A common belief is that any closed, counterclockwise cycle on a pV diagram represents a refrigerator or heat pump. It has been established that this is not the case, but previous papers on this topic have made unnecessary assumptions about the temperatures of the energy reservoirs and about how the system exchanges energy with the reservoirs. Relaxing these assumptions leads to a wide array of unexpected behaviors. In some cases, the same pV cycle can be a heat pump, a cold pump, or a Joule pump simply by changing how the system is connected to the reservoirs. This paper explores the strange world of counterclockwise cycles.

An ideal-gas heat engine follows a closed, clockwise (cw) cycle in a pV diagram, and every closed, clockwise cycle represents some sort of heat engine. That is, over the course of one full cycle, the gas absorbs heat from a hot reservoir, does work on the environment, and exhausts “waste heat” to a cold reservoir.

A refrigerator or heat pump does work on the gas to transfer heat from a cold reservoir to a hot reservoir. An ideal-gas refrigerator or heat pump follows a closed, counterclockwise (ccw) cycle in a pV diagram, but does every closed, counterclockwise cycle represent a refrigerator or heat pump? It is commonly thought that this is the case. Recent statements from seemingly authoritative sources include the following: “If the working substance is taken around a cycle in the pV-plane in the counterclockwise direction, the device is a refrigerator;”1 “If the cycle … is operated counterclockwise, it uses work to transport heat and is therefore acting as a refrigerator or a heat pump;”2 “If the cyclic process … moves counterclockwise then it represents a heat pump …;”3 and “Unlike clockwise cycles, a counterclockwise cycle is a refrigerator or heat pump ….”4 

This common belief may arise from the fact that physics textbooks—be they introductory texts or more advanced texts on thermodynamics—rarely illustrate the pV cycle of anything other than a Carnot-cycle heat pump. Many show only an energy-flow diagram in which work done on the system “lifts” heat from the cold reservoir to the hot reservoir. The text will have already established that any clockwise cycle is a heat engine with thermal efficiency ηηCarnot, so a reader may draw the unwarranted conclusion that any ccw cycle is a refrigerator or heat pump with coefficient of performance KKCarnot.

That not all ccw cycles are refrigerators was discussed at length by da Silva,5 whose exhaustive analysis of the well-known cycles when operated both clockwise and counterclockwise established several important criteria for when a ccw cycle functions as a refrigerator or heat pump. A paper in this journal by Dickerson and Mottman,6 apparently unaware of da Silva's work, made many of the same points. A comment by Bizarro7 on the paper of Dickerson and Mottman added new information. Leff8 then extended the discussion by considering a Carnot heat pump in which the reservoir temperatures differ from the gas temperatures during the isothermal processes so that heat is transferred irreversibly across a non-zero temperature difference.

The impractical but illustrative rectangular pV cycle of Fig. 1 provides a clear example of a ccw cycle that is not a heat pump or refrigerator. We will assume that the device is connected to and exchanges energy with two energy reservoirs, a “hot reservoir” at temperature Thot and a “cold reservoir” at temperature Tcold. (The reservoir temperatures are shown in this and other figures as dashed isotherms. These are not part of the cycle.) The gas temperature decreases during processes 12 and 23, so these must be processes in which heat is exhausted from the gas to a reservoir. The second law of thermodynamics requires heat to flow from hot to cold, so the reservoir receiving this energy must be the cold reservoir at temperature TcoldT3. Processes 34 and 41, with an increasing temperature, must absorb heat from a reservoir. Here, the second law dictates that the reservoir temperature must be ThotT1. Thus, the net result of one complete cycle, during which work is done on the gas, is to remove heat from the hot reservoir and add a larger amount of heat to the cold reservoir—exactly the opposite of what a heat pump does.

Fig. 1.

This rectangular ccw cycle absorbs heat from the hot reservoir and exhausts heat to the cold reservoir. The reservoir temperatures are shown as dashed isotherms. This is not a heat pump or refrigerator.

Fig. 1.

This rectangular ccw cycle absorbs heat from the hot reservoir and exhausts heat to the cold reservoir. The reservoir temperatures are shown as dashed isotherms. This is not a heat pump or refrigerator.

Close modal

The goal of this paper is to show via additional examples that ccw pV cycles have many more interesting and unexpected properties than have been described in previous papers. A clockwise heat-engine cycle is fairly tightly constrained by the fact that the system is turning heat into useful work—ultimately second-law constraints. However, a counterclockwise cycle requires an energy input, so there are fewer constraints as to where that energy goes and how it is used.

Section II of this paper establishes the basic framework for analyzing ccw cycles, then Secs. III and IV apply these ideas to the Carnot cycle and the Kelvin cycle. Section V examines the reversed Brayton cycle, a cycle that is used in many practical heat pumps and refrigerators. Here, the use of a heat exchanger, or regenerator, can have a dramatic impact on how the cycle functions. Finally, Sec. VI analyzes the ccw Stirling cycle as an example of a device that does not rely on adiabatic processes to change the temperature in the absence of heat exchange with the reservoirs.

