Wow. upon review I left out a number of very useful things, without which this problem would be impossible. That’s what I get for trying to only include the relevant parts of the problem.
Okay. Let’s try again.
The goal is to find the drag imposed on a flowing fluid. drag is defined as
dFT/W = – mu times the integral of (du/dy) solved from 0 – L
du/dy is listed in the image above. So if we can solve for du/dy and multiply by negative mu… I’ll have an answer. The problem is, that the question relates to flow under two surfaces at two different heights: h1 and h2. The length, is L.
So, I have to solve the problem for L/2 first (h1) and then from L/2 to L for next (h2). X is the variable used to describe the location along underneath my surfaces.
My method was this, and I’m pretty sure the method is right on.
Look at surface 1 first. at h1.
Substitute in my G and my Q/W. simplify once.
take the integral from Zero to L/2. That adds a variable, X. solve from 0 to L/2.
Simplify.
NEXT
in order to do the second surface, at height h2, I have to change my G equation slightly. (wow, how did I leave all this out)
G becomes a function of h2 not of h1.
Q/W remains the same.
So I go through the same steps, this time integrating from L/2 to L. I add the two equations together… and then supposedly I’m suppose to simplify out to what was the answer in the picture above. I just can’t get it though.
where’s the x? do you mean y?
Wow. upon review I left out a number of very useful things, without which this problem would be impossible. That’s what I get for trying to only include the relevant parts of the problem.
Okay. Let’s try again.
The goal is to find the drag imposed on a flowing fluid. drag is defined as
dFT/W = – mu times the integral of (du/dy) solved from 0 – L
du/dy is listed in the image above. So if we can solve for du/dy and multiply by negative mu… I’ll have an answer. The problem is, that the question relates to flow under two surfaces at two different heights: h1 and h2. The length, is L.
So, I have to solve the problem for L/2 first (h1) and then from L/2 to L for next (h2). X is the variable used to describe the location along underneath my surfaces.
My method was this, and I’m pretty sure the method is right on.
Look at surface 1 first. at h1.
Substitute in my G and my Q/W. simplify once.
take the integral from Zero to L/2. That adds a variable, X. solve from 0 to L/2.
Simplify.
NEXT
in order to do the second surface, at height h2, I have to change my G equation slightly. (wow, how did I leave all this out)
G becomes a function of h2 not of h1.
Q/W remains the same.
So I go through the same steps, this time integrating from L/2 to L. I add the two equations together… and then supposedly I’m suppose to simplify out to what was the answer in the picture above. I just can’t get it though.
by the way, the W in dFT/W = Width.
I’ve done combining and simplifing of the gibberish you gave originally and get
du/dy= -U/h1-3U((h1-h2)/(h1^3+h2^3))(h1-2y)
God, this question reminds me that I really do not miss studying maths.
Mission Accomplished!
The math proof is nasty though. I had to enlist the help of one of my very wise co-workers to get through it.