Centrifugal pumps: characteristics and cavitation

Why it matters

Centrifugal pumps move most of the world's liquids: water supply, coolant loops in machine tools, chemical plants and boiler feed. Selecting one means matching its head–flow curve to the system, predicting what happens when speed or impeller size changes, and making sure the suction side does not cavitate. Cavitation alone destroys more impellers than any other cause.

Key ideas

Construction and energy transfer. Liquid enters the eye of a rotating impeller axially, is flung outward by curved vanes and leaves at the rim with high velocity and pressure. A volute casing (or diffuser vanes) then slows it down, converting kinetic energy into pressure. The pump must be primed (filled with liquid) before starting because the head it develops is proportional to fluid density; running in air gives a negligible pressure rise.

Euler head and velocity triangles. The ideal head given to the liquid is H_e = (V_w2·u₂ − V_w1·u₁)/g. Pumps are designed for radial (shock-free, no pre-whirl) entry, V_w1 = 0, giving H_e = V_w2·u₂/g. At the outlet, V_w2 = u₂ − V_f2/tanβ₂, where β₂ is the blade outlet angle measured from the tangent (backward direction).

• Backward-curved vanes (β₂ < 90°, typically 20–30°): head falls as flow rises, power curve peaks and then falls (non-overloading), best efficiency. Almost all pumps use them.
• Radial vanes (β₂ = 90°): ideal head independent of flow.
• Forward-curved vanes (β₂ > 90°): ideal head rises with flow; unstable and overloading, so rare in pumps. Real impellers deliver less than the Euler head because the relative flow does not follow the vanes exactly (slip), and because of friction and shock losses.

Heads and efficiencies.

• Manometric head H_m: the head actually measured across the pump, H_m = (p_d − p_s)/(ρg) + (V_d² − V_s²)/(2g) + (z_d − z_s). If suction and delivery pipes are the same size and the gauges are at the same level, H_m = (p_d − p_s)/(ρg).
• Manometric efficiency η_mano = g·H_m/(V_w2·u₂).
• Overall efficiency η_o = ρgQH_m ÷ shaft power; it includes manometric, volumetric (leakage) and mechanical losses.
• Minimum starting speed: the impeller must produce a centrifugal head (u₂² − u₁²)/(2g) at least equal to H_m before flow starts.

Characteristics and operating point. At constant speed, the H–Q curve falls with flow, the power curve rises (for backward vanes, then levels off) and efficiency peaks at the best efficiency point (BEP). The system curve is H_sys = static lift + K·Q² (friction and fittings). The pump operates where the two curves intersect. Throttling a delivery valve steepens the system curve and moves the operating point to lower flow. Running far below BEP causes recirculation, heating, radial thrust and vibration; running at zero flow (shut-off or "dead-heading") heats the liquid rapidly and can damage the pump.

Affinity (similarity) laws. For the same pump at a different speed, or geometrically similar pumps: Q ∝ N·D³, H ∝ N²·D², P ∝ N³·D⁵. For a small trim of the impeller diameter in the same casing, Q ∝ D, H ∝ D² and P ∝ D³ are used as approximations. Pump specific speed N_s = N·√Q/H^(3/4) classifies impeller shape: low N_s radial, medium mixed-flow, high N_s axial.

Series and parallel. In series, heads add at the same flow (high-head duties, multistage pumps). In parallel, flows add at the same head (variable demand).

Cavitation and NPSH. Where the pressure at the impeller eye drops below the liquid's vapour pressure, vapour bubbles form and then collapse violently as they move into higher-pressure regions, pitting the vanes, causing noise, vibration and a drop in head and efficiency.

• NPSH available (system property) = absolute pressure head at the pump inlet + velocity head − vapour pressure head, or from the sump: (p_atm − p_v)/(ρg) − suction lift − suction-pipe losses.
• NPSH required (pump property, from the manufacturer's test) rises with flow.
• Avoid cavitation by keeping NPSH_a above NPSH_r with a margin: lower the pump, shorten and enlarge the suction pipe, cool the liquid, reduce speed, or use an inducer.

Formulas

u = π·D·N / 60

• u = blade speed (m/s), D = diameter (m), N = speed (rpm).

H_e = V_w2·u₂ / g (radial entry)

• H_e = Euler head (m), V_w2 = outlet whirl (m/s).

V_w2 = u₂ − V_f2 / tanβ₂, Q = π·D₂·b₂·V_f2

• β₂ = blade outlet angle from the tangent, V_f2 = outlet flow velocity (m/s), b₂ = outlet width (m).

η_mano = g·H_m / (V_w2·u₂), η_o = ρ·g·Q·H_m / P_shaft

N_min = 120·η_mano·V_w2·D₂ / (π·(D₂² − D₁²)) (rpm)

• From (u₂² − u₁²)/(2g) = H_m.

