Hydraulic Machines — Pumps & Turbines | EngineeringHulk

Hydraulic Machines — Pumps & Turbines

Centrifugal pumps, turbine types, specific speed, efficiency, cavitation, NPSH, and similarity laws — with full GATE CE worked examples

Last Updated: April 2026

Key Takeaways
  • Pumps add energy to fluid; turbines extract energy from fluid — both analysed using modified Bernoulli with head terms hp or ht.
  • Centrifugal pump: head H = u₂Vw2/g (Euler’s turbomachinery equation); efficiency η = ρgQH/Pshaft.
  • Specific speed Ns = NQ1/2/H3/4 (pumps) or Ns = NP1/2/(ρ1/2H5/4) (turbines) — determines machine type and geometry.
  • Pelton wheel: impulse turbine for high head (H > 300 m), low flow; tangential flow; jet strikes buckets.
  • Francis turbine: reaction turbine for medium head (30–300 m), medium flow; mixed (radial-axial) flow.
  • Kaplan turbine: reaction turbine for low head (< 40 m), high flow; axial flow; adjustable runner blades.
  • Cavitation occurs when local pressure drops below vapour pressure; characterised by Thoma’s cavitation number σ; prevented by ensuring available NPSH > required NPSH.

1. Classification of Hydraulic Machines

CategoryMachine TypeEnergy TransferExamples
Pumps
(fluid receives energy)
Centrifugal (rotodynamic)Shaft → FluidSingle-stage, multi-stage, axial-flow, mixed-flow
Reciprocating (positive displacement)Shaft → FluidSingle-acting, double-acting piston pump
Rotary (positive displacement)Shaft → FluidGear pump, screw pump, vane pump
Turbines
(fluid gives energy)
Impulse turbineFluid → ShaftPelton wheel (tangential flow)
Reaction turbine (radial/mixed)Fluid → ShaftFrancis turbine (inward radial to axial)
Reaction turbine (axial)Fluid → ShaftKaplan turbine, propeller turbine

1.1 Impulse vs Reaction Turbines

FeatureImpulse Turbine (Pelton)Reaction Turbine (Francis/Kaplan)
Energy conversion at runnerKinetic energy only (jet velocity)Both pressure and kinetic energy
Runner pressureAtmospheric throughoutAbove atmospheric (enclosed casing)
Flow regulationSpear (needle) valve in jet nozzleGuide vanes (wicket gates)
Draft tubeNot usedEssential (recovers pressure energy)
Head rangeHigh (> 300 m)Medium to low (2–300 m)
Degree of reaction0 (all impulse)0.5–0.8 (Francis); ~1.0 (Kaplan)

2. Euler’s Turbomachinery Equation

Euler’s turbomachinery equation is derived from the angular momentum theorem and gives the theoretical (ideal) work done by or on the fluid per unit mass in any rotodynamic machine.

Euler’s equation (work done per unit weight of fluid):

HEuler = (u₂Vw2 – u₁Vw1) / g

where:

  • u₁, u₂ = peripheral (blade tip) velocity at inlet and outlet = ωr₁, ωr₂ (m/s)
  • Vw1, Vw2 = whirl (tangential) component of absolute fluid velocity at inlet and outlet (m/s)
  • g = 9.81 m/s²

For a pump (energy added to fluid): Hp = (u₂Vw2 – u₁Vw1)/g

For a turbine (energy extracted from fluid): Ht = (u₁Vw1 – u₂Vw2)/g

Assumption of no pre-swirl at pump inlet: Vw1 = 0 (radial entry)

→ Hp = u₂Vw2/g (standard assumption for centrifugal pumps)

2.1 Velocity Triangles

At any blade row, three velocities form a triangle:

V = absolute fluid velocity

u = blade (peripheral) velocity = ωr

Vr = relative velocity of fluid with respect to blade

Vector relation: V = u + Vr

Components: Vw = whirl (tangential); Vf = flow (radial/axial)

tan(blade angle β) = Vf / (u – Vw)

3. Centrifugal Pump — Working, Heads & Efficiency

3.1 Working Principle

A centrifugal pump consists of a rotating impeller inside a casing (volute). Fluid enters axially at the eye (centre) of the impeller and is thrown radially outward by centrifugal action. The kinetic energy imparted by the impeller is converted to pressure energy in the volute casing (which acts as a diffuser).

