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8.4: Thermodynamic Cycles, Revisited

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    We concluded Chapter 7 with a discussion of the performance of thermodynamic cycles. We will now investigate what if anything the entropy accounting equation can tell us about the performance of thermodynamic cycles.

    8.4.1 Power Cycles

    Consider a power cycle that operates between a high temperature \(T_{\mathrm{H}}\) and a lower temperature \(T_{\mathrm{L}\). (See Figure \(\PageIndex{1}\).) Recall that a power cycle is a thermodynamic cycle that has a net power output while exchanging energy by heat transfer at its boundaries. Our goal is to examine how changing the boundary temperatures and the entropy production rate influence the cycle performance — the thermal efficiency of the power cycle.

    A power cycle has heat transfer entering the system through a boundary at temperature T_H. Output work exits the system, and output heat transfer exits the system through a boundary at temperature T_L.

    Figure \(\PageIndex{1}\): Power cycle operating between \(T_{\mathrm{H}}\) and \(T_{\mathrm{L}}\).

    The first thing we need to do is see what conservation of energy and entropy accounting tell us about this power cycle. First, we apply the conservation of energy equation: \[\underbrace{ \cancel{\frac{dE_{\text{sys}}}{dt}}^{=0} }_{\text{Steady-state}} = \dot{Q}_{\text{H, in}} - \dot{Q}_{\text {L, out}} - \dot{W}_{\text {net, out}} \quad \rightarrow \quad \dot{W}_{\text {net, out}} = \dot{Q}_{\text{H, in}} - \dot{Q}_{\text {L, out}} \nonumber \]

    Not surprisingly, we find that the net power out is just the difference between the heat transfer rate into the cycle and the heat transfer rate out of the cycle. Next we apply the entropy accounting equation: \[\underbrace{ \cancel{\frac{d S_{\text{sys}}}{dt}}^{=0} }_{\text{Steady-state}} = \frac{\dot{Q}_{\text {H, in}}}{T_{\text{H}}} - \frac{\dot{Q}_{\text {L, out}}}{T_{L}} + \dot{S}_{gen} \quad \rightarrow \quad \frac{\dot{Q}_{\text {L, out}}}{T_{\text{L}}} = \frac{\dot{Q}_{\text {H, in}}}{T_{\text{H}}} + \dot{S}_{gen} \nonumber \] Here we discover that the entropy transfer rate out of the cycle by heat transfer at \(T_{\mathrm{L}}\) equals the entropy transfer rate into the cycle at \(T_{\mathrm{H}}\) plus the rate of entropy generation inside the power cycle.

    If we rewrite Eq. \(\PageIndex{2}\) and solve for the entropy production rate we have \[\dot{S}_{gen} = \frac{\dot{Q}_{\text {L, out}}}{T_{\text{L}}} - \frac{\dot{Q}_{\text {H, in}}}{T_{\text{H}}} \geq 0 \nonumber \] To keep the entropy generation rate in the discussion, we can combine Eqs. \(\PageIndex{1}\) and \(\PageIndex{2}\) by eliminating the heat transfer out of the system, \(\dot{Q}_{\text {L, out}}\), from both equations: \[\begin{aligned} \left.\begin{array}{c} \dot{Q}_{\text {L, out}} = \dot{Q}_{\text{H, in}} - \dot{W}_{\text {net, out}} \\ \dot{Q}_{\text {L, out}} = T_{\text{L}}\left(\dfrac{\dot{Q}_{\text{H, in}}}{T_{\text{H}}} + \dot{S}_{gen}\right) \end{array}\right\} \quad \rightarrow \quad \dot{Q}_{\text{H, in}} - \dot{W}_{\text {net, out}} &= T_{\text{L}} \left(\frac{\dot{Q}_{\text {H, in}}}{T_{\text{H}}} +\dot{S}_{gen}\right) \\ \dot{W}_{\text {net, out}} &= \dot{Q}_{\text{H, in}} - T_{\text{L}} \left(\frac{\dot{Q}_{\text {H, in}}}{T_{\text{H}}} +\dot{S}_{gen}\right) \end{aligned} \nonumber \]

