At steady state, a refrigerator whose coefficient of performance is 3 removes energy by heat transfer from a freezer compartment at 0 degrees C at the rate of 6000 kJ/hr and discharges energy by heat transfer to the surroundings, which are at 20 degrees C. a) Determine the power input to the refrigerator and compare with the power input required by a reversible refrigeration cycle operating between reservoirs at these two temperatures. b) If electricity costs 8 cents per kW-hr, determine the actual and minimum theoretical operating costs, each in $/day

(A) 0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost = 0.234 $/day

Explanation:

Solution

Given that:

The coefficient of performance is = 3

Heat transfer = 6000kJ/hr

Temperature = 20°C

Cost of electricity = 8 cents per kW-hr

Now

The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.

Thus

(A) The coefficient of performance is given below:

COP = Heat transfer from freezer/Power input

3 = 6000/P

P = 6000/3

P = 2000

P = 2000 kJ/hr = 2000 / (60*60) kW

= 2000 (3600) kW

= 0.55 kW

Thus

The ideal coefficient of performance = T_low / (T_high - T_low)

= (0+273) / (20-0)

= 13.65

So,

P ideal = 6000/13.65 = 439.6 kJ/hr

= 439.6 / (60*60) kW

= 0.122 kW

(B) For the actual cost we have the following:

Actual cost = 0.55 kW * 0.08 $/kW-hr = $ 0.044 per hour

= 0.044*24 $/day

= 1.056 $/day

For the theoretical cost we have the following:

Theoretical cost = 0.122 kW * 0.08 $/kW-hr = $ 0.00976 per hour

= 0.00976*24 $/day

= 0.234 $/day