AEE Mechanical Engineering Core · Chapter 5

Hydraulic Turbines and Pumps

What to remember

Turbines turn water energy into shaft work: Pelton (impulse, high head, low flow), Francis (mixed flow, medium head) and Kaplan (axial flow, low head, high flow). Specific speed Ns = N√P/H^(5/4) identifies the type.
Centrifugal pumps add head by raising the velocity of liquid in an impeller and converting it to pressure in the casing. Cavitation is avoided by keeping available NPSH above required NPSH.
Affinity laws for a pump at the same geometry: Q ∝ N, H ∝ N², P ∝ N³. For similar machines: Q ∝ ND³, H ∝ N²D², P ∝ N³D⁵.

1. Hydropower basics

Water power available = ρ g Q H (watts, with Q in m³/s and H the net head in m). With ρ = 1000 kg/m³ and g = 9.81, this is 9.81 Q H kW. Shaft power = overall efficiency × water power.
Worked example: Q = 2 m³/s, H = 50 m, overall efficiency 0.9 gives 9.81 × 2 × 50 × 0.9 = 882.9 kW.
Gross head is reservoir level minus tail-race level. Net head = gross head − head lost in penstock friction and fittings.
Efficiencies: hydraulic (runner work ÷ water energy), mechanical (shaft ÷ runner work), volumetric (water through runner ÷ water supplied), overall = product of these.
APGENCO runs hydro and thermal stations. Hydro units use these turbine types according to site head and flow.

2. Classification of turbines

Basis	Types
Action	Impulse (Pelton): pressure constant, only kinetic energy used. Reaction (Francis, Kaplan, Propeller): pressure falls through the runner
Head	High (Pelton), medium (Francis), low (Kaplan)
Flow direction	Tangential (Pelton), mixed (Francis), axial (Kaplan)
Specific speed	Low (Pelton), medium (Francis), high (Kaplan)

Typical heads: Pelton above about 250 m; Francis about 30 to 300 m; Kaplan below about 60 m. Treat ranges as approximate.

3. Pelton wheel (impulse)

A nozzle forms a jet; spear valve regulates flow. The jet strikes double-hemispherical buckets with a splitter ridge. The wheel runs in air at atmospheric pressure; a casing only stops splashing.
Jet velocity V₁ = Cv √(2gH). With Cv = 1 and H = 100 m, V₁ = 44.3 m/s.
Bucket speed u = πDN/60. Practical u ≈ 0.43 to 0.47 of V₁ (ideal 0.5 V₁).
Work done per kg: (V₁ − u)(1 + k cosφ) u, where φ is the bucket outlet angle (measured from the reversed jet direction) and k is the relative-velocity reduction factor. Maximum hydraulic efficiency (when u = V₁/2) is (1 + k cosφ)/2.
Number of jets: up to about 6. Specific speed rises with the number of jets (Ns ∝ √jets).
Governing: a deflector plate or spear movement; sudden shut-off is avoided by the deflector to prevent water hammer in the penstock.
Pelton uses no draft tube because the runner is above the tail-race at atmospheric pressure.

4. Francis and Kaplan (reaction)

Francis: radial inflow, mixed outflow. Guide vanes (wicket gates) control flow. Water fills the casing (spiral scroll casing), so pressure acts on the runner.
Kaplan: axial flow with adjustable runner blades and adjustable guide vanes (double regulation). It keeps good efficiency at part load. Propeller turbine has fixed blades.
Draft tube: a diverging tube from runner outlet to tail race. It allows the runner to sit above tail-race level and recovers kinetic energy as pressure. Cone angle is kept small (about 8° or less) to avoid flow separation.
Euler's turbine equation: work per unit weight = (V_w1 u₁ − V_w2 u₂)/g.

5. Specific speed and unit quantities

Turbine specific speed Ns = N √P / H^(5/4) (N in rpm, P in kW, H in m). It is the speed of a geometrically similar turbine producing 1 kW under 1 m head.
Unit speed N_u = N/√H; unit discharge Q_u = Q/√H; unit power P_u = P/H^(3/2).
Worked example: N = 500 rpm, P = 1000 kW, H = 100 m gives Ns = 500 × 31.62 / 316.2 = 50.

Turbine	Specific speed (approx.)
Pelton	about 10 to 35
Francis	about 60 to 300
Kaplan / Propeller	about 300 to 1000

Cavitation in turbines: low pressure at runner exit or draft tube vapourises water, causing pitting, noise and efficiency loss. Thoma cavitation factor σ = (H_atm − H_vap − H_s)/H. Prevent by limiting runner setting height above tail water and using suitable materials.

6. Centrifugal pump

Parts: impeller, volute casing (or diffuser), suction pipe with foot valve and strainer, delivery pipe with valve.
Priming: filling suction pipe and casing with liquid before starting, because air cannot generate enough suction.
Heads: static head H_s = suction lift + delivery lift. Manometric head H_m = H_s + friction losses in both pipes + V_d²/2g.
Efficiencies: manometric = H_m ÷ (V_w2 u₂/g); mechanical = impeller power ÷ shaft power; overall = water power ÷ shaft power.
Shaft power = ρ g Q H_m / η_o. Worked example: Q = 0.05 m³/s, H_m = 20 m, η = 0.7 gives 9810 × 0.05 × 20 / 0.7 ≈ 14.0 kW.
Impeller blades: backward-curved (most usual, stable and efficient), radial, forward-curved (rare). Closed impellers are the most efficient.
Multistage: series stages raise head (high-head service). Parallel pumps increase discharge.
Pump specific speed Ns = N √Q / H^(3/4) (Q in m³/s, H in m). Radial impellers have low Ns; axial-flow pumps have high Ns.
NPSH (net positive suction head) available = (p_atm − p_vapour)/ρg − H_suction − suction friction head. Pump is free of cavitation when NPSH available exceeds NPSH required.

7. Similarity and operating curves

Law	Same pump, different speed	Similar pumps
Discharge	Q ∝ N	Q ∝ N D³
Head	H ∝ N²	H ∝ N² D²
Power	P ∝ N³	P ∝ N³ D⁵

Example: doubling the speed multiplies head by 4 and power by 8. Characteristic curves plot head, power and efficiency against discharge. A centrifugal pump should be started with delivery valve closed (minimum power). An axial-flow pump is started with valve open.

8. Reciprocating pump

Positive displacement. Theoretical discharge for single-acting: Q = A L N / 60. Double-acting: 2 A L N / 60 (neglect piston rod area). Example: D = 0.1 m, L = 0.15 m, N = 60 rpm gives A = 0.00785 m² and Q = 0.00785 × 0.15 × 60/60 = 1.18 × 10⁻³ m³/s (single acting).
Slip = theoretical − actual discharge. Percentage slip = slip ÷ theoretical × 100. Negative slip is possible when delivery pipe is short, suction pipe long and speed high.
Air vessels on suction and delivery side reduce acceleration head, reduce friction head and give a nearly uniform discharge.
Indicator diagram shows pressure in the cylinder against piston position.
Comparison: reciprocating gives high head, low discharge, pulsating flow. Centrifugal gives high discharge, moderate head, steady flow.

9. Other devices and selection

Hydraulic ram: uses the water-hammer effect of a large flow falling through a small head to lift a small part of the water to a greater height, without external power.
Hydraulic accumulator stores energy as pressurised liquid. Hydraulic intensifier raises pressure. Hydraulic press works on Pascal's law.
Pumped-storage plants use reversible machines: they pump water up when power is cheap and generate when demand is high.
Selection by site: very high head and small flow suits Pelton; medium head suits Francis; low head with large river flow suits Kaplan. The turbine is chosen so its specific speed matches the site head and generator speed.
Governor: controls turbine speed by changing flow (spear in Pelton, wicket gates in Francis and Kaplan) when electrical load changes.

Exam traps

Pelton is impulse, so no draft tube; Francis and Kaplan are reaction.
Pelton: high head, low specific speed; Kaplan: low head, high specific speed. Do not reverse.
Kaplan has adjustable runner blades; Propeller has fixed blades.
Turbine specific speed uses power P; pump specific speed uses discharge Q.
Priming applies to centrifugal pumps, not reciprocating pumps (they are self-priming).
Start a centrifugal pump with delivery valve closed.
Manometric head includes suction and delivery friction and delivery velocity head.
NPSH available must exceed NPSH required to avoid cavitation.

One-liners

1. Hydropower = ρ g Q H.
2. Pelton jet velocity V₁ = Cv√(2gH).
3. Ideal Pelton bucket speed is half the jet speed.
4. Draft tube recovers kinetic energy at runner outlet.
5. Kaplan: double regulation (guide vanes and runner blades).
6. Specific speed of Francis turbines is medium.
7. Cavitation causes pitting and noise.
8. Unit speed N_u = N/√H.
9. Pump overall efficiency = water power ÷ shaft power.
10. Multistage in series raises head.
11. Slip in reciprocating pump = Q_th − Q_act.
12. Air vessel smooths flow in reciprocating pumps.

Practice questions

The Pelton wheel is classified as a
1. reaction turbine
2. axial-flow reaction turbine
3. mixed-flow turbine
4. impulse turbine
Answer
D. impulse turbine
Water leaves the nozzle at atmospheric pressure; only kinetic energy is used.
Which turbine does NOT need a draft tube?
1. Propeller turbine
2. Francis turbine
3. Pelton wheel
4. Kaplan turbine
Answer
C. Pelton wheel
Pelton runner works in air above the tail race at atmospheric pressure.
A turbine best suited to a low head and very large discharge is the
1. Kaplan turbine
2. Turgo impulse wheel
3. Pelton wheel
4. Francis turbine with small runner
Answer
A. Kaplan turbine
Axial-flow Kaplan runners handle large flow at low head.
A turbine works under a net head of 50 m with a discharge of 2 m³/s and overall efficiency 0.9 (g = 9.81 m/s²). The shaft power is
1. 8829 kW
2. 882.9 kW
3. 98.1 kW
4. 981 kW
Answer
B. 882.9 kW
P = 9.81 × 2 × 50 × 0.9 = 882.9 kW.
The jet velocity from a nozzle under a head of 100 m with Cv = 1 (g = 9.81 m/s²) is about
1. 22.1 m/s
2. 44.3 m/s
3. 31.3 m/s
4. 99.0 m/s
Answer
B. 44.3 m/s
V = √(2 × 9.81 × 100) = √1962 = 44.3 m/s.
For maximum efficiency the ideal Pelton bucket speed is
1. one-third of the jet velocity
2. equal to the jet velocity
3. twice the jet velocity
4. half the jet velocity
Answer
D. half the jet velocity
Work output (V₁ − u)u is maximum at u = V₁/2.
Among the turbines, the highest specific speed is that of the
1. Kaplan turbine
2. Francis turbine
3. single-jet Pelton wheel
4. Pelton wheel
Answer
A. Kaplan turbine
Kaplan: about 300 to 1000; Francis 60 to 300; Pelton 10 to 35.
A turbine runs at 300 rpm and develops 10 000 kW under a head of 100 m. Its specific speed is about
1. 30
2. 300
3. 95
4. 9.5
Answer
C. 95
Ns = 300 × √10 000 / 100^1.25 = 30 000/316.2 ≈ 95.
A turbine runs at 600 rpm under a head of 36 m. Its unit speed is
1. 3.6 rpm
2. 100 rpm
3. 21 600 rpm
4. 16.7 rpm
Answer
B. 100 rpm
N_u = N/√H = 600/6 = 100.
A turbine develops 2160 kW at a head of 36 m. The unit power P/H^(3/2) is
1. 10
2. 60
3. 1.67
4. 77.8
Answer
A. 10
H^1.5 = 216; 2160/216 = 10.
Statements on a Pelton wheel: 1. The runner operates at atmospheric pressure. 2. A draft tube is essential for a Pelton wheel. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
The runner is in air; no draft tube is needed.
Statements: 1. The Kaplan turbine has adjustable runner blades. 2. The Propeller turbine has fixed runner blades. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Kaplan = adjustable blades (double regulation); Propeller = fixed blades.
A draft tube on a reaction turbine is used to
1. increase the jet velocity at nozzle
2. reduce the number of guide vanes
3. raise the specific speed above 1000
4. recover kinetic energy and allow the runner to sit above tail-water level
Answer
D. recover kinetic energy and allow the runner to sit above tail-water level
A gradually widening tube converts exit velocity head into pressure.
Cavitation in a hydraulic turbine mainly results in
1. increase of efficiency
2. pitting of blades, noise and drop in efficiency
3. rise in generator voltage
4. decrease in water temperature only
Answer
B. pitting of blades, noise and drop in efficiency
Vapour bubbles collapse at the surface and erode it.
The flow to a Pelton wheel is regulated by
1. wicket gates
2. runner blade angle
3. draft tube cone
4. a spear (needle) valve in the nozzle
Answer
D. a spear (needle) valve in the nozzle
Spear moves axially to vary jet area.
A turbine has hydraulic efficiency 0.92 and mechanical efficiency 0.95 with no leakage. Its overall efficiency is about
1. 0.920
2. 0.935
3. 0.874
4. 0.970
Answer
C. 0.874
η_o = 0.92 × 0.95 = 0.874.
The purpose of priming a centrifugal pump is to
1. remove air from the suction pipe and casing
2. reduce the delivery head
3. increase the speed of the motor
4. avoid pump overheating by water cooling
Answer
A. remove air from the suction pipe and casing
Air gives negligible suction; casing must be full of liquid.
A centrifugal pump is normally started with its delivery valve
1. half open
2. removed
3. closed
4. fully open
Answer
C. closed
Power demand is minimum at zero discharge.
A pump delivers 0.05 m³/s at a manometric head of 20 m with overall efficiency 0.7 (g = 9.81 m/s²). The shaft power is about
1. 9.8 kW
2. 14.0 kW
3. 1.4 kW
4. 19.6 kW
Answer
B. 14.0 kW
P = 9810 × 0.05 × 20/0.7 = 14 014 W.
A pump has a suction lift of 3 m, a delivery lift of 17 m, total friction loss of 1 m and delivery velocity head 0.5 m. The manometric head is
1. 21.0 m
2. 22.5 m
3. 20.0 m
4. 21.5 m
Answer
D. 21.5 m
H_m = 3 + 17 + 1 + 0.5 = 21.5 m.
A centrifugal pump speed is raised from 1000 rpm to 1500 rpm. The head developed (same pump) changes by a factor of
1. 2.25
2. 1.5
3. 1.0
4. 3.375
Answer
A. 2.25
H ∝ N²: (1.5)² = 2.25.
If the speed of a pump is doubled, the power required changes by a factor of
1. 2
2. 8
3. 4
4. 16
Answer
B. 8
P ∝ N³ for the same pump.
Two geometrically similar pumps run at the same speed. If the impeller diameter of one is double the other, its discharge is
1. 4 times
2. 16 times
3. 8 times
4. 2 times
Answer
C. 8 times
Q ∝ N D³: 2³ = 8.
The specific speed of a pump is defined using
1. power and head, Ns = N√P/H^(5/4)
2. head and diameter only
3. torque and discharge
4. discharge and head, Ns = N√Q/H^(3/4)
Answer
D. discharge and head, Ns = N√Q/H^(3/4)
Pump specific speed uses Q, whereas turbine specific speed uses P.
The NPSH available at the pump suction with atmospheric head 10.3 m, vapour pressure head 0.3 m, suction lift 4 m and suction friction loss 0.5 m is
1. 14.5 m
2. 6.0 m
3. 4.5 m
4. 5.5 m
Answer
D. 5.5 m
NPSHa = 10.3 − 0.3 − 4 − 0.5 = 5.5 m.
Cavitation in a pump is avoided if
1. NPSH available exceeds NPSH required
2. NPSH available equals zero
3. NPSH required exceeds NPSH available
4. the delivery valve is fully closed
Answer
A. NPSH available exceeds NPSH required
Suction pressure must stay above vapour pressure.
The most commonly used blade shape in centrifugal pump impellers is
1. forward curved
2. radial straight only
3. backward curved
4. spiral hollow
Answer
C. backward curved
Backward-curved vanes are stable and efficient.
Pumps connected in series are used mainly to increase
1. specific speed to a very high value
2. head
3. discharge
4. suction lift
Answer
B. head
Heads add; discharge stays the same.
Statements: 1. Pumps in series increase the head. 2. Pumps in parallel increase the discharge. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Series adds head; parallel adds flow.
A single-acting reciprocating pump has bore 0.1 m, stroke 0.15 m and runs at 60 rpm. The theoretical discharge is about
1. 7.07 × 10⁻² m³/s
2. 1.18 × 10⁻³ m³/s
3. 1.18 × 10⁻² m³/s
4. 2.36 × 10⁻³ m³/s
Answer
B. 1.18 × 10⁻³ m³/s
Q = (π/4)(0.1)² × 0.15 × 60/60 = 1.18 × 10⁻³.
For the same bore, stroke and speed, the theoretical discharge of a double-acting reciprocating pump compared with single-acting is
1. same
2. half
3. four times
4. twice as large (rod area neglected)
Answer
D. twice as large (rod area neglected)
Delivery occurs on both sides of the piston.
A reciprocating pump has theoretical discharge 0.0100 m³/s and actual discharge 0.0095 m³/s. The percentage slip is
1. 5%
2. 9.5%
3. 95%
4. 0.5%
Answer
A. 5%
Slip = 0.0005/0.0100 = 5%.
Negative slip in a reciprocating pump can occur when
1. the pump leaks badly at valves
2. the delivery pipe is short, the suction pipe is long and speed is high
3. the delivery pipe is very long
4. the suction pipe is very short and speed is low
Answer
B. the delivery pipe is short, the suction pipe is long and speed is high
Inertia of the water in the suction pipe keeps delivery valve open too long.
The main purpose of an air vessel in a reciprocating pump is to
1. give a nearly uniform discharge and reduce acceleration head
2. increase the theoretical discharge
3. prime the pump
4. increase the stroke length
Answer
A. give a nearly uniform discharge and reduce acceleration head
Air cushion absorbs flow fluctuations.
Statements: 1. A reciprocating pump is self-priming. 2. A centrifugal pump needs priming before starting. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Reciprocating pump can lift air; centrifugal cannot.
A hydraulic ram works on the principle of
1. Pascal's law only
2. centrifugal action
3. Bernoulli's equation with no losses
4. water hammer
Answer
D. water hammer
Sudden valve closure raises pressure, lifting a part of water.
If the number of jets of a Pelton wheel is increased from 1 to 4 (same wheel, same head), the specific speed changes by a factor of
1. 2
2. 4
3. 16
4. 1/2
Answer
A. 2
Ns ∝ √(number of jets): √4 = 2.
In a Francis turbine, water enters the runner and leaves it
1. axially in and radially out
2. radially inward and axially (mixed flow)
3. tangentially, in air
4. radially outward only
Answer
B. radially inward and axially (mixed flow)
Inward radial flow with mixed or axial outflow.
A Kaplan turbine is said to have double regulation because it has
1. two nozzles and two spears
2. two draft tubes
3. adjustable guide vanes and adjustable runner blades
4. two governors only
Answer
C. adjustable guide vanes and adjustable runner blades
Both vanes and blades are adjusted to match load.
The splitter on a Pelton bucket serves to
1. raise the jet velocity
2. reduce bucket speed
3. admit air into the bucket
4. divide the jet into two equal streams
Answer
D. divide the jet into two equal streams
The central ridge splits the jet to balance axial thrust.
A pumped-storage plant uses machines that
1. can run both as pump and as turbine
2. run only as a Pelton turbine
3. run only as a reciprocating pump
4. convert steam to hydraulic power
Answer
A. can run both as pump and as turbine
Reversible pump-turbines pump when power is cheap and generate at peak.
The net head on a turbine is
1. gross head plus friction head
2. gross head minus the head lost in the penstock and fittings
3. tail-race level only
4. reservoir level only
Answer
B. gross head minus the head lost in the penstock and fittings
H_net = H_gross − h_f.
Statements: 1. A turbine of low specific speed is suited to high head. 2. A turbine of high specific speed is suited to low head. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Pelton (low Ns) high head; Kaplan (high Ns) low head.
A pump delivers 0.02 m³/s of water against a head of 30 m (g = 9.81 m/s²). The water power is
1. 0.589 kW
2. 58.86 kW
3. 588.6 kW
4. 5.886 kW
Answer
D. 5.886 kW
P = 9810 × 0.02 × 30 = 5886 W.
The cone angle of a draft tube is kept small mainly to prevent
1. leakage at the runner
2. spear valve blockage
3. increase of jet velocity
4. separation of flow from the walls
Answer
D. separation of flow from the walls
A wide angle causes separation, losses and cavitation.

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