A typical automobile engine Hot and cold reservoir temperature 515 K, 300 K Heat loss of cold reservoir 870 J and useful work 470 J . (1) What is the heat lost to the hot reservoir? (ii) Calculate the entropy changes in the hot reservoir, the cold reservoir, and the engine, respectively. (iii) What is the thermal efficiency of the engine? (iv) Compute the Reversible (Carnot) efficiency. Heat lost Hot reservoir qh = Entrophy change in engine s = Thermal efficiency nt = carnot efficiency nc =

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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A typical automobile engine Hot and cold reservoir temperature 515 K, 300 K Heat loss of cold reservoir
870 J and useful work 470 J . (i) What is the heat lost to the hot reservoir? (ii) Calculate the entropy changes
in the hot reservoir, the cold reservoir, and the engine, respectively. (iii) What is the thermal efficiency of the
engine? (iv) Compute the Reversible (Carnot) efficiency.
Heat lost Hot reservoir qh =
Entrophy change in engine s =
Thermal efficiency nt =
carnot efficiency nc
Transcribed Image Text:A typical automobile engine Hot and cold reservoir temperature 515 K, 300 K Heat loss of cold reservoir 870 J and useful work 470 J . (i) What is the heat lost to the hot reservoir? (ii) Calculate the entropy changes in the hot reservoir, the cold reservoir, and the engine, respectively. (iii) What is the thermal efficiency of the engine? (iv) Compute the Reversible (Carnot) efficiency. Heat lost Hot reservoir qh = Entrophy change in engine s = Thermal efficiency nt = carnot efficiency nc
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