Derive minimum upsurge without damping

Part 1: Derivation (no friction)

Let $z$ be the water level in the tank above the reservoir level (positive upward), $V$ the velocity in the tunnel (length $L$, area $A_t$), $A_s$ the area of the tank, and $Q_t$ the turbine flow after the sudden change.

 reservoir ~~~~~~~~
            \___________ tunnel L, A_t, V ___________ surge tank
                                              | z    |  A_s
                                              |______|__ to turbine Q_t

Continuity at the tank junction: inflow from the tunnel = flow to the tank + flow to the turbine

$$A_tV=A_s\frac{dz}{dt}+Q_t \quad (1)$$

Momentum for the water column in the tunnel (net force = mass x acceleration; the head $z$ acts to decelerate the flow):

$$\rho A_tL\frac{dV}{dt}=-\rho gA_t z\ \Rightarrow\ \frac{L}{g}\frac{dV}{dt}=-z \quad (2)$$

Differentiate (1) with $Q_t$ constant: $A_t\dfrac{dV}{dt}=A_s\dfrac{d^2z}{dt^2}$. Substituting in (2):

$$\frac{d^2z}{dt^2}+\frac{gA_t}{LA_s}\,z=0$$

This is simple harmonic motion with $\omega=\sqrt{\dfrac{gA_t}{LA_s}}$, so the period is $T=2\pi\sqrt{\dfrac{LA_s}{gA_t}}$.

Boundary conditions at $t=0$ (sudden change of discharge from $Q_0$ to $Q_t$, so $\Delta V=\dfrac{Q_0-Q_t}{A_t}$): $z=0$ and $\dfrac{dz}{dt}=\dfrac{A_t\Delta V}{A_s}$. Hence

$$z=\frac{A_t\Delta V}{A_s\,\omega}\sin\omega t,\qquad Z_{max}=\frac{A_t\Delta V}{A_s}\sqrt{\frac{LA_s}{gA_t}}$$ $$\boxed{Z_{max}=\Delta V\sqrt{\frac{L\,A_t}{g\,A_s}}}$$

For load rejection this is the upsurge $+Z_{max}$; for sudden demand the same magnitude is the downsurge $-Z_{max}$ (undamped).

Part 2: Numerical design

Given: reservoir RL 320 m, tailwater RL 200 m, gross head $H_0=120$ m; $Q=100$ m³/s for three units, so each unit takes $33.3$ m³/s; two units are rejected or demanded, so the change is $\Delta Q=66.7$ m³/s. Maximum and minimum tank levels are RL 330 m and RL 310 m, so $Z=330-320=10$ m upwards and $320-310=10$ m downwards. Tunnel $D=7$ m, tunnel head loss $h_f=10\%H_0=12$ m. Penstocks: 3 x 500 m, $D=2.5$ m, $f=0.016$. The friction factor of the tunnel is assumed to be the same as the penstocks, $f=0.016$.

Tunnel velocity and change in velocity

$$A_t=\frac{\pi}{4}(7)^2=38.48\ \text{m}^2,\quad V_0=\frac{100}{38.48}=2.598\ \text{m/s},\quad \Delta V=\frac{2}{3}V_0=1.732\ \text{m/s}$$

Length of the tunnel (from the head loss, Darcy-Weisbach)

$$h_f=\frac{f\,L}{D}\frac{V_0^2}{2g}\ \Rightarrow\ L=\frac{2gD\,h_f}{fV_0^2}=\frac{19.62\times7\times12}{0.016\times6.752}=15256\ \text{m}\ (\approx15.3\ \text{km})$$

Area of the tank from the permissible surge ($Z=10$ m both ways)

$$Z^2=\Delta V^2\frac{L\,A_t}{g\,A_s}\ \Rightarrow\ A_s=\frac{\Delta V^2\,L\,A_t}{g\,Z^2}=\frac{3.001\times15256\times38.48}{9.81\times100}=1796\ \text{m}^2$$

This corresponds to a circular tank of diameter $\sqrt{4A_s/\pi}=47.8$ m.

Thoma check (stability)

$$\alpha=\frac{h_f}{V_0^2}=\frac{12}{6.752}=1.777\ \text{s}^2/\text{m}$$

Penstock velocity $V_p=\dfrac{33.3}{\frac{\pi}{4}(2.5)^2}=6.79$ m/s and its loss $h_p=\dfrac{0.016\times500}{2.5}\dfrac{6.79^2}{19.62}=7.52$ m. Net head $H_0-h_f-h_p=120-12-7.52=100.5$ m.

$$A_{th}=\frac{L\,A_t}{2g\,\alpha\,(H_0-h_f-h_p)}=\frac{15256\times38.48}{19.62\times1.777\times100.5}=168\ \text{m}^2$$

Since $A_s=1796$ m² $>A_{th}=168$ m², the tank is stable.

Period of oscillation: $T=2\pi\sqrt{\dfrac{LA_s}{gA_t}}=1693$ s (about 28.2 min).

Answer: minimum area of the surge tank $A_s\approx1800$ m² (diameter about 48 m) and the permissible tunnel length $L\approx15250$ m (about 15.3 km), with the surge limited to RL 330 m and RL 310 m.