0:18

So, our first fundamental law that we're going to look at is Pascal's Law.

Â And Pascal's Law simply tells us that the pressure

Â in the fluid is the same in all directions.

Â Now there's two major assumptions that you have to pay attention to here.

Â First is that, we're assuming the fluid is at rest.

Â So it has no velocity, has no viscous components to

Â it, has no kinetic energy that we're paying attention to.

Â The second is that the gravitational force is negligible, and

Â we'll talk about this, this limitation in just a moment.

Â But, what Pascal's Law allows us to do is if you

Â look at the system I've got on the slide where I've got

Â two different areas here, and on the left side I've got

Â this small area piston connected to a larger area piston over here.

Â Well, Pascal's Law is telling us that the pressure

Â in that fluid is the same in all locations.

Â And because the pressure's the same.

Â And I know that the pressure is the force divided by the area.

Â Well, now I can, so if my pressure is the force divided by the area.

Â Now I know the pressure's the same.

Â I have different areas, so I can create force amplification.

Â So, this is the, really the key to all fluid power

Â systems where we're creating a force amplifier, or sometimes a flow amplifier.

Â And if you look at this car jack right here, this is a prime example where.

Â I've got a small piston which would be my input right here.

Â And I can apply a meager amount of force to this.

Â Nowhere near what I need to lift a vehicle.

Â But then you look underneath here and I've

Â got a very large diameter piston right here.

Â The pressure's the same in the fluid, both in that small

Â piston and the large piston, but I can amplify my force.

Â And, we're either amplifying force or we're using this to

Â do precise, position control, because you can imagine, if I

Â move some volume with this small diameter cylinder, I'm not

Â going to get very much stroke with a larger diameter cylinder.

Â So we can use this for precise positioning control, and also force amplification.

Â 2:06

So, two limitations I want to highlight here

Â of Pascal's Law that we're going to address

Â later in the class, the first is, assuming that the fluid is not in motion.

Â Which means we don't have any viscous forces.

Â We don't have any kinetic energy losses.

Â And the second is that we're neglecting any gravitational force.

Â So, if I look at a vertical pipe like this one right here, and I say,

Â you know, what is the, the pressure relationship

Â as I'm moving up and down that, that fluid?

Â And I assume that it's at rest.

Â Well, lemme quickly do a force balance right here, so a force balance.

Â On this fluid, I'm going to say my forces going upwards would be P2.

Â Pardon me.

Â Would be, P2 times the area,

Â and then the forces coming down, I've got the pressure on the top cyl, of the cap.

Â P1 times the area.

Â And then I've got the gravitational force which would be rho gh, times the area.

Â And as I look at these terms I can

Â recognize very quickly that I've got areas on both sides.

Â Or in all three terms, so I can.

Â Cross out my areas.

Â And then my pressure differential between the two would be just P, P,

Â in this case, P2 minus P1, which ends up being just rho gh.

Â And if I'm interested in what the pressure drop is or the, the delta P,

Â with respect to a given height, I can divide that out and say my delta P.

Â Which is just P1, er, P2 minus P1, divided by H, is just, rho times G.

Â 4:14

And, I multiply these out and now I can get the pressure difference, which would

Â be right around, 8.7, kilopascals, per meter.

Â And remember we were saying atmospheric pressure

Â is about a 101.3 kilopascals, so this

Â 8.7 per meter is actually very small in the, the grand scheme of things.

Â And quite often in fluid power systems it's, it's unimportant unless you're

Â dealing with, perhaps a wind turbine where you have a 100 meter

Â rotor height, and you're having a large pressure differential as the fluids

Â moving up and down in a, in a hydrostatic drive wind turbine.

Â One place that it does play a role is in pumps,

Â where we need to minimize the pressure drop coming into the pump.

Â So we'll talk about this pressure drop more once we get to pumps.

Â But for now recognized these pressures are relatively small.

Â Compared to the hydraulic pressures that we're

Â often dealing with, which is in the megapascals.

Â 5:09

So, the second law I want to highlight now is the Conservation of Mass.

