Thermodynamics - GATE practice papers - Mechanical for Q. 375

Q.  A cyclic heat engine operates between source temperature of 1000 °C and a sink temperature 50 °C. Calculate least rate of heat rejection when net work output is 2 kW. (Marks: 02)
- Published on 19 Oct 15

a. 2.6799 kW
b. 0.6799 kW
c. 0.7463 kW
d. none of the above

ANSWER: 0.6799 kW
 
The rate of heat rejection will be least for reversible engine,

ηrev = ηmax = 1 – (Tsink / Tsource )

= 1 – (323 / 1273)
= 0.7463

ηmax = Wnet / Q1

Q1 = Wnet / ηmax
= 2 / 0.7463
= 2.6799 kW

Therefore the least heat rejection rate, Q2 is

Q2 = Q1 – Wnet

= 2.6799 – 2
= 0.6799 kW

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