---
title: "Minimum specific energy for triangular channel"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:minimum-specific-energy-triangular-5-4-yc"
source: https://www.bctengineer.com/q/semester_4/hydraulics/minimum-specific-energy-triangular-5-4-yc
site: BCT Engineering
---

# Minimum specific energy for triangular channel

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2076 Bhadra, 6 marks
- Video: [Open channel flow-17 | Critical depth for triangular section | Shubham sarathe #fluid mechanics](https://www.youtube.com/watch?v=xNKDqKcbAEk) — EIQ
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/minimum-specific-energy-triangular-5-4-yc
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

Show that the minimum specific energy ($E_c$) is 5/4 times the critical depth ($y_c$) for triangular channel.

## Short answer

For a triangular channel with side slope z, critical depth yc yields velocity head Vc²/(2g) = A/(2T) = yc/4. Hence minimum specific energy Ec = yc + yc/4 = (5/4)yc.

## Answer

#### Condition for critical flow
Minimum specific energy occurs at critical flow:
$$
\frac{Q^2T}{gA^3}=1\quad\text{or}\quad \frac{Q^2}{gA^2}=\frac{A}{T}\qquad(1)
$$

#### Triangular section
Side slope $z$ (H:V), depth $y$:
$$
A=zy^2,\qquad T=2zy,\qquad \frac{A}{T}=\frac{zy^2}{2zy}=\frac y2
$$

```text
 T = 2zy
 \~~~~~~~/
  \     /  y
   \   /
    \ /
```

#### At critical depth $y=y_c$
From (1):
$$
\frac{Q^2}{gA^2}=\frac{A}{T}=\frac{y_c}{2}\;\Rightarrow\;\frac{V_c^2}{2g}=\frac{Q^2}{2gA^2}=\frac{y_c}{4}
$$
(the velocity head at critical flow is half the hydraulic depth, $\dfrac{A}{2T}=\dfrac{y_c}{4}$).

#### Minimum specific energy
$$
E_c=y_c+\frac{V_c^2}{2g}=y_c+\frac{y_c}{4}=\frac54\,y_c
$$
$$
\boxed{E_c=\frac54\,y_c}
$$
Hence proved. (Compare: rectangular channel $E_c=\tfrac32y_c$; parabolic channel $E_c=\tfrac43y_c$. In general $E_c=y_c+\dfrac{A_c}{2T_c}$.)

## Check yourself

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

1. The critical flow condition used for any channel section is
   A. Q^2 T/(g A^3) = 1
   B. Q^2 A/(g T) = 1
   C. Q T/(g A) = 1
   D. V = 2 sqrt(g y)
   Answer: A. Setting dE/dy = 0 gives Q^2 T/(g A^3) = 1, i.e. Fr = 1. (Key point: Critical flow condition)

2. For a triangular channel with side slope z, A/T equals
   A. z y
   B. 2 y
   C. y
   D. y/2
   Answer: D. A = z y^2 and T = 2 z y, so A/T = y/2. (Key point: Triangular section geometry)

3. At critical flow in a triangular channel, V^2/2g equals
   A. yc/2
   B. yc/4
   C. yc
   D. yc/8
   Answer: B. V^2/(2g) = A/(2T) = yc/4. (Key point: Velocity head at critical)

4. Hence Ec for a triangular channel is
   A. 3/2 yc
   B. 4/3 yc
   C. 5/4 yc
   D. 2 yc
   Answer: C. Ec = yc + yc/4 = 5/4 yc (rectangular gives 3/2, parabolic 4/3). (Key point: Result)

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