Â And so we know in Newtonian physics, mass is always conserved.

Â And, we take one step beyond this and

Â say that hydraulics, oil, water, is relatively incompressible.

Â And if I can say that mass is always conserved and I've got a constant

Â density and incompressible fluid, then I can

Â also say that my flow rate is conserved.

Â So if I look at a junction, which might be some t-junction in our, in our system.

Â It might be a cross where I have a fourth port coming out of this.

Â But some place where I have multiple flows coming in there.

Â Well, from a mass conservation perspective, I

Â could say the mass flow rate coming into

Â each one of these ports, as I add them all up, it better sum to zero.

Â Well, once I assume incompressible flow, now I've got these three flow

Â rates right here and these three flow rates better also sum to zero.

Â So I can say Q1 plus Q2, plus Q3, is equal to zero.

Â And again I've got arrows all pointing inward

Â so this, this all should sum to zero.

Â So the natural question here is, is this a good assumption or not?

Â Is the incompressible assumption good?

Â Well, typically hydraulic oils are somewhere in

Â w, water is well, somewhere two to

Â three percent compressible at the pressures we're

Â dealing with, so these are relatively good assumptions.

Â For the majority of the time, although the compressibility

Â of the fluid which was referred to through the

Â bulk modulus, does become important during some situations and

Â we'll, we'll talk about those later in the class.

Â But for now, incompressible assumption is, is a relatively good one.

Â So where I want to go now is apply this to a hydraulic cylinder,

Â so a cylinder that looks like this where I've got two different areas.

Â On this side, I've got my cap side.

Â This side I've got my rod side.

Â I realize that the cross-sectional area or the effective area

Â that the, the f, fluid is acting on is larger on

Â the cap side than it is on the rod side

Â because we've got this piston sticking out on the rod side.

Â And what I want to do is look at the, the

Â pressure and flow relationships for this, for this hydraulic cylinder.

Â So, first of all, from a pressure standpoint, I can say that.

Â If I sum up, the summation of forces is equal to mass times acceleration.

Â My summation of forces, well, I've got on one side a

Â hydraulic pressure acting, which would just be P1, multiplied by A1.

Â So that's this area here, which we refer to as the cap side area.

Â And then, acting in the opposite direction, I've got the force

Â acting on the rod, and as all, I also have P2.

Â And P2 now is acting on this smaller area right here, which is the entire

Â outer bore, minus the area that the rod is act, is occupying.

Â So we're subtracting F of the rod.

Â 8:17

We also have transformers for the flow relationships,

Â where we can recognize that our flow, flow

Â rate is, Q, which is equal to the

Â cross-sectional area times the velocity of our rod.

Â So I've got the velocity labelled right here and this will, I have to define

Â what's positive and negative and correspond that

Â with the, the flow rate I'm talking about.

Â But we just relate these through the, the cross-sectional area.

Â 9:03

And, one thing you have to be careful of, is that

Â in a hydraulic actuator, in something like this where I've got a

Â different area on the cap side and the rod side, I'm going to

Â have different flow rates coming into this versus going out of it.

Â And it's actually going to be increasing the flow rate, as I go from

Â the rod to the cap, or, and, and reducing in the, in the opposite case.

Â So we have to be, be careful as to

Â amplifying or reducing our flow rates in the system.

Â So, this area ratio, this A2 to A1 captor rod side area

Â is our transformer ratio for both

Â the force amplification and the flow amplification.

Â 9:40

So, to summarize, we've talked about two different

Â laws that we want to apply to fluid power.

Â One is Pascal's law, telling us the

Â pressure is, is the same in all directions.

Â And again, we had some simplifying assumptions that we have to be careful of.

Â And then we said we've got mass conservation applied to fluid power.

Â And if we make the assumption of incompressible

Â flow, now we have a conservation of flow rate.

Â And this allows us to look at flow rates through

Â junctions and then we took one step farther and said let's

Â look at a hydraulic cylinder and look at the transformer

Â ratios to calculate pressure to force and flow to velocity relationships.

Â Thank you.

Â [SOUND]

Â