So we've been talking about analogies between mechanical and hydraulic systems. The fact that a mechanical system has springs, masses, and dampers. While we have the same thing in hydraulic systems thus far we've talked about the compressibility or the spring. We've talked about the resistances or the dampers and now today we're going to talk about the mass or the hydraulic inertia. And to do that, let's talk about a very common, everyday application, which is a water system. And what happens when we turn on and off water valves in a house. So I'm standing here at a kitchen sink that has a faucet with an electronic valve. So this valve closes very quickly and we can hear a water hammer event whenever I stop the flow of water. So you can hear when I stop the flow, and when impact of the water coming up through the, through the pipes. And what we hear is not the closing of the valve, but actually the, the pressure impact force. So now you have seen the Water Hammer Effect in a residential, you know, water type of a system. Let's talk about how we can calculate Hydraulic Inertia. So, I'm going to be looking at an example here. I've got a long, small diameter tube and i'm going to say it's horizontal so I don't have to worry about the gravitational influences. And let me assume that the fluid in there is incompressible and invisible so I don't have to worry about viscous effects and things like that. So, first of all I'm just going to write Newton's second law, and then let me put this into hydraulic terms and we'll start to substitute these in. So, let me first of all say that I can write the force here. And the force is just equal to the change in pressure, or the delta P across this pipe times the area, the cross sectional area. I can say that my mass. It's just equal to the density times the volume of the fluid. And the volume is the area times the length, AL. And the velocity of the fluid. Well, we've got the flow rate. And we've got the cross sectional area. So we can say, this is the flow rate, volumetric flow rate, divided by the cross-sectional area, q over a. So let me take those 3 quantities, substitute them in. And we get this form of the equation. And what you'll notice right away is that I can cancel the areas. And the numerator on both sides and we're left with this. Which looks just like Newton's second law. Where on the left side of, of it I have the change in pressure. the delta P across the, the tube, which is the equivalent of the force. I then have my mass term, which is the hydraulic inertia. Rho l over a, and then I have the change in flow rate with respect to time. This is the acceleration of the fluid. So, this hydraulic inertia term, notice that, I've got length in the numerator, and area in the denominator. So, our hydraulic inertia is greatest for long, small diameter tubes. So, it might be counter, counter intuitive that normally we think of just more mass. Well, we would have more mass with a larger cross sectional area. We actually have more. Higher inertia with a smaller diameter area, but a long tube. So that's where we're going to see the, the largest inertia, like that household, water system. Now, this plays a large roll in certain types of hydraulic applications when we do have long, small diameter tubes. With large changes in flow rate, like opening and closing valves very quickly. So in those types of situations you have to pay attention to. The water hammer effect, or the large pressure gradients that are created by accelerating, and decelerating, and the flows. And that it might be causing just vibrations, or it might be causing some fatigue on the, on your hydraulic conduits. So let's go back to that faucet example and let me take this small long tube and attach a valve to the end of it, a faucet of some sort. And look at what happens as far as what the pressure gradients would be, or the pressure differential across this pipe. So, I'm going to say I'm going to close this valve in about 10 milliseconds. My pipe is 10 meters long, and has a flow rate going through it at 10 liters per minute. Also, I'm going to say it, the radius of my pipe is about 6 millimeters. I'm sorry, Six centimeters, and so let me go, go ahead and, and run through these calculations and see what kind of inertial pressure I have. So I'm going to start off with my, my equations that I just generated a moment ago. and now let me go ahead and crunch through my numbers here. So my Delta P. It's just going to be equal to first of all in the numerator I'm going to have the density of hydraulic fluid which is 1000 kilograms per meter cubed. And then I'm going to multiply it by the length of ten meters. Divide all this by my cross sectional area and my area here I've calculated it out as 1.13. Times ten to the minus fourth. Meters squared. And now I'm, need to multiply this by dq dt. So I've got a flow rate, which is ten liters per minute. I need this in cubic meters per second before I move on with my calculation so I can do a quick conversion of this and this ends up being 1.7. Times ten to the minus fourth. Cubic meters per second. And I know the time that this is occurring across and I'm going to assume that it's just as linearly decreasing across that, that time. Just to simplify my equations. So, my dq term will be 1.7 times ten to the minus fourth. And this would be in units of cubic meters per second. And then I'm going to divide this by the 10 milliseconds, or 0.01 seconds. And, as I crunch through the numbers here, first of all I better check my units. And as you start to cancel out the, the meters and. And what not, you will find that you're going to get Newtons per meters squared, and, therefore, this will be a Pascal. And the number actually ends up being 1.5 times ten to the sixth, or 1.5 Megapascals, which is about 220 PSI. So, we're actually getting a very large water hammer event of very large pressure that's being created by turning off this valve so quickly. And this is for a household re you know, residential system where we normally have 60 psi of water or getting 220 psi due to this hammer event. So it can be quite significant and we need to pay attention to that as were designing our hydraulic conduits. To make sure that we don't fail them due to this. So again, remember that our hydraulic inertia is proportional to the density and the length over the area ratio. And so the longer diameter, small small crests long lengths, small crests. The actual area has a large amount of inertia. And just in general with hydraulic systems, we have springs, masses and dampers. As far as the compliance for the compressibility of our hydraulic fluid. The damping being, the resistance going thorough valves and other components. And now, we have the inertia or the mass. So, we'll take all of these and use this for our modeling, as we move throughout the rest of the course.
