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.

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

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3