Carnot engine working between a source temperature of T2 and sink temperature of T1 has efficiency of 25%. If the sink temperature is reduced by 20o C, the efficiency is increased to 30%. Find the source and the sink temperature.

Maximum efficiency of an engine working between temperatures T2 and T1 is given by the fraction of the heat absorbed by an engine which can be converted into work is known as efficiency of the heat engine.

Mathematically,
Efficiency, η = (T2 – T1) / T2 = 25% = 0.25

Where T2 is the source temperature and T1 is the sink temperature.

T2 – T1 = 0.25 T2
T1 = 0.75 T2
In the second case, if the sink is reduced by 20o C, then
η= T2 – (T1 – 20) / T2 = 30% = 0.3
T2 – T1 + 20 = 0.3 T2 ………………………………. (1)

Putting the value of T1 in equation (1), we will get:
T2 – 0.75 T2 + 20 = 0.3 T2
0.25 T2 + 20 = 0.3 T2
20 = 0.3 T2 – 0.25 T2 = 0.05 T2
T2 = 20 / 0.05 = 400o C
Hence, T1 = 0.75 T2
T1 = 0.75 X 400 = 300o C
Hence, temperature of the source and the sink is 400o C and 300o C respectively.

Category: Second Law of Thermodynamics

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