College Physics

# 15.5Applications of Thermodynamics: Heat Pumps and Refrigerators

College Physics15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
Figure 15.26 Almost every home contains a refrigerator. Most people don’t realize they are also sharing their homes with a heat pump. (credit: Id1337x, Wikimedia Commons)

Heat pumps, air conditioners, and refrigerators utilize heat transfer from cold to hot. They are heat engines run backward. We say backward, rather than reverse, because except for Carnot engines, all heat engines, though they can be run backward, cannot truly be reversed. Heat transfer occurs from a cold reservoir $QcQc size 12{Q rSub { size 8{c} } } {}$ and into a hot one. This requires work input $WW size 12{W} {}$, which is also converted to heat transfer. Thus the heat transfer to the hot reservoir is $Qh=Qc+WQh=Qc+W size 12{Q rSub { size 8{h} } =Q rSub { size 8{c} } +W} {}$. (Note that $QhQh size 12{Q rSub { size 8{h} } } {}$, $QcQc size 12{Q rSub { size 8{c} } } {}$, and $WW size 12{W} {}$ are positive, with their directions indicated on schematics rather than by sign.) A heat pump’s mission is for heat transfer $QhQh size 12{Q rSub { size 8{h} } } {}$ to occur into a warm environment, such as a home in the winter. The mission of air conditioners and refrigerators is for heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ to occur from a cool environment, such as chilling a room or keeping food at lower temperatures than the environment. (Actually, a heat pump can be used both to heat and cool a space. It is essentially an air conditioner and a heating unit all in one. In this section we will concentrate on its heating mode.)

Figure 15.27 Heat pumps, air conditioners, and refrigerators are heat engines operated backward. The one shown here is based on a Carnot (reversible) engine. (a) Schematic diagram showing heat transfer from a cold reservoir to a warm reservoir with a heat pump. The directions of $WW size 12{W} {}$, $QhQh size 12{Q rSub { size 8{h} } } {}$, and $QcQc size 12{Q rSub { size 8{c} } } {}$ are opposite what they would be in a heat engine. (b) $PVPV size 12{ ital "PV"} {}$ diagram for a Carnot cycle similar to that in Figure 15.28 but reversed, following path ADCBA. The area inside the loop is negative, meaning there is a net work input. There is heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ into the system from a cold reservoir along path DC, and heat transfer $QhQh size 12{Q rSub { size 8{h} } } {}$ out of the system into a hot reservoir along path BA.

### Heat Pumps

The great advantage of using a heat pump to keep your home warm, rather than just burning fuel, is that a heat pump supplies $Qh=Qc+WQh=Qc+W size 12{Q rSub { size 8{h} } =Q rSub { size 8{c} } +W} {}$. Heat transfer is from the outside air, even at a temperature below freezing, to the indoor space. You only pay for $WW size 12{W} {}$, and you get an additional heat transfer of $QcQc size 12{Q rSub { size 8{c} } } {}$ from the outside at no cost; in many cases, at least twice as much energy is transferred to the heated space as is used to run the heat pump. When you burn fuel to keep warm, you pay for all of it. The disadvantage is that the work input (required by the second law of thermodynamics) is sometimes more expensive than simply burning fuel, especially if the work is done by electrical energy.

The basic components of a heat pump in its heating mode are shown in Figure 15.28. A working fluid such as a non-CFC refrigerant is used. In the outdoor coils (the evaporator), heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ occurs to the working fluid from the cold outdoor air, turning it into a gas.

Figure 15.28 A simple heat pump has four basic components: (1) condenser, (2) expansion valve, (3) evaporator, and (4) compressor. In the heating mode, heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ occurs to the working fluid in the evaporator (3) from the colder outdoor air, turning it into a gas. The electrically driven compressor (4) increases the temperature and pressure of the gas and forces it into the condenser coils (1) inside the heated space. Because the temperature of the gas is higher than the temperature in the room, heat transfer from the gas to the room occurs as the gas condenses to a liquid. The working fluid is then cooled as it flows back through an expansion valve (2) to the outdoor evaporator coils.

The electrically driven compressor (work input $WW size 12{W} {}$) raises the temperature and pressure of the gas and forces it into the condenser coils that are inside the heated space. Because the temperature of the gas is higher than the temperature inside the room, heat transfer to the room occurs and the gas condenses to a liquid. The liquid then flows back through a pressure-reducing valve to the outdoor evaporator coils, being cooled through expansion. (In a cooling cycle, the evaporator and condenser coils exchange roles and the flow direction of the fluid is reversed.)

