Pressure-time diagram 1200 m from reservoir

Closure time $t_c=1$ s is shorter than the pipe period $2L/c=2\times3000/1400=4.29$ s, so the closure is rapid and the full Joukowsky pressure develops. The pressure at the section is then a trapezoidal wave that repeats every $4L/c$.

Given data

$D=0.6$ m, $L=3000$ m, $v=3.5$ m/s, $c=1400$ m/s (given), $t_c=1$ s, section at $x=1200$ m from the reservoir, i.e. 1800 m from the valve. The static head $H_0$ is not given, so pressure is plotted as $H_0\pm\Delta H$.

(Note: the given $K$ and $E$ would give $c\approx1015$ m/s for a thin-walled elastic pipe; the stated $c=1400$ m/s is used as instructed.)

Pressure rise

$$\Delta p=\rho c v=1000\times1400\times3.5=4.90\ \text{MPa},\qquad \Delta H=\frac{cv}{g}=\frac{1400\times3.5}{9.81}=499.5\ \text{m}$$

Wave travel times

• Pipe period: $4L/c=4\times3000/1400=8.571$ s.
• Wave from valve reaches the section: $(L-x)/c=1800/1400=1.286$ s.
• Wave from valve reaches the reservoir: $L/c=2.143$ s, and returns as a negative wave.

Pressure history at the section (1200 m from reservoir)

Pressure-time diagram (friction neglected)

H0+dH|         ___                            
     |     ////   \\\\                        
H0   |_____           ______           _______
     |                      \\\\\  ////       
H0-dH|                           __           
     +|---|---|---|---|---|---|---|---|---|---
      0   1   2   3   4   5   6   7   8   9    t(s)

dH = 499 m. Sloping parts last 1 s (equal to the closure time). In practice friction damps the wave, and the low-pressure part cannot fall below vapour pressure.

Answer: peak pressure rise = 4.90 MPa (499 m of water); the section sees +499 m from 1.29 s to 3.00 s (falling to $H_0$ by 4.00 s), then −499 m from 5.57 s to 7.29 s, repeating every 8.57 s.