 We starten met de cam-modeling, de modeling van het sedimentation proces in de hopper. Juste een simpel, ideele zetting-basis waar we de trajectories van de partijkels volgden. Maar de, nou, meer of meer de simplified method is dat we naar een soort laminar flow in de hopper gaan kijken. Dus, heel mooie horizontal flowlijnen en de partijkels zet in die horizontal flow. Dus je krijgt een heel mooie straat trajectorie van een partijkel. En met de straat trajectorie zou je estimeren dan de... ...numere partijkels die de bed in tijd reachen en andere partijkels die de domain leveren. En met de zetting van de hele partijkelszijde distributie en verschillende fracties... ...kan je dan integreren met de zetting van alle verschillende zettingen... ...en de lage zetting die de zetting van een soort partijkelszijde distributie ontdekken. Het probleem is dat in de realiteit de flow is niet laminar, maar turbulent. En dat betekent dat als je een partijkel normaal zettingen over een straat trajectorie in de laminar flow... ...door de turbulence, dan kan je de sturen van eddies die de partijkel in de overvloossectorie transporten. Nou, dat is iets wat we willen modelen. Hoe kunnen we dat doen? We gaan dat doen met een continuutie equatie, de attractie en de diffusie equatie. We definen in effect de hele flowdomain in controle volumes. En in every control volume we look at the fluxes, so the rate of exchange within a control volume... ...because the rate of change of the number of particles in a volume is equal to the total of the sum of transport through all the sides of that volume. And we have attraction, the particles are carried with the flow and we have diffusion, which is in fact the mixing due to turbulent eddies. And we will finally arrive at the affection diffusion equation. Well, this is a big step, but I will slowly explain how you will end up with this equation. Does this equation look familiar to you, this affection diffusion equation? Sometimes seen it before in other applications, possibly. Ok, well, how do we arrive at this equation? Because this is in fact the kind of basic transport equation. And like the Navier-Stokes equation is more or less a transport equation like that, where the only difference is that we are transporting momentum and we include pressures, but it's just a transport equation. Well, when we look at the part of the domain, and I'm now looking at just a two-dimensional situation, so to make it not too complicated, so we have horizontal axis, the x and the y axis in vertical. So we look at the certain element, flow element, with the unit width of delta x length and delta y height. En at the left side of the boundary particles are flowing in that area with a certain velocity u, that's the horizontal velocity over the particles, and a certain concentration c. And so the total amount of material entering through the left boundary is u times c, the velocity at that location times the volumetric concentration, just the number of particles. At the other side, of course, particles are leaving. At the moment I'm only looking at horizontal adfections, so I'm neglecting vertical transport. Assume that vertical transport is zero, so no vertical velocities, only horizontal velocities. Well, at the right side you'll see that there's, of course, material leaving as well. And there can be a gradient in that term uc. And so at the right side you can see that what is leaving there, uc plus the gradient of that u times c times a value times the length delta x. So this is, in fact, the total transport at the right side. If there's no gradient, so no difference in the product of u times c, that means that the ddx times uc is zero, no gradient. This term is zero, and you see that at the left side the same amount of particles are coming in, and at the right side the same amount of particles are leaving. So the transport is no difference in transport over two boundaries. So there's no difference in the number of particles in this room. If we are in this room and the number of people leaving, of entering through that door is equal to the number of people leaving through that door, then there's a lot of transport in this room. There's a lot of people walking around, but the number of people is staying the same. There's a gradient in the number of people entering and the number of people leaving. You see that the number of people is increasing or decreasing. And then this is, in fact, the control volume that I'm looking at. Okay, well, if we look at the change of concentration during time, so what is happening, because this is the number of particles of the volume of particles in time changing must be equal to the difference in what's entering here and leaving there. Well, just at the left side you see times height delta y is entering and on the right side delta y times this term is leaving. That is just a volumetric balance equation. Well, we can just subtract this. You can see that the uc term drops out because there's a positive in front of the term in brackets and you see within the brackets the uc minus with a negative sign, so that term drops out. And you arrive finally at this simple affection equation, which states that the rate of change, so the concentration difference in time, so the increase or decrease of concentration in time, is equal to the difference in horizontal transport over a certain distance. So if there's no gradient in horizontal fluxes, because for instance the velocity is the same and the concentration is the same at the left and the right side boundary as well, there's no change in concentration. Yes. There's several possibilities. Let's assume for instance that the concentration is constant over the whole area. So there's a uniform concentration in this area. If then the velocity of the left side is larger than the velocity of the right side, then you see that there will be more particles coming in than leaving. Because it's the product of u times c, which is the flux. So there's a rate of change. Let's take this room again. If the concentration, if people are walking with a larger speed into this room, but we have the same amount of people, so every person has a distance of one meter, but if they're walking with a larger velocity inside of the room and leaving at a lower velocity, but everybody is at the same distance, you'll see that more people will come in. So it's the product from the velocity times concentration which determines the flux. So we can rate of change because the u can be varying over the x or the concentration can be varying over the x. Only if the product of these two, u times c is constant over the x, then you'll see that there's no change of the number of particles within time in that area. And it's important to realize it. And these are just the particles carried along with the flow. Because there's a flow and particles are in the flow and the particles are dragged with the flow, you'll see that the particles will enter into the domain. So this is a very basic affection equation because