0:00

When we looked at our video from the lab,we saw that

Â we can use valves to direct and restrict or control flow.

Â We can change velocities of hydraulic cylinders.

Â We can change the speed of a hydraulic motor or the tork.

Â We can do all sorts of things with valves.

Â Now, we're going to use Bernoulli's equation, to derive what

Â we call the Orpheus flow model, and use that then

Â to predict the pressure drop to flow rate relationship for,

Â a hydraulic valve, and then use that in an example.

Â 0:27

so Bernoulli's equation, is really just a conservation of energy,

Â and this is energy per unit volume, so recognize right away

Â that this is different from when we were talking about

Â Pascal's law, where we were saying our fluid was at rest.

Â Now our fluid is in motion, so I've got a fluid conduit here,

Â with two different points labeled on it, as location one and location two.

Â And I'm going to make assumptions that first this fluid is in-compressible.

Â Second, that I've got a steady flow or you know, fully developed steady flow.

Â So, I've got my valve at its position or it's given

Â area for a certain amount of time that it's at steady state.

Â I've got friction less fluid, and my points are along a stream line.

Â Meaning that I can get from point one to

Â point two, as the fluid moves along the conduit.

Â So, this is Bernoulli's equation, and I've got a few

Â different terms here that I want to pay attention to.

Â 1:42

So we have these three different energy

Â terms, and these are energies per unit volume.

Â Now, when I've, I've got these three terms on the, the left side of my

Â equation for position one, three terms on the

Â right side for position two, and one thing

Â I recognize in many of my fluid power

Â applications is that my, my change in height,

Â my delta h from location one to position

Â two, is relatively negligible, for many common situations.

Â So, because of that I can then get rid of

Â those pardon me, I can then get rid of those points.

Â Let me grab a different color.

Â And, remove these two components of my equation.

Â 2:44

So, now I may apply Bernoulli's Equation to, again, a

Â fluid conduit, but now I have a restriction in it.

Â I'm calling it a knife-edge orifice, or just a, you can imagine it being a

Â plate in the center of my flow field, and has a very sharp edge to it.

Â So, I'm going to take Bernoulli's equation, and I'm going

Â to say, first of all, I'm not going to have much

Â delta H across my, I'm not going to have much delta

Â H across this so I can, knock out these two terms.

Â The second assumption I'm going to make, is I'm going

Â to assume, so first, assume that H1 is equal to H2.

Â The second thing, is that if you look at the, the area that

Â the, the flow is going through, and again I've drawn these two streamlines here.

Â If I look at the area that those are occupying, notice

Â that the area on the upstream is much larger, so the

Â velocity is much lower, so I can say that the velocity

Â is downstream V2 is going to be much greater than V1.

Â To the extent that I can say V 1 becomes relatively

Â negligible, let me set it equal to 0, and so this term here it goes to 0 as well.

Â 3:50

So, once I do that, now I can rewrite this equation gather some

Â terms and I can say that I've got, on let me gather my terms.

Â 2 over rho times the pressure drop, p one minus p 2, in

Â the square root of all of that, this is then equal to V2.

Â So, all I've done is I've taken V2 on the

Â right side of my equation, kept it there, moved all

Â the other terms over to the left side, and then

Â taken the square root of those, so I've got V2 now.

Â And, I know that the, the velocity at this location two right here, well.

Â How do I express that, in something I can measure?

Â It's a little more challenging to measure

Â direct velocity, but I can measure flow rates.

Â So, let me look at a cross-sectional area right here.

Â So I'll refer to that cross-sectional area as A2.

Â And that is just kind of between these stream lines.

Â What's the area that that velocity is occurring at?

Â And I know I can rate velocities with flow rates, through an area.

Â So, I can rewrite this V2, as being a flow rate divided by A2.

Â 5:26

So, if I can measure A nought, and

Â I recognize that these streamlines are reducing in area,

Â what if I can relate, write some sort of

Â relationship there, and call the reduction area discharge coefficient?

Â And that's exactly what I'm going to do.

Â So, let me swap colors here, and I'll write this as just the flow rate.

Â And now I'm going to be the discharge coefficient

Â times the area that I can actually measure, A naught.

Â So, all I'm doing is saying how much does that area reduce as I go from the orifice

Â down to the, that location downstream, that, that second location.

Â So, now I've written this CD, which

Â I'm going to define as a discharge coefficient.

