A winding natural stream with weeds has an average depth of 0.86m and is 7.25m across. The stream channel drops 0.34 m/km. What is the stream’s average velocity?
You should have been provided with a graph based upon actual data. There is no simple formula to apply to this question. The facts that the stream winds and is weed filled affect the answer.
Use the graph provided, or search for one online. I will see what I can find and post it later, if I find anything
Refer to sources that discuss the Mannong Equatiion, such as
http://www.lmnoeng.com/manning.htm
Different constants k apply to different situation.
To calculate the average velocity of the stream, we need to use the formula:
Average Velocity = Discharge / Cross-sectional area
First, let's calculate the cross-sectional area of the stream:
Cross-sectional area = Average depth * Width
Given:
Average depth = 0.86 m
Width = 7.25 m
Cross-sectional area = 0.86 m * 7.25 m
Cross-sectional area = 6.245 m²
Next, we need to calculate the discharge of the stream:
Discharge = Velocity * Cross-sectional area
Given:
Drop in elevation per unit distance (also known as slope or gradient) = 0.34 m/km
In this case, the slope is given in meters per kilometer, so we need to convert it to meters per meter.
1 kilometer = 1000 meters
Slope = 0.34 m/km * (1/1000) km/m
Slope = 0.00034 m/m
Now, we can rearrange the equation to solve for velocity:
Velocity = Discharge / Cross-sectional area
Velocity = (Slope * Cross-sectional area) / Cross-sectional area
Velocity = Slope
Given that the slope is 0.00034 m/m, the stream's average velocity is 0.00034 m/m.