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13.4 Carnot’s perfect heat engine: the second law of thermodynamics  (Page 2/7)

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Carnot’s interesting result implies that 100% efficiency would be possible only if T c = 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 T c / T h 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 PV 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.

Part a of the figure shows a graph of pressure P versus volume V 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 B C D. The path begins at point A, then it moves smoothly down till point B along the direction of the X axis. This is marked as an isotherm at temperature T sub h. Then the curve drops down further, along a different curve, from point B to point C. This is marked as adiabatic expansion. The curve rises from point C to point D along the direction opposite to that of A B. This is also an isotherm but at temperature T sub c. The last part of the curve rises up from point D back to A along a direction opposite to that of B C. This is marked as adiabatic compression. The path C D is lower than path A B. Heat Q sub h enters the system, as shown by a bold arrow to the curve A B. Heat Q sub c leaves the system as shown by a bold arrow near C D. Part b of the diagram shows an internal combustion engine represented as a circle. The hot reservoir is a rectangular section at the top of the circle shown at temperature T sub h. A cold reservoir is shown as a rectangular section in the bottom part of the circle at temperature T sub c. Heat Q sub h enters the heat engine as shown by a bold arrow; work W is produced as output, shown to leave the system, and the remaining heat Q sub c is returned back to the cold reservoir, as shown by a bold arrow toward it.
PV size 12{ ital "PV"} {} diagram for a Carnot cycle, employing only reversible isothermal and adiabatic processes. Heat transfer Q h size 12{Q rSub { size 8{h} } } {} occurs into the working substance during the isothermal path AB, which takes place at constant temperature T h size 12{T rSub { size 8{h} } } {} . Heat transfer Q c size 12{Q rSub { size 8{c} } } {} occurs out of the working substance during the isothermal path CD, which takes place at constant temperature T c size 12{T rSub { size 8{c} } } {} . The net work output W 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 T h size 12{T rSub { size 8{h} } } {} and T c 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.

Maximum theoretical efficiency for a nuclear reactor

A nuclear power reactor has pressurized water at 300 º C 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 [link] ). Steam, produced in the steam generator, is used to drive the turbine-generators. Eventually the steam is condensed to water at 27 º C 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.

Diagram shows a schematic diagram of a pressurized water nuclear reactor and the steam turbines that convert work into electrical energy. There is a pressure vessel in the middle, dome shaped at the ends. This has a nuclear core in it. The core is a small square in the center of the reactor. Control rods are shown as sticks of equal length attached to the core. The pressure vessel has some coolant tubes passing through it and then back to a steam chamber. These coolant tubes contain a coolant liquid that transports the heat from the pressure vessel to the steam chamber. This whole system is enclosed in another dome shaped containment structure of steel. The water supply to steam chamber and the steam outlet are seen to come out of this chamber. This steam is now shown to run two steam turbines, one a high pressure one and another low pressure one. The turbines are nearly triangular and segmented in shape. The steam turbine in turn generates power using a turbine generator, which is attached to the turbine system. The turbines are again housed in another chamber which gets the steam from the steam chamber and return the steam as water back to the steam chamber with pipes. A coolant tower is shown near the turbine system, which is shown to supply cool water in tubes to the turbine system to cool the steam back to water.
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.)
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Source:  OpenStax, College physics: physics of california. OpenStax CNX. Sep 30, 2013 Download for free at http://legacy.cnx.org/content/col11577/1.1
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