Answer:
a)[tex]Q_H=16.4kW[/tex]
b) [tex]w=5.467[/tex]
c) [tex]w=1.2345[/tex]
Explanation:
From the question we are told that:
Interior temperature [tex]TH = 20 °C[/tex]
outside temperature is [tex]TC = 2 °C[/tex]
Rate [tex]H=16.4kW[/tex]
a)
Generally electric resistance heating is the heat transfer from interior building
Therefore
[tex]Q_H=R[/tex]
[tex]Q_H=16.4kW[/tex]
b)
COP =3
Generally the equation for coefficient of performance is mathematically given by
[tex]COP=\frac{Q_H}{w}[/tex]
Therefore
[tex]w=\frac{Q_H}{COP}[/tex]
[tex]w=16.4/3[/tex]
[tex]w=5.467[/tex]
c)
Generally the equation for coefficient of performance is mathematically given by
[tex]COP_r=\frac{TH}{TH-TC}[/tex]
[tex]COP_r=\frac{20+273}{(20+273)-(2+273)}[/tex]
[tex]COP_r=16.2[/tex]
Therefore
[tex]w=1.2345[/tex]
[tex]w=20/16.2[/tex]
[tex]w=1.2345[/tex]