Answer:
[tex]\dot W_{in} = 0.82363 hp[/tex]
Explanation:
Given data:
T_1 = 55 degree F
t_2 - 25 degree F
COP = 2.45
Production rate = 28 lbm/h
we knwo that cooling load is given
[tex]\dot Q_L = \dot m q_L[/tex]
[tex]\dot Q_L = 28 lbm/h \times 16 = 4732 Btu/h[/tex]
Power is determined as
[tex]\dot W_{in} = \frac{\dot Q_L}COP}[/tex]
[tex]\dot W_{in} = \frac{4732}{2.4} = 1931.42 Btu/h[/tex]
[tex]\dot W_{in} =1931.42 Btu \times \frac{1 hp}{2345 Btu/h} = 0.82363 hp[/tex]