We will consider only ideal-gas systems that follow a closed, counterclockwise cycle in the pV plane. As is customary, we will assume that all processes are frictionless and quasi-static. Thus, the system itself—the gas—has no net change in entropy over the course of one full cycle. Any entropy changes are those of the reservoirs.

We will also assume that the environment consists of no more than two reservoirs (it will be found that one reservoir is sufficient in some cases), each with infinite heat capacity and, thus, a fixed temperature. How the system is connected to the reservoirs is an engineering issue, and we will assume that any part of the system's cycle can be connected to either reservoir. The second law of thermodynamics is satisfied if we insist that all heat transfers between the system and a reservoir must be from a higher temperature to a lower temperature.

A device that uses external work to transfer heat from a cold reservoir to a hot reservoir, as in Fig. 2(a), could be either a refrigerator or a heat pump, depending on whether one wishes to keep the cold side cold or the hot side hot. We will refer to all such devices as heat pumps. We are accustomed to situations in which heat is extracted from the cold reservoir during one part of the cycle, and then the same heat plus the work done, to conserve energy, is exhausted to the hot reservoir during a different part of the cycle. However, we will see cycles that both extract heat from and exhaust heat to each of the two reservoirs. This is allowed by the second law if the heat transfers are always from a higher temperature to a lower temperature. Such a cycle is a heat pump if, over the course of a full cycle, there's a net transfer of energy from the cold reservoir to the hot reservoir.

Fig. 2.

Three modes of operation of a counterclockwise cycle. These show the net energy flows for one full cycle of operation.

Fig. 2.

Three modes of operation of a counterclockwise cycle. These show the net energy flows for one full cycle of operation.

Close modal

It is also possible to use the work done on the system to transfer heat from the hot reservoir to the cold reservoir, as in Fig. 2(b). This is what the rectangular ccw cycle of Fig. 1 does. It is analogous to using a water pump to move water from a higher-elevation reservoir to a lower-elevation reservoir rather than letting it flow spontaneously. There may be little need for such a device, but it does not violate any laws of physics. Dickerson and Mottman6 referred to such a device as a “cold pump.” We will follow that terminology because it makes a nice distinction between devices that move energy from the cold reservoir to the hot reservoir (heat pumps) and devices that move energy from the hot reservoir to the cold reservoir (cold pumps).

Bizarro7 added a third possibility, shown in Fig. 2(c), that he called a “Joule pump” in honor of Joule's paddle-wheel demonstration that a system's temperature can be increased by doing work on it. The work done on a Joule pump adds energy to both reservoirs.

The standard textbook analysis of heat engines and heat pumps treats the heat transfers to and from the reservoirs as positive quantities—simply the amount of heat. For a heat engine, for example, the heat transferred from the hot reservoir to the system—usually designated QH or Qin—is a positive number. Similarly, QC or Qout is a positive number for the heat transferred from the system to the cold reservoir. This is reasonable and practical if (1) we know a priori the directions of the heat flows and (2) the heat flow between a reservoir and the system is always in the same direction. Neither of these will be true for some of the cycles we examine, so we will use the more general sign convention, applicable to both the system and the reservoirs, that heat Q is positive when heat is transferred in, negative when heat is transferred out. Summed over the entire cycle, Qsys+Qres=0, because all heat that enters/leaves the system must leave/enter a reservoir.

The work W done on the system in a ccw cycle is positive. The first law of thermodynamics for a complete cycle requires Qsys=W, and thus Qres=W. That is, all energy that enters the system as work is ultimately transferred to the reservoirs as heat.

We begin with the well-known Carnot heat pump. The cycle, shown in Fig. 3, is likely the only explicit heat-pump cycle that many readers will have seen. Let the reservoir temperatures be Tα and Tβ, with, in this case, Tα>Tβ. We will use α and β throughout this paper as labels for the reservoirs, but there is no implication or assumption that Tα is always larger than Tβ. There will be cycles for which it is most convenient to let α label the colder reservoir.

Fig. 3.

The cycle of a reversible Carnot heat pump.

Fig. 3.

The cycle of a reversible Carnot heat pump.

Close modal

A reversible Carnot cycle has an isothermal compression at temperature Tcomp=Tα, during which heat is exhausted to the hot reservoir, and an isothermal expansion at temperature Texp=Tβ, during which heat is absorbed from the cold reservoir. That is, the gas temperatures equal the reservoir temperatures during the isothermal processes.

Now this cannot be literally true. Isothermal processes require a heat transfer, and there can be no heat transfer if the temperature difference is zero. Thus, we suppose that the gas temperature during the 12 compression is actually Tcomp=Tα+dT, allowing heat to flow from the infinitesimally hotter gas to the slightly cooler hot reservoir. This is possible because the adiabatic compression 41 can overshoot the point where Tgas=Tα. Similarly, the gas temperature during the expansion, when heat flows into the gas, must be Texp=TβdT. Needless to say, it takes infinitely long to transfer a finite amount of heat across an infinitesimal temperature difference. These infinitesimal differences do not affect the operation of a cw heat engine and are not even mentioned in most presentations, but they will turn out to be significant for some ccw cycles.

