Fluid power has the highest power density of all conventional power-transmission technologies. Learn the benefits and limitations of fluid power, how to analyze fluid power components and circuits, and how to design and simulate fluid power circuits for applications. In this course, you will be introduced to the fundamental principles and analytical modeling of fluid power components, circuits, and systems. You will learn the benefits and limitations of fluid power compared with other power transmission technologies; the operation, use, and symbols of common hydraulic components; how to formulate and analyze models of hydraulic components and circuits; and how to design and predict the performance of fluid power circuits. This course is supported by the National Science Foundation Engineering Research Center for Compact and Efficient Fluid Power, and is endorsed by the National Fluid Power Association, the leading industry trade group in fluid power.
Theory: Orifice flow
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Del curso dictado por University of Minnesota
Fundamentals of Fluid Power
313 calificaciones
De la lección
Week 3: Components and Concepts: Part 2
This will be a busy week diving into valves and pumps. We will discuss how basic valves function, how to use them in hydraulic circuits, and how to calculate pressure drop for a given flow rate, or vice versa. The videos will directly address the discussion on the forum about seeing hydraulic components working in real world circuits. In our discussion of pumps we will look at many different positive displacement pumps, exploring flow ripple and pump efficiency, look at the supporting components that form a hydraulic power supply, and see how we can make a transmission with a hydraulic pump and a motor. We are now into the heart of this course; we hope you enjoy seeing the components come together into useful circuits.
Conoce a los instructores
James D. Van De Ven, PhDAssistant Professor
Mechanical Engineering
Will Durfee, PhDMorse Alumni Distinguished Teaching Professor
Mechanical Engineering
When we looked at our video from the lab,we saw that
we can use valves to direct and restrict or control flow.
We can change velocities of hydraulic cylinders.
We can change the speed of a hydraulic motor or the tork.
We can do all sorts of things with valves.
Now, we're going to use Bernoulli's equation, to derive what
we call the Orpheus flow model, and use that then
to predict the pressure drop to flow rate relationship for,
a hydraulic valve, and then use that in an example.
so Bernoulli's equation, is really just a conservation of energy,
and this is energy per unit volume, so recognize right away
that this is different from when we were talking about
Pascal's law, where we were saying our fluid was at rest.
Now our fluid is in motion, so I've got a fluid conduit here,
with two different points labeled on it, as location one and location two.
And I'm going to make assumptions that first this fluid is in-compressible.
Second, that I've got a steady flow or you know, fully developed steady flow.
So, I've got my valve at its position or it's given
area for a certain amount of time that it's at steady state.
I've got friction less fluid, and my points are along a stream line.
Meaning that I can get from point one to
point two, as the fluid moves along the conduit.
So, this is Bernoulli's equation, and I've got a few
different terms here that I want to pay attention to.
First term here is the pressure.
The second term is rho-gh.
Well, remember when we were talking about
Pascal's law and the gravitational limitations of it.
Well this is the, the potential energy or the gravitational potential.
So this is potential energy, and then the, the final term is
one half row v squared, this is the kinetic energy of the fluid.
So we have these three different energy
terms, and these are energies per unit volume.
Now, when I've, I've got these three terms on the, the left side of my
equation for position one, three terms on the
right side for position two, and one thing
I recognize in many of my fluid power
applications is that my, my change in height,
my delta h from location one to position
two, is relatively negligible, for many common situations.
So, because of that I can then get rid of
those pardon me, I can then get rid of those points.
Let me grab a different color.
And, remove these two components of my equation.
And, then what we're left with is we
can recognize that, whenever my velocity the fluid
increases, which would happen when our cross-sectional area
gets smaller, my pressure is going to decrease.
So, this is slightly counterintuitive feature, but one thing that
is very important as we get to the, the orifice model.
So, now I may apply Bernoulli's Equation to, again, a
fluid conduit, but now I have a restriction in it.
I'm calling it a knife-edge orifice, or just a, you can imagine it being a
plate in the center of my flow field, and has a very sharp edge to it.
So, I'm going to take Bernoulli's equation, and I'm going
to say, first of all, I'm not going to have much
delta H across my, I'm not going to have much delta
H across this so I can, knock out these two terms.
The second assumption I'm going to make, is I'm going
to assume, so first, assume that H1 is equal to H2.
The second thing, is that if you look at the, the area that
the, the flow is going through, and again I've drawn these two streamlines here.
If I look at the area that those are occupying, notice
that the area on the upstream is much larger, so the
velocity is much lower, so I can say that the velocity
is downstream V2 is going to be much greater than V1.
To the extent that I can say V 1 becomes relatively
negligible, let me set it equal to 0, and so this term here it goes to 0 as well.
So, once I do that, now I can rewrite this equation gather some
terms and I can say that I've got, on let me gather my terms.
2 over rho times the pressure drop, p one minus p 2, in
the square root of all of that, this is then equal to V2.
So, all I've done is I've taken V2 on the
right side of my equation, kept it there, moved all
the other terms over to the left side, and then
taken the square root of those, so I've got V2 now.
And, I know that the, the velocity at this location two right here, well.
How do I express that, in something I can measure?
It's a little more challenging to measure
direct velocity, but I can measure flow rates.
So, let me look at a cross-sectional area right here.
So I'll refer to that cross-sectional area as A2.
And that is just kind of between these stream lines.
What's the area that that velocity is occurring at?
And I know I can rate velocities with flow rates, through an area.
So, I can rewrite this V2, as being a flow rate divided by A2.
So I have gotten one step closer to an equation that
I can use to predict flow rate pressure drop, across an orifice.
Now, one challenge is left here.
How do I measure A2?
It's not something physical in my system, so I'd have to somehow be looking
at the fluid, look at the, the tracer particles in a fluid or something.
It's a difficult thing to measure.
But what I can measure, is I can measure the area of my orifice itself.
And I'll call that area A naught.
So, if I can measure A nought, and
I recognize that these streamlines are reducing in area,
what if I can relate, write some sort of
relationship there, and call the reduction area discharge coefficient?
And that's exactly what I'm going to do.
So, let me swap colors here, and I'll write this as just the flow rate.
And now I'm going to be the discharge coefficient
times the area that I can actually measure, A naught.
So, all I'm doing is saying how much does that area reduce as I go from the orifice
down to the, that location downstream, that, that second location.
So, now I've written this CD, which
I'm going to define as a discharge coefficient.
And our discharge coefficient is typically somewhere between 0.6 and 1.
It would be 0.6 for a knife edge orifice such as I've drawn here.
For a very smooth nozzle, it would approaching 1.
because I wouldn't get any further,
further reduction in the cross sectional area.
So, what we've now developed here is, what we refer to, as the Orifice equation.
Which is Q is equal to CD times A, the discharge
coefficient times the area of the Orifice time the square root of
2 over the density of the fluid, which is rho, times
the pressure drop, from the up stream to the down stream side.
And we use this is all sorts
of hydraulic applications, whenever we're talking about.
Pressure drops across the valve, this is the first approach at doing that.
So, now we got our arvest equation lets, lets examine it a little bit.
So, I'm going to take my equation here and
I'm going to rewrite it, and my purpose here is
I want to look at, whats its the influence on
the diameter of the Orpheus on the, the pressure drop.
So, I'll rewrite it in terms of pressure drops
so, my delta P is simply P1 minus P2.
And then I rearrange the right side of my equation, and then
I can also say that the area of my orifice is, PI
times the diameter squared over four, so I'm going to take this
area here, and I'm going to substitute it in here to the denominator.
And so when I rewrite this, I can say delta P is equal
to the fluid density roll, divided by 2, and then in the numerator, I
get Q squared, over CD squared, and now what happens to this area,
well I get, PI squared, I get diameter, which is now to the fourth.
And then my 4 also gets squared, goes up to the
numerator so I end up with 16 here in the numerator.
Now, at the end of the day what I want you to recognize
is that, we have got this diameter to the fourth power right here.
So, if I maybe have a 2 millimeter orifice
that fluid is moving through, and I reduce that
down to a 1 millimeter orifice, well now I
have a 16 times increase in my pressure drop.
So we have to be very careful as we start
to change areas, as to what the pressure drop relationship is.
It's not linear, it's actually to the fourth power, so very substantial.
So you might say, how do you, how do you change the, the area?
Well, one of our most common valves that we use, is called a needle valve.
And I've got a diagram of it here.
Basically what's happening is I am turning this knob here, and it
is taking a, a conical shape, or a needle if you will.
And pulling it out of a seat or bringing it down into the seat.
What I'm doing is I'm changing the cross-sectional
area, so if you can imagine this is
the part down here that's actually restricting the
area, and the flow path is right through here.
So, I'm restricting that flow path as I turn this knob.
This is one of the common, simple ways to change the area
or the pressure drop flow rate relationship in a, in a hydraulic circuit.
So, now let's look at this in an application.
So, let's say I've got a valve here and I'm going to
pick a valve that, that is just an on/off valve, if you will.
It's got two, two ways, is what we call it.
It's got an input flow and an output flow.
I have a, a manifold block and so this is what would house the valve,
I've got it cut in half, so you can see it a little bit better.
And this valve with thread in here, and inside of this valve I've got
a cylinoid spool, I'll show you a better picture of it in just a moment.
And when I apply a magnetic field with a, a cylinoid, I then pull this
spool and the movement of the spool will either, open a port or block a port.
And provides the valving operation.
So, I've got a, a very simple circuit here, or I've got my hydraulic pump.
I then have this on off valve that I'm discussing that
I'm showing right here, and then I have a hydraulic cylinder.
So, recognize that this is not a very useful circuit because, we can only drive
the cylinder an extension and we can never return it to a retraction state, but it.
I'm, I'm using to provide an example of this flow rate pressure relationship.
So, I'm saying that I've got a, a flow
rate going across this valve of 25 liters per minute.
