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So, thus far in this class we've

Â sent, been considering our hydraulic fluid as incompressible.

Â And we've said that's a reasonable approximation

Â for most of what we've been doing.

Â Now we'll start to explore, when the fluid is compressible.

Â Or when we consider the compressibility effects to be important in our system.

Â So.

Â I've mentioned that it's, that it's compressible, and,

Â just the, the bulk fluid itself is slightly

Â compressible, but when it becomes much more compressible,

Â is when we have entrained air in the fluid.

Â In other words, if we, are agitating this and we get little air bubbles

Â in here, then those air bubbles become

Â very spongy and add to the compressibility.

Â So, we described the, the fluid compressibility through the bulk modulus.

Â That's how we define it.

Â And, I have it here, in a, in a

Â way you can think about the bulk modulus as really,

Â what is the pressure required to cause a given

Â volume of fluid to change volume by a certain amount.

Â So.

Â Lots of, lots of derivatives here.

Â And, a very nice thing about this equation is we

Â can use it now to model the pressure in a given

Â volume, dynamically, given an input flow rate, an output flow

Â rate, and perhaps a moving wall, or a moving, moving volume.

Â So.

Â This is a very useful equation for a variety of, of situations.

Â To give you a feeling of, of what the values are of

Â a bulk modulus, well, first of all, think about the units here.

Â Notice that the denominator of the bulk modulus equation is,

Â unit-less, dV by V, so the numerator has to be

Â the same as, or I'm sorry, the, the, the bulk

Â modulus has to be the same as the numerator or pressure.

Â So we're in units of pressure here.

Â And, typically hydraulic oils are 1.5 to 1.9 gigapascal.

Â Water is typically 2.2 gigapascal.

Â And again, this changes greatly with the amount of entrained air.

Â It's also a function of pressure.

Â So, to give you a feeling of what that looks like,

Â here's a plat of the bulk modulus on the x-axis of pressure.

Â And then the, the different curves are for,

Â different amounts of trained air so when I say

Â that hydraulic oil might have 1.6 gigapascals of

Â bulk modulus, that is what it plateaus to, but,

Â as we add air to that and as we decrease the pressure the bulk modulus drops and

Â so to do a really good job of modeling

Â this, we would ideally have a pressured dependent and.

Â And trained air ration dependent bulk modulus term.

Â 2:29

Let me throw out some, some common numbers.

Â Maybe for a 5,000 psi system which would be 35,000 five gigapascals.

Â So I've got minus 35 MPa, so that's ten to the sixth.

Â Divided by lets say 1.6 gigapascals 10 to the 9th.

Â I'm ending up with compressibility of right around 2.2%.

Â So minus 2.2%.

Â So, just to give you a feeling, you know, our, our oils that we're dealing with are

Â usually 1 to 3, maybe 5% compressible, depending on

Â how much air is in that, in that oil.

Â So, often reasonable assumption that it's incompressible.

Â But in some cases this does become important.

Â Now before I move on let me just mention the sign here notice

Â that we have a negative sign and a definition of the bulk modulus.

Â This is due just to how we define a

Â change in volume you notice that, a dV is considered

Â a, a reduction in volume so, that's a little bit

Â arbitrary, we can, we can take care of that later.

Â What I was mentioning, why is this important.

Â So, first of all, it's important for

Â systems where we need to precise positioning.

Â So, consider that I have a hydraulic cylinder and

Â I'm trying to very precisely position this end effector and

Â perhaps there's a lot of mass attached to this

Â and, I don't want to have much compliance to my system.

Â I want it to be very stiff and, that can be an issue if

Â our fluid is compressible, especially if there's

Â a lot of air trapped in the system.

Â So we have to be aware of that.

Â Second, I'll talk about more in just a moment, is resonance of the system.

Â The fact that we can get dynamic resonances

Â of, square root of k over m, sort of term.

Â And having that, [SOUND] influence our system,

Â especially with high frequency systems, often several hydraulic

Â systems are running at very high frequencies,

Â getting, getting into the resonant frequency of the.

Â The hydraulics.

Â And then, compressible energy losses.

Â You might say well w, what do you mean by that?

Â If we look at a hydraulic pump as we were talking

Â about before and we were looking at the piston cylinder interface.

Â Now, in that piston cylinder we have.

Â When the piston comes up to the top dead center

Â position, we still have some dead volume in the chamber.

Â Well that fluid was compressed previously, and now

Â we're going to open a valve to tank pressure.

Â And so the energy that went into compressing that fluid, is exhaust

Â in the tank, and then we're going to do that cycle after cycle.

