Steam turbines: impulse and reaction staging

Impulse and reaction stages, velocity triangles, blade work and diagram efficiency, optimum blade-speed ratios, degree of reaction and pressure/velocity compounding, with impulse and 50 % reaction stage examples.

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Why it matters

The Rankine cycle tells you how much enthalpy is available; the turbine blading decides how much of it becomes shaft work. Velocity triangles, blade-speed ratio and degree of reaction explain why a turbine needs dozens of stages, why the high-pressure end uses impulse stages and the rest reaction stages, and where axial thrust and leakage losses come from. The same velocity-triangle reasoning carries over to gas turbines, compressors, turbochargers and pumps.

Key ideas

Energy conversion in a stage. A stage is a row of fixed blades (nozzles or guide vanes) followed by a row of moving blades. In the fixed row, enthalpy drops and steam velocity rises. In the moving row, the change in the steam's tangential (whirl) momentum pushes the blades. The power comes from the change in whirl velocity, not from the absolute speed alone.

Impulse stage. The whole stage enthalpy drop occurs in the nozzles; pressure is constant across the moving blades, and the relative velocity changes only by friction (Vr₂ = k Vr₁, with k ≈ 0.85–0.95). Symmetric blades have β₁ = β₂. With no pressure difference across the rotor, the blade tips need no tight clearance and partial admission is possible.

Reaction stage. Steam expands in both fixed and moving blades; the moving blades act as nozzles, so the relative velocity increases (Vr₂ > Vr₁). Degree of reaction R = enthalpy drop in moving blades ÷ stage enthalpy drop. The common Parsons (50 %) reaction stage has identical fixed and moving blade profiles and symmetric velocity triangles (α₁ = β₂, β₁ = α₂, V₁ = Vr₂, Vr₁ = V₂). Pressure difference across the rotor causes tip leakage and axial thrust, balanced with a dummy piston or opposed flows.

Blade-speed ratio and optimum efficiency.

  • Single-stage impulse: maximum diagram (blade) efficiency cos²α₁ (with k = 1, symmetric blades) at ρ = u/V₁ = cos α₁ / 2.
  • 50 % reaction: maximum 2 cos²α₁ / (1 + cos²α₁) at ρ = cos α₁.
  • So for the same steam speed, a reaction stage needs roughly twice the blade speed, and therefore extracts less enthalpy per stage. More stages are needed, but each is more efficient.

Why compounding? Expanding from boiler to condenser pressure in one nozzle would give steam velocities above 1000 m/s, needing blade speeds near 500 m/s — beyond rotor stress limits and wasteful in leaving loss. Compounding splits the drop.

  • Pressure compounding (Rateau): several impulse stages in series, each with its own nozzles and a share of the pressure drop. Efficient, but long.
  • Velocity compounding (Curtis): one nozzle drop, then two or three rows of moving blades with fixed guide blades between them that only redirect the steam. Absorbs a large drop in a short length but is less efficient. Optimum ρ = cos α₁/(2n) for n moving rows. Often used as the first (control) stage of large turbines with partial-arc admission.
  • Pressure–velocity compounding: Curtis stages in series.
  • Reaction blading: many 50 % stages with gradual pressure drop; dominant in the intermediate- and low-pressure sections.

Losses. Nozzle friction (nozzle efficiency), blade friction (k), leaving loss (V₂²/2), tip leakage, disc friction and windage, partial-admission loss, wetness loss in LP stages, and gland leakage. Stage efficiency = nozzle efficiency × diagram efficiency.

Formulas

V₁ = √(2 Δh_nozzle + V₀²) — nozzle exit velocity (m/s), Δh in J/kg; multiply the ideal value by the velocity coefficient if one is given. V_w1 = V₁ cos α₁, V_f1 = V₁ sin α₁ — whirl and flow components; α measured from the blade-plane (tangential) direction. Vr₁ = √((V_w1 − u)² + V_f1²), tan β₁ = V_f1 / (V_w1 − u) u = π D N / 60 — blade speed (m/s); D: mean blade-ring diameter (m), N: rpm. w = u ΣV_w — work per kg (J/kg). ΣV_w = V_w1 + V_w2 when the exit whirl opposes the blade motion (the usual case), = V_w1 − V_w2 if in the same direction. ΣV_w = (V_w1 − u) + Vr₂ cos β₂ — impulse or reaction, with β measured from the tangential direction. F_t = ṁ ΣV_w — tangential force (N); F_a = ṁ (V_f1 − V_f2) — axial thrust from flow change (N), plus pressure-difference thrust for reaction stages. η_b = 2 u ΣV_w / V₁² — impulse blade (diagram) efficiency. η_b,max = cos²α₁ at u/V₁ = cos α₁ / 2 (impulse, symmetric blades, k = 1); with friction η_b,max = (cos²α₁ / 2)(1 + k). R = Δh_moving / Δh_stage w = u (2 V₁ cos α₁ − u) — 50 % reaction stage with symmetric triangles. η_b = w / (V₁² − Vr₁² / 2) — 50 % reaction stage (energy input = V₁²/2 + (Vr₂² − Vr₁²)/2). η_b,max = 2 cos²α₁ / (1 + cos²α₁) at u/V₁ = cos α₁ — 50 % reaction.

