Pelton, Francis and Kaplan turbines

Heads and efficiencies, impulse versus reaction turbines, Pelton, Francis and Kaplan construction and velocity triangles, draft tubes, cavitation and specific speed, with Pelton and Francis design examples.

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Why it matters

Hydraulic turbines convert the energy of water stored at height into shaft power, at efficiencies above 90 %. Choosing between Pelton, Francis and Kaplan machines for a site, sizing the runner, and predicting performance at part load are core tasks for hydro-power engineers, and the same velocity-triangle and specific-speed reasoning carries straight over to pumps, fans and compressors.

Key ideas

Heads and efficiencies. The gross head is the difference between headrace and tailrace levels; the net head H is what reaches the turbine inlet after penstock losses. Power supplied by the water is ρgQH.

  • Hydraulic efficiency η_h = power delivered to the runner ÷ ρgQH.
  • Mechanical efficiency η_m = shaft power ÷ runner power (bearing and disc friction).
  • Volumetric efficiency η_v = water actually acting on the runner ÷ total supplied (leakage).
  • Overall efficiency η_o = shaft power ÷ ρgQH ≈ η_h × η_m × η_v.

Impulse versus reaction. In an impulse turbine all the available head is turned into jet kinetic energy in a nozzle; the runner works at atmospheric pressure and only changes the jet's direction (Pelton). In a reaction turbine, part of the pressure energy is converted inside the runner, which runs full of water under pressure in a closed casing (Francis, Kaplan). The degree of reaction is the fraction of the energy transfer that comes from the pressure change in the runner.

Pelton wheel (high head, low flow, low specific speed).

  • One or more nozzles with spear valves make jets at V₁ = C_v·√(2gH), C_v ≈ 0.97–0.99.
  • Double-cup buckets split the jet and turn it through about 160–165°; a full 180° would throw water onto the next bucket.
  • Best efficiency when bucket speed u ≈ 0.45–0.47 V₁ (ideally 0.5 V₁ with no friction).
  • Jet ratio m = D/d (wheel pitch diameter ÷ jet diameter), typically 10 to 24; number of buckets is often estimated as about 15 + m/2.
  • Governing: the spear valve changes the jet area; a deflector diverts the jet quickly on load rejection to avoid water hammer from closing the spear too fast.

Francis turbine (medium head, medium flow).

  • Water enters a spiral casing, passes adjustable guide vanes (wicket gates) that set the inlet angle α₁, flows radially inward through the runner and leaves axially into a draft tube.
  • Usually designed for radial (or axial) discharge, V_w2 = 0, so the work per unit weight is V_w1·u₁/g and no leaving whirl is wasted.
  • Governing is by rotating the guide vanes.

Kaplan turbine (low head, large flow, high specific speed).

  • Axial flow through a propeller-type runner with typically four to eight blades; u₁ = u₂ at any radius.
  • Both guide vanes and runner blades are adjustable (double regulation), so efficiency stays high over a wide range of load; a fixed-blade propeller turbine has a peaked efficiency curve.

Draft tube. A diverging tube from the reaction-turbine outlet to the tailrace. It lets the turbine be set above tailwater without losing that head, and it recovers much of the exit kinetic energy by reducing velocity, creating a pressure below atmospheric at the runner outlet. That low pressure is where cavitation risk is greatest; the Thoma cavitation number σ = (H_atm − H_v − H_s)/H must exceed a critical value given by the manufacturer or data book, which limits the suction height H_s.

Specific speed. N_s = N·√P/H^(5/4) is the speed of a geometrically similar turbine that would develop unit power under unit head. It fixes the turbine type for a site (with N in rpm, P in kW, H in m; take exact ranges from your data book): Pelton about 10–35 per jet, Francis about 60–300, Kaplan and propeller about 300–1000. Unit quantities (Q_u = Q/√H, N_u = N/√H, P_u = P/H^1.5) predict how one turbine behaves under a different head.

Formulas

P = η_o · ρ · g · Q · H

  • P = shaft power (W), Q = flow (m³/s), H = net head (m).

V₁ = C_v · √(2gH), u = φ · √(2gH), u = π·D·N / 60

  • V₁ = jet velocity (m/s), φ = speed ratio (–), D = pitch-circle diameter (m), N = speed (rpm).

P_runner = ρ·Q·(V₁ − u)·(1 + k·cosφ_b)·u (Pelton)

  • φ_b = bucket outlet angle measured from the reversed jet direction (180° − deflection), k = V_r2/V_r1 (friction factor).

W/g = (V_w1·u₁ ± V_w2·u₂) / g, η_h = (V_w1·u₁ ± V_w2·u₂) / (g·H)

  • Euler head (m). For radial or axial discharge, V_w2 = 0.

tanα₁ = V_f1 / V_w1, tanθ = V_f1 / (V_w1 − u₁)

  • α₁ = guide-vane angle, θ = runner inlet blade angle for shockless entry.

