Question 15.1: For a theoretical single-stage cycle with a condenser pressu......

Use the pressure–enthalpy diagram for refrigerant 22, Chart 4:

(a)  COP = $\frac{i_{3}  –  i_{1}}{i_{4}  –  i_{3}}  =  \frac{104.6  –  43.1}{125.4  –  104.6}  =  \frac{61.5}{20.8}  =  2.96$

(b)  $t_{4}$ = 168 F

(c)  $t_{1}$ = 116 F

(d)  $t_{2}$ = 2F

The COP in this case is determined almost entirely by the difference between
evaporating and condensing temperatures. For the highest COP the cycle should be operated at the lowest possible condensing temperature and the highest possible evaporating temperature.