this is only horizontal affection. But you can imagine that if I will apply the whole same reasoning in vertical direction, that there will be a term ddy times u times c as well entering in the formula. But I'm not going to repeat that. So this is affection. Now we're going to look at the influence of turbulence. What's happening with turbulence? Well, let's look again at this control volume. At the left boundary I've plotted a lot of particles. And at the right side there's no particles at all. En you have a kind of an eddy, a turbulent eddy, which is more a stirring up the flow. Well, what happens when there's a kind of an eddy busy here at the left side of the boundary? You'll see that at the left side of the boundary there are more particles than at the right side of the boundary. Because there's less particles, here the concentration in the domain is lower than the concentration just outside of the domain. Well, what happens with a turbulent eddy? You'll see that when there's an eddy, only one particle is moving out of the domain, while more particles, because there's more busy at the left side, more particles are getting in. So you see that there's no net flow into the boundary and the same amount of water is entering here and leaving at the other side. Because just imagine it as a kind of a circular movement. In fact it's not exactly circular, but just to make it not too difficult. But you need a certain gradient and concentration to get transport. If there are the same number of particles at the left side and the right side of the boundary, then there will be no net transport, because the same amount of particles from the left, which are carried in to the boundary, will leave again at the other side. So you need a gradient, a difference in concentration over a horizontal distance to get real transport over the boundary. En dat's a diffusion process. No gradient, no net transport. Let's again look at this room again. And there's one large eddy. Eddy is going through the door at the left side and leaving to that door at the other side. There are 20 people waiting for this door and only two people waiting for him at the inner side of the room at this door. When the eddy enters, a lot of people are sucked in the room, 20 people are sucked in the room. At that door, only two are leaving out and we have 18 people extra in the room. If the same number of people waiting there in and out of the room, then the same people are leaving and coming in. So there's no net change in the number of people in this room. Quite simple in fact. Well, how do we write this in terms of an equation? The transport through the left wall is equivalent with the concentration gradient. So you need a concentration difference over that wall, that is the dc dx. And that epsilon, that is a kind of a diffusion constant that takes into account how effective that eddy is. There's a very small turbulence and the epsilon has a low value. So there's almost no mixing. If it's a very turbulent field, then the epsilon is very effective. There's a lot of mixing. It's kind of a mixing coefficient which takes into account the effect of turbulence. So at the left side, these number of particles are entering. This control volume. Well, of course, at the right side, you can look at the same transport as well. And at the right side, you will see that there is a term epsilon times dc dx. But we have to look at the gradient in the concentration gradient itself to look at the net transport. From the previous slide, we have seen that it is the gradient in u times c which determines the net transport. So now we have to look at the gradient, d dx, of this transport due to turbulence. So that's why you see the d dx over the transport. In de previous slide, you see the d dx over uc. uc was the transport here. The d dx was the contribution in the horizontal direction. The same holds for the turbulence. This is the transport at the left side and the d dx times that term is the difference. Well, then you can look again at what's the influence of the turbulence on the rate of the increase of concentration in the control volume. And you see that again, d dx times the volume, d dx times delta y, that's the number of particles. The volume of particles in the control volume is equivalent to the rate of change of this transport due to diffusion of turbulence. Well, then we have identified both influences and we can add them up because finally, when we just divide by delta y, delta x, we are left with this equation and if we sum them up, you see that the total, we see the first effect part of the affection diffusion equation, only in the horizontal direction. Well, in the vertical direction, exactly the same is happening and the direction in the third dimension, which is not taken of course as well, so it's easily, you can see that we can build, we can, the total equation in two dimensions will be this equation which was the equation where I started with. And if I include the third coordinate, so the coordinate out of this plane, then the only thing which changes is that there's a dz times wc, u is horizontal, v is vertical and w is the flow velocity to the third dimension and we have an extra term which takes the influence of turbulence in the account and there's a dz times epsilon times dw dz entering into the equation. So that's all in fact and it's important to know that we're looking at the concentration of particles, so the u and the v in this equation are the velocities of the particles, we're not interested because we're looking at the transport equation of the volume of particles, we have to look at the concentration of particles and the velocity of particles. That's important to note and there can be a difference, as you know, between the velocity of the fluid and the velocity of particles. The simplest example that I can give you is just when I drop one particle in a stagnant fluid the velocity is zero and I have a vertical velocity of the particle and so normally these two velocities will not be the same. Ok. So when we start with this general equation we see a lot of in this shape it's difficult to solve it. We can solve it numerically but not analytically. There's no analytical solution. We can simplify it a little bit with some assumptions. One of the first approximations is that the horizontal diffusion the transport due to horizontal diffusion is small compared with the horizontal affection. It means that because we have a horizontal flow through the settling basin that's the main transport mechanism and of course there will be horizontal transport due to turbulence but it is more or less small compared to the affection. So which terms does that which terms do we compare them in this equation? What is the horizontal affection term? This one. This horizontal affection. Horizontal diffusion is this one. This one is small compared to that one. We just drop it out. That's the first approximation. The second approximation for the ideal settling basin is that we have a stationary flow