 Ch 5. Second Law of Thermodynamics Multimedia Engineering Thermodynamics Heat Engine The SecondLaw CarnotCycle Carnot HeatEngine CarnotRefrigerator
 Chapter 1. Basics 2. Pure Substances 3. First Law 4. Energy Analysis 5. Second Law 6. Entropy 7. Exergy Analysis 8. Gas Power Cyc 9. Brayton Cycle 10. Rankine Cycle Appendix Basic Math Units Thermo Tables Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Meirong Huang Kurt Gramoll ©Kurt Gramoll THERMODYNAMICS - CASE STUDY SOLUTION A heat engine is operating between a refrigerator's outside fin coil unit and the kitchen air. Determine the maximum power generated by this heat engine. Problem Description (1) Determine the heat dissipated from the refrigerator The coefficient of performance(COP) of a refrigerator is expressed as       COPR = QL /Wnet,in = (QH - Wnet,in)/Wnet,in Rearranging this equation gives       QH = (1 + COPR)Wnet,in = (1 + 3.0)(1.0) = 4.0 kW (2) Determine power generated by the heat engine The thermal efficiency of a heat engine is expressed as       ηth = Wnet,out/QH where       Wnet,out = power generated by the heat engine       QH = heat transferred to the heat engine Rearranging this equation yields       Wnet,out =  ηth QH where QH is the heat transferred from the heat-exchange pipes to the heat engine, which is the heat dissipated from the refrigerator, hence       QH = 4.0 kW The thermal efficiency of a Carnot heat engine is       ηth,rev = 1- (273 + 20)/(273 + 30) = 3.3% The work output from this Carnot heat engine is       Wnet,out =  ηth,rev QH = (3.3%)(4.0) = 0.12 kW The result shows the power generated from this Carnot heat engine is only enough to light a 100 W bulb. Hence, there is little benefit to use the heat dissipated by a refrigerator.