Question

A reversible engine is supplied with heat from two constant temperature sources at 900K and 600K and rejects heat to a constant temperature sink at 300 K. The engine develops work equivalent to 90KJ/s rejects heat at the rate of 56KJ/s. Estimate: 1. Heat supplied by each source and 2. Thermal efficiency of the engine .
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Answer 1

Explanation:

A reversible engine is supplied with heat from two constant temperature sources at 900K and 600K and rejects heat to a constant temperature sink at 300 K. The engine develops work equivalent to 90KJ/s rejects heat at the rate of 56KJ/s. Estimate: 1. Heat supplied by each source and 2. Thermal efficiency of the engine .

Answer 2

Answer:

Heat supplied by each source:

Q₁ = 102kJ/s

Q₂ = 44kJ/s

The thermal efficiency of the engine is 61.64%.

Explanation:

Given,

T₁ = 900K, T₂ = 600K, T₃ = 300K

W(engine) = 90kJ/s

Q(rejects) = 56kJ/s

As we know,

$Q_{1} + Q_{2} = W_{engine} + Q_{rejects}\\\\Q_{1} + Q_{2} = 90 + 56 = 146kJ/s --(i)\\\\\eta_{1} = \frac{T_{1} -T_{3}}{T_{1}} = \frac{900-300}{900} = \frac{600}{900} = \frac{2}{3} \\\\\eta_{1} = \frac{T_{2} -T_{3}}{T_{2}} = \frac{600-300}{600} = \frac{300}{600} = \frac{1}{2}$

$W_{1} = \eta_{1}Q_{1} = \frac{2}{3} Q_{1} \\\\W_{2} = \eta_{2}Q_{2} = \frac{1}{2} Q_{2} \\ \\Total \ Work = W_{1} + W_{2} \\\\Total \ Work = \frac{2}{3} Q_{1} + \frac{1}{2} Q_{2}\\\\\frac{2}{3} Q_{1} + \frac{1}{2} Q_{2} = 90\ \ \ -(ii)\\\\Solving\ equation \ (i) \ and \ (ii), \ we \ get\\\\Q_{1} = 102kJ/s\\\\and, \ Q_{2} = 44kJ/s$

The thermal efficiency of the engine is

$Efficiency \ of \ engine = \frac{Total \ Work\ Done}{Total\ heat\ absorbed} = \frac{90}{146} \\\\Efficiency \ of \ engine = 0.6164$

∴ The efficiency of the engine is 61.64%

To learn more about thermal efficiency, click on the link below:

https://brainly.in/question/17848821

To learn more about heat supplied, click on the link below:

https://brainly.in/question/33140895

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