---
title: "Economic diameter; penstock thickness (1.25 m)"
subject: "Hydropower Engineering"
chapter: 6
key: "semester_7/hydropower_engineering:penstock-economic-diameter-thickness-1-25m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/penstock-economic-diameter-thickness-1-25m
site: BCT Engineering
---

# Economic diameter; penstock thickness (1.25 m)

Hydropower Engineering · Chapter 6: Water Conveyance Structures

- Asked in: 2078 Bhadra, 2+4 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_7/hydropower_engineering/penstock-economic-diameter-thickness-1-25m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6

## Question

What is the economic diameter of a penstock? A steel penstock with an internal diameter of 1.25 m supplies water at a head equivalent to 18 kg/cm². There is a possibility of a 20% increase in pressure due to transient conditions. The design stress and efficiency of the joint may be assumed to be 1025 kg/cm² and 85% respectively. Compute the thickness of the penstock required.

## Short answer

Economic diameter minimizes the sum of annual fixed charges on the pipe and annual cost of power lost by friction. With 20% transient rise, design pressure is 21.6 kg/cm²; t = pD/(2σηj) = 21.6×125/(2×1025×0.85) = 1.55 cm = 15.5 mm, so provide 16 mm or more with corrosion allowance.

## Answer

#### Economic diameter
The economic diameter of a penstock is the diameter for which the sum of the annual fixed charges on the pipe and the annual cost of the power lost by friction is least. A smaller pipe costs less but loses more head; a larger pipe costs more but loses less head. At the minimum, $\dfrac{d}{dd}(\text{total annual cost}) = 0$.

#### Thickness
Design pressure including 20 % transient rise:

$$
p = 1.2 \times 18 = 21.6\ \text{kg/cm}^2
$$

With $D = 1.25\ \text{m} = 125\ \text{cm}$, $\sigma = 1025\ \text{kg/cm}^2$, $\eta_j = 0.85$:

$$
t = \frac{pD}{2\sigma\eta_j} = \frac{21.6 \times 125}{2 \times 1025 \times 0.85} = 1.55\ \text{cm}
$$

**Answer: required thickness $= 15.5$ mm. Provide 16 mm (about 17 to 18 mm if 1 to 2 mm corrosion allowance is added).**

## Check yourself

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

1. What is the condition at the economic diameter?
   A. d(total annual cost)/dd = 0, the minimum of fixed charges plus power loss cost
   B. Fixed charges equal zero
   C. Head loss equals the static head
   D. Total cost is maximum
   Answer: A. The total annual cost curve is minimum there. (Key point: Economic diameter)

2. What design pressure is used with 18 kg/cm2 and 20% transient rise?
   A. 18 kg/cm2
   B. 21.6 kg/cm2
   C. 3.6 kg/cm2
   D. 38 kg/cm2
   Answer: B. p = 1.2 x 18 = 21.6 kg/cm2. (Key point: Design pressure)

3. Which result follows from t = pD/(2 sigma eta_j)?
   A. 1.55 mm
   B. 31 mm
   C. 15.5 mm
   D. 7.7 mm
   Answer: C. t = 21.6 x 125 /(2 x 1025 x 0.85) = 1.55 cm = 15.5 mm. (Key point: Thickness)

4. What does the joint efficiency 0.85 do in the formula?
   A. It multiplies the pressure
   B. It reduces thickness
   C. It is replaced by corrosion allowance
   D. It sits in the denominator, so lower efficiency requires a thicker wall
   Answer: D. Welded joints are weaker, so eta_j reduces effective strength and thickness rises. (Key point: Joint efficiency)

## Related in Hydropower Engineering

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## Related in other subjects

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