Question

A heat engine working between two temperatures could theoretically convert one-eighth of the heat supplied to the work. If the lowered temperature is reduced by 95°C , the theoretical efficiency would be doubled. Find the source and sink temperature

Answer 1

Efficiency $η = \frac{(T_{1} – T_{2})}{T_{1}} = \frac{1}{8}$ (1)

T₁ is the source temperature

T₂ is the sink temperature

$η = \frac{T_{1} – (T_{2} - 95)}{T_{1}} = 2\times \frac{1}{8} = \frac{1}{4}$

4T₁ – 4T₂ + 380 = T₁

3T₁ – 4T₂ + 380 = 0

Substituting the value of T₂ from equation (1)

8T₁ – 8T₂ = T₁

7T₁ – 8T₂ = 0

7T₁ = 8T₂

$\frac{7}{2}$ T₁ = 4T₂

3T₁ – $\frac{7}{2}$ T₁ + 380 = 0

$\frac{1}{2}$ T₁ = 380

T₁ = 760 K or 487 °C

T₂ = 665 K or 392 °C

Answer: source temperature 487 °C, sink temperature 392 °C