Ask Question
9 March, 09:41

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.

+4
Answers (1)
  1. 9 March, 11:02
    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?
Find 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 you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers