Question

Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase efficiency.

1. By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
2. by increasing the temperature of the hot reservoir from 300°C to 350°C by keeping the cold reservoir temperature constant.

Which is the suitable method?

Answer 1

Heat engine operates at initial temperature = 100°C + 273 = 373 K

Final temperature = 300°C + 273 = 573 K

At melting point = 273 K

Efficiency η = ${\displaystyle 1-\frac{{\text{T}}_2}{{\text{T}}_1}}$

= ${\displaystyle 1-\frac{273}{573}}$

= 0.3491

η = 34.9%

(a) By decreasing the cold reservoir, efficiency

T₁ = 300°C + 273 = 573 K

T₂ = 50°C + 273 = 323 K

Efficiency η = ${\displaystyle 1-\frac{{\text{T}}_2}{{\text{T}}_1}}$

= ${\displaystyle 1-\frac{323}{573}}$

= 0.436

η = 43.6%

(b) By increasing the temperature of hot reservoir, efficiency

T₁ = 350°C + 273 = 623 K

T₂ = 100°C + 273 = 373 K

Efficiency η = ${\displaystyle 1-\frac{{\text{T}}_2}{{\text{T}}_1}}$

= ${\displaystyle 1-\frac{373}{623}}$

= 0.401

η = 40.1%

Method (a) More efficiency than method (b).