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 Incoming links: Lumped thermal model

Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2019 / Module NE1 / 3 / c / ii / Solution 2026-10-20

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.

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