    Rearranging gives the following relation for the net power out of the power cycle: \[\dot{W}_{\text {net, out}} = \dot{Q}_{\text {H, in}}\left(1-\frac{T_{\text{L}}}{T_{\text{H}}}\right) - \left(T_{\text{L}} \dot{S}_{gen}\right) \nonumber \] Solving for the thermal efficiency of the power cycle gives the following: \[\eta=\frac{\dot{W}_{\text {net, out}}}{\dot{Q}_{\text{H, in}}} = \left[1-\frac{T_{\text{L}}}{T_{\text{H}}}\right] - \underbrace{\left(\frac{T_{\text{L}} \dot{S}_{gen}}{\dot{Q}_{\text {H, in}}}\right)}_{\text {Always} \geq 0} \leq 1 \nonumber \] Now we are in a position to examine the performance of our power cycle.

    Investigating Power Cycle Performance

    Using Eq. \(\PageIndex{5}\), answer the following questions about power cycle performance before continuing your reading:

    (a) If the temperatures and input heat transfer rate are fixed, how can we increase the thermal efficiency?

    (b) Does a reversible or an irreversible power cycle give the best performance?

    (c) What is the theoretical maximum value of the thermal efficiency if the temperatures and input heat transfer rate are fixed? (This is known as the Carnot efficiency.) Is it surprising that your answer only depends on temperatures?

    (d) How can we increase the numerical value of the maximum thermal efficiency?

    (e) A typical steam power plant is a closed-loop, periodic cycle where water circulates in a closed loop formed by a boiler, a steam turbine, a condenser, and a boiler feedpump. The high-temperature heat transfer of energy into the cycle occurs in the boiler, and the low-temperature heat transfer of energy out of the cycle occurs in the condenser.

    • What are the physical limits on the values of \(T_{\mathrm{H}}\), the high temperature at which the power cycle receives energy by heat transfer?
    • What are the physical limits on the values of \(T_{\mathrm{L}}\), the low temperature at which the power cycle rejects energy by heat transfer?

    If you did not try to answer the questions above, please go back and try them before reading further. This is a great opportunity to learn something new on your own. Don’t miss out!

    Now let's review what you should have discovered from your investigation of power cycle performance. If the input heat transfer rate and the boundary temperatures are fixed, the only way to improve the performance of this power cycle is to reduce the entropy production rate. In the limit of an internally reversible process, the thermal efficiency reaches a maximum value called the Carnot efficiency: \[\eta_{\max} = 1-\frac{T_{L}}{T_{H}}<1 \quad\quad \begin{array}{c} \text { Carnot } \\ \text{ Efficiency } \end{array} \nonumber \] where \(T_{L}\) and \(T_{H}\) are both thermodynamic temperatures. Thus, the efficiency of an ideal power cycle that exchanges energy by heat transfer with the surroundings at two temperatures only depends on the temperatures. [Carefully note that Eq. \(\PageIndex{6}\) is only valid for a power cycle with heat transfer at two temperatures. A different equation must be developed for power cycles that have heat transfer at more than two temperatures.] This is exactly the conclusion that Sadi Carnot summarized in the Carnot Principles given earlier in Section 8.1.1.

    Stop for a moment and consider the significance of the Carnot efficiency. This is a fairly astounding result! It implies that no matter what kind of power cycle you construct, no matter what working fluid you select, and no matter how much money you spend, your power cycle efficiency can never exceed that predicted by the Carnot efficiency. And furthermore, I can tell you right now with a very simple calculation the maximum possible efficiency you can ever achieve if you will just give me the two temperatures your cycle operates between.

    Example — Efficiencies of some ideal power cycles

    Calculate the Carnot efficiency for the following power cycles that operate between two temperatures. Note that typically the actual efficiency of a power cycle is approximately \(25 \text{-} 50 \%\) of the ideal value.

    (a) A steam power plant receives heat transfer in the boiler at \(1000^{\circ} \mathrm{F} \ \left(1460^{\circ} \mathrm{R}\right)\) and rejects energy by heat transfer at \(140^{\circ} \mathrm{F} \ \left(600^{\circ} \mathrm{R}\right)\).

    (b) The first American nuclear-powered merchant ship was the N. S. Savannah. The nuclear--powered propulsion system used nuclear fission to provide the heat source to boil water in the reactor. The water in the boiler received energy by heat transfer at a mean temperature of \(508^{\circ} \mathrm{F} \ \left(968^{\circ} \mathrm{R}\right)\) and the heat transfer was rejected in the condenser at a temperature of \(347^{\circ} \mathrm{F} \ \left(807^{\circ} \mathrm{R}\right)\).

