Economic diameter factors; penstock wall thickness (3 m)

Factors governing economic diameter

The economic diameter is found by balancing the annual cost of the penstock against the annual value of the energy lost in friction. It depends on:

• Discharge and head: a larger $Q$ or a lower head needs a larger diameter, because the same loss is a larger share of the head.
• Cost of steel and fabrication: cost of plates, welding, transport, erection, painting and anchor blocks; the pipe cost varies roughly with $d^2$ (weight $\propto d \cdot t$ and $t \propto d$).
• Annual charges: interest, depreciation, maintenance and insurance, as a percentage of the penstock cost.
• Value of energy lost: tariff or unit cost of power, and the plant load factor; a high value of power favours a bigger pipe.
• Friction loss: pipe roughness (friction factor), length and velocity; the loss varies as $Q^2/d^5$.
• Turbine and overall efficiency: the loss of power is $\gamma Q h_f \eta$.
• Permissible velocity: limits for erosion (silt), water hammer and noise; usually 3 to 6 m/s (up to 8 m/s for high heads).
• Other factors: surge/water-hammer pressures, bends and fittings, transport limits on plate size, and the number of penstocks.

Wall thickness of the penstock

Design pressure head $= 1.4 \times 220 = 308$ m of water. Since 10 m of water $\approx 1\ \text{kg/cm}^2$,

$$p = 30.8\ \text{kg/cm}^2$$

For a thin-walled pipe (hoop stress), with $D = 300$ cm, $\sigma_{st} = 1400\ \text{kg/cm}^2$, $\eta_j = 0.95$:

$$t = \frac{pD}{2\sigma_{st}\eta_j} = \frac{30.8 \times 300}{2 \times 1400 \times 0.95} = 3.47\ \text{cm}$$

Answer: thickness $\approx 34.7$ mm; in practice provide about 36 to 37 mm (34.7 mm plus a 1 to 2 mm corrosion allowance).