Question

If 4.5×105kg of emergency cooling water at 10 ∘C are dumped into a malfunctioning nuclear reactor whose core is producing energy at the rate of 200 MW, and if no circulation or cooling of the water is provided, how long will it be before half the water has boiled away?

Answer 1

Answer:

$\Delta t= 2962.395\,s\,(49.373\,min)$

Explanation:

Let assume that cooling water works under a pressure of 1 atmosphere. The time required to boil half of the water is determined by the First Law of Thermodynamics:

$\dot Q \cdot \Delta t = m \cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}]$

$\Delta t = \frac{m\cdot [c_{p,w}\cdot (T_{2}-T_{1})+h_{fg}]}{\dot Q}$

$\Delta t = \frac{(2.25\times 10^{5}\,kg)\left[\left(4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (100\,^{\textdegree}C - 10\,^{\textdegree}C)+2256.5\,\frac{kJ}{kg} \right]}{200000\,kW}$

$\Delta t= 2962.395\,s\,(49.373\,min)$