Pelton, Francis and Kaplan turbines

Heads and efficiencies, Euler turbine equation, Pelton, Francis and Kaplan turbines with velocity triangles, draft tube, cavitation and specific-speed selection.

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Why it matters

Hydraulic turbines convert the head of stored water into shaft power at 90 %+ efficiency, and choosing between Pelton, Francis and Kaplan is a classic design decision driven by head, flow and speed. The same Euler equation and velocity-triangle method carries over to pumps, compressors and the turbine side of a turbocharger. GATE uses these machines to test velocity triangles, efficiencies and specific speed.

Key ideas

  • Heads. Gross head H_g is the difference between headrace and tailrace levels. Net head H = H_g − penstock losses is what the turbine receives at inlet.
  • Efficiencies.
    • Hydraulic η_h = power developed by the runner / water power (ρgQH).
    • Mechanical η_m = shaft power / runner power.
    • Volumetric η_v = water actually passing through the runner / water supplied.
    • Overall η_o = shaft power / ρgQH = η_h·η_m (·η_v).
  • Euler turbine equation. Work per unit mass = V_w1·u₁ − V_w2·u₂. Best efficiency is usually obtained with radial or axial (whirl-free) discharge, V_w2 = 0, so the exit kinetic energy is minimal.
  • Impulse vs reaction. In an impulse turbine (Pelton) all available head is converted to jet velocity in the nozzle, and the runner works at atmospheric pressure. In a reaction turbine (Francis, Kaplan) part of the pressure drop occurs in the runner, which runs full of water inside a casing. The degree of reaction R = (pressure-energy change in runner)/(total energy change in runner).
  • Pelton wheel.
    • High head (roughly above 250 m), low flow; tangential flow; one to six jets.
    • Jet velocity V₁ = C_v·√(2gH), with C_v ≈ 0.97–0.99. Flow is regulated by a spear (needle) valve; a deflector diverts the jet on sudden load loss to avoid water hammer.
    • Double-hemispherical buckets with a central splitter turn the jet through about 165°.
    • Theoretical optimum bucket speed u = V₁/2; in practice speed ratio u/√(2gH) ≈ 0.43–0.47.
    • Jet ratio m = D/d (wheel pitch diameter / jet diameter) is typically about 12–14 or more; it sets the number of buckets.
  • Francis turbine.
    • Medium head (roughly 30–300 m), mixed flow: water enters radially inward through a spiral casing and adjustable guide vanes (wicket gates) and leaves axially.
    • Guide-vane angle α₁ sets the inlet whirl; the runner blade inlet angle β₁ must match the relative velocity.
    • Flow ratio V_f1/√(2gH) ≈ 0.15–0.3; speed ratio u₁/√(2gH) ≈ 0.6–0.9.
  • Kaplan turbine.
    • Low head (roughly below 30–60 m), large flow, axial flow.
    • A few (3–8) runner blades whose pitch is adjusted together with the guide vanes (double regulation), so the efficiency curve is flat over a wide load range. A propeller turbine has fixed blades.
    • Flow area = π/4·(D_o² − D_b²), with D_o tip and D_b hub diameter.
  • Draft tube. A gradually diverging pipe from the reaction-turbine exit to the tailrace. It lets the runner be set above tailrace level without losing that height, and it recovers exit kinetic energy by diffusion. It lowers the runner-exit pressure below atmospheric, which raises the risk of cavitation.
  • Cavitation. If the local pressure falls to the vapour pressure, vapour bubbles form and collapse violently, causing pitting, noise, vibration and loss of efficiency — worst at runner outlets and blade tips. It is controlled through the Thoma cavitation factor σ: the plant σ must exceed a critical σ_c for that specific speed (take σ_c from design charts).
  • Specific speed and unit quantities.
    • Specific speed N_s = N·√P/H^(5/4) is the speed of a geometrically similar turbine producing 1 kW under 1 m head.
    • It is the main selection parameter. In metric units (N rpm, P kW, H m), approximate ranges are: Pelton about 10–35 per jet, Francis about 50–300, Kaplan about 300–1000. Textbooks give slightly different limits.
    • Unit speed N_u = N/√H, unit discharge Q_u = Q/√H and unit power P_u = P/H^1.5 let you predict performance at another head for the same turbine.

Formulas

P_water = ρ·g·Q·H P_shaft = η_o·ρ·g·Q·H Power (W); ρ (kg/m³), Q (m³/s), H net head (m).

W = V_w1·u₁ ± V_w2·u₂ (J/kg) η_h = W/(g·H) Euler work per kg; + if V_w2 opposes u₂ (Pelton), − if it is in the same direction. u = π·D·N/60 (m/s).

V₁ = C_v·√(2gH) Pelton jet velocity; C_v nozzle velocity coefficient.

