A large electrical power plant generates 1050 MW of electricity with an efficien
ID: 1481864 • Letter: A
Question
A large electrical power plant generates 1050 MW of electricity with an efficiency of 36.0%. (a) Calculate the heat input to the plant, Qh, in one day. J (b) How much heat Qc is exhausted in one day? J (c) If the cooling towers exhaust the heat in the form of 35.0°C air into the local air mass, which increases in temperature from 18.0°C to 20.0°C, what is the total increase in entropy due to this heat transfer? J/K (d) How much energy becomes unavailable to do work because of this increase in entropy, assuming an 18.0°C lowest temperature? (Part of Qc could, of course, be utilized to operate heat engines or for simple space heating, but it rarely is.) J
Explanation / Answer
a) from the defintion of power = eenrgy/time
so eenrgy = power* time
in a day
Energy = 1050 e6 * 24* 60*60
E = 9.072 e 13 Joules
36% of this = 0.36 * 9.072 e 13 = 3.26 e+13 Joules
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b . exhausted energy = input energy - used energy
Ee = 9.072 e 13 - 3.26e 13
Ee = 5.81 e 13 J
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c.
Entropy change Delta S = Q/T1 - Q/T2.
Q is the exhaust energy
T1 is average temperature of the local air mass =18 +273 = 291 K.
T2 is the temperature of the exhaust tower = 293 K.
DeltaS = (9.072e 13 /291 - 3.26e 13/293)
delta S = 2 e 11 J/K.
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d)
Unavailable energy dS = 291*2 e1 = 5.83 e 13 Joules
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