Penstock thickness and economic diameter

Thickness of penstock

A penstock is a thin cylinder under internal pressure. Consider a unit length of pipe of internal diameter $D$ cut along a diameter. The bursting force due to pressure $p$ is resisted by hoop tension in the two walls:

$$\begin{aligned} p \times D \times 1 &= 2\,\sigma\, t \times 1 \times \eta_j \\ \Rightarrow\ t &= \frac{p\,D}{2\,\sigma\,\eta_j} + t_c \end{aligned}$$

where $p = \gamma_w (H + h_{wh})$ is the design pressure (static head plus water hammer rise, typically 15–30%), $\sigma$ = allowable (design) stress of steel, $\eta_j$ = joint efficiency (0.85–0.95), $t_c$ = corrosion allowance (1.5–3 mm). A minimum thickness for handling is also used, e.g. $t_{min} = \dfrac{D + 500}{400}$ mm ($D$ in mm), not less than 6 mm.

Economic diameter of penstock

The economic diameter gives minimum total annual cost = annual cost of the penstock + annual value of energy lost by friction.

1. Pipe weight $\propto \pi D t$, and $t \propto pD$, so the annual cost of the penstock is

$$C_1 = k_1 D^2$$

2. Head loss (Darcy):

$$h_f = \frac{f L V^2}{2 g D} = \frac{8 f L Q^2}{\pi^2 g D^5}$$

Power lost $= 9.81\,Q\,h_f\,\eta$; annual value of lost energy

$$C_2 = \frac{k_2}{D^5}$$

3. Total annual cost and minimum:

$$\begin{aligned} C &= k_1 D^2 + \frac{k_2}{D^5} \\ \frac{dC}{dD} &= 2k_1 D - \frac{5k_2}{D^6} = 0 \\ D_{eco} &= \left(\frac{5k_2}{2k_1}\right)^{1/7} \end{aligned}$$

At the optimum, $2C_1 = 5C_2$, i.e. annual pipe cost is 2.5 times the annual cost of energy lost.

4. Graphical method: compute $C_1$ and $C_2$ for several trial diameters, plot them and their sum; the minimum of the total curve gives $D_{eco}$.

5. Empirical formulas for preliminary design ($P$ in kW, $H$ in m, $D$ in m):

• USBR (Bier): $D = 0.176\,(P/H)^{0.466}$
• Sarkaria: $D = 0.71\,P^{0.43}/H^{0.65}$
• Fahlbusch: $D = 0.52\,H^{-0.17}(P/H)^{0.43}$