Question

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.

Answer 1

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:

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

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

$COP_{HP} =2$

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