Steam turbines: impulse and reaction staging

Impulse and reaction steam-turbine stages, compounding, velocity triangles, blade work, efficiency and thrust, optimum blade-speed ratios and degree of reaction.

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Why it matters

Steam turbines generate most of the world's thermal and nuclear electricity and drive ship propellers, compressors and boiler-feed pumps. How the enthalpy drop is split between nozzles and moving blades (impulse vs reaction), and how it is divided over many stages (compounding), determines blade speed, efficiency, axial thrust and machine size. The velocity-triangle method is the same one used for hydraulic turbines and compressors.

Key ideas

  • Energy conversion. Steam expands from high to low enthalpy. In nozzles (or fixed blades) enthalpy turns into kinetic energy; moving blades then turn the jet's momentum into torque. Work per kg follows the Euler equation, u·(V_w1 ± V_w2), with u the mean blade speed.
  • Impulse stage. The whole stage pressure drop occurs in the nozzles; pressure is constant across the moving blades. Moving blades are symmetrical with a constant-area passage. With friction, the relative velocity falls slightly: V_r2 = k·V_r1, with blade velocity coefficient k ≈ 0.85–0.95.
  • Reaction stage. Pressure drops in both fixed and moving blades. Moving-blade passages converge like nozzles, so the relative velocity increases (V_r2 > V_r1) and the reaction of the accelerating steam adds to the impulse force. The degree of reaction R = enthalpy drop in moving blades / enthalpy drop in the stage. The Parsons turbine uses R = 50 % with identical fixed and moving blade profiles, so the triangles are symmetrical: α₁ = β₂, β₁ = α₂, V₁ = V_r2, V_r1 = V₂.
  • Why compounding? A single de Laval impulse stage expanding steam from boiler to condenser produces jet speeds over 1000 m/s and needs blade speeds near half of that — impractical rotor speeds (about 30 000 rpm) and stresses, and a large leaving loss. The drop is therefore split:
    • Pressure compounding (Rateau): several impulse stages in series, each with its own nozzles and a share of the pressure drop. This gives moderate jet speeds and good efficiency; common in HP and IP sections.
    • Velocity compounding (Curtis): one nozzle expansion, then two or three rows of moving blades with fixed guide blades between them that only redirect the flow. Compact, but lower efficiency; used as the first (control) stage or in small drives.
    • Pressure-velocity compounding: a combination of the two.
    • Reaction turbines need many stages, because the drop per stage is small; their blade height grows along the turbine as the steam volume increases.
  • Optimum blade-speed ratio ρ = u/V₁.
    • Single impulse stage with symmetric blades: η_b,max = (cos²α₁/2)·(1 + k) at ρ = cos α₁/2; with k = 1, η_b,max = cos²α₁.
    • Two-row Curtis stage: optimum ρ = cos α₁/4.
    • Parsons (50 %) stage: optimum ρ = cos α₁, with η_b,max = 2cos²α₁/(1 + cos²α₁).
    • So for the same jet speed, reaction blading runs faster, and for the same blade speed it takes a smaller enthalpy drop per stage.
  • Efficiencies. Nozzle efficiency = actual jet KE / isentropic enthalpy drop. Blade (diagram) efficiency = blade work / energy supplied to the blades. Stage efficiency = nozzle efficiency × blade efficiency (for an impulse stage). In multistage turbines, friction reheats the steam and makes later stages produce a little more work. The reheat factor (about 1.03–1.05) = Σ stage isentropic drops / overall isentropic drop.
  • Axial thrust. The change in axial (flow) velocity, plus any pressure difference across the moving blades, gives a thrust. It is small in impulse stages but large in reaction turbines, which need balance pistons or double-flow designs.
  • Other practical points. Partial admission (nozzles round only part of the circumference) in HP stages; governing by throttling or by nozzle-group control; losses from leaving velocity, tip leakage, disc friction and wetness in the last LP stages, which also cause blade erosion.

Formulas

V₁ = √(2·Δh_nozzle) (V in m/s, Δh in J/kg), or with nozzle efficiency V₁ = √(2·η_n·Δh_s) Nozzle exit velocity; inlet velocity neglected.

V_w1 = V₁·cos α₁, V_f1 = V₁·sin α₁, tan β₁ = V_f1/(V_w1 − u) Inlet triangle; angles from the plane of rotation; u = π·D·N/60 (m/s).

W = u·(V_w1 + V_w2) (J/kg) Blade work per kg when V_w2 points opposite to u (use − if in the same direction).

F_a = ṁ·(V_f1 − V_f2) Axial thrust (N) on an impulse stage with no pressure difference across the moving blades.

η_b = 2·W/V₁² Impulse blade efficiency.

ρ_opt = cos α₁/2, η_b,max = (cos²α₁/2)·(1 + k) Single-row impulse stage, symmetric blades (β₁ = β₂).