The quality of a heat pump is judged by how much heat transfer $QhQh size 12{Q rSub { size 8{h} } } {}$ occurs into the warm space compared with how much work input $WW size 12{W} {}$ is required. In the spirit of taking the ratio of what you get to what you spend, we define a heat pump’s coefficient of performance ($COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$) to be

$COPhp=QhW.COPhp=QhW. size 12{ ital "COP" rSub { size 8{"hp"} } = { {Q rSub { size 8{h} } } over {W} } } {}$
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Since the efficiency of a heat engine is $Eff=W/QhEff=W/Qh size 12{ ital "Eff"=W/Q rSub { size 8{h} } } {}$, we see that $COPhp=1/EffCOPhp=1/Eff size 12{ ital "COP" rSub { size 8{"hp"} } =1/ ital "Eff"} {}$, an important and interesting fact. First, since the efficiency of any heat engine is less than 1, it means that $COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$ is always greater than 1—that is, a heat pump always has more heat transfer $QhQh size 12{Q rSub { size 8{h} } } {}$ than work put into it. Second, it means that heat pumps work best when temperature differences are small. The efficiency of a perfect, or Carnot, engine is $EffC=1−Tc/ThEffC=1−Tc/Th size 12{ ital "Eff" rSub { size 8{C} } =1 - left (T rSub { size 8{c} } /T rSub { size 8{h} } right )} {}$; thus, the smaller the temperature difference, the smaller the efficiency and the greater the $COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$ (because $COPhp=1/EffCOPhp=1/Eff size 12{ ital "COP" rSub { size 8{"hp"} } =1/ ital "Eff"} {}$). In other words, heat pumps do not work as well in very cold climates as they do in more moderate climates.

Friction and other irreversible processes reduce heat engine efficiency, but they do not benefit the operation of a heat pump—instead, they reduce the work input by converting part of it to heat transfer back into the cold reservoir before it gets into the heat pump.

Figure 15.29 When a real heat engine is run backward, some of the intended work input $W W { left (W right )} {}$ goes into heat transfer before it gets into the heat engine, thereby reducing its coefficient of performance $COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$. In this figure, $W ' W ' {W'} {}$ represents the portion of $W W {W} {}$ that goes into the heat pump, while the remainder of $W W {W} {}$ is lost in the form of frictional heat $Q f Q f { left (Q rSub { {f} } right )} {}$ to the cold reservoir. If all of $WW size 12{W} {}$ had gone into the heat pump, then $QhQh size 12{Q rSub { size 8{h} } } {}$ would have been greater. The best heat pump uses adiabatic and isothermal processes, since, in theory, there would be no dissipative processes to reduce the heat transfer to the hot reservoir.

### Example 15.5The Best COP hp of a Heat Pump for Home Use

A heat pump used to warm a home must employ a cycle that produces a working fluid at temperatures greater than typical indoor temperature so that heat transfer to the inside can take place. Similarly, it must produce a working fluid at temperatures that are colder than the outdoor temperature so that heat transfer occurs from outside. Its hot and cold reservoir temperatures therefore cannot be too close, placing a limit on its $COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$. (See Figure 15.30.) What is the best coefficient of performance possible for such a heat pump, if it has a hot reservoir temperature of $45.0ºC45.0ºC size 12{"45" "." 0°C} {}$ and a cold reservoir temperature of $−15.0ºC−15.0ºC size 12{-"15" "." 0°C} {}$?

Strategy

A Carnot engine reversed will give the best possible performance as a heat pump. As noted above, $COPhp=1/EffCOPhp=1/Eff size 12{ ital "COP" rSub { size 8{"hp"} } =1/ ital "Eff"} {}$, so that we need to first calculate the Carnot efficiency to solve this problem.

Solution

Carnot efficiency in terms of absolute temperature is given by:

$EffC=1−TcTh.EffC=1−TcTh. size 12{ ital "Eff" rSub { size 8{C} } =1 - { {T rSub { size 8{c} } } over {T rSub { size 8{h} } } } } {}$
15.38

The temperatures in kelvins are $Th=318 KTh=318 K size 12{T rSub { size 8{h} } ="318"" K"} {}$ and $Tc=258 KTc=258 K size 12{T rSub { size 8{c} } ="258"" K"} {}$, so that

$EffC=1−258 K318 K=0.1887.EffC=1−258 K318 K=0.1887. size 12{ ital "Eff" rSub { size 8{C} } =1 - { {"258"" K"} over {"318 K"} } =0 "." "1887"} {}$
15.39

