 Solution

ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2019/paper-ne1/3/c/ii/solution

The average moderator temperature is $725^{\circ} C$ . First the mean coolant temperature in the lumped thermal model is

\overline{T}_{C} = \frac{600 + 700}{2} = 650^{\circ} C .

At a steady state, the same thermal power crosses both thermal resistances. If

R_{1} = R_{2}

, then

\overline{T}_{F} - \overline{T}_{M} = \overline{T}_{M} - \overline{T}_{C} .

Thus

\overline{T}_{M} = \frac{800 + 650}{2} = 725^{\circ} C .

In particular

R_{1} = R_{2} = (800 - 650) / (2 P) = 3.17797 \times 1 0^{- 7} K W^{- 1}

at the specified state.

 New to topics? Read the docs here!