---
title: "Geometric properties from channel offsets"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:open-channel-cross-section-from-table-properties"
source: https://www.bctengineer.com/q/semester_4/hydraulics/open-channel-cross-section-from-table-properties
site: BCT Engineering
---

# Geometric properties from channel offsets

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2078 Chaitra, 1+1+1+1+1 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/open-channel-cross-section-from-table-properties
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Plot the open channel cross-section from the following table given below and calculate the following geometric properties: area, conveyance factor, equivalent hydraulic radius, hydraulic depth and section factor for uniform flow. Take manning's n = 0.02.

| Distance from left bank (m) | Water depth (m) |
|---|---|
| 0.0 | 0.0 |
| 1.0 | 3.0 |
| 3.0 | 3.0 |
| 5.0 | 5.0 |
| 7.0 | 3.0 |
| 10.0 | 0.0 |

## Short answer

By segment summation: area A=28.00 m², wetted perimeter P=15.06 m, hydraulic radius R=1.859 m. Conveyance K= (1/0.02)×28×1.859^{2/3}=2117 m³/s, hydraulic depth D_h=2.80 m, section factor AR^{2/3}=42.33 m^{8/3}.

## Answer

#### Plotting the section
Take distance as $x$ and depth as measured below the water surface (the depth is $0$ at both banks, so the water surface is level with the ground at $x=0$ and $x=10$ m). Top width $T=10$ m, maximum depth $5$ m at $x=5$ m.

```text
 0    2    4    6    8   10  (m)
 ~~~~~~~~~~~~~~~~~~~~~~~~~~  water surface
 \                        /
  \__ 3 m ___      ___   /
          5 m  \  /   /
                \/
```
(points: (0,0), (1,3), (3,3), (5,5), (7,3), (10,0))

#### Area and wetted perimeter by segments
Area of each strip = width × mean depth; length = $\sqrt{\Delta x^2+\Delta y^2}$.

| Segment | $\Delta x$ (m) | Depth (m) | Area (m²) | Length (m) |
|---|---|---|---|---|
| 1 | 1 | 0 → 3 | 1.50 | 3.162 |
| 2 | 2 | 3 → 3 | 6.00 | 2.000 |
| 3 | 2 | 3 → 5 | 8.00 | 2.828 |
| 4 | 2 | 5 → 3 | 8.00 | 2.828 |
| 5 | 3 | 3 → 0 | 4.50 | 4.243 |
| **Total** | | | **28.00** | **15.062** |

#### (i) Area
$A=28.00$ m².

#### (ii) Equivalent hydraulic radius
$$
R=\frac{A}{P}=\frac{28.00}{15.062}=1.8590\ \text{m}
$$

#### (iii) Conveyance factor
$$
K=\frac{1}{n}AR^{2/3}=\frac{1}{0.02}\times28.00\times(1.8590)^{2/3}=2116.7\ \text{m}^3/\text{s}
$$
(so that $Q=K\sqrt{S_0}$; the section factor is $AR^{2/3}=42.333$ m$^{8/3}$.)

#### (iv) Hydraulic depth
$$
D_h=\frac{A}{T}=\frac{28.00}{10}=2.80\ \text{m}
$$

#### (v) Section factor for uniform flow
$$
AR^{2/3}=28.00\times(1.8590)^{2/3}=42.333\ \text{m}^{8/3}
$$

**Answer: $A=28.0$ m², $K=2117$, $R=1.859$ m, $D_h=2.8$ m, $AR^{2/3}=42.33$ m$^{8/3}$.**

## Check yourself

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

1. How is the area computed from the tabulated points?
   A. Sum of strips: width x mean depth
   B. Maximum depth x top width
   C. Top width x minimum depth
   D. P x R
   Answer: A. Each strip has area = delta x times average of its end depths; the total is 28.0 m^2. (Key point: Area method)

2. How is wetted perimeter computed?
   A. Top width only
   B. Sum of sloping segment lengths sqrt(dx^2 + dy^2)
   C. Sum of depths
   D. 2 x maximum depth
   Answer: B. Each segment length is sqrt(dx^2 + dy^2); the total is 15.06 m. (Key point: Wetted perimeter)

3. What is the hydraulic depth?
   A. A/P = 1.86 m
   B. Maximum depth = 5 m
   C. A/T = 28/10 = 2.8 m
   D. Mean of the table depths
   Answer: C. Hydraulic depth is area divided by top width. (Key point: Hydraulic depth)

4. What is the section factor for uniform flow?
   A. A sqrt(D) 
   B. A R^(2/3) = 42.33 m^(8/3)
   C. A/T
   D. R^(2/3)/n
   Answer: B. A R^(2/3) = 28 x 1.859^(2/3) = 42.33; dividing by n gives conveyance K. (Key point: Section factor)

## Related in Hydraulics

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