Thus, from the discussion above,

$COP hp = 1 Eff = 1 0 . 1887 = 5 . 30 , COP hp = 1 Eff = 1 0 . 1887 = 5 . 30 , size 12{ ital "COP" rSub { size 8{"hp"} } = { {1} over { ital "Eff"} } = { {1} over {0 "." "1887"} } =5 "." "30",} {}$
15.40

or

$COP hp = Q h W = 5 . 30 , COP hp = Q h W = 5 . 30 , size 12{ ital "COP" rSub { size 8{"hp"} } = { {Q rSub { size 8{h} } } over {W} } =5 "." "30",} {}$
15.41

so that

$Q h = 5.30 W . Q h = 5.30 W . size 12{Q rSub { size 8{h} } =5 "." "30"" W" "." } {}$
15.42

Discussion

This result means that the heat transfer by the heat pump is 5.30 times as much as the work put into it. It would cost 5.30 times as much for the same heat transfer by an electric room heater as it does for that produced by this heat pump. This is not a violation of conservation of energy. Cold ambient air provides 4.3 J per 1 J of work from the electrical outlet.

Figure 15.30 Heat transfer from the outside to the inside, along with work done to run the pump, takes place in the heat pump of the example above. Note that the cold temperature produced by the heat pump is lower than the outside temperature, so that heat transfer into the working fluid occurs. The pump’s compressor produces a temperature greater than the indoor temperature in order for heat transfer into the house to occur.

Real heat pumps do not perform quite as well as the ideal one in the previous example; their values of $COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$ range from about 2 to 4. This range means that the heat transfer $QhQh size 12{Q rSub { size 8{h} } } {}$ from the heat pumps is 2 to 4 times as great as the work $WW size 12{W} {}$ put into them. Their economical feasibility is still limited, however, since $WW size 12{W} {}$ is usually supplied by electrical energy that costs more per joule than heat transfer by burning fuels like natural gas. Furthermore, the initial cost of a heat pump is greater than that of many furnaces, so that a heat pump must last longer for its cost to be recovered. Heat pumps are most likely to be economically superior where winter temperatures are mild, electricity is relatively cheap, and other fuels are relatively expensive. Also, since they can cool as well as heat a space, they have advantages where cooling in summer months is also desired. Thus some of the best locations for heat pumps are in warm summer climates with cool winters. Figure 15.31 shows a heat pump, called a “reverse cycle” or “split-system cooler” in some countries.

Figure 15.31 In hot weather, heat transfer occurs from air inside the room to air outside, cooling the room. In cool weather, heat transfer occurs from air outside to air inside, warming the room. This switching is achieved by reversing the direction of flow of the working fluid.

### 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 $QcQc size 12{Q rSub { size 8{c} } } {}$ occurs from a cold environment compared with how much work input $WW size 12{W} {}$ 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 $(COPref)(COPref) size 12{ ital "COP" rSub { size 8{"ref"} } } {}$ of an air conditioner or refrigerator to be

$COP ref = Q c W . COP ref = Q c W . size 12{ ital "COP" rSub { size 8{ ital "ref"} } = { {Q rSub { size 8{c} } } over {W} } "." } {}$
15.43

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

$COP ref = COP hp − 1 COP ref = COP hp − 1 size 12{ ital "COP" rSub { size 8{"ref"} } = ital "COP" rSub { size 8{"hp"} } - 1} {}$
15.44

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 $COPrefCOPref size 12{ ital "COP" rSub { size 8{"ref"} } } {}$ ranging from 2 to 6. These numbers are better than the $COPhpCOPhp size 12{ ital "COP" rSub { size 8{"hp"} } } {}$ 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 $COPCOP size 12{ ital "COP"} {}$ rating system called the “energy efficiency rating” ($EER EER size 12{ ital "EER"} {}$ ) 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. $EER s EER s size 12{ ital "EER"} {}$ 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 $EER s EER s size 12{ ital "EER"} {}$ ranging from 6 to 12. Although not the same as the $COPsCOPs size 12{ ital "COP"} {}$ just described, these $EER s EER s size 12{ ital "EER"} {}$ are good for comparison purposes—the greater the $EER EER size 12{ ital "EER"} {}$, the cheaper an air conditioner is to operate (but the higher its purchase price is likely to be).

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

$EER = Q c / t 1 W / t 2 , EER = Q c / t 1 W / t 2 , { ital "EER"= { {Q rSub { {c} } /t rSub { {1} } } over {W/t rSub { size 8{2} } } } ,} {}$
15.45

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