The need for infinitesimal temperature differences, to provide non-zero driving forces, as well as assumptions such as quasi-static processes and reservoirs with infinite heat capacity raise fundamental questions about the nature or even the existence of thermodynamically reversible processes. However, an analysis of the foundations of thermodynamics is not the purpose of this paper. The interested reader is referred to papers by Norton,9 Valente,10 and Leff and Mungan.11 

The heat transfers of the Carnot cycle are calculated in every introductory physics text. Qcomp=nRTαln(V1/V2) during the compression, negative because heat is transferred out of the gas, and Qexp=nRTβln(V4/V3) during the expansion. The reservoir heat flows are Qα=Qcomp>0 and Qβ=Qexp<0. The adiabatic processes require V1/V2=V4/V3, and thus the coefficient of performance is
KCarnot=QαW=QαQα+Qβ=TαTαTβ=11Tβ/Tα.
(1)
Alternatively, perfect reversibility with ΔS=0 requires
ΔS=ΔSgas+ΔSres=0+ΔSα+ΔSβ=QαTα+QβTβ=0.
(2)
Thus, Qβ/Qα=Tβ/Tα, which, if used in Eq. (1), gives the same result for KCarnot.

The first change we can make is not to insist that the reservoir temperatures match the gas temperatures during the isothermal processes. The idea of an endoreversible heat engine was introduced in this journal in a classic paper by Curzon and Ahlborn.12 An endoreversible heat engine is one in which the system—the gas—undergoes a reversible cycle with ΔSgas=0, but the heat transfers with the reservoirs are irreversible processes across finite temperature differences. Since then, several papers13–15 have considered endoreversible refrigerators and heat pumps.

Note that the two isothermal processes of a reversible Carnot cycle are established by having close thermal contact between the system and the reservoirs. In contrast, an endoreversible engine or heat pump requires some sort of temperature regulation to maintain a constant temperature during the isothermal processes. That is an engineering issue, not a physics issue, and similar issues will be relevant to numerous cycles herein.

Figure 4(a) shows an endoreversible Carnot cycle with reservoir temperatures Tα<Tcomp and Tβ>Texp. We have assumed, for now, that Tα>Tβ. This is a heat pump—previously analyzed by Leff8—in which heat is absorbed from the cold reservoir by the even colder gas during the isothermal expansion and exhausted to the hot reservoir by an even hotter gas during the compression. The isothermal processes must be slow enough to justify the quasi-static assumption, but they no longer require infinitely long duration to complete.

Fig. 4.

Two endoreversible Carnot cycles. (a) A heat pump. (b) A cold pump.

Fig. 4.

Two endoreversible Carnot cycles. (a) A heat pump. (b) A cold pump.

Close modal
The entropy change of this endoreversible heat pump is
ΔS=QαTα+QβTβ=nRln(V1/V2)(TcompTαTexpTβ)>0.
(3)
The inequality, and thus compliance with the second law, follows from the design stipulation that Tα<Tcomp and Tβ>Texp. That is, all heat flows are from hotter to colder.
The coefficient of performance is calculated as earlier if we use the actual gas temperatures Tcomp and Texp in the expressions for Qα and Qβ rather than the reservoir temperatures Tα and Tβ. We find that
K=QαW=QαQα+Qβ=TcompTcompTexp=11Texp/Tcomp.
(4)
It follows from Eq. (3) that Texp/Tcomp<Tβ/Tα and, thus, referring to Eq. (1), K<KCarnot. As expected, the coefficient of performance is less than that of a reversible Carnot heat pump that operates at the reservoir temperatures.

Previous papers on endoreversible Carnot refrigerators and heat pumps have assumed (explicitly or tacitly) that the isothermal compression exhausts heat to the hotter of the two reservoirs; that is, Tα>Tβ. However, there is no physical requirement that Tα>Tβ. Suppose, as in Fig. 4(b), that Tα<Tβ. That is, the compression exhausts heat to the colder reservoir at Tα, while the expansion absorbs heat from the hotter reservoir at Tβ. This violates no laws; energy is conserved, and the heat transfers—out of the system during the compression and into the system during the expansion—are all from hotter to colder. However, now we have a cold pump, rather than a heat pump, with work being done to move energy from the hotter reservoir to the colder reservoir. Leff's conclusion8 that “The irreversible CCW Carnot cycle is a traditional refrigerator…” is, in general, not correct. It is a valid statement only if Tα>Tβ.

