--- title: "Surge and water hammer: 25 m³/s, 15 m surge tank" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:surge-numerical-25cumec-2-penstocks-15m-tank" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-25cumec-2-penstocks-15m-tank site: BCT Engineering --- # Surge and water hammer: 25 m³/s, 15 m surge tank Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2079 Baishakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-25cumec-2-penstocks-15m-tank - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question The design discharge through the tunnel of a hydropower project is 25 m³/sec and is conveyed by two penstocks to the turbine. The length and diameter of the tunnel are 4 km and 8 m respectively, friction factor of the tunnel is 0.016 and length of each penstock is 500 m, diameter and friction factor of the penstock are 2 m and 0.04 respectively and velocity of wave in penstock = 1600 m/sec. If a surge tank of 15 m diameter has been provided at the end of the tunnel, find the following for full load rejection: (i) maximum up-surge, (ii) maximum down-surge, (iii) water hammer pressure, (iv) time of oscillation of wave. ## Short answer Maximum upsurge is 5.29 m, downsurge 5.16 m. The water hammer pressure head is 649.0 m for rapid closure, and the oscillation time period is 237.9 s. ## Answer #### Method used Let the tunnel have area $A_t$, length $L$, diameter $D$ and friction factor $f$; the tank has area $A_s$. - Tunnel velocity $V = Q/A_t$; steady friction loss $h_f = \dfrac{fLV^2}{2gD}$. - Frictionless amplitude: $Z_0 = V\sqrt{\dfrac{L A_t}{g A_s}}$. - Upsurge with friction (full load rejection), $Z$ measured above reservoir level, from the equation of motion and continuity: $$ \frac{Z_{max}}{h_f}=\frac{Z_0^2}{2h_f^2}\left[1-e^{-2h_f (Z_{max}+h_f)/Z_0^2}\right] $$ - The following down-surge is found by integrating the reversed-flow equations step by step (below reservoir level). - Period of mass oscillation: $T = 2\pi\sqrt{\dfrac{L A_s}{g A_t}}$. #### Surge in the tunnel $A_t = 50.27\ \text{m}^2$, $A_s = \frac{\pi}{4}(15)^2 = 176.71\ \text{m}^2$, $L = 4000$ m. - $V = 25/50.27 = 0.497$ m/s - $h_f = \dfrac{0.016 \times 4000 \times 0.497^2}{2(9.81)(8)} = 0.101$ m - $Z_0 = 0.497\sqrt{\dfrac{4000 \times 50.27}{9.81 \times 176.71}} = 5.36$ m #### (i) Maximum up-surge $Z_{max} = 5.29$ m above reservoir level. #### (ii) Maximum down-surge First downswing $= 5.16$ m below reservoir level. #### (iii) Water hammer pressure Flow per penstock $= 25/2 = 12.5$ m³/s; $v = 12.5/3.142 = 3.979$ m/s. Closure time for full load rejection is taken as less than $2L/a = 2(500)/1600 = 0.625$ s (rapid closure): $$ h = \frac{a v}{g} = \frac{1600 \times 3.979}{9.81} = 649.0\ \text{m} $$ $p = \rho g h = 6.37\ \text{MPa}$ ($\approx 64.9\ \text{kg/cm}^2$). #### (iv) Time of oscillation $$ T = 2\pi\sqrt{\frac{4000 \times 176.71}{9.81 \times 50.27}} = 237.9\ \text{s} \approx 4.0\ \text{min} $$ **Answer: up-surge 5.29 m; down-surge 5.16 m; water hammer head 649.0 m (6.37 MPa); $T = 237.9$ s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives the frictionless surge amplitude Z0 in the tunnel-surge tank system? A. Z0 = V sqrt(L At / (g As)) B. Z0 = V sqrt(L As / (g At)) C. Z0 = V sqrt(g At / (L As)) D. Z0 = V^2 / (2g) Answer: A. Z0 comes from equating the kinetic energy of the tunnel water to the potential energy gained in the tank, giving V sqrt(L At/(g As)). (Key point: Frictionless amplitude) 2. Which formula gives the period of mass oscillation of the surge tank? A. T = 2 pi sqrt(L At / (g As)) B. T = 2 pi sqrt(L As / (g At)) C. T = 4 L / a D. T = 2 L / a Answer: B. The mass oscillation period is 2 pi sqrt(L As/(g At)); 4L/a and 2L/a belong to the pressure wave in the penstock. (Key point: Period of oscillation) 3. Which expression gives the steady friction head loss hf used in the upsurge equation? A. hf = f L v^2 / (2 g d) for the penstock B. hf = a V / g C. hf = f L V^2 / (2 g D) for the tunnel D. hf = Z0^2 / L Answer: C. Only the tunnel friction acts on the tank oscillation, so hf = f L V^2/(2gD) with tunnel values (0.101 m here). (Key point: Steady friction loss in tunnel) 4. With a = 1600 m/s and v = 3.979 m/s per penstock, what is the water hammer head? A. About 5.3 m, from v^2 / 2g B. About 162 m, from a v / (4g) C. About 1298 m, from 2 a v / g D. About 649 m, from a v / g Answer: D. h = a v / g = 1600 x 3.979 / 9.81 = 649 m for rapid closure. (Key point: Water hammer pressure) ## Close variants set in other exams - [Surge and water hammer: 60 m³/s, 30 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-60cumec-3-penstocks-30m-tank) — Hydropower Engineering, chapter 6, asked 2× - [Surge and water hammer: 45 m³/s, 20 m surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-45cumec-3-penstocks) — Hydropower Engineering, chapter 6, asked 2× ## Related in Hydropower Engineering - [Surge tank: 7.5 m³/s through 2.5 m tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-7-5cumec-2-5m) — Hydropower Engineering, chapter 6 - [Surge numerical: 13 m tank, 1.8 km tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-13m-tank-1-8km-tunnel) — Hydropower Engineering, chapter 6 - [Surge tank: tunnel 5 m dia, 28 cumec](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-5m-28cumec) — Hydropower Engineering, chapter 6 - [Pumped storage plant: power, turbine, setting, surge](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apumped-storage-turbine-surge-numerical) — Hydropower Engineering, chapter 7 - [Surge numerical: 4.5 m tunnel, 2500 m, 25 m³/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-4-5m-2500m-25cumec) — Hydropower Engineering, chapter 6, asked 2× - [Hydraulic design of surge tank and pressure tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-tank-hydraulic-design-pressurised-tunnel) — Hydropower Engineering, chapter 6 ## Related in other subjects - [Water hammer in penstock and surge tank](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-hammer-surge-tank) — Power Plant Equipment, chapter 1, asked 2× - [Surge tank objectives and economic penstock diameter factors](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asurge-tank-objectives-economic-diameter) — Hydro Power, chapter 5