This problem demonstrates the continuity equation in integral form.
As you may recall our governing equation is d/dt over the control volume.
Rho, dV, which is volume.
Plus over all the control surfaces, rho, V dot n, were both V and n are vectors, and
n is the unit normal vector, dA equals 0.
So lets take a problem here.
So we have an entrance it is a 2 foot wide channel and it has a uniform velocity flow.
So we will make it look like this.
Down stream the velocity profile is no longer uniform, and in fact it is a profile, such
that u=3y-y^2.
So if the down stream height of the channel is 1.5 ft, and the velocity, uniform velocity
here is 2 ft/s.
What we are looking for is the height at the entrance of the channel.
So lets start with our governing equation.
Since this is steady state.
This is going to be 0, and since this is water the fluid is incompressible.
So with an incompressible flow what we can assume is that Q1=Q2.
Now with our uniform velocity.
We can say V1A1 there is our Q1.
Equals the intgral over A2 of our udA.
The reason we have to keep this in the integral.
Is that u is a function of the are.
In particular it is a function of y, and y is in the upward direction.
So the first thing that we will do.
Is we will define our dA in terms of dy.
So since the width is constant dA is wdy.
Now lets replace this with our Q1=Q2.
So we have V1 time x times this h that we are trying to find.
Equals w and in the integral from 0 to 1.5, because that is the height of that down stream
channel.
3y-y^2dy, and if you notice the widths cancel out.
So we did not even need to know that it was 2 feet wide.
So lets do this integral.
We have V1 times h.
Equals 3/2y^2-1/3y^3.
Integrated from 0 to 1.5, and that equals 2.25 ft^2/s.
So our h is that 2.25 ft^2/s divided by our uniform velocity, which is 2 ft/s.
Our height at the entrance of the channel is 1.125 ft.