hi I'm Mike Crowley and to day at Fluid Mechanics i'm going to explain water hammer in pipes. Water hammer is a special transient flow case. Transient flow and the study of transition flow which is called surge analysis is concerned with dynamically changing flow velocities in pipe. Water hammer occurs when there is a sudden or rapid change in the flow velocity. It's usually associated with a valve slamming closed or rapid closing of a valve. It can lead to very high pressure transients which can cause the pipe to fail often is associated with a banging noise which leads to the term water hammer. Basically you have a long column of water and you're rapidly stopping it. It bangs against the valve and it causes a banging noise. In this lesson I will explain the theory behind water hammer i'll show you how to calculate the pressure transients that are induced due to water hammer. I will explain this shortly at Fluid Mechanics. So let me explain what is happening and how to calculate the induced pressures. So if I draw a sketch of a tank, connected to a pipe. This is a header tank. A pipe line connected to it. And in this tank we have a head of fluid and that is pushing the fluid along the pipe. Its going to have an initial velocity Ui and it's going into a open tank, at this end here. So this is our initial conditions, constant velocity Ui initially along the pipe into a tank. And there's a head of fluid, which is pushing the flow along. The pressure at the inlet to the pipe is... The pressure equals. Rho, which is a density, Gravity times h The head. Now knowing the pressure at the inlet to the pipe and if you know the other conditions along the pipe. You know the length of the pipe, diameter of the pipe, the viscosity of the fluid, it is possible to calculate what the flow rate is along the pipe. Now in this video I'm not going to explain how to do that, but it but it's not very difficult job to calculate the velocity along the pipe. So then what happens, In the water hammer case, we have a sudden closure of valve at the end of the pipe, So that some instance in time the end of the pipe is closed off. I'm just going to show a blockage on the end of the pipe, there to show that the pipe has been closed now into the instantanes you do that, you still got flow coming into the starts of the pipe. But at this end of the pipe here the flow has stopped, because it has got nowhere to go. So what actually happens is it sets up a pressure fronter or a wave front which travels up the pipe and i will show it at this position here. And this pressure or way front travels up the pipe at velocity C. And C is the velocity the sonic velocity in the pipe. So on this side of the way front here. The velocity and U equals 0. And on this side of the pipe, the velocity is still the initial velocity. Now that is a little bit theoretical, because it assumes you had an instantaneous closure valve. But no matter how far you close it, it will take some time to close the valve. And in that case what happens instead of just being a one plane in the pipe the the change of velocity will occur over a section of pipe, so this is probably a bit more realistic and basically what we're saying is that over this length, here there will be a pressure change, Delta P. Where on this side the velocity is U, the initial velocity. And on this side of the wave front the velocity is zero. So the velocity will be changing across this wave front now the length of this wave front from here to here, is to do with how long it takes to close the valve. So if the valve was closed instantaneously it would be just be a plane but if it takes a fraction of a second basically it's how far that wave front travels in the time. So the time it takes to close the valve, times the sonic velocity will determine what the length of that wave front is. Now across the wavefront the velocity is going from the initial velocity down to zero velocity there's a change in momentum or change in velocity across that wave front. That wave front can only change momentum, or the velocity can only change if there's a force applied to the fluid, okay. We've now got to look at Newton's second law to work out what force is applied to the fluid as it goes across the wavefront. Newton's second law is force equals mass times acceleration now in our case we're not talking about forces were talking about pressures and we're not talking about mass and acceleration. We're talk about changes in momentum. So for us the the force that's acting across that wave front there is the the differential pressure, DP across the wavefront acting on the area of the pipe. So I will put down A for the area of the pipe. So now we need to look at what is the momentum change across that wavefront well the wavefront is traveling up the pipe at velocity C so at any instance in time we can actually work out how much fluid is traveling through that wavefront okay and the amount of fluid that's traveling through that wavefront. Is basically how fast it's going up the pipe times the area of the of the area wavefront times the density of the fluid. So the mass flow rate part of it is. The velocity of the wavefront C times the area of the pipe A times the density of the fluid rho okay. So the fluid that's actually go through that wavefront in terms of kilograms per second, going up through the wavefront is C, A, rho. So in other other words the velocity of the wavefront that's the sonic velocity of the wavefront, the area of the pipe and the density of the fluid. And that's the mass flow rate going through that wavefront. And how much is the velocity changing? Well it's going from Ui down to zero. So in other words it's going from the initial velocity down to zero. So the momentum change is Ui. So we can take out A from both sides of that equation there. so we've basically got delta p equals C rho Ui. Or more generally we say that the pressure for a sudden closure of a valve is C rho U, okay. Now that that equation there is called the Joukowsky equation and it's a famous equation, and that determines what the maximum pressure rise you can get to water hammer is. The maximum pressure rises is the sonic velocity, the speed of sound in the fluid the density of the fluid times the change in speed of the fluid. So its initial speed going down to zero. So let's try and apply this equation to say a 50-mmr copper pipe. And say we had a 50mm copper pipe with