A nuclear power plant generates 3000 MW of heat energy from nuclear reactions in

Question

A nuclear power plant generates 3000 MW of heat energy from nuclear reactions in the reactor's core. This energy is used to boil water and produce high-pressure steam at 280 C. The steam spins a turbine, which produces 1200 MW of electric power, then the steam is condensed and the water is cooled to 25C before starting the cycle again.

1) What is the maximum possible thermal efficiency of the power plant?

2) What is the plant's actual efficiency?

3) Cooling water from a river flows through the condenser (the low-temperature heat exchanger) at the rate of 1.2×108L/h (30 million gallons per hour). If the river water enters the condenser at 18C, what is its exit temperature?

Explanation / Answer

1)-maximum possible thermal efficiency of the power plant:

1 - (25+273) / (280 + 273) = 47%

2)-the plant's actual efficiency is:

1200 MW / 3000 MW = 40%

3)- exit temperature is:

Qh / t = (0.4)*(3000MW) = 1200 MW

in other side 1.2x10^8L/hr*3600s=33333.33L/s
so:

Tf=( Qh / (mc) )+Ti = [ (1.2x10^9 ) / (33333.33*4190) ] + (18 + 273)

Tf = 299.59 K = 26.5°C

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