Question

An actual refrigerator operates with a COP that is half the Carnot COP. It removes 10 kW of heat from a cold reservoir at 250 K and dumps the waste heat into the atmosphere at 300 K. (a) Determine the net work consumed by the refrigerator. (b) What-if Scenario: How would the answer change if the cold storage were to be maintained at 200 K without altering the rate of heat transfer

Answer 1

Answer:

Explanation:

Given

$\left ( COP\right )_{actual}=0.5\left ( COP\right )_{ideal}$

$T_L=250 K$

$T_H=300 K$

ideal COP$=\frac{T_L}{T_H-T_L}=\frac{250}{300-250}=5$

$\left ( COP\right )_{actual}=2.5$

Also $COP=\frac{Desired\ effect}{Power\ supplied}$

$2.5=\frac{10}{W_{in}}$

$W_{in}=4 kW$

(b)if $T_L=200 K$

$\left ( COP\right )_{actual}=0.5\times \frac{200}{300-200}=1$

thus $W_{in}=10 kW$