 Consider a flow of water being split by a wedge as shown in the figure below. Both the wedge and the stream of water are very wide, by which I mean into and out of the screen here. They extend very far into the screen and out of the screen, which means that we are treating it as a two-dimensional problem. If the horizontal force required to keep this wedge stationary is 124 newtons, that's per meter into and out of the page, what is the angle of the wedge? This time around, I recognize that I have a force in the x direction, which means that I'm going to care about my x component of our conservation of momentum, and I don't have a control volume defined this time, so I will define my own. I'm going to call my inlet state 1, I'm going to call this top outlet state 2, and the bottom outlet state 3. Furthermore, I was told an average velocity at state 1, and I'm going to assume that both states 2 and 3 have the same average velocity. Since I have incompressible flow, or I'm assuming incompressible flow, that means that the volumetric flow rate at state 1 has to equal the volumetric flow rate at state 2 plus the volumetric flow rate at state 3, because all of the mass flow rate that's entering has to leave for this to be steady state, which by the way I can also assume. What that means for our purposes here is that the cross-sectional area at state 2 and state 3 are each going to be half of the cross-sectional area of state 1, because they have the same average velocity and the volumetric flow rate entering must then leave because I have the same mass flow rate entering and leaving and I have incompressible flow. Therefore, whatever the height is in my control volume at state 1, half of that height is present at 2 and 3 each. Remember that for later. Next I will write out my x component of our conservation of momentum and then I can begin to simplify by neglecting terms that aren't relevant and simplifying my surface forces into whatever is actually appearing. Again, I have fx as my label for my force in the x direction and I'm going to define my x-axis as being positive to the right. That means that fx is in the negative direction and again to try to avoid confusion I will give fx a different name. Let's go with Brian. Okay, so the Brian force is to the left so I'm going to plug that in as a surface force. By the way, I'm plugging it in as a surface force because it's acting on the surface of my control line. So negative Brian. Then do I have any body forces? Well remember the only body force we consider is gravitational acceleration for now. Therefore, I only have to consider if gravity is relevant or not. I'm going to assume that gravitational acceleration is in the y direction and that I don't have any body forces at all in the x direction. So Brian plus 0, excuse me, negative Brian plus 0 is equal to then we have our volume and surface terms. Because I have steady state analysis this entire term is going to disappear and then how many interfaces do I have water crossing the boundary in the x direction? How many ingress and egress points? How many orifices are present for the water to enter or exit my control volume in the x direction? That's right, there are three. I have state one and state two and state three. So I'm going to write that out as three separate integrals. Okay, next I can begin to consider what's actually happening in each one of those three integrals. I've assumed incompressible flow which means that our density is constant so it can come out of the integral. Furthermore, the velocity or rather the x component of velocity is constant. It doesn't change with respect to area because I'm assuming uniform flow everywhere. So states one, two and three are all uniform flow. Basically, I'm saying those velocities of six meters per second are on average at that state point. With that assumption, the velocity comes out and I can collapse each of these three integrals into the magnitude of velocity times area. So that's going to be rho one times u one times the x component of velocity of instate one times a one. Now, is that a positive quantity or negative quantity? That's right, it's negative. Why is it negative? Because the area vector and the velocity vector are in opposite directions. The area vector always points out. The velocity vector here is pointing in which means that they are in opposite directions so that collapses to a negative value. At state two, I'm going to have rho two times u two times the magnitude average velocity of two times area two. Is that a positive or negative quantity? That's right, it's positive quantity. Why? Because the area vector and velocity vector are in the same direction. Lastly, rho three u three times average velocity at state three and then multiply by a three. And is that a positive or negative quantity? That's right, positive for the same reasons as state two. The velocity vector and the area vector are in the same direction, therefore it's a positive quantity. All the densities are going to be the same because of incompressible flow. And I know that the density of water is going to be about 998 kilograms per cubic meter because I am assuming that the water is at standard temperature and pressure. That value comes from table A one or A three in your appendix. Okay, so I know rho one, I know rho two, I know rho three. I know the area at state one because it is a rectangular area that is one meter wide because we are describing the Brian force as 124 newtons and that's per meter into and out of the page. Therefore the cross-sectional area at state