---
title: "Most economic penstock diameter (cost numerical)"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:penstock-economic-diameter-cost-numerical"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/penstock-economic-diameter-cost-numerical
site: BCT Engineering
---

# Most economic penstock diameter (cost numerical)

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2076 Ashwin, 8 marks
- Video: [How to calculate Economic diameter of penstock](https://www.youtube.com/watch?v=b9TGJ3_cPZA) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/penstock-economic-diameter-cost-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

A penstock carries 8 m³/s of water at a head of 25 m. The cost of the pipe line in place is given by US\$250hd² per meter length, where h = head and d = diameter of the pipe. Annual fixed charges are 8% of the pipe line cost. The estimated head loss in friction is $\frac{0.025Q^2}{12.1d^5}$ per m length of the pipe. Efficiency of the turbine is 80% and the selling price of the power is US\$500 per kW per annum. Calculate the most economic diameter of the penstock.

## Short answer

Minimise total annual cost per metre: fixed charge 500d² plus power-loss cost 4151/d⁵. Setting derivative zero gives d⁷ = 20.755, hence most economic diameter is 1.54 m.

## Answer

#### Given
$Q = 8\ \text{m}^3/\text{s}$, $h = 25$ m, $\eta = 0.8$.

- Pipe cost per metre: $C = 250\,h\,d^2 = 250(25)d^2 = 6250\,d^2$ (US\$)
- Annual fixed charge: $0.08 \times 6250\,d^2 = 500\,d^2$ (US\$/m/year)

#### Head loss and power loss per metre
$$
h_f = \frac{0.025\,Q^2}{12.1\,d^5} = \frac{0.025 \times 64}{12.1\,d^5} = \frac{0.1322}{d^5}\ \text{m/m}
$$

Power lost per metre length:

$$
P = \gamma Q h_f \eta = 9.81 \times 8 \times 0.8 \times \frac{0.1322}{d^5} = \frac{8.302}{d^5}\ \text{kW}
$$

Annual cost of lost power at US\$500 per kW per year:

$$
C_{loss} = 500 \times \frac{8.302}{d^5} = \frac{4151}{d^5}
$$

#### Total annual cost and minimum
$$
C_T = 500\,d^2 + \frac{4151}{d^5}
$$

$$
\frac{dC_T}{dd} = 1000\,d - \frac{20755}{d^6} = 0 \;\Rightarrow\; d^7 = 20.755
$$

$$
d = (20.755)^{1/7} = 1.542\ \text{m}
$$

Check: at $d = 1.542$ m the fixed charge is US\$1189 and the power-loss cost is US\$476, total US\$1665 per metre per year.

**Answer: most economic diameter $\approx 1.54$ m.**

## Check yourself

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

1. What is the annual fixed charge per metre?
   A. 0.08 x 6250 d^2 = 500 d^2
   B. 0.08 x 250 d = 20 d
   C. 6250 d^2
   D. 8 x 6250 d^2
   Answer: A. Cost per metre is 250 h d^2 = 6250 d^2; 8% gives 500 d^2. (Key point: Annual fixed charge)

2. How is power lost per metre found?
   A. P = Q hf only
   B. P = gamma Q hf eta = 8.302/d^5 kW
   C. P = gamma h d^2
   D. P = 9.81 x 8 x 25 kW
   Answer: B. hf = 0.1322/d^5 so P = 9.81 x 8 x 0.8 x 0.1322/d^5 = 8.302/d^5. (Key point: Power lost per metre)

3. How is the most economic diameter obtained?
   A. Set 500 d^2 = 0
   B. Set the two terms to their maximum
   C. Differentiate C = 500 d^2 + 4151/d^5, set to zero: d^7 = 20.755
   D. Choose d where hf = 25 m
   Answer: C. dC/dd = 1000 d - 20755/d^6 = 0 gives d^7 = 20.755. (Key point: Optimisation)

4. What is the economic diameter?
   A. About 0.54 m
   B. About 2.9 m
   C. About 20.8 m
   D. About 1.54 m
   Answer: D. d = 20.755^(1/7) = 1.542 m. (Key point: Result)

## Related in Hydropower Engineering

- [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
- [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 of penstock (analysis)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aeconomic-diameter-penstock-mathematical) — Hydropower Engineering, chapter 6, asked 2×
- [Pelton wheel 70 kW: flow rate and diameter](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-wheel-70kw-100m-400rpm) — Hydropower Engineering, chapter 7
- [Surge numerical: 4.5 m tunnel, 2500 m, 25 m³/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-tunnel-4-5m-2500m-25cumec) — Hydropower Engineering, chapter 6, asked 2×

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