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Steam turbines and nozzles

Mechanical Engineering (B.E./B.Tech) · Thermodynamics and thermal engineering · free study notes for exams, placements and interviews

In short

A steam nozzle converts the enthalpy (pressure energy) of steam into a high-velocity jet; a steam turbine then converts that jet's kinetic energy into shaft rotation through rows of moving blades. Together they are the expansion end of every thermal power station, and their analysis is pure applied thermodynamics plus velocity triangles. In India this is the working heart of NTPC, Tata Power, Adani Power and JSW stations, of BHEL's and Siemens Energy's turbine shops, of sugar, paper and refinery cogeneration sets, and of nuclear plants under NPCIL. Universities examine nozzle sizing, critical pressure ratio and blade-work derivations every year, and GATE ME regularly asks a numerical on optimum blade speed ratio or maximum blade efficiency.

Explained

Flow through a steam nozzle

Nozzle efficiency, friction and supersaturated flow

Impulse and reaction turbines

Velocity diagrams and blade work (single-stage impulse)

Compounding, multi-stage machines and the reheat factor

Losses, governing, and plant practice in India

Key points

Formulas and examples

C₂ = √(2(h₁ − h₂) + C₁²) ; with C₁ ≈ 0, C₂ = 44.72 √Δh [C in m/s, Δh in kJ/kg]
ṁ = A C / v [ṁ kg/s, A m², C m/s, v m³/kg - specific volume at that section]
p*/p₁ = [2/(n + 1)]^{n/(n−1)} ; n = 1.3 superheated, 1.135 dry saturated, n = 1.035 + 0.1x wet
ṁ_max = A_throat √{ n (p₁/v₁) [2/(n+1)]^{(n+1)/(n−1)} } (choked flow)
η_N = (h₁ − h₂′)/(h₁ − h₂) = (C_actual/C_ideal)² ; velocity coefficient k = C_actual/C_ideal = √η_N
ΔC_w = C_w1 + C_w2 = C₁cos α₁ + C₂cos α₂ (exit whirl opposing) ; axial thrust F_a = ṁ(C_a1 − C_a2)
Blade work W = ṁ U ΔC_w [W in watts if ṁ kg/s, U m/s, ΔC_w m/s] ; blade efficiency η_b = 2 U ΔC_w / C₁²
Impulse: ρ_opt = cos α₁/2, η_b,max = (cos²α₁/2)(1 + K C) where C = cos β₂/cos β₁ ; ideal η_b,max = cos²α₁
n moving rows (velocity compounded): ρ_opt = cos α₁/(2n) ; work in rows ratio 3:1 for ideal 2 rows
50% reaction: ρ_opt = cos α₁, η_b,max = 2cos²α₁/(1 + cos²α₁) ; ΔC_w = 2C₁cos α₁ − U
Blade friction loss = ṁ (V₁² − V₂²)/2 ; Reheat factor RF = Σ Δh_s,stage / Δh_s,overall
Nozzle/blade height: ṁ v = π D h ε C_a , ε = fraction of periphery admitted

Example 1 - Critical pressure, throat area and exit area of a convergent-divergent nozzle. Steam at 10 bar, dry saturated (h₁ = 2,778 kJ/kg) expands to 1.5 bar; ṁ = 0.5 kg/s; inlet velocity negligible; nozzle efficiency 100% up to the throat and 90% for the divergent part (isentropic drops from steam tables/Mollier chart).

  1. n = 1.135 → critical ratio = 0.577 → p\* = 5.77 bar.
  2. Isentropic drop to throat ≈ 103 kJ/kg → C_t = 44.72√103 ≈ 454 m/s; at the throat v ≈ 0.317 m³/kg, x ≈ 0.962.
  3. A_throat = ṁ v / C = 0.5 × 0.317 / 454 = 3.49 × 10⁻⁴ m² = 3.49 cm² → throat diameter = √(4 × 3.49/π) ≈ 2.1 cm.
  4. Isentropic drop 10 bar → 1.5 bar ≈ 330 kJ/kg, so the divergent part's isentropic share is 330 − 103 = 227 kJ/kg, of which 0.9 × 227 = 204 kJ/kg is realised; total actual drop = 103 + 204 = 307 kJ/kg → C₂ = 44.72√307 ≈ 784 m/s.
  5. Actual exit state: x₂ ≈ 0.90 at 1.5 bar (friction reheat raises it) → v₂ ≈ 0.90 × 1.159 ≈ 1.043 m³/kg.
  6. A₂ = 0.5 × 1.043 / 784 = 6.65 × 10⁻⁴ m² = 6.65 cm² → exit diameter ≈ 2.9 cm. (Divergence check: A₂/A_t ≈ 1.9.)

Example 2 - Single-stage impulse turbine, full velocity diagram. Nozzle angle α₁ = 20°, C₁ = 600 m/s, U = 250 m/s, blade exit angle β₂ = 25°, K = V₂/V₁ = 0.9, ṁ = 5 kg/s.