Q₂/Q₁ = N₂/N₁, H₂/H₁ = (N₂/N₁)², P₂/P₁ = (N₂/N₁)³ (same pump)

N_s = N·√Q / H^(3/4)

• N in rpm, Q in m³/s, H in m (per stage).

NPSH_a = (p_atm − p_v)/(ρg) − z_s − h_fs

• p_v = vapour pressure (Pa), z_s = height of pump inlet above sump level (m), h_fs = suction-pipe head loss (m).

Worked examples

Example 1 — impeller design point (standard). An impeller of 0.3 m outer diameter and 20 mm outlet width runs at 1450 rpm. The blade outlet angle is 30° (backward), the outlet flow velocity is 2.5 m/s, entry is radial and the manometric efficiency is 80 %. Find the Euler head, manometric head, discharge and power given to the water.

1. u₂ = πD₂N/60 = π × 0.3 × 1450/60 = 22.78 m/s.
2. V_w2 = u₂ − V_f2/tanβ₂ = 22.78 − 2.5/0.5774 = 22.78 − 4.33 = 18.45 m/s.
3. H_e = V_w2·u₂/g = 18.45 × 22.78/9.81 = 42.83 m.
4. H_m = 0.8 × 42.83 = 34.26 m.
5. Q = πD₂b₂V_f2 = π × 0.3 × 0.02 × 2.5 = 0.04712 m³/s.
6. Power delivered to water (useful): ρgQH_m = 9810 × 0.04712 × 34.26 = 15.84 kW; power transferred by the impeller ρQ·V_w2·u₂ = 19.80 kW. Answer: H_e ≈ 42.8 m, H_m ≈ 34.3 m, Q ≈ 0.0471 m³/s, useful power ≈ 15.8 kW.

Example 2 — speed change and suction check (GATE level). A pump delivers 0.05 m³/s against 30 m at 1450 rpm with an overall efficiency of 75 %. (a) Find its shaft power, its specific speed, and the duty at 1750 rpm (efficiency unchanged). (b) It lifts water (vapour pressure 7.4 kPa at the operating temperature, from steam tables; ρ = 1000 kg/m³) from an open sump at 101.3 kPa; the pump inlet is 3 m above the sump level and suction losses are 0.8 m. NPSH_r = 4.5 m. Check for cavitation and find the maximum suction lift.

1. Shaft power: P = ρgQH/η = 9810 × 0.05 × 30/0.75 = 19.62 kW.
2. N_s = N√Q/H^(3/4) = 1450 × 0.2236/12.82 = 25.3 (a radial-flow impeller).
3. At 1750 rpm, ratio r = 1.207: Q = 0.05 × 1.207 = 0.0603 m³/s; H = 30 × 1.207² = 43.7 m; P = 19.62 × 1.207³ = 34.5 kW.
4. NPSH_a = (p_atm − p_v)/(ρg) − z_s − h_fs = (101 300 − 7400)/9810 − 3 − 0.8 = 9.57 − 3.8 = 5.77 m.
5. NPSH_a (5.77 m) > NPSH_r (4.5 m), so no cavitation, with a 1.27 m margin.
6. Maximum suction lift: z_s,max = 9.57 − 0.8 − 4.5 = 4.27 m. Answer: (a) 19.6 kW, N_s ≈ 25, 0.060 m³/s at 43.7 m needing 34.5 kW; (b) NPSH_a ≈ 5.8 m, safe; maximum lift ≈ 4.3 m.

Common mistakes

• Using gauge pressure in NPSH; it must be absolute pressure minus vapour pressure.
• Measuring the blade outlet angle from the radial direction instead of the tangent, which changes tanβ₂.
• Applying H ∝ N² while keeping Q fixed; flow changes too, and the new operating point must be found on the system curve.
• Using the turbine specific speed formula (with power) for a pump.
• Thinking a throttled discharge valve reduces pump head; it moves the operating point back along the curve, usually raising head.
• Forgetting priming, or forgetting that suction lift is limited to well under 10 m of water.

For GATE ME

Expect Euler-head and manometric-efficiency calculations from velocity triangles, power and overall efficiency, affinity-law scaling, specific speed, minimum starting speed, operating point from given pump and system curves, pumps in series or parallel, and NPSH or maximum suction height. Conceptual MCQs test vane shapes, cavitation remedies and priming. Practise reading a pump curve and intersecting it with a system curve.

Quick check

1. Why must a centrifugal pump be primed?
2. If pump speed is doubled, how do flow, head and power change?
3. What is NPSH available, in words?
4. Which vane shape gives a non-overloading power characteristic?
5. Two identical pumps in parallel: what is common to both?

Answers: 1. the head developed is proportional to fluid density, so in air the pressure rise is too small to lift liquid; 2. ×2, ×4, ×8; 3. the margin of absolute total head at the pump inlet above the vapour-pressure head; 4. backward-curved; 5. the head across each pump.