Question

A Bell-Coleman refrigerator operates between pressure limits of 1 bar and 8 bar. Air is drawn from the cold chamber at 9°C, compressed and then it is cooled to 29°C before entering the expansion cylinder. Expansion and compression follow the law pv^1.35 = constant. Calculate the theoretical C.O.P. of the system. 

For air take γ = 1.4, c_p = 1.003 kJ/kg K

Answer 1

shows the working cycle of the refrigerator. 

Given : p₂ = 1.0 bar ; 

p₁ = 8.0 bar ; 

T₃ = 9 + 273 = 282 K ; 

T₄ = 29 + 273 = 302 K.

Considering polytropic compression 3-4, we have

Again, considering polytropic expansion 1-2, we have

Heat extracted from cold chamber per kg of air 

= c_p (T₃ – T₂) = 1.003 (282 – 176.6) = 105.7 kJ/kg. 

Heat rejected in the cooling chamber per kg of air 

= c_p (T₄ – T₁) = 1.003 (482.2 – 302) = 180.7 kJ/kg. 

Since the compression and expansion are not isentropic, difference between heat rejected and heat absorbed is not equal to the work done because there are heat transfers to the surroundings and from the surroundings during compression and expansion. 

To find the work done, the area of the diagram ‘1-2-3-4’ is to be considered :

Work done = \(\cfrac{n}{n-1}\) (p₄V₄ – p₃V₃) – \(\cfrac{n}{n-1}\)(p₁V₁ – p₂V₂)

= \(\cfrac{n}{n-1}\) R[(T₄ – T₃) – (T₁ – T₂)]

The value of R can be calculated as follows

\(\cfrac{c_p}{c_v}\) = γ