Everything in this lesson is the depth identity from the first teach card, used in anger: 1 mm over 1 ha is 10 m³, exactly. Write it on the inside of your hard hat.

Run time. $t = \dfrac{d\,A}{Q}$ — t equals d A over Q. $d$ is the application depth wanted on this set, $A$ the area the set covers, and $Q$ the flow rate reaching it. Depth times area is a volume; a volume divided by a flow rate is a time. In practical units, 25 mm over 4 ha is 1,000 m³, and at 50 L/s — which is 0.05 m³/s — that is 20,000 seconds, or about five and a half hours. The arithmetic is trivial; the unit bridges are where the errors live.

System capacity. $Q = \dfrac{A\,ET_p}{E_a\,f}$ — Q equals A E-T-p over E-a f. This is the sizing calculation behind every pump, well and mainline on the farm. $A$ is the whole area under the system, $ET_p$ the peak crop water use in mm/day — the worst fortnight, not the seasonal average — $E_a$ the application efficiency as a bare fraction, and $f$ the operating time fraction: the hours a day the system can actually run, over 24.

Look at what sits on the bottom. Both terms are fractions below one, so both make the pump bigger. And $f$ is where optimism gets expensive: sizing on 24 hours a day leaves no margin for a move, a power interruption or a failed fitting, and the peak-use fortnight is precisely when the system cannot afford a day off. Design practice is 20 to 22 hours.

Run the same relation backwards and it answers the only question that matters on an existing farm: can this system keep up? Convert the flow into a depth per day and compare it against peak use. If the delivery is smaller, the shortfall does not average out — it draws the soil reserve down a little further every day, and the reserve is finite. The reserve buys DAYS; it does not supply litres.