3.2 Head Definitions

Head TermSymbolDefinition
Gross (total) static headHsVertical distance from lower reservoir surface to upper reservoir surface = Hss + Hsd
Suction headHssVertical distance from pump centreline to lower reservoir surface (positive if pump is above sump)
Delivery headHsdVertical distance from pump centreline to upper reservoir surface
Manometric headHmTotal head added to fluid by pump = Hs + hf,total + Vd²/(2g) – Vs²/(2g)
Euler (theoretical) headHEulerHead computed from Euler’s equation (velocity triangles); no losses
Shut-off headH0Head developed at zero discharge (Q = 0)

3.3 Pump Efficiency

Manometric efficiency:

ηm = Manometric head / Euler head = Hm / HEuler = gHm / (u₂Vw2)


Volumetric efficiency:

ηv = Q / (Q + qleakage)


Mechanical efficiency:

ηmech = Power delivered to fluid / Shaft power = ρgQHm / Pshaft


Overall efficiency:

η = ρgQHm / Pinput = ηm × ηv × ηmech


Power input to pump:

Pinput = ρgQHm / η (W)

3.4 Priming

A centrifugal pump must be primed (filled with liquid) before starting because it cannot develop sufficient suction to lift liquid from a dry state — it can only pressurize liquid already present. Self-priming pumps have special features to handle this automatically.

4. Pump Characteristic Curves & System Curve

4.1 Pump Characteristic Curves

The pump manufacturer provides characteristic curves at a fixed speed N, showing how H, efficiency η, and power P vary with discharge Q:

  • H–Q curve: Head decreases as discharge increases (centrifugal pump). At Q = 0: H = H₀ (shut-off head). As Q increases, H decreases.
  • η–Q curve: Efficiency peaks at the best efficiency point (BEP) and falls on both sides.
  • P–Q curve: Power increases with Q for centrifugal pumps — motor must be sized for maximum expected flow.

4.2 System Curve

The system curve represents the head required to push flow through the pipe network at various flow rates:

Hsystem = Hstatic + K Q²

where Hstatic = static head (elevation difference + pressure difference between reservoirs); K Q² = total friction and minor head losses = Σ(RiQ²)

Operating point: Intersection of pump H–Q curve and system curve → gives actual Q and H at which pump operates.

5. Pumps in Series and Parallel

ConfigurationHeadDischargeUse Case
Pumps in SeriesHtotal = H₁ + H₂ + … (heads add)Qtotal = Q (same flow through each)When high head is needed for a given flow — same pipe, boosting pressure
Pumps in ParallelHtotal = H (same head across each)Qtotal = Q₁ + Q₂ + … (flows add)When high flow rate is needed — multiple pumps feeding the same main

Graphical construction for combined curve:

Series: For each Q, add heads → combined H–Q curve is shifted upward

Parallel: For each H, add discharges → combined H–Q curve is shifted rightward

New operating point = intersection of combined pump curve with same system curve

6. Reciprocating Pump

A reciprocating pump uses a piston or plunger moving back and forth in a cylinder to displace fluid. Unlike centrifugal pumps, it is a positive displacement pump — it delivers a fixed volume per cycle regardless of the head.

Single-acting pump (one delivery stroke per revolution):

Qth = A × L × N / 60

where A = piston area (m²); L = stroke length (m); N = speed (rpm)


Double-acting pump (two delivery strokes per revolution):

Qth = (2A – Arod) × L × N / 60

where Arod = rod cross-sectional area (m²)


Actual discharge: Qact = ηv × Qth

Slip: Slip = Qth – Qact; % slip = (Slip/Qth) × 100

Negative slip (Qact > Qth) can occur with inertia effects in delivery pipe.


Work done per stroke: W = (pd – ps) × A × L

Power: P = ρg(Hs + Hd) × Qact / η

6.1 Air Vessel

An air vessel is a closed chamber connected to the suction or delivery pipe of a reciprocating pump. It acts as an accumulator — absorbing the pulsating flow from the pump and delivering a steady stream to the pipe. Benefits: reduces acceleration head, allows higher pump speeds, prevents separation in suction pipe, and saves power.

7. Specific Speed — Pumps & Turbines

Specific speed is a dimensionless (or dimensional, in common practice) parameter that characterises the geometry and type of a hydraulic machine operating at its best efficiency point. It allows comparison and selection of machine types independently of size.