so there's no change in time of the whole situation in the basin. In fact we assume that after some time of course when we start loading there will be a change in concentration in the domain but after some time we have a kind of a stationary situation and nothing will change anymore within time. So which terms drops out at the equation then when it's stationary? There's no change in time anymore. Sorry? Yeah, DCDT, that's right. Okay, so stationary flow. Well, there's another approximation because here we're looking at vertical and you see in the equations are vertical velocities at the value v and the second term of the equation. Well, that v is the sum, is in fact the particle velocity and the particle velocity can be the settling velocity of the particle itself plus for instance the flow velocity. If we have a vertical flow velocity then the particle will settle with a larger speed than when there's a stagnant when the fluid velocity in vertical direction is zero. Assume, it can't even be the case that we have a vertical velocity coming up and when that vertical velocity of the fluid is the same as the settling velocity of a particle then that v is zero because it's just in balance. Well, what we assume now is that there's no vertical flow velocity in the domain because all more or less the average flow velocity is horizontal. Flow is coming from left to right and the only vertical velocity of the particles is due to the settling velocity itself. Well, and the second thing is that we assume that the settling velocity is not a function of the concentration which in fact we know it's not right because we have a hindered settling effect that has been explained just yesterday. Well, it makes it easy because we can then write this, because when we look at this previous term we see that the v times c is within the, within brackets so we have to look at the gradient of the product but if the velocity is constant we have no gradient so we can just get it out of the bracket and that's why we can just remove this one and write only the concentration gradient times the settling velocity, yes. Now u is the actual u and v are the actual particle velocities and if the vertical fluid velocity is zero so in fact you can see that we have a when we have a velocity field going down that is for instance the v water that's the vertical velocity and then we have a particle which is settling with a settling speed relatively to the moving water then the actual particle velocity v is v water plus the settling velocity well, in the transport equation we are using this one but if I assume that the water velocity is zero in vertical direction this one is zero and then v equals Ws, ja. Goed question dat je asked it, I could explain it a little bit better. Ok. Then we have an extra simplification and then we are looking at this term because here you see that there is still a possibility that the u at the horizontal velocity can change over the horizontal coordinate, we assume now that the u is constant as well in the whole domain so u is just uniform and just one value no matter where you are in the domain, in the basin so no velocity distribution or whatever well, that's easy as well because then we can take this one at the equation as well and we can write it like that and u is now the average horizontal flow velocity in the domain well, we are almost there the last simplification is in the third term and because you see that within the brackets is still a variation of the diffusion coefficient possible so the diffusion coefficient can still be a function of the vertical coordinate which is in fact always the case because very near to the bottom very near to the bed, the eddies are quite small so there is less turbulent mixing then when I am higher up in the basin then the eddies are much bigger so in reality that diffusion coefficient that mixing coefficient is indeed a function of the vertical coordinate low, small at the bottom small at the water surface and theoretically the largest value will be around in the middle of the basin from the bed well, this is another simplification the diffusion is just a constant average value in the domain and then I finally arrive at this simplified affection diffusion equation which you might have had in one of your mathematics courses on separation of variables or partial differential equations did you ever follow the courses on partial differential equations because this is a very popular equation to solve in that field you can solve it by separation of variables so you assume that the c is a function of the product of a function of something only variable in x something only in y and you introduce it in this equation and then you get two normal partial normal differential equations and you get a solution in a kind of series like solution I'm not going to do it but only look at what will be the result of such a calculation what we see then if we apply such a, when we solve the equation we see that when we for instance have a situation where we have a very nice homogeneous concentration over the height in the intern section and the particles are transported to the right, you see that due to the settling effect the concentration will be the total concentration will be lower at the end because the particles are just well, being removed from the flow and settling in the bed and at the same time we will see that there will be less particles near the water surface and more particles near the bed just due to the settling so that will be the solution and how can we then determine the overflow loss, well we know that at the entrance section the total volume of particles is u times the height times the inflowing concentration and there is a total sand flux coming in at the left side at the right side the total volume of particles is the horizontal flow velocity which is a constant so I can put it outside of the integral and integrate the concentration over height so the integrated concentration over height times the horizontal velocity is the number of volume of particles ja, ja, ja that is one of the assumptions that was disassumption this term you see that the horizontal velocity is a function of the can be a function of the X coordinate and so that's why g dx times the term and now we have said we simplify it and we say the horizontal flow velocity is everywhere the same in horizontal direction and in vertical direction so uniform flow over the whole domain and that's why we can just put the u as a constant in front of this d dx this is one of the simplifications which is not in fact not the case but it's just a simplified model otherwise we cannot solve it ja, ja when we have a stationary situation the volume in is equal to volume out so that is, well that's then the case so Q in is Q out but then still the flow velocity can vary for instance in the inflow section you can have a very low velocity over a large height en at the out, at the entrance, at the overflow section you can have a high velocity over a small depth so there