A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejects energy by heat transfer to a cold reservoir at temperature TC. Determine the energy rejected, in kJ, and TC, in K.

Question

Eugene

Engineering

4 April, 17:58

A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejects energy by heat transfer to a cold reservoir at temperature TC. Determine the energy rejected, in kJ, and TC, in K.

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Imani Ashley · Answer 1

thermal efficiency = 40%

efficiency = (T₂ - T₁) / T₂ where T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.

(600 - T₁) / 600 =.4

600 - T₁ = 240

T₁ = 360K

Energy converted into work = 50 x. 4 = 20 kJ

heat rejected = 50 - 20 = 30 kJ