A residential air conditioning system operates at steady sta

A residential air conditioning system operates at steady state, as shown in the figure. Refrigerant 22 circulates through the components of the system. Property data at key locations are given on the figure. If the evaporator removes energy by heat transfer from the room air at a rate of 600 Btu/min. determine the rate of heat transfer between the compressor and the surroundings, in Btu/min, and the coefficient of performance.

Solution

b) Coefficient of performance = ( Desired Effect ) / ( Work input required );

Here the desired effect is cooling and its magnitude is 600 units

While the work input required to produce this cooling effect is 200Btu/min.

Therefore COP = 600 / 200 = 3

COP = 3

Heat transfer between the compressor and surroundings

Q₁₂ = U₁₂ + W₁₂ = U₂ - U₁ + V₁ (P₂ - P₁)

Here the work transfer is flow work i.e. VdP

H = U +PV ; U = H - PV

U₁ = H₁ - P₁V_{1 ;} U₂ = H₂ - P₂V₂

H₁ = 109.9 btu/lb , H₂ = 130 btu/lb from tables

V₁ = V₂ = V of saturated vapour =0.458 ft³/ lbm ; P₁ = 120 units ; P₂ = 225 units;

Therefore U₁ = 109.9 - (120 * 0.458 ) = 54.94units ; U₂ = 130 - ( 225 * 0.458 ) = 26.95 units;

Work transfer = V ( P₂- P₁ ) = 0.458 * ( 225 - 120 ) = 48. 09 Units

Therefore heat transfer = (U₂ - U₁) + Work transfer = ( 26.95 - 54.94 ) + 48.09 = 20.1 Units

Heat transfer to the surroundings during the compression process = 20.1 Units