η_b = W / (V₁²/2 + (V_r2² − V_r1²)/2) Reaction stage blade efficiency; for Parsons this equals W/(V₁² − V_r1²/2).

ρ_opt = cos α₁, η_b,max = 2·cos²α₁/(1 + cos²α₁) 50 % reaction (Parsons) stage.

R = Δh_moving / Δh_stage Degree of reaction.

Worked examples

Example 1 (standard) — single impulse stage. Given: steam leaves the nozzle at V₁ = 600 m/s at α₁ = 20°; blade speed u = 250 m/s; symmetric blades (β₂ = β₁); blade velocity coefficient k = 0.9; ṁ = 1 kg/s. Find the blade angle, work, blade efficiency and axial thrust.

  1. V_w1 = 600 cos 20° = 563.8 m/s; V_f1 = 600 sin 20° = 205.2 m/s.
  2. V_w1 − u = 313.8 m/s → tan β₁ = 205.2/313.8 → β₁ = 33.2°.
  3. V_r1 = √(313.8² + 205.2²) = 375.0 m/s; V_r2 = 0.9 × 375.0 = 337.5 m/s.
  4. V_r2 cos β₂ = 337.5 × cos 33.2° = 282.4 m/s → V_w2 = 282.4 − 250 = 32.4 m/s, opposite to u.
  5. W = u(V_w1 + V_w2) = 250 × (563.8 + 32.4) = 149 060 J/kg.
  6. η_b = 2W/V₁² = 298 120/360 000 = 0.828.
  7. V_f2 = 337.5 × sin 33.2° = 184.7 m/s → axial thrust = 1 × (205.2 − 184.7) = 20.5 N. Answer: β₁ ≈ 33.2°, W ≈ 149 kJ/kg (149 kW per kg/s), η_b ≈ 82.8 %, thrust ≈ 20.5 N. Here ρ = 0.417, a little below the optimum cos 20°/2 = 0.470.

Example 2 (GATE level) — Parsons (50 % reaction) stage. Given: α₁ = β₂ = 20°, absolute steam velocity leaving the fixed blades V₁ = 250 m/s, blade speed u = 180 m/s, ṁ = 10 kg/s. Find the work, power, blade efficiency and the maximum possible blade efficiency.

  1. Symmetry: V_r2 = V₁ = 250 m/s and β₂ = 20°.
  2. V_w1 = 250 cos 20° = 234.9 m/s.
  3. V_w2 = V_r2 cos β₂ − u = 234.9 − 180 = 54.9 m/s (opposite to u).
  4. W = u(V_w1 + V_w2) = 180 × 289.8 = 52 170 J/kg → power = 10 × 52 170 = 521.7 kW.
  5. V_r1² = V₁² + u² − 2V₁u cos α₁ = 62 500 + 32 400 − 84 572 = 10 328 → V_r1 = 101.6 m/s.
  6. Energy supplied per kg = V₁² − V_r1²/2 = 62 500 − 5164 = 57 336 J/kg.
  7. η_b = 52 170/57 336 = 0.910.
  8. Maximum at u = V₁ cos α₁: η_b,max = 2cos²20°/(1 + cos²20°) = 1.766/1.883 = 0.938. Answer: W ≈ 52.2 kJ/kg, P ≈ 522 kW, η_b ≈ 91.0 %, η_b,max ≈ 93.8 % (at u ≈ 235 m/s).

Common mistakes

  • Assuming the absolute velocity is unchanged across moving blades; it is the relative velocity that is constant (frictionless impulse) or increases (reaction).
  • Using 1 + k without symmetric blades — the general factor is (1 + k·cos β₂/cos β₁).
  • Mixing angle references (plane of rotation vs axial direction).
  • Using the impulse optimum (cos α/2) for a reaction stage (cos α).
  • Forgetting to convert kJ/kg to J/kg in V = √(2Δh).
  • Saying a reaction turbine has no nozzles — its fixed blades act as nozzles.

For GATE ME

Expect: impulse-stage velocity triangles, blade work, blade efficiency and axial thrust; optimum blade-speed ratio and maximum efficiency for impulse and 50 % reaction stages; degree of reaction; power from ṁ·Δh·η; nozzle velocity from enthalpy drop; and conceptual questions on compounding and reheat factor. Practise drawing combined inlet and outlet triangles on a common base.

Quick check

  1. In an impulse stage, where does the pressure drop occur?
  2. What is the optimum blade-speed ratio for a single-row impulse stage with α₁ = 20°?
  3. A turbine has ṁ = 8 kg/s and an actual enthalpy drop of 400 kJ/kg. What is its power?
  4. What is the degree of reaction of a Parsons turbine?
  5. Why is compounding used? Answers: 1. Entirely in the nozzles. 2. cos 20°/2 = 0.47. 3. 3200 kW. 4. 50 %. 5. To keep rotor speed and jet velocities practical and reduce leaving losses by dividing the enthalpy drop over several stages or rows.

