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Q38PE

Expert-verifiedFound in: Page 551

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Suppose you have an ideal refrigerator that cools an environment at −20.0ºC and has heat transfer to another environment at 50.0ºC . What is its coefficient of performance?**

The refrigerator’s performance coefficient is 3.61.

**Coefficient of Performance is the ratio of useful heat produced by the pump to the energy intake of the pump. It is the reciprocal of the efficiency of the pump.**

Temperature of cold environment \({T_c} = - 20\;^\circ {\rm{C}} = {\rm{253}}\;{\rm{K}}\)

The temperature of hot environment \({T_h} = 50\;^\circ {\rm{C}} = {\rm{323}}\;{\rm{K}}\)

The efficiency of the Carnot engine \(\eta = 1 - \frac{{{T_c}}}{{{T_h}}}\)

Heat pump’s performance coefficient \({\beta _{hp}} = \frac{1}{\eta }\)

Refrigerator’s performance coefficient \({\beta _{ref}} = {\beta _{hp}} - 1\)

Here,

\(\eta \) - efficiency of the engine.

\({T_c}\)- the temperature of the cold environment.

\({T_h}\)- temperature of the hot environment

\({\beta _{hp}}\) - heat pump’s performance coefficient.

\({\beta _{ref}}\) - refrigerator’s performance coefficient.

Maximum efficiency can be calculated by calculating Carnot efficiency as

\(\begin{aligned}{}\eta & = 1 - \frac{{{T_c}}}{{{T_h}}}\\ & = 1 - \frac{{253\;{\rm{K}}}}{{323\;{\rm{K}}}}\\ &= 0.217\end{aligned}\)

Heat pump’s performance coefficient is

\(\begin{aligned}{}{\beta _{hp}} &= 1/\eta \\ &= 1/0.217\\ &= 4.61\end{aligned}\)

Refrigerator’s performance coefficient is

\(\begin{aligned}{}{\beta _{ref}} &= {\beta _{hp}} - 1\\ &= 4.61 - 1\\ &= 3.61\end{aligned}\)

Therefore, the refrigerator’s performance coefficient is \(3.61\).

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