Newton’s Second Law in Terms of Momentum, Linear Momentum, Force Classification, and Free Body Diagram (FBD)

Okay, so I'm gonna today talk about the conservation of momentum, which is applied to a particular control volume that we would select. Okay, all right. The first equation that we are gonna write down is gonna be the Newton's second law. Okay, so basically what it says is the rate of change of linear momentum is equal to the summation of forces. Okay, as you note right now that, this is a vector equation, not a scalar equation. As you know, scalar equations do not have direction. The vector equations have directions, so they will have components in the directions that you're interested in, such as X, Y, Z, or R, theta, Z in the polar coordinates. Okay, so basically this is Newton's second law. Now I'm gonna actually insert what linear momentum is, and you may remember from your physics courses that linear momentum is equal to mass times the velocity. Okay, so this formula then becomes summation of the forces becomes D Dt. Remember, this is a material derivative because this is applied to a particular point in the flow. Okay, and mV. All right, next is I'm going to remind you the Reynolds transport theorem and see whether we can play with that. Okay, so DB Dt, B was just a placeholder. Del Del t of triple integral over the control volume, b, lowercase b, d volume plus control surface, b rho velocity dotted with n dA. Right, this was my Reynolds transport theorem, and I can apply this to conservation of mass, momentum, energy, by replacing this B value in here. Okay, all right. So what I want to accomplish today is I'm gonna be strategic and insert this capital B is equal to this m V, okay, because if you take a look at it, if I simply go out and insert this right over here, what will happen is, on the left-hand side of the equation, is gonna get, I'm gonna get D Dt of m times V, which is the linear momentum. And this and I'm gonna go and say that this is equal to sum of the forces that will be giving me something useful for me to work with. Okay, so let me repeat what I said. This capital B is equal to m times velocity vector, and if you remember from our earlier discussions, the ratio between the capital B and lowercase b is per unit mass. Okay, and I'm gonna insert here m velocity vector, and from here I'll get the lowercase b as velocity vector. Right, you can see, ms will cancel out. Okay, all right. So now I seem to know everything that I need to proceed. Okay, so I'm gonna say that D m V Dt, which is the linear momentum. I'm gonna use Newton's second law and say that this is going to be equal to the summation of the forces now. And then I'm gonna write the right-hand side of the equation. So it's gonna be Del Del t, and again, triple control volume, rho velocity vector, d volume. Be careful about these symbols, they look close to each other. Right, and control surface, rho velocity vector, velocity dotted with n dA. Okay, so then actually, I'm gonna go ahead and box this equation up, as this is my main equation for the conservation of momentum. What I would like you to note that this is a vector equation, as I mentioned before, and it's all components in the x direction, y direction, and z direction. So now I want to go ahead and write these three equations if I'm using the Cartesian coordinates. Okay, for your notes, for completeness purposes. So the summation of the forces in the x direction will be equal to Del Del t, triple integral over the control volume, rho Vx d volume, plus over control surface, rho Vx V dotted with n dA. Okay, you can have the similar equation in the Y direction as well. So you'll get Del Del t of the control volume, rho Vy d volume, plus over the control surface, rho Vy V dotted with n dA. And the third one will be summation of the forces in the Z direction, will be equal to Del Del t, one two three control volume, rho Vz d volume, plus over the control surface, rho Vz V dotted with n dA. Okay. Okay, now I have the three components that I need. If I go up here, as I mentioned, this is a vector equation. When we looked at the conservation of mass, we have seen that that is a scalar equation. And typically what happens is, that is the left-hand side was zero. Okay, is equal to, let's say the first term is plus five, the second term has to be minus five. As simple as that. Okay, in here it's a bit more complicated. Okay, the left-hand side is a summation of the forces. So if you think about it, this is still the Newton's second law. Okay, so this is it. This is the fluid equivalent of F is equal to ma. The right-hand side is ma, actually, if you think about it. Okay, let's say that this is vector one, this is vector two, this is vector three. What happens is, let's say the vector two is like this. Okay, I'll say that vector three is this. Right, and then this is what it means. This one on summation of the forces will be the summation of the vector two, plus the summation of the vector three. Okay, so this is quite different than the conservation of mass. I would like you to realize that. Okay. Okay, let's talk about the special cases of linear momentum. The first thing I want to highlight here is the first assumption or special case that we have evaluated when we're doing the conservation of mass was the steady flow, and I will do the same. Okay, what did steady flow mean is another recommendation. Did it mean that the velocity is constant throughout the flow? No, it doesn't. Okay, and I gave the example. I was doing an experiment in my lab, and I take a break. I go to the lunch. I came back from the lunch, and nothing has changed, okay, in terms of the fluids flow characteristics-wise and properties-wise. So that means that the rate of change of things over time will be zero. Okay, steady means that. Okay, and if I go up in here, if I look as I look at the z equation, you will see that the first term is looking at how does this triple integral change over time. For a steady case, the answer would be it doesn't change, okay, because del of del t of this triple integral over the control volume, velocity vector times Rho times d volume will be equal to zero. I'll talk about more this equation. Okay, so you can see that I number the number two vector turns out to be zero. Okay, and what it basically means then, my summation of the forces will be just this last term. So it will be over the control surface, Rho, velocity vector, velocity dotted with n, dA. So this will be the equation that I obtain after I go through this process. Okay, and I gave the example of vector one, vector three. If I use the same analogy, so what it says is vector three and vector one is identical. So if one is three, one is one, because two is zero, right? So vector one and vector three are the same. That's what it means. So when I was doing the conservation of mass, you may remember that I did steady plus incompressible flow, right? So I don't really need to do it in here, because in that