Carnot Cycles with Irreversible Heat Transfer

This Demonstration shows a Carnot cycle operating either as a heat engine or a heat pump, with finite temperature differences between the hot and cold reservoirs and the high and low temperatures of the Carnot cycle, respectively. The entropy changes for the reservoirs ( and ) and the overall entropy change are calculated. When the temperature differences between the reservoirs and the engine/pump are nonzero, the total entropy change is positive. The entropy change of the engine/pump, which is at steady state, is zero. All energies and entropy changes are per unit time, since these are continuous processes, but the time scale is arbitrary. The cycle efficiency is calculated for the heat engine, and the coefficient of performance () is calculated for the heat pump. As the temperature difference between the reservoirs and the engine/pump increases, the efficiency and the coefficient of performance decrease. For the heat engine, is held constant as the temperature differences change. For the heat pump, is held constant as the temperature differences change.


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where and are user-defined differences in reservoir and Carnot cycle temperatures (K). The subscripts and refer to either the hot or cold reservoirs, while the subscripts and refer to the hot or cold sides Carnot cycle. is temperature (K), is the change in entropy (J/K), and is heat gained or lost (J).
For a Carnot heat engine, heat is transferred from a hot reservoir to a cold reservoir and the engine does work (). For a real process, and . The efficiency of a Carnot engine, in terms of the engine temperatures, is
For a Carnot heat pump, heat is transferred from the cold reservoir to the hot reservoir and work is added. For a real process, and to have a reasonable rate of heat transfer. The coefficient of performance () is
[1] J. R. Elliott and C. T. Lira, Introductory Chemical Engineering Thermodynamics, New York: Pearson Education, 2012 pp. 157–158.
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