Water hammer in 20 m steel pipeline with valve

For instantaneous closure (frictionless), the valve creates a pressure wave of magnitude $\Delta p=\rho cV$. Upstream of the valve the wave is a pressure rise; downstream of the valve it is a pressure drop. The waves reflect at the constant-head tanks (head unchanged, sign reversed) and at the closed valve (sign unchanged).

Given data

$D=0.075$ m, $t=6$ mm, $Q=0.022$ m³/s, $K=2\times10^9$ N/m², $E=204\times10^9$ N/m². The valve is 10 m from the upstream tank (6 m head) and 10 m from the downstream tank (2 m head).

Step 1: Velocity and celerity

$$A=\frac{\pi}{4}(0.075)^2=0.004418\ \text{m}^2,\qquad V=\frac{Q}{A}=4.980\ \text{m/s}$$ $$c=\sqrt{\frac{K/\rho}{1+KD/(Et)}}=\sqrt{\frac{2\times10^6}{1+\dfrac{2\times10^9\times0.075}{204\times10^9\times0.006}}}=1334.8\ \text{m/s}$$

Step 2: Magnitude of the pressure wave

$$\Delta p=\rho cV=1000\times1334.8\times4.980=6.65\ \text{MPa},\qquad \Delta H=\frac{cV}{g}=677.6\ \text{m}$$

• Upstream of the valve (towards the upstream tank): a wave of +677.6 m.
• Downstream of the valve (towards the downstream tank): a wave of −677.6 m.

Step 3: Steady pressure head

In a frictionless pipe, the steady head at any point upstream of the valve is the tank head minus the velocity head:

$$h_s=6-\frac{V^2}{2g}=6-1.26=4.74\ \text{m}$$

(The 2 m downstream head is only a boundary level; the remaining head is lost at the valve.)

Step 4: Timing

The valve is $L_v=10$ m from the tank, so the wave period is $4L_v/c=40/c=29.97$ ms. For a point $x$ m from the upstream tank:

• First rise: $(10-x)/c$
• Back to steady (reflection from the tank): $(10+x)/c$
• Drop of $\Delta H$ (reflection from the valve): $(30-x)/c$
• Back to steady: $(30+x)/c$; the cycle repeats every 29.97 ms.

Step 5: Pressure-time curves

The steady pressure head is $4.74$ m (a horizontal line). The unsteady (total) head switches between three levels:

 head
 +H_hi  |      ____            ____
        |     |    |          |
 h_s    |_____|    |__    ____|   (steady line)
        |           |    |
 H_lo   |           |____|
        +---------------------------------> t
          rise  back  fall  back  next rise

H_hi = 682 m, h_s = 4.74 m, H_lo = −673 m. (The shape is the same at all three points; only the timing differs, as shown in the table. Both the positive and the negative pulse are $2x/c$ wide, so the pulses are narrowest at 0.5 m and widest at 5 m.)

The theoretical low head (−673 m) is below absolute vacuum, so in reality the water column separates (cavitation) and the pressure is limited to the vapour pressure (about −10 m gauge).

Answer: wave speed 1335 m/s; pressure wave = ±678 m of water (±6.65 MPa), positive upstream and negative downstream of the valve; steady head 4.74 m at the points.