Thermodynamics, Carnot Engine

Q. Thermodynamics, Carnot Engine.

Question .
A reversible heat engine receives heat from two thermal reserviors at 870K and 580K,and rejects 50KW of heat to a sink at 290K.If the engine output is 85KW,make calculations for the energy efficiency and heat supplied by each reservior.

Answer  
Let Q1 = heat supplied by the first reservoir 
and Q2 = heat received by the 2nd reservoir
=> Q1 + Q2 = work done + heat rejected 
=> Q1 + Q2 = 85 + 50 = 135 kW ... (1)

Efficiency of the engine for heat received from the first source
= (870 - 290) / (870) = 2/3
Efficiency of the engine for heat received from the 2nd source
= (580 - 290) / (580) = 1/2

=> (2/3)Q1 + (1/2)Q2 = 85
i.e., 4Q1 + 3Q2 = 510 ... (2)

Solving eqns. (1) and (2),
Q1 = 105 kW and
Q2 = 30 kW

Efficiency 
= W / (Q1 + Q2) x 100 
= 85 / (105 + 30) x 100
= 62.96 %.