Ask Question
1 January, 11:01

A heat engine is operating on a Carnot cycle and has a thermal efficiency of 75 percent. The waste heat from this engine is rejected to a nearby lake at 15 ⁰C at a rate of 14 kW. Determine the power output of the engine and the temperature of the heat source.

+3
Answers (1)
  1. 1 January, 11:14
    0
    Power Output = 42 KW

    T₁ = 1152 K = 879°C

    Explanation:

    The thermal efficiency of a Carnot's Engine is given by:

    η = 1 - Q₂/Q₁

    where,

    η = efficiency = 75% = 0.75

    Q₂ = Heat Rejection Rate = 14 KW

    Q₁ = Heat Absorption Rate = ?

    Therefore,

    0.75 = 1 - 14 KW/Q₁

    14 KW = (1 - 0.75) (Q₁)

    Q₁ = 14 KW/0.25

    Q₁ = 56 KW

    Thus,

    Power Output = Q₁ - Q₂

    Power Output = 56 kW - 14 KW

    Power Output = 42 KW

    The thermal efficiency of a Carnot's Engine is also given by:

    η = 1 - T₂/T₁

    where,

    η = efficiency = 75% = 0.75

    T₂ = Temperature of Heat Sink (Lake) = 15°C + 273 = 288 K

    T₁ = Temperature of Heat Sink = ?

    Therefore,

    0.75 = 1 - 288 K/T₁

    288 K = (1 - 0.75) (T₁)

    T₁ = 288 K/0.25

    T₁ = 1152 K = 879°C
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A heat engine is operating on a Carnot cycle and has a thermal efficiency of 75 percent. The waste heat from this engine is rejected to a ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers