--- title: "Pelton wheel design (310 m, 5 m³/s)" subject: "Hydropower Engineering" chapter: 7 key: "semester_7/hydropower_engineering:pelton-design-310m-5cumec-specific-speed" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-design-310m-5cumec-specific-speed site: BCT Engineering --- # Pelton wheel design (310 m, 5 m³/s) Hydropower Engineering · Chapter 7: Hydro-electric Machines - Asked in: 2071 Chaitra, 7+1 marks - Video: [Pelton wheel problems | Solved problems in Pelton Wheel Turbine | Fluid mechanics and Machinery](https://www.youtube.com/watch?v=-SxzvLuk6mc) — Amizhdhagam - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-design-310m-5cumec-specific-speed - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7 ## Question Design a Pelton wheel turbine for a hydropower plant having net head of 310 m and discharge of 5 m³/s. Take the efficiency of the turbine as 90%. What will be the specific speed of such turbine? ## Short answer Power is 13.68 MW; use 2 jets, synchronous speed 333.3 rpm, wheel diameter 2.06 m, jet diameter 0.204 m; turbine specific speed is about 30, about 21 per jet. ## Answer #### Step 1: Power $$ P = \eta\gamma Q H = 0.90 \times 9.81 \times 5 \times 310 = 13{,}685\ \text{kW} $$ #### Step 2: Jet velocity and bucket speed Assuming a nozzle velocity coefficient $C_v = 0.98$ (not given): $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 310} = 76.43$ m/s Speed ratio $K_u = u/\sqrt{2gH} = 0.46$ (usual range 0.43 to 0.47) with $\sqrt{2gH} = 77.99$ m/s: $u = 0.46 \times 77.99 = 35.87$ m/s #### Step 3: Number of jets and speed For a single jet, the empirical (Siervo and Lugaresi) specific speed is $N_s \approx 85.49/H^{0.243} = 21.2$. For $z$ jets the wheel specific speed is $N_s\sqrt{z}$. Take **$z = 2$ jets** (horizontal shaft; 50 Hz assumed): $$ N = \frac{N_s\sqrt{z}\,H^{5/4}}{\sqrt{P}} = 333.5\ \text{rpm} $$ Synchronous speed for 50 Hz: $N = 6000/p$. Choose $p = 18$ poles, **$N = 333.3$ rpm**. #### Step 4: Wheel (pitch circle) diameter $$ D = \frac{60u}{\pi N} = \frac{60 \times 35.87}{\pi \times 333.3} = 2.056\ \text{m} $$ #### Step 5: Jet diameter Discharge per jet $q = 5/2 = 2.5\ \text{m}^3/\text{s}$: $$ d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 2.5}{\pi \times 76.43}} = 0.2041\ \text{m} $$ Jet ratio $m = D/d = 10.07$ (acceptable range 10 to 20). #### Step 6: Bucket details - Number of buckets $Z = 15 + \dfrac{D}{2d} = 15 + 5.04 \approx 20$ - Bucket width $B \approx 3.5d = 0.71$ m; depth about $1.0d$; length about $2.5d$. #### Step 7: Specific speed of the wheel $$ N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{333.3\sqrt{13{,}685}}{310^{1.25}} = 30.0 $$ (about 21.2 per jet) **Answer: $P = 13.68$ MW; 2 jets; $N = 333.3$ rpm; $D = 2.06$ m; $d = 0.204$ m; specific speed of the turbine $N_s \approx 30$ (kW, m, rpm), i.e. 21 per jet.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the available turbine power for 310 m, 5 m3/s and 90% efficiency? A. About 13.7 MW B. About 27 MW C. About 15.2 MW D. About 12.3 MW Answer: A. P = 0.9 x 9.81 x 5 x 310 = 13,685 kW. (Key point: Power) 2. How is the final speed of 333.3 rpm chosen? A. Speed with u = V1 B. Synchronous speed 6000/p for 50 Hz with 18 poles, near the speed computed from Ns for 2 jets C. Highest speed the buckets can take D. Speed that makes D = 1 m Answer: B. The computed 333.5 rpm is rounded to a synchronous speed, 6000/18 = 333.3 rpm. (Key point: Speed choice) 3. What specific speed does the wheel have, and what is it per jet? A. About 21 for the wheel, about 30 per jet B. About 30 for the wheel, about 21 per jet C. About 60 for the wheel, 30 per jet D. About 100 for the wheel, 50 per jet Answer: B. Ns = N sqrt(P)/H^(5/4) = 30, which is 21.2 x sqrt(2) for two jets. (Key point: Final specific speed) 4. How is the number of buckets estimated? A. Z = 2 pi D/B B. Z = 15 D/d C. Z = D/d D. Z = 15 + D/(2d) Answer: D. Z = 15 + D/(2d) = 15 + 5.04, about 20 buckets. (Key point: Bucket count) ## Related in Hydropower Engineering - [Pelton turbine design (312.5 m, 5 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-312-5m-5cumec) — Hydropower Engineering, chapter 7 - [Pelton turbine design (6867 kW, 350 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-6867kw-350m) — Hydropower Engineering, chapter 7 - [Multi-jet Pelton wheel design (200 m, 4 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-multijet-200m-4cumec) — Hydropower Engineering, chapter 7 - [Pelton unit speed, discharge, power](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-unit-quantities-190rpm) — Hydropower Engineering, chapter 7 - [Francis design (90 m, 60 m³/s, 94%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-60cumec-94) — Hydropower Engineering, chapter 7 - [Francis design (90 m, 40 m³/s, 86%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-40cumec) — Hydropower Engineering, chapter 7 ## Related in other subjects - [Pelton turbine design (5 m³/s, 310 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-310m) — Hydro Power, chapter 7 - [Design: 310 m head plant, HV breaker](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2080-310m-hvcb) — Power Plant Design, chapter 4, asked 3× - [Design of Pelton wheel for given power and head](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-design-numerical) — Fluid Machines, chapter 3 - [Design: 300 m head plant, GCB rating](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2081-300m-gcb) — Power Plant Design, chapter 4, asked 11×