---
title: "Economic penstock diameter by empirical formulas"
subject: "Hydro Power"
chapter: 5
key: "semester_6/hydro_power:penstock-economic-diameter-formulas"
source: https://www.bctengineer.com/q/semester_6/hydro_power/penstock-economic-diameter-formulas
site: BCT Engineering
---

# Economic penstock diameter by empirical formulas

Hydro Power · Chapter 5: Components of Hydropower System

- Asked in: 2078 Chaitra, 8 marks
- Video: [Penstock Thickness and Economic Diameter Calculation | Formulae Explained #Thickness #economic](https://www.youtube.com/watch?v=4UqCfkXvVI8) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/penstock-economic-diameter-formulas
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5

## Question

A penstock is to convey 6 m³/s discharge with a design head of 60 m. Determine the economic diameter of the penstock based on i) USBR formula, ii) Sarkaria formula, and iii) formula based on graphical solution. Take n = 85%.

## Short answer

Power = 9.81 × 6 × 60 × 0.85 = 3002 kW. Using USBR: D = 0.176(3002/60)^0.466 = 1.09 m; Sarkaria: D = 0.71×3002^0.43/60^0.65 = 1.55 m; graphical formula: D = 1.12×6^0.45/60^0.12 = 1.53 m. Adopt about 1.5 m.

## Answer

**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
| Formula | $D$ (m) | $V$ (m/s) |
|---|---|---|
| USBR (Bier) | 1.09 | 6.43 |
| Sarkaria | 1.55 | 3.17 |
| Graphical-solution formula | 1.53 | 3.24 |

**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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the power computed here for the empirical formulas?
   A. P = 9.81 Q H eta, about 3002 kW
   B. P = 9.81 Q H, about 3532 kW
   C. P = Q H eta, about 306 kW
   D. P = 9.81 Q eta / H, about 0.85 kW
   Answer: A. 9.81 x 6 x 60 x 0.85 = 3001.9 kW. (Key point: Power)

2. Which formula and result is the USBR (Bier) one?
   A. D = 0.71 P^0.43 / H^0.65, giving 1.09 m
   B. D = 0.176 (P/H)^0.466, giving 1.09 m
   C. D = 0.176 (P/H)^0.466, giving 1.55 m
   D. D = 1.12 Q^0.45 / H^0.12, giving 1.09 m
   Answer: B. USBR gives D = 0.176 x 50.03^0.466 = 1.09 m. (Key point: USBR formula)

3. Why is the USBR result of 1.09 m questioned?
   A. It gives a velocity below 1 m/s
   B. It is larger than the Sarkaria value
   C. It gives a high velocity of about 6.4 m/s
   D. It ignores the discharge
   Answer: C. V = 6/(pi x 1.09^2/4) = 6.43 m/s, too high; Sarkaria and graphical give about 3 m/s. (Key point: Velocity check)

4. Which diameter is finally adopted?
   A. About 1.09 m, from USBR
   B. About 3 m
   C. About 0.85 m
   D. About 1.5 m, since Sarkaria (1.55 m) and graphical (1.53 m) agree
   Answer: D. Those two agree and give a reasonable velocity of about 3 m/s. (Key point: Adopted diameter)

## Related in Hydro Power

- [Economic diameter of penstock](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-economic-diameter) — Hydro Power, chapter 5, asked 5×
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## Related in other subjects

- [Economic diameter of penstock by graphical method](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aeconomic-diameter-penstock-graphical) — Hydropower Engineering, chapter 6
- [Economic diameter; penstock thickness (1.25 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-economic-diameter-thickness-1-25m) — Hydropower Engineering, chapter 6
- [Economic diameter factors; penstock wall thickness (3 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aeconomic-diameter-and-wall-thickness-factors) — Hydropower Engineering, chapter 6
- [Most economic penstock diameter (cost numerical)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-economic-diameter-cost-numerical) — Hydropower Engineering, chapter 6