91Ó°ÊÓ

The initial power of the reactor, $P_0=400\mathrm{MW}$

The neutron generation time, $t_{gen}=30\mathrm{ms}$

The multiplication factor, $k=1.0003$

The reactor power is increase for $t=5\min$.

Power means the average speed of electricity supply during one measurement interval, converted to an hourly rate of electricity supply in kWhper hour, which is equal to the product of the measured energy for one measurement interval, inkWhper measurement interval, times the number of measurement intervals in one hour.

The expression to calculate the power output after time t (according to problem 19) is given as follows.

$\mathrm{P}\left(\mathrm{t}\right)={\mathrm{P}}_0{\mathrm{k}}^{\mathrm{t}/{\mathrm{t}}_{\mathrm{gen}}}$ ……. (1)

Calculate the output power of the reactor.

Substitute 400 MW for $P_0$, 30 ms for $t_{gen}$ , 1.0003 for k and 5 min for t into equation (1).

$$\begin{gathered} P\left(5\min \right)=400{\left(1.0003\right)}^{5×60/30×{10}^{-3}} \\ =400{\left(1.0003\right)}^{300×{10}^3/30} \\ =400{\left(1.0003\right)}^{{10}^4} \\ =400×20.076 \end{gathered}$$

Solve further as,

$$\begin{gathered} P\left(5\min \right)=8030.60\mathrm{MW} \\ =8.03/{10}^3\mathrm{MW} \end{gathered}$$

Hence, the output power of the reactor after 5 min is $8.03/{10}^3\mathrm{MW}$.