So in this video we're going to discuss two fundamental laws of fluid power. We're going to apply them to a couple of different applications, one being a t-junction, one being a hydraulic cylinder. And we're going to use transformer ratios in a hydraulic cylinder to look at force pressure relationships and flow rate velocity relationships. So, our first fundamental law that we're going to look at is Pascal's Law. And Pascal's Law simply tells us that the pressure in the fluid is the same in all directions. Now there's two major assumptions that you have to pay attention to here. First is that, we're assuming the fluid is at rest. So it has no velocity, has no viscous components to it, has no kinetic energy that we're paying attention to. The second is that the gravitational force is negligible, and we'll talk about this, this limitation in just a moment. But, what Pascal's Law allows us to do is if you look at the system I've got on the slide where I've got two different areas here, and on the left side I've got this small area piston connected to a larger area piston over here. Well, Pascal's Law is telling us that the pressure in that fluid is the same in all locations. And because the pressure's the same. And I know that the pressure is the force divided by the area. Well, now I can, so if my pressure is the force divided by the area. Now I know the pressure's the same. I have different areas, so I can create force amplification. So, this is the, really the key to all fluid power systems where we're creating a force amplifier, or sometimes a flow amplifier. And if you look at this car jack right here, this is a prime example where. I've got a small piston which would be my input right here. And I can apply a meager amount of force to this. Nowhere near what I need to lift a vehicle. But then you look underneath here and I've got a very large diameter piston right here. The pressure's the same in the fluid, both in that small piston and the large piston, but I can amplify my force. And, we're either amplifying force or we're using this to do precise, position control, because you can imagine, if I move some volume with this small diameter cylinder, I'm not going to get very much stroke with a larger diameter cylinder. So we can use this for precise positioning control, and also force amplification. So, two limitations I want to highlight here of Pascal's Law that we're going to address later in the class, the first is, assuming that the fluid is not in motion. Which means we don't have any viscous forces. We don't have any kinetic energy losses. And the second is that we're neglecting any gravitational force. So, if I look at a vertical pipe like this one right here, and I say, you know, what is the, the pressure relationship as I'm moving up and down that, that fluid? And I assume that it's at rest. Well, lemme quickly do a force balance right here, so a force balance. On this fluid, I'm going to say my forces going upwards would be P2. Pardon me. Would be, P2 times the area, and then the forces coming down, I've got the pressure on the top cyl, of the cap. P1 times the area. And then I've got the gravitational force which would be rho gh, times the area. And as I look at these terms I can recognize very quickly that I've got areas on both sides. Or in all three terms, so I can. Cross out my areas. And then my pressure differential between the two would be just P, P, in this case, P2 minus P1, which ends up being just rho gh. And if I'm interested in what the pressure drop is or the, the delta P, with respect to a given height, I can divide that out and say my delta P. Which is just P1, er, P2 minus P1, divided by H, is just, rho times G. And, if we're talking about this in the SI system, a typical density of a hydraulic fluid might be, 870 kilogrou, I'm sorry, 870 kilograms per meter cubed. So 870, kilograms, per meter cubed. I've got, gravity around ten meters per second squared. So 9.8, meters per second squared. And, I multiply these out and now I can get the pressure difference, which would be right around, 8.7, kilopascals, per meter. And remember we were saying atmospheric pressure is about a 101.3 kilopascals, so this 8.7 per meter is actually very small in the, the grand scheme of things. And quite often in fluid power systems it's, it's unimportant unless you're dealing with, perhaps a wind turbine where you have a 100 meter rotor height, and you're having a large pressure differential as the fluids moving up and down in a, in a hydrostatic drive wind turbine. One place that it does play a role is in pumps, where we need to minimize the pressure drop coming into the pump. So we'll talk about this pressure drop more once we get to pumps. But for now recognized these pressures are relatively small. Compared to the hydraulic pressures that we're often dealing with, which is in the megapascals. So, the second law I want to highlight now is the