 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 solar pond is built to test the possibility of generating power. The maximum thermal efficiency and the maximum power generation need to be determined. A Heat Engine can Work between a Solar Pond and the Ambient Air (1) Determine the maximum thermal efficiency Assuming the power plant is a Carnot heat engine, which works between a source at temperature 85 oC, and a sink at temperature 35 oC.       ηth,rev = 1 - (35+273)/(85+273) = 14% (2) Determine the maximum power generated by the pond The definition of the thermal efficiency of a heat engine is       ηth = Wnet,out/Qin Rearranging this equation gives       Wnet,out = Qin ηth If the efficiency of this heat engine equals the Carnot efficiency, and the heat absorbed by the salt water is totally input to the heat engine, the output net work can achieve its maximum value.       Wnet,max = QHηth,rev The bottom area of the pond is       Apond = (10) (10) = 100 m2 Heat absorbed by the salt water in the pond bottom area is       QH = (0.3) (100) = 30 kW Hence, the maximum net work output by the heat engine equals       Wnet,out,max = Qin ηth,rev = ( 30)(0.14) = 4.2 kW