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Penstock thickness (1.20 m, 17.5 kg/cm²)

Short answer

Design pressure p = 1.20 × 17.5 = 21 kg/cm². Thickness t = pD/(2 σ η_j) = (21 × 120)/(2 × 1050 × 0.87) = 1.379 cm = 13.79 mm, so use 14 mm.

Calculate the thickness of a penstock of internal diameter 1.20 m which conveys water at a pressure head of 17.5 kg/cm². There is a possibility of a 20% increase in pressure due to the water hammer. Take design stress = 1050 kg/cm² and joint efficiency = 87%.

Given: D=1.20D = 1.20 m =120= 120 cm, pressure p0=17.5p_0 = 17.5 kg/cm², water hammer rise 20%, design stress σ=1050\sigma = 1050 kg/cm², joint efficiency ηj=0.87\eta_j = 0.87.

Formula (hoop stress in a thin-walled pipe):

t=p D2 σ ηjt = \frac{p\,D}{2\,\sigma\,\eta_j}

Design pressure

p=1.20×17.5=21 kg/cm2p = 1.20 \times 17.5 = 21\ \text{kg/cm}^2

Thickness

t=21×1202×1050×0.87=25201827=1.379 cm=13.79 mm\begin{aligned} t &= \frac{21 \times 120}{2 \times 1050 \times 0.87} \\ &= \frac{2520}{1827} = 1.379\ \text{cm} = 13.79\ \text{mm} \end{aligned}

Answer: penstock thickness t=13.8t = 13.8 mm, say 14 mm (a corrosion allowance of about 2 mm may be added, giving 16 mm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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