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Impulse turbine blade work and diagram efficiency

Short answer

For a single-stage impulse turbine, work per kg is w = U(C1 cosα1 - U)(1 + kc) with c = cosβ2/cosβ1. Diagram efficiency ηd = 2ρ(cosα1 - ρ)(1+kc), with ρ=U/C1. Maximum occurs at U = C1 cosα1/2, giving ηd,max = (cos^2α1)/2 (1+kc). For a symmetrical frictionless blade, ηd,max = cos^2α1.

For a single-stage impulse turbine draw the velocity diagrams and derive expressions for the work done per kg of steam and the diagram (blade) efficiency. Show that the efficiency is maximum when the blade speed is half of the whirl component of the nozzle exit velocity, and find the maximum value for a symmetrical blade with no friction.

Velocity diagrams

Steam leaves the nozzle at C1C_1 at angle α1\alpha_1 to the blade direction. The relative velocity at inlet is Vr1V_{r1} (angle β1\beta_1) and at outlet Vr2V_{r2} (angle β2\beta_2); the absolute outlet velocity is C2C_2.

 Inlet                      Outlet
      C1                       Vr2  C2
     /|\                        \   /|
    / | \ Ca1                    \ / | Ca2
   /a1| b1\Vr1                    \b2|
  +---U---+----                 ---U--+--
  |<--Cw1 ----->|            |<--Vw2-->|

Work done

Take the blade speed UU as positive. The whirl of the relative velocity at inlet is Vw1=C1cos⁡α1−UV_{w1} = C_1\cos\alpha_1 - U. At outlet the relative velocity has whirl Vw2=Vr2cos⁡β2V_{w2} = V_{r2}\cos\beta_2 in the direction opposite to UU. With blade friction, Vr2=kVr1V_{r2} = k V_{r1}, and Vr1cos⁡β1=Vw1V_{r1}\cos\beta_1 = V_{w1}.

The change of whirl velocity is Vw1+Vw2V_{w1} + V_{w2}, so the work per kg is:

w=U(Vw1+Vw2)=U(C1cos⁡α1−U)(1+kcos⁡β2cos⁡β1)w = U (V_{w1} + V_{w2}) = U (C_1\cos\alpha_1 - U)\left(1 + k\frac{\cos\beta_2}{\cos\beta_1}\right)

Diagram efficiency

ηd=wC12/2=2U(C1cos⁡α1−U)(1+kc)C12,c=cos⁡β2cos⁡β1\eta_d = \frac{w}{C_1^2/2} = \frac{2U (C_1\cos\alpha_1 - U)(1 + kc)}{C_1^2}, \quad c = \frac{\cos\beta_2}{\cos\beta_1}

With blade speed ratio ρ=U/C1\rho = U/C_1:

ηd=2ρ(cos⁡α1−ρ)(1+kc)\eta_d = 2\rho(\cos\alpha_1 - \rho)(1 + kc)

Condition for maximum

dηddρ=2(1+kc)(cos⁡α1−2ρ)=0  ⇒  ρ=cos⁡α12\frac{d\eta_d}{d\rho} = 2(1 + kc)(\cos\alpha_1 - 2\rho) = 0 \;\Rightarrow\; \rho = \frac{\cos\alpha_1}{2}

that is, U=12C1cos⁡α1=12Cw1U = \tfrac12 C_1\cos\alpha_1 = \tfrac12 C_{w1}. The second derivative is negative, so this is a maximum.

Maximum efficiency

ηd,max=cos⁡2α12(1+kc)\eta_{d,max} = \frac{\cos^2\alpha_1}{2}(1 + kc)

For a symmetrical blade (β1=β2\beta_1 = \beta_2, so c=1c = 1) with no friction (k=1k = 1):

ηd,max=cos⁡2α1\boxed{\eta_{d,max} = \cos^2\alpha_1}

For α1=20∘\alpha_1 = 20^\circ this is 0.8830.883. With friction (k<1k < 1) the value is lower. Because the whole enthalpy drop in one nozzle gives a very high C1C_1, the optimum blade speed U≈0.47C1U \approx 0.47C_1 is still very high. This is why a single-stage impulse turbine runs at a very high rotor speed, and why velocity compounding or pressure compounding is used.

✓ Answer checked by Claude Opus 5.5 (high effort)

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