So, let me figure out what the pressure's drop, drop is going to be.
And again this is a valve that I can buy off the shelf
and because of that, can go to the data sheet for this valve.
See if that add link will work.
And here's the data sheet for the valve.
So this is 2-Way, Normally Closed Spool valve.
I can scroll down through here, and I mentioned I'd give
you a little better picture of it, over here on the
right is a little bit better picture where you can see
this is the, the spool component that is, is actually moving.
I've got my cylinoid up here which is a bunch of wires
that are wrapped around and when I apply electrical current it creates,
a magnetic field which then pulls this, this first component upwards moving
that spool, and we're opening or closing the valve by doing so.
When it's says it's normally closed that means the valve, is blocked
in the off position, when I energize the spool, the valve is open.
So, this is the kind of data that you get from a
manufacturer's data sheet, and it tells you how to order the valve.
Now, what we're most interested in here, is this plot right here.
If you notice, on the X axis, we've got the flow rate.
On the y axis, we've got the pressure drop, and then we've got a curve.
Well this curve ends up basically just being the
orifice equation and so, I'm going to toggle back to
my slide, and here is that curve I've blown it
up a little bit so we can see it better.
So what I'm going to do, is I'm going to
try to pick off this 25 liters per minute that my
actual flow rate will be for this application, and you'll see
that I only go up to about 19 liters per minute.
And I'm off the chart.
So am I out of, out of luck here?
Well no, I can use my, my knowledge of the orifice equation, to calculate a, a, a,
a value for this valve that will allow me
to find the pressure drop at higher flow rates.
So, let me now do that.
So, I'm going to take my orifice equation.
And I'm going to lump some terms now, because I
don't know what the discharge coefficient of this valve is.
It's a very complex flow, we can't call it a knife-edge orifice.
I've got an area, I can't measure that in because I haven't torn apart this
valve, or maybe I have in this case, but normally we haven't torn about the
valve, and I don't really care a whole lot where the fluid density is, so let me take
this term right here and, just call that a factor some, some coefficient.
So, I'm going to lump all this and say my coefficient K, is just equal to
the discharge coefficient times the area, times 2
over rho and the, the root of that.
So, I've simply lumped those together, and now
I want to figure out what that value is.
So, I'm going to go to this data sheet and, calculate this K value.
So as I do so, I can recognize, let me, let me rewrite my equation.
Q is equal to K times the square root of delta P.
I take this one step farther so that I can get the values that I have on my, my data
sheet here, and I can say K is equal to, Q over the square root of delta P.
So, let me pull those values right off my data sheet.
So I'm going to maybe look at, in a certain situation
where I've got, right around 15 liters per minute of flow.
And as I move over to the y axis, I've got
somewhere right around, I'll call that 3.8 bar, a pressure drop.
And remember a bar is 100 kpa, so we can already convert
this into, to units, that, that we're used to, used to working in.
So, I'll plug these numbers into, to my equation.
And if I plug in a flow rate of 15 liters per minute, convert this into
cubic meters per second, I can, well let me go ahead and do that.
So this 15 meters per minute ends up being 2.5 x 10 to the minus 4th.
And this is in cubic meters per second, and remember 3.8
bar is equal to, this would be, 380 kilo pascals.
So, just a little bit over 31/2 atmospheres of, of pressure drop.
So I plug in these two numbers and I, I crunch through it and I get a K value
of 4.05 times 10 to the, 10 to the minus 7th in this case.
So I've got a k value.
I only need one step farther now to actually figure out what my pressure drop
is going to be, so at 25 liters per minute, which is my given in this case.
And I can convert that into cubic meters per second
this is 4.2 times 10 to the minus 4th cubic meters.
Now I can figure out what my actual pressure drop will be.
And my Delta P is going to be equal to, in this case Q squared
over, I'm getting into my Q squared over K squared.
And in this case I end up with about 1,000 kilo
pascals, 1,075 kpa, a pressure drop.
So, I went from the data sheet where I had 15 liters per minute,
and I said I want to push that up to 25 liters per minute.
I went from 380 kilo pascal of pressure drop, to about 1000 or about a megapascal.
So, now this pressure drop is very significant,as the
fluid is going through my valve, to a hydraulic cylinder.
And the pressure drop might be to the point that it's unacceptable,
in which case I'd have to go to a, a larger size valve.
So this would allow me to say, is
this valve adequate for, for what I'm doing here?
So, in summary, I took, first of all,
Bernoulli's Equation, and I reduced some terms from it.
Turned it and used that to create the, the orifice equation.
And we said the orifice equation is applicable to all sorts
of valves that we would experience in, in fluid power systems.
We then use that to calculate the, the flow rate going
to a hydraulic cylinder through this on off valve, as an example.
Now, you might wonder, where does the, the logo for the course come from.
Well it is from the, the orifice equation, and
so, you can see it right there, the orifice
equation, and there it's related to a, to a
check valve that we'll explore in our next video lecture.
Thank you.
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