A pump curve says what the machine can OFFER at each flow. The pipework has its own curve, saying what it DEMANDS at each flow, and the pump has no vote on it.

$H = H_{st} + kQ^{2}$, read aloud H equals H-static plus k Q squared. $H$ is the system head in metres: the head the pipework demands at flow $Q$. $H_{st}$ is the static head in metres: the lift, plus any pressure difference between the two ends written as a height. $Q$ is the flow, in litres per second in this chapter. $k$ is the system resistance coefficient, in metres of head per (L/s)²: one number that folds every pipe, valve, strainer and coil into a single rate of climb. Reference pages quote $k$ in strict SI, with flow in m³/s. That figure is a million times larger and describes the same pipe.

The two halves behave differently, and that is the point of drawing the curve. The static head is there at zero flow and unchanged at full flow. The friction term follows the SQUARE of the flow, so half the flow leaves a quarter of the friction. The honest way to get $k$ is to measure one duty point and read the curve backwards: $k = \dfrac{H - H_{st}}{Q^{2}}$. And $k$ is not fixed for ever. Throttle a valve or foul a strainer and it rises.

Now the crossing. Fit the pump's own curve as $H = H_0 - cQ^{2}$, where $H_0$ (H-nought) is the shutoff head the pump makes at zero flow, in metres, and $c$ is its droop coefficient, in the same units as $k$. The pump can only run where offer equals demand. Set the two curves equal and solve: $Q_{op} = \sqrt{\dfrac{H_0 - H_{st}}{k + c}}$, read aloud Q-op equals the square root of H-nought minus H-static, over k plus c. $Q_{op}$ is the operating flow, the unknown. Put it back into either curve for the head. If both curves give the same head, the arithmetic is right.

One warning worth carrying. A pump does not run at its nameplate duty. It runs where the curves cross, and if the system was estimated generously the crossing sits further out than anyone planned: more flow, more power, less suction margin.