can be a change in velocity and that in reality that will be the case but in this ideal settling basin we have said ok we have an entrance section and we have an overflow section and we're not going to, well to model what's happening over there that's too complicated the only thing that we're concentrating on is the middle section we define it as the settling section and in the settling section we say that we assume that the horizontal flow velocity is equal everywhere is uniform ok so the when we calculate the overflow when we integrate the concentration over height then we know the volume of particles leaving and then the removal efficiency is then just the amount of particles settling and there's a difference between coming in and out over the amount of particles going in because we have an analytical solution of the partial differential equation we can really calculate this effect and the result of these calculations can be shown in a graph where we see the removal ratio that is the efficiency of the number of particles settling in the bed as a function of the turbulence well in this case in the horizontal scale you see the influence of the turbulence where from when we go to the left you see that the increase in turbulence is larger or the influence of turbulence is larger the right side there's no turbulence and at the right side you can see that the effect for when there's so here you see the effect of that influence of the here you see the effect of the turbulence on the total ratio when there's no turbulence at all when we go to the right side you see that all the lines they are approaching the normal values of removal ratio of 0.1 until 0.9 when the influence of turbulence is increasing you see that these lines are dropping and less particles are indeed settling when the line is dropping you see that less particles are removed from the flow and this is the whole stirring up behavior of the turbulence well this whole analysis was already done in 1946 of 1948 or something like that this is quite old theory and that's why because they didn't have strong computers or no computers at all at the moment why they derived an analytical solution now we can just solve the whole equation just with a numerical method ok so if we go back again to the whole settling process so far we didn't look at the effect of the flow velocity on the sedimentation itself even we took into account the influence of turbulence the influence of turbulence stirs up the particles and that means that particles are longer in suspension ja, that's possible in theory but what we see in practice is that the effect of the flow velocity near the bed will have a reducing effect on the sedimentation velocity itself for instance imagine that I have a glass of water and sand particles and I mix it up and I just put it down then I don't have any flow velocity horizontal flow velocity in that container and the particles will just settle and you see the bed coming up in the function of time then you have the sedimentation and you can calculate the sedimentation velocity just with this equation where E is zero so there is no pickup by turbulence because there is no horizontal flow in that glass another extreme situation is when you look at a pipeline a pipeline transport of particles then you have a very large horizontal flow velocity all the time particles try to settle on the bottom of the pipeline because they are just settling but at the same time the turbulence within the pipeline is stirring the particles up again from the pipeline bottom and then you'll see that E is more or less equal to S en er is no sedimentation velocity at all and that is another extreme situation a pipeline transport above the critical velocity you don't have any sedimentation well in between you have a in fact an area where there is influence for instance if I make a graph and I put on the horizontal horizontal flow velocity and flow velocity is zero and this is the sedimentation velocity is when I stir up a mixture I'll get a certain sedimentation velocity in the glass well for a certain flow velocity you'll see that it won't settle anymore because the turbulence in the pipeline is just stirring up the particles at the whole time and they are settling but there is no net sedimentation I have a point over here that is the U critical in the horizontal pipeline transport well and there is then a kind of a relation and when we see from tests that we did you see that it looks like that so for low velocities you don't see too much influence of the velocity near the bed on the sedimentation velocity and after some time this will drop down quite dramatically this value is a little bit depending on the particle size but for instance for 150 microns this will be around 0.5 meters per second and this will be around well maybe 2.5 meters per second or something like that to get an idea well in hopper sedimentation we're always in well this region because if we would be around here or there there won't be any sedimentation at all and then we're loading two fine particles or the hopper size is just too small for the discharge that we're putting in the hopper and so for practical well the practical applications this is the interesting area where we will look at for hopper sedimentation so we have always will have a situation where we have sedimentation because otherwise we cannot load the hopper as we are always left of that critical velocity but there can be an influence of that flow velocity near the bed sometimes they use its cower or erosion but in fact it's not erosion it's more kind of a hindered settling near the bed the sedimentation velocity decreasing due to the effect of the horizontal flow but if we're talking about scour or we're talking about erosion then the bed will go down scour is well just when you have scour is more or less a type of erosion for instance a bridge pier or windmill pile in the seabed and you have wave and currents along that pile you see that there always a kind of a hole in the seabed is created due to the higher flow velocities near that obstacle that's called scour and very often people use this hindered sedimentation due to bed shear stress they call it scour as well but it's not really a good expression because well finally the bed will still come up there will be always net sedimentation in the hopper in practical situations of course in theory it can be possible that you first start loading the hopper with a very low velocity and you have a settling bed the bed goes up and then after some times you increase the velocity enormously and decrease the concentration and then you can blow all the sand out of the hopper again of course then you would have real erosion in practical situations we'll never do that ok we'll have a short break and then continue at a quarter to twelve let's continue so the sedimentation we know now that the sedimentation velocity is influenced by the flow velocity near the bed it's reduced in fact normally we never get the situation where e is