Optimization of hydraulic tunnel diameter

The diameter of a power tunnel is chosen so that the net benefit (value of the energy saved minus the cost of the tunnel) is highest. A small diameter is cheap but gives a high velocity, a high head loss and less energy; a large diameter costs more but loses less head.

Step by step procedure

1. Select several trial diameters $D_1, D_2, \dots$ (e.g. 3.0, 3.5, 4.0, 4.5, 5.0 m) for the design flow $Q$ and the tunnel length $L$.
2. Head loss for each: $h_f=\dfrac{f\,L}{D}\dfrac{V^2}{2g}$ (or Manning), with $V=\dfrac{4Q}{\pi D^2}$, so $h_f\propto\dfrac{Q^2L}{D^5}$. Add the minor losses.
3. Power and energy loss: $P_{loss}=\rho g Q h_f\eta$, and the annual energy loss is $E_{loss}=P_{loss}\times8760\times PF$ (PF = plant factor; the variation of $Q$ over the flow-duration curve should be included).
4. Annual cost of the lost energy: $C_E=E_{loss}\times\text{tariff}$ (loss of income). Add the cost of the lost capacity.
5. Capital cost of the tunnel for each diameter: excavation, support, lining and grouting, usually of the form $C_c=k\,D^{m}$ (with $m$ about 1.2-2). Convert it to an annual cost: $C_A=C_c\times CRF+\text{O\&M}$, where CRF is the capital recovery factor for the given interest rate and life.
6. Total annual cost: $C_T(D)=C_A(D)+C_E(D)$.
7. Optimum diameter: the diameter at which $C_T$ is minimum, i.e. where $\dfrac{dC_A}{dD}+\dfrac{dC_E}{dD}=0$; or the diameter at which the incremental benefit/cost ratio (extra energy value / extra annual cost) just exceeds 1. Plot $C_A$, $C_E$ and $C_T$ against $D$ and read off the minimum.

If $C_c=kD^{m}$ and $C_E=a\,D^{-5}$, then

$$\frac{d}{dD}\left(k\,CRF\,D^{m}+aD^{-5}\right)=0\ \Rightarrow\ D_{opt}=\left(\frac{5a}{m\,k\,CRF}\right)^{\frac{1}{m+5}}$$

8. Check the practical limits: velocity (about 2.5-4 m/s for lined pressure tunnels, less for unlined ones), the minimum construction diameter (about 2.5-3 m for TBM or drill and blast), geology and surge conditions.