 We're now going to derive a very important equation in fluid mechanics. That equation is the Bernoulli equation. So what the Bernoulli equation will provide is it will provide a useful relationship between pressure, velocity, and elevation. So we'll get an equation that will have pressure, we'll have velocity in it, and we will have elevation z. And this will be, we'll come up with a function for it, but it will then be written in terms of a constants. We'll have some function of all three of these variables is equal to a constant. So that's what we're going to get out of the Bernoulli equation. And the place where we're going to begin with this derivation is we're going to begin with a control volume analysis using the linear momentum equation for a little piece of fluid between two streamlines. So let's begin with that. So what we have here is we have a little differential control volume and it's indicated in red. And so this is a little differential chunk of fluid and it is bounded by a couple of streamlines. So there's one on the top and one on the bottom. And what we're examining, we're examining the change in the fluid as it's going from this point here to this point here. So the inlet to the exit. And we'll notice that as the fluid goes between the inlet and the exit, pressure goes from p to p plus dp, and the velocity goes from vs to vs plus ds and the area can change as well. So we can either have the streamlines opening up, getting wider or getting narrower and converging. And what we're assuming as well, and this will be one of the assumptions here, is that the density is not changing. So we're dealing with an incompressible flow. So let me write out the assumptions for what we're looking at. Okay, so we're dealing with a steady flow, frictionless flow, no shear forces and incompressible. So the density is not changing. So what we're going to do, we want to apply control volume analysis and we're going to use two different techniques. One, we're going to use a continuity equation, mass conservation. And the other one that we're going to use is F equals mA or the linear momentum equation. So let's go about applying those two equations to this system. So that is the continuity equation. Now to begin with, one of the assumptions we made was that we're dealing with steady flow. So right off the bat, this term disappears. And then what we're left with is the mass flux across the two control surfaces. If we look back at our schematic, where is mass crossing the boundaries? Well, it's bounded by streamlines. We have a streamline up here and we have a streamline here. So the only place the mass is crossing the boundary is going to be on the inlet and we have the inlet there and then on the exit. So those are the two control surfaces where we have mass crossing the boundary and those are the two that we will work with. So for mass crossing on the inlet boundary, if you recall what we have, remember the area vector is always pointed in the outward direction. And here we have a velocity vector. Well, it's V in the S direction. I'll drop that way. So that's going to give us a negative term for the first part of the mass conservation. And then on the exit side, there we have a situation where we would have, again, dA. Actually, that should be A naught. dA, I wrote A plus dA. But that would be the area vector, A plus dA. And then we will have VS and then we had VS plus dS. But we know that the velocity in the area are pointing in the same direction and consequently this here will be a plus. So the velocity was VS plus dVS and then the area was A plus dA. So that's what we get from continuity. We can then simplify area range. And we get, so we get this relationship here. And what we're going to do, we're just going to park that and we will come back and use that in a few minutes as we look at the momentum equation. So let's now move on to looking at applying the momentum equation. And what we're going to do, we're going to apply the momentum equation in the direction of a streamline. So that means we're applying momentum in the direction of S. So that's what we get for the momentum equation. Now what we have on the left hand side, we have the surface forces and we have body forces. So surface forces, there is no viscous shear, but there is pressure. And then we have B, FB, that's the body force. And then on the right hand side, we have the time rate of change term. So that's going to disappear. And then we have the mass flux crossing the control boundary and exchange of momentum. So let's begin with the first term here on the left hand side. We'll begin with the pressure force. So let's take a look at that. Okay. So let's do some explaining here. First of all, the first term here, this is the left hand face of our little chunk of fluid. So remember, we had a control volume that kind of looked like this. So that is the left hand. And then this here is going to be the right hand face, which is the second term. And then we have this third term. What is that representing? Well, what that represents, we're going to take the pressure, average pressure in the middle, which we'll say is P plus DP over 2. And let's assume that the streamlines are getting wider. Well, we need to apply the force due to that expansion in the streamlines. And that's what this term is doing. And so the amount that the streamlines may be increasing could be DA. And so it's just the pressure at the midpoint, multiplied by that change in area term. And so if the streamlines are converging, that would then be negative and you'd have a force in the left hand direction in the negative S. But here we're assuming that they're getting wider and so you would have a force in the positive S direction. So that's what that is referring to. So there we have an expression for the surface forces. Let's now move on and take a look at the body forces. Before we do that, let me just rewrite this a little bit to clean it up. Okay, so what I've done here is I've just expanded out some of the terms and if you go through the math, you'll find that some of them do cancel out. It's not the cleanest, but that's where we will stop with that part. And next, what