Answer:[tex]P_{in}=0.25\ Hp[/tex]
Explanation:
Given
Operating temperatures are
[tex]T_L=70^{\circ}F\approx 294.261\ K[/tex]
[tex]T_H=93^{\circ}F\approx 307\ K[/tex]
[tex]COP[/tex] is [tex]27\% [/tex] of Carnot refrigerator
[tex]COP_{carnot}=\dfrac{T_L}{T_H-T_L}[/tex]
[tex]COP_{carnot}=\dfrac{294.261}{307-294.261}[/tex]
[tex]COP_{carnot}=\dfrac{294.261}{12.74}[/tex]
[tex]COP_{carnot}=23.09\approx 23.1[/tex]
actual [tex]COP=0.27 \times COP_{carnot}[/tex]
[tex]COP_{actual}=6.237[/tex]
Also, [tex]COP=\dfrac{\text{Desired effect}}{\text{Power in}}[/tex]
[tex]6.237=\frac{4000}{P_{in}}[/tex]
[tex]P_{in}=\frac{4000}{6.237}=641.33\ Btu/h[/tex]
[tex]P_{in}=0.25\ Hp[/tex]