- Practice · 8 marks
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 at angle to the blade direction. The relative velocity at inlet is (angle ) and at outlet (angle ); the absolute outlet velocity is .
Inlet Outlet
C1 Vr2 C2
/|\ \ /|
/ | \ Ca1 \ / | Ca2
/a1| b1\Vr1 \b2|
+---U---+---- ---U--+--
|<--Cw1 ----->| |<--Vw2-->|
Work done
Take the blade speed as positive. The whirl of the relative velocity at inlet is . At outlet the relative velocity has whirl in the direction opposite to . With blade friction, , and .
The change of whirl velocity is , so the work per kg is:
Diagram efficiency
With blade speed ratio :
Condition for maximum
that is, . The second derivative is negative, so this is a maximum.
Maximum efficiency
For a symmetrical blade (, so ) with no friction ():
For this is . With friction () the value is lower. Because the whole enthalpy drop in one nozzle gives a very high , the optimum blade speed 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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