Gradual valve closure with retardation law

Data. Normal net head at the valve $= 300$ m with $Q = 2.5$ m³/s; $D = 1$ m, $L = 1.6$ km, $f = 0.02$; the valve is closed gradually over $T = 8$ s with retardation $a \propto t^{5/4}$. The pipe is treated as rigid and the water as incompressible (rigid-column theory); minor losses are neglected.

Initial velocity and reservoir head

$$V_0 = \frac{Q}{\pi D^2/4} = 3.183\ \text{m/s},\qquad h_{f0} = f\frac{L}{D}\frac{V_0^2}{2g} = 0.02(1600)\frac{3.183^2}{19.62} = 16.53\ \text{m}$$

The head at the upstream end of the pipe (reservoir level) is the valve head plus the friction loss: $H_R = 300 + 16.53 = 316.53$ m.

Velocity during closure

$a = kt^{5/4}$, and the velocity falls to zero at $t = T$: $\displaystyle V_0 = \int_0^T kt^{5/4}\,dt = \frac{4}{9}kT^{9/4}$, so $k = \dfrac{9V_0}{4T^{9/4}}$. Then

$$V(t) = V_0 - \int_0^t a\,dt = V_0\left[1 - \left(\frac{t}{T}\right)^{9/4}\right]$$

At $t = 6$ s: $\left(\dfrac{6}{8}\right)^{2.25} = 0.5235$, so

$$V = 3.183(1 - 0.5235) = 1.517\ \text{m/s}$$

Head at the valve at $t = 6$ s

Retardation $a = \left|\dfrac{dV}{dt}\right| = \dfrac{9}{4}V_0\dfrac{t^{5/4}}{T^{9/4}} = 2.25(3.183)\dfrac{6^{1.25}}{8^{2.25}} = 0.6248\ \text{m/s}^2$.

Equation of motion of the water column (deceleration raises the head at the valve):

$$H_{valve} = H_R - f\frac{L}{D}\frac{V^2}{2g} + \frac{L}{g}a$$ $$H_{valve} = 316.53 - 3.75 + \frac{1600(0.6248)}{9.81} = 316.53 - 3.75 + 101.91 = 414.7\ \text{m}$$

Answer: At $t = 6$ s, the velocity in the pipe is $1.52$ m/s and the head at the valve is about $415$ m. This is about 115 m above the normal 300 m: 101.9 m comes from the deceleration, and 12.8 m because the friction loss has dropped from 16.53 m to 3.75 m.