---
title: "Economic diameter of penstock (analysis)"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:economic-diameter-penstock-mathematical"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/economic-diameter-penstock-mathematical
site: BCT Engineering
---

# Economic diameter of penstock (analysis)

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2081 Bhadra, 5 marks; 2072 Chaitra, 4 marks (asked 2 times)
- Video: [Analytical Method of Penstock Pipe Optimization](https://www.youtube.com/watch?v=7CEgquj9IAI) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/economic-diameter-penstock-mathematical
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

How can the economical diameter of a penstock be determined using mathematical analysis (explain with mathematical expression the optimization of a penstock)?

## Short answer

The economical diameter minimizes annual total cost: pipe cost ∝ D² + lost energy cost ∝ Q³/D⁵. Setting dC/dD=0 yields D_econ = (5K₂/(2K₁))^(1/7)·Q^(3/7); cost of lost energy equals 2/5 of pipe cost. Practical formulas give D ∝ Q^(3/7); check velocity 3–6 m/s.

## Answer

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.

## Check yourself

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

1. How does annual pipe cost vary with D?
   A. C1 = K1 D^2, since thickness is proportional to D and weight to D t
   B. C1 proportional to 1/D^5
   C. C1 independent of D
   D. C1 proportional to D^(1/2)
   Answer: A. t = gamma H D/(2 sigma eta) is proportional to D and steel weight pi D t is proportional to D^2. (Key point: Pipe cost vs diameter)

2. How does annual cost of lost energy vary?
   A. C2 = K2 Q^3 / D^5
   B. C2 = K2 Q D^2
   C. C2 = K2 Q^2 D
   D. C2 independent of Q
   Answer: A. h_f is proportional to Q^2/D^5 and power loss is proportional to Q h_f, giving Q^3/D^5. (Key point: Energy loss cost)

3. What does dC/dD = 0 give?
   A. D_econ = (5K2/2K1)^(1/7) Q^(3/7)
   B. D_econ proportional to Q^2
   C. D_econ proportional to Q^(1/2)
   D. D_econ = K1/K2
   Answer: A. 2K1 D = 5 K2 Q^3/D^6 gives D^7 = 5K2Q^3/(2K1), so D is proportional to Q^(3/7). (Key point: Economic diameter result)

4. At the optimum, how are energy-loss cost and pipe cost related?
   A. C2 = (2/5) C1
   B. C2 = C1
   C. C2 = 5 C1
   D. C2 = 0
   Answer: A. From 2K1D^2 = 5K2Q^3/D^5 we get C2 = K2Q^3/D^5 = (2/5)K1D^2 = (2/5)C1. (Key point: Cost relation at the optimum)

## Related in Hydropower Engineering

- [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 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
- [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
- [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
- [Optimization of hydraulic tunnel diameter](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Atunnel-optimization-diameter) — Hydropower Engineering, chapter 6, asked 2×

## Related in other subjects

- [Penstock thickness and economic diameter](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apenstock-thickness-economic-diameter) — Hydro Power, chapter 5
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