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Air conditioners and refrigerators

Air conditioners and refrigerators are designed to cool something down in a warm environment. As with heat pumps, work input is required for heat transfer from cold to hot, and this is expensive. The quality of air conditioners and refrigerators is judged by how much heat transfer Q c size 12{Q rSub { size 8{c} } } {} occurs from a cold environment compared with how much work input W size 12{W} {} is required. What is considered the benefit in a heat pump is considered waste heat in a refrigerator. We thus define the coefficient of performance     (COP ref ) size 12{ ital "COP" rSub { size 8{"ref"} } } {} of an air conditioner or refrigerator to be

COP ref = Q c W . size 12{ ital "COP" rSub { size 8{ ital "ref"} } = { {Q rSub { size 8{c} } } over {W} } "." } {}

Noting again that Q h = Q c + W size 12{Q rSub { size 8{h} } =Q rSub { size 8{c} } +W} {} , we can see that an air conditioner will have a lower coefficient of performance than a heat pump, because COP hp = Q h / W size 12{ ital "COP" rSub { size 8{"hp"} } =Q rSub { size 8{h} } /W} {} and Q h size 12{Q rSub { size 8{h} } } {} is greater than Q c size 12{Q rSub { size 8{c} } } {} . In this module’s Problems and Exercises, you will show that

COP ref = COP hp 1 size 12{ ital "COP" rSub { size 8{"ref"} } = ital "COP" rSub { size 8{"hp"} } - 1} {}

for a heat engine used as either an air conditioner or a heat pump operating between the same two temperatures. Real air conditioners and refrigerators typically do remarkably well, having values of COP ref size 12{ ital "COP" rSub { size 8{"ref"} } } {} ranging from 2 to 6. These numbers are better than the COP hp size 12{ ital "COP" rSub { size 8{"hp"} } } {} values for the heat pumps mentioned above, because the temperature differences are smaller, but they are less than those for Carnot engines operating between the same two temperatures.

A type of COP size 12{ ital "COP"} {} rating system called the “energy efficiency rating” ( EER size 12{ ital "EER"} {} ) has been developed. This rating is an example where non-SI units are still used and relevant to consumers. To make it easier for the consumer, Australia, Canada, New Zealand, and the U.S. use an Energy Star Rating out of 5 stars—the more stars, the more energy efficient the appliance. EER s size 12{ ital "EER"} {} are expressed in mixed units of British thermal units (Btu) per hour of heating or cooling divided by the power input in watts. Room air conditioners are readily available with EER s size 12{ ital "EER"} {} ranging from 6 to 12. Although not the same as the COP s size 12{ ital "COP"} {} just described, these EER s size 12{ ital "EER"} {} are good for comparison purposes—the greater the EER size 12{ ital "EER"} {} , the cheaper an air conditioner is to operate (but the higher its purchase price is likely to be).

The EER { ital "EER"s} {} of an air conditioner or refrigerator can be expressed as

EER = Q c / t 1 W / t 2 , { ital "EER"= { {Q rSub { {c} } /t rSub { {1} } } over {W/t rSub { size 8{2} } } } ,} {}

where Q c {Q rSub { {c} } } {} is the amount of heat transfer from a cold environment in British thermal units, t 1 {Q rSub { {c} } } {} is time in hours, W {W} {} is the work input in joules, and t 2 is time in seconds.

Problem-solving strategies for thermodynamics

  1. Examine the situation to determine whether heat, work, or internal energy are involved . Look for any system where the primary methods of transferring energy are heat and work. Heat engines, heat pumps, refrigerators, and air conditioners are examples of such systems.
  2. Identify the system of interest and draw a labeled diagram of the system showing energy flow.
  3. Identify exactly what needs to be determined in the problem (identify the unknowns) . A written list is useful. Maximum efficiency means a Carnot engine is involved. Efficiency is not the same as the coefficient of performance.
  4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns) . Be sure to distinguish heat transfer into a system from heat transfer out of the system, as well as work input from work output. In many situations, it is useful to determine the type of process, such as isothermal or adiabatic.
  5. Solve the appropriate equation for the quantity to be determined (the unknown).
  6. Substitute the known quantities along with their units into the appropriate equation and obtain numerical solutions complete with units.
  7. Check the answer to see if it is reasonable: Does it make sense? For example, efficiency is always less than 1, whereas coefficients of performance are greater than 1.
Practice Key Terms 2

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Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
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