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Economic diameter; penstock thickness (1.25 m)

Short answer

Economic diameter minimizes the sum of annual fixed charges on the pipe and annual cost of power lost by friction. With 20% transient rise, design pressure is 21.6 kg/cm²; t = pD/(2σηj) = 21.6×125/(2×1025×0.85) = 1.55 cm = 15.5 mm, so provide 16 mm or more with corrosion allowance.

What is the economic diameter of a penstock? A steel penstock with an internal diameter of 1.25 m supplies water at a head equivalent to 18 kg/cm². There is a possibility of a 20% increase in pressure due to transient conditions. The design stress and efficiency of the joint may be assumed to be 1025 kg/cm² and 85% respectively. Compute the thickness of the penstock required.

Economic diameter

The economic diameter of a penstock is the diameter for which the sum of the annual fixed charges on the pipe and the annual cost of the power lost by friction is least. A smaller pipe costs less but loses more head; a larger pipe costs more but loses less head. At the minimum, ddd(total annual cost)=0\dfrac{d}{dd}(\text{total annual cost}) = 0.

Thickness

Design pressure including 20 % transient rise:

p=1.2×18=21.6 kg/cm2p = 1.2 \times 18 = 21.6\ \text{kg/cm}^2

With D=1.25 m=125 cmD = 1.25\ \text{m} = 125\ \text{cm}, σ=1025 kg/cm2\sigma = 1025\ \text{kg/cm}^2, ηj=0.85\eta_j = 0.85:

t=pD2σηj=21.6×1252×1025×0.85=1.55 cmt = \frac{pD}{2\sigma\eta_j} = \frac{21.6 \times 125}{2 \times 1025 \times 0.85} = 1.55\ \text{cm}

Answer: required thickness =15.5= 15.5 mm. Provide 16 mm (about 17 to 18 mm if 1 to 2 mm corrosion allowance is added).

✓ Answer checked by Claude Opus 5.5 (high effort)

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