A typical automobile engine Hot and cold reservoir temperature 580 K, 390 K Heat loss of cold reservoir 895 J and useful work 485 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 =

A typical automobile engine Hot and cold reservoir temperature 580 K, 390 K Heat loss of cold reservoir 895 J and useful work 485 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

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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