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Lapatulllka [165]
3 years ago
12

The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal re

servoirs are now used to power a reversible refrigeration system and then a reversible heat pump, determine the Coefficient of Performance of these two systems.
Physics
1 answer:
inna [77]3 years ago
5 0

Answer:

Explanation:

The thermal efficiency of a Power cycle \eta = \dfrac{Q_H -Q_c}{Q_H}

where;

\eta = 50\% = 0.5

Q_H = Heat \ flow \ from \ higher \ temperature

Q_c = Heat \ flow \ from \ lower \ temperature

0.5 = \dfrac{Q_H -Q_c}{Q_H}

0.5 Q_H = Q_H - Q_c --- (1)

Q_c = 0.5 Q_H         ---- (2)

The coefficient of performance is:

COP_R = \dfrac{Q_c}{Q_H -Q_c}

let replace the value of Q_c = 0.5 Q_H   in the above equation then;

COP_R = \dfrac{0.5Q_H}{Q_H -0.5 Q_H}

COP_R = \dfrac{0.5Q_H}{0.5 Q_H}

COP_R = 1

The

On the other hand,  the heat pump

COP_{HP} = \dfrac{Q_H}{Q_H -Q_c}

By replacing equation (1) into the above equation; we have:

COP_{HP} = \dfrac{Q_H}{0.5Q_{H}}

COP_{HP} = \dfrac{1}{0.5}

COP_{HP} =2

t

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