larger than s because then we just blow out all the sand out of the hopper so normally during hopper sedimentation e will always be smaller than s but the risk can be an influence of that flow velocity well how can we determine the sediment pickup how can we determine the influence well first we have to know the flow velocity in the hopper especially near the bed because the flow velocity near the bed will influence that sedimentation and secondly we like to have then a relationship between that erosion or hindered sedimentation which is in fact a better definition the bed shear stress the particle size distribution the concentration if you know first the flow velocity and then the relationship then we can determine the value of e well one of the problem is that the situation in a hopper is quite different than we encounter in nature like in river flows in under waves in sand transport due to two effects especially that the concentration near the bed in a hopper is much higher than you will normally see in natural conditions you can never have a 30% concentration near the bed with a very low flow velocity in nature maybe when you have a very large avalanche in the water where a lot of material is more or less eroded then something can happen like that but normally you will only have very small concentrations near the bed so we really have to look at the specific situations in dredging to find suitable relationships well the first methods of the first models assume a uniform or logarithmic velocity profile so when we start loading we have a certain water depth the discharge is distributed over the height so we have an average flow velocity and during time the sand bed will come up there is less space available for flow so the flow velocity is increasing in an average and most models are most simple models are based on these kind of assumptions so you just assume that the total height of the hopper is used by the flow and when the sand bed goes up there is less space available well then we can for instance look at the vertical of the horizontal flow velocity and you can compute a certain critical speed for a certain critical flow velocity particles will not settle anymore or you can compute then a critical diameter which is not settling anymore depending on the horizontal flow velocity so a larger flow will result in a larger particle and that still can settle this is in fact the camp approach well and you can for a certain flow velocity you can then determine a certain particle size that will not settle anymore and it means that if we then look at the integral of a camp there is a certain size p s and that will not settle anymore due to that hindered sedimentation due to the effect of the bed shear stress so this is more or less a camp approach and this is the medium approach to take the flow velocity into account in practice we see that in fact there is not too much influence of this flow velocity en that is due to two effects so if you try to get if you are using this method to calculate the critical particle size that does not settle anymore and introduce it in the camp integral and look at the results that you will get you will see that there is only very little influence on the settling behavior and why is that because of that uniform flow assumption or logarithmic flow assumption and you assume that the whole flow is distributed over the whole height you will see that the horizontal flow velocity is only very small so maybe 10 centimeters per second that kind of value you get and when you look at the influence you will see that you are around here somewhere so a very low flow velocity will have a very small influence on that hindered sedimentation effect so you see effect during the whole stage you won't see any influence apart from the very last moment where you have a very limited water depth and then you get very high flow velocities but then your hoper is almost already completely filled with sand and so there is no influence well, when we look at the camp model and see what is really happening in the model we see something like that the mixture is entering from left to right here we see the flow velocity we see a very nice uniform flow velocity that is what we assumed everywhere the flow velocity is around about the same value and here we enter the concentration nicely distributed over the total height of the hopper and in the camp model this is just the application of the camp model what I did here this is the numerical simulation and you see that due to settling the whole surface of this mixture will go down it's just settling ok but this is in fact quite strange because what we see over here is we have a length of only 40 meters a height of 2 meters and we see that the tip the top of this concentrated mixture here is the concentration that is 0.2 that is the mixture of about 1 about 1300 kilogram per cubic meter blue is around 1000 kilogram per cubic meter so we have a quite large difference in density en we have a sloping surface of that dense fluid and that is strange if you have ever been near a border of a lake you see a water surface which is a density of 1000 kilogram per cubic meter above the lake there is air about 1 kilogram per cubic meter and the horizontal you see that the lake has a very nice horizontal surface now we have a large difference in density as well and we see a very sloping surface it's impossible in practice this cannot happen because it's just a heavy fluid if you have two fluids one heavy fluid and a less heavier fluid on top why should there be a sloping surface now because this is not sand this is a mixture of sand and water not settled sand what you see over here is not settled sand it's a mixture of particles and water but still a fluid 1.3 20% concentration is around 1.3 ton per cubic meter or 1300 kilogram per cubic meter water is 1000 so it's a very heavy compared with fluids and why is this in the model this is happening why because we have assumed that there is a horizontal flow velocity and we didn't couple the concentration resulting from that assumed flow velocity assumed velocity field so there's no coupling we have seen the density calculated by the diffusion equation the adfraction diffusion equation with the actual flow but of course there should be a coupling because this surface won't be in this horizontal direction due to gravity and now I'll show you the same experiment numerical experiment I will release at the left side the same amount of concentration particles with the same velocity so at the left side of the hopper we have indeed a uniform flow velocity with a uniform concentration but in the remaining part of the hopper there is an influence of this concentration distribution on the actual flow velocities that will develop and then we'll get this situation it's totally different you see that due to the high density difference the mixture drops to the bottom and now it starts filling up like a heavy fluid and we don't see any uniform flow velocity in the hopper at all we