case there was a density I was able to take out of the integrals. Left-hand side was zero, so I divided everything by the density, so I cancelled out the density. I don't have anything like this in here, but maximum what I'm going to obtain, if looking at this boxed equation is, if the density is out, it's going to be summation of the forces divided by density. It's not that special. Okay, so then I'm going to go and make my second assumption as steady, and I'm going to add one more to it, which is called the uniform flow. Okay, so also I see some confusion for the definition of uniform flow, so let's discuss that for a minute here. Uniform flow does not mean that the velocity is constant. Okay. Uniform flow means that I have, let's say that I have this shape, right? In the example, one inlet, two outlets. Okay, remember, vectors are defined at a point, not at an area. Okay, when I say velocity at the inlet is equal to five meter per second, then I need to be much more specific and say at the inlet, at this particular point, the velocity is five. This uniform flow assumption will enable me to say that, hey, the velocity at the inlet, regardless of which particular point that I take, is the same. It doesn't change. Okay, however, it doesn't say that this velocity will be equal to this velocity. Alright, so now if I go ahead with the steady and uniform flow assumption, what I'm going to have is the equation will be the summation over the exits. What it will enable me to do, you can look at the video from the conservation of mass, how I obtained the summation sign. Because integrals, as you know by definition, are we take small areas and we them add them up. So I'm kind of going back to the definition of the integral. Okay, Rho exit, V exit, Vn exit, A exit. Okay, and I'll talk about this Vn exit in a moment. Summation over the inlets, Rho inlet, V inlet, Vn inlet, A inlet. Okay, now let's look at this n terms. This what, what I mean by Vn is, I'm going to underline here as well, up there in the boxed equation, that is what I mean by Vn. So this is the velocity at the inlet, okay, the velocity at the inlet and the velocity at the exit, but it doesn't have any components in x, y, z. It's in the normal direction. Okay, at the exits, if you think about it, let's say this exit, you can see that the normals, so I'm drawing the normals over here in the figure, you can see that they are aligned with the velocity. So that's why I'm going to get the positive sign in front of the summation. At the inlet, if you think about the normal, normal is pointing away from the surface, right, but the velocity is to the into. So I'm going to get a negative sign. That's why this negative is. So I would like you to be careful, and although this is not an official way of representing this, I'd like you to think that there is no negative sign associated with either Vn inlet or Vn exit. Granted, I do know that inlet is a negative. That's why I have this circled-up negative sign over there. Okay, so now we talked about this as well, so I'm going to actually like take these three terms and see what this multiplication of these three terms are. Rho, V, A. Okay, the multiplication of V times A, if you remember, V times A was the volumetric flow rate. And if I have V times A times density, that will be giving me the mass flow rate. Okay, so I can simplify this equation further. It's not even simplification, just for mathematical purposes. I'm going to have my exits as m dot exit times V exit, minus over the inlets, m dot inlet, Vi. Okay, so this is another version of this particular formula as well. The last thing I want to talk about is the left-hand side of the equation. I don't really need to talk about the left-hand side of the equation. Any of these assumptions are not practical to the assumption number two, a special case number two, but in any cases, I have to look at the forces. Okay, and from our experience, what we noticed is there are three common types of forces that we have to account for in all of the cases, or we have an assumption for stating that this type of force doesn't exist. Okay, and those three types of forces are: First one is going to be the surface forces, and a typical one is the pressure force. Okay, if you think about the pressure force, it's P times A. So I need to know my pressure value. If I multiply by area, I'm gonna get a force, so I have to account for it. Okay, and I want to talk about it in here because I see issues here as well. So let me write this particular example again, or redraw again. I have an inlet here, I have an exit here, and I have an exit over here. Okay, as we discussed in the fluid statics section as well, pressure always pushes it in. If I have a balloon like this, in any particular shape, it always is perpendicular to the surface and pushes it in. So when I pick my control volume in a traditional way like that, what will happen is I have to indicate the direction of the pressures to be pushing into the surface. So if I draw with longer arrows, so this is gonna be the P1 A1, assuming this is section 1. And, careful, this is P2 A2 pointing it in, and this is P3 A3 pointing in as well, assuming this is section number 3. This is section number 2, and this is section number 1. If I change the direction of the flow, let's say that 2 and 3 become an inlet and 1 becomes an exit, what happens to the direction of the pressure forces? Nothing. It still stays the same. Note that the direction of the pressure forces is nothing to do with the direction of the flow. I see these issues often from the students that didn't quite understand the topic well. Okay, the second force are viscous forces. Okay, viscous forces are similar to friction in your solid mechanics courses. Okay, when I have a pipe, and the flow is flowing in it, there will be some kind of drag on it, right? It's going to slow down. If I don't have a pump, let's say I just turn off the pump, what happens is the flow is going to stop. Right? That is the force stopping the flow in my pipe. Okay? It's called the viscous forces, and we touched upon this in the first introductory concepts videos. So it's going to be the viscosity times du dy. Remember that if I multiply this by area, I'm gonna get myself viscous force that I should be incorporating into my analysis. And this viscous force will be opposite to the direction of the flow. So now, this one has something to do with the direction of the flow. Okay? If you want to get rid of this particular force because you weren't given enough information, then I can just simply say this is inviscid flow. Okay? Inviscid flow means that this is zero, so then I don't have to worry about viscous forces. Okay, and the third one is, I want to call it body forces. But the most common one that we know of is the weight of the control volume. Okay, the weight of the control volume. There's also magnetic forces sometimes that act on my flow. That's a body force, but that's not as common as the weight.