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Q25CQ

Expert-verifiedFound in: Page 549

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Why do refrigerators, air conditioners, and heat pumps operate most cost-effectively for cycles with a small difference between T _{h} and T_{c}? (Note that the temperatures of the cycle employed are crucial to its COP.)**

Refrigerators, air conditioners, and heat pumps work efficiently if the temperature difference is small because the work required for their performance is less.

The efficiency of heat pumps, refrigerators and air conditioners are determined by the coefficient of performance. It is calculated as follows:

\(\begin{array}{l}CO{P_{hp}} = \frac{{{Q_h}}}{W}\\CO{P_{ref}} = \frac{{{Q_c}}}{W}\\W = {Q_h} - {Q_c}\end{array}\)

Here, \(CO{P_{hp}}\)is the coefficient of performance of the heat pump, \({Q_h}\)is amount of heat transfer into the warmer space and \(W\) is the work input. \(CO{P_{ref}}\) is the coefficient of performance of refrigerator and \({Q_c}\) is the heat in the environment. The expression for the coefficient of performance of an air conditioner is same as that of a refrigerator.

The expressions of coefficient of performance indicates that it is inversely proportional to the work output. In a Carnot engine,

\(\begin{array}{l}{Q_c} = {T_c}\\{Q_h} = {T_h}\\W = {T_h} - {T_c}\end{array}\)

Here, \({T_c}\)is the temperature of the environment and \({T_h}\) is the temperature at the system. So, if this difference is small, the work input required for the performance will be smaller as well. Therefore, the coefficient of performance increases.

Thus, air conditioners, refrigerators, and heat pumps operate most cost-efficiently while working with a small temperature difference between the system and its surroundings.

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