Question 4 of 6: PWR Heat Generation — fuel inventory, fission heat release and core power densities

Question 4: PWR Heat Generation — fuel inventory, fission heat release and core power densities (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. A 157-assembly pressurised water reactor core with 264 rods per 17 × 17 assembly and the geometry, material and neutronic data below; Avogadro’s number is taken from page 16 of the paper as $N_{A}=0.602\times10^{24}$ atoms per mole.

Find. The uranium dioxide inventory, the fission heat release rate, the fuel and core power densities, the average rod surface heat flux, and the thermal power obtained independently from the coolant enthalpy rise.

Approach. The fuel inventory is pure geometry, the pellet volume times the rod count times the density. The heat release follows the three reference equations on page 16 of the paper: nuclei per unit volume from the density and molecular mass, fissile nuclei from the enrichment, and volumetric heat release from the fission rate times the energy per fission. The power densities and heat flux are then that power divided by the appropriate mass, volume or area, and the coolant enthalpy rise gives a wholly independent check on the answer.

1. Part (a) — mass of uranium dioxide. Each rod holds a stack of pellets of diameter $D_{p}$ over the effective length $L$: $$V_{rod}=\frac{\pi}{4}D_{p}^{2}L=\frac{\pi}{4}(8.19\times10^{-3})^{2}\times3.658=1.9271\times10^{-4}\ \text{m}^{3}$$ Over $n=41\,448$ rods the fuel volume is $V=41\,448\times1.9271\times10^{-4}=7.987\ \text{m}^{3}$, so $$m_{UO_{2}}=\rho V=10\,400\times7.987=\boxed{83\,069\ \text{kg}}$$ Of this, the uranium fraction $238/270$ accounts for 73 224 kg of uranium metal, of which 2.8 % or about 2050 kg is U-235. As a geometric check, the clad bore is $9.5-2\times0.57=8.36\ \text{mm}$ against a pellet of 8.19 mm, leaving a radial gap of 0.085 mm for fission-gas release and pellet swelling.
2. Part (b) — nuclei per unit volume. Taking the dominant isotopes as the question directs, the molecular mass of uranium dioxide is $M=238+2\times16=270\ \text{g}\cdot\text{mol}^{-1}$. The reference equation for the number of nuclei per cubic centimetre is $$N=\frac{N_{A}\rho}{M}=\frac{0.602\times10^{24}\times10.400}{270}=2.3188\times10^{22}\ \text{molecules}\cdot\text{cm}^{-3}$$ where the density has been expressed in g/cm³ to match the units of $N_{A}$ and $M$. Each molecule contains one uranium atom.
3. Part (b) continued — fissile nuclei and heat release. The enrichment gives the fissile fraction, $$N_{f}=\gamma N=0.028\times2.3188\times10^{22}=6.493\times10^{20}\ \text{U-235 nuclei}\cdot\text{cm}^{-3}$$ and the volumetric heat release is the fission rate per unit volume multiplied by the energy released per fission: $$q^{*}=N_{f}\sigma_{f}\phi E_{f}=6.493\times10^{20}\times380\times10^{-24}\times4.5\times10^{13}\times32\times10^{-12}=355.3\ \text{W}\cdot\text{cm}^{-3}$$
4. Part (b) continued — total power. Multiplying by the fuel volume, and noting that $1\ \text{m}^{3}=10^{6}\ \text{cm}^{3}$, $$\dot{Q}=q^{*}V=355.3\times7.987\times10^{6}=2.838\times10^{9}\ \text{W}=\boxed{2838\ \text{MW}}$$ which is a thoroughly typical thermal rating for a four-loop pressurised water reactor of this size.
5. Part (c) — average fuel power density. Referred to the mass of uranium dioxide, $$\frac{\dot{Q}}{m_{UO_{2}}}=\frac{2.838\times10^{6}\ \text{kW}}{83\,069\ \text{kg}}=\boxed{34.16\ \text{kW}\cdot\text{kg}^{-1}}$$ This is the figure that governs fuel burnup and, with the thermal conductivity of uranium dioxide, the centreline temperature of the pellet.
6. Part (d) — average core power density. The core power density is defined on the whole core envelope, not on the fuel alone, so it uses the equivalent core cylinder: $$V_{core}=\frac{\pi}{4}D_{c}^{2}L=\frac{\pi}{4}(3.040)^{2}\times3.658=26.55\ \text{m}^{3}$$ and therefore $$\frac{\dot{Q}}{V_{core}}=\frac{2838}{26.55}=\boxed{106.9\ \text{MW}\cdot\text{m}^{-3}}$$ The fuel occupies only $7.987/26.55=30$ % of the core volume, the balance being clad, coolant, guide tubes and structure, which is why the core figure is so much lower than the volumetric heat release inside the pellets.
7. Part (e) — average rod heat flux. All of the fission heat must cross the outside surface of the cladding, whose total area is $$A=\pi D_{r}Ln=\pi\times9.5\times10^{-3}\times3.658\times41\,448=4525\ \text{m}^{2}$$ so the average surface heat flux is $$q^{\prime\prime}=\frac{\dot{Q}}{A}=\frac{2.838\times10^{6}}{4525}=\boxed{627\ \text{kW}\cdot\text{m}^{-2}}$$ Equivalently the average linear heat rate is $2.838\times10^{6}/(41\,448\times3.658)=18.7\ \text{kW}\cdot\text{m}^{-1}$. Since the real axial and radial flux distributions peak well above the average, the hot channel sees roughly twice these figures, and it is the peak rather than the average that must be kept clear of the critical heat flux.
8. Part (f) — thermal power from the coolant flow. The coolant is single-phase liquid throughout, so an energy balance on the primary side gives the core power directly from the enthalpy rise. At 15.5 MPa the steam tables give $h_{in}=h(15.5\ \text{MPa},286\,{}^{\circ}\text{C})=1263.6\ \text{kJ}\cdot\text{kg}^{-1}$ and $h_{out}=h(15.5\ \text{MPa},325\,{}^{\circ}\text{C})=1484.4\ \text{kJ}\cdot\text{kg}^{-1}$, a rise of $220.8\ \text{kJ}\cdot\text{kg}^{-1}$. Hence $$\dot{Q}_{c}=\dot{m}_{c}(h_{out}-h_{in})=12\,600\times220.8=2.782\times10^{6}\ \text{kW}=\boxed{2783\ \text{MW}}$$
9. Part (f) continued — reconciling the two routes. The neutronic route gave 2838 MW and the coolant route 2783 MW, a difference of 2.0 %. The two need not agree exactly: a few per cent of the fission energy is deposited outside the fuel, in the moderator and structure by gamma and neutron heating, and the coolant figure is also sensitive to the assumed uniform flux. The agreement to 2 % is the real confirmation that the neutronic data are self-consistent. Note also that the saturation temperature at 15.5 MPa is $344.8\,{}^{\circ}\text{C}$, so the core outlet at $325\,{}^{\circ}\text{C}$ retains 19.8 K of subcooling — the whole reason the primary circuit is pressurised to 15.5 MPa in the first place.