Question 9.3: Assume a pressurized water reactor has the parameters specif......
Assume a pressurized water reactor has the parameters specified in the example at the end of Chapter 8.3. Assume the core has a thermal resistance or R_ƒ = 0.50 °C/MW(t), If the reactor fuel and moderator temperature coefficients are α_ƒ = -3.2 \cdot 10^{-5} °C^{-1} and α_m = -1.4 \cdot 10^{-5} °C^{-1}
a. Determine the isothermal temperature coefficient.
b. Determine the power coefficient.
Part a: From Eq. (9.29) the isothermal temperature coefficient is just
\alpha_{T}=\alpha_{f}+\alpha_{m}=-3.2\cdot10^{-5}-1.4\cdot10^{-5}=-4.6\cdot10^{-5}~/^{o}\,C
Part b: Following Eq. (9.33) the power coefficient is
\alpha_{P}=R_{f}\alpha_{f}+(2W c_{p})^{-1}(\alpha_{f}+\alpha_{c})
Using data from Chapter 8 example
{W}c_{p}=27.2\cdot10^{3} \ \ \mathrm{kg}/\mathrm{s} \ \ 6.4\cdot10^{3} \ \ \mathrm{J/kg}=174\cdot10^{6 \ \ }\mathrm{J/s}=174 \ \mathrm{MW}
Thus
\alpha_{P}=0.50(-3.2\cdot10^{-5})+(2\cdot174)^{-1}(-3.2\cdot10^{-5}-1.4\cdot10^{-5})
=-(1.6+0.013).10^{-5}=-1.61\cdot10^{-5}\,/\,\mathrm{MW}
Thus the fuel provides much more of the power coefficient than does the coolant.