A centrifugal pump transfers energy to liquid through a rotating impeller. This page uses the Golding pump as a starting point for an ideal Euler-head calculation: enter impeller diameter, speed and outlet whirl velocity, then compare theoretical head, delivered head and fluid power. The animation illustrates a centrifugal impeller; it is not a dimensioned reproduction of a particular historic pump.
Golding Centrifugal Pump Interactive Calculator
Evaluate an ideal zero-inlet-whirl Euler head from impeller diameter, speed and an entered outlet whirl component, then apply an entered hydraulic efficiency.
Equation Used
- Water density 1000 kg/m3 and gravitational acceleration 9.81 m/s2.
- Zero inlet whirl and user-supplied outlet whirl component Cw2.
- Entered efficiency is hydraulic efficiency from Euler head to delivered head.
- No blade/slip model, measured pump curve, NPSH, cavitation, leakage, disk friction, bearings, shaft, motor or safety model.
Ideal velocity-triangle estimate only. The power result is ideal Euler fluid power, not brake/shaft input power. Outlet whirl is an independent input, and no blade geometry, slip, pump curve, NPSH, cavitation, leakage, mechanical loss or motor model is included.
How the impeller adds energy
Liquid enters near the impeller axis and travels outward through the rotating passages. The impeller changes the liquid’s angular momentum. A casing collects the discharge, and diffusion can convert some velocity head into pressure head.
The Euler equation relates energy transfer to peripheral speed and the tangential component of absolute liquid velocity. With no inlet whirl, only the outlet term remains. This calculator asks for that outlet component directly; it does not derive it from the curved blades in the animation.
Parts to identify
- Impeller: the rotating element that transfers energy to the liquid.
- Inlet or eye: the region where liquid enters the impeller.
- Casing and discharge: collect liquid leaving the impeller and connect it to the delivery pipe.
- Shaft and seal: support rotation and control leakage where the shaft crosses the casing.
Actual impeller passages, shrouds, seals and clearances depend on the pump. The picture does not specify their dimensions.
Using the head calculation
The calculation helps explain energy transfer in centrifugal pumping. Increasing diameter or speed raises peripheral velocity; increasing the entered outlet whirl raises Euler head. These comparisons can help interpret a pump’s velocity triangle.
For drainage, irrigation, circulation or other pumping duties, use the manufacturer’s measured curves to find the operating point where pump head meets the system requirement. An entered flow rate and an Euler-head estimate alone cannot establish that a pump will deliver that flow.
Ideal Euler-head estimate
Impeller-tip speed is u₂=πD₂N/60. With zero inlet whirl, theoretical Euler head is H_E=u₂Cw₂/g. Applying entered hydraulic efficiency gives delivered head H=η_hH_E, and the ideal pressure rise is ρgH.
The displayed power is ρgQH_E. It is ideal Euler fluid power, not brake or shaft input, because mechanical and volumetric losses are not modeled.
Example: head and fluid power
Use diameter 600 mm, speed 800 rpm, outlet whirl 2.51 m/s, flow 300 L/s and hydraulic efficiency 70%.
- Peripheral speed: π × 0.6 × 800 / 60 = 25.13 m/s.
- Euler head: 25.13 × 2.51 / 9.81 = 6.43 m.
- Delivered head: 0.70 × 6.43 = 4.50 m.
- Pressure equivalent: 1000 × 9.81 × 4.50 = 44.16 kPa.
- Ideal Euler fluid power: 1000 × 9.81 × 0.300 × 6.43 = 18.92 kW.
The useful head-based fluid power is 70% of 18.92 kW, or 13.25 kW. The 18.92 kW result is the ideal impeller-to-fluid energy transfer for the entered flow. Shaft and motor input require additional losses and cannot be selected from this result alone.
If hydraulic efficiency changes to 60%, delivered head becomes 3.86 m and pressure equivalent becomes 37.85 kPa. Ideal Euler power remains 18.92 kW because diameter, speed, whirl and flow have not changed.
Scope and limitations
A real pump must be evaluated from manufacturer curves at the required flow, head, speed and impeller diameter. Selection also requires fluid properties, suction conditions, NPSH margin, cavitation and operating-range checks, efficiency, shaft power, motor/service factor, materials, seals and applicable safety requirements.
Frequently asked questions
Why does hydraulic efficiency not change Ideal Euler Power?
Euler head depends on impeller speed and outlet whirl. Hydraulic efficiency reduces delivered head relative to that ideal value; the ideal fluid-power result remains ρgQH_E.
Does the drawing calculate outlet whirl?
No. Outlet whirl is an independent input. Blade angles, passage area and slip are not calculated.
Why does changing flow leave head unchanged?
Flow multiplies the fluid-power result. The simplified model keeps the entered outlet whirl constant; it has no measured head–flow curve.
Is Pressure Rise the difference between two pressure gauges?
It is the pressure equivalent ρgH of delivered head. A real total-head comparison also accounts for inlet and outlet elevations and velocity heads.
Can I use the result as a motor rating?
No. Mechanical and volumetric losses and motor efficiency are absent. Use pump and drive data at the required operating point.
Does the animation show actual rpm?
No. It runs at a fixed slowed playback speed so the impeller can be seen. The Speed input changes the calculation.
Does this calculate cavitation risk?
No. Suction pressure, liquid vapor pressure and required NPSH are not inputs.
References & Further Reading
- Wikipedia contributors. Centrifugal pump. Wikipedia
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