Once the chart has told you what one kilogram is worth, the size of the machine is a division. $\dot{m} = \dfrac{\dot{Q}}{\Delta h}$ — read aloud m-dot equals Q-dot over delta-h. $\dot{m}$ is the refrigerant mass flow in kilograms per second; $\dot{Q}$ is the cooling capacity in kilowatts; and $\Delta h$ is the refrigerating effect in kJ per kilogram — the enthalpy one kilogram gains crossing the evaporator, which is $h_1 - h_4$ from the previous lesson.

The units do the explaining. A kilowatt IS a kilojoule per second, so kJ/s divided by kJ/kg leaves kg/s and nothing else. That also means no thousands-conversion is needed anywhere in this relation — which is precisely why the ×1000 slip is so tempting and so easy to spot afterwards.

Rotate it and it answers the two other questions a designer asks. Solved for capacity, $\dot{Q} = \dot{m}\,\Delta h$: a compressor of known displacement tells you what the machine can deliver. Solved for the effect, $\Delta h = \dfrac{\dot{Q}}{\dot{m}}$: a measured duty and a measured flow tell you what the refrigerant is really achieving, which is how a starved evaporator gets caught.

Worth carrying: a refrigerant with a large refrigerating effect needs FEWER kilograms per second for the same duty, and therefore a smaller compressor and smaller pipe. That one line explains most of the refrigerant selection argument of the last forty years — and it is why ammonia, with an enormous $\Delta h$, still runs the world's cold stores despite everything else about it.