  1. C_w1 = 600 cos 20° = 600 × 0.9397 = 563.8 m/s; C_a1 = 600 sin 20° = 600 × 0.3420 = 205.2 m/s.
  2. Relative inlet: V_w1 = 563.8 − 250 = 313.8 m/s → V₁ = √(313.8² + 205.2²) = √(98,470 + 42,107) = √140,577 ≈ 374.9 m/s; β₁ = tan⁻¹(205.2/313.8) = 33.2°.
  3. V₂ = 0.9 × 374.9 = 337.4 m/s; V_w2 = 337.4 cos 25° = 337.4 × 0.9063 = 305.8 m/s; C_a2 = 337.4 sin 25° = 142.6 m/s.
  4. ΔC_w = V_w1 + V_w2 = 313.8 + 305.8 = 619.6 m/s (equivalently C_w1 + C_w2 with C_w2 = 305.8 − 250 = 55.8 m/s → 563.8 + 55.8 = 619.6 m/s ✓).
  5. Power = ṁ U ΔC_w = 5 × 250 × 619.6 = 774,500 W ≈ 774.5 kW.
  6. η_b = 2UΔC_w/C₁² = 2 × 250 × 619.6/600² = 309,800/360,000 = 86.1%.
  7. Axial thrust = 5 × (205.2 − 142.6) = 313 N. Blade friction loss = 5 × (374.9² − 337.4²)/2 = 5 × (140,577 − 113,839)/2 ≈ 66.8 kW.

Example 3 - Optimum blade speed ratio and maximum efficiency. For the same blading (α₁ = 20°, β₂ = 25°, β₁ = 33.2°, K = 0.9):

  1. ρ_opt = cos 20°/2 = 0.9397/2 = 0.470 → for C₁ = 600 m/s the best blade speed is U = 281.9 m/s (the 250 m/s above is slightly below optimum).
  2. C = cos β₂/cos β₁ = 0.9063/0.8368 = 1.083 → η_b,max = (cos²20°/2)(1 + 0.9 × 1.083) = (0.8830/2)(1.9747) = 0.4415 × 1.9747 = 87.2%.
  3. Ideal symmetric frictionless limit = cos²20° = 88.3% - so friction costs only about 1 percentage point here.
  4. If the same jet drove a two-row Curtis stage, ρ_opt = cos 20°/4 = 0.235 → U ≈ 141 m/s, i.e. half the rotor speed for the same jet.

Example 4 - 50% reaction (Parsons) stage. Fixed and moving blades identical, α₁ = 20°, blade mean diameter 1.2 m, rotor speed 3,000 rpm, ṁ = 12 kg/s, stage working at the optimum ratio.

  1. U = πDN/60 = π × 1.2 × 3,000/60 = 188.5 m/s.
  2. ρ_opt = cos 20° = 0.9397 → C₁ = U/0.9397 = 200.6 m/s.
  3. ΔC_w = 2C₁cos α₁ − U = 2 × 200.6 × 0.9397 − 188.5 = 376.9 − 188.5 = 188.4 m/s.
  4. Power = 12 × 188.5 × 188.4 = 426,100 W ≈ 426 kW per stage.
  5. η_b,max = 2cos²20°/(1 + cos²20°) = 2 × 0.8830/1.8830 = 1.7660/1.8830 = 93.8%.
  6. Stage isentropic drop needed: work/kg = UΔC_w = 188.5 × 188.4 = 35,520 J/kg ≈ 35.5 kJ/kg → with 93.8% blade efficiency and 95% nozzle efficiency the stage needs about 39.9 kJ/kg of isentropic drop, hence the large number of stages in a reaction machine.

Example 5 - Reheat factor and internal efficiency. A turbine expands from 40 bar, 400 °C to 0.1 bar in four stages; the stage isentropic drops are 265, 272, 279 and 285 kJ/kg; the overall isentropic drop (single expansion line) is 1,069 kJ/kg; stage efficiency 0.85.

  1. Σ stage drops = 265 + 272 + 279 + 285 = 1,101 kJ/kg.
  2. RF = 1,101/1,069 = 1.030.
  3. Internal (turbine) efficiency = 0.85 × 1.030 = 0.876 → 87.6%.
  4. Actual work = 0.876 × 1,069 = 936 kJ/kg; at ṁ = 100 kg/s, output ≈ 93.6 MW.
FeatureImpulse stage50% reaction stage
Pressure dropOnly in nozzlesIn fixed and moving rows
Pressure on moving bladesConstantFalls
Blade shapeSymmetrical, constant areaAerofoil, converging passage
AdmissionPartial possibleMust be full
Optimum U/C₁cos α₁/2cos α₁
Max blade efficiencycos²α₁2cos²α₁/(1 + cos²α₁)
Stages for a given dropFewerMany more
Axial thrustSmallLarge - needs balancing
Typical useControl/first stage, small setsHP and IP sections of large sets
CompoundingPressure dropVelocity levelNote
Velocity (Curtis)One nozzle set, all at onceVery high jet, absorbed in 2-3 rowsCompact, cheaper, lower efficiency
Pressure (Rateau)In each nozzle ringModerate throughoutEfficient, long machine
Pressure-velocityLarge per wheelHigh per wheelShorter than Rateau
Reaction (Parsons)Small, in every rowLowHighest stage efficiency, most stages

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