an initial speed of 1m/s and what we want to do, is find out when we suddenly closed the valve how much pressure rise we're going to get for a 15mm copper pipe. Well let's just put down some details first of all of this copper pipe, so the diameter of the copper pipe is 15mm and the initial velocity U equals 1m/s 1m/s in a 15mm pipe is actually equivalent to 8.7 l/min Okay. So when we look at this equation and we try to apply it, If we were talking about water in a copper pipe, that's what I'm talking about here, we know the initial velocity that's going to be 1m/s we know the density of water that's normally a 1000 kg/m^3 the thing we're not sure about is, what's the sonic velocity. And that's what I'm going to talk about next. So to find the sonic velocity in a fluid you need to apply Hooke's law to it. If we assume that the pipe is perfectly rigid and does not flex okay, you can apply this equation which is Hooke's law which basically says, the specific speed is equal to the square root of the bulk modulus, divided by the density of the fluid. Now for water let's just calculate that. For water we got C equals the square root of. The bulk modulus of water, is 2.19x10^9 Pa and the density is a 1000, so if you calculate that you get a speed of 1480 m/s. Now that's assuming that the pipe is perfectly rigid, but pipes aren't perfectly rigid they actually flex and that actually affects the stiffness of the system. And as it gets less stiff the sonic speed comes down. So there's a modification you can do to this equation to take into account the stiffness of the pie. Basically you modify Hooke's law equation, so that C equals the square root of, one on, rho k plus D on. So what's this equation saying? Basically what this equation is saying is that the sonic speed is the density. Same as there, one on K, that's the bulk modulus + D. D is the diameter of the pipe, E is the Youngs modulus of the pipe material. And then little e is the wall thickness, Okay. This part of the equation here is taking into account the stiffness of the actual pipe itself. If the pipe was perfectly rigid then effectively what that's saying is that you have infinite young's modulus, for the material and if that number was infinitely large then this term would would drop down to zero okay, and if thats zero, if you put zero in there you'll effectively come back to this this original equation here. So basically that's that's how its modified, so as this becomes less stiff then this term in the equation becomes more important and it actually reduces the speed. So if we actually now put in some numbers for that. Now for a standard 15 mm copper pipe, I believe the wall thickness is 0.7mm and for copper E, young's modulus is 120x10^9 Pa. Okay, so if i put those numbers into that equation, will get Okay and if you work that out. You get C equals 1254 m/s. So the velocity has come down from, for a copper pipe from 1484 a perfectly rigid copper pipe down to 1254 m/s. Actually copper pipe is very stiff but it all depends on the pipe your choosing. So if you're talking about the pipes that take water to your house. The plastic pipes that nowadays they use in the road. Typically you'd find the wave speed in one of those would be around about a 1000 m/s, but if you took a very flexible pipe likes a garden hose pipe you know you could be talking in terms of 100s m/s the other thing to bear in mind about the wave speed though is the bulk modulus. Water in particular is very stiff okay. So that's 2.19x10^9 now that's true as long as there's no air in the in the water. But often you get little small air bubbles in the water and they can have quite a significant effect on the bulk modulus. and bring down to speed quite significantly. So that can be quite an important factor but anyway we'll carry on with the calculation. So we now want to work out what the pressure rises due to this closure of this 15mm pipe with a 1m/s flow in it, and we close the end of the valve. We have the numbers now to apply to the Joukowsky equation so the pressure rise looking at the Joukowsky equation is going to be C which is 1254 times the density of water which is normally a 1000 kg/m^3 times the velocity which in our particular example is one and if we work that out, that comes out 12.54x10^5. I'm going to put in x10^5 because 1x10^5 is 1 bar. ok, So that's equals 12.5 bar. So that's the pressure rise you'd get in that particular case, maximum. I happen to know that the pressure rating of a copper pipe, of this specification is 58 bar. So the safety factor for that particular case is 58 bar divided by 12.5bar. Which equals 4.64. So the safety factor is 4.64 Another way of looking at that is if we actually had a much higher velocity. If the initial velocity was 4.64m/s for we would have actually then got 58 bar. Now for copper pipe, that will be going some. So normally for copper pipes when you close the end of the valve you don't have a problem from a burst point of view. But just be a little bit careful with that because the burst pressure is not the only thing that's important when you're designing a hydraulic system you have all the fittings on the end of the pipes, there highly likely to be ripped off you go to excessive pressures you have all the bracketry on the walls, and things like that. If you've got movement in the pipes you might affect that. If you have bends in pipes they can tend to flex. So there are other things to take into account. So in summary to calculate the pressure rise due to a sudden closure and water hammer what you need to know is the initial flow conditions, an and the initial flow velocity. You need to understand and work out what the the sonic speed is and I have shown you in the in the lesson how to calculate that. And you need to know the density of the fluid. From that you can apply the Joukowsky equation and basically the maximum pressure rise is the the product of the velocity, the wave speed, the density. If you have any questions on this then please leave a comment on my website blog and I will endeavor to answer any questions there. I cannot answer any general questions directly by email but I will if you leave a question on the blog try and answer it there. If you need any more detailed advice particularly need advice on surge analysis on a consultancy type basis. Then please contact me directly. That's it today from fluid mechanics thank you for listening.