7.1 Specific Speed for Pumps

Ns (pump) = NQ1/2 / H3/4

where N = rotational speed (rpm); Q = discharge (m³/s or litres/s); H = head (m)

Units depend on which unit system is used — always state units when quoting Ns.

Pump type selection by Ns (rpm, m³/s, m):

  • Ns < 0.5 (or < 50 in rpm-L/s-m system): Centrifugal, radial flow — high head, low discharge
  • 0.5 < Ns < 2.0: Mixed flow centrifugal pump
  • Ns > 2.0: Axial flow pump — low head, very high discharge

7.2 Specific Speed for Turbines

Ns (turbine) = NP1/2 / (ρ1/2 H5/4)

Dimensional form commonly used in India (N in rpm, P in kW, H in m):

Ns = N√P / H5/4

Turbine selection by Ns (rpm, kW, m):

  • Ns = 4 – 30: Pelton wheel (high head, low Ns)
  • Ns = 51 – 255: Francis turbine (medium head)
  • Ns = 255 – 860: Kaplan / propeller turbine (low head, high Ns)

7.3 Physical Meaning of Specific Speed

Specific speed is the speed at which a geometrically similar turbine (or pump) of unit size would run while developing unit power (or delivering unit discharge) under unit head. It is the same for all geometrically similar machines regardless of size — it characterises the shape (geometry) of the runner.

8. Affinity Laws (Similarity Laws) for Pumps

The affinity laws relate the performance of a pump at different speeds (or impeller sizes) using dimensional analysis and dynamic similarity.

For speed change (same pump, same impeller diameter D):

Q₁/Q₂ = N₁/N₂

H₁/H₂ = (N₁/N₂)²

P₁/P₂ = (N₁/N₂)³


For size change (same speed N, different impeller diameter D):

Q₁/Q₂ = (D₁/D₂)³

H₁/H₂ = (D₁/D₂)²

P₁/P₂ = (D₁/D₂)⁵


General (both N and D change — model to prototype):

Q ∝ ND³; H ∝ N²D²; P ∝ N³D⁵

These laws allow manufacturers to predict performance at different operating speeds without testing at every speed, and allow engineers to scale model test results to prototype dimensions.

9. Pelton Wheel (Impulse Turbine)

The Pelton wheel is an impulse turbine in which one or more high-velocity water jets strike double-cupped (bucket-shaped) vanes attached to the rim of a wheel. All the available head is converted to kinetic energy in the jet before striking the buckets — there is no pressure change across the runner.

9.1 Velocity Triangle Analysis

Jet velocity: V₁ = Cv√(2gH) (Cv ≈ 0.97–0.99)

Bucket velocity: u = πDN/60

Relative velocity at inlet: Vr1 = V₁ – u

Relative velocity at outlet: Vr2 = k Vr1 (k = 0.85–0.95, accounts for bucket friction)

Whirl component at outlet: Vw2 = u – Vr2cosβ

where β = deflection angle of bucket (usually 160°–170°; β = 180° – β₂ where β₂ is bucket exit angle)


Work done per unit weight of fluid (Euler):

W = (u/g)(V₁ – u)(1 + k cosβ)


Hydraulic efficiency:

ηh = 2(u/V₁)(1 + k cosβ)(1 – u/V₁)


Maximum efficiency condition:

u/V₁ = 0.5 → u = V₁/2 (bucket speed = half jet speed)

ηh,max = (1 + k cosβ)/2 ≈ 0.90–0.95 for k ≈ 1, β ≈ 165°

9.2 Design Features

  • Jet deflector: Deflects jet away from buckets during load rejection to prevent overspeed.
  • Spear (needle) valve: Regulates jet area (and hence flow) smoothly — flow control without surge.
  • Number of jets: Single jet standard; multi-jet (up to 6) for high power — jets equally spaced around the wheel.
  • Jet ratio (m): m = D/(djet) — typically m = 12 to 25. Higher m → lower Ns → more suitable for higher heads.

10. Francis Turbine (Reaction Turbine)

The Francis turbine is an inward-flow reaction turbine — water enters radially inward through adjustable guide vanes (wicket gates) and exits axially. Both pressure energy and kinetic energy decrease across the runner. The turbine is enclosed in a spiral casing (scroll casing) that maintains uniform velocity distribution around the runner periphery.

