Centrifugal Pump Basics & Performance Curves

Impeller and volute anatomy, the falling head–flow curve, why pumps are rated in head (metres) not pressure, and the efficiency eta = rho g Q H / P_shaft.

TurbomachineryMechanical EngineeringFree preview
⏱️ About 18 min

Turn on a tap and the water climbs four storeys to your apartment as if by magic. It is not magic — it is a <strong>centrifugal pump</strong>, the workhorse of modern fluid handling and the subject of this module. A centrifugal pump does something deliberately counter-intuitive: it does <em>not</em> 'suck' water upward. Instead a spinning <strong>impeller</strong> flings fluid outward, a spiral <strong>volute</strong> catches it and slows it down, and that deceleration turns speed into pressure. This lesson introduces the anatomy, explains why engineers describe a pump's output in <strong>head</strong> (metres of fluid) rather than pressure, and shows how to read a pump's <strong>performance curve</strong> and turn it into brake power and efficiency. By the end you will understand the single most useful fact about a pump: what it actually <em>does</em> to the fluid passing through it.

💡
The big idea: A centrifugal pump adds energy to a fluid as <strong>head</strong> <em>H</em> — energy per unit weight, measured in <em>metres of the fluid</em>. Fluid enters axially at the <strong>impeller eye</strong>, is accelerated tangentially and radially by the rotating vanes, and is decelerated in the <strong>volute</strong>, where kinetic energy is converted to pressure rise. The <strong>performance curve</strong> <em>H(Q)</em> is <em>falling</em>: head is highest at shutoff (<em>Q = 0</em>) and drops as flow increases. Rating pumps in head rather than pressure is deliberate — a pump that delivers 25 m of head delivers 25 m whether the fluid is water, gasoline, or brine, because <em>H = Δp/(ρg)</em> cancels density. The <strong>hydraulic (water) power</strong> delivered to the fluid is <em>P_hyd = ρgQH</em>, and the pump's <strong>efficiency</strong> is <em>η = P_hyd/P_shaft = ρgQH/P_shaft</em>, where <em>P_shaft = ωM</em> is the brake (shaft) power the motor supplies. Crucially, head is density-independent but power is <em>not</em>: the denser the fluid, the more power the same head costs.
🎯 By the end, you'll be able to
  • Describe the anatomy of a centrifugal pump (impeller, vanes, volute, eye) and the energy conversion from kinetic to pressure
  • Explain why pump output is rated in head H (metres of fluid) rather than pressure, and relate head to pressure rise via H = Δp/(ρg)
  • Read a falling H–Q performance curve and identify the shutoff head and the best-efficiency point
  • Compute hydraulic power P_hyd = ρgQH and pump efficiency η = ρgQH/P_shaft
  • Recognise that pump head is density-independent but pump power is not — denser fluids cost more power at the same head
📎 Helpful to know first
  • The Angular Momentum Equation — a Turbomachinery Preview
  • The General Energy Equation — the First Law for a Control Volume

Anatomy: impeller, volute, and the eye

A centrifugal pump has three working parts worth naming. Fluid enters axially — along the shaft direction — at the centre of the impeller, a passage called the eye. The impeller is a disc carrying curved vanes; as it spins, the vanes sweep the fluid around and fling it radially outward by centrifugal action, accelerating it to a high tangential and radial speed. Surrounding the impeller is the volute, a spiral casing whose flow area grows steadily toward the discharge. That growing area is the point: it lets the fast fluid leaving the impeller decelerate, and by Bernoulli that deceleration is a pressure rise. The pump has turned shaft work into fluid speed (in the impeller) and then fluid speed into pressure (in the volute).

Where is the pressure lowest? Right at the eye, where the fluid is accelerating into the impeller. Remember that fact — it is the whole story of cavitation in Lesson 3. The eye is the point of minimum pressure in the entire machine, lower even than the pressure at the suction flange where the inlet pipe bolts on, because the fluid must accelerate from the flange into the impeller channels. Keep that in the back of your mind.