one is a rectangle that is one meter wide and six centimeters tall. Then state two and state three are each going to have half of that cross-sectional area. So just for our purposes here, I can write this as one meter times six centimeters. And then this as 0.5 times one meter times six centimeters. And that also applies to top training page on iPad. No, very slowly. A three which is one half times one meter times six centimeters. Cool. For state one, the average velocity at state one is the same as the x component of velocity at state one. So these are both this velocity up here which is given as six meters per second. And for state two, the average velocity is still six meters per second. And for state three, the average velocity is still six meters per second. However, the x component of velocity at states two and three is no longer six meters per second because that is at an angle. So the x component of that velocity is going to have a trigonometric term involved. This is theta over two. And I can describe cosine of theta over two as the x component of that velocity which I will describe as u divided by six meters per second. Therefore, u two is going to be six meters per second times cosine of theta over two. And for state three, the x component of that velocity is also going to be cosine of theta over two multiplied by six meters per second because it's the same angular offset from the x direction just in the opposite direction. So here I will plug in six meters per second multiplied by cosine of theta over two and that also applies over here. And this is all equal to the negative bryan force which means I have everything I need to solve for theta. Just to make the algebra a little bit easier, I'm going to be referring to all of the instances of six meters per second as v bar. And I'm going to describe all of the areas in terms of a one. So at that point I have negative bryan force is equal to rho times negative rho one times v bar one squared times a one. I'm going to make a little bit more room here. Plus rho equal to one times cosine of theta over two times v bar one times v bar one times a two which is one half of a one. Alright that is one half times a one. And then at state three I have plus rho times cosine of theta over two times v bar one times one half times a one. And again that velocity squared because one of them is appearing here and here and the other one is appearing here. Does that make sense? So at state two u two is v bar one times cosine of theta over two and then I'm multiplying that by the quantity v bar one times a two. Therefore it simplifies down to density times cosine of theta over two times v bar one squared and the area I'm writing in terms of a one to reduce how many things I have to keep track of. Following that logic. Okay, cool. Then I can factor out v bar one squared and density and a one. So this is going to become v bar one squared times a one times density times the quantity negative one plus cosine of theta over two times one half plus cosine of theta over two times one half. Cosine of theta over two times one half plus cosine of theta over two times one half is going to become one times cosine of theta over two. Therefore that entire term inside the parentheses is going to be cosine of theta over two minus one. And again that's multiplied by v bar one squared times a one times density of water. And I'm running out of room on this page so I will create a new page. So I can say cosine of theta over two is equal to negative Brian divided by v bar one squared times a one times row. Keep running one after that kind of a necessary and then I'm adding one to that. Therefore, theta is going to equal two times the arc cosine of one minus Brian over v bar one squared times a one times row. And that's going to give us an answer. So I will take two times the arc cosine of one minus a Brian force was 124 Newtons. And I will point out that I am not writing that as Newtons per one meter because we've already accounted for the fact that it's one meter wide in the area. So this is like we're analyzing a one meter wide section so I don't actually care about Newtons per meter right now. Does that make sense? Then I'm dividing by six meters per second I think. Correct. Six meters per second. And then I square that. Then I'm multiplying by the cross sectional area at say one which was six centimeters by one meter. So six centimeters times one meter. And then the density of water at standard temperature and pressure was 998 kilograms per cubic meter. And then I want a unitless proportion inside of the arc cosine so that I can end up with a theta term at the end. So I'm going to want to take one minus unitless proportion which means I need all of these dimensions to cancel. So I will break apart the Newton and do its components and I will convert from centimeters into meters. And at that point everything should cancel. Newton cancels Newtons, kilograms cancels kilograms, seconds squared cancels seconds squared, centimeters cancels centimeters, and I have square meters, meters, meters, cancelling cubic meters and meters. Yay I have a unitless proportion. Then theta can be calculated, if my calculator will turn on, and I will remember that my calculator is in degrees. So two times arc cosine of one minus, and then two parentheses just to be safe, 124 times 100 divided by six squared times six times 998 and one more parentheses. And we get two times the arc cosine of 12,698 divided by 13,473. I appreciate your accuracy calculator. We get about 39 degrees. So theta is 39.06 degrees.