Equations (3) and (4) are still correct, because we made no assumption about which was the hotter reservoir, but we must be cautious in interpreting the coefficient of performance K. First, this is now the efficiency of doing work to deliver energy to the cold reservoir—an appropriate figure of merit for a cold pump. The second law, Eq. (3), still requires Texp/Tcomp<Tβ/Tα, but now Tβ/Tα>1. However, Texp/Tcomp<1 in a ccw cycle, so this second-law limit on Texp/Tcomp is not relevant. Unlike the heat pump, there is no second-law upper bound on the coefficient of performance of a cold pump. Said another way, there is no ideal, reversible cold pump with ΔS=0 to establish a comparison cycle that cannot be bettered.

Two cases are of particular interest. Figure 5(a) shows an endoreversible cold pump that, at first glance, looks like the reversible Carnot heat pump of Fig. 3. However, there is an important difference. Here, the isothermal compression exhausts heat not to the hotter reservoir but to the colder reservoir at temperature Tα=Texp. Similarly, the isothermal expansion absorbs heat from the hotter reservoir at Tβ=Tcomp. Thus, there is a net heat transfer from the hotter reservoir to the colder reservoir, making this a cold pump.

Fig. 5.

Two examples of endoreversible Carnot cold pumps.

Fig. 5.

Two examples of endoreversible Carnot cold pumps.

Close modal
The coefficient of performance, from Eq. (4), is
K=11Texp/Tcomp=11Tα/Tβ.
(5)
Here, Tα is the temperature of the colder reservoir, so the expression on the right is the coefficient of performance of a reversible Carnot heat pump that operates between a hotter reservoir at temperature Tβ and a colder reservoir at temperature Tα. That is, switching the connections to the reservoirs of a reversible Carnot heat pump produces an irreversible cold pump with the same coefficient of performance.

Figure 5(b) shows that nothing prevents Tα from being less than Texp. Likewise, Tβ can exceed Tcomp. Now there is no constraint on the coefficient of performance. Referring to Eq. (4), we see that K as TexpTcomp. The quantity of heat delivered to the cold reservoir is fixed at Qα=nRTcompln(V1/V2), but the amount of work required to do so—the area enclosed within the cycle —approaches zero as TexpTcomp.

For example, a Carnot cold pump with isothermal processes at 400 and 410 K could absorb heat from a 500 K reservoir and exhaust heat to a 300 K reservoir with coefficient of performance K=41. Endoreversible heat engines do not have these types of unexpected behaviors because reversing the order of the reservoirs would violate the second law.

What if, as in Fig. 6, the reservoirs have the same temperature, Tα=Tβ, with this common temperature falling between Texp and Tcomp? These could be two physically separate reservoirs that just happen to have the same temperature, but they need not be distinct. If not, we see that a ccw Carnot cycle can run from a single reservoir. Heat |Qβ| is extracted from this reservoir during the expansion, then a larger amount of heat Qα=|Qβ|+W is exhausted to the same reservoir during the compression. The reservoir is an energy source during one part of the cycle but an energy sink during a different part. The net result of one full cycle is to add energy W to the single reservoir. This is not a heat pump, nor a cold pump, nor a Joule pump of the type envisioned by Bizarro,7 although it could be considered an alternative type of Joule pump.

Fig. 6.

A Carnot-cycle Joule pump that operates with one reservoir.

Fig. 6.

A Carnot-cycle Joule pump that operates with one reservoir.

Close modal

This might seem to be all the possibilities for a Carnot cycle, but that is not the case. The  Appendix looks at a situation in which each of the isothermal processes, the compression and the expansion, is connected first to one reservoir and then, later in the process, to the other reservoir.

A key finding from all these examples is that the pV cycle of the gas does not, by itself, tell us whether we have a heat pump, a cold pump, or a Joule pump. We also need to know the reservoir temperatures and how the different processes in the cycle are connected to the reservoirs.

The cycle shown in Fig. 7 is, effectively, half of a ccw Carnot cycle. Leff8 called it a Kelvin cycle because it roughly models a process described by Lord Kelvin. Its significance, in Leff's analysis, is that it operates from a single reservoir. Heat is absorbed from a reservoir at temperature Tβ, but no higher-temperature reservoir is needed because the gas reaches higher temperatures by adiabatic compression and then exhausts heat to the reservoir at temperature Tβ. This cycle is neither a heat pump nor a cold pump, but, as Leff noted, a type of Joule pump that transforms work into an increased thermal energy of the reservoir. Note that the reversed, cw heat-engine Kelvin cycle does need two reservoirs.

Fig. 7.

The counterclockwise Kelvin cycle with the isothermal expansion at the reservoir temperature. This is neither a heat pump nor a cold pump.

Fig. 7.

The counterclockwise Kelvin cycle with the isothermal expansion at the reservoir temperature. This is neither a heat pump nor a cold pump.

Close modal

The gas cools during the isobaric compression by being in thermal contract with the reservoir at temperature Tβ. However, there is a problem. The gas temperature approaches Tβ asymptotically from above, so it can only reach temperature Tβ+dT. However, the subsequent isothermal expansion needs the gas temperature to be TβdT so that the gas can absorb heat from the reservoir. That is, the end points of the isobaric and isothermal processes are on opposite sides of the reservoir temperature.

