Question

A cold storage plant of 20 tonne of refrigerator capacity operates between 200 K and 300 K. The power required to run the plant if the plant has half COP of a Carnot cycle (Tonne of refrigerator = 3.5 kW) is
1. 50 kW
2. 60 kW
3. 70 kW
4. 80 kW

Answer 1

Correct Answer - Option 3 : 70 kW

Concept: 

The COP of refrigerator is given as,

Ideal \(COP = \frac{{{T_L}}}{{{T_H} - {T_L}}}\) and Actual \(COP = \frac{Q}{W_{input}}\)

Calculation:

Given:

Q̇ = 20 × 3.5 = 70 kW, TL = 200 K, TH = 300 K

Ideal \(COP = \frac{{{200}}}{{{300} - {200}}}\) = 2

Actual COP = \(\frac{1}{2}\) × Ideal COP = \(\frac{1}{2}\) × 2 = 1

Actual COP = 1

Also, Actual COP = \(\frac{{\dot Q}}{{\dot W}} = \frac{{70}}{{\dot W}}\)

∴ \(\dot {W}\) = 70 kW.