--- title: "Parsons reaction turbine blade efficiency" subject: "Turbo Machines" chapter: 2 key: "semester_7/turbo_machines:reaction-turbine-parsons-efficiency" source: https://www.bctengineer.com/q/semester_7/turbo_machines/reaction-turbine-parsons-efficiency site: BCT Engineering --- # Parsons reaction turbine blade efficiency Turbo Machines · Chapter 2: Velocity Vector Diagram - Asked in: year not given, 6 marks - Video: [Don't Struggle with Turbine Efficiency! Parson's Reaction Turbine Simplified](https://www.youtube.com/watch?v=LGFyWFBg090) — Tech-Shodhak Rishi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/reaction-turbine-parsons-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-2 ## Question What is the degree of reaction? For a Parsons (50% reaction) turbine stage, draw the velocity diagrams and derive the expression for the blade (diagram) efficiency. Show that the efficiency is maximum when $U/C_1 = \cos\alpha_1$ and find the maximum value. ## Short answer Degree of reaction R = Δh_rotor / Δh_stage. For a 50% reaction (Parsons) stage, work w = U(2C1 cosα1 - U). Blade efficiency ηb = 2ρ(2 cosα1 - ρ)/(1+2ρ cosα1 - ρ^2) with ρ=U/C1. Maximum at ρ = cosα1 yields ηb,max = 2 cos^2α1/(1+cos^2α1). ## Answer #### Degree of reaction The **degree of reaction** $R$ is the ratio of the enthalpy drop in the moving blades to the enthalpy drop in the whole stage: $$ R = \frac{h_2 - h_3}{h_1 - h_3} = \frac{\Delta h_{rotor}}{\Delta h_{stage}} $$ $R = 0$ for impulse; $R = 0.5$ for a Parsons turbine, where fixed and moving blades are identical and mirror images of each other. #### Velocity diagrams (50% reaction) Because fixed and moving blades are identical: $\alpha_1 = \beta_2$ and $\beta_1 = \alpha_2$, so $V_{r2} = C_1$ and $C_2 = V_{r1}$. ```text Inlet Outlet C1 Vr2=C1 C2=Vr1 /|\ \ /| / | \ Vr1 \ / | Ca /a1|b1\ \b2/ | +---U---+ +--U--+ ``` #### Work done per kg $$ C_{w1} = C_1\cos\alpha_1, \quad V_{w2} = V_{r2}\cos\beta_2 = C_1\cos\alpha_1 $$ $$ w = U(C_{w1} + V_{w2} - U) = U(2C_1\cos\alpha_1 - U) $$ #### Blade efficiency Energy supplied = enthalpy drop in the fixed blades ($C_1^2/2$, entry velocity neglected) + enthalpy drop in the moving blades $(V_{r2}^2 - V_{r1}^2)/2$. This equals the work done plus the exit kinetic energy, $w + C_2^2/2$. With $C_2^2 = V_{r1}^2 = C_1^2 - 2UC_1\cos\alpha_1 + U^2$: $$ \text{Supply} = w + \frac{C_2^2}{2} = \frac{C_1^2}{2} + UC_1\cos\alpha_1 - \frac{U^2}{2} $$ $$ \eta_b = \frac{2U(2C_1\cos\alpha_1 - U)}{C_1^2 + 2UC_1\cos\alpha_1 - U^2} = \frac{2\rho(2\cos\alpha_1 - \rho)}{1 + 2\rho\cos\alpha_1 - \rho^2}, \quad \rho = \frac{U}{C_1} $$ #### Condition for maximum Putting $d\eta_b/d\rho = 0$ gives $\rho = \cos\alpha_1$, i.e. $U = C_1\cos\alpha_1$. $$ \boxed{\eta_{b,max} = \frac{2\cos^2\alpha_1}{1 + \cos^2\alpha_1}} $$ For $\alpha_1 = 20^\circ$, $\eta_{b,max} = 2(0.883)/(1.883) = 0.938$. This is higher than the single-stage impulse maximum $\cos^2\alpha_1 = 0.883$, and the optimum blade speed is about twice as high, so reaction stages are used in multistage turbines at moderate speed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Degree of reaction is the ratio of: A. Enthalpy drop in moving blades to that in the whole stage B. Enthalpy drop in fixed blades to that in moving blades C. Work done to kinetic energy supplied D. Blade speed to steam speed Answer: A. R = rotor enthalpy drop / stage enthalpy drop; 0 for impulse and 0.5 for Parsons. (Key point: Degree of reaction) 2. Because fixed and moving blades are identical in a Parsons stage: A. Vr2 = Vr1 and C2 = C1 B. alpha1 = beta1 C. Vr2 = C1 and C2 = Vr1 D. U = C1 Answer: C. Mirror-image blades give alpha1 = beta2 and beta1 = alpha2, hence Vr2 = C1 and C2 = Vr1. (Key point: Parsons velocity triangles) 3. At what blade speed ratio is Parsons efficiency maximum? A. U/C1 = cos(alpha1)/2 B. U/C1 = 2 cos(alpha1) C. U/C1 = sin(alpha1) D. U/C1 = cos(alpha1) Answer: D. Setting d(eta)/d(rho) = 0 gives rho = cos(alpha1), twice the impulse optimum. (Key point: Optimum speed ratio) 4. The maximum blade efficiency of a Parsons stage is: A. 2cos^2(a1)/(1 + cos^2(a1)) B. cos^2(a1) C. cos^2(a1)/2 D. 1 - cos^2(a1) Answer: A. Substituting rho = cos(a1) gives 2cos^2 a1/(1 + cos^2 a1), e.g. 0.938 at 20 degrees, higher than impulse. 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