boxed{\dot Q_{air,coal} = 162.17 + 41.67 = 203.8\ \text{MW}} \] Both accounts close on the fuel input: for the CANDU plant \(1000 + 2146.8 + 41.7 = 3188.5\) MW, and for the coal plant \(1000 + 1499.0 + 203.8 = 2702.8\) MW.
7. Interpret the comparison. For identical electrical output the nuclear station discharges 2147 MW to the cooling water against the fossil station's 1499 MW — 43% more thermal load on the receiving river or lake. Two effects compound: the lower steam cycle efficiency (0.33 against 0.41, because a CANDU's saturated steam conditions are far below a fossil unit's 538 °C superheat) means much more heat must be rejected per unit of work, and the reactor has no stack, so essentially all of its loss must also go into water. In cooling-water terms, at a 10 °C temperature rise the CANDU plant needs \(2147/(4.19 \times 10) \approx 51\ \mathrm{m^3/s}\) against about 36 m³/s for the coal plant — which is why Canadian CANDU stations sit on the Great Lakes or the Bay of Fundy, and why discharge-plume modelling and thermal-limit permitting under the Fisheries Act are a routine part of their design.