Suppose a power reactor has negative values of α_ƒ and α_c , the fuel and coolant temperature coefficients. Using the thermal hydraulic model developed in Chapter 8:
a. Show that if very slow changes take place in the coolant inlet temperature and mass flow rate, but no control poisons are added or subtracted, the power will undergo a quasi-static change of
dP = \frac{\left|\alpha_{f}+\alpha_{c}\right|\left({\frac{P}{2W^{2}c_{p}}}d W-d T_{i}\right)}{\left|\left(R+\frac{1}{2W c_{p}}\right)\alpha_{f}+\frac{1}{2W c_{p}}\alpha_{c}\right|}
b. If the flow rate increases, does the power increase or decrease? Why?
c. If the inlet temperature increases, does the power increase or decrease? Why?
a. For negative temperature coefficients, we begin by writing Eq. (9.25) as
d\rho=-\left|\alpha_{f}\right|d\overline{{{T}}}_{f}-\left|\alpha_{c}\right|d\overline{{{T}}}_{c}
Use the quasi- steady state temperatures of Eqs. (8.40) and (8.41). Taking the differentials:
\ d{\overline{{{T}}}}_{c}=d\left(\frac{P}{2W c_{p}}\right)+d T_{i}=\frac{1}{2W c_{p}}d P-\frac{P}{2W^{2}c_{p}}d W+d T_{i}
d\overline{{{T}}}_{c}=d\left(\,R+\frac{P}{2W c_{p}}\,\right)+d T_{i}=\left(\,R+\frac{1}{2W c_{p}}\,\right)d P-\frac{P}{2W^{2}c_{p}}\,d W+d T_{i}
Combining equations:
d\rho=-|\alpha_{f}|\left\{\left ({ R}+\frac{1}{2W{ c}_{p}}\right)d P-\frac{P}{2W^{2}{ c}_{p}}dW+d T_{i}\right\}_{f}-\left|\alpha_{c}\right|\left \{\frac{1}{2W c_{p}}d P-\frac{P}{2W^{2}{c}_{p}}dW+d T_{i}\right \}
For very slow changes, we assume that the feedback instantaneously compensates for the reactivity, thus dρ=0. and we have
Solving for P, we obtain the desired equation.
b. The Power increases since increasing the flow rate decreases both fuel and moderator temperature, causing increased feedback reactivity.
c. The power decreases, since raising the inlet temperature causes increases in fuel and moderator temperature, causing negative reactivity from the feedback.