A nuclear-fueled electric power plant utilizes a so-called \"boiling water react

Question

A nuclear-fueled electric power plant utilizes a so-called "boiling water reactor". In this type of reactor, nuclear energy causes water under pressure to boil at 280 °C (the temperature of the hot reservoir). After the steam does the work of turning the turbine of an electric generator, the steam is converted back into water in a condenser at 41.0 °C (the temperature of the cold reservoir). To keep the condenser at 41.0 °C , the rejected heat must be carried away by some means - for example, by water from a river. The plant operates at 76 % of its Carnot efficiency, and the electrical output power of the plant is 1.20 x 109 watts. A river with a water flow rate of 1.04 x 105 kg/s is available to remove the rejected heat from the plant. Find the number of Celsius degrees by which the temperature of the river rises.

Explanation / Answer

efficiency of the plant is eta = (3/4)[1-(T2/T1)]

T1 = 280+273 = 553 K

T2 = 41+273 = 314 K

eta = (3/4)*(1-(314/553)) =0.324

heat taken is Q_in = Q_out/eta

given that Q_out = 1.2*10^9 Watts

eta = 0.324

Q_in = (1.2*10^9)/0.324 = 3.7*10^9 Watts

heat removed is Q= Q_in - Q_out = (3.7-1.2)*10^9 = 2.5*10^9 W

but heat removed and given to water is

Q = m*S*dT

S is the specific heat capacity of water = 4186 J/Kg-k

change in temperature is dT = Q/(m*S) = (2.5*10^9)/(1.04*10^5*4186)

dT = 5.74 K or deg C

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