Q = π·D₁·B₁·V_f1 (Francis, neglecting blade thickness); Q = (π/4)·(D_o² − D_b²)·V_f (Kaplan)

  • B₁ = runner width at inlet (m), D_o, D_b = outer and hub diameters (m).

N_s = N · √P / H^(5/4)

  • N in rpm, P in kW, H in m (metric convention; ranges depend on the unit system used).

Worked examples

Example 1 — Pelton wheel (standard). A Pelton wheel works under a net head of 400 m with 0.5 m³/s, at 750 rpm. Take C_v = 0.98, speed ratio 0.46, bucket deflection 165° and k = 0.9. Find the jet diameter, wheel diameter, power developed by the runner and hydraulic efficiency.

  1. √(2gH) = √(2 × 9.81 × 400) = 88.59 m/s.
  2. V₁ = C_v·√(2gH) = 0.98 × 88.59 = 86.82 m/s; u = 0.46 × 88.59 = 40.75 m/s.
  3. Jet diameter: d = √(4Q/(πV₁)) = √(4 × 0.5/(π × 86.82)) = 0.0856 m.
  4. Wheel diameter: D = 60u/(πN) = 60 × 40.75/(π × 750) = 1.038 m; jet ratio m = 12.1.
  5. φ_b = 180° − 165° = 15°. Power: ρQ(V₁ − u)(1 + k·cosφ_b)·u = 1000 × 0.5 × 46.07 × (1 + 0.9 × 0.9659) × 40.75 = 1.755 × 10⁶ W.
  6. η_h = 1.755 × 10⁶/(1000 × 9.81 × 0.5 × 400) = 1.755/1.962 = 0.894. Answer: d ≈ 85.6 mm, D ≈ 1.04 m, runner power ≈ 1.75 MW, η_h ≈ 89.4 %.

Example 2 — Francis runner (GATE level). A Francis turbine has a runner outer diameter of 1.2 m, runs at 300 rpm under a net head of 60 m, with guide-vane angle 15° at inlet and radial discharge. Hydraulic efficiency is 90 %. Find the runner inlet blade angle, and the specific speed if it passes 3 m³/s at an overall efficiency of 85 %.

  1. u₁ = πD₁N/60 = π × 1.2 × 300/60 = 18.85 m/s.
  2. Radial discharge, so η_h·g·H = V_w1·u₁: V_w1 = 0.9 × 9.81 × 60/18.85 = 28.10 m/s.
  3. Flow component: V_f1 = V_w1·tan15° = 28.10 × 0.2679 = 7.53 m/s.
  4. Blade angle: tanθ = V_f1/(V_w1 − u₁) = 7.53/(28.10 − 18.85) = 0.814, so θ = 39.1°.
  5. Shaft power: P = 0.85 × 9810 × 3 × 60 = 1.501 × 10⁶ W = 1501 kW.
  6. N_s = N√P/H^(5/4) = 300 × √1501/60^1.25 = 300 × 38.74/167.0 = 69.6. Answer: θ ≈ 39.1°; P ≈ 1.50 MW; N_s ≈ 70 (a slow-speed Francis runner).

Common mistakes

  • Using gross head in place of net head.
  • Taking the bucket deflection (165°) as φ in (1 + k·cosφ); φ is 180° minus the deflection.
  • Using the pump specific speed N√Q/H^(3/4) for a turbine, or mixing kW and W in N_s.
  • Forgetting that V_w2 = 0 only for radial or axial discharge; otherwise include it with the correct sign.
  • Assuming the draft tube is only a pipe; it recovers kinetic energy and sets the cavitation limit.
  • Claiming reaction turbines are always more efficient; peak efficiencies are similar, and Pelton wheels keep efficiency better at part load.

For GATE ME

Expect numericals on Pelton jet and wheel dimensions, power and hydraulic efficiency with friction; Francis and Kaplan velocity triangles giving blade angles, power and flow; specific speed and choosing a turbine type; unit quantities and scaling of performance to a new head; and overall efficiency chains. Conceptual MCQs test impulse versus reaction, degree of reaction, functions of the draft tube and guide vanes, and governing. Practise the triangles until you can draw them without hesitation.

Quick check

  1. Which turbine suits a head of 500 m and a small flow?
  2. What does radial discharge mean for the work equation of a Francis runner?
  3. Why do Pelton buckets deflect the jet by about 165° rather than 180°?
  4. What two parts are adjusted in a Kaplan turbine during governing?
  5. Give two functions of a draft tube.

Answers: 1. Pelton wheel; 2. V_w2 = 0, so work per unit weight is V_w1·u₁/g; 3. so the leaving water clears the following bucket; 4. guide vanes and runner blades; 5. allows the turbine to be set above tailwater without losing head, and recovers exit kinetic energy.

Interview questions

All Fluid Mechanics and Fluid Power interview questions

Try answering each one aloud before you open it.

