Question 4 of 6: Power Plant Heat Discharge and Cooling Towers

Question 4: Power Plant Heat Discharge and Cooling Towers (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 II, cooling tower: electrical output 600 MW; heat rejected 1500 MJ/s; turbine exhaust 30 °C; cooling water 15 °C in, 25 °C out; ambient air 30 °C dry bulb at 40 % relative humidity; evaporative loss read from the page-12 chart.

Find. Part I: for each plant the heat rejected to cooling water and to atmosphere, the cooling-water volume flow and the water used per unit generated. Part II: the cooling-water flow, the evaporative loss in three forms and the water consumption per kWh.

Approach. Work backwards from the electrical output through the generator, the cycle and the heat source to find the fuel heat, then attribute each loss to the medium the question specifies. Everything that is neither electricity nor an air-side loss must end up in the cooling water. For Part II, size the water circuit from the condenser duty, then read the evaporative loss from the page-12 psychrometric chart.

1. Part I — Setting up the energy chain. The definitions given run in series: fuel heat \(\dot Q_f\) enters the reactor or boiler, which delivers \(\dot Q_s = \eta_b \dot Q_f\) to the steam; the cycle converts \(\dot W = \eta_{cycle}\dot Q_s\) to shaft work; the generator delivers \(P_e = \eta_{elec}\dot W\). Working backwards from 1000 MW, the shaft power is the same for both plants: $$\dot W = \frac{P_e}{\eta_{elec}} = \frac{1000}{0.96} = 1041.7\ \text{MW}$$
2. Part I (a) — CANDU: heat discharged to the cooling water. $$\dot Q_s = \frac{1041.7}{0.33} = 3156.6\ \text{MW},\qquad \dot Q_f = \frac{3156.6}{0.99} = 3188.5\ \text{MW}$$ The condenser rejects everything the steam brought that did not become shaft work, \(3156.6 - 1041.7 = 2114.9\ \text{MW}\); and because the question states the reactor is water cooled, the reactor's own 31.9 MW loss joins it: $$\boxed{\dot Q_{water,\ CANDU} = 2114.9 + 31.9 = 2146.8\ \text{MW}}$$
3. Part I (a) concluded — Coal plant: heat discharged to the cooling water. $$\dot Q_s = \frac{1041.7}{0.41} = 2540.7\ \text{MW},\qquad \dot Q_f = \frac{2540.7}{0.94} = 2702.8\ \text{MW}$$ Here the boiler loss leaves up the stack, so only the condenser duty reaches the water: $$\boxed{\dot Q_{water,\ coal} = 2540.7 - 1041.7 = 1499.0\ \text{MW}}$$ The nuclear station therefore imposes 43 % more thermal load on the receiving water for the same saleable megawatts.
4. Part I (b) — Heat lost to the atmosphere. The generator and its auxiliaries are air cooled in both cases and lose \(1041.7 - 1000 = 41.7\ \text{MW}\). The coal plant adds its boiler loss of \(2702.8 - 2540.7 = 162.2\ \text{MW}\): $$\boxed{\dot Q_{air,\ CANDU} = 41.7\ \text{MW};\qquad \dot Q_{air,\ coal} = 162.2 + 41.7 = 203.8\ \text{MW}}$$ As a check the first law must close for each plant: \(1000 + 2146.8 + 41.7 = 3188.5\) ✓ and \(1000 + 1499.0 + 203.8 = 2702.8\) ✓.
5. Part I (c) — Cooling-water flow rate at a 10 °C rise. With \(\dot Q = \dot m c_p \Delta T\) and \(\rho = 1000\ \text{kg/m}^3\), $$\dot V = \frac{\dot Q}{\rho\,c_p\,\Delta T}$$ $$\dot V_{CANDU} = \frac{2\,146\,800}{1000 \times 4.19 \times 10} = 51.2\ \text{m}^3\text{/s}, \qquad \dot V_{coal} = \frac{1\,498\,980}{41\,900} = 35.8\ \text{m}^3\text{/s}$$ $$\boxed{\dot V_{CANDU} = 51.2\ \text{m}^3\text{/s};\qquad \dot V_{coal} = 35.8\ \text{m}^3\text{/s}}$$ For scale, 51 m³/s is roughly the mean flow of a substantial river — which is precisely why once-through cooling drives station siting onto the Great Lakes or the seaboard.