The energy bookkeeping is the angular-momentum relation you previewed in Module 6: a pump adds swirl (it raises r·V_t, the per-unit-mass angular momentum), and the torque is M = ṁ(r₂V_{t2} − r₁V_{t1}) with shaft power Ẇ = ωM. That is the deep theory. In day-to-day pump work you will almost never compute r·V_t by hand; instead you will read a manufacturer's measured performance curve. But it helps to know that the curve is that physics, packaged empirically.

A centrifugal pump in cross-section. Fluid enters axially along the shaft direction at the impeller eye on the left, is flung radially outward by curved vanes on the spinning impeller, collects in the spiral volute casing whose flow area grows toward the discharge, and leaves upward through the discharge pipe at the top. The impeller rotates at angular speed omega. The eye is the lowest-pressure point in the pump. discharge ω inlet (eye) lowest pressure volute collects & decelerates flow volute (spiral casing) growing area: KE → pressure

A centrifugal pump in cross-section. Fluid enters axially at the impeller eye on the left (the lowest-pressure point), is flung radially outward by curved vanes on the spinning impeller, collects in the spiral volute casing whose area grows toward the discharge, and leaves upward through the discharge pipe. The impeller rotates at angular speed omega.

Centrifugal-pump anatomy. Fluid enters axially at the eye, the vanes fling it outward, and the volute's growing area decelerates the flow — converting the impeller's kinetic energy into a pressure rise. The eye is the lowest-pressure point in the pump, which is why cavitation nucleates there (Lesson 3).

Why head, not pressure

Ask a pump engineer 'how much does this pump lift?' and the answer comes in metres, not pascals. A pump is said to deliver, say, H = 25 m of head. Head is energy per unit weight of fluid — the height to which the pump could lift that fluid against gravity — and it is related to the pressure rise by

H = Δp / (ρg),

where Δp is the pressure rise, ρ the fluid density, and g gravity. The density sits in the denominator, so it cancels: a pump that delivers 25 m of head delivers 25 m of water, 25 m of gasoline, 25 m of any liquid. That makes head a universal, fluid-independent rating of what the pump does to the flow. If pumps were rated in pressure, the same machine would have a different number for every fluid — useless for comparison.

This is the first of two facts that look like one and trip people up. Head is independent of density; pressure is not. A 25-m-head pump lifting water produces a pressure rise of ρgH = (1000)(9.81)(25) ≈ 245 kPa; lifting gasoline (ρ = 750) the same pump produces only (750)(9.81)(25) ≈ 184 kPa. Same head, different pressure — because pressure depends on density and head does not. Hold that distinction; the very next section turns it into a power calculation where density reappears with a vengeance.

\[ H=\frac{\Delta p}{\rho g}\qquad\bigl(\text{head}=\text{pressure rise divided by specific weight}\bigr)\qquad P_{\text{hyd}}=\rho\,g\,Q\,H \]
Head H is the pressure rise divided by the specific weight ρg — so head is independent of the fluid density, while the pressure rise is not. The hydraulic (water) power the pump delivers to the fluid is P_hyd = ρgQH, which (unlike head) explicitly contains the density.
⚠️ Head is density-free; power is not

This is the misconception to stamp out early. Students hear 'head is independent of density' and wrongly conclude that everything about the pump is density-independent. It is not. Head H is density-independent — a 25 m pump is a 25 m pump. But the power the pump consumes is P_hyd = ρgQH, and ρ is right there in the formula. Pumping a denser fluid at the same head and flow costs more power. Pumping mercury at 25 m of head takes about 13× the power of pumping water at 25 m. The head is the same; the motor must be far larger. Head tells you what the pump does to the flow; power tells you what the flow costs to move.

The performance curve (a falling H–Q)

Every pump comes with a measured performance curve — a plot of the head it produces against the flow it delivers. For a centrifugal pump the curve falls: head is largest at shutoff (the discharge valve closed, Q = 0) and decreases as you open the valve and let flow through. A simple model of the shape is a downward parabola,

H(Q) = H₀ − aQ²,

where H₀ is the shutoff head (the head at zero flow) and a sets how steeply the curve drops. Real curves are not perfect parabolas, but this captures the essential behaviour and is exactly the model the interactive below uses.