The same issue arises with the counterclockwise Stirling cycle analyzed by Bizarro,7 which was not recognized in his paper but was pointed out by Dickerson and Mottman16 in their reply to Bizarro. Dickerson and Mottman argued that this slight difference in the temperatures prevents the cycle from functioning, but that is incorrect. If a gas expands while being in contact with an ever-so-slightly cooler reservoir, something has to happen. It cannot be an isothermal expansion, or any expansion that requires heat, because heat cannot be absorbed from the cooler reservoir. However, lacking a heat source, the gas can expand adiabatically. The adiabatic expansion lowers the temperature as it follows a pV curve that is steeper than the isotherm, but it only needs to move an infinitesimal distance along this curve until the temperature has dropped to TβdT. At that point, heat will begin to flow from the reservoir to the gas, and the process can continue as an isothermal expansion.

In other words, a cycle such as this requires an infinitesimal “adiabatic corner” in which the gas temperature changes from one side of the reservoir temperature to the other. This infinitesimal adiabatic process has no effect on the analysis. The author has been unable to find any cw heat-engine cycles that need an adiabatic corner; it appears to be another peculiarity of ccw cycles.

Although the counterclockwise Kelvin cycle can operate from one reservoir, there are richer and more complex behaviors if it uses two. For example, Fig. 8(a) shows a hotter reservoir at Tβ>Texp and a colder reservoir at Tα<Texp. The isothermal expansion absorbs heat Qβ from the hotter reservoir, while the isobaric compression exhausts heat Qα to the colder reservoir, so this is a cold pump.

Fig. 8.

Variations on the Kelvin cycle. (a) This cycle is always a cold pump. (b) This cycle can be a heat pump or a joule pump, depending on the location of point 2.

Fig. 8.

Variations on the Kelvin cycle. (a) This cycle is always a cold pump. (b) This cycle can be a heat pump or a joule pump, depending on the location of point 2.

Close modal

However, suppose, as in Fig. 8(b), that the colder reservoir is at temperature Tα=Texp (plus dT as needed), from which the isothermal expansion absorbs heat Q34, and the hotter reservoir is at temperature Tβ such that Tα<Tβ<T1. The isobaric compression can exhaust heat to the hotter reservoir until reaching point 2, where the Tβ isotherm crosses the isobar, and then, by changing the connections, exhaust heat to the colder reservoir. How to change the connections to the reservoirs is an engineering issue, not a physics issue. The net heat exchange of the colder reservoir is Qα=Q23Q34, and this device is a heat pump—heat is transferred from the colder reservoir to the hotter reservoir—if Qα<0. This will be true if Q34>|Q23|; that is, if more heat is absorbed from the colder reservoir during the isothermal expansion than is exhausted to the colder reservoir during the isobaric process 23.

This cycle can be analyzed in detail using the known expressions for heat and work during the three processes, the ideal-gas law, and the adiabatic relationship between points 1 and 4. Doing so finds that
Qα=γγ1nR[(TβTα)Tαln(T1Tα)],
(6)
where γ=CP/CV is the specific-heat ratio, and T1 is the temperature at point 1. Qα is negative, making the cycle a heat pump, if Tβ<Tβ*, where
Tβ*=Tα[1+ln(T1Tα)].
(7)

We see that Tβ*>Tα, and the identity lnxx1 for x>0 can be used to show that Tβ*<T1. That is, there is always a range of hotter reservoir temperatures Tα<Tβ<Tβ* for which Fig. 8(b) is a heat pump. The device is a Joule pump if Tβ*<Tβ<T1 because then net heat is exhausted to both reservoirs.

As a numerical example, suppose Tα=Texp=300K and T1=400K. Equation (7) gives Tβ*=386K, so the device of Fig. 8(b) is a heat pump if the hotter reservoir temperature is in the range 300K<Tβ<386K. It is a Joule pump if 386K<Tβ<400K.

The Carnot and Kelvin cycles are of academic but not practical interest. In contrast, the reverse Brayton cycle is actually used in applications ranging from air cooling in jet aircraft to the liquefaction of natural gas. In addition, the Brayton cycle allows the use of a heat exchanger that can dramatically change how the cycle functions.

The counterclockwise Brayton cycle, with two adiabats and two isobars, is often drawn as in Fig. 9(a) with the reservoir temperatures matching the gas temperatures at points 2 and 4 on the cycle. This is a heat pump, absorbing heat from the cold reservoir at Tβ and exhausting heat to the hot reservoir at Tα. However, a heat pump with these reservoirs would require an infinitely long cycle time because the isobaric processes approach the reservoir temperatures asymptotically.

Fig. 9.

(a) Narrow and (b) wide Brayton cycles that function, respectively, as heat pumps and cold pumps.

Fig. 9.

(a) Narrow and (b) wide Brayton cycles that function, respectively, as heat pumps and cold pumps.

