--- title: "Surge tank area, up/downsurge, invert level" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:surge-numerical-tunnel-5m-4000m-pressure-shaft" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/surge-numerical-tunnel-5m-4000m-pressure-shaft site: BCT Engineering --- # Surge tank area, up/downsurge, invert level Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2069 Chaitra, 3+3+3 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-tunnel-5m-4000m-pressure-shaft - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question A power station is fed by a 4000 m long concrete lined tunnel of 5.0 m dia and 600 m long pressure shaft of 4.0 m dia operating under a gross head of 250 m. If the design discharge of the plant is 60 m³/sec and the friction factors in tunnel and pressure shaft are 0.014 and 0.012 respectively, i) compute the sectional area required for mass oscillation in a surge tank; ii) find the maximum upsurge and downsurge levels; iii) if the headwater level is 1048 m, find out the invert level of the headrace tunnel at the surge tank. ## Short answer Thoma area is 28.9 m2, equivalent diameter 6.07 m. Full-load rejection gives maximum upsurge level 1095.36 m and downsurge level 1006.10 m, so up 47.36 m and down 41.90 m from 1048 m headwater. Invert level of headrace tunnel at surge tank is about 1000.1 m. ## Answer #### Given data and losses $A_t = \frac{\pi}{4}(5)^2 = 19.635\ \text{m}^2$, $V_t = 60/19.635 = 3.056$ m/s. Shaft: $A_p = 12.566\ \text{m}^2$, $V_p = 4.775$ m/s. - Tunnel loss: $h_{ft} = \dfrac{0.014 \times 4000 \times 3.056^2}{2(9.81)(5)} = 5.33$ m - Shaft loss: $h_{fp} = \dfrac{0.012 \times 600 \times 4.775^2}{2(9.81)(4)} = 2.09$ m - Net head $H_0 = 250 - 5.33 - 2.09 = 242.58$ m #### i) Surge tank area (Thoma) Head-loss coefficient of the tunnel, $\alpha = h_{ft}/V_t^2 = 5.33/3.056^2 = 0.5708\ \text{s}^2/\text{m}$. $$ A_{th} = \frac{L A_t}{2g\alpha H_0} = \frac{4000 \times 19.635}{2(9.81)(0.5708)(242.58)} = 28.9\ \text{m}^2 $$ Equivalent diameter $= 6.07$ m. Adopting $A_s = A_{th} = 28.9\ \text{m}^2$ (a larger area, e.g. $1.5A_{th}$, gives more safety). #### ii) Maximum upsurge and downsurge $Z_0 = 3.056\sqrt{\dfrac{4000 \times 19.635}{9.81 \times 28.9}} = 50.85$ m. With friction ($h_{ft} = 5.33$ m), for full load rejection: - Maximum upsurge $= 47.36$ m above reservoir level, i.e. level $1048 + 47.36 = 1095.36$ m. - Maximum downsurge (first downswing) $= 41.90$ m below reservoir level, i.e. level $1048 - 41.90 = 1006.10$ m. For comparison, with $A_s = 1.5A_{th} = 43.4\ \text{m}^2$: upsurge 38.05 m and downsurge 32.88 m. #### iii) Invert level of the headrace tunnel at the tank The tunnel crown must stay below the lowest water level in the tank to prevent air entering the tunnel. Assume a minimum water cover of 1.0 m over the crown. - Lowest tank level $= 1006.10$ m - Crown $= 1006.10 - 1.0 = 1005.10$ m - Invert $= 1005.10 - 5.0 = 1000.10$ m **Answer: $A_{th} = 28.9\ \text{m}^2$; upsurge level $= 1095.36$ m; downsurge level $= 1006.10$ m; invert level $\approx 1000.1$ m a.s.l.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the net head H0 used in the Thoma area obtained? A. Gross head minus tunnel and shaft friction losses: 250 - 5.33 - 2.09 = 242.58 m B. Gross head plus friction losses C. Gross head only, 250 m D. Gross head minus the upsurge Answer: A. Friction losses reduce the gross head to the net head H0 = 242.58 m. (Key point: Net head) 2. Which formula gives the Thoma surge tank area? A. Ath = 2 g alpha H0 / (L At) B. Ath = L At / (2 g alpha H0), with alpha = hft / Vt^2 C. Ath = Q / V D. Ath = pi D^2 / 4 using tunnel diameter Answer: B. Thoma's stability criterion gives the minimum area for stable oscillation, here 28.9 m2. (Key point: Thoma area) 3. How is the tunnel invert level at the tank fixed? A. Add the upsurge to the reservoir level B. Use the headwater level minus gross head C. Take lowest tank level, subtract minimum water cover over the crown, then subtract the tunnel diameter D. Use the upsurge level minus tunnel diameter Answer: C. Crown must stay below the lowest tank level to stop air entering: 1006.10 - 1 - 5 = 1000.10 m. (Key point: Invert level) 4. What effect does adopting 1.5 Ath have on the surges? A. They increase B. They are unchanged C. Only the downsurge changes D. They decrease (upsurge 38.05 m, downsurge 32.88 m vs 47.36 and 41.90 m) Answer: D. A larger area lowers Z0 = V sqrt(L At/(g As)) and so lowers the surges and gives more stability. 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