W = (V₁ − u)·(1 + k·cos φ)·u Pelton work per kg; φ = 180° − bucket deflection angle; k = V_r2/V_r1 (friction factor; k = 1 if frictionless).

η_h,max = (1 + k·cos φ)/2 at u = V₁/2 Ideal maximum Pelton hydraulic efficiency.

Q = π·D₁·B₁·V_f1 (Francis) Q = (π/4)·(D_o² − D_b²)·V_f (Kaplan) Flow through runner; B₁ inlet width (m); blade thickness neglected.

N_s = N·√P / H^(5/4) Specific speed; N (rpm), P shaft power (kW), H (m).

N_u = N/√H, Q_u = Q/√H, P_u = P/H^(3/2) Unit quantities.

Worked examples

Example 1 (standard) — Pelton wheel. Given: net head H = 300 m, Q = 0.5 m³/s, C_v = 0.98, speed ratio 0.46, bucket deflection 165°, relative-velocity factor k = 0.9, N = 500 rpm. Find the wheel diameter, jet diameter, power developed and hydraulic efficiency.

  1. √(2gH) = √(2 × 9.81 × 300) = 76.72 m/s.
  2. Jet velocity V₁ = 0.98 × 76.72 = 75.19 m/s; bucket speed u = 0.46 × 76.72 = 35.29 m/s.
  3. Wheel diameter: D = 60u/(πN) = 60 × 35.29/(π × 500) = 1.348 m.
  4. Jet diameter: d = √(4Q/(πV₁)) = √(4 × 0.5/(π × 75.19)) = 0.0920 m (jet ratio D/d ≈ 14.6).
  5. φ = 180° − 165° = 15°; W = (75.19 − 35.29)(1 + 0.9 cos 15°)(35.29) = 39.90 × 1.8693 × 35.29 = 2632 J/kg.
  6. Power developed by the runner = ρQW = 1000 × 0.5 × 2632 = 1.316 × 10⁶ W.
  7. η_h = W/(gH) = 2632/(9.81 × 300) = 0.894. Answer: D ≈ 1.35 m, d ≈ 92 mm, runner power ≈ 1.32 MW, η_h ≈ 89.4 %.

Example 2 (GATE level) — Francis runner. Given: outer diameter D₁ = 1.0 m, inlet width B₁ = 0.12 m, N = 300 rpm, Q = 1.2 m³/s, guide-vane angle α₁ = 10°, radial discharge, net head H = 32 m, overall efficiency 85 %. Find the runner blade inlet angle, hydraulic efficiency, shaft power and specific speed.

  1. V_f1 = Q/(πD₁B₁) = 1.2/(π × 1.0 × 0.12) = 3.183 m/s.
  2. V_w1 = V_f1/tan α₁ = 3.183/tan 10° = 18.05 m/s.
  3. u₁ = πD₁N/60 = π × 1.0 × 300/60 = 15.71 m/s.
  4. tan β₁ = V_f1/(V_w1 − u₁) = 3.183/(18.05 − 15.71) = 3.183/2.344 → β₁ = 53.6°.
  5. Radial discharge → V_w2 = 0, so W = V_w1·u₁ = 18.05 × 15.71 = 283.6 J/kg.
  6. η_h = 283.6/(9.81 × 32) = 0.903.
  7. Shaft power P = 0.85 × 1000 × 9.81 × 1.2 × 32 = 320.2 kW.
  8. N_s = 300 × √320.2/32^1.25 = 300 × 17.89/76.11 = 70.5 → in the Francis range. Answer: β₁ ≈ 53.6°, η_h ≈ 90.3 %, P ≈ 320 kW, N_s ≈ 70.5.

Common mistakes

  • Using gross head instead of net head.
  • Taking u = V₁/2 when the problem gives a speed ratio based on √(2gH).
  • Using P in W instead of kW in N_s, or using H^(3/2) (unit power) instead of H^(5/4).
  • For Francis, forgetting that radial discharge means V_w2 = 0, not V_f2 = 0.
  • For Kaplan, using the full tip area instead of the annulus area between hub and tip.
  • Confusing hydraulic and overall efficiency when going from runner power to shaft power.

For GATE ME

Expect: Pelton jet velocity, bucket speed and efficiency; Francis inlet triangles and blade angles with radial discharge; Kaplan flow from hub and tip diameters; power from ρgQHη; specific speed and turbine selection; unit quantities when the head changes; and the role of the draft tube and cavitation. Practise one complete triangle-based problem for each turbine type.

