--- title: "Pelton turbine design (1.8 m³/s, 605 m)" subject: "Hydro Power" chapter: 7 key: "semester_6/hydro_power:pelton-design-605m-1-8" source: https://www.bctengineer.com/q/semester_6/hydro_power/pelton-design-605m-1-8 site: BCT Engineering --- # Pelton turbine design (1.8 m³/s, 605 m) Hydro Power · Chapter 7: Hydro-Electric Machines - Asked in: 2070 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/pelton-design-605m-1-8 - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7 ## Question Design a pelton turbine with following data: Overall efficiency = 87%, Discharge (Qd) = 1.8 m³/s, Net head (H) = 605 m. ## Short answer Single-jet Pelton wheel: power 9.29 MW at 500 rpm (12 poles). Jet diameter 147 mm, pitch circle diameter 1.91 m, jet ratio 13.07. Specific speed 16.1. 22 buckets; bucket width 0.73 m, depth 0.18 m. ## Answer A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets. Of the two nearest synchronous speeds (500 and 600 rpm), 500 rpm is chosen because it gives a jet ratio nearer 12. **Data:** $Q = 1.8$ m³/s, $H = 605$ m, $\eta_o = 0.87$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47). #### 1. Power developed $$ P = \eta_o \rho g Q H = 0.87 \times 9.81 \times 1.8 \times 605 = 9294.3\ \text{kW} $$ #### 2. Jet velocity and bucket (peripheral) velocity $$ \begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 605} = 108.95\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 108.95 = 106.77\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 108.95 = 50.12\ \text{m/s} \end{aligned} $$ #### 3. Jet diameter (single jet) $$ \begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 1.8}{\pi \times 106.77}} = 0.1465\ \text{m} \approx 147\ \text{mm} \end{aligned} $$ #### 4. Speed of the wheel Trial with jet ratio $m = 12$: $D = 12d = 1.758$ m, so $$ N = \frac{60u}{\pi D} = \frac{60 \times 50.12}{\pi \times 1.758} = 544.4\ \text{rpm} $$ Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/12 = 500$ rpm (12 poles). #### 5. Wheel (pitch circle) diameter and jet ratio $$ \begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 50.12}{\pi \times 500} = 1.914\ \text{m} \\ m &= \frac{D}{d} = \frac{1.914}{0.1465} = 13.07 \end{aligned} $$ $m$ lies within 10 to 14, so a **single-jet** wheel is satisfactory. #### 6. Specific speed check $$ N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{500 \times 96.41}{3000.5} = 16.1 $$ This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice. #### 7. Number of buckets (Tygun's formula) $$ z = 15 + \frac{D}{2d} = 15 + \frac{13.07}{2} = 21.53 \approx 22 $$ #### 8. Bucket size (usual proportions) - Bucket width $B \approx 5d = 0.733$ m - Bucket depth $T \approx 1.2d = 0.176$ m - Bucket radial length $L \approx 2.5d = 0.366$ m #### Summary of design | Item | Value | |---|---| | Power | 9.29 MW | | Speed | 500 rpm | | Number of jets | 1 | | Jet diameter $d$ | 147 mm | | Wheel diameter $D$ | 1.91 m | | Jet ratio $m$ | 13.07 | | Specific speed $N_s$ | 16.1 | | Number of buckets | 22 | | Bucket width × depth | 0.73 m × 0.18 m | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the power developed by the Pelton turbine calculated? A. P = rho g Q H with no efficiency B. P = eta x Q x H / g C. P = overall efficiency x rho g Q H = 0.87 x 9.81 x 1.8 x 605 = 9294.3 kW D. P = eta x 0.5 rho Q V1^2 with V1 = sqrt(2gH) only, ignoring head Answer: C. Output power is the hydraulic power times the overall efficiency. (Key point: Power developed) 2. How are the jet velocity V1 and bucket velocity u found? A. V1 = sqrt(2gH) with no coefficient, and u = V1 B. V1 = phi sqrt(2gH) and u = Cv sqrt(2gH) C. V1 = sqrt(gH) and u = 2 V1 D. V1 = Cv sqrt(2gH) = 106.77 m/s and u = phi sqrt(2gH) = 50.12 m/s Answer: D. The jet velocity uses the velocity coefficient Cv (0.98), the bucket velocity uses the speed ratio phi (0.46), both times sqrt(2gH). (Key point: Jet and bucket velocities) 3. How is the jet diameter d found for a single jet? A. From Q = (pi/4) d^2 V1, so d = sqrt(4Q/(pi V1)) = 0.1465 m (147 mm) B. From Q = d V1, so d = Q/V1 C. From Q = pi d^2 u, using the bucket velocity D. From P = pi d^2 V1, using the power Answer: A. The discharge equals jet area times jet velocity. (Key point: Jet diameter) 4. How are the wheel speed and diameter fixed? A. Use any speed and accept any m, because only d matters B. Pick a synchronous speed N = 120f/p (500 rpm), then D = 60u/(pi N) = 1.914 m, giving m = D/d = 13.07, within 10 to 14 C. Take D = 12d always and N follows with no synchronous restriction D. Choose N from Ns = 100 and ignore m Answer: B. The speed must suit the generator, and the resulting jet ratio D/d must lie in 10 to 14 for good efficiency. (Key point: Speed and jet ratio) 5. What do the specific speed check and bucket count show? A. Ns = 16.1 is above 30, so several jets are needed; z = 15 + m = 26 buckets B. Ns must equal 100 exactly; z = D/d buckets C. Ns = N sqrt(P)/H^(5/4) = 16.1, inside the single-jet range 8.5 to 30; z = 15 + m/2 = 21.53, so 22 buckets D. Ns is irrelevant to Pelton choice; z = 2m = 26 buckets Answer: C. The Pelton wheel suits low specific speed (about 8.5 to 30 for a single jet); bucket number follows Tygun's formula. 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