Problem 92 - Algebra and Trigonometry

Question

Flow Speed in a Channel The speed of water flowing in a channel, such as a canal or river bed, is governed by the Manning Equation, $$ V=1.486 \frac{A^{2 / 3} S^{1 / 2}}{p^{2 / 3} n} $$ Here \(V\) is the velocity of the flow in \(\mathrm{ft} / \mathrm{s}\) ; \(A\) is the crosssectional area of the channel in square feet; \(S\) is the downward slope of the channel; \(p\) is the wetted perimeter in feet (the distance from the top of one bank, down the side of the channel, across the bottom, and up to the top of the other bank); and \(n\) is the roughness coefficient (a measure of the roughness of the channel bottom). This equation is used to predict the capacity of flood channels to handle runoff from heavy rainfalls. For the canal shown in the figure, \(A=75 \mathrm{ft}^{2}, S=0.050\) , \(p=24.1 \mathrm{ft},\) and \(n=0.040\) (a) Find the speed at which water flows through the canal. (b) How many cubic feet of water can the canal discharge per second? \([\text { Hint: Multiply } V \text { by } A \text { to get the volume of the }\) flow per second.]

Step-by-Step Solution

Answer

(a) The flow speed is approximately 17.742 ft/s. (b) The canal can discharge around 1330.65 cubic feet per second.

Step 1: Identify Known Values

We are given the following values: \(A = 75 \, \text{ft}^2\), \(S = 0.050\), \(p = 24.1 \, \text{ft}\), and \(n = 0.040\). These will be used to calculate the flow speed using the Manning equation.

Step 2: Substitute Values into Manning Equation

Substitute the known values into the Manning equation: \[ V = 1.486 \frac{A^{2/3} S^{1/2}}{p^{2/3} n} \] This becomes \[ V = 1.486 \frac{75^{2/3} \times 0.050^{1/2}}{24.1^{2/3} \times 0.040} \]

Step 3: Calculate Each Component

Calculate each component separately:- \(A^{2/3} = 75^{2/3} \approx 18.849\)- \(S^{1/2} = 0.050^{1/2} \approx 0.2236\)- \(p^{2/3} = 24.1^{2/3} \approx 8.836\)

Step 4: Compute Flow Speed (V)

Now, substitute these calculated values back into the formula: \[ V = 1.486 \times \frac{18.849 \times 0.2236}{8.836 \times 0.040} \] Simplifying further, we find: \[ V \approx 1.486 \times \frac{4.2158}{0.35344} \approx 17.742 \, \text{ft/s} \]

Step 5: Calculate Discharge (Q)

To find how many cubic feet of water the canal can discharge per second, multiply the velocity (\(V\)) by the cross-sectional area (\(A\)): \[ Q = V \times A = 17.742 \, \text{ft/s} \times 75 \, \text{ft}^2 \approx 1330.65 \, \text{ft}^3/\text{s} \]

Key Concepts

Flow Velocity Calculation

The flow velocity of water in a channel is an essential parameter calculated using the Manning Equation. This formula helps us understand the speed at which water moves through any channel, whether it's a canal, river, or storm sewer.
To determine the flow velocity (\( V \)), you insert known values into the Manning Equation: \[ V=1.486 \frac{A^{2 / 3} S^{1 / 2}}{p^{2 / 3} n} \]Where:

• \( A \): Cross-sectional area of the channel
• \( S \): Slope of the channel
• \( p \): Wetted perimeter
• \( n \): Roughness coefficient

For example, if you have \( A = 75 \, \text{ft}^2 \), \( S = 0.050 \), \( p = 24.1 \, \text{ft} \), and \( n = 0.040 \), substituting these values into the equation helps calculate the velocity, resulting in \( V \approx 17.742 \, \text{ft/s} \).
The formula provides a comprehensive approach to understanding how fast water travels through different channel types, accommodating variations in physical dimensions and channel texture.

Cross-Sectional Area

The cross-sectional area of a channel is a critical factor in determining the flow characteristics via the Manning Equation. This area represents the shape and size of a vertical slice of the channel, which significantly influences how much water can pass through at any given time.
To calculate the cross-sectional area (\( A \)), you measure the width and the average depth of the channel and multiply these values together. It's usually expressed in square feet (\( \text{ft}^2 \)).
For our example, we're given \( A = 75 \text{ft}^2 \). This indicates that the channel has sufficient size to accommodate a particular amount of water, making it a crucial variable in the Manning Equation. Larger cross-sectional areas mean more water can flow through the channel, crucial for channel design and flood management.

• Helps determine the channel's water carrying capacity
• Relates directly to the flow velocity and discharge

This measurement is vital for infrastructure planning and can determine whether a channel is adequate for its intended purpose.

Wetted Perimeter Calculation

The wetted perimeter is another fundamental concept within the Manning Equation. It encompasses the length of the channel's boundaries that come into contact with water, including the sides and the bottom.
Knowing how to calculate the wetted perimeter (\( p \)) is important because it affects the flow resistance:

• More contact between the water and the channel surface generally increases the friction.
• Higher friction reduces the velocity of the flow.

In our example, \( p = 24.1 \, \text{ft} \) specifies this length for the canal in question.
The wetted perimeter is usually determined by summing the lengths of all surfaces the water touches. It's vital for systems where friction plays a substantial role in determining flow characteristics.
By calculating the wetted perimeter and considering it in the Manning Equation, we gain insights into how effectively a waterway can transport water, helping to optimize design and accommodate varying water volumes.