Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temperature reservoir at 1000 K (727 °C) and a large, cold reservoir at 308 K (35°C). If it withdraws 1.2 MJ/s from the high-temperature reservoir, what would be the rate of loss of entropy from that reservoir and what would be the rate of gain by the low-temperature reservoir? (4 marks) b. Express the work done by the engine in watts. (3 marks) What would be the total entropy gain of the system? d. Determine Carnot efficiency and recalculate the a, b, and c, accordingly.

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ogorwyne ogorwyne · Answer 1 · 2022-04-05T17:37:42+02:00

With a high temperature of 1000k and a cold temperature of 308K, The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

[tex]\frac{w}{Q1} = \frac{35}{100}[/tex]

W = [tex]1.2*\frac{35}{100}*1000kj/s[/tex]

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

[tex]\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ[/tex]

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

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