Economic penstock diameter by empirical formulas

Given: $Q = 6$ m³/s, $H = 60$ m, overall efficiency $\eta = 85\%$ (the "n" in the question is read as efficiency).

Power

$$\begin{aligned} P &= 9.81\,Q\,H\,\eta = 9.81 \times 6 \times 60 \times 0.85 \\ &= 3001.9\ \text{kW} \end{aligned}$$

i) USBR formula (Bier)

$D = 0.176\,(P/H)^{0.466}$, with $P$ in kW and $H$ in m:

$$\begin{aligned} \frac{P}{H} &= \frac{3001.9}{60} = 50.03 \\ D &= 0.176 \times 50.03^{0.466} = 0.176 \times 6.19 = 1.09\ \text{m} \end{aligned}$$

Velocity $V = \dfrac{6}{\pi(1.09)^2/4} = 6.43$ m/s.

ii) Sarkaria formula

$D = 0.71\,P^{0.43}/H^{0.65}$ ($P$ in kW, $H$ in m):

$$\begin{aligned} D &= \frac{0.71 \times 3001.9^{0.43}}{60^{0.65}} = \frac{0.71 \times 31.28}{14.31} \\ &= 1.55\ \text{m} \end{aligned}$$

Velocity $V = \dfrac{6}{\pi(1.55)^2/4} = 3.17$ m/s.

iii) Formula based on graphical solution

The empirical fit to the graphical (cost-curve) solution: $D = 1.12\,Q^{0.45}/H^{0.12}$ ($Q$ in m³/s, $H$ in m):

$$\begin{aligned} D &= \frac{1.12 \times 6^{0.45}}{60^{0.12}} = \frac{1.12 \times 2.240}{1.635} \\ &= 1.53\ \text{m} \end{aligned}$$

Velocity $V = \dfrac{6}{\pi(1.535)^2/4} = 3.24$ m/s.

Summary

Answer: $D_{USBR} = 1.09$ m, $D_{Sarkaria} = 1.55$ m, $D_{graphical} = 1.53$ m. The Sarkaria and graphical values agree closely and give a reasonable penstock velocity of about 3 m/s, so a diameter of about 1.5 m is adopted. (The USBR velocity form $V = 0.125\sqrt{2gH} = 4.29$ m/s gives $D = 1.33$ m.)