Â And our discharge coefficient is typically somewhere between 0.6 and 1.

Â It would be 0.6 for a knife edge orifice such as I've drawn here.

Â For a very smooth nozzle, it would approaching 1.

Â because I wouldn't get any further,

Â further reduction in the cross sectional area.

Â So, what we've now developed here is, what we refer to, as the Orifice equation.

Â Which is Q is equal to CD times A, the discharge

Â coefficient times the area of the Orifice time the square root of

Â 2 over the density of the fluid, which is rho, times

Â the pressure drop, from the up stream to the down stream side.

Â And we use this is all sorts

Â of hydraulic applications, whenever we're talking about.

Â Pressure drops across the valve, this is the first approach at doing that.

Â 6:48

So, now we got our arvest equation lets, lets examine it a little bit.

Â So, I'm going to take my equation here and

Â I'm going to rewrite it, and my purpose here is

Â I want to look at, whats its the influence on

Â the diameter of the Orpheus on the, the pressure drop.

Â So, I'll rewrite it in terms of pressure drops

Â so, my delta P is simply P1 minus P2.

Â 7:09

And then I rearrange the right side of my equation, and then

Â I can also say that the area of my orifice is, PI

Â times the diameter squared over four, so I'm going to take this

Â area here, and I'm going to substitute it in here to the denominator.

Â And so when I rewrite this, I can say delta P is equal

Â to the fluid density roll, divided by 2, and then in the numerator, I

Â get Q squared, over CD squared, and now what happens to this area,

Â well I get, PI squared, I get diameter, which is now to the fourth.

Â 8:01

So, if I maybe have a 2 millimeter orifice

Â that fluid is moving through, and I reduce that

Â down to a 1 millimeter orifice, well now I

Â have a 16 times increase in my pressure drop.

Â So we have to be very careful as we start

Â to change areas, as to what the pressure drop relationship is.

Â It's not linear, it's actually to the fourth power, so very substantial.

Â So you might say, how do you, how do you change the, the area?

Â Well, one of our most common valves that we use, is called a needle valve.

Â And I've got a diagram of it here.

Â Basically what's happening is I am turning this knob here, and it

Â is taking a, a conical shape, or a needle if you will.

Â And pulling it out of a seat or bringing it down into the seat.

Â What I'm doing is I'm changing the cross-sectional

Â area, so if you can imagine this is

Â the part down here that's actually restricting the

Â area, and the flow path is right through here.

Â So, I'm restricting that flow path as I turn this knob.

Â This is one of the common, simple ways to change the area

Â or the pressure drop flow rate relationship in a, in a hydraulic circuit.

Â So, now let's look at this in an application.

Â So, let's say I've got a valve here and I'm going to

Â pick a valve that, that is just an on/off valve, if you will.

Â It's got two, two ways, is what we call it.

Â It's got an input flow and an output flow.

Â I have a, a manifold block and so this is what would house the valve,

Â I've got it cut in half, so you can see it a little bit better.

Â And this valve with thread in here, and inside of this valve I've got

Â a cylinoid spool, I'll show you a better picture of it in just a moment.

Â And when I apply a magnetic field with a, a cylinoid, I then pull this

Â spool and the movement of the spool will either, open a port or block a port.

Â And provides the valving operation.

Â So, I've got a, a very simple circuit here, or I've got my hydraulic pump.

Â I then have this on off valve that I'm discussing that

Â I'm showing right here, and then I have a hydraulic cylinder.

Â So, recognize that this is not a very useful circuit because, we can only drive

Â the cylinder an extension and we can never return it to a retraction state, but it.

Â I'm, I'm using to provide an example of this flow rate pressure relationship.

Â So, I'm saying that I've got a, a flow

Â rate going across this valve of 25 liters per minute.

Â So, let me figure out what the pressure's drop, drop is going to be.

Â And again this is a valve that I can buy off the shelf

Â and because of that, can go to the data sheet for this valve.

Â 10:32

See if that add link will work.

Â And here's the data sheet for the valve.

Â So this is 2-Way, Normally Closed Spool valve.

Â I can scroll down through here, and I mentioned I'd give

Â you a little better picture of it, over here on the

Â right is a little bit better picture where you can see

Â this is the, the spool component that is, is actually moving.