Conservation of Mass. And so we know in Newtonian physics, mass is always conserved. And, we take one step beyond this and say that hydraulics, oil, water, is relatively incompressible. And if I can say that mass is always conserved and I've got a constant density and incompressible fluid, then I can also say that my flow rate is conserved. So if I look at a junction, which might be some t-junction in our, in our system. It might be a cross where I have a fourth port coming out of this. But some place where I have multiple flows coming in there. Well, from a mass conservation perspective, I could say the mass flow rate coming into each one of these ports, as I add them all up, it better sum to zero. Well, once I assume incompressible flow, now I've got these three flow rates right here and these three flow rates better also sum to zero. So I can say Q1 plus Q2, plus Q3, is equal to zero. And again I've got arrows all pointing inward so this, this all should sum to zero. So the natural question here is, is this a good assumption or not? Is the incompressible assumption good? Well, typically hydraulic oils are somewhere in w, water is well, somewhere two to three percent compressible at the pressures we're dealing with, so these are relatively good assumptions. For the majority of the time, although the compressibility of the fluid which was referred to through the bulk modulus, does become important during some situations and we'll, we'll talk about those later in the class. But for now, incompressible assumption is, is a relatively good one. So where I want to go now is apply this to a hydraulic cylinder, so a cylinder that looks like this where I've got two different areas. On this side, I've got my cap side. This side I've got my rod side. I realize that the cross-sectional area or the effective area that the, the f, fluid is acting on is larger on the cap side than it is on the rod side because we've got this piston sticking out on the rod side. And what I want to do is look at the, the pressure and flow relationships for this, for this hydraulic cylinder. So, first of all, from a pressure standpoint, I can say that. If I sum up, the summation of forces is equal to mass times acceleration. My summation of forces, well, I've got on one side a hydraulic pressure acting, which would just be P1, multiplied by A1. So that's this area here, which we refer to as the cap side area. And then, acting in the opposite direction, I've got the force acting on the rod, and as all, I also have P2. And P2 now is acting on this smaller area right here, which is the entire outer bore, minus the area that the rod is act, is occupying. So we're subtracting F of the rod. Minus P2 A2. Recognizing that A1 and A2 are not equal. And again this is equal to mass times acceleration. So, I can now use this as my transformer equation if you will to transform force into pressure or vice versa, through the areas of the, the cap and rods via my cylinder. We also have transformers for the flow relationships, where we can recognize that our flow, flow rate is, Q, which is equal to the cross-sectional area times the velocity of our rod. So I've got the velocity labelled right here and this will, I have to define what's positive and negative and correspond that with the, the flow rate I'm talking about. But we just relate these through the, the cross-sectional area. So, on my cap side, I can say that the flow rate going into the cap side, Q1, divided by the area on the cap side, A1. This is just equal to the rod velocity. And I also know on the rod side, that I've got a flow rate going out of this, in this case, of Q2. Divided by A2. And, one thing you have to be careful of, is that in a hydraulic actuator, in something like this where I've got a different area on the cap side and the rod side, I'm going to have different flow rates coming into this versus going out of it. And it's actually going to be increasing the flow rate, as I go from the rod to the cap, or, and, and reducing in the, in the opposite case. So we have to be, be careful as to amplifying or reducing our flow rates in the system. So, this area ratio, this A2 to A1 captor rod side area is our transformer ratio for both the force amplification and the flow amplification. So, to summarize, we've talked about two different laws that we want to apply to fluid power. One is Pascal's law, telling us the pressure is, is the same in all directions. And again, we had some simplifying assumptions that we have to be careful of. And then we said we've got mass conservation applied to fluid power. And if we make the assumption of incompressible flow, now we have a conservation of flow rate. And this allows us to look at flow rates through junctions and then we took one step farther and said let's look at a hydraulic cylinder and look at the transformer ratios to calculate pressure to force and flow to velocity relationships. Thank you. [SOUND]