--- title: "Parsons turbine stage work and efficiency" subject: "Fluid Machines" chapter: 5 key: "semester_5/fluid_machines:parsons-reaction-turbine-numerical" source: https://www.bctengineer.com/q/semester_5/fluid_machines/parsons-reaction-turbine-numerical site: BCT Engineering --- # Parsons turbine stage work and efficiency Fluid Machines · Chapter 5: Steam turbine and Hydraulic machine - Asked in: year not given, 8 marks - Video: [Parsons Turbine](https://www.youtube.com/watch?v=j_NAsDbLbfA) — Sidramappa Alur - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/fluid_machines/parsons-reaction-turbine-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-5 ## Question In a stage of a Parsons (50 % reaction) turbine the mean blade diameter is 1.0 m and the speed is 3000 rpm. The fixed and the moving blades have the same exit angle of 20 degrees (measured from the direction of blade motion). Steam leaves the fixed blades at 250 m/s. The steam flow rate is 10 kg/s. Draw the velocity diagrams and determine (a) the inlet angle of the moving blades, (b) the power developed in the stage, and (c) the blade efficiency of the stage. ## Short answer Blade speed u = 157.1 m/s. For symmetrical Parsons triangles, moving blade inlet angle θ = tan⁻¹(V sin20° / (V cos20° – u)) = 47.7°. Work per kg = u (V cos20° + (V cos20° – u)) = 49.13 kJ/kg, power = 10 kg/s × 49.13 = 491 kW. Blade efficiency = 49.13 / (49.13 + V₂²/2) = 88.0%. ## Answer **Given:** $D=1.0$ m, $N=3000$ rpm, $\alpha_1=\phi=20^\circ$, $V_1=250$ m/s, $m=10$ kg/s. For a Parsons turbine the moving blades are the mirror image of the fixed blades, so the velocity triangles are symmetrical: $V_{r2}=V_1$, $V_{r1}=V_2$ and the exit angle of moving blades $\phi=\alpha_1=20^\circ$. #### Blade speed $$ u=\frac{\pi DN}{60}=\frac{\pi\times1.0\times3000}{60}=157.1\ \text{m/s} $$ #### Inlet triangle (leaving fixed blades, entering moving blades) $$ V_{w1}=V_1\cos20^\circ=234.9\ \text{m/s},\qquad V_{f1}=V_1\sin20^\circ=85.5\ \text{m/s} $$ $$ V_{r1}=\sqrt{(V_{w1}-u)^2+V_{f1}^2}=\sqrt{77.8^2+85.5^2}=115.6\ \text{m/s} $$ #### (a) Inlet angle of moving blades $$ \tan\theta=\frac{V_{f1}}{V_{w1}-u}=\frac{85.5}{77.8}\ \Rightarrow\ \theta=47.7^\circ $$ #### Outlet triangle ```text Inlet (moving blade) Outlet V1 Vf V2 Vf \ | Vr1 \ | Vr2 = V1 20 deg\ |/ \ | / -------*----> u ----*-----> u (Vr2 backward) ``` $V_{r2}=V_1=250$ m/s at $\phi=20^\circ$. Exit whirl (backward): $$ V_{w2}=V_{r2}\cos20^\circ-u=234.9-157.1=77.8\ \text{m/s} $$ Exit velocity of steam: $V_2=\sqrt{V_{w2}^2+V_f^2}=115.6$ m/s, (equal to $V_{r1}$, as expected). #### (b) Power Work per kg of steam: $$ W=u\,(V_{w1}+V_{w2})=157.1\times312.8=49.13\ \text{kJ/kg} $$ $$ P=mW=10\times49.13=491\ \text{kW} $$ #### (c) Blade efficiency Energy supplied $=$ work done $+$ kinetic energy carried away by steam leaving the stage $\left(\frac12V_2^2\right)$: $$ \frac12V_2^2=\frac{115.6^2}{2}=6.69\ \text{kJ/kg} $$ $$ \eta_b=\frac{W}{W+\frac12V_2^2}=\frac{49.13}{49.13+6.69}=88.0\% $$ The kinetic energy leaving this stage is not wasted when the next stage follows; it is used again by the next fixed blades. **Answer:** (a) $\theta=47.7^\circ$; (b) $P=491$ kW (49.13 kJ/kg); (c) $\eta_b=88.0\%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the outlet triangle of a Parsons stage be taken as the mirror image of the inlet triangle? A. Because the moving blades are identical to the fixed blades, giving Vr2 = V1 and Vr1 = V2 B. Because the blade speed equals the steam speed C. Because there is no pressure drop in the moving blades D. Because the axial velocity is zero Answer: A. For 50 % reaction the moving blades are the same shape as the fixed blades, so the triangles are symmetrical: Vr2 = V1 and V2 = Vr1. (Key point: Parsons velocity triangles) 2. What is the mean blade speed for D = 1.0 m and N = 3000 rpm? A. 314.2 m/s B. 78.5 m/s C. 157.1 m/s D. 250 m/s Answer: C. u = pi D N / 60 = pi x 1.0 x 3000 / 60 = 157.1 m/s. (Key point: Blade speed) 3. How is the inlet angle of the moving blades found, and what is its value? A. tan(theta) = Vf1/Vw1 = 20 degrees B. tan(theta) = Vf1/(Vw1 - u) = 85.5/77.8, theta = 47.7 degrees C. tan(theta) = Vf1/(Vw1 + u), theta = 20.0 degrees D. tan(theta) = (Vw1 - u)/Vf1, theta = 42.3 degrees Answer: B. The relative velocity at inlet has whirl component Vw1 - u = 77.8 m/s and flow component 85.5 m/s, giving theta = 47.7 degrees. (Key point: Inlet angle of moving blades) 4. With work W = u(Vw1 + Vw2) = 49.13 kJ/kg and exit kinetic energy V2^2/2 = 6.69 kJ/kg, what are the power (m = 10 kg/s) and blade efficiency? A. 491 kW and 88.0 % B. 491 kW and 12.0 % C. 49.1 kW and 88.0 % D. 558 kW and 88.0 % Answer: A. P = 10 x 49.13 = 491 kW, and efficiency = W/(W + V2^2/2) = 49.13/55.82 = 88.0 %. 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