see a very high flow velocity near the bed over here we see the mixture going down a kind of circulation is taking place this is when we have an actual coupling between the sediment transport equation and the momentum equation so the actual flow velocities in the hopper during loading are not uniform I'll show it again so the big density wave bounces at the other side of the hopper the wave density wave is moving back again there is still no sedimentation there is still fluid as you can see there is no bed developing but the heavy fluid is more or less with the very nice horizontal surfings more or less filling up the domain with the heavy fluid this is just high flow velocity this is velocity so we have one meter per second near the bed the flow velocity and here you see kind of a circular well, kind of an eddy developing and this is the whole filling up with the mixture and now here you have a uniform horizontal inflow section so the horizontal flow velocity is equal and then due to the density the mixture accelerates to the vertical direction due to gravity you see here a nice high velocity then it more or less bends off due to the bed it has to go through this bend you get a high velocity and then it distributes again and you are filling up this hopper with the heavy fluid you see the surface of the fluid just going up during time I'll show you another animation a little bit later well, these are the model tests that we carried out to check this theory done by Deltar a laboratory a little bit further on the Rotterdamseweg you can see a top view with the diffuser that we used to introduce the mixture so we had a nice two-dimensional flow distribution but you can see all those vertical pipes at the bottom end of the pipe there was a device to measure the flow velocity in the hopper so we could measure the direction of the flow in two directions and at the same time we measured the concentration on about 14 different locations so we had I think 13 locations where we measured velocities and about 40 positions where we measured density how this hole is developing here another view seen from the entrance section total length about 12 meters 3 meter wide, 2,5 meter deep we tested other inflow mechanisms as well this is kind of a pipe the previous one is a very nice diffuser where we have a better distribution over the horizontal axis than this pipe but there was during these model tests a little influence on this type of inflow construction well this is then the actual flow pattern already seen on the simulations this is what we really measured as well and this is the measured flow velocity in time I have to explain this a little bit what you see over here are two signals at the bottom you see a measured flow velocity horizontal flow velocity and what we did was we started with a measurement device near the original bed of the hopper it was measuring the horizontal flow velocity then we started filling the hopper and then the sand bed comes up and after some time the instrument is buried in the sand and so we stayed we kept the velocity measurement on the same vertical position but because of the bed moving up and then you'll see that we are more or less measuring through the whole density current until the moment that is buried in the sand and that is this moment here you see that the flow velocity is increasing and then it drops to zero and what we then did we lifted the instrument again it started measuring and then it is buried again but with this method you are more or less measuring through the whole density current because you are making a kind of a profile with the assumption that the flow velocity is more or less constant in time of course and now what you can see is very interesting you see that at the start of this test when the instrument was still quite deep you see quite high flow velocities then the velocity goes down and then the velocity goes up again and so first you see the highest velocities you see at the beginning of the loading process and at the end of the loading process well at the end of the loading process that's due to the fact that you only have a limited space available because the hopper is almost completely filled the high velocity at the start of the filling process is the density current what you saw on the animation that the flow velocity near the bed had the highest value very nicely and what you see over here is the vertical position of the instrument so we started here at about 1.2 meters then we lifted it up again and up again and up again and the blue line is the measured send bed level on that location so you see every time when the velocity drops to zero you see that the send bed is at that location and then they waited for some small time it was buried maybe a few centimeters under the bed and then they lifted again it started measuring here it drops to zero again here it drops to zero again every time you see when the send bed the measured send bed crosses the vertical position of the instrument you see that the velocity goes to zero because it's in the send bed so what you can see is that the actual flow velocity what we measured the maximum flow velocity is this red line this is another example while in the original uniform flow distribution when we would assume that the flow would be uniform and nicely distributed over the whole hopper you would expect this flow velocity we see that the uniform flow velocity is completely different from what we actually measured and in fact what we calculate as well when we have a model where the sediment transport equations are coupled to the momentum equations so that are the shortcomings from the camp approach because we prescribe the flow field but in reality we have density currents we have an influence of the bed shear stress on sedimentation we have the inflow and outflow zone which are in fact not modeled on the camp model because we only are concentrating on the settling zone but nevertheless you'll see that the camp metal does give a quite reasonable estimate for the sedimentation and loading time and that's due to the effect that the influences of the overflow positions and the inflow conditions are more or less second order effects the most important parameters are that loading hopper loading parameter so the discharge over the width and length of the hopper and the settling velocity of the particles these are the two most dominant effects in hopper sedimentations and these two are available are introduced in the camp approach if you'd like to have, well, more if you'd like to do some more fundamental research and what's really the influence of the loading positions or should we start with hopper filled with hopper water or not that kind of influences you need a better model and well, that's why we I derived a two dimensional model where in the camp model we had uncoupled the sediment transport equations and the momentum equations and now we are going to couple it and you couple what you then do is to solve the Navier-Stokes equations in the hopper