Question 9.5: A 3000 MW(t) pressurized water reactor has the following spe......
A 3000 MW(t) pressurized water reactor has the following specifications:; core thermal resistance 0.45 °C/MW(t), coolant flow, 68 × 10^6 kg/hr; coolant specific heat 6.4 × 10³ J/kg °C . The fuel temperature coefficient is
\frac{1}{k} \frac{∂k}{∂\overline {T_ƒ}} = -\frac{7.2 \cdot 10^{-4}}{\sqrt{273 + \overline {T_ƒ} }} \ \ (°C)^{-1}
and the coolant temperature coefficient by
\frac{1}{k}\frac{∂k}{∂\overline{T}_{c}}=\Bigl(30+1.5\overline{{{T}}}_{c}-0.010\overline{{{T}}}_{c}^{2}\Bigr)\cdot10^{-6}\quad({}^{o}C)^{-1}
a. Over what temperature range is the core overmoderated?
b. What is the value of the temperature defect? Assume room temperature of 35 °C and an operating coolant inlet temperature of 290 °C.
c. What is the value of the power defect?
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\{-720\cdot2\,x\bigg |_{17.5}^{23.7}+\left.\left(30T+{\frac{1.5}{2}}T^{2}-{\frac{0.010}{3}}T^{3}\right)\right|_{35}^{290}\right\}.10^{-6}
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}