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A food processing plant is installing a new freezer that has a CoP of 4.9. The freezer is rated at an input power of 92 hp. At what rate (Btu/hr) should this freezer be able to remove heat from the products inside the freezer?

Respuesta :

Answer:

Heat freezer will remove 450.8 Btu/hr heat

Explanation:

We have given CoP of the freezer = 4.9

Input power is given of the freezer = 92 hp

We have to find the rate at which freezer remove the heat

We know that CoP is given by

[tex]CoP=\frac{haet\ removed}{input\ power}[/tex]

[tex]4.9=\frac{haet\ removed}{92}[/tex]

So heat removed = 4.9 ×92 = 450.8 Btu/hr

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