Water hammer in 200 mm pipe

For abrupt (instantaneous) closure, the pressure rise is given by Joukowsky's equation and does not depend on the pipe diameter.

Pressure rise

$$\begin{aligned} \Delta p &= \rho c V = 1000 \times 1400 \times 2 = 2.8 \times 10^6\ \text{N/m}^2 = 2.8\ \text{MPa} \\ \Delta H &= \frac{cV}{g} = \frac{1400 \times 2}{9.81} = 285.4\ \text{m of water} \end{aligned}$$

Period of oscillation

One full cycle of the pressure wave takes four trips along the pipe:

$$T = \frac{4L}{c} = \frac{4L}{1400} = \frac{L}{350}\ \text{s}$$

The pipe length L is not given, so T is left in terms of L. For example, for $L = 1000$ m, $T = 4 \times 1000/1400 = 2.86$ s.

Answer: Pressure rise = 2.8 MN/m² (≈ 285.4 m head); period $T = 4L/c = L/350$ s (2.86 s if L = 1000 m).