Question 3 of 8: Nuclear Plant Output

Question 3: Nuclear Plant Output (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. One pressurised water reactor unit circulates 16 460 kg/s of primary coolant at 15 MPa, heating it from 290 °C to 318 °C in the core; the two once-through steam generators raise slightly superheated steam at 6 MPa, and the steam cycle and station auxiliaries have stated efficiencies.

Find. The reactor and steam generator heat output rates, the steam flow to the turbine, the gross and net electrical output, and the heat rejected to the cooling water, together with a labelled system sketch.

Approach. Take the reactor duty from the enthalpy rise of compressed liquid water across the core, add the pump work that also enters the coolant, apply the steam generator efficiency, divide the resulting duty by the water-to-steam enthalpy rise for the flow, and close an energy balance on the steam cycle for the rejected heat.

1. Part (b) — do not use the 4.19 constant for the primary coolant. At 15 MPa the water is a compressed liquid near 300 °C, where its specific heat is far above the room-temperature value. Reading the compressed-liquid table, $h(15\ \text{MPa},\,290\ ^\circ\text{C}) = 1284.7$ and $h(15\ \text{MPa},\,318\ ^\circ\text{C}) = 1441.8\ \text{kJ/kg}$, so $$\Delta h = 1441.8 - 1284.7 = 157.1\ \text{kJ/kg},$$ an effective $c_p = 157.1/28 = 5.61\ \text{kJ/kg K}$ — a third higher than 4.19. Using the page-17 constant here would understate the reactor power by a quarter.
2. Reactor heat output. The core delivers $$\dot{Q}_{\text{reactor}} = M \Delta h = 16\,460 \times 157.1 = \boxed{2\,586\,000\ \text{kW} = 2586\ \text{MW(th)}} .$$ The published rating of an Oconee unit is 2568 MW(th), so the calculation lands 0.7 % from the real machine — a free confirmation that the compressed-liquid reading, and not the 4.19 shortcut, is the intended route.
3. Steam generator heat output. The four reactor coolant pumps do 16 MW of work on the coolant, and that work is degraded to heat inside the primary loop, so it must be carried to the steam generators along with the fission heat. Applying the 98 % steam generator efficiency to the sum, $$\dot{Q}_{sg} = \eta_{sg}(\dot{Q}_{\text{reactor}} + P_{\text{pump}}) = 0.98 \times (2\,586\,000 + 16\,000) = \boxed{2\,549\,500\ \text{kW} = 2550\ \text{MW}} .$$ The 2 % shortfall, about 52 MW, is casing and pipework loss to the containment atmosphere.
4. Part (c) — enthalpy rise across the steam generator. At 6 MPa the saturation temperature is 275.6 °C, so the 220 °C feedwater is indeed subcooled and the 300 °C outlet is superheated, exactly as the question states — a once-through generator, with no drum and no recirculation. From the compressed-liquid and superheat tables, $$h_{fw} = 944.6\ \text{kJ/kg},\qquad h_{\text{steam}} = 2885.5\ \text{kJ/kg},$$ $$\Delta h_{sg} = 2885.5 - 944.6 = 1940.9\ \text{kJ/kg} .$$
5. Steam mass flow. Dividing the steam generator duty by that rise, $$M_{\text{steam}} = \frac{\dot{Q}_{sg}}{\Delta h_{sg}} = \frac{2\,549\,500}{1940.9} = \boxed{1314\ \text{kg/s}} ,$$ that is 4730 t/h, split between the two steam generators.
6. Part (d) — turbine power. The steam cycle efficiency is defined on the heat supplied to it, so $$P_{\text{gross}} = \eta_{cy}\dot{Q}_{sg} = 0.36 \times 2\,549\,500 = \boxed{917\,800\ \text{kW} = 918\ \text{MW(e)}} .$$
7. Part (e) — net electrical output. Auxiliaries take 7 % of that gross figure: $$P_{\text{aux}} = 0.07 \times 917\,800 = 64\,200\ \text{kW},\qquad P_{\text{net}} = 0.93 \times 917\,800 = \boxed{853\,600\ \text{kW} = 854\ \text{MW(e)}} .$$ The overall station efficiency is then $853.6/2586 = 33.0$ %, which is the figure the real units achieve and is a second free check on the whole chain.
8. Part (f) — heat rejected to the cooling water. The steam cycle receives $\dot{Q}_{sg}$ and delivers $P_{\text{gross}}$ as electricity; everything else leaves through the condenser. The 16 MW of pump work is already inside that balance because it entered the primary coolant, whereas the remaining 48 MW of auxiliary load is dissipated to atmosphere and never reaches the condenser. Hence $$\dot{Q}_{\text{rej}} = \dot{Q}_{sg} - P_{\text{gross}} = 2\,549\,500 - 917\,800 = \boxed{1\,631\,700\ \text{kW} = 1632\ \text{MW}} .$$ As a check on scale, the cooling water carrying it away at a 10 K rise must flow at $1\,631\,700/(4.19 \times 10) = 38\,940\ \text{kg/s}$, that is 38.9 m³/s — the once-through circulating water demand of a large PWR, and the reason such stations sit on a lake or a large river.