A home refrigerator is listed to run at 60% of ideal when the internal temperatu

Question

A home refrigerator is listed to run at 60% of ideal when the internal temperature is 4 °C and the room temperature is 25 °C. What is the observed efficiency? If the refrigeration compartment gains energy (from opening the door and from leaks) at a rate of 500 KJ/hr., how much electrical energy (in Watts) must the motor run at to keep the temperature inside the refrigerator constant (Note 1 Watt = 1J/s)? The back of the refrigerator is observed to be warm. Calculate the amount of heat released from the refrigerator into the surrounding room.

Explanation / Answer

part a:

let W be the energy input to the refrigerator.

Qc is the temperature removed from the cold reservoir.

Qh is the heat
coefficient of performance=Qh/W=hot temperature/(hot temperature-cold temperature)

=(273+25)/(25-4)=14.19

so efficiency=14.19*0.6=8.5143

part b:

from part a,

coefficient of performance=14.19

==>Qh/W=14.19

==>Qh=14.19*W

to keep Qc constant, and to take 500 kJ/hr into account, let W be increased to W1.

then (Qh+500)/W1=14.19

==>Qh+500=14.19*W1

==>14.19*W+500=14.19*W1

==>W1-W=500/14.19=35.236 kJ/hr=9.7878 watts

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