Worked examples

Example 1 (standard, single-stage impulse). Steam leaves the nozzles at V₁ = 600 m/s, α₁ = 20°. Blade speed u = 250 m/s; blades are symmetric (β₂ = β₁) with friction factor k = 0.9. Find the blade inlet angle, work per kg, diagram efficiency and axial thrust per kg/s.

  1. V_w1 = 600 cos 20° = 563.8 m/s; V_f1 = 600 sin 20° = 205.2 m/s.
  2. Vr₁ = √((V_w1 − u)² + V_f1²) = √(313.8² + 205.2²) = 375.0 m/s; β₁ = tan⁻¹(205.2/313.8) = 33.2°.
  3. Vr₂ = 0.9 × 375.0 = 337.5 m/s at β₂ = 33.2°.
  4. ΣV_w = (V_w1 − u) + Vr₂ cos β₂ = 313.8 + 337.5 × 0.8370 = 313.8 + 282.4 = 596.2 m/s.
  5. w = u ΣV_w = 250 × 596.2 = 149 060 J/kg = 149.1 kJ/kg.
  6. η_b = 2w/V₁² = 2 × 149 060/360 000 = 0.828.
  7. V_f2 = 337.5 sin 33.2° = 184.7 m/s; axial thrust = 205.2 − 184.7 = 20.5 N per kg/s.

β₁ ≈ 33.2°, w ≈ 149 kJ/kg, η_b ≈ 82.8 %, axial thrust ≈ 20.5 N per kg/s (Optimum check: u/V₁ = 0.417 against cos 20°/2 = 0.470; the maximum with k = 0.9 would be 0.5 × 0.883 × 1.9 = 0.839.)

Example 2 (GATE level, 50 % reaction stage). A Parsons stage has V₁ = 200 m/s at α₁ = 20° and u = 140 m/s, with symmetric triangles. Find the work per kg, the blade efficiency and the enthalpy drop in each blade row (repeating stage, so the incoming velocity equals V₂ = Vr₁).

  1. V_w1 = 200 cos 20° = 187.94 m/s; V_f1 = 68.40 m/s.
  2. Vr₁ = √((187.94 − 140)² + 68.40²) = √(2298.2 + 4678.6) = 83.53 m/s.
  3. w = u (2 V₁ cos α₁ − u) = 140 × (375.88 − 140) = 140 × 235.88 = 33 023 J/kg ≈ 33.0 kJ/kg.
  4. η_b = w / (V₁² − Vr₁²/2) = 33 023 / (40 000 − 3488.6) = 33 023/36 511 = 0.904.
  5. Fixed row: Δh = (V₁² − V₂²)/2 = (40 000 − 6977.3)/2 = 16 511 J/kg ≈ 16.5 kJ/kg; the moving row drops the same, so the stage total equals w (33.0 kJ/kg).
  6. Optimum for comparison: u/V₁ = cos 20° = 0.940 gives η_b,max = 2 × 0.8830/1.8830 = 0.938.

w ≈ 33.0 kJ/kg, η_b ≈ 90.4 %, ≈ 16.5 kJ/kg per blade row

At their optimum blade-speed ratios, an impulse stage (k = 1) does w = 2u² while a 50 % reaction stage does w = u², so for the same blade speed a reaction stage takes only about half the enthalpy drop — that is why reaction turbines have many stages.

