0:00

So we've been talking about analogies between mechanical and hydraulic systems.

Â The fact that a mechanical system has springs, masses, and dampers.

Â While we have the same thing in hydraulic systems thus far

Â we've talked about the compressibility or the spring.

Â We've talked about the resistances or the dampers and

Â now today we're going to talk about the mass or the hydraulic inertia.

Â And to do that,

Â let's talk about a very common, everyday application, which is a water system.

Â And what happens when we turn on and off water valves in a house.

Â So I'm standing here at a kitchen sink that has a faucet with

Â an electronic valve.

Â So this valve closes very quickly and

Â we can hear a water hammer event whenever I stop the flow of water.

Â So you can hear when I stop the flow,

Â and when impact of the water coming up through the, through the pipes.

Â And what we hear is not the closing of the valve, but

Â actually the, the pressure impact force.

Â 0:57

So now you have seen the Water Hammer Effect in a residential, you know,

Â water type of a system.

Â Let's talk about how we can calculate Hydraulic Inertia.

Â So, I'm going to be looking at an example here.

Â I've got a long, small diameter tube and i'm going to say it's horizontal so

Â I don't have to worry about the gravitational influences.

Â And let me assume that the fluid in there is incompressible and invisible so

Â I don't have to worry about viscous effects and things like that.

Â So, first of all I'm just going to write Newton's second law, and

Â then let me put this into hydraulic terms and we'll start to substitute these in.

Â So, let me first of all say that I can write the force here.

Â And the force is just equal to the change in pressure, or

Â the delta P across this pipe times the area, the cross sectional area.

Â I can say that my mass.

Â It's just equal to the density times the volume of the fluid.

Â And the volume is the area times the length, AL.

Â And the velocity of the fluid.

Â 1:52

Well, we've got the flow rate.

Â And we've got the cross sectional area.

Â So we can say, this is the flow rate, volumetric flow rate,

Â divided by the cross-sectional area, q over a.

Â So let me take those 3 quantities, substitute them in.

Â And we get this form of the equation.

Â And what you'll notice right away is that I can cancel the areas.

Â And the numerator on both sides and we're left with this.

Â Which looks just like Newton's second law.

Â Where on the left side of, of it I have the change in pressure.

Â the delta P across the, the tube, which is the equivalent of the force.

Â I then have my mass term, which is the hydraulic inertia.

Â Rho l over a, and then I have the change in flow rate with respect to time.

Â This is the acceleration of the fluid.

Â So, this hydraulic inertia term,

Â notice that, I've got length in the numerator, and area in the denominator.

Â So, our hydraulic inertia is greatest for long, small diameter tubes.

Â So, it might be counter, counter intuitive that normally we think of just more mass.

Â Well, we would have more mass with a larger cross sectional area.

Â We actually have more.

Â Higher inertia with a smaller diameter area, but a long tube.

Â So that's where we're going to see the,

Â the largest inertia, like that household, water system.

Â Now, this plays a large roll in certain types of hydraulic applications when we

Â do have long, small diameter tubes.

Â With large changes in flow rate, like opening and closing valves very quickly.

Â So in those types of situations you have to pay attention to.

Â The water hammer effect, or the large pressure gradients that are created

Â by accelerating, and decelerating, and the flows.

Â And that it might be causing just vibrations, or

Â it might be causing some fatigue on the, on your hydraulic conduits.

Â So let's go back to that faucet example and let me take this small long tube and

Â attach a valve to the end of it, a faucet of some sort.

Â And look at what happens as far as what the pressure gradients would be,

Â or the pressure differential across this pipe.

Â 3:42

So, I'm going to say I'm going to close this valve in about 10 milliseconds.

Â My pipe is 10 meters long, and

Â has a flow rate going through it at 10 liters per minute.

Â Also, I'm going to say it, the radius of my pipe is about 6 millimeters.

Â I'm sorry, Six centimeters, and so let me go, go ahead and,

Â and run through these calculations and see what kind of inertial pressure I have.

Â 4:43

Meters squared.

Â And now I'm, need to multiply this by dq dt.

Â So I've got a flow rate, which is ten liters per minute.

Â I need this in cubic meters per second before I move on with my calculation so

Â I can do a quick conversion of this and this ends up being 1.7.

Â Times ten to the minus fourth.

Â Cubic meters per second.

Â And I know the time that this is occurring across and I'm going to assume that it's

Â just as linearly decreasing across that, that time.

Â Just to simplify my equations.

Â So, my dq term will be 1.7 times ten to the minus fourth.

Â 5:20

And this would be in units of cubic meters per second.

Â And then I'm going to divide this by the 10 milliseconds, or 0.01 seconds.

Â And, as I crunch through the numbers here, first of all I better check my units.

Â And as you start to cancel out the, the meters and.

Â And what not, you will find that you're going to get Newtons per meters squared,

Â and, therefore, this will be a Pascal.

Â And the number actually ends up being 1.5 times ten to the sixth, or

Â 1.5 Megapascals, which is about 220 PSI.

Â So, we're actually getting a very large water hammer event of

Â very large pressure that's being created by turning off this valve so quickly.

Â And this is for a household re you know, residential system where we normally have

Â 60 psi of water or getting 220 psi due to this hammer event.

Â So it can be quite significant and

Â we need to pay attention to that as were designing our hydraulic conduits.

Â To make sure that we don't fail them due to this.

Â So again, remember that our hydraulic inertia is proportional to the density and

Â the length over the area ratio.

Â And so the longer diameter, small small crests long lengths, small crests.

Â The actual area has a large amount of inertia.

Â And just in general with hydraulic systems, we have springs,

Â masses and dampers.

Â As far as the compliance for the compressibility of our hydraulic fluid.

Â The damping being, the resistance going thorough valves and other components.

Â And now, we have the inertia or the mass.

Â So, we'll take all of these and use this for

Â our modeling, as we move throughout the rest of the course.

Â