A vapour-compression cycle is four processes, and the pressure-enthalpy chart numbers the corners in the direction the refrigerant travels. Fix the subscript convention now, because the whole lesson is bookkeeping: $h_1$ is the vapour leaving the evaporator at the compressor suction; $h_2$ is the hot gas the compressor discharges into the condenser; $h_3$ is the liquid leaving the condenser; and $h_4$ is that same liquid after the expansion valve, entering the evaporator. All four are specific enthalpies — energy per kilogram of refrigerant, in kJ/kg.

Two differences do all the work. The refrigerating effect is $h_1 - h_4$: what one kilogram picks up crossing the evaporator, which is the useful cooling. The compressor work is $h_2 - h_1$: what one kilogram costs to lift to the high side, which is the electricity bill. Divide the first by the second and the kilograms cancel:

$\mathrm{COP} = \dfrac{h_1 - h_4}{h_2 - h_1}$ — read aloud C O P equals h-one minus h-four, over h-two minus h-one. Because the mass flow cancels, this single ratio grades a machine of any size, which is exactly why the chart is drawn per kilogram in the first place.

One nugget that catches people out. Run the same cycle for HEATING and the useful output moves to the condenser: $\mathrm{COP}_{hp} = \dfrac{h_2 - h_4}{h_2 - h_1}$, which is always exactly ONE more than the refrigeration COP — the compressor's work ends up in the heated space too. Same hardware, same numbers, one extra unit of heat for free. It is the strongest argument in building services, and it is also the commonest wrong answer to a cooling question.