--- title: "Impulse turbine blade work and diagram efficiency" subject: "Turbo Machines" chapter: 2 key: "semester_7/turbo_machines:impulse-turbine-diagram-efficiency" source: https://www.bctengineer.com/q/semester_7/turbo_machines/impulse-turbine-diagram-efficiency site: BCT Engineering --- # Impulse turbine blade work and diagram efficiency Turbo Machines · Chapter 2: Velocity Vector Diagram - Asked in: year not given, 8 marks - Video: [Velocity Triangle of Impulse Turbine](https://www.youtube.com/watch?v=fxn3lfHehLc) — BEST MECHANICAL ENGINEERING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/impulse-turbine-diagram-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-2 ## Question 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. ## 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. ## Answer #### Velocity diagrams Steam leaves the nozzle at $C_1$ at angle $\alpha_1$ to the blade direction. The relative velocity at inlet is $V_{r1}$ (angle $\beta_1$) and at outlet $V_{r2}$ (angle $\beta_2$); the absolute outlet velocity is $C_2$. ```text Inlet Outlet C1 Vr2 C2 /|\ \ /| / | \ Ca1 \ / | Ca2 /a1| b1\Vr1 \b2| +---U---+---- ---U--+-- |<--Cw1 ----->| |<--Vw2-->| ``` #### Work done Take the blade speed $U$ as positive. The whirl of the relative velocity at inlet is $V_{w1} = C_1\cos\alpha_1 - U$. At outlet the relative velocity has whirl $V_{w2} = V_{r2}\cos\beta_2$ in the direction opposite to $U$. With blade friction, $V_{r2} = k V_{r1}$, and $V_{r1}\cos\beta_1 = V_{w1}$. The change of whirl velocity is $V_{w1} + V_{w2}$, so the work per kg is: $$ w = U (V_{w1} + V_{w2}) = U (C_1\cos\alpha_1 - U)\left(1 + k\frac{\cos\beta_2}{\cos\beta_1}\right) $$ #### Diagram efficiency $$ \eta_d = \frac{w}{C_1^2/2} = \frac{2U (C_1\cos\alpha_1 - U)(1 + kc)}{C_1^2}, \quad c = \frac{\cos\beta_2}{\cos\beta_1} $$ With blade speed ratio $\rho = U/C_1$: $$ \eta_d = 2\rho(\cos\alpha_1 - \rho)(1 + kc) $$ #### Condition for maximum $$ \frac{d\eta_d}{d\rho} = 2(1 + kc)(\cos\alpha_1 - 2\rho) = 0 \;\Rightarrow\; \rho = \frac{\cos\alpha_1}{2} $$ that is, $U = \tfrac12 C_1\cos\alpha_1 = \tfrac12 C_{w1}$. The second derivative is negative, so this is a maximum. #### Maximum efficiency $$ \eta_{d,max} = \frac{\cos^2\alpha_1}{2}(1 + kc) $$ For a symmetrical blade ($\beta_1 = \beta_2$, so $c = 1$) with no friction ($k = 1$): $$ \boxed{\eta_{d,max} = \cos^2\alpha_1} $$ For $\alpha_1 = 20^\circ$ this is $0.883$. With friction ($k < 1$) the value is lower. Because the whole enthalpy drop in one nozzle gives a very high $C_1$, the optimum blade speed $U \approx 0.47C_1$ 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Work per kg for the impulse stage is: A. U (Vw1 - Vw2) B. C1^2/2 C. U (Vw1 + Vw2), the change of whirl times blade speed D. Vr1 - Vr2 Answer: C. Relative whirl reverses direction on the blade, so the change of whirl is Vw1 + Vw2. (Key point: Work done) 2. At what blade speed is the diagram efficiency maximum? A. U = C1 B. U = half of C1 cos(alpha1), i.e. half the whirl component of nozzle velocity C. U = 2 C1 cos(alpha1) D. U = C1 cos(alpha1) Answer: B. d(eta)/d(rho) = 0 gives rho = cos(alpha1)/2, so U = Cw1/2. (Key point: Optimum blade speed) 3. For a symmetrical frictionless blade the maximum diagram efficiency is: A. cos^2(alpha1)/2 B. 2 cos^2(alpha1) C. 1 D. cos^2(alpha1) Answer: D. eta_max = (cos^2 a1 /2)(1 + kc) with k = c = 1 gives cos^2 a1. (Key point: Maximum efficiency) 4. Why are velocity or pressure compounding used with impulse turbines? A. The optimum blade speed is low relative to steam velocity, so a single stage would need very high rotor speed B. Efficiency is always above 100% otherwise C. Nozzles cannot expand steam D. Blades cannot turn the flow Answer: A. Optimum U is about 0.47 C1, so single-stage rotors would spin extremely fast; compounding splits the drop over stages. 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