Question 9.5: A 3000 MW(t) pressurized water reactor has the following spe......

Part a: As we learned in chapter 4 a liquid cooled and moderated reactor’s temperature coefficient is positive if it is overmoderated. This is based on the isothermal temperature coefficient. Thus for this problem

$\alpha_{T}=\alpha_{f}+\alpha_{m}=-\frac{7.2\cdot10^{-4}}{\sqrt{273+T}}+\left(30+1.5T-0.010T^{2}\right)\cdot10^{-6}$

Determining the temperature at which this coefficient vanishes analytically, involves solving a quadric equation. Instead we simply plot the coefficient.

Part b. We apply Eq. (9.34) to $\alpha _T$

taking $x = \sqrt{273 + T}$ and hence dT = 2xdx in the first integral then gives

$D_{T}=\left\{-89.28+7650+62156-81215\right\}\cdot10^{-6}=-11.5\cdot10^{-3}$

Part c. The power defect is determined by substituting the temperature dependent coefficients into Eq. (9.35)

$D_{P}=\int_{{T}_{i}}^{\overline{T}_{f}(P)}\,\alpha_{f}d\overline{{{T}}}_{f}+\int_{{T}_{i}}^{\overline{T}_{c}(P)}\,\alpha_{c}d\overline{{{T}}}_{c}$

where $T_{i}=T_{i}=290\ {}^{\circ}{ C},\,\,R_{f}=0.45\ {}^{\circ}{ C}/\mathrm{MW}(t), W=68\times10^{6}\,\mathrm{kg}/\mathrm{hr}\, × 3600^{-1}\,\mathrm{hr}/\mathrm{s}= 18.9\times10^{3}\mathrm{kg /s} \ \ { c}_{p}\,=6.4  ×  10^{3}~\mathrm{J/kg} \ \ ^{\circ }C$ . Thus

$\overline{{{T}}}_{c}=\frac{1}{2W c_{p}}P+T_{i}$

$={\frac{1}{2\cdot18.8\cdot10^{3}\cdot6.4\cdot10^{3}\cdot10^{-6}}}\,3000+290=302.5 \ \ ^{\circ} C$

$\overline{{{T}}}_{f}=R_{f}P+\overline{{{T}}}_{c}=0.45 \cdot 3000+302.5=1,652.5\ ^{\circ}C$

Thus

Evaluating the integrals:

$D_{P}=\left\{-29,088+377+5,555-10,972\right\}\cdot10^{-6}=-34.1\cdot10^{-3}$