Economic diameter of penstock (analysis)

The economic diameter of a penstock is the diameter at which the total annual cost (cost of the pipe plus cost of the energy lost in head loss) is least. A larger diameter needs more steel but loses less head.

Mathematical analysis

1. Cost of the penstock. The thickness is $t=\dfrac{p\,D}{2\sigma\eta}=\dfrac{\gamma H D}{2\sigma\eta}$ (neglecting the corrosion allowance), where $H$ is the head including water hammer, $\sigma$ is the allowable stress and $\eta$ is the joint efficiency. The weight of steel per unit length is $W=\pi D\,t\,\gamma_s\propto D^2$. With the unit cost $c_s$ per unit weight and the annual charge rate $r$ (interest + depreciation + maintenance):

$$C_1=r\,c_s\,\pi\gamma_s\frac{\gamma H}{2\sigma\eta}\,L\,D^2=K_1\,D^2$$

2. Cost of the energy lost. The head loss in a pipe of length $L$ carrying $Q$ is

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

The power lost is $P=\gamma Q h_f\eta_{pl}$, and the annual value of the lost energy (with plant factor PF and unit rate $c_e$) is

$$C_2=\gamma Q\,\eta_{pl}\,c_e\,(8760\,PF)\frac{8fLQ^2}{\pi^2gD^5}=\frac{K_2\,Q^3}{D^5}$$

3. Total annual cost and optimisation.

$$C=K_1D^2+\frac{K_2Q^3}{D^5}$$ $$\frac{dC}{dD}=2K_1D-\frac{5K_2Q^3}{D^6}=0\ \Rightarrow\ D_{econ}=\left(\frac{5K_2}{2K_1}\right)^{1/7}Q^{3/7}$$

Thus $D_{econ}\propto Q^{3/7}$. At the optimum, the cost of the energy lost is $\frac{2}{5}$ of the cost of the pipe: $C_2=\frac{2}{5}C_1$.

4. Practical use. Several empirical formulas relate $D$ to $Q$ and $H$; some of them (e.g. Bondschu's, $D=\sqrt[7]{5.2Q^3/H}$) show the same trend $D\propto Q^{3/7}$ found above. After the analysis, the diameter is rounded to a standard plate/pipe size, and the velocity (about 3-6 m/s), the head loss and the water hammer are checked.