---
title: "Critical depth in four channel sections"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:critical-depth-four-channel-sections-6-cumecs"
source: https://www.bctengineer.com/q/semester_4/hydraulics/critical-depth-four-channel-sections-6-cumecs
site: BCT Engineering
---

# Critical depth in four channel sections

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

- Asked in: 2069 Poush, 8 marks
- Video: [Critical Depth and Specific Energy](https://www.youtube.com/watch?v=GPQYjxmruxM) — Seema Shah-Fairbank
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/critical-depth-four-channel-sections-6-cumecs
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

Calculate the critical depth for a discharge of 6 cumecs in the following section of channel: i) Circular having diameter 1.5 m. ii) Rectangular having bed width 3 m. iii) Trapezoidal having bed width 2.5 m and side slope 2:1. iv) Triangular having side slope 1:1.

## Short answer

Using A^3/T = Q^2/g: circular D=1.5 m, yc = 1.26 m; rectangular b=3 m, yc = 0.74 m; trapezoidal b=2.5 m, z=2, yc = 0.69 m; triangular z=1, yc = 1.49 m.

## Answer

**Condition for critical flow:**

$$
\frac{Q^2}{g} = \frac{A^3}{T}
$$

For $Q = 6\ \text{m}^3/\text{s}$: $\dfrac{Q^2}{g} = \dfrac{36}{9.81} = 3.670\ \text{m}^5$.

#### i) Circular, $D = 1.5\ \text{m}$

With $\theta$ (rad) the angle subtended at the centre by the water surface:

$$
A = \frac{D^2}{8}(\theta - \sin\theta),\quad T = D\sin\frac{\theta}{2},\quad y = \frac{D}{2}\left(1 - \cos\frac{\theta}{2}\right)
$$

Trial and error (bisection) on $A^3/T = 3.670$ gives $\theta = 4.651\ \text{rad}$:

$$
A = 1.589\ \text{m}^2,\quad T = 1.093\ \text{m},\quad \frac{A^3}{T} = \frac{4.011}{1.093} = 3.670 \checkmark
$$

$$
y_c = 0.75\,(1 - \cos 2.3254) = 1.264\ \text{m}
$$

#### ii) Rectangular, $b = 3\ \text{m}$

$$
q = \frac{6}{3} = 2\ \text{m}^2/\text{s},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{4}{9.81}\right)^{1/3} = 0.742\ \text{m}
$$

#### iii) Trapezoidal, $b = 2.5\ \text{m}$, $z = 2$ (2H : 1V)

$$
A = (2.5 + 2y)y,\quad T = 2.5 + 4y
$$

Solve $\dfrac{[(2.5 + 2y)y]^3}{2.5 + 4y} = 3.670$. Trial: $y = 0.691$ gives $A = 2.683$, $T = 5.265$, $A^3/T = 19.32/5.265 = 3.670$.

$$
y_c = 0.691\ \text{m}
$$

#### iv) Triangular, $z = 1$ (1H : 1V)

$A = y^2$, $T = 2y$, so $\dfrac{y^6}{2y} = \dfrac{Q^2}{g}$:

$$
y_c = \left(\frac{2Q^2}{g\,z^2}\right)^{1/5} = \left(\frac{2 \times 36}{9.81}\right)^{1/5} = 1.490\ \text{m}
$$

| Section | $y_c$ (m) |
|---|---|
| Circular, $D = 1.5$ m | 1.264 |
| Rectangular, $b = 3$ m | 0.742 |
| Trapezoidal, $b = 2.5$ m, $z = 2$ | 0.691 |
| Triangular, $z = 1$ | 1.490 |

## Check yourself

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

1. Which condition is solved for critical depth in any section?
   A. Q^2/g = A^3/T
   B. Q = A R^(2/3) S^(1/2)/n
   C. E = y + V^2/2g is maximum
   D. Q^2/g = A T^3
   Answer: A. Fr = 1 gives Q^2 T/(g A^3) = 1, i.e. Q^2/g = A^3/T = 3.670 m^5 here. (Key point: General critical condition)

2. For b = 3 m and Q = 6 m^3/s, how is yc found?
   A. yc = q/g = 0.204 m
   B. yc = (q^2/g)^(1/3) with q = 2 m^2/s, giving 0.742 m
   C. yc = Q/b = 2 m
   D. yc = (Q^2/g)^(1/3) = 1.54 m
   Answer: B. q = 6/3 = 2, yc = (4/9.81)^(1/3) = 0.742 m. (Key point: Rectangular section)

3. For a triangular section with A = y^2 and T = 2y, A^3/T = Q^2/g reduces to:
   A. y^3 = Q^2/g
   B. y^4 = Q^2/g
   C. y^5 = 2Q^2/g
   D. y^6 = Q^2/(2g)
   Answer: C. y^6/(2y) = Q^2/g gives y^5 = 2Q^2/g, so yc = (72/9.81)^(1/5) = 1.49 m. (Key point: Triangular section)

4. How are yc values found for circular and trapezoidal sections?
   A. Direct closed-form formula
   B. Trial and error on A^3/T until it equals Q^2/g
   C. Using yc = 2/3 E
   D. Using Manning's equation
   Answer: B. A and T depend non-linearly on y, so the depth is found by trial until A^3/T = 3.670. (Key point: Circular and trapezoidal method)

5. Which section has the smallest critical depth for the same Q = 6 m^3/s?
   A. Circular, D = 1.5 m
   B. Triangular
   C. Rectangular
   D. Trapezoidal b = 2.5 m, z = 2 (0.691 m)
   Answer: D. The wide flaring trapezoid has large top width and area, giving the smallest yc of 0.691 m. (Key point: Ordering of results)

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