Water hammer in 250 m penstock

Sudden closure means the valve closes in less than $2L/c$, so the full Joukowsky pressure rise occurs.

Pressure rise

$$\begin{aligned} \Delta p &= \rho c V = 1000 \times 1450 \times 2.5 = 3.625 \times 10^6\ \text{N/m}^2 = 3.625\ \text{MPa} \\ \Delta H &= \frac{cV}{g} = \frac{1450 \times 2.5}{9.81} = 369.5\ \text{m of water} \end{aligned}$$

Period of oscillation

The pressure wave travels to the reservoir and back twice in one full cycle (high pressure phase + low pressure phase):

$$\begin{aligned} T &= \frac{4L}{c} = \frac{4 \times 250}{1450} = 0.690\ \text{s} \end{aligned}$$

(The time for one round trip, $2L/c = 0.345$ s, is the critical closure time.)

Answer: Pressure rise = 3.625 MN/m² (≈ 369.5 m head); period of oscillation = 0.69 s.