Consider a refrigerator whose dimensions are 1.8 m × 1.2 m × 0.8 m and whose walls are 3 cm thick. The refrigerator consumes 600 W of power when operating and has a COP of 2.5. It is observed that the motor of the refrigerator remains on for 5 minutes and then is off for 15 minutes periodically.

Question

Consider a refrigerator whose dimensions are [latex] 1.8 m imes 1.2 m imes 0.8 m [/latex] and whose walls are 3 cm thick. The refrigerator consumes 600 W of power when operating and has a COP of 2.5. It is observed that the motor of the refrigerator remains on for 5 minutes and then is off for 15 minutes periodically. If the average temperatures at the inner and outer surfaces of the refrigerator are 6°C and 17°C, respectively, determine the average thermal conductivity of the refrigerator walls. Also, determine the annual cost of operating this refrigerator if the unit cost of electricity is $0.08/kWh.

Accepted Answer

A refrigerator consumes 600 W of power when operating, and its motor remains on for 5 min and then off for 15 min periodically. The average thermal conductivity of the refrigerator walls and the annual cost of operating this refrigerator are to be determined.

Assumptions 1 Quasi-steady operating conditions exist. 2 The inner and outer surface temperatures of the refrigerator remain constant.

Analysis The total surface area of the refrigerator where heat transfer takes place is

[latex] A_{ ext {total }}=2[(1.8 imes 1.2)+(1.8 imes 0.8)+(1.2 imes 0.8)]=9.12 m ^{2} [/latex]

Since the refrigerator has a COP of 2.5, the rate of heat removal from the refrigerated space, which is equal to the rate of heat gain in steady operation, is

[latex] \dot{Q}=\dot{W}_{e} imes COP =(600 W ) imes 2.5 = 1500 W [/latex]

But the refrigerator operates a quarter of the time (5 min on, 15 min off). Therefore, the average rate of heat gain is

[latex] \dot{Q}_{ ext {ave }}=\dot{Q} / 4=(1500 W ) / 4 = 375 W [/latex]

Then the thermal conductivity of refrigerator walls is determined to be

[latex] \dot{Q}_{ ext {ave }}=k A \frac{\Delta T_{ ext {ave }}}{L} \longrightarrow k=\frac{\dot{Q}_{ ext {ave }} L}{A \Delta T_{ ext {ave }}}=\frac{(375 W )(0.03 m )}{\left(9.12 m ^{2} ight)(17-6){ }^{\circ} C }= 0 . 1 1 2 W / m \cdot{ }^{\circ} C [/latex]

The total number of hours this refrigerator remains on per year is

[latex] \Delta t=365 imes 24 / 4 = 2190 h [/latex]

Then the total amount of electricity consumed during a one-year period and the annular cost of operating this refrigerator are

[latex] ext { Annual Electricity Usage }=\dot{W}_{e} \Delta t=(0.6 kW )(2190 h / yr ) =1314 kWh / yr [/latex]

[latex] ext { Annual cost } =(1314 kWh / yr )(\$ 0.08 / kWh )=\$ 1 0 5 . 1 / yr [/latex]

Question 1.131: Consider a refrigerator whose dimensions are 1.8 m × 1.2 m ×...

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