Three real heat engines have the same thermal efficiency and are connected in series. The first engine absorbs 2400 kJ of heat from a thermal reservoir at 1250 K and the third engine ejects its waste of 300 kJ to a sink at 150 K. determine the work output from each engine.

Data given:
Q₁ = 2400 KJ
Q₄ = 300 KJ
For engine E₁,
Efficiency η₁ = (Q₁ – Q₂) / Q₁
= 1 – (Q₂ / Q₁)
I.e. Q₂ / Q₁ = 1 – η₁
Similarly for engine E₂,
Q₃ / Q₂ = 1 – η₂
For engine E₃,
Q₄ / Q₃ = 1 – η₃
Therefore,
(1 – η₁) (1 – η₂) (1 – η₃)
= (Q₂ / Q₁) X (Q₃ / Q₂) X (Q₄ / Q₃)
= Q₄ / Q₁ = 300 / 2400 = 1 / 8

For same thermal efficiency of each engine:
(1 – η)³ = 1 / 8
1 – η = 1 / 2
η = 1 / 2 = 50%
Now, Q₂ / Q₁ = 1 – 1 / 2 = 1 / 2
Q₁ = 2 Q₂
Q₂ = Q₁ / 2 = 2400 / 2 = 1200 kJ
And
Q₃ / Q₂ = 1 / 2
Q₃ = Q₂ / 2 = 1200 / 2 = 600 kJ
Hence work output from first engine (W₁) = Q₁ – Q₂
= 2400 – 1200 = 1200 kJ
From second engine (W₂) = Q₂ – Q₃
= 1200 – 600 = 600 kJ
From third engine (W₃) = Q₃ – Q₄
= 600 – 300 = 300 kJ
Total work done W = W₁ + W₂ + W₃
= 1200 + 600 +300 = 2100 kJ

Hence the total work output of the engines in series is equal to 2100 kJ