    Why would anyone settle for such a low efficiency? What compensating advantage did a nuclear-fueled system have for a ship?

    (c) Your neighbor has a solar-powered power cycle that receives energy by heat transfer from a solar collector on her roof. The maximum temperature of the solar collector is \(95^{\circ} \mathrm{C} \ (368 \mathrm{~K})\) and the power cycle rejects energy by heat transfer to a pond in her backyard at \(15^{\circ} \mathrm{C} \ (288 \mathrm{~K})\).

    How could you justify spending money to build a system with such a low efficiency? (Remember the actual efficiency is typically \(25 \text{-} 50 \%\) of the ideal value.)

    Now how can we improve the ideal efficiency of our power cycle? Since the ideal efficiency only depends on the temperatures, we should increase \(T_{\mathrm{H}}\) and decrease \(T_{\mathrm{L}}\). Although this is a perfectly reasonable idea, it turns out in practice that physical considerations limit our ability to arbitrarily change these temperatures.

    The high temperature \(T_{\mathrm{H}}\) is limited by two factors. First, it is limited by the temperature of the heat source. For example, a combustion process found in a coal-fired boiler will typically produce a higher temperature than will a flat-plate solar collector. Second, it is limited by the material properties of the physical components of the cycle.

    Material limitations are probably the most significant ones in limiting the efficiency of practical power cycles. At some point the steel used to manufacture a boiler tube for a fossil-fueled power plant will melt. However, we can never even get close to this temperature because the water is pressurized and the strength of the tubes which depends on temperature must be sufficient to resist the water pressure. (An interesting phenomenon called creep is of great interest to boiler designers. Just like a stretched rubber band will slowly deform with time, boiler tubes that are pressurized will also slowly stretch or creep with time. Extensive testing programs are used to ensure that boilers do not fail due to creep.) When designing gas-turbine power plants, operating temperatures in the combustor are limited by the ability of the first row of turbine blades to withstand the hot gases coming out of the combustor. This discussion about how materials limit performance underscores the importance of materials science in engineering. (Classroom design that only occurs on paper seems separated from real materials, but any engineer who has had to build what they designed understands the crucial role materials play in engineering.)

    The low temperature \(T_{\mathrm{L}}\) is not typically subject to material property limitations; however, it is still limited. When building a power cycle that must reject energy by heat transfer, the temperature \(T_{\mathrm{L}}\) must be greater than the temperature of the ultimate energy sink. On the surface of the Earth the ultimate energy sink is the atmosphere, a lake, a river, or an ocean. These all have temperatures that are close to what is commonly called ambient temperature. To obtain a temperature significantly below ambient temperature requires the use of a refrigeration system and is not practical. Thus, \(T_{\mathrm{L}}\) is limited to the lowest temperature that naturally occurs in the surroundings of the power cycle.

    8.4.2 Refrigeration and Heat Pump Cycles

    Now we want to look at the performance of refrigeration and heat pump cycles. As we stated earlier, these are sometimes called "reversed" cycles because all of the heat transfer and work transfer interactions are reversed from that of a power cycle. Again we will restrict our discussion to cycles that exchange energy by heat transfer with the surroundings at only two temperatures, \(T_{H}\) and \(T_{L}\).

    A system has a net input of work, an input of heat transfer at a boundary of a lower temperature, and an output of heat transfer at a boundary of higher temperature.

    Figure \(\PageIndex{2}\): Refrigeration or Heat Pump Cycle operating between \(T_{\mathrm{H}}\) and \(T_{\mathrm{L}}\).

    Figure \(\PageIndex{2}\) shows a schematic that we will use to model the behavior of both a refrigeration cycle and a heat pump cycle. Remember that from a performance standpoint the only difference is which heat transfers we are interested in. For a refrigeration cycle, we are interested in the heat transfer into the cycle at a low temperature \(T_{\text{L}}\). For a heat pump, we are interested in the heat transfer out of the cycle at the high temperature \(T_{\text{H}}\).

