Pumps and piping systems

Understand it in 30 seconds

A pump adds mechanical energy to a fluid so it can rise in elevation or overcome pipe and valve resistance. The energy added per unit weight is expressed as total head $H$ , with the SI unit metre.

For a centrifugal pump, available head generally decreases as flow $Q$ increases. The head required by a piping system rises with flow because friction losses increase. Their intersection is the actual operating point [1–3].

What is total pump head?

Total head $H$ is the mechanical energy added by the pump per unit weight of fluid, expressed as a length. Comparing pressure, elevation and velocity between suction and discharge gives Equation (1).

Pressure rise alone is not total head when elevation or pipe area changes. For transfer between tanks, first find the static requirement from elevation and pressure differences, then add the flow-dependent head loss.

Eq. (1)Total head added by a pump

H=\left(\frac{p_2-p_1}{\rho g}\right)+\left(z_2-z_1\right)+\left(\frac{V_2^2-V_1^2}{2g}\right)+h_{\mathrm L}

The operating point is the intersection

The pump curve $H_\mathrm{p}(Q)$ gives the head a pump can develop at each flow for a specified speed and impeller diameter. Manufacturer test curves commonly include efficiency, input power and NPSH required [1].

The system curve $H_\mathrm{sys}(Q)$ gives the head required to pass each flow through the piping. The flow settles where available pump head equals required system head. The intersection of the blue and orange curves in the figure is this operating point [2,3].

Eq. (2)Operating-point condition

H_\mathrm{p}(Q_\mathrm{op})=H_\mathrm{sys}(Q_\mathrm{op})

Constructing the system curve

For a simple single-path system, required head separates into static head $H_\mathrm{static}$ , which remains at zero flow, and head loss $h_\mathrm L(Q)$ , which rises with flow. If friction and minor-loss coefficients are approximated as constant, velocity is proportional to flow and losses are approximately proportional to $Q^2$ [2,3].

Throttling a valve increases resistance and steepens the curve, moving the operating point toward lower flow. Changing liquid levels or vessel pressures changes static head and shifts the curve's zero-flow intercept.

Eq. (3)Required head for a simple system

H_\mathrm{sys}(Q)=H_\mathrm{static}+h_\mathrm L(Q)\approx H_\mathrm{static}+bQ^2

Symbols and units

Use $\mathrm{m^3/s}$ for flow and $\mathrm m$ for head in SI calculations. A curve coefficient changes numerically with the unit chosen for flow, so always keep the equation and units together [4].

Symbols and SI units for pumps and piping systems
Symbol	Meaning	SI unit
$Q$	Volumetric flow rate	$\mathrm{m^3/s}$
$Q_\mathrm{op}$	Operating-point flow rate	$\mathrm{m^3/s}$
$H$	Total head	$\mathrm m$
$H_\mathrm{p}$	Head developed by the pump	$\mathrm m$
$H_\mathrm{sys}$	Head required by the system	$\mathrm m$
$H_\mathrm{static}$	Static head from elevation and pressure	$\mathrm m$
$P_\mathrm h$	Hydraulic power delivered to the fluid	$\mathrm W$
$P_\mathrm{in}$	Pump shaft input power	$\mathrm W$
$\eta$	Pump efficiency	—

From head to required power

Once operating flow and head are known, Equation (4) gives the hydraulic power transferred to the fluid. Internal hydraulic losses, leakage, disc friction and mechanical losses mean that shaft input power exceeds hydraulic power.

Efficiency $\eta$ varies across the operating range. Read it at the calculated operating point rather than applying the catalogue peak efficiency everywhere. Locating the operating point near the best efficiency point (BEP) matters for both energy use and reliability [1,2].

Eq. (4)Hydraulic power

P_\mathrm h=\rho gQH

Eq. (5)Required input power

P_\mathrm{in}=\frac{P_\mathrm h}{\eta}=\frac{\rho gQH}{\eta}

Worked example: pump-system operating point

For water transfer, approximate the pump curve by $H_\mathrm{p}=30-1200Q^2$ and the system curve by $H_\mathrm{sys}=10+6800Q^2$ , with $Q$ in $\mathrm{m^3/s}$ and $H$ in $\mathrm m$ . Take $\rho=1000\ \mathrm{kg/m^3}$ and pump efficiency $\eta=0.75$ . Find the operating point and input power.

Equating the heads gives $Q_\mathrm{op}=0.050\ \mathrm{m^3/s}$ , or 50 L/s. Substitution into either curve gives 27 m. Hydraulic power is about 13.2 kW and required input power is about 17.7 kW.

Eq. (6)Operating flow

30-1200Q^2=10+6800Q^2\quad\Rightarrow\quad Q_\mathrm{op}=0.050\ \mathrm{m^3/s}

Eq. (7)Operating head

H_\mathrm{op}=30-1200(0.050)^2=27.0\ \mathrm m

Eq. (8)Hydraulic and input power

P_\mathrm h=(1000)(9.80665)(0.050)(27)=13.2\ \mathrm{kW},\qquad P_\mathrm{in}=\frac{13.2}{0.75}=17.7\ \mathrm{kW}

Calculator

Calculate the intersection of simplified parabolas $H_\mathrm p=H_0-aQ^2$ and $H_\mathrm{sys}=H_\mathrm{static}+bQ^2$ . Use manufacturer test curves, not this approximation, for actual pump selection.

CALCULATOR

Calculate the pump-system operating point

Shut-off head H₀

m

Pump-curve coefficient a

s²/m⁵

Static head Hstatic

m

System coefficient b

s²/m⁵

Density ρ

kg/m³

Pump efficiency η

%

Operating flow Qop50L/s

Operating head Hop27m

Hydraulic power Ph13.24kW

Input power Pin17.65kW

This learning tool uses parabolic approximations. Use manufacturer test curves and allowable operating ranges for equipment selection.

Conditions and limits

The simple system curve used here represents steady, single-phase flow along one path from one suction source to one receiving point. Branched networks, multiple pressure zones, flow-actuated valves, non-Newtonian fluids and two-phase flow require a simultaneous network model.

Pump curves change with speed, impeller diameter, liquid viscosity and wear. Selection also requires efficiency, allowable operating range, NPSH required, suction conditions, motor rating, materials and temperature; the intersection alone is not sufficient.

Common mistakes

Before finding the operating point, mark the analysis boundary, liquid levels, diameters, valves and reference pressures on a system sketch.

Use maximum flow as operating flow: Actual flow is set by the pump-curve and system-curve intersection.
Use pressure rise alone as total head: Include elevation and velocity differences in the energy equation.
Assume throttling changes the pump curve: At fixed speed, throttling mainly steepens the system curve and shifts the operating point.
Assume constant efficiency: Efficiency varies with flow and must be read at the operating point.
Add NPSH to total head: NPSH is a separate suction-side cavitation criterion.

Applications and related topics

Operating-point analysis applies to throttling, variable-speed control, impeller trimming, series and parallel pumps, and energy evaluations of piping modifications. For variable demand, evaluate the entire operating range rather than one design point.

Review Bernoulli, continuity, pipe-flow pressure loss and Reynolds number first. Next, evaluate cavitation and NPSH from suction pressure and vapour pressure to connect system performance with safe pump operation.

Prerequisites: Bernoulli equation, continuity, pipe-flow pressure loss, Reynolds number
Next: Cavitation and NPSH; pumps in series and parallel

References

Sources accessed 26 August 2026. Independent expert review has not yet been completed.