During a nuclear accident, 450 m3 of emergency cooling water at 19 C are injecte

Question

During a nuclear accident, 450 m3 of emergency cooling water at 19 C are injected into a reactor vessel where the reactor core is producing heat at the rate of 200 MW.

Part A

If the water is allowed to boil at normal atmospheric pressure, how long will it take to boil the reactor dry?

Express your answer using two significant figures.

8.4•103

During a nuclear accident, 450 m3 of emergency cooling water at 19 C are injected into a reactor vessel where the reactor core is producing heat at the rate of 200 MW.

Part A

If the water is allowed to boil at normal atmospheric pressure, how long will it take to boil the reactor dry?

Express your answer using two significant figures.

8.4•103

Explanation / Answer

P = 200 MW = 200 x 106 W = 2 x 108 W = 2 x 108 J/s

The density of water is 1000 kg/m3.

Mass = 450 x 1000 = 4.5 x 105 kg.

Let’s use the following equation to determine the amount of heat energy that is required to increase the water’s temperature from 190C to 1000C.

Q = m x S x (Tfinal – Tinitial)

The specific heat of water is 4186 J/(kg C)

Q = 4.5 x 105 x 4186 x 81 = 1.5258 x 1011 J

The heat of vaporization of water is 2.26 * 106 J/kg.

Q = 4.5 x 105 x 2.26 x 106 = 1.017 x 1012 J

Total energy = 1.5258 x 1011 + 1.017 x 1012 = 1.16958 x 1012 J

Power = Energy ÷ Time

Time = Energy ÷ Power

Time = (1.16958 x 1012) ÷ (2 x 108) = 5847.89 seconds

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