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The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal reservoirs are now used to power a reversible refrigeration system and then a reversible heat pump, determine the Coefficient of Performance of these two systems.

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9 votes

Answer:

Step-by-step explanation:

The thermal efficiency of a Power cycle
\eta = (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 = (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 = (Q_c)/(Q_H -Q_c)

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


COP_R = (0.5Q_H)/(Q_H -0.5 Q_H)


COP_R = (0.5Q_H)/(0.5 Q_H)


COP_R = 1

The

On the other hand, the heat pump


COP_(HP) = (Q_H)/(Q_H -Q_c)

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


COP_(HP) = (Q_H)/(0.5Q_(H))


COP_(HP) = (1)/(0.5)


COP_(HP) =2

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