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By Luqman Ismat © 2025

Engineering API Solutions • Hydraulics Calculations • Thermal Systems • Pump Design

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Centrifugal Pumps Capacity Modulation

Compare throttling, bypass and speed control using the pump and system together.

Source review: September 8, 2026

Control changes the operating point

The operating point is where pump head and required system head agree. A change in valve position, speed or network demand can move that point. Compare useful flow delivered to the process, not merely total flow through the pump.

  • Discharge throttling increases resistance and dissipates head across a valve. The resulting power change depends on the pump curve.
  • Bypass control diverts part of the pumped flow. It can serve minimum-flow or process needs, but the energy spent circulating bypass flow does not necessarily serve the useful load.
  • Speed control changes the pump curve. It can reduce unnecessary head and flow; evaluate drive losses and the pump's permissible operating range.

These distinctions and the system approach are described in the DOE pumping-system sourcebook. No control method has a universal efficiency ranking or payback period.

What the affinity laws do and do not predict

For the same impeller and approximately unchanged efficiency, corresponding points scale with speed ratio r: Q₂ = rQ₁, H₂ = r²H₁ and shaft power P₂ ≈ r³P₁. These are pump-performance relationships. The new operating point must still satisfy the system curve.

On a friction-dominated system this can closely resemble cubic power scaling. With static elevation or pressure head, that head remains when flow decreases. DOE Tip Sheet 11 calls for constructing the system curve when static head is significant. At sufficiently low speed, the pump may not overcome the required static head.

Worked example: static head changes the answer

This is a constructed teaching example, not a manufacturer curve. Let Q be in m³/s and H in metres. At full speed, choose Hp = 40 − 20,000Q²; let the system be Hs = 10 + 10,000Q². Assume constant liquid density of 1000 kg/m³.

Scaling the pump curve gives Hp,r = 40r² − 20,000Q². Equating pump and system heads gives Q = √[(40r² − 10)/30,000]. Positive flow requires r > 0.5 in this model.

Calculated intersections; hydraulic power = ρgQH, using g = 9.80665 m/s²
Speed ratioFlow (m³/s)Head (m)Hydraulic power (W)
10.03162320.006202.3
0.80.02280415.203399.1

At 80% speed, flow is 72.1% of its original value, not 80%. Hydraulic power is 54.8%, compared with the naive cubic estimate of 51.2%. Electrical power additionally depends on pump, motor and drive efficiencies at each duty point.

Evaluate the actual duty

Collect operating hours by flow and head requirement, measured electrical input, and manufacturer curves at candidate speeds. Check minimum flow, allowable operating region and suction requirements. Estimate energy from the duty profile, then compare installed cost and maintenance changes against annual savings. A universal payback estimate would omit these inputs.

The pump-duty calculator estimates head and power at an entered flow. It does not find the pump/system intersection. Use the NPSH calculator for a separate reservoir suction-head estimate, with a manufacturer-supplied requirement.

References

  • US DOE: Adjustable Speed Pumping Applications, Tip Sheet 11Accessed 2026-09-08
  • US DOE: Improving Pumping System Performance, second editionAccessed 2026-09-08