Interview questions

All Fluid Mechanics and Turbomachinery interview questions

Try answering each one aloud before you open it.

  1. 1.What is an impulse steam turbine?Concept

    An impulse steam turbine is a type of turbine where the steam expands in nozzles and the high-velocity jets of steam are directed onto the turbine blades. The blades change the direction of the steam flow, which results in a change in momentum and thus imparts a force on the blades, causing them to rotate.

  2. 2.What is a reaction steam turbine?Concept

    A reaction steam turbine is a type of turbine where the steam expands both in the stationary and moving blades. The pressure drop occurs over both sets of blades, and the reaction force generated by the steam's acceleration through the moving blades causes the rotor to turn.

  3. 3.Explain the difference between impulse and reaction staging in steam turbines.Concept

    In impulse staging, the steam expands in nozzles and the kinetic energy is converted into mechanical work on the rotor blades. In reaction staging, the steam expands in both the fixed and moving blades, and the pressure drop occurs in both, providing a reaction force that contributes to the rotor's motion. Impulse turbines typically have fewer stages and are used for high-pressure applications, while reaction turbines have more stages and are used for lower pressure applications.

  4. 4.Why are impulse turbines often used in high-pressure stages?Application

    Impulse turbines are used in high-pressure stages because they can efficiently handle the high energy levels and pressure drops. The design allows for the conversion of high-pressure steam into high-velocity jets, which can be effectively managed by the impulse blades. This makes them suitable for the initial stages of steam turbines where the steam pressure is highest.

  5. 5.What happens if the steam velocity is too low relative to the blade speed in an impulse turbine?Application

    Blade efficiency depends on the speed ratio ρ = u/V₁. For a single-row stage the optimum is about cos α₁/2. If V₁ is too low for the blade speed, ρ rises above the optimum, the whirl change across the blades falls, and blade work and efficiency drop. In the limit, the blades run as fast as the steam's whirl and do no work. If V₁ is too high, ρ is too low and a large leaving velocity is wasted. Designers match the stage enthalpy drop to the blade speed, which is one reason for compounding.

  6. 6.How does the blade design differ between impulse and reaction turbines?Concept

    Impulse moving blades are roughly symmetrical, with equal inlet and outlet angles and a constant-area passage. They only turn the steam, so pressure and relative speed stay nearly constant (apart from friction). Reaction moving blades are aerofoil-shaped with a converging passage, like a nozzle. Pressure falls across them and the relative velocity increases, which adds a reaction force. In a Parsons (50 %) turbine, fixed and moving blades have the same profile. Reaction blades need tight tip clearances because a pressure difference across them drives leakage.

  7. 7.What is the role of nozzles in an impulse turbine?Concept

    Nozzles in an impulse turbine are responsible for converting the thermal energy of steam into kinetic energy. They accelerate the steam to a high velocity, directing it onto the turbine blades. This high-velocity steam is what imparts the force needed to turn the turbine rotor.

  8. 8.Steam leaves the nozzles of an impulse stage at 400 m/s with a mass flow rate of 5 kg/s. How much kinetic energy is supplied to the blades per second, and what limits the stage power?Numerical

    The jets carry kinetic energy at ½·ṁ·V₁² = 0.5 × 5 × 400² = 400 000 W, or 400 kW. That is the energy supplied to the blades, not the stage output. The power actually developed is ṁ·u·(V_w1 ± V_w2), which equals the blade efficiency times 400 kW. The blade efficiency depends on the speed ratio u/V₁, the nozzle angle and blade friction, and the leaving kinetic energy and friction are lost, so the output is always less than 400 kW.

  9. 9.Why is it important to maintain a pressure drop across the blades in a reaction turbine?Application

    Maintaining a pressure drop across the blades in a reaction turbine is crucial because it allows for the continuous acceleration of steam through the moving blades. This acceleration generates a reaction force that contributes to the rotor's motion. Without a pressure drop, the reaction force would be insufficient, reducing the turbine's efficiency and power output.

  10. 10.Steam leaves the moving blades of a turbine stage with a relative velocity of 250 m/s at 20° to the plane of rotation, directed opposite to the blade motion. The blade speed is 150 m/s. What is the absolute exit velocity?Numerical

    The absolute velocity is the vector sum V₂ = V_r2 + u. Since V_r2 points backward at 20° to the blade motion, V₂² = V_r2² + u² − 2·V_r2·u·cos 20° = 62 500 + 22 500 − 70 477 = 14 523, so V₂ ≈ 120.5 m/s. Equivalently, the whirl component is 250 cos 20° − 150 = 84.9 m/s opposite to u, and the flow component is 250 sin 20° = 85.5 m/s. You can never simply add the speeds arithmetically unless the two vectors are parallel.

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