in two dimensions I introduced the mixture model so no multi phase flow so more or less you regard the fluid as a mixture not a separate so you don't solve the separate equations for the solid phase and the fluid phase it's in hydrodynamic model so not hydrostatic and you have the coupling between the momentum equations and sediment transport equations and what I've already showed you and that's why you get the buoyancy effect density currents only will develop when you have that coupling between the momentum equations and sediment transport equations well, to solve this Reynolds average model the runs you need to model the turbulence and I did it with a K-option turbulence model I'll go a bit quickly through the details I don't know why it's not working again other important condition is that of course you need to have a movable bed in the hopper of course you like to really fill up the hopper so the bed should move to the computational domain you need erosion sedimentation boundary condition at the hopper and you like to introduce not one particle size but the whole particle size distribution and when you look at that you get the total model is more or less built up by 3 different blocks you have the navistokes equation solved you have the turbulence model and the sediment transport equations well, here you recognize the camp model but for every fraction so for instance if I divide the particle size distribution in 10 different particle sizes I'm solving 10 different sediment transport equations in fact 10 kind of camp models but the only thing is that here we don't assume that the flow velocity is uniform now the flow velocity is coming from the navistokes equations but the navistokes equation is influenced by the density distributions coming out of the sediment transport equations and the turbulence model is giving the added viscosity and the turbulence model is again influenced by the sediment transport equation as well because if you have a high density near the bed and a low density above the bed a turbulent addy will be damped to that vertical gradient density stratification influence that will the same in atmospheric flows well, when you solve it you can calculate the flow velocities whereas you see simulations with different levels of the bed with different velocity profiles and here again you see that that the maximum flow velocity measured during time is highest when we start loading and highest when we stop loading the same phenomena that we have measured in the calculations concentration distribution and here you see a nice animation of a loading of a hopper where you see in fact all effects at the same time density currents building up sand bed building up you can see that the loading position is varying and this as well more or less a simulation of what could happen during the loading process we have more or less a complete two-dimensional simulation of the loading process in a hopper and this model is now well used I made it from my PhD and after my PhD I more or less made it well full proof and it is now well currently used in Boscas and van Orten when they tried to optimize hopper sizes for new hoppers they are really using this model ok een ander, een heel simpel model te laten zien waar je de eerste oplossing voor de overflow los kan geven is based op een heel simpel balans van sediment waar we zeggen ok de overflow los is in fact de functie van de inflowing sediment flux en de dus wat wat is de amount of particles going in over de amount of particles settling in the bed wel we recognize this equation again de sedimentation velocity is the sedimentation minus erosion of hindered sedimentation over this the one minus the project times the concentration near the bed if I looked in at the ratio between the sedimentation flux so the amount of particles settling in the bed over the amount of particles going in we can derive this formula where you see that the ratio is equivalent with the ingoving concentration and here you see the erosion term and here you see the hopper load parameter had a camp influence effect so this is in fact the camp influence the overflow rate and this is kind of a correction factor to take into account that the concentration in the hopper is not is varying well for a simple situation where the erosion or hindered sedimentation is zero so there is no influence on the flow velocity near the bed you can simplify it even further and then you see that we can define a kind of a parameter hopper loading parameter S star which is the ratio between the ingoving scent flux over the scent flux settling in the bed and you can write it as a function of something of the concentration and bed porosity times the kind of dimensionless hopper load parameter so the hopper load parameter over the settling velocity we can look at the experiments if we can really find such a relationship because now I like to find a relationship between overflow loss and this settling flux parameter the S star and what I did was first go to the next graph put them in a graph here you see all the tests that we did the model tests here you see the S star parameter and vertically the overflow loss cumulative measured during the experiments and indeed we see that there is a relationship between the overflow loss and the S star parameter and you see that the relationship between S star and overflow loss is better then I only would take the H star and the H star is the dimensionless overflow rate in more or less the camp parameter well that's encouraging but you have to be quite careful with this approach because this is based on model tests and there isn't a scale effect during the tests because we scale down the hopper you'll see that the erosion or the hindered sedimentation process is not on the right scale compared with the full size hoppers so in model tests you'll always find that the hindered sedimentation is less compared with the actual situation it's very difficult because in fact you are dealing with two contradicting scale laws you have the fruit scale which is necessary to get the whole sedimentation process and the settling process and the density currents on the right scale and the situation near the bed where you have that hindered settlement is more or less more related to a Reynolds type of scale and these two scale laws are more or less fighting with each other so if you have the one scale the whole flow is not right the only thing that you can do is really measure and then you can validate the models on lab scale and if the whole physical process is right in your equations then you can use it to scale up the whole calculation but it's difficult or a little bit dangerous to scale up just directly the model results with this approach so this can be used for a first approximate but you have to be careful ok now I'll give you a simple example how to calculate an overflow loss just using the camp method and we do it in fact we're going to solve that integral but now with a simple numerical approximation let's assume that