  1. 1.What is a Pelton turbine and where is it typically used?Concept

    A Pelton turbine is an impulse-type water turbine used for high-head, low-flow applications. It is typically used in hydroelectric power plants where water is available at high pressure but low flow rates. The turbine converts the kinetic energy of water jets into mechanical energy by striking the buckets on the wheel.

  2. 2.Explain the working principle of a Francis turbine.Concept

    A Francis turbine is a mixed-flow (inward radial entry, axial exit) reaction turbine for medium heads. Water from a spiral casing passes adjustable guide vanes, which set the inlet whirl and control flow, then flows inward through a runner that is completely full of water under pressure. Both kinetic energy and pressure energy are given up in the runner; the runner is designed for nearly radial or axial discharge so that V_w2 ≈ 0 and the work per unit weight is V_w1·u₁/g. A draft tube at the exit recovers kinetic energy and lets the runner sit above tailwater.

  3. 3.Describe the main features of a Kaplan turbine.Concept

    A Kaplan turbine is a propeller-type reaction turbine with adjustable blades, designed for low-head, high-flow applications. It features a runner with three to six blades that can be adjusted to optimize efficiency for varying flow conditions. This adaptability makes it suitable for rivers and tidal power plants.

  4. 4.Why is a Pelton turbine not suitable for low-head applications?Application

    Jet velocity is √(2gH), so at low head the jets are slow and, for a given power, the flow must be very large. That means many or very large jets, and with bucket speed tied to about 0.46 of jet speed, a huge wheel turning very slowly, which needs an expensive multi-pole generator or gearbox. In specific-speed terms the site demands a high N_s, while a Pelton wheel is a low-N_s machine (about 10–35 per jet). A Francis or Kaplan turbine handles large flows at low head in a far smaller, faster unit.

  5. 5.What happens if the blade angle of a Kaplan turbine is not adjusted properly?Application

    A Kaplan turbine is double-regulated: the runner blade angle is linked to the guide-vane opening through a cam (combinator) so that the relative flow meets the blades without shock at every load. If the blade angle does not match the guide-vane setting, the inlet relative velocity strikes the blades at the wrong angle, causing flow separation, shock losses, outlet whirl that is not recovered, and a sharp efficiency drop. It also raises vibration and cavitation risk, so mis-adjustment shortens runner life as well.

  6. 6.How does the efficiency of a Francis turbine compare to that of a Pelton turbine?Application

    At their design points both reach similar peak efficiencies, typically around 90–95 % for large units, each within its own head range. The difference is at part load: a Pelton wheel, regulated by spear valves and with several jets that can be switched off, keeps a flat efficiency curve down to low loads, whereas a Francis turbine's efficiency falls more steeply away from design flow and it can suffer draft-tube surging. So the choice is driven by head and specific speed first, and by the expected load pattern second.

  7. 7.In what scenario would you choose a Kaplan turbine over a Francis turbine?Application

    A Kaplan turbine is chosen over a Francis turbine in low-head, high-flow scenarios, such as in river or tidal power plants. Its adjustable blades allow it to maintain high efficiency across varying flow conditions, making it ideal for environments where water flow rates fluctuate significantly.

  8. 8.Calculate the power output of a Pelton turbine given a water flow rate of 2 m³/s and a net head of 300 m. Assume 85% overall efficiency.Numerical

    Shaft power P = η_o·ρ·g·Q·H = 0.85 × 1000 × 9.81 × 2 × 300 = 5 003 100 W, i.e. about 5.0 MW. The water power available is ρgQH = 5.886 MW, so about 0.88 MW is lost in the nozzle, buckets, bearings and leaving kinetic energy.

  9. 9.Determine the specific speed of a Francis turbine running at 150 rpm with a flow rate of 10 m³/s under a net head of 50 m, assuming an overall efficiency of 90%.Numerical

    Turbine specific speed uses power, N_s = N√P/H^(5/4) with P in kW. Shaft power P = 0.9 × 9810 × 10 × 50 = 4 414 500 W = 4414.5 kW, so √P = 66.44. H^(5/4) = 50^1.25 = 132.96. N_s = 150 × 66.44/132.96 ≈ 75, which lies in the slow-speed Francis range. The pump form N√Q/H^(3/4) should not be used for turbines.

  10. 10.What are the advantages of using a reaction turbine like the Francis or Kaplan over an impulse turbine like the Pelton?Application

    At low and medium heads, reaction turbines handle large flows in a compact, fast-running unit because the runner is full of water and uses its whole periphery, giving higher specific speed and a cheaper generator. A draft tube lets them use the head between the runner and tailwater and recover exit kinetic energy. Kaplan turbines with adjustable blades also keep high efficiency over a wide load range. The drawbacks are cavitation risk, the need for careful setting relative to tailwater, and (for Francis) poorer part-load efficiency than a Pelton wheel.

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