Question

A simple vapour compression plant produces 5 tonnes of refrigeration. The enthalpy values at inlet to compressor, at exit from the compressor, and at exit from the condenser are 183.19, 209.41 and 74.59 kJ/kg respectively. Estimate : 

(i) The refrigerant flow rate, 

(ii) The C.O.P., 

(iii) The power required to drive the compressor, and 

(iv) The rate of heat rejection to the condenser.

Answer 1

Total refrigeration effect produced = 5 TR (tonnes of refrigeration)

= 5 × 14000 = 70000 kJ/h or 19.44 kJ/s (\(\because\) 1 TR = 14000 kJ/h)

Given : h₂ = 183.19 kJ/kg ; h₃ = 209.41 kJ/kg ;

h₄ (= h₁) = 74.59 kJ/kg (Throttling process)

(i) The refrigerant flow rate, m :

Net refrigerating effect produced per kg = h₂ – h₁

= 183.19 – 74.59 = 108.6 kJ/kg

∴ Refrigerant flow rate, m = \(\cfrac{19.44}{108.6}\) = 0.179 kg/s.

(ii) The C.O.P. :

(iii) The power required to drive the compressor, P :

P = m (h₃ – h₂) = 0.179 (209.41 – 183.19) = 4.69 kW.

(iv) The rate of heat rejection to the condenser :

The rate of heat rejection to the condenser

= m (h₃ – h₄) = 0.179 (209.41 – 74.59) = 24.13 kW.