Where on the curve should a pump operate? Each pump has a best efficiency point (BEP): a particular flow at which its efficiency peaks. The manufacturer's curve usually overlays iso-efficiency contours, and the BEP is the design target. Running a pump far from its BEP — too far left (low flow, near shutoff) or too far right (high flow) — wastes energy and can damage the pump through recirculation or cavitation. Good pump selection means matching the operating point to the BEP, which is the entire subject of Lesson 2.

\[ H(Q)=H_{0}-a\,Q^{2}\qquad\bigl(\text{centrifugal pump: head falls with flow}\bigr)\qquad H_{0}=\text{shutoff head (}Q=0\text{)} \]
A simple performance-curve model: head falls as a downward parabola from the shutoff head H0. Real curves deviate from a pure parabola, but the falling trend — high head at shutoff, lower head at high flow — is universal for centrifugal pumps. The best-efficiency point (BEP) is the design target.

Brake power and efficiency

The useful power the pump actually delivers to the fluid — the rate at which it raises the fluid's energy — is the hydraulic power (sometimes called water power):

P_hyd = ρgQH.

This is the energy per unit weight (H) times the weight flow rate (ρgQ). It is the output of the pump as a fluid machine. The input is the brake (shaft) power P_shaft = ωM that the motor supplies — the angular speed times the torque. The difference between input and output is lost to friction in the bearings and seals, to internal recirculation, and to fluid friction in the impeller and volute. The ratio is the pump's efficiency:

η = P_hyd / P_shaft = ρgQH / P_shaft.

A well-designed centrifugal pump runs at η ≈ 0.70–0.85 near its BEP; small or off-design pumps can be much worse. Notice that efficiency also carries the density through P_hyd: at the same head and flow, a denser fluid means more hydraulic power out, so for the same shaft power the efficiency number changes. In practice η is read off the manufacturer's curve for the fluid being pumped.

\[ P_{\text{hyd}}=\rho g Q H\qquad P_{\text{shaft}}=\omega M\qquad \eta=\frac{P_{\text{hyd}}}{P_{\text{shaft}}}=\frac{\rho g Q H}{P_{\text{shaft}}} \]
Hydraulic (water) power P_hyd = ρgQH is the rate of energy delivered to the fluid; brake power P_shaft = ωM is what the motor supplies; the ratio is the pump efficiency η. Head is density-independent, but both P_hyd and (via it) the efficiency depend on density — pumping a denser fluid costs more power at the same head.
🎮 Interactive: pump curve, system curve, and the operating point LIVE
Predict first: Start with the default pump (shutoff head H0 = 40 m). Notice the pump curve falls to the right while the system curve rises. Their crossing is the operating point. Now drag the static-head slider up: the system curve lifts, the operating point slides left (less flow) and up (more head). Finally tick 'Oversized pump': a dashed second pump curve appears, and shading shows the head wasted by throttling the bigger pump back to the same flow.

An interactive pump-curve and system-curve plot. Sliders set the static head, the system resistance coefficient, and the pump shutoff head. The falling pump curve and the rising system curve are drawn on the same head-versus-flow axes; their intersection, the operating point, is marked with its flow and head read out, and a best-efficiency band is shaded. A toggle overlays an oversized pump curve and shades the head wasted by throttling.

