Water hammer sketch, 2300 m pipeline

Data. $L = 2300$ m, $D = 0.15$ m, wall thickness $t = 2.8$ mm, $Q = 2200$ l/min $= 0.03667$ m³/s, $K = 2\times10^9$ Pa, $E = 2.08\times10^{11}$ Pa. The valve is closed suddenly and completely.

Wave speed and pressure rise

Cross-section area $A = \dfrac{\pi}{4}(0.15)^2 = 0.017671$ m², so $V_0 = \dfrac{Q}{A} = 2.075$ m/s.

$$c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}} = \sqrt{\frac{2\times10^9/1000}{1 + \dfrac{(2\times10^9)(0.15)}{(2.08\times10^{11})(0.0028)}}} = \frac{1414.2}{\sqrt{1 + 0.5151}} = 1148.9\ \text{m/s}$$ $$\Delta p = \rho cV_0 = 1000(1148.9)(2.075) = 2.384\ \text{MPa},\qquad \Delta H = \frac{\Delta p}{\rho g} = 243.0\ \text{m}$$

Time for the wave to travel the pipe: $L/c = 2.002$ s; $2L/c = 4.004$ s; period $4L/c = 8.007$ s.

(i) Pressure variation at the valve end

$\Delta p$ rises at $t = 0$ and stays at $+2.38$ MPa above the static pressure until $t = 2L/c = 4.00$ s. Then it falls to $-2.38$ MPa until $t = 4L/c = 8.01$ s, then rises again. (Square wave, period 8.01 s.)

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(top = 1.0, bottom = -1.0; '-' is zero line)

(ii) Pressure variation at 57.5 m from the upstream tank

The point is $d = 2300 - 57.5 = 2242.5$ m from the valve. Wave arrival times:

• $t_1 = d/c = 1.952$ s: pressure rises by $+2.38$ MPa
• $t_2 = (2L - d)/c = 2.052$ s: the wave reflected at the tank (negative) returns, pressure falls back to the static value
• $t_3 = (2L + d)/c = 5.956$ s: the negative wave reflected at the valve arrives, pressure falls to $-2.38$ MPa
• $t_4 = (4L - d)/c = 6.056$ s: the positive wave reflected at the tank arrives, pressure returns to the static value

So at this point the pressure rise is a short pulse: $+2.38$ MPa for only $0.100$ s (from 1.95 to 2.05 s), then the static value (up to 5.96 s), then $-2.38$ MPa for 0.100 s (5.96 to 6.06 s), then static until $t = 4L/c + t_1$, and the cycle repeats.

(not to scale in time; pulses are narrow)
 +2.38 |       __
       |      |  |
     0 +------+--+----------__----------> t (s)
       |    1.95 2.05    5.96  6.06      8.0
 -2.38 |                   |__|

(The pulses are only about 0.1 s wide because the point is close to the tank.)

Answer: $c = 1149$ m/s; $\Delta p = 2.38$ MPa. At the valve: $\pm2.38$ MPa square wave of half-period 4.00 s. At 57.5 m from the tank: short $\pm2.38$ MPa pulses of about 0.1 s each cycle of 8.01 s.