--- title: "Height and area of simple surge shaft" subject: "Hydro Power" chapter: 5 key: "semester_6/hydro_power:surge-shaft-height-area" source: https://www.bctengineer.com/q/semester_6/hydro_power/surge-shaft-height-area site: BCT Engineering --- # Height and area of simple surge shaft Hydro Power · Chapter 5: Components of Hydropower System - Asked in: 2080 Chaitra, 6 marks - Video: [Design of Surge Tank(Shaft) | Water Hammer Effect | Hydropower Engineering | IOE , TU , PU](https://www.youtube.com/watch?v=r1J4tINkknw) — Parash Joshi - Civil Construction and Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/surge-shaft-height-area - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5 ## Question How do you determine the height and area of simple circular surge shaft? Discuss with neat sketch and mathematical equation. ## Short answer The area is set by the Thoma criterion: A_th = (L A_t V0^2)/(2 g h_f H0) with a safety factor of 1.5–1.8. Height is determined by maximum upsurge Z_max = V0 sqrt(L A_t/(g A_s)) and downsurge, adding freeboard and submergence to prevent overflow and air entry. ## Answer A **simple surge shaft** is a vertical circular shaft at the end of a pressure tunnel. Its **area** is fixed for stable oscillations and its **height** must contain the maximum upsurge (on load rejection) and maximum downsurge (on load acceptance) without overflowing or letting air into the penstock. ```text ___ Max upsurge level | | -- --------------------- +Z_max Reservoir| | ~~~~~~~~~~|~~~|~~ static level (SWL) | | | -- steady level (SWL - h_f) | | | -- Min downsurge level -Z_max | | A | |_______| |______________ tunnel (A_t, L, h_f) \ penstock -> turbine ``` #### Notation $A_t$ = tunnel area, $L$ = tunnel length, $V_0$ = steady tunnel velocity, $Q_0 = A_t V_0$, $h_f$ = tunnel friction loss, $A_s$ = surge-shaft area, $g$ = 9.81 m/s². #### 1. Area of the shaft (Thoma criterion) For the oscillations to be stable (damped) after a small load change, the area must be larger than the Thoma area: $$ A_{th} = \frac{L\,A_t}{2 g\, c\, H_0}, \qquad c = \frac{h_f}{V_0^2} $$ i.e. $$ A_{th} = \frac{L\,A_t\,V_0^2}{2 g\, h_f\, H_0} $$ where $H_0$ = net head on the turbine. A factor of safety of 1.5–1.8 is applied: $A_s = (1.5 \text{ to } 1.8)A_{th}$; diameter $D_s = \sqrt{4A_s/\pi}$. #### 2. Maximum surge (upsurge) – full load rejection Neglecting friction, the dynamic equation of tunnel water and continuity of the shaft give simple harmonic oscillation: $$ \begin{aligned} \frac{L}{g}\frac{dV}{dt} &= -z, \qquad A_s\frac{dz}{dt} = A_t V \\ \Rightarrow\ \frac{d^2 z}{dt^2} &+ \frac{g A_t}{L A_s} z = 0 \end{aligned} $$ Solving with $V = V_0$ at $z = 0$: $$ Z_{max} = V_0\sqrt{\frac{L\,A_t}{g\,A_s}}, \qquad T = 2\pi\sqrt{\frac{L\,A_s}{g\,A_t}} $$ With friction, the upsurge is reduced, approximately (Jaeger/Johnson): $$ Z_{up} \approx Z_{max}\left(1 - \tfrac{2}{3}k + \tfrac{1}{9}k^2\right), \quad k = \frac{h_f}{Z_{max}} $$ #### 3. Maximum downsurge – sudden load acceptance For a sudden full-load demand, the downsurge below the static level is about $$ Z_{down} \approx h_f + (0.5 \text{ to } 1.0) Z_{max} $$ (often taken equal to $Z_{max}$ below the minimum steady level for a conservative design). #### 4. Height of the shaft $$ \begin{aligned} \text{Top level} &= \text{Max reservoir level} + Z_{up} + \text{freeboard (1–2 m)} \\ \text{Bottom level} &= \text{Min reservoir level} - Z_{down} - \text{submergence (2–3 m)} \end{aligned} $$ Height of shaft = top level − bottom level. The bottom must stay below the minimum downsurge so that air never enters the tunnel or penstock. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The Thoma criterion is used to find A. the minimum shaft area for stable (damped) oscillations B. the maximum upsurge height C. the tunnel friction loss D. the shaft freeboard Answer: A. A shaft area greater than the Thoma area gives stable oscillations after small load changes. (Key point: Thoma area) 2. Which expression gives the Thoma area? A. A_th = 2 g c H0 / (L A_t) B. A_th = L A_t / (2 g c H0), with c = h_f / V0^2 C. A_th = L A_t V0 / g D. A_th = h_f H0 / (L A_t) Answer: B. A_th = L A_t / (2 g c H0), where c = h_f/V0^2. (Key point: Thoma formula) 3. What area is finally adopted for the shaft? A. Exactly the Thoma area B. Half the Thoma area C. 1.5 to 1.8 times the Thoma area D. 10 times the Thoma area Answer: C. A factor of safety of 1.5-1.8 is applied to A_th. (Key point: Safety factor) 4. For full load rejection with friction neglected, the maximum upsurge is A. Z_max = V0 sqrt(g A_s / (L A_t)) B. Z_max = V0^2 / (2 g) C. Z_max = L V0 / g D. Z_max = V0 sqrt(L A_t / (g A_s)) Answer: D. Solving the SHM equation gives Z_max = V0 sqrt(L A_t/(g A_s)). 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