Lecture 29 -- Limits to efficiency. Now one might expect that there should be some limitations on the efficiency of a Carnot engine, beyond the constraints of conservation of energy. For example, you might or might not be persuaded that the following is true: It is impossible to convert heat to work with perfect efficiency, that is, with Ql = 0. This statement is called the Kelvin formulation of the second law of thermodynamics. If you are not persuaded that this postulate is an always-true statement about the world, you probably will be persuaded by the following statement: It is impossible for heat to flow from a cold object to a hot object spontaneously, i.e., without any work being done. This statement is called the Clausius formulation of the second law. The reason for the qualification &quot;spontaneously&quot;, is the existence of heat pumps; if we do work, we can pump heat from a cold place to a hot place. Remarkably, these statements are equivalent -- one implies the other. We now prove this by using the <a href="/keyword/carnot-cycle/" >carnot cycle</a> . First, assume that Kelvin was wrong; that is, suppose we could have a heat engine with no heat exhaust, Ql = 0, and hence W = -Qh . Then (see Fig. 4) we could couple this engine to a <a href="/keyword/carnot-cycle/" >carnot cycle</a> operated as a heat pump; the combined engine would have the effect of pumping heat from cold to hot with no input of work. This would violate the Clausius postulate, so if Kelvin was wrong then Clausius must also be wrong. Now assume Clausius was wrong; that is, suppose we could have a perfect refrigerator, that pumps heat from cold to hot with no work input. Then (see Fig. 5) we can couple this device with a <a href="/keyword/carnot-cycle/" >carnot cycle</a> operated as an engine; the combined engine would have the effect of extracting work from the hot reservoir with no heat exhaust. This would violate the Kelvin postulate, so if Clausius is wrong then Kelvin must also be wrong. Carnot theorem. So far, we have made no mention of the properties of the working fluid, other than our calculation (your calculation actually, assigned as homework) of the Carnot efficiency for an ideal gas. Now, using the same techniques of combining <a href="/keyword/carnot-cycle/" >carnot cycle</a> s, we prove the following remarkable statements, as a consequence of the second law (Kelvin or Clausius): Any heat engine must have an efficiency less than that of the reversible Carnot engine operating between the same two reservoirs; 67 Figure 4: If Kelvin is wrong, then Clausius is wrong too. Figure 5: If Kelvin is wrong, then Clausius is wrong too. 68 Any heat pump must have a coefficient of performance less than that of the reversible Carnot engine operating between the same two reservoirs. Suppose we have a more efficient engine (which may be irreversible!) than the Carnot engine, operating between two reservoirs. Couple this engine to a Carnot engine with the same magnitude of work, run in reverse as a heat pump (see Fig. 6). The waste stream Ql for the assumed moreefficient engine, is smaller in magnitude than the heat Ql pumped by the Carnot engine for the same work (which is, by reversibility, the same as the waste stream would have been for the Carnot engine operated in the forward direction). Thus the combined device pumps heat from cold to hot with no work input, which violates the Clausius postulate. Figure 6: No heat engine is more efficient than Carnot. Next, suppose we have a more efficient heat pump (which again may be irreversible) than the Carnot engine run in reverse as a heat pump. Drive this heat pump with a Carnot engine with the same magnitude of waste heat Ql , as the heat pumped from the cold reservoir by the heat pump (see Fig. 7). Because the heat pump is assumed more efficient than Carnot, the work required to drive the pump will be less than the Carnot engine produces. Thus the combined system produces work with no waste heat, which violates the Kelvin postulate. As a corollary of this theorem, we see that: All Carnot engines operating between the same two reservoirs have the same efficiency, regardless of working fluid. If we had Carnot engines of two different efficiencies, we could use the more efficient one to drive the less efficient one as a heat pump, and the combined system would pump heat from cold to hot with no work input. 69 Figure 7: No heat pump is more efficient than Carnot. This allows us to say that the efficiency of any <a href="/keyword/carnot-cycle/" >carnot cycle</a> is the same as that of a <a href="/keyword/carnot-cycle/" >carnot cycle</a> with an ideal gas as working fluid; therefore, = 1 - |Ql |/Qh = 1 - Tl /Th As a consequence, we may write that for any <a href="/keyword/carnot-cycle/" >carnot cycle</a> Qh /Th + Ql /Tl = 0 (This may be regarded as a constraint on possible equations of state: any valid EOS must be such that if I compute the heat and work effects in a <a href="/keyword/carnot-cycle/" >carnot cycle</a> , the efficiency works out to be the Carnot efficiency.) We will make use of this result in the next lecture. 70