6. Part I (d) — Cooling water per unit generated. Convert to a per-kilowatt-hour basis by multiplying by 3600 s/h and dividing by 10⁶ kW: $$\text{CANDU: } \frac{51.24 \times 3600}{1\,000\,000} = 0.184\ \text{m}^3\text{/kWh}\qquad \text{Coal: } \frac{35.78 \times 3600}{1\,000\,000} = 0.129\ \text{m}^3\text{/kWh}$$ $$\boxed{0.184\ \text{m}^3\text{/kWh (CANDU)};\qquad 0.129\ \text{m}^3\text{/kWh (coal)}}$$ That is about 184 litres of lake water pushed through the condenser for every kilowatt-hour of CANDU electricity — a useful figure for arguing a thermal-discharge permit under provincial water-quality regulation.
7. Part II (a) — Cooling-water flow round the tower circuit. The condenser rejects 1500 MJ/s into water that rises from 15 °C to 25 °C: $$\dot m_w = \frac{\dot Q}{c_p \Delta T} = \frac{1\,500\,000}{4.19 \times 10} = 35\,800\ \text{kg/s}$$ $$\boxed{\dot V_w = 35.8\ \text{m}^3\text{/s}}$$ The 30 °C turbine exhaust sits 5 °C above the 25 °C water leaving the condenser, a normal terminal difference, but the 15 °C cold water is 5.1 °C below the 20.1 °C ambient wet bulb found in part (b). A wet tower cannot cool water below the wet bulb, so the printed operating conditions are not mutually consistent; none of parts (a)–(e) depends on the approach, so the data are used as given and the inconsistency is simply noted.
8. Part II (b) — Evaporative loss from the page-12 chart. The chart is entered with the atmospheric dry-bulb temperature on the horizontal axis and the wet-bulb temperature on the vertical axis, the humidity lines locating the point. At 30 °C dry bulb and 40 % relative humidity the wet bulb is 20.1 °C, and that point falls on the contour marked $$\boxed{0.375\ \text{m}^3\text{/GJ of heat rejected}}$$ Sanity check: if every joule left as latent heat, the loss would be \(10^6/h_{fg} = 10^6/2430 = 412\ \text{kg}\), i.e. 0.412 m³/GJ. The chart value is 91 % of that ceiling, which is what hot, moderately dry air should give — most of the duty leaves as vapour and only a tenth as sensible heating of the air.
9. Part II (c) — Evaporative loss as a volume flow. The heat rejected is 1500 MJ/s = 1.5 GJ/s, so $$\dot V_{evap} = 0.375 \times 1.5$$ $$\boxed{\dot V_{evap} = 0.5625\ \text{m}^3\text{/s}}$$ i.e. 562.5 litres every second leaving the tower as a plume.
10. Part II (d) — Percentage loss of the circulating water. $$\frac{\dot V_{evap}}{\dot V_w} \times 100 = \frac{0.5625}{35.80}\times 100$$ $$\boxed{1.57\ \%}$$ The rule of thumb is about 1 % of circulating flow per 6 °C of cooling range; with a 10 °C range that predicts 1.7 %, so 1.57 % is consistent. This evaporation is what concentrates dissolved solids in the circuit, and it is the reason blowdown and make-up are needed — make-up must cover evaporation plus drift plus blowdown, so it is appreciably more than 1.57 %.
11. Part II (e) — Water consumption per unit generated. Referring the loss to the 600 MW sent out: $$\frac{0.5625\ \text{m}^3\text{/s} \times 1000\ \text{L/m}^3 \times 3600\ \text{s/h}}{600\,000\ \text{kW}}$$ $$\boxed{3.375\ \text{L/kWh generated}}$$ Compare Part I: a once-through plant circulates 129 litres per kWh but consumes essentially none of it, whereas this tower circulates 215 litres and truly consumes 3.4. That trade — a fifty-fold cut in thermal discharge to the receiving water, bought with real consumptive water use — is the central decision in choosing between once-through and evaporative cooling.