How Does Channel Slope Affect Water Depth?

[dreamliner]

Homework Statement

A rectangular channel has a width of 2 meters, and an slope of 1:80. The Manning's number for the entire channel is n = 0,014. In this situation the depth is y= 1,50m. The flow of water is normal.

Further down the slope decreases to 1:150 and the width increases to b=3 meters. The waterflow Q and the MAnning's number is the same as in the situation above. Find the depth of water, y.

Homework Equations

Hydraulic radius R = A/P
Manning formula for gravity flow: Q= 1/n*A*R*S₀^1/2
Froude number U/(√(g*y))

The Attempt at a Solution

I have the water flow Q= 17,04m^3/s from before. Since I also have Manning's number, width and slope, I'm thinking Manning's formula for gravity flow.

However, not having the new depth prevents me from calculating the hydraulic radius - which is a part of Manning's formula so it would end up looking like this:

17,04= 1/0,014*(3*y)*?^(2/3)*(1/150)^(1/2). (? being the place where the number for hydraulic radius would go).
So I'm kind of stuck at this point. Any suggestions would be appreciated.

 

[SteamKing]

It seems like you are missing the wetted perimeter for the channel. The problem doesn't state it explicitly, but assume the channel cross section is still rectangular downstream. Can you figure out the formula for the perimeter given a width of 3.0 m and a depth of y?

 

[dreamliner]

We just touched on the subject before end of class today, so I'm not sure if this is correct, but using the formula for hydraulic radius R=A/P - where P is the wetted perimeter:

R= A/P can be re-written as (w*y)/(y+y+w)= (w*y)/(2y+w) giving P=2y+3.

Looking at it I'm not sure if this will work since there is a slope in the channel...

 

[SteamKing]

But the slope is in the direction of water flow, isn't it? The area and wetted perimeter of the channel are properties of the cross section of the channel, which is normal to the direction of flow.

 

[rezeew]

I would suggest using the Manning's formula for gravity flow to solve for the hydraulic radius. The formula can be rearranged to solve for the depth, y, as follows:

y = (Q * n * b * (1/150)^(1/2))^(3/8) / (2 * 1.5 * (1/150)^(1/2))

Substituting the values given in the question, we get:

y = (17.04 * 0.014 * 3 * (1/150)^(1/2))^(3/8) / (2 * 1.5 * (1/150)^(1/2))

y = 0.896 meters

Therefore, the depth of the channel at the new slope and width is 0.896 meters.