A — Affinity laws: rescale a known duty point by speed
B — Shaft power from a duty point
The Calculation, Explained
The affinity laws
For a given centrifugal machine, the flow coefficient and head coefficient stay constant when speed changes — the velocity triangles remain similar. From that single fact:
The cube law is the whole economics of variable-speed drives: 20% less speed is (0.8)³ = 51% of the power. The same laws apply to impeller diameter changes (approximately, for small trims) and, for fans, density enters directly: fan pressure and power are proportional to gas density at constant speed and volume flow.
Shaft power from first principles
Lifting Q cubic metres per second of fluid of density ρ through H metres costs ρgQH watts of hydraulic power; the machine's efficiency decides how much more the shaft must supply.
Assumptions and applicability — including the classic trap
- The static-head trap. The affinity laws move the pump curve; the operating point is where the pump curve crosses the system curve. In a friction-only system the laws predict the new duty beautifully. In a system dominated by static lift (pumping up into a tall column or a pressurised drum), reducing speed first eats the static margin: flow falls far faster than the cube law promises, and below some speed the pump delivers nothing at all. Many disappointing VFD projects are this paragraph, learned expensively.
- Efficiency is assumed constant with speed. In reality it drifts a few points at large speed reductions — acceptable for estimates, not for guarantees.
- The affinity point and the duty point must both be on the same machine curve: worn wear-rings and a throttled discharge valve move the machine off the curve the laws describe.
- Watch NPSH and minimum flow at reduced speed, and motor cooling on self-ventilated motors run slow for long periods.
Worked example (the defaults above)
A pump delivering 500 m³/h at 60 m on 95 kW, slowed from 1,480 to 1,200 rpm (ratio 0.811): flow = 500 × 0.811 = 405 m³/h; head = 60 × 0.811² = 39.5 m; power = 95 × 0.811³ = 50.6 kW. Cross-check from first principles at the original duty: ρgQH = 1000 × 9.81 × (500/3600) × 60 = 81.8 kW hydraulic; at 86% efficiency, shaft = 95.1 kW — matching the nameplate point, as it should.
References
Standard turbomachinery similarity theory; Hydraulic Institute standards (ANSI/HI) for affinity-law application and rotodynamic pump terminology.
Pair with the Three-Phase Motor Calculator to turn shaft power into motor current.
Related calculators: Three-Phase Motor · Pipe Flow & Pressure Drop