8.12

Applications of Thermodynamics: Heat Pumps and Refrigerators

Chapter contents: Chapter 8: Thermal Physics

Photograph of various expensive refrigerators displayed in a home appliance store.
Figure 8.12.1. 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 operate by causing heat transfer from cold to hot. Imagine a heat-engine cycle run backward: the system absorbs heat Qc from a cold reservoir, and work W is done on it so that it can release heat Qh into a hot reservoir. Over a complete cycle, the work input adds to the energy released as heat, so the first law gives Qh=Qc+W. (Note that Qh, Qc, and W are positive, with their directions indicated on schematics rather than by sign.) A heat pump’s mission is for heat transfer Qh to occur into a warm environment, such as a home in the winter. The mission of air conditioners and refrigerators is for heat transfer Qc 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.)
Part a of the figure shows a heat pump, drawn as a circle. Work W, indicated by a bold orange arrow, is put in to to the pump to transfer heat Q sub c, indicated by a bold orange arrow, out of a cold temperature reservoir T sub c, drawn as a blue rectangle, and pumps heat Q sub h, indicated by a larger bold orange arrow, into high temperature reservoir T sub h. Part b of the figure shows a P V diagram for a Carnot cycle. The pressure P is along the Y axis and the volume V is along the X axis. The graph shows a complete cycle A D C B A. The path begins at point A, then it drops sharply down and slightly to the right until point D. This is marked as an adiabatic expansion. Then the curve drops down more gradually, still to the right, from point D to point C. This is marked as an isotherm at temperature T sub c, during which heat Q sub c enters the system. The curve then rises from point C to point B along the direction opposite to that of A D. This is an adiabatic compression. The last part of the curve rises up from point B back to A. This is marked as an isotherm at temperature T sub h, during which heat Q sub h leaves the system. The path D C is lower than path B A. Heat entering and leaving the system is indicated by bold orange arrows, with Q sub h larger than Q sub c.
Figure 8.12.2. 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 W, Qh, and Qc are opposite what they would be in a heat engine. (b) PV diagram for a Carnot cycle similar to that in Figure 8.12.3 but reversed, following path ADCBA. The area inside the loop is negative, meaning there is a net work input. There is heat transfer Qc into the system from a cold reservoir along path DC, and heat transfer Qh 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+W. Heat transfer is from the outside air, even at a temperature below freezing, to the indoor space. You only pay for W, and you get an additional heat transfer of Qc 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 8.12.3. A working fluid such as a non-CFC refrigerant is used. In the outdoor coils (the evaporator), heat transfer Qc occurs to the working fluid from the cold outdoor air, turning it into a gas.
The diagram shows a diagram of a heat pump. There are four components connected by pipes. They are a condenser (1), an expansion valve (2), an evaporator (3), and a compressor (4), connected in that order. The evaporator coils are outside; all of the other components are inside. Heat Q sub c is absorbed from the outside air at the evaporator, and heat Q sub h is emitted inside from the condenser.
Figure 8.12.3. A simple heat pump has four basic components: (1) condenser, (2) expansion valve, (3) evaporator, and (4) compressor. In the heating mode, heat transfer Qc 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 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. (To switch to cooling, a reversing valve changes the refrigerant flow so that the indoor coil acts as the evaporator and the outdoor coil acts as the condenser.)
The quality of a heat pump is judged by how much heat transfer Qh occurs into the warm space compared with how much work input 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 (COPhp) to be
COPhp=QhW.
(8.12.1)
For a reversible heat-engine cycle run backward as a heat pump, the efficiency of the forward cycle is Eff=W/Qh, we see that COPhp=1/Eff, an important and interesting fact. First, since the efficiency of any heat engine is less than 1, it means that COPhp is always greater than 1—that is, a heat pump always has more heat transfer Qh 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/Th); thus, the smaller the temperature difference, the smaller the efficiency and the greater the COPhp (because COPhp=1/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. This is part of what it means for a process to be irreversible. We can model these losses by letting only a portion W′ of the work input W reach a reversible heat pump, with the rest dissipated outside the hot reservoir. For the same heat delivery Qh to the hot reservoir at the same reservoir temperatures, greater work input W is needed.
A diagram of a heat pump (shown as a circle). Work W, indicated by a large, wavy orange arrow, is the total work put into the pump. Part of this work is done against friction and is lost in the form of frictional heat, Q sub f, to the cold reservoir. The portion of work that is used by the heat pump is represented by W prime. The pump transfers heat Q sub h, indicated by a large orange arrow, into the hot reservoir, a tan-colored rectangle, at temperature T sub h. Frictional heat Q sub f, indicated by a wavy orange arrow, is transferred to the cold reservoir, a blue rectangle at temperature T sub c. Heat Q sub c, indicated by a smaller wavy orange arrow, is transferred into the pump from the cold reservoir. Heat Q sub h is formed from a combination of W prime and Q sub c.
Figure 8.12.4. This figure models losses in a heat pump. Only a portion W′ of the work input W reaches the reversible heat pump; the remainder is dissipated as heat Qf into the cold reservoir. For the same total work input and reservoir temperatures, these losses reduce the heat delivered to the hot reservoir and therefore reduce the coefficient of performance.
Example 8.12.1

The 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 COPhp. (See Figure 8.12.5.) What is the best coefficient of performance possible for such a heat pump, if it has a hot reservoir temperature of 45.0°C and a cold reservoir temperature of −15.0°C?

Strategy

A Carnot engine reversed will give the best possible performance as a heat pump. As noted above, COPhp=1/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.
The temperatures in kelvins are Th=318K and Tc=258K, so that
EffC=1−258K318K=0.1887.
Thus, from the discussion above,
COPhp=1Eff=10.1887=5.30,
or
COPhp=QhW=5.30,
so that
Qh=5.30W.

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.
A heat pump absorbs heat Qc from outdoor air at -5 degrees Celsius into refrigerant at Tc = -15 degrees Celsius. Electrical work W drives the pump, which releases heat Qh from refrigerant at Th = 45 degrees Celsius to indoor air at 20 degrees Celsius.
Figure 8.12.5. 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 COPhp range from about 2 to 4. This range means that the heat transfer Qh from the heat pumps is 2 to 4 times as great as the work W put into them. Their economical feasibility is still limited, however, since 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. The mild winters of the San Francisco Bay Area, home to College of Alameda, are well suited to efficient heat-pump heating. 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 8.12.6 shows a heat pump, called a “reverse cycle” or “split-system cooler” in some countries.
A residential heat pump.
Figure 8.12.6. 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 Qc occurs from a cold environment compared with how much work input 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) of an air conditioner or refrigerator to be
COPref=QcW.
(8.12.2)
Noting again that Qh=Qc+W, we can see that an air conditioner will have a lower coefficient of performance than a heat pump, because COPhp=Qh/W and Qh is greater than Qc. In this module’s Problems and Exercises, you will show that
COPref=COPhp−1
(8.12.3)
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 COPref ranging from 2 to 6. These numbers are better than the COPhp 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 rating system called the “energy efficiency rating” (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. EERs 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 EERs ranging from 6 to 12. Although not the same as the COPs just described, these EERs are good for comparison purposes—the greater the EER, the cheaper an air conditioner is to operate (but the higher its purchase price is likely to be).
The EER of an air conditioner or refrigerator can be expressed as
EER = Q c / t 1 W / t 2 ,
(8.12.4)
where Q c is the amount of heat transfer from a cold environment in British thermal units, t 1 is time in hours, W is the work input in joules, and t 2 is time in seconds.
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