College Physics for AP® Courses

15.4Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated

College Physics for AP® Courses15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated

Learning Objectives

By the end of this section, you will be able to:

• Identify a Carnot cycle.
• Calculate maximum theoretical efficiency of a nuclear reactor.
• Explain how dissipative processes affect the ideal Carnot engine.

Figure 15.22 This novelty toy, known as the drinking bird, is an example of Carnot's engine. It contains methylene chloride (mixed with a dye) in the abdomen, which boils at a very low temperature—about $100ºF 100ºF$. To operate, one gets the bird's head wet. As the water evaporates, fluid moves up into the head, causing the bird to become top-heavy and dip forward back into the water. This cools down the methylene chloride in the head, and it moves back into the abdomen, causing the bird to become bottom heavy and tip up. Except for a very small input of energy—the original head-wetting—the bird becomes a perpetual motion machine of sorts. (credit: Arabesk.nl, Wikimedia Commons)

We know from the second law of thermodynamics that a heat engine cannot be 100% efficient, since there must always be some heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ to the environment, which is often called waste heat. How efficient, then, can a heat engine be? This question was answered at a theoretical level in 1824 by a young French engineer, Sadi Carnot (1796–1832), in his study of the then-emerging heat engine technology crucial to the Industrial Revolution. He devised a theoretical cycle, now called the Carnot cycle, which is the most efficient cyclical process possible. The second law of thermodynamics can be restated in terms of the Carnot cycle, and so what Carnot actually discovered was this fundamental law. Any heat engine employing the Carnot cycle is called a Carnot engine.

What is crucial to the Carnot cycle—and, in fact, defines it—is that only reversible processes are used. Irreversible processes involve dissipative factors, such as friction and turbulence. This increases heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ to the environment and reduces the efficiency of the engine. Obviously, then, reversible processes are superior.

Carnot Engine

Stated in terms of reversible processes, the second law of thermodynamics has a third form:

A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. Furthermore, all engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures.

Figure 15.23 shows the $PVPV size 12{ ital "PV"} {}$ diagram for a Carnot cycle. The cycle comprises two isothermal and two adiabatic processes. Recall that both isothermal and adiabatic processes are, in principle, reversible.

Carnot also determined the efficiency of a perfect heat engine—that is, a Carnot engine. It is always true that the efficiency of a cyclical heat engine is given by:

$Eff=Qh−QcQh=1−QcQh.Eff=Qh−QcQh=1−QcQh. size 12{ ital "Eff"= { {Q rSub { size 8{h} } - Q rSub { size 8{c} } } over {Q rSub { size 8{h} } } } =1 - { {Q rSub { size 8{c} } } over {Q rSub { size 8{h} } } } } {}$
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What Carnot found was that for a perfect heat engine, the ratio $Qc/QhQc/Qh size 12{Q rSub { size 8{c} } /Q rSub { size 8{h} } } {}$ equals the ratio of the absolute temperatures of the heat reservoirs. That is, $Qc/Qh=Tc/ThQc/Qh=Tc/Th size 12{Q rSub { size 8{c} } /Q rSub { size 8{h} } =T rSub { size 8{c} } /T rSub { size 8{h} } } {}$ for a Carnot engine, so that the maximum or Carnot efficiency $EffCEffC size 12{ ital "Eff" rSub { size 8{c} } } {}$ is given by

$EffC=1−TcTh,EffC=1−TcTh, size 12{ ital "Eff" rSub { size 8{c} } =1 - { {T rSub { size 8{c} } } over {T rSub { size 8{h} } } } } {}$
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where $ThTh size 12{T rSub { size 8{h} } } {}$ and $TcTc size 12{T rSub { size 8{c} } } {}$ are in kelvins (or any other absolute temperature scale). No real heat engine can do as well as the Carnot efficiency—an actual efficiency of about 0.7 of this maximum is usually the best that can be accomplished. But the ideal Carnot engine, like the drinking bird above, while a fascinating novelty, has zero power. This makes it unrealistic for any applications.

Carnot's interesting result implies that 100% efficiency would be possible only if $Tc=0 KTc=0 K size 12{T rSub { size 8{c} } =0" K"} {}$ —that is, only if the cold reservoir were at absolute zero, a practical and theoretical impossibility. But the physical implication is this—the only way to have all heat transfer go into doing work is to remove all thermal energy, and this requires a cold reservoir at absolute zero.

It is also apparent that the greatest efficiencies are obtained when the ratio $Tc/ThTc/Th size 12{T rSub { size 8{c} } /T rSub { size 8{h} } } {}$ is as small as possible. Just as discussed for the Otto cycle in the previous section, this means that efficiency is greatest for the highest possible temperature of the hot reservoir and lowest possible temperature of the cold reservoir. (This setup increases the area inside the closed loop on the $PVPV size 12{ ital "PV"} {}$ diagram; also, it seems reasonable that the greater the temperature difference, the easier it is to divert the heat transfer to work.) The actual reservoir temperatures of a heat engine are usually related to the type of heat source and the temperature of the environment into which heat transfer occurs. Consider the following example.

