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1. Problem statement
A power plant generates electricity from the difference in sea temperature. At the surface of the sea the temperature is 27°C. At the bottom of the sea the temperature is 6°C.
a) What is the highest efficiency that this plant can have?
b) If the power plant generates 210 W of power, how much heat per time must be released by the hot sea water? How much heat per time is released into the cold sea water?
2. Equations
The Carnot efficiency: efficiency = 1-(Tc/Th)
P[supplied] = P[output]/efficiency
P[wasted] = P[supplied] – P[output]
3. Attempt at a solution
The idea here is that the power plant uses heat from the hot sea water to run a heat engine. The heat engine turns the heat into mechanical work to generate electricity. The heat engine is not efficient enough, so some heat energy is wasted. This wasted heat energy is used to heat the cold sea water.
In a) I used The Carnot efficiency: efficiency = 1-(Tc/Th)
Tc = temperature in kelvin of cold reservoir = 273.15+6
Th = temperature in kelvin of hot reservoir = 273.15+27
Efficiency = 0.0699 = 7%
In b) to find heat per time released by the hot sea water I used:
P[supplied] = P[output]/efficiency
P[output] = 210 W
Efficiency = 1-(Tc/Th) = 0.0699
P[supplied] = 3043.478 W = 3 KW
To find heat per time released into the the cold sea water I used:
P[wasted] = P[supplied] – P[output]
P[wasted] = 3043.478 W – 210 W = 2833.478 W = 2.8 KW
A power plant generates electricity from the difference in sea temperature. At the surface of the sea the temperature is 27°C. At the bottom of the sea the temperature is 6°C.
a) What is the highest efficiency that this plant can have?
b) If the power plant generates 210 W of power, how much heat per time must be released by the hot sea water? How much heat per time is released into the cold sea water?
2. Equations
The Carnot efficiency: efficiency = 1-(Tc/Th)
P[supplied] = P[output]/efficiency
P[wasted] = P[supplied] – P[output]
3. Attempt at a solution
The idea here is that the power plant uses heat from the hot sea water to run a heat engine. The heat engine turns the heat into mechanical work to generate electricity. The heat engine is not efficient enough, so some heat energy is wasted. This wasted heat energy is used to heat the cold sea water.
In a) I used The Carnot efficiency: efficiency = 1-(Tc/Th)
Tc = temperature in kelvin of cold reservoir = 273.15+6
Th = temperature in kelvin of hot reservoir = 273.15+27
Efficiency = 0.0699 = 7%
In b) to find heat per time released by the hot sea water I used:
P[supplied] = P[output]/efficiency
P[output] = 210 W
Efficiency = 1-(Tc/Th) = 0.0699
P[supplied] = 3043.478 W = 3 KW
To find heat per time released into the the cold sea water I used:
P[wasted] = P[supplied] – P[output]
P[wasted] = 3043.478 W – 210 W = 2833.478 W = 2.8 KW
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