Pumps

Pumps add energy to liquids. Learn centrifugal vs positive-displacement pumps, head, flow and hydraulic power, pump and system curves, efficiency, affinity laws, NPSH and cavitation, and series/parallel operation.

Pumps use roughly 20 % of the world's electrical motor energy. They move water to cities, coolant through engines, fuel to jet engines and blood through heart-lung machines.

Choosing a pump means matching its head–flow curve to the system's needs, keeping it efficient, and avoiding cavitation.

An analogy

A pump is like a lift for water: 'head' is how many metres it can lift the water, and flow is how many people (litres) it carries per second.

Types of pumps

Centrifugal (rotodynamic) pumps spin an impeller to add velocity, converted to pressure in the casing — high flow, moderate head, smooth delivery. Positive-displacement pumps (gear, piston, diaphragm, screw) trap and push fixed volumes — high pressure, viscous fluids, metering.

Pump curve and operating point

A centrifugal pump's head falls as flow rises. The system curve rises (static head + friction ∝ Q²). They intersect at the operating point. Select pumps so this lies near the best-efficiency point (BEP).

Affinity laws and variable speed

Changing speed shifts the curve: flow ∝ N, head ∝ N², power ∝ N³. Reducing speed 20 % cuts power ≈ 49 % — why variable-speed drives save so much energy compared with throttling valves.

Cavitation, NPSH and specific speed

If suction pressure falls near vapour pressure, bubbles form and implode on the impeller — noise, vibration and erosion. Keep NPSH_a ≥ NPSH_r + margin. Specific speed N_s = N√Q/H^¾ classifies impeller geometry (radial, mixed, axial). Parallel pumps add flow; series pumps add head.

Formulas

• Hydraulic power: P_hyd = ρ g Q H
• Shaft power: P_shaft = ρ g Q H / η
• Affinity laws: Q ∝ N, H ∝ N², P ∝ N³
• NPSH available: NPSH_a = (p_s − p_v)/ρg + z_s − h_Ls

Worked example: Pump power

Given: Q = 0.02 m³/s, H = 30 m; η = 70 %

Find: Shaft power.

• P_hyd = 998 × 9.81 × 0.02 × 30 = 5.87 kW
• P_shaft = 5.87/0.7 = 8.39 kW

Answer: ≈ 8.4 kW (choose an 11 kW motor)

Worked example: Affinity laws

Given: Pump at 1450 rpm: 20 L/s, 30 m, 8.4 kW; Speed raised to 1750 rpm

Find: New performance.

• Ratio 1.207
• Q = 24.1 L/s
• H = 30 × 1.457 = 43.7 m
• P = 8.4 × 1.758 = 14.8 kW

Answer: 24 L/s, 44 m, 14.8 kW

Common mistakes

• Wrong: Controlling flow with a throttling valve on a fixed-speed pump. — Right: Use a VFD where flow varies — much lower energy use.
• Wrong: Ignoring suction conditions. — Right: Check NPSH_a vs NPSH_r to prevent cavitation.

Applications

• Water & wastewater: Booster stations, borehole pumps and sewage lifts.
• Process industry: Chemical transfer, dosing (PD metering pumps) and boiler feed pumps.
• Automotive: Coolant, fuel, oil and brake-fluid pumps.

Key takeaways

• P = ρgQH/η.
• Operating point = pump curve ∩ system curve.
• Affinity: Q ∝ N, H ∝ N², P ∝ N³.
• Protect against cavitation with NPSH margin.

Going further

• Euler pump equation: Head = (u₂V_t2 − u₁V_t1)/g — impeller exit blade angle and tip speed set the ideal curve.
• Minimum-flow protection: Running at shut-off overheats the liquid; recirculation valves protect pumps.
• Pump system optimisation: ISO 14414 energy audits often find 20–50 % savings by right-sizing, trimming impellers and adding VFDs.