Figure 15.23 $PVPV size 12{ ital "PV"} {}$ diagram for a Carnot cycle, employing only reversible isothermal and adiabatic processes. Heat transfer $QhQh size 12{Q rSub { size 8{h} } } {}$ occurs into the working substance during the isothermal path AB, which takes place at constant temperature $ThTh size 12{T rSub { size 8{h} } } {}$. Heat transfer $QcQc size 12{Q rSub { size 8{c} } } {}$ occurs out of the working substance during the isothermal path CD, which takes place at constant temperature $TcTc size 12{T rSub { size 8{c} } } {}$. The net work output $WW size 12{W} {}$ equals the area inside the path ABCDA. Also shown is a schematic of a Carnot engine operating between hot and cold reservoirs at temperatures $ThTh size 12{T rSub { size 8{h} } } {}$ and $TcTc size 12{T rSub { size 8{c} } } {}$. Any heat engine using reversible processes and operating between these two temperatures will have the same maximum efficiency as the Carnot engine.

Example 15.4

Maximum Theoretical Efficiency for a Nuclear Reactor

A nuclear power reactor has pressurized water at $300ºC300ºC size 12{"300"°C} {}$. (Higher temperatures are theoretically possible but practically not, due to limitations with materials used in the reactor.) Heat transfer from this water is a complex process (see Figure 15.24). Steam, produced in the steam generator, is used to drive the turbine-generators. Eventually the steam is condensed to water at $27ºC27ºC size 12{"27"°C} {}$ and then heated again to start the cycle over. Calculate the maximum theoretical efficiency for a heat engine operating between these two temperatures.

Figure 15.24 Schematic diagram of a pressurized water nuclear reactor and the steam turbines that convert work into electrical energy. Heat exchange is used to generate steam, in part to avoid contamination of the generators with radioactivity. Two turbines are used because this is less expensive than operating a single generator that produces the same amount of electrical energy. The steam is condensed to liquid before being returned to the heat exchanger, to keep exit steam pressure low and aid the flow of steam through the turbines (equivalent to using a lower-temperature cold reservoir). The considerable energy associated with condensation must be dissipated into the local environment; in this example, a cooling tower is used so there is no direct heat transfer to an aquatic environment. (Note that the water going to the cooling tower does not come into contact with the steam flowing over the turbines.)

Strategy

Since temperatures are given for the hot and cold reservoirs of this heat engine, $EffC=1−TcThEffC=1−TcTh size 12{ ital "Eff" rSub { size 8{C} } =1- { {T rSub { size 8{c} } } over {T rSub { size 8{h} } } } } {}$ can be used to calculate the Carnot (maximum theoretical) efficiency. Those temperatures must first be converted to kelvins.

Solution

The hot and cold reservoir temperatures are given as $300ºC300ºC size 12{"300"°C} {}$ and $27.0ºC27.0ºC size 12{"27" "." 0°C} {}$, respectively. In kelvins, then, $Th=573 KTh=573 K$ and $Tc=300 KTc=300 K size 12{T rSub { size 8{c} } ="300"" K"} {}$, so that the maximum efficiency is

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

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Discussion

A typical nuclear power station's actual efficiency is about 35%, a little better than 0.7 times the maximum possible value, a tribute to superior engineering. Electrical power stations fired by coal, oil, and natural gas have greater actual efficiencies (about 42%), because their boilers can reach higher temperatures and pressures. The cold reservoir temperature in any of these power stations is limited by the local environment. Figure 15.25 shows (a) the exterior of a nuclear power station and (b) the exterior of a coal-fired power station. Both have cooling towers into which water from the condenser enters the tower near the top and is sprayed downward, cooled by evaporation.

Figure 15.25 (a) A nuclear power station (credit: BlatantWorld.com) and (b) a coal-fired power station. Both have cooling towers in which water evaporates into the environment, representing $QcQc size 12{Q rSub { size 8{c} } } {}$. The nuclear reactor, which supplies $QhQh size 12{Q rSub { size 8{h} } } {}$, is housed inside the dome-shaped containment buildings. (credit: Robert & Mihaela Vicol, publicphoto.org)

Since all real processes are irreversible, the actual efficiency of a heat engine can never be as great as that of a Carnot engine, as illustrated in Figure 15.26(a). Even with the best heat engine possible, there are always dissipative processes in peripheral equipment, such as electrical transformers or car transmissions. These further reduce the overall efficiency by converting some of the engine's work output back into heat transfer, as shown in Figure 15.26(b).

Figure 15.26 Real heat engines are less efficient than Carnot engines. (a) Real engines use irreversible processes, reducing the heat transfer to work. Solid lines represent the actual process; the dashed lines are what a Carnot engine would do between the same two reservoirs. (b) Friction and other dissipative processes in the output mechanisms of a heat engine convert some of its work output into heat transfer to the environment.
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