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Pumps and turbines

Advanced and Design Fluid Mechanics and Hydraulics Reviewed 24 Oct 2026 2 min read

Pumps add energy to water; turbines take it out. Both are rotodynamic machines in civil works: centrifugal pumps lift water for supply and drainage, and turbines in hydropower stations convert the head of a reservoir into electricity. The same basic relations describe both, with energy flowing in opposite directions.

A pump is chosen by matching its characteristic curve, head against discharge at a given speed, to the system curve, the static lift plus friction losses that rise roughly with the square of the flow. The intersection is the operating point, which should lie near the best efficiency point. The power needed equals the weight of water lifted per second times the head, divided by efficiency. Pumps in parallel add flow; pumps in series add head.

Diagram: Indiacon original, not to scale.

Affinity laws scale a pump's performance with speed: discharge varies with speed, head with its square and power with its cube. That is why variable-speed drives save so much energy in systems with varying demand. Specific speed, a combination of speed, discharge and head, classifies impellers into radial, mixed and axial flow types and, for turbines, separates Pelton wheels for high head, Francis turbines for medium head and Kaplan turbines for low head.

Cavitation sets the limit. If pressure at the impeller eye falls to the vapour pressure, bubbles form and collapse violently, pitting the metal and cutting performance. The net positive suction head available at the site must exceed that required by the pump with a margin, which controls how high above the water a pump can sit.

Tip

Plot the system curve for minimum and maximum static lift; the operating point moves between them.

Key formulas
  • Pump power: P = ρ g Q H / η
  • Affinity laws: Q ∝ N, H ∝ N², P ∝ N³
  • Turbine output: P = η ρ g Q H
Common mistake

Choosing a pump far from its best efficiency point at the usual operating flow.

Solved example

Problem. A pump delivers 0.1 m³/s against 40 m at 75 percent efficiency, running at 1 450 rpm. Find the power. If the speed is reduced to 1 200 rpm, estimate the new discharge, head and power.

  1. P = 1 000 × 9.81 × 0.1 × 40 / 0.75 = 52 320 W ≈ 52.3 kW.
  2. Speed ratio = 1 200 / 1 450 = 0.8276.
  3. Q₂ = 0.1 × 0.8276 = 0.0828 m³/s; H₂ = 40 × 0.8276² = 27.4 m.
  4. P₂ = 52.3 × 0.8276³ = 52.3 × 0.567 = 29.6 kW.

Answer. 52.3 kW at 1 450 rpm; at 1 200 rpm about 0.083 m³/s, 27.4 m and 29.6 kW

Quick quiz

1How does pump power vary with speed?

With its cube.

2Which turbine suits high head and low flow?

Pelton wheel.

3What is cavitation?

Formation and violent collapse of vapour bubbles where pressure falls to vapour pressure.

Code note
  • CPHEEO Manual on Water Supply and Treatment (India): Central government manual for planning, design and operation of public water supply.

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • P. N. Modi, S. M. Seth, Hydraulics and Fluid Mechanics including Hydraulic Machines, Standard Book House, latest edition.
  • V. T. Chow, Open-Channel Hydraulics, McGraw Hill, 1959.
  • K. Subramanya, Flow in Open Channels, McGraw Hill Education (India), latest edition.

Free and official sources

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