Question

A trapezoidal open channel with a base width of 4 m and side slope 1 horizontal to 2 vertical carries water at a depth of 2 m. The flow area of section is

• A. 16 m$^2$
• B. 12 m$^2$
• C. 10 m$^2$
• D. 8 m$^2$

Hint:

The most common error in trapezoidal channel problems is misinterpreting the side slope.
Always be sure what 'm' represents. The standard formula $A=(B+my)y$ assumes $m$ is the horizontal component for a 1-unit vertical component.
If the slope is given as "X horizontal to Y vertical," then $m = X/Y$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to calculate the cross-sectional area of flow for a trapezoidal channel with given dimensions.

Step 2: Key Formula or Approach:
The formula for the cross-sectional area ($A$) of a trapezoidal channel is:
\[ A = (B + my)y \] where:
$B$ = Base width
$y$ = Depth of flow
$m$ = Side slope, defined as the ratio of horizontal to vertical (1 vertical to m horizontal).

Step 3: Detailed Explanation:
First, carefully determine the side slope parameter, $m$.
The problem states the side slope is "1 horizontal to 2 vertical".
Our standard definition of $m$ is "1 vertical to m horizontal". We need to convert the given slope to this format.
Given slope: $\frac{H}{V} = \frac{1}{2}$.
So, $m = 1/2 = 0.5$.
Now, identify the other given values:
- Base width ($B$) = 4 m
- Depth of flow ($y$) = 2 m
Substitute these values into the area formula:
\[ A = (B + my)y \] \[ A = (4 + (0.5)(2)) \times 2 \] \[ A = (4 + 1) \times 2 \] \[ A = 5 \times 2 = 10 \text{ m}^2 \]

Step 4: Final Answer:
The flow area of the section is 10 m$^2$.