PWR Heat Generation: 22-Mec-B3 Energy Conversion and Power Generation, December 2019, Q4

Question 4: PWR Heat Generation (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.

Check: the average neutron flux is printed on the source page as 4.5 × 10¹⁷ neutron/m²s. That is the value used throughout; it is the standard order for a commercial PWR core, and it is confirmed by part (c) independently reproducing the same thermal power from the coolant balance.

Given. The geometry and the reactor-physics data that fix the fission rate, plus the primary-coolant conditions:

Quantity	Symbol	Value
Assemblies × rods per assembly	$N$	157 × 264 = 41 448 rods
Fuel pellet diameter	$d_p$	8.19 mm
Fuel rod effective length	$L$	3.658 m
Uranium dioxide density	$\rho$	10 400 kg/m³
Average U-235 enrichment	$e$	2.8 %
Effective fission cross-section	$\sigma_f$	380 barns = 3.80 × 10⁻²⁶ m²
Average neutron flux	$\phi$	4.5 × 10¹⁷ n/m²s
Energy per fission	$E_f$	32 pJ
Coolant inlet / outlet	$T$	286 / 325 °C at 15.5 MPa
Coolant flow rate	$\dot{m}$	12 600 kg/s

Find. The mass of uranium dioxide in the core, the heat release rate the fission data imply, and the thermal power the coolant balance gives — the last being an independent check on the first two.

The three length scales of the core. Only the pellet stack carries fuel: the cladding and the pellet-clad gap are excluded from the volume in part (a), and the lattice pitch and core diameter play no part in the heat-generation calculation at all — they are there so that the power density and linear rating can be checked.

Approach. Get the fuel volume from the pellet stack alone, convert it to a number of U-235 nuclei through the molecular mass and Avogadro's number, multiply by the microscopic cross-section, the flux and the energy per fission to get power, and finally close a separate energy balance on the primary coolant using compressed-liquid enthalpies rather than a constant specific heat.

Part (a) — count only the fuel itself. The fuel is the pellet stack; the Zircaloy cladding and the gap beneath it are not uranium dioxide, so the rod outside diameter and the 0.57 mm clad thickness are distractors here. Per rod, $V_{rod} = \frac{\pi}{4}d_p^2 L = \frac{\pi}{4}(0.00819)^2 \times 3.658 = 1.9271 \times 10^{-4}\ \text{m}^3$, so with 41 448 rods the core holds $V = 7.987$ m³ and $$m = \rho V = 10\,400 \times 7.987 = 83\,069\ \text{kg} = \boxed{83.1\ \text{tonnes of UO}_2}$$ That is the right order for a three-loop PWR, which typically carries 80–90 t.
Part (b) — convert mass to a count of fissile nuclei. The question directs us to the dominant isotopes, so uranium dioxide is taken as $^{238}\text{U}^{16}\text{O}_2$ and $M = 238 + 2(16) = 270$ kg/kmol. The core therefore holds $n = 83\,069/270 = 307.7$ kmol, and $N_U = 307.7 \times 10^3 \times 6.022 \times 10^{23} = 1.853 \times 10^{29}$ uranium atoms — one per molecule. At 2.8 % enrichment $$N_{235} = 0.028 \times 1.853 \times 10^{29} = \boxed{5.188 \times 10^{27}\ \text{atoms}}$$
Part (b) — fission rate from the reaction-rate density. The macroscopic reaction rate in a volume containing $N$ target nuclei exposed to a flux $\phi$ is $R = N\sigma_f\phi$. Substituting, $$R = 5.188 \times 10^{27} \times 3.80 \times 10^{-26} \times 4.5 \times 10^{17} = 8.871 \times 10^{19}\ \text{fissions/s}$$ The units check: nuclei × m² × (neutrons per m² per second) leaves reactions per second.
Part (b) — heat release rate. Each fission deposits 32 pJ, so $$\dot{Q}_{fission} = R\,E_f = 8.871 \times 10^{19} \times 32 \times 10^{-12} = 2.839 \times 10^{9}\ \text{W} = \boxed{2839\ \text{MW(th)}}$$ Two sanity checks on the geometry data that were not otherwise used: over the equivalent core volume $\frac{\pi}{4}(3.040)^2 \times 3.658 = 26.55$ m³ that is a power density of 107 kW/L, and spread over the 41 448 rods it is an average linear rating of 18.7 kW/m. Both are textbook values for a commercial PWR, so the fission chain is sound.
Part (c) — do not use a constant specific heat here. The coolant is compressed liquid at 15.5 MPa, only 20 K below its saturation temperature of 344.8 °C, where the specific heat has risen far above its cold-water value. Taking $c_p = 4.19$ kJ/kg°C from the page-17 constant sheet would give $12\,600 \times 4.19 \times 39 = 2059$ MW, which is 27 % below the fission answer — the tell that the approximation has failed. The compressed-liquid table must be read instead: $h = 1263.6$ kJ/kg at 286 °C and $h = 1484.4$ kJ/kg at 325 °C.
Part (c) — thermal power from the coolant balance. With the correct enthalpy rise $\Delta h = 1484.4 - 1263.6 = 220.8$ kJ/kg (an effective specific heat of 5.66 kJ/kg°C), $$\dot{Q}_{coolant} = \dot{m}\,\Delta h = 12\,600 \times 220.8 = 2.782 \times 10^{6}\ \text{kW} = \boxed{2783\ \text{MW(th)}}$$ This is 98.0 % of the 2839 MW obtained from the fission parameters. The 2 % gap is exactly what one expects: a small part of the fission energy escapes the fuel as neutrino energy and as gamma energy deposited in the reflector and vessel rather than in the coolant, and the flux and cross-section are quoted to two significant figures. Two entirely independent routes agreeing to 2 % is the confirmation the question is looking for.

