--- title: "Turbine setting; three single-jet Pelton wheels" subject: "Hydropower Engineering" chapter: 7 key: "semester_7/hydropower_engineering:turbine-setting-pelton-three-jets" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/turbine-setting-pelton-three-jets site: BCT Engineering --- # Turbine setting; three single-jet Pelton wheels Hydropower Engineering · Chapter 7: Hydro-electric Machines - Asked in: 2070 Ashad, 2+6 marks - Video: [Pelton Wheel Turbine](https://www.youtube.com/watch?v=ArTxn7obZHc) — BEST MECHANICAL ENGINEERING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/turbine-setting-pelton-three-jets - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7 ## Question What do you mean by setting of a turbine? The pipe line 1200 meter supplies water to 3 single jet Pelton wheels. The head above the nozzle is 360 m. The velocity coefficient for the nozzle is 0.98 and the coefficient of friction for the pipe line is 0.02. The turbine efficiency is 0.85. The specific speed of the turbine is 15.3 rpm and loss head is 18 meter in the pipeline due to friction. If the operating speed of each turbine is 560 rpm, determine (i) total power developed, (ii) discharge, (iii) diameter of each jet and diameter of pipe line. ## Short answer Setting fixes turbine elevation relative to tailwater; Pelton wheel above maximum flood level. Using specific speed formula, total power = 5507 kW. From efficiency, discharge Q = 1.834 m³/s. Jet velocity (82.36 m/s) gives jet diameter 0.097 m; Darcy–Weisbach yields pipe diameter 0.82 m. ## Answer #### Setting of a turbine **Setting** means fixing the elevation of the turbine (runner outlet or centre line) relative to the tailwater level. For a reaction turbine it is limited by cavitation, $H_s \le H_{atm} - H_v - \sigma_c H$. For a Pelton wheel the wheel centre is set above the maximum flood level in the tailrace (clearance 1 to 2 m) so that the runner never dips in water. #### Data Net head at nozzle $H = 360$ m, $C_v = 0.98$, $\eta = 0.85$, $N = 560$ rpm, $N_s = 15.3$ (for each single-jet wheel), $L = 1200$ m, $h_f = 18$ m, $f = 0.02$ (Darcy). #### (i) Total power developed Power of one wheel from specific speed (kW, m, rpm): $$ P = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{15.3 \times 360^{1.25}}{560}\right)^2 = 1835.5\ \text{kW} $$ $$ P_{total} = 3 \times 1835.5 = 5506.6\ \text{kW} $$ #### (ii) Discharge $$ Q = \frac{P_{total}}{\eta \gamma H} = \frac{5506.6}{0.85 \times 9.81 \times 360} = 1.834\ \text{m}^3/\text{s} $$ Per wheel (one jet): $q = 0.6115\ \text{m}^3/\text{s}$. #### (iii) Jet diameter and pipe diameter Jet velocity $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 360} = 82.36$ m/s. $$ d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 0.6115}{\pi \times 82.36}} = 0.0972\ \text{m} $$ Pipeline: $h_f = \dfrac{fLv^2}{2gD}$ with $v = \dfrac{4Q}{\pi D^2}$: $$ D^5 = \frac{16 f L Q^2}{\pi^2 \, 2g \, h_f} = \frac{16 \times 0.02 \times 1200 \times 1.834^2}{\pi^2 \times 2 \times 9.81 \times 18} $$ $$ D = 0.820\ \text{m},\quad v = 3.47\ \text{m/s} $$ **Answer: total power $= 5507$ kW; $Q = 1.834$ m³/s; jet diameter $= 97$ mm; pipe diameter $= 0.82$ m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the setting of a turbine mean? A. Fixing the number of runner blades B. Fixing the elevation of the runner outlet or centre line relative to the tailwater level C. Setting the governor gate opening D. Choosing the generator speed Answer: B. Setting is the elevation of the turbine (runner outlet or centre line) with respect to the tailwater level. (Key point: Meaning of setting) 2. How is a Pelton wheel set relative to the tailrace? A. Above the maximum flood level in the tailrace, with 1 to 2 m clearance B. At the level determined by the Thoma coefficient C. At the level of the nozzle exactly D. Below minimum tailwater so it is submerged Answer: A. The Pelton wheel centre is set above the maximum tailrace flood level so the runner never dips in water. (Key point: Pelton setting) 3. How is the power of each single-jet wheel found from the specific speed? A. P = Ns N / H^(5/4) B. P = eta gamma Q H with Q assumed C. P = Ns sqrt(H) / N D. P = (Ns H^(5/4) / N)^2 Answer: D. From Ns = N sqrt(P)/H^(5/4), P = (Ns H^(5/4)/N)^2, giving 1835.5 kW per wheel. (Key point: Power from specific speed) 4. After finding the total discharge, how is the jet diameter obtained? A. Using the specific speed directly B. Using the total Q with jet velocity 82.36 m/s for one jet C. Using the per-wheel discharge q = Q/3 and V1 = Cv sqrt(2gH), d = sqrt(4q/(pi V1)) D. Using the pipe velocity 3.47 m/s Answer: C. Each wheel has one jet carrying q = 0.6115 m3/s at V1 = 82.36 m/s, so d = 97 mm. 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