Penstock plate thickness (1.5 m, 250 m head)

Given: $D = 1.5$ m $= 1500$ mm, static head $H = 250$ m, water hammer allowance 30%, design stress $\sigma = 120$ N/mm², joint efficiency $\eta_j = 0.85$.

Formula (thin cylinder, hoop stress):

$$t = \frac{p\,D}{2\,\sigma\,\eta_j}$$

Design pressure

$$\begin{aligned} H_d &= 1.30 \times 250 = 325\ \text{m} \\ p &= \gamma_w H_d = 9810 \times 325 = 3.188 \times 10^6\ \text{N/m}^2 \\ &= 3.188\ \text{N/mm}^2 \end{aligned}$$

Thickness

$$\begin{aligned} t &= \frac{3.188 \times 1500}{2 \times 120 \times 0.85} \\ &= \frac{4782.4}{204} = 23.44\ \text{mm} \end{aligned}$$

Answer: required plate thickness $t = 23.44$ mm, say 24 mm (adding a corrosion allowance of about 2 mm, a 26 mm plate would be used in practice).