Runner head: Hr = (u₁Vw1 – u₂Vw2)/g (Euler)

Hydraulic efficiency:

ηh = Runner work / Available head = (u₁Vw1 – u₂Vw2) / (gH)

For no-swirl exit (Vw2 = 0, ideal condition):

ηh = u₁Vw1/(gH)

Overall efficiency: η = ρgQHnet × η / Pshaft

Typical overall efficiency: 85–93% at design point

10.1 Draft Tube

A draft tube is a diverging conduit connecting the turbine runner exit to the tailwater. It serves two functions:

  • Recovers kinetic energy: The expanding tube decelerates the exit flow, recovering velocity head as pressure head — this increases the effective net head across the turbine.
  • Allows runner installation above tailwater: The draft tube creates a sub-atmospheric pressure at the runner exit (suction effect), allowing the turbine to be set above the tailwater level without losing the head between runner exit and tailwater.

Head recovery by draft tube:

Hdt = Hs + (V₃² – V₄²)/(2g) – hf,dt

where Hs = height of runner above tailwater; V₃ = velocity at draft tube entry; V₄ = velocity at draft tube exit

11. Kaplan Turbine (Propeller Turbine)

The Kaplan turbine is an axial-flow reaction turbine with adjustable runner blades (like a ship’s propeller). Water flows axially through the runner. The blade angle can be adjusted as guide vane opening changes, maintaining high efficiency over a wide range of flows — making it ideal for run-of-river plants with highly variable flow.

Key features:

  • Axial flow: water enters and leaves the runner along the axis of rotation
  • Adjustable runner blades: optimise blade angle for each flow condition → high efficiency across wide range
  • Double regulation: both guide vanes and runner blades adjust simultaneously
  • Typical efficiency: 88–93% at design point

Runner velocity triangle (axial entry at mean radius r):

u = ωr; Vf = axial flow velocity = Q/(πR² – πrh²) where R = runner radius, rh = hub radius

Euler work: W = u(Vw1 – Vw2)/g; for ideal no-swirl exit Vw2 = 0

12. Turbine Comparison & Selection

FeaturePelton WheelFrancis TurbineKaplan Turbine
TypeImpulseReaction (mixed flow)Reaction (axial flow)
Head range (m)150 – 200030 – 7002 – 40
DischargeLow to mediumMediumVery high
Specific speed Ns4 – 30 (low)51 – 255 (medium)255 – 860 (high)
Flow direction through runnerTangentialRadial to axialAxial
Runner pressureAtmosphericAbove atmosphericAbove atmospheric
Draft tubeNoYesYes
Part-load efficiencyGoodModerateExcellent (adjustable blades)
Typical installationMountain hydropower, very high headDam-based hydropowerRun-of-river, tidal, low-head dam
Peak efficiency85 – 92%85 – 93%88 – 93%

13. Cavitation & NPSH

13.1 Cavitation Mechanism

Cavitation occurs when the local pressure in a flowing liquid drops to or below the vapour pressure pv at that temperature. Vapour bubbles form and then collapse violently when they move into higher-pressure regions, generating intense local shock waves, noise, vibration, surface pitting (erosion), and loss of efficiency.

13.2 Thoma’s Cavitation Number (σ)

σ = (Hatm – Hs – Hv) / H

where:

  • Hatm = atmospheric head = patm/(ρg) ≈ 10.33 m of water
  • Hs = suction head (height of runner above tailwater, positive if runner is above tailwater)
  • Hv = vapour pressure head = pv/(ρg)
  • H = net head across turbine (m)

Condition to avoid cavitation: σ > σc

where σc = critical Thoma number (from turbine model tests)

If σ < σc: cavitation occurs; turbine must be set lower (Hs reduced)

13.3 Net Positive Suction Head (NPSH) for Pumps

NPSHavailable = (patm – pv)/(ρg) – Hs – hf,suction

where Hs = height of pump above sump; hf,suction = head loss in suction pipe

Condition to avoid cavitation: NPSHavailable > NPSHrequired

NPSHrequired is specified by the pump manufacturer — obtained from tests.