    Relations to predict the coefficient of performance (COP) for refrigeration cycles and heat pump cycles can be developed following an approach like that used in the previous section to develop the thermal efficiency for a power cycle. Only the final results for these cycles will be presented here:

    Refrigeration cycle: \[\mathrm{COP}_{\text {Ref}} \equiv \frac{\dot{Q}_{\text{L}}}{\dot{W}_{\text {net, in}}} = \frac{T_{\text{L}}}{\left[\left(T_{\text{H}}-T_{\text{L}}\right) + T_{\text{L}}\left(\dfrac{T_{\text{H}} \dot{S}_{gen}}{\dot{Q}_{\text{L}}}\right)\right]} \nonumber \]

    Heat pump cycle: \[\mathrm{COP}_{\mathrm{HP}} \equiv \frac{\dot{Q}_{\text{H}}}{\dot{W}_{\text {net, in}}} = \frac{T_{\text{H}}}{\left[\left(T_{\text{H}}-T_{\text{L}}\right) + T_{\text{H}}\left(\frac{T_{\text{L}} \dot{S}_{gen}}{\dot{Q}_{\text{H}}}\right)\right]} \nonumber \] As before, we will now investigate the ideal performance of these cycles.

    Performance of Refrigeration Cycles and Heat Pump Cycles —

    Please answer the following questions before continuing. Because of the similarities between these two cycles it makes sense to consider their performance in parallel.

    Refrigeration Cycles [Use Eq. \(\PageIndex{7}\)] Heat Pump Cycles [Use Eq. \(\PageIndex{8}\)]
    (a) Assuming that the two boundary temperatures and the heat transfer rate into the cycle are fixed, how can you increase the \(\mathrm{COP}_{\text{Ref}}\)? (a) Assuming that the two boundary temperatures and the heat transfer rate out of the cycle are fixed, how can you increase the \(\mathrm{COP}_{\text{HP}}\)?
    (b) Assuming that the two boundary temperatures and the heat transfer rate into the cycle is fixed, what is the maximum possible \(\mathrm{COP}_{\text {Ref}}\)? (b) Assuming that the two boundary temperatures and the heat transfer rate out of the cycle is fixed, what is the maximum possible \(\mathrm{СОР}_{\text{HP}}\)?
    (c) Based on this result, why do you think that cryogenic refrigerators are very expensive to run? (c) Based on this result, why do you think that electric heat pumps are an easier sell in Knoxville, TN than they are in Minneapolis, MN?

    Again, please try your hand at answering these questions before you proceed.

    Before we examine the ideal performance of a refrigeration cycle it will pay us to look at a typical refrigerator or freezer and see how this familiar device relates to our refrigeration cycle. Figure \(\PageIndex{3}\) shows the main components of a refrigerator or freezer.

    A refrigerator is represented as an insulated box, whose interior temperature is T_cold. Heat transfers from the interior of the refrigerator into a refrigeration cycle through a boundary of temperature T_L. The cycle also has a net input of work, and an output heat transfer through a boundary of temperature T_H from the cycle to the surrounding room. Heat from the room "leaks" into the refrigerator through the insulation.

    Figure \(\PageIndex{3}\): How a refrigeration cycle is used to keep the inside of a refrigerator or freezer cold.

    A simple refrigerator consists of an insulated box and a refrigeration cycle. The low-temperature boundary of the refrigeration cycle is inside the insulated box, and typically forms a portion of the inner wall of the insulated box. The power input \(\dot{W}_{\text {net, in}}\) is achieved by plugging the cycle into a wall outlet. The heat transfer of energy out of the refrigeration cycle occurs at a high temperature \(T_{\mathrm{H}}\) boundary surface that is physically outside the insulated box. Before a refrigerator is turned on, \(T_{\text {cold}}=T_{\mathrm{L}}=T_{\mathrm{H}}=T_{\text {room}}\). After the refrigerator is turned on, \(T_{\mathrm{H}}\) starts to increase above \(T_{\text {room}}\) and \(T_{\mathrm{L}}\) starts to drop below \(T_{\text {room}}\). The system including the "cold space" reaches steady-state when the "heat" leaking through the thermal insulation equals the \(\dot{Q}_{\text {L, in}}\) entering the refrigeration cycle. At steady-state, \(T_{\text {cold}}>T_{\mathrm{L}}\) and \(T_{\mathrm{H}}>T_{\text {room}}\). Obviously, if we had a perfect thermal insulator then all we would have to do is cool the "cold space" to the appropriate temperature and shut down the cycle. Unfortunately there is no perfect thermal insulator, and your refrigerator cycles on and off to maintain a specified interior temperature.