Common mistakes

  • Measuring blade and nozzle angles from the axial direction in one step and the tangential direction in the next. Fix the convention first; here angles are from the tangential (blade-plane) direction.
  • Subtracting whirl components when the exit whirl opposes blade motion. They add.
  • Using kJ/kg with velocities in m/s; V²/2 is in J/kg.
  • Assuming impulse blades have no relative-velocity loss. With friction, Vr₂ < Vr₁.
  • Saying reaction stages need lower blade speed. Their optimum u/V₁ is about twice the impulse value.
  • Thinking velocity compounding is more efficient. It is compact, not efficient.

For GATE ME

Expect velocity-triangle numericals: work, power, blade efficiency, axial thrust and blade angles for single-stage impulse and 50 % reaction stages; optimum blade-speed ratios and maximum efficiencies; nozzle exit velocity from enthalpy drop; and conceptual questions on compounding and degree of reaction. Practise drawing combined inlet and exit triangles on one base and checking which way each whirl component points.

Quick check

  1. What is the optimum u/V₁ for a single-row impulse stage with α₁ = 20°?
  2. What is the degree of reaction of an impulse stage?
  3. An adiabatic nozzle has an isentropic enthalpy drop of 125 kJ/kg and negligible inlet velocity. Find V₁.
  4. Which compounding gives the shortest turbine for a large enthalpy drop?
  5. In a 50 % reaction stage with symmetric blades, which velocities are equal?

Answers: 1. cos 20°/2 ≈ 0.47. 2. Zero. 3. √(2 × 125 000) = 500 m/s. 4. Velocity (Curtis) compounding. 5. V₁ = Vr₂ and Vr₁ = V₂ (α₁ = β₂, β₁ = α₂).

Interview questions

All Thermal Engineering 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 main differences between impulse and reaction turbines.Concept

    In an impulse stage the whole stage pressure drop occurs in the fixed nozzles; pressure is constant across the moving blades, and relative velocity only falls by friction. In a reaction stage, steam also expands in the moving blades, which act as nozzles, so relative velocity rises and a pressure difference exists across the rotor. As a result reaction stages need tighter tip clearances, produce axial thrust that must be balanced, and have an optimum blade-speed ratio about twice that of impulse stages (cos α against cos α/2), so they extract less enthalpy per stage but with higher efficiency.

  4. 4.Why are impulse stages often used at the high-pressure end of a steam turbine?Application

    At the high-pressure end the volume flow is small, so blades are short and tip leakage in a reaction stage, which has a pressure drop across its moving blades, would be a large fraction of the flow. An impulse stage has no pressure difference across the rotor, so it tolerates larger clearances and allows partial-arc admission for governing. A velocity-compounded (Curtis) impulse stage also takes a large enthalpy drop at once, quickly lowering the pressure and temperature that the rest of the casing must withstand.

  5. 5.What happens if the pressure drop per stage in a reaction turbine is made too large?Application

    Steam velocities rise, so for a fixed blade speed the blade-speed ratio moves away from its optimum (about cos α for 50 % reaction) and diagram efficiency falls. The larger pressure difference across the moving blades also raises tip leakage and axial thrust. That is why reaction turbines use many stages with small, roughly equal enthalpy drops.

  6. 6.How does the efficiency of impulse and reaction stages compare?Application

    With ideal blades, a single-row impulse stage has a maximum diagram efficiency of cos²α₁ at u/V₁ = cos α₁/2, while a 50 % reaction stage reaches 2cos²α₁/(1 + cos²α₁) at u/V₁ = cos α₁. For α₁ = 20° that is about 88 % against 94 %. Reaction stages are therefore more efficient per stage but extract less enthalpy each, so they need more stages; impulse stages are preferred where a large drop must be taken compactly or leakage must be avoided.

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

    In an impulse turbine, nozzles play a crucial role by converting the thermal energy of steam into kinetic energy. The nozzles accelerate the steam to a high velocity, directing it onto the turbine blades. This high-velocity steam jet imparts momentum to the blades, causing them to rotate and produce mechanical work.

  8. 8.Steam expands in a nozzle from 10 MPa to 1 MPa with an enthalpy drop of 500 kJ/kg. Find the exit velocity, neglecting the inlet velocity.Numerical

    From the steady-flow energy equation for an adiabatic nozzle, V₂ = √(2Δh + V₁²). With V₁ ≈ 0 and Δh = 500 000 J/kg, V₂ = √(1 000 000) = 1000 m/s. A pressure ratio of 0.1 is far below the critical ratio (about 0.55 for superheated steam), so this nozzle must be convergent–divergent; a nozzle velocity coefficient below 1 would lower the actual value.

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