The cycle on the last page had a perfect compressor: it followed a line of constant entropy straight up the chart. No real machine does. Friction, valve losses and heat from the motor windings all add entropy, and at a fixed discharge pressure more entropy means a point further RIGHT: a higher enthalpy. The real discharge always sits above the ideal one.

$\eta_s = \dfrac{h_{2s} - h_1}{h_2 - h_1}$, read aloud eta-s equals h-two-s minus h-one, over h-two minus h-one. $h_1$ is the suction enthalpy, the vapour entering the compressor. $h_{2s}$ (h-two-s, s for isentropic) is the ideal discharge enthalpy, read where the constant-entropy line from point 1 meets the discharge pressure. $h_2$ is the actual discharge enthalpy. All three are in kJ/kg. $\eta_s$ (eta-s) is the isentropic efficiency, a bare fraction, typically 0.6 to 0.8 for refrigeration machines.

Now the trap, and it is the whole lesson. This ratio runs the opposite way to a turbine's. A turbine takes work OUT, so the ideal drop is the most it could ever deliver and the actual drop goes on top. A compressor puts work IN, so the ideal rise is the LEAST it could ever need and the ideal goes on top. Either way the efficiency lands below 1. Carry the turbine's form across and an ordinary compressor scores about 130 %.

Read backwards, the same line finds the real discharge point: $h_2 = h_1 + \dfrac{h_{2s} - h_1}{\eta_s}$. Divide the ideal rise by the efficiency, never multiply. Then everything downstream uses the REAL $h_2$: the compressor power is $\dot{W} = \dot{m}\,(h_2 - h_1)$, with $\dot{m}$ the refrigerant circulated in kg/s and $\dot{W}$ in kW, and the real COP is the refrigerating effect over the actual rise. Because only the denominator grew, the real COP is exactly $\eta_s$ times the chart's ideal one.