--- title: "Single-stage impulse turbine blade work and efficiency" subject: "Fluid Machines" chapter: 5 key: "semester_5/fluid_machines:de-laval-turbine-velocity-diagram-numerical" source: https://www.bctengineer.com/q/semester_5/fluid_machines/de-laval-turbine-velocity-diagram-numerical site: BCT Engineering --- # Single-stage impulse turbine blade work and efficiency Fluid Machines · Chapter 5: Steam turbine and Hydraulic machine - Asked in: year not given, 8 marks - Video: [single stage impulse turbine solved numerical](https://www.youtube.com/watch?v=mAsn4Kc3Fvw) — E - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/fluid_machines/de-laval-turbine-velocity-diagram-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-5 ## Question In a single-stage impulse steam turbine (De Laval type) the steam leaves the nozzle at 900 m/s at an angle of 20 degrees to the plane of the wheel. The mean blade speed is 350 m/s and the blades are symmetrical (equiangular). The relative velocity is reduced by 10 % in passing over the blades. The steam flow is 2 kg/s. Draw the velocity diagram and calculate (a) the blade angles, (b) the tangential force on the blades and the power developed, (c) the axial thrust, and (d) the blade efficiency. Also find the blade speed for maximum efficiency. ## Short answer From velocity triangles: blade angles θ = φ = 31.8°. Tangential force m(Vw1+Vw2) = 1884 N, power = 659 kW, axial thrust = 61.6 N. Blade efficiency = 81.4%. Optimum blade speed for maximum efficiency = V1 cosα / 2 = 423 m/s. ## Answer **Given:** $V_1=900$ m/s, $\alpha=20^\circ$, $u=350$ m/s, blades symmetrical ($\theta=\phi$), $k=V_{r2}/V_{r1}=0.9$, $m=2$ kg/s. #### Inlet triangle $$ V_{w1}=V_1\cos20^\circ=845.7\ \text{m/s},\qquad V_{f1}=V_1\sin20^\circ=307.8\ \text{m/s} $$ $$ V_{r1}=\sqrt{(V_{w1}-u)^2+V_{f1}^2}=\sqrt{495.7^2+307.8^2}=583.5\ \text{m/s} $$ #### (a) Blade angles $$ \tan\theta=\frac{V_{f1}}{V_{w1}-u}=\frac{307.8}{495.7}\ \Rightarrow\ \theta=31.8^\circ,\qquad \phi=\theta=31.8^\circ $$ #### Outlet triangle ```text Inlet Outlet V1 Vf1 V2 Vf2 \ | \ | 20deg \| Vr1 Vr2 \ | --------*---------> u <-------*------- u |<-Vw1->| |Vw2| ``` $$ V_{r2}=0.9\times583.5=525.2\ \text{m/s},\quad V_{f2}=V_{r2}\sin\phi=277.0\ \text{m/s} $$ $$ V_{w2}=V_{r2}\cos\phi-u=446.2-350=96.2\ \text{m/s (opposite to blade motion)} $$ The exit whirl is opposite to the blade motion, so it adds to the inlet whirl: $$ V_{w1}+V_{w2}=845.7+96.2=941.9\ \text{m/s} $$ #### (b) Force and power $$ F_t=m(V_{w1}+V_{w2})=2\times941.9=1884\ \text{N} $$ $$ P=F_tu=1884\times350=659\ \text{kW} $$ #### (c) Axial thrust $$ F_a=m(V_{f1}-V_{f2})=2\times(307.8-277.0)=61.6\ \text{N} $$ #### (d) Blade efficiency $$ \eta_b=\frac{\text{work done}}{\text{kinetic energy supplied}}=\frac{2u(V_{w1}+V_{w2})}{V_1^2}=\frac{2\times350\times941.9}{900^2}=81.4\% $$ (Check: $\frac12mV_1^2=810$ kW and $P/\text{KE}=81.4\%$.) #### Blade speed for maximum efficiency For a single-stage impulse turbine with equiangular blades, $\eta_b$ is a maximum at $$ u_{opt}=\frac{V_1\cos\alpha}{2}=\frac{900\times\cos20^\circ}{2}=423\ \text{m/s} $$ and the maximum efficiency (with no friction) is $\cos^2\alpha=88.3\%$. The given 350 m/s is below this value, so the efficiency is lower than the maximum. **Answer:** (a) $\theta=\phi=31.8^\circ$; (b) $F_t=1884$ N, $P=659$ kW; (c) $F_a=61.6$ N; (d) $\eta_b=81.4\%$; $u_{opt}=423$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With V1 = 900 m/s at 20 degrees to the wheel plane, what are the whirl and flow components? A. Vw1 = V1 sin20, Vf1 = V1 cos20 B. Vw1 = V1 cos20 = 845.7 m/s, Vf1 = V1 sin20 = 307.8 m/s C. Both equal 900 m/s D. Vw1 = 307.8 m/s, Vf1 = 845.7 m/s Answer: B. The whirl component lies along the blade motion (cos), and the flow component is axial (sin). (Key point: Inlet velocity components) 2. What does equiangular (symmetrical) blades mean for the blade angles? A. Outlet angle is 90 degrees B. Inlet angle equals outlet angle (31.8 degrees each) C. Outlet angle is twice the inlet angle D. Both angles are 20 degrees Answer: B. Symmetrical blades have theta = phi. (Key point: Equiangular blades) 3. Why are the whirl components added in F = m (Vw1 + Vw2)? A. Exit whirl is opposite to blade motion, so the change in whirl momentum is the sum B. Both are in the same direction as u C. It reduces the axial thrust D. Because friction is zero Answer: A. Vw2 = 96.2 m/s is reversed relative to blade motion, so F = 2 x 941.9 = 1884 N. (Key point: Tangential force) 4. For a single-stage impulse turbine with equiangular blades, the blade speed for maximum efficiency is: A. V1 B. V1 cos(alpha) / 2 C. V1 sin(alpha) / 2 D. 2 V1 cos(alpha) Answer: B. u_opt = 900 x cos20 / 2 = 423 m/s, higher than the given 350 m/s so efficiency is below maximum. (Key point: Optimum blade speed) ## Related in Fluid Machines - [Parsons turbine stage work and efficiency](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Aparsons-reaction-turbine-numerical) — Fluid Machines, chapter 5 - [Kaplan turbine discharge and blade angles](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Akaplan-turbine-numerical) — Fluid Machines, chapter 3 - [Convergent-divergent steam nozzle design](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Asteam-nozzle-numerical) — Fluid Machines, chapter 5 - [Pelton wheel power with penstock losses](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-numerical-penstock) — Fluid Machines, chapter 3 ## Related in other subjects - [Impulse stage: blade angles, work and thrust](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Asingle-stage-impulse-steam-numerical) — Turbo Machines, chapter 2 - [Impulse turbine blade work and diagram efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aimpulse-turbine-diagram-efficiency) — Turbo Machines, chapter 2 - [Parsons stage: work, power and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aparsons-stage-numerical) — Turbo Machines, chapter 2 - [Axial gas turbine stage design calculation](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aaxial-turbine-stage-gas-numerical) — Turbo Machines, chapter 2