The pump_curve widget plots a falling pump curve H = H0 − aQ² and a rising system curve H_sys = h_static + KQ² on the same axes. Sliders move the static head h_static, the system resistance K, and the pump's shutoff head H0; the operating point is solved live from the quadratic intersection, with a best-efficiency band marked. An 'oversized pump' toggle overlays a larger pump and shades the head burned off by throttling.
📝 Worked example: A centrifugal pump moving water (ρ = 1000 kg/m³) delivers a flow Q = 0.020 m³/s against a head of H = 25 m. The drive motor supplies a brake (shaft) power P_shaft = 6.3 kW. Find the hydraulic power delivered to the water and the pump efficiency.
  1. Hydraulic (water) power: P_hyd = ρgQH.
  2. P_hyd = (1000)(9.81)(0.020)(25) = (1000)(9.81)(0.50) = 4905 W ≈ 4.9 kW.
  3. Efficiency: η = P_hyd / P_shaft = 4905 / 6300 = 0.779.
  4. η ≈ 78%. The remaining 22% of the shaft power is lost to bearing and seal friction, internal recirculation, and fluid friction in the impeller and volute.
✓ P_hyd ≈ 4.9 kW; η ≈ 78%.
✏️ Practice: A pump delivers a head H = 40 m at a flow Q = 0.010 m³/s of water (ρ = 1000 kg/m³). Compute the hydraulic power delivered to the water. Give your answer in kW.
kW
Solution
  1. P_hyd = ρgQH.
  2. P_hyd = (1000)(9.81)(0.010)(40) = (1000)(9.81)(0.40) = 3924 W ≈ 3.9 kW.
✏️ Practice: A pump rated for H = 20 m of head runs at Q = 0.030 m³/s. Compute the hydraulic power when the fluid is gasoline (ρ = 750 kg/m³). Give your answer in kW.
kW
Solution
  1. P_hyd = ρgQH — note the density is now gasoline, not water.
  2. P_hyd = (750)(9.81)(0.030)(20) = (750)(9.81)(0.60) = 4414.5 W ≈ 4.4 kW.
  3. Same head and flow, but the lighter fluid costs less power. Had this been water the power would be (1000)(9.81)(0.030)(20) ≈ 5.9 kW — about 34% more. Head is density-independent; power is not.

Check your understanding

1. A centrifugal pump's output is rated in head H (metres of fluid) rather than pressure because:
Because H = Δp/(ρg) divides pressure by the specific weight, head is independent of density. A 25 m pump lifts water, gasoline, or brine the same 25 m. Rating in pressure would give a different number per fluid and make comparison impossible. (The pressure rise ρgH, and the power ρgQH, do depend on density — head alone does not.)
2. For a centrifugal pump, the head–flow performance curve H(Q):
Centrifugal-pump head falls with flow: a simple model is H = H0 − aQ², with the maximum (shutoff) head H0 at Q = 0. The best-efficiency point (BEP) sits partway down the curve and is the design target. Lesson 2 shows how this falling curve meets the rising system curve at the operating point.
3. A pump delivers the same head H and flow Q to two liquids, water (ρ = 1000) and a denser brine (ρ = 1100). Compared with water, the brine requires:
Head is density-independent, so the pump delivers the same H to both fluids. But P_hyd = ρgQH is proportional to density: the brine needs 10% more power than water at the same head and flow. Head tells you what the pump does; power tells you what it costs — and power depends on density.
✅ Key takeaways
  • A centrifugal pump accelerates fluid at the impeller (eye → vanes) and decelerates it in the volute, converting shaft work to a pressure rise; the eye is the lowest-pressure point.
  • Output is rated in head H (metres of fluid): H = Δp/(ρg) cancels density, so a 25 m pump lifts any liquid 25 m — head is density-independent, pressure is not.
  • The performance curve H(Q) falls with flow (model H = H0 − aQ²); the best-efficiency point (BEP) is the design target.
  • Hydraulic power P_hyd = ρgQH and efficiency η = ρgQH/P_shaft; power (unlike head) depends on density — denser fluids cost more power at the same head.
➡️ You can now read a pump curve and turn head and flow into power and efficiency. But a pump curve on its own tells you only what the pump <em>can</em> do — not what it <em>will</em> do once bolted into a real piping system. The pipes push back: they demand a certain head at every flow. The next lesson builds that <strong>system curve</strong> and finds the single flow where the pump's supply meets the system's demand — the <strong>operating point</strong> — and shows why oversizing a pump is a quietly expensive mistake.
Want to test yourself on this? Try the Mechanical Aptitude test →
🎓 Go deeper: external courses & trusted references