Maximum suction head to avoid cavitation:

Hs,max = (patm – pv)/(ρg) – NPSHreq – hf,suction

≈ 10.33 – Hv – NPSHreq – hf,suction

In practice: Hs,max ≈ 6–7 m for most centrifugal pumps at sea level with cold water

13.4 Effects of Altitude and Temperature on NPSH

  • Higher altitude: patm decreases → NPSHavailable decreases → maximum suction lift decreases.
  • Higher water temperature: pv increases → NPSHavailable decreases → pump must be placed closer to (or below) the sump.

14. Worked Examples (GATE CE Level)

Example 1 — Centrifugal Pump Power and Efficiency (GATE CE type)

Problem: A centrifugal pump delivers 0.06 m³/s of water against a total head of 25 m. The shaft power input is 18 kW. Find (a) the overall efficiency and (b) the power delivered to water.

Given: Q = 0.06 m³/s; Hm = 25 m; Pinput = 18,000 W

(a) Power delivered to water:

Pwater = ρgQHm = 1000 × 9.81 × 0.06 × 25 = 14,715 W = 14.72 kW

(b) Overall efficiency:

η = Pwater/Pinput = 14,715/18,000 = 0.8175 = 81.75%

Answer: Pwater = 14.72 kW; η = 81.75%

Example 2 — Pelton Wheel Efficiency (GATE CE 2021 type)

Problem: A Pelton wheel has a jet velocity V₁ = 45 m/s and bucket velocity u = 20 m/s. The side leakage angle β₂ (deflection from straight) is 165° so the exit angle for the relative velocity is β = 15° from horizontal. Friction factor k = 0.9. Find the hydraulic efficiency.

Given: V₁ = 45 m/s; u = 20 m/s; k = 0.9; bucket exit angle β = 15° (from axis of jet direction)

cosβ = cos15° = 0.9659

Hydraulic efficiency formula:

ηh = 2(u/V₁)(1 + k cosβ)(1 – u/V₁)

u/V₁ = 20/45 = 0.4444

1 – u/V₁ = 0.5556

1 + k cosβ = 1 + 0.9 × 0.9659 = 1 + 0.8693 = 1.8693

ηh = 2 × 0.4444 × 1.8693 × 0.5556

= 2 × 0.4444 × 1.039 = 2 × 0.4617 = 0.9234 = 92.3%

Answer: Hydraulic efficiency ηh = 92.3%

Example 3 — Specific Speed of a Turbine (GATE CE type)

Problem: A turbine runs at 300 rpm under a head of 80 m and produces 5000 kW. Find the specific speed and identify the turbine type.

Given: N = 300 rpm; H = 80 m; P = 5000 kW

Specific speed:

Ns = N√P / H5/4

H5/4 = (80)5/4 = (80)¹ × (80)0.25 = 80 × 800.25

800.25 = (80)1/4 = (2⁴ × 5)1/4 = 2 × 50.25 = 2 × 1.4953 = 2.9907

H5/4 = 80 × 2.9907 = 239.26

√P = √5000 = 70.71

Ns = 300 × 70.71 / 239.26 = 21,213 / 239.26 = 88.67

Ns ≈ 89 (rpm, kW, m units)

Range 51–255 → Francis turbine

Answer: Ns = 88.7 → Francis turbine

Example 4 — Affinity Laws: Speed Change (GATE CE type)

Problem: A centrifugal pump operates at 1200 rpm delivering 0.05 m³/s against a head of 30 m, consuming 18 kW. If the speed is increased to 1500 rpm (same impeller), find the new discharge, head, and power.

Given: N₁ = 1200 rpm; Q₁ = 0.05 m³/s; H₁ = 30 m; P₁ = 18 kW; N₂ = 1500 rpm

Speed ratio: N₂/N₁ = 1500/1200 = 1.25

New discharge:

Q₂ = Q₁ × (N₂/N₁) = 0.05 × 1.25 = 0.0625 m³/s

New head:

H₂ = H₁ × (N₂/N₁)² = 30 × (1.25)² = 30 × 1.5625 = 46.875 m ≈ 46.9 m

New power:

P₂ = P₁ × (N₂/N₁)³ = 18 × (1.25)³ = 18 × 1.9531 = 35.16 kW

Answer: Q₂ = 62.5 L/s; H₂ = 46.9 m; P₂ = 35.2 kW

Example 5 — NPSH and Maximum Suction Head

Problem: A centrifugal pump is to be installed to lift water at 25°C (pv = 3.17 kPa absolute) from a sump. Atmospheric pressure = 101.3 kPa. NPSHrequired = 4.0 m. Head loss in suction pipe = 0.5 m. Find the maximum permissible suction head.

