Consider a refrigeration system using a watercooled condense

Consider a refrigeration system using a water-cooled condenser with a specific heat of 1.0 Btu/lbm degree F. The condenser uses water at 60 degree F and a rate of 1.33 lbm/s. If the system absorbs heat from a source at 27 degree F and at a rate of 20,000 Btu/hr, it is estimated to have a COP of 1.8. a. Determine the electricity needed to be input to the system. b. What is the temperature of the water that leaves the condenser? c. Compare the efficiency of this system to the possible maximum.

Solution

a) Given COP = 1.8 = heat absorbed / Work input. Work input is electricity needed to be input to the system.

heat absorbed = 20,000 Btu/hr. so work input = 20,000/1.8 = 11,111.11 Btu/hr

b) Heat released in condenser = heat absorbed + work input = 20,000 + 11,111.11 = 31,111.11 Btu/hr

heat released in condenser should be taken by water.

31,111.11 = Condenser water mass * specific heat * change in temperature of condenser water = 1.33*3600*1*Delta T

Delta T = 6.5 ^oF. so,temperature of water leaving condenser = 60+6.5 = 66.5 ^oF

C) We can have maximum possible efficiency if process is carnot cycle.

change in entropy of refrigerant during evaporation is = 20000/270.372 = 73.972

let temparature of refrigerant during condenstion is T₁ = 31111.11/ 73.972 = 420.57 K

COP of carnot cycle = 270.372 / (420.57-270.372 ) = 1.8001

So given refrigeration system is slightly operating under maximum efficiency