Quick check

  1. Which turbine suits a 500 m head and small flow?
  2. What is the condition for maximum hydraulic efficiency of a Pelton wheel (ideal)?
  3. A Kaplan plant has Q = 3 m³/s, H = 20 m, η_o = 0.9. What is the shaft power?
  4. A turbine runs at 150 rpm, giving 500 kW under 30 m. What is N_s?
  5. Why is a draft tube used? Answers: 1. Pelton. 2. u = V₁/2. 3. 0.9 × 1000 × 9.81 × 3 × 20 = 529.7 kW. 4. 150 × √500/30^1.25 = 47.8. 5. To recover exit kinetic energy and allow the runner to be set above tailrace without losing head.

Interview questions

All Fluid Mechanics and Turbomachinery 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 hydroelectric power generation. It is typically used in locations where water is available at high pressure and low flow rates, such as mountainous regions. The turbine converts the kinetic energy of water jets into mechanical energy by striking the buckets on the wheel's periphery.

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

    A Francis turbine is a mixed-flow reaction turbine for medium heads, roughly 30–300 m. Water from the spiral casing passes through adjustable guide vanes, which give it whirl and partly convert pressure to velocity. It then enters the runner radially inward and leaves axially. Pressure keeps falling inside the runner, so the runner extracts both pressure and kinetic energy, giving torque by the Euler equation, ideally with whirl-free discharge. A draft tube recovers exit kinetic energy, and the guide vanes regulate flow as the load changes.

  3. 3.Describe the main differences between Kaplan and Francis turbines.Concept

    Kaplan turbines are axial-flow reaction turbines, while Francis turbines are mixed-flow reaction turbines. Kaplan turbines have adjustable blades and are suitable for low-head, high-flow applications, whereas Francis turbines have fixed blades and are used for medium-head applications. Kaplan turbines are more efficient in variable flow conditions due to their adjustable blades, while Francis turbines are more efficient in stable flow conditions.

  4. 4.Why is a draft tube used in a Francis turbine?Application

    A draft tube is used in a Francis turbine to recover the kinetic energy of the water exiting the turbine. It allows the water to exit at a lower velocity, reducing energy losses and increasing the overall efficiency of the turbine. The draft tube also helps maintain a low pressure at the turbine exit, which is essential for the proper functioning of the turbine.

  5. 5.What happens if the jet of a Pelton turbine is not correctly aligned or regulated?Application

    The jet must be tangential to the pitch circle and hit the bucket splitter centrally. If it is misaligned, water splashes off one half, the bucket deflection is uneven, side thrust and vibration appear, and efficiency falls. Flow is regulated with a spear (needle) valve, which changes jet area while keeping jet velocity, and so the velocity ratio, nearly constant. A jet deflector handles sudden load rejection without causing water hammer in the penstock.

  6. 6.How does cavitation affect the performance of a Kaplan turbine?Application

    Cavitation occurs when the local pressure in the turbine falls below the vapor pressure of water, leading to the formation of vapor bubbles. In a Kaplan turbine, cavitation can cause damage to the blades, reduce efficiency, and lead to vibrations and noise. To minimize cavitation, the turbine must be designed to operate within specific pressure and flow conditions, and the installation must ensure proper submergence of the turbine.

  7. 7.Calculate the specific speed of a turbine with a power output of 500 kW, a head of 30 m, and a rotational speed of 150 rpm. Which type is it?Numerical

    N_s = N·√P / H^(5/4), with N in rpm, P in kW and H in m. √500 = 22.36 and 30^1.25 = 70.21, so N_s = 150 × 22.36/70.21 ≈ 47.8. That is at the low end of the Francis range, just above multi-jet Pelton values. At this head, a slow-speed Francis runner would normally be chosen.

  8. 8.What is the role of guide vanes in a Francis turbine?Application

    Guide vanes in a Francis turbine control the flow of water entering the runner. They adjust the angle and velocity of the incoming water to optimize the energy transfer to the runner blades. By regulating the flow, guide vanes help maintain efficient operation across varying load conditions and prevent cavitation by ensuring proper flow alignment.

  9. 9.Explain why Kaplan turbines are preferred for low-head applications.Application

    Kaplan turbines are preferred for low-head applications because they are designed to handle high flow rates efficiently. Their adjustable blades allow them to maintain high efficiency even when the flow conditions vary. This adaptability makes them suitable for locations where water levels fluctuate, ensuring consistent power generation.

  10. 10.A Pelton wheel receives a jet at 50 m/s and its buckets move at 20 m/s. The buckets deflect the jet through 165° with no friction. Find the hydraulic efficiency.Numerical

    Take φ = 180° − 165° = 15° and k = 1. Work per kg = (V − u)(1 + cos φ)·u = 30 × 1.966 × 20 = 1179.5 J/kg. The jet's kinetic energy per kg is V²/2 = 1250 J/kg. So η_h = 2u(V − u)(1 + cos φ)/V² = 1179.5/1250 ≈ 0.944, or 94.4 %. The ideal maximum would occur at u = V/2 = 25 m/s.

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