Given:
patm = 101,300 Pa; pv = 3,170 Pa
NPSHreq = 4.0 m; hf,s = 0.5 m

Atmospheric head:

Hatm = patm/(ρg) = 101,300/(1000 × 9.81) = 10.33 m

Vapour pressure head:

Hv = pv/(ρg) = 3,170/(1000 × 9.81) = 0.323 m

Maximum suction head:

Hs,max = Hatm – Hv – NPSHreq – hf,s

= 10.33 – 0.323 – 4.0 – 0.5 = 5.507 m ≈ 5.5 m

Answer: Maximum suction head = 5.5 m. Pump must be installed within 5.5 m above the sump water surface.

Example 6 — Reciprocating Pump Discharge (GATE CE type)

Problem: A double-acting reciprocating pump has a piston diameter of 200 mm and a piston rod diameter of 40 mm. The stroke length is 300 mm and the pump speed is 60 rpm. Volumetric efficiency = 95%. Find the actual discharge.

Given:
D = 200 mm = 0.2 m → A = π(0.2)²/4 = 0.03142 m²
drod = 40 mm = 0.04 m → Arod = π(0.04)²/4 = 0.001257 m²
L = 300 mm = 0.3 m; N = 60 rpm; ηv = 0.95

Theoretical discharge (double-acting):

Qth = (2A – Arod) × L × N/60

= (2 × 0.03142 – 0.001257) × 0.3 × 60/60

= (0.06284 – 0.001257) × 0.3 × 1

= 0.061583 × 0.3 = 0.018475 m³/s

Actual discharge:

Qact = ηv × Qth = 0.95 × 0.018475 = 0.01755 m³/s = 17.55 L/s

Answer: Qact = 17.55 litres/s

15. Common Mistakes

Mistake 1 — Confusing Specific Speed Formulas for Pumps vs Turbines

Error: Using Ns = NQ1/2/H3/4 (pump formula) for a turbine, or Ns = N√P/H5/4 (turbine formula) for a pump.
Root Cause: Pumps are characterised by discharge Q (output), turbines by power P (output) — different quantities are used because they represent different performance outputs.
Fix: Pump specific speed uses Q (what the pump delivers). Turbine specific speed uses P (what the turbine produces). The H exponent differs (3/4 for pump, 5/4 for turbine). Always identify the machine type first.

Mistake 2 — Using the Affinity Law H ∝ N Instead of H ∝ N²

Error: Writing H₂ = H₁ × (N₂/N₁) instead of H₂ = H₁ × (N₂/N₁)².
Root Cause: Mixing up the three affinity laws. Q scales linearly with N, but H scales with N² and P scales with N³.
Fix: Memorise as a sequence: Q ∝ N¹; H ∝ N²; P ∝ N³. The exponent increases by one each time: 1, 2, 3.

Mistake 3 — Forgetting That u = V₁/2 Gives Maximum Pelton Wheel Efficiency

Error: Assuming the Pelton wheel runs at its most efficient when bucket speed equals jet speed (u = V₁).
Root Cause: At u = V₁, the relative velocity is zero — no work is done at all. Maximum efficiency occurs when u = V₁/2 (half jet speed) — this is derived by differentiating the hydraulic efficiency expression with respect to u.
Fix: Optimal condition: u/V₁ = 0.5 → u = V₁/2. At this condition, ηh,max = (1 + k cosβ)/2.

Mistake 4 — Including the Draft Tube in the Net Head Calculation Incorrectly

Error: Not accounting for draft tube suction head Hs in the net head for a Francis turbine, computing Hnet = headwater level – tailwater level without the draft tube correction.
Root Cause: The draft tube allows the runner to be set above tailwater while still utilising the full gross head. The net head = gross head + draft tube recovery – draft tube losses. For preliminary calculations, Hnet ≈ gross head is acceptable, but detailed design must account for draft tube efficiency.
Fix: For GATE problems, use H = gross head (difference between headwater and tailwater levels) unless the problem specifically asks about draft tube effects. When turbine setting height Hs is given, check cavitation using Thoma’s σ.

