Ask Question
9 December, 10:56

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.

+3
Answers (1)
  1. 9 December, 11:41
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers