Hydropower: Resources, Plant Layout, Head and Flow, Pelton, Francis and Kaplan Turbines, Small Hydro

This chapter belongs to the Engineering Sciences (XE) paper and sits in Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. It answers the heading “Hydropower: Hydro resources, hydro power plants, turbines (Pelton, Kaplan, Francis), small hydro systems”. It teaches the hydro resource and the power equation, flow-duration curves and the main plant types, the layout of a plant with head losses in the penstock, the Pelton, Francis and Kaplan turbines and their selection by specific speed, cavitation and the draft tube, and small hydro schemes. The numericals are plant power from head, flow and efficiency, penstock head loss, specific speed, Pelton wheel speed and bucket power; every constant is stated in the question.

1. Hydro resource, head and flow

Water of volume flow rate Q falling through a height H carries power ρgQH. The gross head Hg is the level difference between the water surface at the intake and the tailwater; the net head H = Hg − hf is what remains at the turbine after the friction and other losses hf in the waterway. The electrical output is P = η ρ g Q H, where the overall efficiency η = ηh ηm ηg multiplies the hydraulic, mechanical and generator efficiencies. Worked example. Q = 20 m³/s, Hg = 100 m with a 5 m loss, so H = 95 m; η = 0.88; ρ = 1000 kg/m³, g = 9.81 m/s²: P = 0.88 × 1000 × 9.81 × 20 × 95 = 16.4 × 10⁶ W, 16.4 MW. Each factor is linear, so a plant is classed by its head (low, medium, high) and flow together: the same power can come from a large flow at low head or a small flow at high head.

The flow of a river varies through the year, so the resource is described by a flow-duration curve: stream flow plotted against the fraction of time for which it is equalled or exceeded. The area under it is the energy available; the design flow of a run-of-river plant is chosen from it, balancing the extra energy of a larger turbine against its cost and the days it would run part-loaded. The capacity factor is the actual annual energy divided by (rated power × 8760 h) and is always below 1 because of flow variation, maintenance and demand.

Main plant types
Plant typeIdeaConsequence
Run-of-riveruses the natural flow with little or no storageoutput follows the river; low environmental footprint
Storage (reservoir)a dam stores water over a seasonoutput follows demand; large submergence and sediment issues
Pumped storagewater is pumped up when power is cheap and run back down at peakstores energy rather than creating it; a net consumer over a cycle

2. Plant layout and penstock losses

Along the water path come: the dam or weir that raises the level and stores or diverts water; the intake with a trash rack to stop debris; a canal or tunnel (a forebay where the water joins the penstock); the penstock, the steep pressure pipe to the turbine; a surge tank that absorbs pressure surges when the flow is changed suddenly (water hammer) on long conduits; the powerhouse with turbine, governor and generator; and the draft tube (reaction turbines) and tailrace returning the water. Isolating valves and spillways protect the structure in floods.

Friction in the penstock reduces the net head. By the Darcy-Weisbach relation hf = f L v²/(2gD), with v = Q/(πD²/4). Worked example. L = 500 m, D = 1.5 m, Q = 6 m³/s, friction factor f = 0.02 (given): v = 6/1.767 = 3.395 m/s; hf = 0.02 × 500 × 3.395²/(2 × 9.81 × 1.5) = 3.92 m. Because hf grows as Q², there is a best flow for a given pipe: the power ρgQ(Hg − kQ²) is largest when hf = Hg/3, so the net head at the maximum is 2Hg/3. A designer uses a larger pipe to keep hf to a few percent of Hg, since the pipe is a major cost.

⚠️ Gross head is not the head the turbine sees
Use the net head H = Hg − hf in P = ηρgQH and in specific speed. Putting Hg into the power formula overstates the output; and the maximum-power rule hf = Hg/3 is a result about the best flow for a fixed pipe, not a design target, since it would waste a third of the head.

3. Pelton, Francis and Kaplan turbines

The three main hydro turbines
TurbineAction and flowHead and flowFeatures
Peltonimpulse; free jets strike split buckets at atmospheric pressure; tangentialhigh head, small flowregulated by a spear valve and jet deflector; no draft tube
Francisreaction; pressure falls across the runner; radial inflow, mixed outflowmedium head, medium flowguide vanes regulate; draft tube; most widely used
Kaplanreaction; axial flow; adjustable runner bladeslow head, large flowkeeps good efficiency at part load by adjusting the blade angle with the guide vanes; draft tube

Selection by specific speed. The specific speed Ns = N√P/H^(5/4), with N in rpm, P in kW (the shaft power) and H in m, is the speed of a geometrically similar turbine that gives 1 kW under 1 m of head; it collects speed, power and head into one number that describes the shape of the runner. Typical ranges (metric units, indicative): Pelton about 10 to 60 (single jet at the lower end), Francis about 60 to 300, Kaplan about 300 to 1000. Low Ns means a high-head, small-flow machine, and high Ns a low-head, large-flow one. Worked example. N = 300 rpm, P = 10 000 kW, H = 100 m: Ns = 300 × 100/100^1.25 = 30 000/316.2 = 94.9, in the Francis range.

Pelton basics. The jet speed is V₁ = Cv√(2gH), with Cv ≈ 0.98 for the nozzle. The best bucket speed u is a little under V₁/2, for example 0.46V₁ in practice; for a frictionless 180° deflection, u = V₁/2 gives the whole kinetic power of the jet: P = ρQ u (V₁ − u)(1 + 1). With Q = 0.15 m³/s, V₁ = 80 m/s and u = 40 m/s, P = 1000 × 0.15 × 40 × 40 × 2 = 480 kW, equal to ½ρQV₁² = 480 kW. A real bucket turns the jet by less than 180° and has friction, so the power is less. Wheel speed: with H = 400 m, Cv = 0.98, u = 0.46V₁ and wheel diameter 1.4 m, V₁ = 86.8 m/s, u = 39.9 m/s and N = 60u/(πD) = 545 rpm.

4. Draft tube, cavitation and part-load behaviour

Draft tube and cavitation. The draft tube of a reaction turbine is a diverging passage that recovers the exit kinetic energy and lets the runner sit above the tailwater, using the suction head. Setting the runner too high drops the local pressure towards the vapour pressure and causes cavitation, which pits blades and loses efficiency. The Thoma coefficient σ = (Ha − Hv − Hs)/H, with Ha the atmospheric head, Hv the vapour-pressure head and Hs the setting above the tailwater, must exceed the critical value for the turbine; a larger Hs lowers σ and invites cavitation. Pelton runners run in air and do not cavitate in this way.

Part-load behaviour. Demand and flow change, so the efficiency-against-load curve matters. A Pelton wheel with a spear valve, especially a multi-jet one, and a double-regulated Kaplan turbine keep a high efficiency over a wide range of load. A Francis turbine, with fixed runner blades, falls off noticeably at part load, and a fixed-blade propeller turbine falls sharply. The overall efficiency is the product of the hydraulic, mechanical (bearing) and generator efficiencies, η = ηh ηm ηg; for example 0.92 × 0.98 × 0.97 = 0.875, so about one-eighth of the water power is lost even at best.

5. Small hydro systems

Small hydro is hydropower at a modest scale, generally run-of-river with a diversion weir and a short penstock and no large reservoir, so it has far less submergence and a smaller environmental footprint than a large dam. The size limit for “small” differs between countries, so classes are best remembered by name in descending order: small, mini, micro and pico. Equipment is standardised and often built locally; plants are placed on canals, drops in irrigation channels and mountain streams.

  • Turbine choice follows head and flow: Pelton or Turgo for high head and small flow; Francis for medium head; Kaplan, propeller or tubular types for low head; the cross-flow (Banki) turbine is simple, cheap, tolerant of changing flow and common in micro hydro at low to medium head.
  • Design flow is picked from the flow-duration curve: a higher design flow gives more energy but a larger, costlier turbine that runs part-loaded on more days.
  • Generators and control: small plants often use induction or synchronous generators; an isolated micro-hydro plant with a fixed-speed generator holds frequency with an electronic load controller that diverts surplus power to a ballast (dump) load, instead of a costly mechanical governor.
🎯 Why head matters more than flow for cost
For a given power, a higher head needs a smaller flow and so a smaller, faster turbine, a narrower penstock and a smaller generator for the same output, which is why high-head sites are often cheaper per kW, while low-head sites need big, slow machines and large civil works.

Key takeaways

  • P = ηρgQH with the net head H = Hg − hf; hf = fLv²/(2gD), and for a fixed pipe the maximum power occurs when hf = Hg/3.
  • The flow-duration curve shows time for which flow is equalled or exceeded; design flow and energy come from it. Run-of-river follows the river; storage follows demand; pumped storage stores energy.
  • Pelton: impulse, high head, free jets at atmospheric pressure, spear valve. Francis: reaction, medium head, guide vanes. Kaplan: axial, low head, adjustable blades, good part-load efficiency.
  • Ns = N√P/H^(5/4) (rpm, kW, m): low for Pelton, medium for Francis, high for Kaplan. Pelton jet V₁ = Cv√(2gH) and bucket speed near 0.45V₁; ideal 180° deflection gives all the jet power at u = V₁/2.
  • A draft tube recovers exit kinetic energy but a high setting risks cavitation (Thoma σ = (Ha − Hv − Hs)/H). Small hydro is run-of-river; cross-flow, Pelton, Kaplan-type machines suit it by head, and micro plants use a load controller.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A site has a very high head of several hundred metres and a small flow of water. Which turbine suits it best?

    1. Kaplan turbine
    2. Pelton wheel
    3. Propeller turbine
    4. Cross-flow turbine at the lowest head
    Show answer

    Answer: B — Pelton wheel

    A high head with small flow gives a low specific speed, which is the Pelton range. Kaplan and propeller machines are for low head with large flow.

  2. Which turbine has an axial flow runner with adjustable blades and is used for low heads?

    1. Pelton wheel
    2. Turgo turbine
    3. Francis turbine
    4. Kaplan turbine
    Show answer

    Answer: D — Kaplan turbine

    The Kaplan turbine has axial flow and runner blades whose pitch can be adjusted with the guide vanes. The Francis has fixed runner blades and radial inflow, and Pelton and Turgo are impulse machines for high head.

  3. Which statement distinguishes an impulse turbine such as the Pelton wheel from a reaction turbine such as the Francis?

    1. In the Pelton wheel the whole pressure drop occurs in the nozzle, and the jet strikes the buckets at atmospheric pressure
    2. In the Pelton wheel the water pressure falls steadily across the runner while the runner passages are full of water
    3. The Francis turbine runs in air, free of the tailwater, with free jets striking its blades at atmospheric pressure
    4. Only the Francis turbine needs a governor, because the Pelton wheel keeps its speed without regulation
    Show answer

    Answer: A — In the Pelton wheel the whole pressure drop occurs in the nozzle, and the jet strikes the buckets at atmospheric pressure

    In an impulse turbine the head is converted to jet velocity in the nozzle and the buckets only change its direction at atmospheric pressure. In a reaction machine the runner passage is full and the pressure drops across it.

  4. A turbine has N = 300 rpm, delivers P = 10 000 kW at a net head H = 100 m, and specific speed Ns = N√P/H^(5/4). If specific speed ranges of about 10 to 60 (Pelton), 60 to 300 (Francis) and 300 to 1000 (Kaplan) are taken, the suitable turbine is

    1. Pelton wheel, since Ns is below 60
    2. Kaplan turbine, since Ns is above 300
    3. Francis turbine, since Ns is about 95
    4. A cross-flow turbine only, because the speed is too low for the three main types
    Show answer

    Answer: C — Francis turbine, since Ns is about 95

    Ns = 300 × √10 000/100^1.25 = 300 × 100/316.2 = 94.9, within the Francis range. The Kaplan answer would follow if H were mistakenly dropped from the denominator.

  5. What is the main function of the draft tube of a reaction turbine?

    1. It increases the jet speed at the runner inlet by throttling the flow before the guide vanes
    2. It regulates the flow to the runner by opening and closing with the load
    3. It absorbs the pressure surges that arise in the penstock when the load changes
    4. It recovers exit kinetic energy and lets the runner sit above the tailwater
    Show answer

    Answer: D — It recovers exit kinetic energy and lets the runner sit above the tailwater

    The diverging draft tube slows the water, converting velocity head into pressure recovery, and uses the suction head below the runner. Surge absorption is the job of a surge tank, and regulation of flow is done by the guide vanes.

  6. A Francis turbine runner is moved higher above the tailwater level, with head and flow unchanged. What is the main risk?

    1. Cavitation, because the Thoma coefficient falls and the outlet pressure nears the vapour pressure
    2. The specific speed rises and the runner turns into a Pelton wheel, which is better suited to the new setting
    3. The gross head increases with the setting height, so the turbine is overloaded and trips
    4. The penstock friction loss rises with the setting height
    Show answer

    Answer: A — Cavitation, because the Thoma coefficient falls and the outlet pressure nears the vapour pressure

    σ = (Ha − Hv − Hs)/H falls as Hs grows, and the lower pressure at the outlet drops toward the vapour pressure, so cavitation sets in. Setting height does not alter the specific speed formula and has no effect on penstock friction.

  7. A penstock of fixed size feeds a turbine from a reservoir of fixed gross head Hg, and friction loss is proportional to Q². The flow that gives the maximum power is the one at which the friction head loss is

    1. Hg/2
    2. Hg/4
    3. Hg/3
    4. Hg, so that the whole head is used for friction
    Show answer

    Answer: C — Hg/3

    P ∝ Q(Hg − kQ²); setting dP/dQ = 0 gives Hg = 3kQ², so hf = kQ² = Hg/3. At hf = Hg the net head is zero and so is the power.

  8. The purpose of a surge tank on a long penstock is to

    1. filter debris and floating matter out of the water before it enters the turbine
    2. absorb the pressure surges caused by sudden changes in flow (water hammer)
    3. raise the net head at the turbine
    4. store enough water to carry the plant through the dry season
    Show answer

    Answer: B — absorb the pressure surges caused by sudden changes in flow (water hammer)

    A surge tank gives a free surface that lets the water column move and limits the pressure rise when the turbine gates close quickly. Debris is stopped by the trash rack and seasonal storage is the job of the reservoir.

  9. A micro-hydro scheme on a small stream has a low-to-medium head, wide variations of flow, and needs a simple, cheap turbine that is easily made locally. Which turbine is the most natural choice?

    1. A cross-flow (Banki) turbine
    2. A large Francis turbine with a full draft tube
    3. A double-regulated Kaplan turbine
    4. A multi-jet Pelton wheel of high specific speed
    Show answer

    Answer: A — A cross-flow (Banki) turbine

    The cross-flow turbine is simple, tolerates variable flow with good part-load efficiency, and is widely used in micro hydro. A Francis or double-regulated Kaplan is costly and complex, and a Pelton is for high head.

  10. Which statement about a flow-duration curve of a river is correct?

    1. It plots the head available at the dam against the time of day, so that peak hours can be seen
    2. It plots the flow in chronological order through the year, month by month, so that floods and droughts appear in sequence
    3. It plots the turbine efficiency against the flow, so that the best operating point can be found
    4. It plots the flow against the fraction of time for which that flow is equalled or exceeded, and the area under it gives the available volume of water
    Show answer

    Answer: D — It plots the flow against the fraction of time for which that flow is equalled or exceeded, and the area under it gives the available volume of water

    Ordering the flows from largest to smallest against the exceedance fraction gives the curve, and its area is the volume per year. A chronological plot is a hydrograph, not a duration curve.

  11. Which of the following statements about the Pelton wheel are correct?

    1. It is an impulse turbine
    2. The jet strikes the buckets at atmospheric pressure
    3. The flow is regulated by a spear valve in the nozzle
    4. It needs a draft tube to discharge to the tailwater
    Show answer

    Answer: A — It is an impulse turbine; B — The jet strikes the buckets at atmospheric pressure; C — The flow is regulated by a spear valve in the nozzle

    The impulse action, the atmospheric-pressure jet and the spear-valve regulation are standard properties of the Pelton wheel. The wheel runs in air above the tailwater, so it has no draft tube.

  12. Which of the following statements about reaction turbines are correct?

    1. Francis and Kaplan turbines are fitted with a draft tube
    2. A Kaplan turbine keeps good efficiency at part load by adjusting its runner blade angle
    3. A Kaplan turbine has fixed runner blades and is regulated only by the guide vanes
    4. The specific speed of a Francis turbine is higher than that of a Pelton wheel
    Show answer

    Answer: A — Francis and Kaplan turbines are fitted with a draft tube; B — A Kaplan turbine keeps good efficiency at part load by adjusting its runner blade angle; D — The specific speed of a Francis turbine is higher than that of a Pelton wheel

    Both reaction machines use a draft tube, the Kaplan is double-regulated with adjustable runner blades, and Ns rises from Pelton to Francis to Kaplan. A Kaplan with fixed blades would be a propeller turbine.

  13. Which of the following statements about hydropower schemes are correct?

    1. The output of a hydro plant is proportional to the product of flow and net head
    2. A small run-of-river plant has little storage and its output follows the river flow
    3. An isolated micro-hydro plant needs no means of controlling frequency, since river flow is constant
    4. The design flow of a plant is chosen with the help of the flow-duration curve
    Show answer

    Answer: A — The output of a hydro plant is proportional to the product of flow and net head; B — A small run-of-river plant has little storage and its output follows the river flow; D — The design flow of a plant is chosen with the help of the flow-duration curve

    P = ηρgQH, run-of-river output follows the river, and the flow-duration curve fixes the design flow. Load varies in a micro-hydro plant, so a load controller or governor is needed to hold the frequency.

  14. A hydro plant has a flow of 20 m³/s, a gross head of 100 m and a head loss of 5 m in the waterway. The overall efficiency is 0.88, water density 1000 kg/m³ and g = 9.81 m/s². The electrical power output in MW, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 16.4

    Net head = 100 − 5 = 95 m; P = 0.88 × 1000 × 9.81 × 20 × 95 = 16.40 × 10⁶ W = 16.4 MW. Using 100 m gives 17.3 MW. Verified in Python (16.402).

  15. A penstock of length 500 m and diameter 1.5 m carries 6 m³/s of water. Using the Darcy-Weisbach equation hf = fLv²/(2gD) with friction factor f = 0.02 and g = 9.81 m/s², the head loss in metres, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 3.92

    v = Q/(πD²/4) = 6/1.7671 = 3.3953 m/s; hf = 0.02 × 500 × 3.3953²/(2 × 9.81 × 1.5) = 115.28/29.43 = 3.917, i.e. 3.92 m. Verified in Python.

  16. A turbine runs at N = 300 rpm, delivers P = 10 000 kW (shaft power) under a net head H = 100 m. Its specific speed Ns = N√P/H^(5/4) (rpm, kW, m), rounded to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 95

    Ns = 300 × √10 000/100^1.25 = 300 × 100/316.23 = 94.87, i.e. 95. Verified in Python.

  17. A Pelton wheel works under a net head of 400 m. The nozzle coefficient is Cv = 0.98, so the jet speed is V₁ = Cv√(2gH) with g = 9.81 m/s². The bucket speed is 0.46V₁ and the wheel (pitch-circle) diameter is 1.4 m. The wheel speed in rpm, rounded to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 545

    V₁ = 0.98 × √(2 × 9.81 × 400) = 0.98 × 88.59 = 86.82 m/s; u = 0.46 × 86.82 = 39.94 m/s; N = 60u/(πD) = 60 × 39.94/(π × 1.4) = 544.8, i.e. 545 rpm. Verified in Python.

  18. A frictionless Pelton bucket deflects the jet through 180°. The jet speed is 80 m/s, the jet discharge is 0.15 m³/s, water density is 1000 kg/m³, and the bucket speed is 40 m/s. Using P = ρQ u (V₁ − u)(1 + cos 0°), the power developed, in kW, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 480

    P = 1000 × 0.15 × 40 × (80 − 40) × (1 + 1) = 1000 × 0.15 × 40 × 40 × 2 = 480 000 W = 480 kW, which equals the jet kinetic power ½ρQV₁² = 0.5 × 1000 × 0.15 × 6400 = 480 kW, so the ideal efficiency is 100 %.