 Water enters a pipe flowing at a uniform velocity u. As a result of passing through the pipe, the velocity becomes non-uniform and the flow at the exit follows a profile as shown in the figure below. So that profile is u is equal to u max times 1 minus little r squared over big r squared. Note that little r is going to be the actual axis up and down, measured from the zero, that's the radius, and big r is the constant that represents the overall radius of the pipe. So u is a function of little r and that is constant times 1 minus r squared over constant squared. If the radius of the pipe, big r, is three inches and the maximum velocity recorded at the center of the pipe outlet, u max, is 10 feet per second. Determine the inlet velocity that's uppercase u. Okay, so this is a conservation of mass problem. So we are going to start with our Reynolds transport theorem, simplified for the conservation of mass. So dM dt of our system, got a little bit ahead of myself there, which is zero, because we're analyzing a control volume is equal to, and then I'm saying d dt integral of density d volume plus the integral across the control surface density velocity vector d area vector. And then I will pose the question, can we eliminate that second term? The answer is yes. Why is it yes? Because we have steady state analysis in this problem. We weren't explicitly told that, but we can see that there's no opportunity for water to build up in the pipe or to be draining out of the pipe. It's just that regardless of when we look at the problem, the conditions are the same. If you've been running water through this pipe for 10 hours or two hours or a million and a half years, time has no effect on the problem. So I am going to add steady state to my assumptions. That's steady state. And then I am going to assume incompressible flow because we are analyzing water and for our purposes, all liquids are incompressible unless we have enough information to deduce otherwise. Incompressible implies to us that the density does not change at all. And then I was told that I have a uniform flow at the inlet. So that's not really an assumption that I need to make. I think that's enough information to move on. So because we have steady state analysis, this entire second term is zero because nothing can change with respect to time. And remember, this entire integral is the thing that is being derivated with respect to time. It's D that thing with respect to time. So I have zero is equal to the integral across my control surface of density, velocity vector, the area vector, and then density is constant. So it comes out, then I divide both sides by density and I have zero is equal to the integral across the control surface of the velocity vector times the area vector or rather the integral of the velocity vector with respect to area. And that's across the entire control surface, which is going to be my entire outside surface as crudely drawn here. There are only two orifices, that is two points at which fluid is crossing the boundary. So I am going to split that integral across those two and write this as zero is equal to the area across the inlet, which doesn't have a state point. Let's call that one. And over here we will call this two. And remember that the area vector is always defined in the outward direction. So this is across area one, that would be velocity one, the a one. And I'm adding the integral across a two, the velocity two, the a two. And then the first integral is going to simplify because that's uniform flow. So I'm going to collapse the vectors and write that as the velocity that's day one on average times area that's day one. And then if my vectors are in opposite directions, I have to add a negative. So because v one is to the right and the area at state one area vector is to the left, they're in opposite directions. So this collapses to a minus or rather a negative quantity. But I cannot simplify the integral on the right in such a manner because the velocity changes as a function of where it is within the area. So I have to leave that as an area integral a two v two a two. And then I will bring v bar one a one to the left. And I will recognize that I already have a name for the average velocity at the inlet. That's big you. So I'm going to write that as you times a one. And that's equal to the integral across a to the vector two times the area differential. Okay, now how does that simplify? Well, my velocity vector has a profile. And I was told that that velocity, which is really the velocity in the x direction. So I really should be writing you to follow the nomenclature as established in the problem. But that velocity is you max times the quantity one minus little r squared over bigger r squared. So I'm going to be plugging in you max, which is a constant remember, times one minus little r squared divided by big R, which is again a constant squared. In order to be able to evaluate this integral, then I'm going to have to write d a in terms of r. So I am going to be representing that circle in terms of radius. So quick little recap of calculus when I'm talking about integrating across a circle where the theta position doesn't matter. Then what I'm saying is I'm evaluating a whole bunch of little circular slices. This is called an annulus from the inside to the outside. And this annulus has an area that's a d a and that annulus is area because it's infinitesimally thin that is equal to two times pi times the radius that would be the circumference of that circle times dr. Does that make sense? So we're taking the circumference of this circle and we are multiplying by the thickness of the ring. So we're saying that it's such a thin circle that's basically a rectangle. So we're taking the height of the rectangle, which is the thickness, which is dr multiplied by the width of the rectangle that's wrapped around the circle. I'm making a lot of hand gestures that you can't see, but just imagine in your mind me holding an apple pencil and then moving my hand around in a circle over the image as though that's helpful. Hopefully you guys got a good foundation in that from your calculus classes, but when we're evaluating an integral for a circular section and the only thing that matters is radius, we split it into an infinite number of circular rings called annulus or annuli because it's plural. And the d a term is two pi r, which is the circumference of that annulus times dr, which is the thickness, the height, whatever the size from the inlet to outlet direction. And we are integrating from an r position of zero to big R. So I will plug that in over here. d a is really two pi r dr. And we are integrating from zero to big R. So integral from zero to big R of u max times one minus little r squared over big r squared. That's not really a good two. Let's try that again. Two times two times pi times r dr. And then it's just a matter of evaluating the integral. So I will split that integral. And I will, here, let's split it after I bring everything inside the parentheses. So that's really just u max times two pi r dr minus zero big R u max times two pi r dr times a little r squared over big R squared. So I split the integral and evaluate them separately. And for that we apparently are out of room. I will write big u 181. And I will copy and paste this onto a new page. So I am probably best off just getting rid of that. And look, we have an integral zero to big R. And look, we have another integral from zero to big R. And then I will bring out the constants. So I have u max times two times pi. And then this becomes the integral of R with respect to R from zero to big R. Then I'm subtracting, bringing out the constants, constants, the constants, I'm bringing out the constants like you did it again. Then I'm bringing out the constants guys, times the integral from zero to big R. Guess that's also big R squared in the denominator. And we have R, R squared. So that's R cubed dr. Then I am evaluating the integral, that would be the integral of R with respect to R is one half R squared, evaluated from zero to big R, which means that I have one half times the quantity, big R squared minus zero, zero squared to zero. So this is just going to be u max times two times pi times one half times big R squared minus the integral of x cubed with respect to x would be one quarter x to the fourth. So it's going to be one over four times a little r to the fourth evaluated from zero to big R, which is going to be one quarter times big R to the fourth minus zero to the fourth. So I have one quarter times R to the fourth. So you max times two times pi divided by big R squared times one quarter times big R to the fourth. Well, that's cool. Some stuff is going to start canceling. I'm excited. R squared is going to bring this down to R squared and then two divided by four is one half. And over here, the two and the one half start canceling. So I have u max times pi times R squared minus u max times two times pi two divided by four is one half. So I'll write that as e max times pi divided by two just to be a little bit more straightforward times big R squared. So then I could factor out the stuff that's in common, which would be u max times pi times R squared. So u max times pi times big R squared times the quantity one minus one half one minus one half. Let's try that one more time. One minus one half. There we go. And one minus one half is one half. So this would be u max times pi times big R squared times one half. And don't forget the left hand side of this equation was u. It's the thing that we're looking for times area. And the area at the inlet is going to be pi times big R squared because the area of a circle is going to be pi times radius squared and the radius is three inches. So I have u times pi times big R squared. Hey, look, I have pi appearing on both sides and R squared appearing on both sides, which means this collapses all the way down to u is equal to u max divided by two. I think we can handle that calculation. u max was 10 feet per second. Therefore, the answer to the question is 10 divided by two, which I'm pretty sure is five. But just for good measure, let's have a calculator double check our math for us. 10 divided by two. Haha, it's five. Look, we did it. Five feet per second. So that wasn't too bad, was it? I mean, I would say that's downright fun. Ain't no party like a simplification of the Reynolds transport theorem to the conservation of mass and then simplifying it all the way down to just a constant multiplied by one of the numbers that you were given.