The Nuclear Reactor — Question 6

Question 6

Coolant flow converts a heat rate into a temperature rise

A heat exchanger transfers 12.0 MW to a steady single-phase coolant stream with mass flow $100\,\mathrm{kg/s}$ and specific heat $4.00\,\mathrm{kJ/(kg\,K)}$. Ignore heat losses and kinetic or potential-energy changes.

1. Calculate the outlet-minus-inlet temperature rise.

2. If flow decreases by 20 percent at fixed thermal power, find the new temperature rise and its fractional increase.

Question 6 - Solution

Coolant flow converts a heat rate into a temperature rise

(a) Method and calculation

$$P=\dot m(h_{\mathrm{out}}-h_{\mathrm{in}})\simeq\dot m c_p\Delta T.$$$$\boxed{\Delta T=12.0\times10^6/[100(4000)]=30.0\,\mathrm K}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta T\prime=12.0\times10^6/[80(4000)]=37.5\,\mathrm K}.$$$$\boxed{\Delta T\prime/\Delta T=1/.80=1.25}\quad\text{a 25 percent increase}.$$

Physical check

The calculation assumes constant heat capacity and no boiling. The inverse flow dependence follows from energy conservation, not from a change in generated power.