Mistake 5 — Applying NPSH Incorrectly When Pump is Below Sump

Error: Using a positive Hs in the NPSH formula when the pump is installed below the sump (flooded suction), leading to an artificially reduced NPSHavailable.
Root Cause: Sign convention for Hs: Hs is positive when the pump is above the sump (suction lift) and negative when the pump is below the sump (flooded suction — reduces cavitation risk). Many students always use a positive Hs.
Fix: NPSHavailable = Hatm – Hv – Hs – hf,s. If pump is below sump: Hs is negative → NPSHavailable increases (flooded suction improves cavitation safety). If pump is above sump: Hs is positive → NPSHavailable decreases.

16. Frequently Asked Questions

Q1. Why does a centrifugal pump need priming but a reciprocating pump does not?

A centrifugal pump develops head by centrifugal action — it imparts angular momentum to the fluid. For this to work, the impeller must be surrounded by a liquid (not air), because the density difference between air and water means the centrifugal pump cannot generate enough pressure difference to push air out of the suction pipe and draw water up from the sump. If the pump runs dry (air-filled), it develops negligible head and no suction is created. In contrast, a reciprocating pump is a positive displacement device — the piston physically compresses the fluid in the cylinder and forces it out regardless of its density. Even with air in the cylinder, each stroke displaces a volume and creates pressure, so reciprocating pumps are self-priming by nature. They can pump air initially and gradually evacuate the suction line, drawing water up. Self-priming centrifugal pumps incorporate a recirculation system that retains a reservoir of liquid to prime the impeller automatically before operation begins.

Q2. What determines the choice between a Francis and a Kaplan turbine for a low-head hydropower project?

The primary factor is specific speed Ns — which itself is determined by the head and desired speed. For heads below about 40–50 m, Kaplan turbines have higher specific speed and higher efficiency at part load than Francis turbines, making them the preferred choice for run-of-river plants where flow varies significantly throughout the year. The key advantage of the Kaplan turbine is its double regulation: both the guide vanes (wicket gates) and the runner blade angle are adjusted simultaneously, maintaining near-peak efficiency even when operating at 50–60% of design flow. A Francis turbine has fixed runner geometry — its efficiency drops sharply at part load. In India, run-of-river projects on Himalayan tributaries (where monsoon and dry-season flows differ by a factor of 5 to 10) typically use Kaplan turbines for this reason. Francis turbines dominate storage-type hydro schemes (dams with reservoirs) where flow can be regulated to keep the turbine near its design point.

Q3. How does the concept of specific speed relate to geometrically similar turbine families?

Specific speed Ns = N√P/H5/4 is the same for all geometrically similar turbines (those with the same shape of flow passages, blade angles, and proportions) regardless of their actual size — it is a shape parameter, not a size parameter. This means that all Pelton wheels with single jets belong to a narrow band of Ns values (4–30), all Francis turbines to their band (51–255), and all Kaplan turbines to their band (255–860), regardless of whether they are 1 MW or 500 MW machines. The physical reason is that Ns is derived from dimensional analysis of the governing physics (flow rate, speed, head, and power) — once the Ns value is fixed, the ratios of all key dimensions (impeller diameter to setting diameter, blade angles, volute shape) are determined by geometric similarity. This is why turbine manufacturers use model tests at convenient small scales (typically 1:5 to 1:20) and scale up results using dimensional similarity — the efficiency and flow characteristics of the prototype are predicted from the model’s Ns and appropriate scale correction formulas.

Q4. What is the engineering significance of the Thoma cavitation number σ in turbine design?

Thoma’s cavitation number σ = (Hatm – Hs – Hv)/H quantifies the margin of safety against cavitation in a turbine installation. The numerator represents the available pressure head above vapour pressure at the runner, and the denominator normalises this by the total operating head. The critical value σc is determined from model tests for each turbine design — if the installed σ falls below σc, cavitation begins on the runner blades, causing pitting and vibration. Since σ increases as Hs decreases (runner set lower relative to tailwater), the Thoma number dictates the maximum permissible setting height of the turbine above tailwater. For a given turbine design and head H, the setting height limit is Hs ≤ Hatm – Hv – σcH. In practice, high-head Francis turbines have low σc and can be set well above tailwater, while low-head Kaplan turbines have high σc and are typically set below tailwater level (Hs negative, requiring a deeper powerhouse excavation) to prevent cavitation.