Question 8.10: Consider the PWR design at the end of Section 8.3. Suppose t......
Consider the PWR design at the end of Section 8.3. Suppose that by varying the enrichment in the fuel assemblies and distributing the control poisons in a non-uniform pattern the designers are able to reduce the radial and axial peaking factors to F_r = 1.30 and F_z = 1.46 . Redesign the reactor by solving parts c through g of the pressurized water reactor example using these peaking factors.
With the new peaking factors stipulated, the solution follows the example in the text fairly closely. For the core volume we evaluate Eq. (8.26) at the point of the maximum linear heat rate to obtain {{V}}=A_{c e l l}{{P}}F_{r}F_{z}\,/\,{q}^{\prime}\Big|_{\mathrm{max}} where A_{cell} = p^2 .With the new peaking factors:
V=p^{2}P F_{r}F_{z}\ /q^{\prime}\vert_{\mathrm{max}}=1.536^{2}\cdot3000\cdot10^{6}\cdot1.30\cdot1.46/400
=3.36\cdot10^{7}{\mathrm{~cm}}^{3}=33.6~{\mathrm{m}}^{3}
For a height to diameter ratio of one, V=\pi(H/2)^{2}\,H . Hence
(c) H=(4{ V}/\,\pi)^{1/3}=(4\cdot3.36\cdot10^{7}\;/\,\pi)^{1/3}=350\mathrm{~cm}=3.50\mathrm{~cm}
The core averaged power density is just
(d) \overline{{{P}}}^{\prime \prime \prime}={{{P}}}/V=3000\cdot10^{6}\mathrm{~}\mathrm{~}/ {3}.36\cdot10^{7}=89.3\mathrm{~}\mathrm{W/cm^{3}}=89.3\mathrm{~}\mathrm{~}\mathrm{~}\mathrm{M}\mathrm{W/m^{3}~}
The number of fuel elements is determined by dividing the core cross sectional area by the area of a lattice cell;
(e) N={\frac{\pi R^{2}}{A_{ce l l}}}=\frac{\pi(H/2)^{2}}{p^{2}}=\frac{\pi(350/2)^{2}}{1.536^{2}}=40,780
Since the maximum outlet temperature is limited to 330 °C we determine the mass flow rate from Eq. (8.42); taking c_p = 6.4 \cdot 10^3 J/kg°C as the specific heat of water at the operating coolant temperature yields:
(f) W=\frac{1}{c_{p}}\frac{P F_{r}}{\left(T_{0}|_{\mathrm{max}}-T_{i}\right)}= {\frac{1}{6.4\cdot10^{3}}}{\frac{3000\cdot10^{6}\cdot1.30}{(330-290)}}= 15.2 \cdot 10^3 kg/s = 15.2 \cdot 10^6 gm/s
We determine the mean coolant velocity from {W} = \rho{A}_{flow} \overline{\mathrm{v}} where {A}_{flow} = N \cdot (p^2 – \pi a^2) and we take the density of the pressurized water at 300 °C as 0.676 gm/cm³. Thus
(g) \overline{\mathrm{v}}={\frac{W}{\rho N(p^{2}-\pi a^{2})}}={\frac{15.2\cdot10^{6}}{0.676\cdot40,780\cdot(1.536^{2}-\pi0.509^{2})}}= 357 cm/s = 3.57 m/s
Note that using the same lattice, (i.e. fuel diameter and pitch)the core dimensions, the number of fuel pins, and the mass flow rate all decrease dramatically, but the average flow speed changes very little. If the size of the core had been kept the same, the power of the reactor could have been increased substantially.