Answer:
It is enough
Explanation:
To develop the problem it is necessary to take into account the concepts related to the coefficient of performance of a pump.
The two ways in which the performance coefficient can be expressed are given by:
[tex]COP_p = \frac{T_H}{T_H-T_L}[/tex]
Where,
[tex]T_H =[/tex]High Temperature
[tex]T_L =[/tex] Low Temperature
And the other way is,
[tex]COP_p = \frac{\dot{Q}}{W}[/tex]
Where [tex]\dot{Q}[/tex] is heat rate and W the power consumed.
We have all our terms in Celsius, so we calculate the temperature in Kelvin
[tex]T_H = 22+273 = 295K[/tex]
[tex]T_L = 2+273 = 275k[/tex]
The rate at which heat is lost is:
[tex]\dot{Q} = 110000kJ/h[/tex]
The power consumed by the heat pump is
[tex]\dot{W} = 5kW[/tex]
And the coefficient of performance is
[tex]COP_p = \frac{T_H}{T_H-T_L}[/tex]
[tex]COP_p = \frac{295}{295-275}[/tex]
[tex]COP_p = 14.75[/tex]
With this value we can calculate the Power required,
[tex]COP_p = \frac{\dot{Q}}{W}[/tex]
[tex]14.75 = \frac{110000}{W}[/tex]
[tex]W = \frac{110000}{3600*14.75}[/tex]
[tex]W = 2.07kW[/tex]
The power consumed is consumed is 5kW which is more than 2.07kW so this heat pump powerful enough.
