A small water reactor recently installed at Podunk College is operating at the boiling point of water due to the malfunctioning of the cooling system. The operators observe that the water boils away at the rate of 10 L/min. If they assume that all of the reactor energy is absorbed in the water, the power developed by the reactor is approximately

Question

Quizplus · Accepted Answer

We can use the formula for power developed by a nuclear reactor:

P = ηρVq

where η is the efficiency of energy transfer, ρ is the density of water, V is the volume of water in the reactor, and q is the heat produced per unit time.

Assuming η = 1 and ρ = 1000 kg/m^3, and using the fact that 1 L = 0.001 m^3, we can find V:

10 L/min = 0.01 m^3/min
V = 0.01 m^3/min x 60 min/hr = 0.6 m^3/hr

Now, the heat produced per unit time is equal to the rate of boiling away of water times the latent heat of vaporization of water, which is 2.26 x 10^6 J/kg:

q = 10 L/min x 0.001 m^3/L x 1000 kg/m^3 x 2.26 x 10^6 J/kg = 22,600 W

Substituting these values into the formula for power, we get:

P = ηρVq = 1 x 1000 kg/m^3 x 0.6 m^3/hr x 22,600 W = 378 kW

Therefore, the best choice is D, 378 kW.

A Small Water Reactor Recently Installed at Podunk College Is