In actual air-conditioning applications for R-12 and R-22, and operating at a condenser temperature of 40°C and an evaporator temperature of 5°C, the heat rejection factor is about

Solution (By Examveda Team)

T1 = 5°C or 273 + 5 = 278K
T2 = 40°C or 273 + 40 = 313K
$$\eqalign{ & {\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = \frac{{{\text{T1}}}}{{{\text{T2}} - {\text{T1}}}} \cr & = \frac{{278}}{{313 - 278}} \cr & = \frac{{278}}{{35}} \cr & {\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = 7.942 \cr & {\text{Heat rejection factor (HRF)}} \cr & = 1 + \frac{1}{{{\text{C}}{\text{.O}}{\text{.P}}{\text{.}}}} \cr & = 1 + \frac{1}{{7.942}} \cr & = \frac{{8.942}}{{7.942}} \cr & = 1.12 \cr} $$