we will do is we will move on to take a look at the body force. So what we have here is the component of gravity acting in the direction of the streamline. And through trigonometry, what we can do if we draw our gravity, we said the gravity vector was acting in that direction. If we want to evaluate the components of gravity along the streamline, what we need to do is use trigonometry here. So I'm going to set up a right angle triangle right here and Gs, that would then be equal to minus sine theta multiplied by G, the gravitational constant where in this diagram, this angle here is theta. And that's basically the angle that our streamline is making. And so with that, we can rewrite the body force in the following manner. And this here is a representation for the volume. And what we do, we have our little chunk of fluid. And remember, we're assuming that maybe it's getting larger. So we have A here, A plus DA divided by 2. And then the length of this differential element is DS. So what I'm doing is I'm taking the average area at the middle and assuming that that's a cylinder and then getting the volume. So that gives us one part of the body force. But there's another trig relationship that we can use. And that enables us to do a little bit of a simplification of DS and sine theta. And it turns out that if you look at the direction of DS, what we have, this is DS here. Remember that was theta. And this was DZ. We said that the z direction was in the vertical. So what we can do, we can write out sine theta DS is equal to DZ. And that enables us to simplify because we have a sine theta there and we have a DS there and we can substitute for a DZ. So we're going to go through and do that. And what we end up with for the body force for our differential element is the following equation. So we will let that one rest. And then what we're going to do, let's go back and look at our momentum equation. So we've tackled the surface force. We've tackled the body force. We now need to take a look at this last term. So let's address that. And so here, what we need to do, we need to look at the velocity component on the left hand face multiplied by the mass flux. So remember, here we have mass flux coming into. So it's going to be a negative mass flux. So with momentum, we get this expression, the S, that's the scalar magnitude of the velocity in the S direction. And that is a minus because again, if we look at the the unit normal, the area would be pointing in that direction, but the velocity is coming in in that direction. That's why we get the minus sign there because it's the dot product. And then for the right hand surface and this one, the mass flux term turns out to be positive. Okay. Now we can make a little bit of a simplification. If we go back and take a look at what we got from the continuity equation, which I think was here. And so if you take a look at this term right here and then we go back to what we just had for momentum, that is the same as this term here. And consequently, we can make a substitution from continuity and we will make that substitution. And that will enable us to simplify this equation a little bit. If you go through and you then do the expansion, so multiply D plus VS by that term, you'll notice the first term is going to cancel out. So what we end up with, we end up with this equation here and we will substitute this back into momentum. And so we'll do that in the next slide. So substituting everything back into momentum, we have our surface force, the body force, as well as the momentum flux. We end up with this expression. Okay. So that doesn't look like a very friendly expression, but what we're going to do, we're going to do two simplifying assumptions. The first one, we're going to neglect products of differentials because those will be really small. So we're neglecting things on what we would say, we're neglecting things of order delta S squared. So things that are a differential times a differential. Where do those occur in the equation? And we have one here. So we'll neglect that and we have another one here. So we neglect that. Those are the only two places where those occur. So we'll neglect those two terms because they're really small. And the other thing we're going to do, we're going to divide the entire equation by dA. Or sorry, by row A, not dA. So we divide the entire equation by row A and what do we get for the first term? We have minus dP divided by row minus we get rid of the row in the A, then we get a minus GdZ. And then on the right hand side, we have Vs dVs and that can be rewritten as the differential of Vs squared over 2. And so we can rewrite this. So that is an equation that we get. It's still in differential form. Remember one of the assumptions we had, we said that row was a constant. We're assuming that we're dealing with an incompressible flow. And therefore that simplifies matters and it enables us to be able to integrate. So we can then integrate and I'm not going to go between two conditions. I'm just going to integrate it. I'm going to equate it to a constant but you could integrate it between two different points within the flow field. But let's go ahead and integrate. And when we integrate that, we get this equation here. And we have the constant of integration because we're evaluating this between two different points. So this is the Bernoulli equation. And remember I said that we were after an equation that represented pressure, velocity and elevation. This is the Bernoulli equation. It can be used quite extensively within fluid mechanics. However, we have restrictions. And so let's write those out. We've derived it assuming that we had steady flow. No friction. So that means that we're dealing with a, what we call an inviscid fluid. Or we're applying it at points where viscosity is not, where viscous shear is not that important. Irrotational flow. It applies for flow along a streamline. And then the fourth one is we're dealing with incompressible flow. So that's the Bernoulli equation. What we're going to do in the next couple of segments, we're going to take a look at how we can use the Bernoulli equation to solve problems in fluid mechanics.