Close modal

Interestingly, the isotherms that match the corner temperatures reverse order if the cycle gets too wide relative to its height, as in Fig. 9(b). Now the isobaric expansion takes in energy from the hotter reservoir and the isobaric compression exhausts energy to the colder reservoir. This is a cold pump. Thus, not every counterclockwise Brayton cycle is a heat pump—the width of the cycle matters.

The transition between heat pump and cold pump occurs when points 2 and 4 fall on the same isotherm, in which case the cycle can operate from a single reservoir. The adiabatic expansion 23 satisfies T3=r(1γ)/γT2, where r=p2/p3=pmax/pmin is the pressure ratio that appears in the expression for the efficiency of a Brayton heat engine.17 The expansion 34 is isobaric, so
T4=V4V3T3=V4V3r(1γ)/γT2.
(8)
T4 matches T2, which places both corners on the same isotherm, if V4/V3=r(γ1)/γ. The device is a heat pump for smaller values of the volume-expansion ratio and a cold pump for larger values. A typical value of r is 5, in which case, the transition ratio is V4/V3=1.58 for a diatomic gas such as air with γ=1.40.

Figure 10(a) shows a narrow Brayton cycle with reservoir temperatures that do not match the corner temperatures. This is still a heat pump, but now the isobaric processes are completed in a finite amount of time because each process ends before the gas temperature reaches the reservoir temperature. However, we are not compelled to absorb heat from the cold reservoir and exhaust heat to the hot reservoir. If we switch the connections to the reservoirs, as in Fig. 10(b), the cycle becomes a cold pump that absorbs heat from the hot reservoir at temperature Tβ and exhausts heat to the cold reservoir at temperature Tα. (This cold-pump arrangement also works for Fig. 9(a) when the reservoir temperatures do match the corner temperatures.) In addition, as with the Carnot cycle, we could allow the reservoir temperatures to match and operate this as a Joule pump with a single reservoir.

Fig. 10.

More generally, a narrow Brayton cycle can be either (a) a heat pump or (b) a cold pump.

Fig. 10.

More generally, a narrow Brayton cycle can be either (a) a heat pump or (b) a cold pump.

Close modal

Figure 11 is the wide Brayton cycle from Fig. 9(b) with reservoir temperatures matching the corner temperatures, but now—similar to what we did with the Carnot (see the  Appendix) and Kelvin cycles—we let the isobaric compression exhaust heat first to the hotter reservoir in process 12, then to the colder reservoir in process 23. Similarly, the isobaric expansion absorbs heat from the colder reservoir in process 45, then from the hotter reservoir in process 56. This is a situation in which each of the reservoirs is both a sender and a receiver of heat at different points in the cycle.

Fig. 11.

This Brayton cycle can be a heat pump, a cold pump, or a joule pump, depending on the value of the pressure ratio.

Fig. 11.

This Brayton cycle can be a heat pump, a cold pump, or a joule pump, depending on the value of the pressure ratio.

Close modal
We assume that this is a wide Brayton cycle with V4/V3>r(γ1)/γ and corner temperatures Tβ>Tα. Straightforward calculations of the heat transfers find
Qα=Q23Q45=nCP(Tβ+(r(1γ)/γ2)Tα),Qβ=Q12Q56=nCP(Tα+(r(γ1)/γ2)Tβ).
(9)
The device is a heat pump if Qα<0, a cold pump if Qβ<0, and a Joule pump if neither are negative. Examination of Eq. (9) shows that Qα can be negative only if Tβ/Tα<2, which covers most realistic cases. In that case, the pressure ratios that lead to each of these devices are
Coldpump:r<(2Tα/Tβ)γ/(γ1),Joulepump:(2Tα/Tβ)γ/(γ1)<r<(2Tβ/Tα)γ/(1γ),Heatpump:r>(2Tβ/Tα)γ/(1γ).
(10)
A cold pump or a Joule pump are the only possibilities if Tβ/Tα ≥ 2, with r for the Joule pump then having no upper limit.

As an example, for reservoir temperatures Tα=300 and Tβ=400K, and for a diatomic gas with γ=1.40, the device is a cold pump for pressure ratio r<2.18, a heat pump r>4.13, and a Joule pump for values in between. Thus, a wide Brayton cycle that is a cold pump with “normal” connections to the reservoirs can actually be a heat pump with a change to the engineering of how the device is connected to the reservoirs.

Real-world Brayton cycles are flow systems, and flow systems often use heat exchangers. To see how a heat exchanger can be useful, notice that processes 23 and 56 in Fig. 11 are both isobaric processes with the same specific heat and the same temperature difference TβTα. The heat nCP(TαTβ) exhausted in 23 is exactly the heat absorbed in 56, so these processes can be enabled by an internal exchange of energy rather than by connections to external reservoirs. Figure 12 shows how.

Fig. 12.

A Brayton cycle with a heat exchanger.

Fig. 12.

A Brayton cycle with a heat exchanger.