we have a hopper with a certain length a certain width before he was in that now it was the old situation now there's a section of about 30 meters or 50 meters inserted so they cut it in two and they build a piece in between this is the old situation we have a certain discharge this is still the same and let's assume that we have this particle size distribution so this is the cumulative distribution of the size of particles that we're going to dredge so we see that about 100% is around 2-3 millimeters or everything is smaller than that ok what we can do make just a simple spreadsheet is assume that we calculate is with the length the width in the discharge Q we can calculate that v0 that's just Q divided over L times B that's the overflow rate so we see that the overflow rate is about 8 millimeters per second that can be seen as an average vertical flow velocity to the hopper well, from the particle size distribution that I showed you in the previous graph I distributed in 10 different sections over here and every particle size has about 10% of the total volume of particles so I cut the whole size in 10 different fractions with different particle sizes so 10% has a diameter of 0.01 millimeter and 10% has 3.3 millimeter and so on well, for every diameter particle size I can calculate the settling velocity w0 and just using the formulas that I gave you yesterday you can just introduce it in your spreadsheet you can use that empirical equation so, for every size I have a settling velocity you see that for the largest fraction the 3.3 millimeter will settle with a velocity of about 0.3 meter per second 30 centimeters per second while the very fine size is only settling at about 0.6 millimeters per second is a very large difference in settling velocities ok, well, the camp integral integrates this settling velocity over the integral over vsws over v0 over times vsdp, so the integrated so we can, for every fraction we can calculate the ratio between w the settling velocity of 1 particle and that overflow rate is just this column divided by that column we get a new column again and this determines then the removal ratio of that particle size so what does that mean this column means take for instance column number or row number 3 you see here a value of 0.87 appearing for a size of about 100 microns 0.11 millimeter that means that from this fraction so these particles about 80 percent of the particles will settle in the hopper while for the smaller particles the 50 micron particles only 19 percent of 20 percent are settling and from the very the smallest size that was the 60 microns only, well, 007 the fraction will settle this 0.7 percent will settle and you see values of 1 in all other rows why is that, why is there a 1 it always settles it always settles of course this value removal ratio can never be 41 because it cannot be 41 400 percent settling of the number of particles flowing in so therefore we have to reduce that value to 1 no, that means that from everything above around 100 microns everything will settle that means that this part 10 percent of this size will settle et cetera et cetera et cetera and we can just sum up the total removal ratio of the whole particle size distribution and we see that 80 percent of the total inflowing particle size distribution will settle in the hopper well, the overflow loss is 1 minus the removal ratio per definition that means that we have about 20 percent of overflow loss and this is an example where we don't have we didn't take into account the hindered settling effect because we are basing the equation not on the WS value and the settling velocity of more than one particle but only on one particle so the next graph I think I have in it, yeah we see that we assume a inflow concentration of 0.17 the v0 is of course the same value that's just the q over b times the width but now you see that the settling velocity is now calculated including the hindered settling effect with the formulas of yesterday with the 1 minus c to the power n, n is a function of the particle Reynolds number you see that the settling velocities are a little bit lower compared with the settling velocities of the previous slide because of the hindered settling effect the same well the same we just take the ratio again between the settling velocity and the v0 only now we take the hindered settling into account so we get vS over v0 and well we see that the removal ratio can again be derived we reduce or we limit the value to 1 in this column larger than 1 I just put the 1 we sum up the whole removal ratio again and we find out that the removal ratio is a little bit less we see now that the overflow loss is increasing to almost 30% due to hindered settlement so there is a very simple application that is in fact how the camp model works al taken into account the hindered erosion effect or turbulence is the most simple application but just to show you how it works and if you do it like that you'll see that the actual situation will not be far off from what you calculate over here so it's a very good approximation well you can effect the same procedure you can using this using this method you can calculate which part of the inflowing particle size distribution will settle in the hopper and at the same time you can calculate which part of the fraction is moving out of the hopper and that's why by this method you can then calculate the particle size distributions in the hopper and in the overflow when you for instance look at this graph you'll see that this is the particle size distribution on the sea bed and what we are sucking up pumping into the hopper you'll see this is the particle size distribution which is actually in the hopper itself which is settled and this is the particle size distribution that you will see in the overflow and you see very clearly that the particle size distribution in the hopper is coarser than what we are sucking in because we more or less are getting rid of the fine particles and that's why hopper is in fact a very nice instrument to improve the particle size distribution because you see that more or less all the fine material that we have over here is gone especially this amount of material you don't like on the reclamation site the fines so the hopper is more or less acting as a very nice big sieve getting rid of all the fines so that you can get the coarser material on site of course there are limitations you have to know where you can leave all the fines in the borrow area of course you have to look at the environmental aspects that's why what's one of the reasons why hoppers are so efficient for reclamation purposes ok well something about optimal loading time but I'll continue with that on the next presentation or the next course will be in the beginning of January so the remaining course will be on this a little bit on the hopper sedimentation process and then the other topic that I'd like to talk about you next year the teaching process so final one so that's it for this year and well have a nice