    The ideal COP for a refrigeration cycle that exchanges energy by heat transfer at only two boundary temperatures occurs for an internally reversible cycle: \[ \mathrm{COP}_{\text{Ref, max}} = \frac{T_{\text{L}}}{T_{\text{H}} - T_{\text{L}}} = \frac{T_{\text{L}} / T_{\text{H}}}{1 - T_{\text{L}} / T_{\text{H}}} \quad\quad \begin{array}{c} \text{Ideal} \\ \text{Refrigerator Cycle} \end{array} \nonumber \] The high temperature for a typical refrigeration system is the lowest naturally occurring temperature in the surroundings, typically room temperature. Thus, for a fixed \(T_{\mathrm{H}}\), the ideal COP of a refrigeration cycle decreases as the cold temperature drops.

    Interactive Element

    What would the ideal COP be for a cryogenic refrigeration cycle that operates between \(T_{\text{H}}=300 \mathrm{~K}\) and \(T_{\text{L}}=2 \mathrm{~K}\)? Does this explain why materials that become superconductors at \(150 \mathrm{~K}\) instead of \(2 \mathrm{~K}\) are of such interest?

    A heat pump is used to heat a room to temperature T_room. The heat pump cycle has a net input of work and heat transfer from the cooler outdoors into the cycle, through a boundary at temperature T_L. The cycle outputs heat into the room, through a boundary at temperature T_H between the cycle and the room.

    Figure \(\PageIndex{4}\): How a heat pump cycle is used to heat a house.

    Figure \(\PageIndex{4}\) shows a schematic of a heat pump and a house that it is heating. As you can see, the heat-pump cycle rejects energy by heat transfer into the conditioned space at a boundary with high temperature \(T_{\text{H}}\) and receives energy from the outdoors at a boundary with a low temperature \(T_{\text{L}}\) outside the conditioned space. At steady state conditions, "heat" leaks out of the room through the walls and there is a heat transfer of energy into the room (and out of the cycle) at \(T_{\mathrm{H}}>T_{\text {room}}\). The cycle also receives a heat transfer of energy into the cycle (and out of the surroundings) at \(T_{\text{L}}<T_{\text {outdoors}}\) as well as a net power input. (On very cold days, the outside heat transfer surface may be below freezing and frost up. In cold climates, heat pumps must have a defrost capability just like the freezer compartment in a refrigerator.)

    The ideal COP for a heat pump cycle that exchanges energy by heat transfer at only two boundary temperatures occurs for an internally reversible cycle: \[\mathrm{COP}_{\text{HP, max} = \frac{T_{\text{H}}}{T_{\text{H}}-T_{\text{L}}} = \frac{1}{1-T_{\text{L}} / T_{\text{H}}} \quad\quad \begin{array} \text{Ideal} \\ \text {Heat Pump Cycle} \nonumber \] The high temperature for a residential heat pump system is something above the air temperature needed for human comfort. The low temperature, because it must be below the ambient air temperature outdoors, is at the mercy of the daily weather and average climatic conditions. Some heat pump installations use ground water the as source of energy at \(T_{L}\) to minimize these fluctuations.

    Interactive Element

    Where would a heat pump have a higher COP on average—in Knoxville or Minneapolis? How much would they differ?

    For any installation, the ideal COP will always decrease as the outdoor temperature drops. This means that when you need it the most a heat pump has the worst performance. Because of this, most heat pump installations include a set of electric resistance heating coils to provide auxiliary heating of the air on the coldest days. In fact, as \(T_{\text{L}}\) drops the performance of an expensive electric heat pump approaches that of a less expensive electric resistance furnace. So why do people buy heat pumps? If your alternative is an electric resistance furnace, an electric heat pump is much cheaper to operate. Also the same equipment can be used to air condition the house. (What would you have to do physically to change a heat pump system into an air-conditioner?)

    This page titled 8.4: Thermodynamic Cycles, Revisited is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Donald E. Richards (Rose-Hulman Scholar) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.