Close modal

First, after the adiabatic compression, the gas is cooled from temperature T1 to temperature T2=Tβ by exhausting heat to the hot reservoir. Similarly, the gas is warmed from temperature T4, after the adiabatic expansion, to temperature T5=Tα by absorbing heat from the cold reservoir.

In addition, the gas passes through a counterflow heat exchanger. Heat exchanger theory is fairly messy. Hot gas enters on the left at temperature Tβ and cold gas enters on the right at temperature Tα. Inside, the counterflowing gases are in thermal contact and exchange heat. The two output temperatures depend on the geometry (length, pipe diameter), the rate of heat exchange, the gas specific heats, and the mass flow rates. In our case, with equal specific heats and equal mass flow rates, an ideal heat exchanger simply reverses the two input temperatures:18 the gas entering at Tβ cools to Tα, while the gas entering at Tα warms to Tβ. We assume idealizations for other aspects of the cycle, so we will assume an ideal heat exchanger.

The revised pV diagram of Fig. 13, with the heat exchanger in place, shows that heat is extracted from the cold reservoir during portion 45 of the isobaric expansion and heat is exhausted to the hot reservoir during portion 12 of the isobaric compression. However, the heat exhausted during compression 23 is not transferred to a reservoir but is redistributed internally by the heat exchanger to become the heat absorbed during the expansion 56.

Fig. 13.

A Brayton cycle that had been a cold pump becomes a heat pump if a heat exchanger is used to redistribute heat internally.

Fig. 13.

A Brayton cycle that had been a cold pump becomes a heat pump if a heat exchanger is used to redistribute heat internally.

Close modal
As a result, what had been a cold pump or a Joule pump is now a heat pump! It extracts heat from the cold reservoir and exhausts heat to the hot reservoir. We will leave it as an exercise for the reader to show, using the heat transfers calculated earlier, that the coefficient of performance is
K=QαW=11r(1γ)/γTα/Tβ<KCarnot.
(11)

Carnot and Brayton cycles use two adiabatic processes to change the gas temperature, so they exchange heat with the reservoirs during only two of the four processes that form the cycle. One might wonder if adiabatic processes are essential for a heat pump or a cold pump. The Stirling cycle of Fig. 14, with heat exchange during all four processes, is important for showing that this is not the case. We will assume that the reservoir temperatures match the isothermal-process temperatures (±dT as needed). Heat is absorbed from the cold reservoir during the isothermal expansion and exhausted to the hot reservoir during the isothermal compression. However, in addition, unlike a Carnot heat cycle, there is also heat exchange during the two isochoric processes. The isochoric cooling 23 can come about only if the system is in contact with and exhausts heat to the cold reservoir. Similarly, the isochoric heating 41 requires contact with and absorbs heat from the hot reservoir. Thus, each reservoir is both a source and a sink of heat. It is not clear what this device does.

Fig. 14.

In this ccw Stirling cycle, each reservoir is both a source and a sink of heat.

Fig. 14.

In this ccw Stirling cycle, each reservoir is both a source and a sink of heat.

Close modal

Note that this is another cycle that needs “adiabatic corners.” The gas temperature approaches the cold reservoir temperature Tβ asymptotically during isochoric process 23, reaching Tβ+dT, but the gas temperature needs to be TβdT to absorb heat from the cold reservoir during the isothermal expansion. A similar situation prevails at point 1.

Calculation of the heat transfers during the four processes is straightforward, and we find that the net heat transfers of the reservoirs are
Qα=(Q12+Q41)=nRTαln(V1/V2)nCV(TαTβ),Qβ=(Q23+Q34)=nCV(TαTβ)nRTβln(V1/V2).
(12)
The device is a heat pump if Qβ<0, a cold pump if Qα<0, and a Joule pump if neither are negative. This leads to
Coldpump:V1V2<exp[CVRTαTβTα],Joulepump:exp[CVRTαTβTα]<V1V2<exp[CVRTαTβTβ],Heatpump:V1V2>exp[CVRTαTβTβ].
(13)
Equation (13) bounds on the compression ratio V1/V2 were found by da Silva,5 and, as a requirement for the device to function as a heat pump, by Mungan.19 Thus, depending on the value of the compression ratio, the counterclockwise Stirling cycle can be a heat pump or a cold pump or a Joule pump.

Alternatively, the device could be configured to exhaust heat to the cold reservoir at temperature Tβ during isothermal compression 12, then absorb heat from the hot reservoir at temperature Tα during isothermal expansion 34. In this case, the hot reservoir is always a source of heat and the cold reservoir is always a sink, so the device is a cold pump that is similar to the rectangular-cycle cold pump of Fig. 1.

