Answer:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,
![COP_{HP}=\frac{T_H}{T_H-T_L}](https://tex.z-dn.net/?f=COP_%7BHP%7D%3D%5Cfrac%7BT_H%7D%7BT_H-T_L%7D)
Therefore, the temperature relationship, ![T_H=1.15\;T_L](https://tex.z-dn.net/?f=T_H%3D1.15%5C%3BT_L)
Then, we should apply the values in the COP.
![=\frac{1.15\;T_L}{1.15-1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1.15%5C%3BT_L%7D%7B1.15-1%7D)
![=7.67](https://tex.z-dn.net/?f=%3D7.67)
The number of heat rejected by the heat pump must then be calculated.
![Q_H=COP_{HP}\times W_{nst}](https://tex.z-dn.net/?f=Q_H%3DCOP_%7BHP%7D%5Ctimes%20W_%7Bnst%7D)
![=7.67\times5=38.35](https://tex.z-dn.net/?f=%3D7.67%5Ctimes5%3D38.35)
We must then calculate the refrigerant mass flow rate.
![m=0.264\;kg/s](https://tex.z-dn.net/?f=m%3D0.264%5C%3Bkg%2Fs)
![q_H=\frac{Q_H}{m}](https://tex.z-dn.net/?f=q_H%3D%5Cfrac%7BQ_H%7D%7Bm%7D)
![=\frac{38.35}{0.264}=145.27](https://tex.z-dn.net/?f=%3D%5Cfrac%7B38.35%7D%7B0.264%7D%3D145.27)
The
value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.
![T_L=\frac{T_H}{1.15}](https://tex.z-dn.net/?f=T_L%3D%5Cfrac%7BT_H%7D%7B1.15%7D)
![=\frac{64+273}{1.15}=293.04](https://tex.z-dn.net/?f=%3D%5Cfrac%7B64%2B273%7D%7B1.15%7D%3D293.04)
![=19.89\°C](https://tex.z-dn.net/?f=%3D19.89%5C%C2%B0C)
the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15