Â I've got my cylinoid up here which is a bunch of wires

Â that are wrapped around and when I apply electrical current it creates,

Â a magnetic field which then pulls this, this first component upwards moving

Â that spool, and we're opening or closing the valve by doing so.

Â When it's says it's normally closed that means the valve, is blocked

Â in the off position, when I energize the spool, the valve is open.

Â So, this is the kind of data that you get from a

Â manufacturer's data sheet, and it tells you how to order the valve.

Â Now, what we're most interested in here, is this plot right here.

Â If you notice, on the X axis, we've got the flow rate.

Â On the y axis, we've got the pressure drop, and then we've got a curve.

Â Well this curve ends up basically just being the

Â orifice equation and so, I'm going to toggle back to

Â my slide, and here is that curve I've blown it

Â up a little bit so we can see it better.

Â So what I'm going to do, is I'm going to

Â try to pick off this 25 liters per minute that my

Â actual flow rate will be for this application, and you'll see

Â that I only go up to about 19 liters per minute.

Â And I'm off the chart.

Â 11:56

So am I out of, out of luck here?

Â Well no, I can use my, my knowledge of the orifice equation, to calculate a, a, a,

Â a value for this valve that will allow me

Â to find the pressure drop at higher flow rates.

Â So, let me now do that.

Â So, I'm going to take my orifice equation.

Â And I'm going to lump some terms now, because I

Â don't know what the discharge coefficient of this valve is.

Â It's a very complex flow, we can't call it a knife-edge orifice.

Â I've got an area, I can't measure that in because I haven't torn apart this

Â valve, or maybe I have in this case, but normally we haven't torn about the

Â valve, and I don't really care a whole lot where the fluid density is, so let me take

Â this term right here and, just call that a factor some, some coefficient.

Â So, I'm going to lump all this and say my coefficient K, is just equal to

Â the discharge coefficient times the area, times 2

Â over rho and the, the root of that.

Â So, I've simply lumped those together, and now

Â I want to figure out what that value is.

Â So, I'm going to go to this data sheet and, calculate this K value.

Â So as I do so, I can recognize, let me, let me rewrite my equation.

Â Q is equal to K times the square root of delta P.

Â I take this one step farther so that I can get the values that I have on my, my data

Â sheet here, and I can say K is equal to, Q over the square root of delta P.

Â 13:25

So, let me pull those values right off my data sheet.

Â So I'm going to maybe look at, in a certain situation

Â where I've got, right around 15 liters per minute of flow.

Â And as I move over to the y axis, I've got

Â somewhere right around, I'll call that 3.8 bar, a pressure drop.

Â And remember a bar is 100 kpa, so we can already convert

Â this into, to units, that, that we're used to, used to working in.

Â So, I'll plug these numbers into, to my equation.

Â And if I plug in a flow rate of 15 liters per minute, convert this into

Â cubic meters per second, I can, well let me go ahead and do that.

Â So this 15 meters per minute ends up being 2.5 x 10 to the minus 4th.

Â And this is in cubic meters per second, and remember 3.8

Â bar is equal to, this would be, 380 kilo pascals.

Â 15:27

And in this case I end up with about 1,000 kilo

Â pascals, 1,075 kpa, a pressure drop.

Â So, I went from the data sheet where I had 15 liters per minute,

Â and I said I want to push that up to 25 liters per minute.

Â I went from 380 kilo pascal of pressure drop, to about 1000 or about a megapascal.

Â So, now this pressure drop is very significant,as the

Â fluid is going through my valve, to a hydraulic cylinder.

Â And the pressure drop might be to the point that it's unacceptable,

Â in which case I'd have to go to a, a larger size valve.

Â So this would allow me to say, is

Â this valve adequate for, for what I'm doing here?

Â 16:10

So, in summary, I took, first of all,

Â Bernoulli's Equation, and I reduced some terms from it.

Â Turned it and used that to create the, the orifice equation.

Â And we said the orifice equation is applicable to all sorts

Â of valves that we would experience in, in fluid power systems.

Â We then use that to calculate the, the flow rate going

Â to a hydraulic cylinder through this on off valve, as an example.

Â Now, you might wonder, where does the, the logo for the course come from.

Â Well it is from the, the orifice equation, and

Â so, you can see it right there, the orifice

Â equation, and there it's related to a, to a

Â check valve that we'll explore in our next video lecture.

Â Thank you.

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