--- title: "Pelton turbine design (2 m³/s, 605 m, Ns 100)" subject: "Hydro Power" chapter: 7 key: "semester_6/hydro_power:pelton-design-605m-ns100" source: https://www.bctengineer.com/q/semester_6/hydro_power/pelton-design-605m-ns100 site: BCT Engineering --- # Pelton turbine design (2 m³/s, 605 m, Ns 100) Hydro Power · Chapter 7: Hydro-Electric Machines - Asked in: 2071 Magh, 9 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-ns100 - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7 ## Question Design a Pelton turbine with the following data: Design discharge = 2 m³/s, Net head = 605 m, Overall efficiency = 87%, Specific speed = 100. Assume the ratios and coefficients from their standard range. ## Short answer Given specific speed 100 is impossible for a Pelton wheel at this head. Using synchronous speed 500 rpm (12 poles) yields a single-jet design: power 10.33 MW, jet diameter 154 mm, wheel diameter 1.91 m, jet ratio 12.40, actual Ns 16.9, 22 buckets, bucket width 0.77 m, depth 0.19 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. The specific speed given in the data is checked first. **Data:** $Q = 2$ 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 $$ P = \eta_o \rho g Q H = 0.87 \times 9.81 \times 2 \times 605 = 10327.0\ \text{kW} $$ #### 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) $$ \begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 2}{\pi \times 106.77}} = 0.1544\ \text{m} \approx 154\ \text{mm} \end{aligned} $$ #### 4. Speed of the wheel Trial with jet ratio $m = 12$: $D = 12d = 1.853$ m, so $$ N = \frac{60u}{\pi D} = \frac{60 \times 50.12}{\pi \times 1.853} = 516.5\ \text{rpm} $$ Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/12 = 500$ rpm (12 poles). **Use of the given $N_s = 100$:** if the speed is taken from the given specific speed, $$ N = \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{100 \times 3000.5}{101.62} = 2953\ \text{rpm} \approx 3000\ \text{rpm} $$ which gives $D = 60u/(\pi N) = 0.319$ m and a jet ratio $m = D/d = 2.07$, far below the minimum of about 10. A single jet Pelton has $N_s$ of only about 8.5 to 30; even 6 jets raise this only by $\sqrt{6}$ (to about 70). Reaching 100 would need about $(100/16.9)^2 \approx 35$ jets, which is impossible. So $N_s = 100$ cannot be met by a Pelton wheel at this head; the design below uses the practical speed of 500 rpm and the standard proportions, and the actual $N_s$ is reported. #### 5. Wheel (pitch circle) diameter and jet ratio $$ \begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 50.12}{\pi \times 500} = 1.914\ \text{m} \\ m &= \frac{D}{d} = \frac{1.914}{0.1544} = 12.40 \end{aligned} $$ $m$ lies within 10 to 14, so a **single-jet** wheel is satisfactory. #### 6. Specific speed check $$ N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{500 \times 101.62}{3000.5} = 16.9 $$ 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) $$ z = 15 + \frac{D}{2d} = 15 + \frac{12.40}{2} = 21.20 \approx 22 $$ #### 8. Bucket size (usual proportions) - Bucket width $B \approx 5d = 0.772$ m - Bucket depth $T \approx 1.2d = 0.185$ m - Bucket radial length $L \approx 2.5d = 0.386$ m #### Summary of design | Item | Value | |---|---| | Power | 10.33 MW | | Speed | 500 rpm | | Number of jets | 1 | | Jet diameter $d$ | 154 mm | | Wheel diameter $D$ | 1.91 m | | Jet ratio $m$ | 12.40 | | Specific speed $N_s$ | 16.9 | | Number of buckets | 22 | | Bucket width × depth | 0.77 m × 0.19 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 = eta x 0.5 rho Q V1^2 with V1 = sqrt(2gH) only, ignoring head D. P = overall efficiency x rho g Q H = 0.87 x 9.81 x 2 x 605 = 10327.0 kW Answer: D. 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 = Cv sqrt(2gH) = 106.77 m/s and u = phi sqrt(2gH) = 50.12 m/s B. V1 = sqrt(2gH) with no coefficient, and u = V1 C. V1 = phi sqrt(2gH) and u = Cv sqrt(2gH) D. V1 = sqrt(gH) and u = 2 V1 Answer: A. 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 = d V1, so d = Q/V1 B. From Q = (pi/4) d^2 V1, so d = sqrt(4Q/(pi V1)) = 0.1544 m (154 mm) C. From Q = pi d^2 u, using the bucket velocity D. From P = pi d^2 V1, using the power Answer: B. The discharge equals jet area times jet velocity. (Key point: Jet diameter) 4. Why is 500 rpm used rather than the speed from the given Ns? A. Ns = 100 gives 500 rpm exactly, so the data is satisfied B. 3000 rpm is better because a larger jet ratio is not needed C. 500 rpm is used only because it is the lowest speed available D. Ns = 100 would need about 2953 rpm, giving m = 2.07 far below 10; 500 rpm (12 poles) gives m = 12.40, which is suitable Answer: D. The speed implied by Ns = 100 would make the wheel tiny (D = 0.319 m), so the practical synchronous speed is used. (Key point: Speed and jet ratio) 5. Why can a Pelton wheel not reach Ns = 100 at this head? A. A single jet gives Ns of only about 8.5 to 30, and even many jets (about 35 needed) cannot reach 100; the actual Ns is 16.9 B. Pelton wheels always have Ns above 100, so the data is fine C. The head of 605 m is too low for a Pelton wheel D. Ns can be raised to 100 simply by using 2 jets Answer: A. Specific speed rises only with the square root of the number of jets, so 100 is unreachable; the design reports the actual 16.9. (Key point: Feasibility of Ns = 100) 6. How many buckets are used, and by which formula? A. z = D/d = 12.4, so 12 buckets B. Tygun's formula z = 15 + D/(2d) = 15 + 12.40/2 = 21.2, so 22 buckets C. z = 15 x m = 186 buckets D. z = 15 - m/2, so 9 buckets Answer: B. Tygun's formula gives about 21.2, rounded up to 22. 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