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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

Respuesta :

Answer:

Explanation:

Given

[tex]\left ( COP\right )_{actual}=0.5\left ( COP\right )_{ideal}[/tex]

[tex]T_L=250 K[/tex]

[tex]T_H=300 K[/tex]

ideal COP[tex]=\frac{T_L}{T_H-T_L}=\frac{250}{300-250}=5[/tex]

[tex]\left ( COP\right )_{actual}=2.5[/tex]

Also [tex]COP=\frac{Desired\ effect}{Power\ supplied}[/tex]

[tex]2.5=\frac{10}{W_{in}}[/tex]

[tex]W_{in}=4 kW[/tex]

(b)if [tex]T_L=200 K[/tex]

[tex]\left ( COP\right )_{actual}=0.5\times \frac{200}{300-200}=1[/tex]

thus [tex]W_{in}=10 kW[/tex]

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