Pumps and system curves

Pumps, fans and compressors are the largest users of motor energy in industry, and most of them run for thousands of hours a year. That makes them the place where the choice of drive matters most, and the reason this chapter exists.

The pump curve

A centrifugal pump adds pressure to the fluid by spinning it. Pressure is expressed as head $H$ , the height of water column the pump could hold up (1 bar ≈ 10.2 m of water). The more flow $Q$ it delivers, the less head it can add. A simple model that fits real curves well:

H_\text{pump} = r^2 H_0 - a\,Q^2

where $H_0$ is the shut-off head (valve closed, zero flow) at rated speed and $r = n / n_\text{rated}$ is the speed ratio. The Lab II pump delivers 120 m³/h at 25 m at 1 475 rpm. The booklet does not give its shut-off head; this site, like the reference model, uses $H_0 = 32.5$ m.

FoundationStart here if this is new to you

A pump is like a fan blowing into a straw. If you block the straw, the fan builds up its highest pressure but nothing flows. Open it wide and air flows freely, but with little pressure behind it. Every pump lives somewhere between those two extremes.

The system curve

The pipes, valves and height the water must climb decide how much head a given flow needs:

H_\text{system} = H_s + k\,Q^2

$H_s$ is the static head: the height difference (and any pressure difference) between suction and delivery. It is needed even at zero flow.
$k Q^2$ is the friction in pipes and fittings. Closing a valve increases $k$ .

The operating point

The pump settles where the two curves cross: the flow at which the head the pump gives equals the head the system needs.

Try it: pump curve meets system curve

Pump speed100 %

Valve opening100 %

Static head H_s5.0 m

The height the water must be lifted, whatever the flow. Lab II uses 5 m.

Pump at this speed
Pump at 100 %
System

Operating point: 120.0 m³/h at 25.0 m.

Flow: 120.0 m³/h
Head: 25.0 m
Electrical power (η = 75 %): 10.90 kW
Compared with the rated point: 100 %
Energy over 6,000 h per year: 65.4 MWh
Energy per m³ pumped: 91 Wh/m³

Try three things. Close the valve: the system curve steepens and the operating point slides left, but the pump keeps its speed and still draws a lot of power. Lower the speed instead: the pump curve drops, and the same flow reduction costs far less power. Then raise the static head to 20 m and lower the speed again: the flow collapses much faster.

The affinity laws

For the same pump at a different speed, with geometrically similar flow patterns:

\frac{Q_2}{Q_1} = \frac{n_2}{n_1}, \qquad \frac{H_2}{H_1} = \left(\frac{n_2}{n_1}\right)^2, \qquad \frac{P_2}{P_1} = \left(\frac{n_2}{n_1}\right)^3

The torque follows from $P = T\omega$ : $T \propto n^2$ , the quadratic torque profile (course eq. 2.1). A fan that needs 40 N·m at 750 rpm needs $40 \times (1500/750)^2 = 160$ N·m at 1500 rpm, not 80. At 80 % speed the torque falls to 64 % and the power to 51.2 %.

Tutorial 2.1. A pump driven by a 4-pole motor runs at $N_1 = 1\,475$ rpm and delivers 120 m³/h at 25 m, absorbing 11.5 kW. The flow can be cut by 20 %, so the speed falls by 20 % too: $N_2 = 1\,180$ rpm. Then $Q = 96$ m³/h, $H = 25 \times 0.64 = 16$ m (36 % less head) and $P = 11.5 \times 0.512 = 5.89$ kW (48.8 % less power). At 8 000 h a year and €0.15/kWh that saves $(11.5 - 5.89) \times 8000 \times 0.15 \approx$ €6 700 a year.

Two cautions from the same tutorial. Commanding the VFD to exactly $N_2$ gives the right flow only if the real curves match the design; wear, tolerances and changing conditions shift them, so a closed loop on flow or pressure is more robust. And the laws lose accuracy at very low speed and with a high static head, which is the next point.

Static head breaks the cube law

The Lab II network has 5 m of static head. At 80 % speed the pump curve becomes $0.64 \times 32.5 - aQ^2$ and crosses the system curve at 91 m³/h and 16.5 m, not at the 96 m³/h the affinity laws promise. As the speed falls further, a larger and larger share of the pump's head goes into merely lifting the water to the static height.

Predict first

The Lab II pump runs at 40 % speed. How much water does it deliver?

40 % of 120, so 48 m³/h16 % of 120, so 19 m³/hAlmost none, about 10 m³/h

Power follows from flow and head, with the pump efficiency $\eta$ :

P = \frac{\rho\, g\, Q\, H}{\eta}

with $Q$ in m³/s. At the Lab II rated point: $1000 \times 9.81 \times (120/3600) \times 25 / 0.75 = 10.9$ kW.

ExplorerGo deeper: derivations and open questions

Where the affinity laws come from. For geometrically similar pumps, the flow coefficient $Q / (n D^3)$ and the head coefficient $gH / (n^2 D^2)$ are the same at corresponding points. Keep the diameter $D$ and change $n$ : the three laws follow. What happens if you trim the impeller diameter instead of changing speed?

Best efficiency point. Efficiency is not constant along the curve: it peaks at the best-efficiency point (BEP) and falls on both sides. Corresponding points under the affinity laws keep roughly the same efficiency. Why does that make speed control efficient only if the system curve passes near the origin?

Cavitation. At high flow the pressure at the impeller eye can fall below the vapour pressure of water. Bubbles form and collapse, eroding the impeller. Look up NPSH (net positive suction head). Which way does running faster push the risk?