Explanation Express the entropy balance for the two constant pressure devices. S in − S out + S gen = Δ S system S in − S out + S gen = Δ S cylinder,source + Δ S cylinder,sink + Δ S heat engine (I) Here, net entropy transfer by heat and mass is S in − S out , entropy generation is S gen , change in entropy is Δ S system , change in entropy due to cylinder source is Δ S cylinder,source , change in entropy due to cylinder sink is Δ S cylinder,sink and change in entropy due to heat engine is Δ S heat engine . Since there is no net entropy transfer by heat and mass, entropy generation and change in entropy due to heat engine, take these values as 0 . S in − S out = 0 S gen = 0 Δ S heat engine = 0 Substitute 0 for S in − S out , S gen and Δ S heat engine respectively in Equation (I). 0 + 0 = Δ S cylinder,source + Δ S cylinder,sink + 0 Δ S cylinder,source + Δ S cylinder,sink = 0 [ m c p ln T 2 T 1 A − m R ln P 2 P 1 ] cylinder,source + [ m c p ln T 2 T 1 B − m R ln P 2 P 1 ] cylinder,sink = 0 (II) Here, mass is m , constant pressure specific heat of air at room temperature is c p , initial temperature of source is T 1 A , initial temperature of sink is T 1 B , common final temperature is T 2 , initial pressure is P 1 , final pressure is P 2 and universal gas constant is R . As the pressures are constant for both of the devices, take initial and final pressure as 0 . P 1 = P 2 = 0 Substitute 0 for P 1 and P 2 in Equation (II). [ m c p ln T 2 T 1 A − m R ln 0 0 ] cylinder,source + [ m c p ln T 2 T 1 B − m R ln 0 0 ] cylinder,sink = 0 [ m c p ln T 2 T 1 A ] cylinder,source + [ m c p ln T 2 T 1 B ] cylinder,sink = 0 ln T 2 T 1 A T 2 T 1 B = 0 T 2 2 = T 1 A T 1 B T 2 = T 1 A T 1 B (III) Calculate the heat rejected out by the source by using the energy balance. E in − E out = Δ E system − Q source,out + W b,in = Δ U Q source,out = W b,in − Δ U Q source,out = m c p ( T 1 A − T 2 ) (IV) Here, the total energy entering the system is E in , the total energy leaving the system is E out , work done inside is W b,in , the change in the total energy of the system is Δ E system , change in internal energy is Δ U and the heat rejected out by the source is Q source,out

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 8.8, Problem 116RP

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The maximum work that can be produced by the heat engine and the final temperatures of the devices.

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Chapter 8.8, Problem 115RP

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The maximum work that can be produced by the heat engine and the final temperatures of the devices.

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