Because Q23=Q41 in the Stirling cycle, we could, as with the Brayton cycle, use an ideal heat exchanger—called a regenerator in the parlance of Stirling heat engines and refrigerators—to transfer the heat exhausted in isochoric process 23 to isochoric process 41. All heat exchange with the reservoirs then occurs along the isotherms and is reversible, so a ccw Stirling cycle with a heat exchanger becomes equivalent to a Carnot heat pump and has the Carnot coefficient of performance. This has been previously noted by Mungan.19 

Closed, counterclockwise cycles of an ideal gas offer a rich array of unexpected behaviors. Some are heat pumps, extracting heat from a cold reservoir and exhausting heat to a hot reservoir. Some are cold pumps that do the opposite, extracting heat from the hot reservoir and exhausting heat to the cold reservoir. Some are joule pumps that transform the work done on the system into increased thermal energy of both reservoirs. Some can operate from a single reservoir by extracting heat from this reservoir during part of the cycle and then exhausting heat to the same reservoir during another part of the cycle.

Especially noteworthy is that the function of a device—heat pump, cold pump, or Joule pump—cannot be determined by analyzing only the cycle. The analysis has to include not only the cycle itself but also the reservoir temperatures as well as how various parts of the cycle are connected to the reservoirs. Some ccw cycles can function as any of these three devices simply by changing the connections to the reservoirs. This is in stark contrast with cw cycles that are always heat engines.

The author is grateful to Bob Dickerson and John Mottman for useful discussions over the years.

The author has no conflicts to disclose.

This  Appendix explores a variation of the Carnot cycle in which—similar to the wide Brayton cycle of Fig. 11—each of the isothermal processes is connected first to one reservoir and then, later in the process, to the other reservoir.

The isothermal compression of a ccw Carnot cycle can exhaust heat to either the hotter or the colder reservoir if both happen to be colder than Tcomp. Likewise, the expansion can absorb heat from either reservoir if both are hotter than Texp. Thus, we can envision the counterclockwise cycle shown in Fig. 15 where the two reservoir temperatures are between Texp and Tcomp. We have shown the case with Tα>Tβ, but the only requirement is that Tα and Tβ both lie between Texp and Tcomp.

Fig. 15.

A Carnot cycle in which each isothermal process is connected to both reservoirs.

Fig. 15.

A Carnot cycle in which each isothermal process is connected to both reservoirs.

Close modal

During the compression, at temperature Tcomp, the system first exhausts heat Q12 to the reservoir at Tα as it moves along the isotherm from 1 to 2, then heat Q23 to the reservoir at Tβ as it moves from 2 to 3. These are allowed because both reservoirs are colder than Tcomp. How to change the connections to the reservoirs is an engineering issue, not a physics issue. The total heat transfer is Qcomp=Q12+Q23, with all three heat transfers are negative.

Similarly, heat Q45 is absorbed from the reservoir at Tβ in the isothermal expansion process 45, followed by the absorption of heat Q56 from the reservoir at Tα during the process 56. The total heat transfer is Qexp=Q45+Q56.

What is this device? Whether the device is a heat pump or a cold pump or a Joule pump is determined not by how heat is transferred into and out of the system but by how heat is moved between the reservoirs. Heat that enters/leaves the system must leave/enter a reservoir, so from the reservoirs' perspective Qα=Q12Q56 and Qβ=Q23Q45.

Suppose that points 1, 3, 4, and 6—the corners of the cycle—are fixed but that points 2 and 5 are under our control. A sequential application of Q=nRTln(Vj/Vi) for the isothermal process ij leads to
Qα=nR(Tcompln(V1/V2)Texpln(V6/V5)),Qβ=nR(Tcompln(V2/V3)Texpln(V5/V4)).
(A1)
A bit of algebra finds that the device is a Joule pump, adding heat to both reservoirs (Qα>0 and Qβ>0), if
V3(V5V4)Texp/Tcomp<V2<V1(V5V6)Texp/Tcomp.
(A2)

That is, first pick point 5 anywhere on the expansion isotherm between points 4 and 6. For that choice of point 5, the device is a Joule pump if point 2 on the compression isotherm is within the range given by Eq. (A2). The lower limit is greater than V3, and the upper limit is less than V1, and the use of V6/V4=V1/V3 for a Carnot cycle establishes that the lower limit is always less than the upper limit. Thus, this is always a physically realizable range.

If point 2 is farther left, in the range
V3<V2<V3(V5V4)Texp/Tcomp,
(A3)
then Qα>0, but Qβ<0. Now, we have a device that moves heat from the reservoir at Tβ to the reservoir at Tα, but it could be either a heat pump or a cold pump depending, respectively, on whether Tα>Tβ or Tα<Tβ.
If point 2 is on the right side of the compression isotherm, in the range
V1(V5V6)Texp/Tcomp<V2<V1,
(A4)
then Qα<0, but Qβ>0. That is, this device moves heat from the reservoir at Tα to the reservoir at Tβ. Whichever function this device performs (heat pump or cold pump) if point 2 is in the range given by Eq. (A3), it performs the opposite (cold pump or heat pump) if point 2 is in the range given by Eq. (A4).

Thus, the cycle shown in Fig. 15 can be a heat pump or a cold pump or a Joule pump with no change in the reservoirs or the cycle simply by changing how different parts of the cycle are connected to the reservoirs.

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