Â And this is happening, you know, for 3600 RPMs.

Â This is happening 60 times a second.

Â So it ends up being a significant amount of energy loss over

Â time if we have a large amount of dead volume in our cylinders.

Â So, again, different areas where the small amount of

Â compressibility does add up over time and does become important.

Â So, let me talk a little bit more about resonance, as

Â one of the, the key areas where this, this becomes important.

Â So, think about a long, slender cylinder, something like

Â this, relatively small diameter, long stroke, and I'm going to,

Â [SOUND] just do an analysis of what the residents

Â frequency or what the national frequency of this system is.

Â I'm going to simplify this ever so slightly first of all by

Â saying that, my piston hub length is basically minuscule and so

Â the length of the oil column of one side plus the

Â oil, other oil length is the total length of the cylinder.

Â Second, I'm going to say that, my piston rod is very small, or that

Â I have a double-ended cylinder, and that the area on each side is the same.

Â So they're just some simplifications to make my math a little bit easier.

Â 5:58

Now, I recognize I'm going to treat this system as a spring.

Â I can write traditional spring equation.

Â Just F is equal to k times X or delta X.

Â In this case our X is just the, the length of the.

Â The, the system.

Â So, delta L in this case.

Â So, now let me substitute this in.

Â So I'll solve for k.

Â And this will just be delta F over L.

Â And this is now, the bulk modulus times the area divided by the length.

Â Notice that I did remove my, my negative sign here because

Â again, it's arbitrary decision of what direction is a positive delta L.

Â So I'm removing that right now because it'll end up under square root.

Â And that's why, that's why I've removed it here.

Â 7:19

So now I have this, this form for my, my resident frequency, and

Â we might say well, where is it going to be the worse case?

Â Is it going to be, all the way at one end, is it

Â going to be a quarter of the way, is it going to be centered?

Â Where is it?

Â Well, recognize if I move the piston to one end,

Â now I have a very stiff spring on one side.

Â And so that drives up my, my spring rate.

Â But, if I end up with the piston hub in

Â the dead center, that happens to be the case where

Â we have the, the lowest natural frequency or the, the

Â worst operation, if you will, when L1 is equal to L2.

Â 7:50

And in that case, the natural frequency simplifies quote

Â a bit, [SOUND] and we're left with this equation

Â right here, just, 2 times the radical of the

Â bulk modulus times the area divided by L over m.

Â So again, notice that just like our spring rate,

Â when we get to this natural frequency equation, equation.

Â The, slenderness ratio, that's A over L, becomes

Â the, the driving parameter for the, the resonant frequency.

Â So let's now do an example calculation, for this hydraulic cylinder here.

Â Now, let me say I'm going to add a chunk of mass onto

Â the end of this, 10 kilograms for, say, and I've measured the.

Â The cross sectional area of it 1.2 times 10 to minus 4th

Â meter squared and, the length of the stroke here is 20 centimeters.

Â So let me take those numbers, [SOUND] and plug this this into the equation that I

Â have here and, now I've got 2 times the bulk modules of maybe 1.6 gigapascals.

Â [BLANK_AUDIO]

Â Multiplied by the area of 1.2 times 10 to the

Â minus 4th meters squared, divided by the length

Â of .2 meters, and, the mass of 10 kilograms.

Â And I'll take the square root of all of that.

Â 9:17

And if I convert this into hertz, this ends

Â up being about 68, or 98 hertz, I'm sorry.

Â [BLANK_AUDIO]

Â So, about 100 hertz natural frequency for this system.

Â When we get into, several hydraulic systems, we're going to find that that

Â really is within the range of what, many hydraulic systems operate at.

Â Sometimes they get up and do a couple of 100 hertz.

Â Very rarely we'll get into 1000 hertz.

Â So we are getting into the resonant frequency of,

Â some hydraulic systems, so we have to be careful how

Â we're, how we're operating when we get to those

Â regions, and there's some special things that we can do.

Â One thing that I have neglected with this calculation is the additional

Â hydraulic oil that would be between the valve and the, the cylinder itself.

Â So there would be additional compressibility which

Â would further lower this, this natural frequency.

Â So [SOUND] in some situations, a long slender cylinder

Â like this, we can have fairly low, natural frequencies.

Â So, in this video we've talked about the compressibility of hydraulic oils.

Â And, how we model that with the bulk modulus equation, and then talked

Â about a few specific cases where

Â it's important, including, this case of resonance.

Â [BLANK_AUDIO]

Â