Height and area of simple surge shaft

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.

            ___      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.