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So, thus far in this class we've

sent, been considering our hydraulic fluid as incompressible.

And we've said that's a reasonable approximation

for most of what we've been doing.

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

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

So.

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

just the, the bulk fluid itself is slightly

compressible, but when it becomes much more compressible,

is when we have entrained air in the fluid.

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

in here, then those air bubbles become

very spongy and add to the compressibility.

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

That's how we define it.

And, I have it here, in a, in a

way you can think about the bulk modulus as really,

what is the pressure required to cause a given

volume of fluid to change volume by a certain amount.

So.

Lots of, lots of derivatives here.

And, a very nice thing about this equation is we

can use it now to model the pressure in a given

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

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

So.

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

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

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

Notice that the denominator of the bulk modulus equation is,

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

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

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

So we're in units of pressure here.

And, typically hydraulic oils are 1.5 to 1.9 gigapascal.

Water is typically 2.2 gigapascal.

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

It's also a function of pressure.

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

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

And then the, the different curves are for,

different amounts of trained air so when I say

that hydraulic oil might have 1.6 gigapascals of

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

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

so to do a really good job of modeling

this, we would ideally have a pressured dependent and.

And trained air ration dependent bulk modulus term.

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Let me throw out some, some common numbers.

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

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

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

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

So minus 2.2%.

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

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

how much air is in that, in that oil.

So, often reasonable assumption that it's incompressible.

But in some cases this does become important.

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

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

This is due just to how we define a

change in volume you notice that, a dV is considered

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

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

What I was mentioning, why is this important.

So, first of all, it's important for

systems where we need to precise positioning.

So, consider that I have a hydraulic cylinder and

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

perhaps there's a lot of mass attached to this

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

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

our fluid is compressible, especially if there's

a lot of air trapped in the system.

So we have to be aware of that.

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

The fact that we can get dynamic resonances

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

And having that, [SOUND] influence our system,

especially with high frequency systems, often several hydraulic

systems are running at very high frequencies,

getting, getting into the resonant frequency of the.

The hydraulics.

And then, compressible energy losses.

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

If we look at a hydraulic pump as we were talking

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

Now, in that piston cylinder we have.

When the piston comes up to the top dead center

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

Well that fluid was compressed previously, and now

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

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

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

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

This is happening 60 times a second.

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

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

So, again, different areas where the small amount of

compressibility does add up over time and does become important.

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

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

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

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

[SOUND] just do an analysis of what the residents

frequency or what the national frequency of this system is.

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

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

the length of the oil column of one side plus the

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

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

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

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

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

I can write traditional spring equation.

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

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

The, the system.

So, delta L in this case.

So, now let me substitute this in.

So I'll solve for k.

And this will just be delta F over L.

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

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

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

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

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

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

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

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

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

Where is it?

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

now I have a very stiff spring on one side.

And so that drives up my, my spring rate.

But, if I end up with the piston hub in

the dead center, that happens to be the case where

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

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

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And in that case, the natural frequency simplifies quote

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

right here, just, 2 times the radical of the

bulk modulus times the area divided by L over m.

So again, notice that just like our spring rate,

when we get to this natural frequency equation, equation.

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

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

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

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

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

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

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

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

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

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Multiplied by the area of 1.2 times 10 to the

minus 4th meters squared, divided by the length

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

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

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And if I convert this into hertz, this ends

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

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So, about 100 hertz natural frequency for this system.

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

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

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

Very rarely we'll get into 1000 hertz.

So we are getting into the resonant frequency of,

some hydraulic systems, so we have to be careful how

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

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

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

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

So there would be additional compressibility which

would further lower this, this natural frequency.

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

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

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

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

about a few specific cases where

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

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