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Carnot engine (application of the second law of thermodynamics) - problems and solutions 1. An engine operates between 1200 Kelvin and 300 Kelvin. What is the efficiency of this engine ? Known : High temperature (TH) = 1200 K Low temperature (TL) = 300 K Wanted : Efficiency (e) Solution : T −TL e= H TH 1200− 300 900 e= = =0.75 1200 1200 75 e= =75 100 2. An engine operates between 727oC and 127oC. The engine's heat input is 6000 Joule. What is the efficiency of the engine and work done by the engine each cycle. Known : High temperature (TH) = 727oC + 273 = 1000 K Low temperature (TL) = 127oC + 273 = 400 K Heat input (QH) = 6000 Joule Wanted : Efficiency (e) and work done by the engine (W) Solution : Efficiency of the Carnot engine : T −TL e= H TH 1000− 400 600 e= = =0.60 1000 1000 60 e= =60 100 Work is done by the Carnot engine : W = e QH W = (0.6)(6000) W = 3600 Joule 3. An engine operates between 527oC and 127oC. The engine's heat input is 10,000 Joule. How much heat is discharged as waste heat from this engine ? Known : High temperature (TH) = 527oC + 273 = 800 K Low temperature (TL) = 127oC + 273 = 400 K Heat input (QH) = 10,000 Joule Wanted : Heat output (QL) Solution : Efficiency of the carnot engine :

© 2018 | San Alexander

https://physics.gurumuda.net

TH−TL TH 800 −400 400 e= = =0.5 800 800 Work is done by Carnot engine : W = e Q1 W = (0.5)(10,000) W = 5000 Joule e=

Heat output (QL) = Heat input (QH) – Work (W) QL = Q H – W QL = 10,000 – 5,000 QL = 5000 Joule 4. What is the efficiency of the engine and work done by the engine according to diagram below. Known : High temperature (TH) = 800 K Low temperature (TL) = 300 K Heat input (QH) = 1000 Joule Wanted : Efficiency (e) and work (W) done by the engine Solution : Efficiency of Carnot engine : T −TL e= H TH 800 −300 500 = =0.625 800 800 62.5 e= =62.5 100 e=

Work is done by the Carnot engine : W = e QH W = (0.625)(1000) W = 625 Joule

© 2018 | San Alexander