Quantity	Result
(a) Fuel volume per rod	192.7 cm³
(a) Total fuel volume	7.987 m³
(a) Mass of uranium dioxide	83 069 kg = 83.1 t
(b) Molecular mass of UO₂	270 kg/kmol
(b) Uranium atoms in the core	1.853 × 10²⁹
(b) U-235 atoms in the core	5.188 × 10²⁷
(b) Fission rate	8.871 × 10¹⁹ /s
(b) Total heat release rate	2839 MW(th)
(c) Coolant enthalpy rise at 15.5 MPa	220.8 kJ/kg
(c) Thermal power from coolant balance	2783 MW(th)
Agreement between (b) and (c)	98.0 %

Concept & Theory

Topic. Core heat generation from fission parameters, and its confirmation by the primary-coolant energy balance.

Key relations. $V = N\frac{\pi}{4}d_p^2 L$; $N_{235} = e\,(m/M)N_A$; $R = N_{235}\sigma_f\phi$; $\dot{Q} = R\,E_f$; $\dot{Q} = \dot{m}(h_{out} - h_{in})$ with compressed-liquid enthalpies.

Why this works. Reactor power is the product of how many fissile nuclei are present, how large a target each presents, how many neutrons pass per unit area per second, and how much energy each fission releases. The microscopic cross-section is literally an area, so $N\sigma_f\phi$ has units of reactions per second directly and no further conversion is needed. The coolant balance is a completely separate physical route to the same number, which is why the paper asks for both: agreement to a few per cent validates the geometry, the enrichment and the flux at once. That check only works if the enthalpy rise is taken from the compressed-liquid table, because near the saturation line at 15.5 MPa water's specific heat is about 5.7 rather than 4.2 kJ/kg°C.

Common pitfall. Three pitfalls. Using the rod outside diameter instead of the pellet diameter inflates the fuel mass by 35 %. Taking the molecular mass from U-235 rather than the dominant U-238 shifts the nuclide count by more than 1 %. And using $c_p = 4.19$ for the primary coolant understates the thermal power by a quarter, which then looks like an error in the fission calculation when it is not.

Source. Lamarsh & Baratta, Introduction to Nuclear Engineering, §3.5 and ch. 8 (heat removal from nuclear reactors); El-Wakil, Powerplant Technology, ch. 10; compressed-liquid data from Granet & Bluestein's bound steam tables.

Question 4 of 8: PWR Heat Generation

Question 4: PWR Heat Generation (10 marks)