Question 8.11: An unachievable ideal would be a reactor with a perfectly fl......

With the idealized 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}\cdot1.00\cdot1.00/400

=1.77\cdot10^{7}\mathrm{~cm}^{3}=17.7\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\cdot1.77\cdot10^{7}\;/\,\pi)^{1/3}=282\mathrm{~cm}=2.82\mathrm{~cm}

The core averaged power density is just

(d)  \overline{{{P}}}^{\prime \prime \prime}={{{P}}}/V=3000\cdot10^{6}\mathrm{~}\mathrm{~}/ 1.77\cdot10^{7}=169\mathrm{~}\mathrm{W/cm^{3}}=169\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(282/2)^{2}}{1.536^{2}}=26,474

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}\cdot1.00}{(330-290)}}= 11.7 \cdot 10^3  kg/s = 11.7 \cdot 10^6  gm/s

We determine the mean coolant velocity from {W} = \rho{A}_{flow} \overline{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 v={\frac{W}{\rho N(p^{2}-\pi a^{2})}}={\frac{11.7\cdot10^{6}}{0.676\cdot26,474\cdot(1.536^{2}-\pi0.509^{2})}}= 423  cm/s = 4.23  m/s