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7–91 A refrigerator is to remove heat from the cooled space at a rate of 300 kJ/min to maintain its temperature at 28°C. If the air surrounding the refrigerator is at 25°C, determine the minimum power input required for this refrigerator. Answer: 0.623 kW

Respuesta :

Answer:

minimum power input required for this refrigerator is 0.623 kW

Explanation:

given data

rate = 300 kJ/min

maintain temperature T1 = -8°C = 265 K

air surrounding the refrigerator T2 = 25°C = 298 K

solution

we get here first COP that is

COP = [tex]\frac{T1}{T2-T1}[/tex]    

COP = [tex]\frac{265}{298-265}[/tex]  =  8.03

and we know here

COP = [tex]\frac{Q2}{W}[/tex]  

8.03 = [tex]\frac{300}{60W}[/tex]  

W = 0.623 kW  

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