In the previous lecture, we were discussing about the fundamentals of fully developed flow and developing flow in channels and let us recapitulate what we studied a little bit before proceeding further. So we considered fully developed flow in a parallel plate channel and considering that we started with the continuity equation and the momentum equations to describe the velocity profile, the fully developed velocity profile which is parabolic in nature. Then we calculated the dimensionless wall shear stress and dimensionless pressure drop and because the wall shear stress and pressure drop are related, their dimensionless forms are also related. The dimensionless wall shear stress is the Fanning's friction factor Cf and the dimensionless pressure drop is the Darcy's friction factor f and we derived that f is equal to 4Cf. This was what we discussed in the previous lecture and we will try to extend the discussions of the previous lecture for flow through circular channels, right. In microfluidics, rectangular channels are more common than circular channels but circular channels are also important in the context of micro needles, I mean which may have circular section. So it is important to understand the fundamental mathematical paradigm by which we describe the circular flow through circular tubes, the fully developed flow through circular tubes which is given by the Hagen-Poisouli law. So to understand that, let us appeal to this slide which is there. So in this slide you can see that we have drawn a cylindrical tube, I mean it is better not to call it a pipe because we are considering small dimension. So we may call it a tube, although like geometry wise there is no great difference, tube of radius r and we have Vr, Vtheta and Vz as the 3 components of the velocity. So here we are using the cylindrical polar coordinate system. The flow is steady so that del Vz del t equal to 0. The flow is fully developed so del Vz del z is equal to 0, it is just like del u del x equal to 0 for a rectangular channel, here the axial direction is the z direction, so the del Vz del z equal to 0 and because of axial symmetry there is no variation with respect to theta. So if you consider all these, then you will see that if you write the momentum balance, the z momentum equation, then the left hand side becomes 0, just like the left hand side in the x momentum equation becomes 0 for flow through a rectangular channel. Similar thing the left hand side which is the acceleration term becomes 0, then right hand side there is a balance between the pressure gradient term and the viscous term. Now the r momentum equation, so far we have discussed about the z momentum equation. In the r momentum equation you have basically all the terms because Vr is equal to 0, all terms are 0 that means del p del r which is the only term remaining that term is also 0. So because del p del r is 0 that means p is a function of z only, p is not a function of r. So that boils down to this equation, I am just going through it, I mean in a very straight forward manner because the same thing just in a different coordinate system we have discussed in details. So I just do not want to spend unnecessary time in describing the same physics for another case where only change is the description of the coordinate system but physically it is the same. So you can see that the z momentum equation the pressure is a function of z only and Vz is a function of r only. The only difference between the rectangular channel or the parallel plate channel and the circular pipe is the fact that the viscous term you express instead of d2 Vz dr2 it is 1 by r d dr r d Vz dr that is just a representation in the cylindrical coordinate system. So this is because Vz is a function of r, this is a function of r only, p is a function of z, so this is a function of z only. So function of r is equal to a function of z that may be possible only if it is a constant. So we take it up from here and go to the board to address this problem. So 1 by r d dr of r d Vz dr is equal to 1 by mu dp dz. What we do? So then each is equal to say c. So we can write d of r d Vz dr is equal to cr dr. We can now integrate it. So if we integrate it now r d Vz dr is equal to cr square by 2 plus c1. So you can see that d Vz dr is equal to cr by 2 plus c1 by r. Now if you consider the channel or the tube the center line is r equal to 0. So at r equal to 0 you must have d Vz dr finite. But this 1 by r dependence is killing that finiteness until and unless c1 is equal to 0. So d Vz dr must be finite at r tends to 0 which implies c1 equal to 0. So d Vz dr so Vz is equal to cr square by 4 plus c2. So we will apply the other boundary condition that at small r equal to capital R what is Vz? Vz must be equal to 0. So that means c2 is equal to minus cr square by 4. So that means you have Vz is equal to c by 4 into small r square minus capital R. So we can now express Vz in terms of V average. So what is V average? Integral Vz into 2 pi r dr 0 to r divided by pi r square right integral Vz df by the area. So c by 4 into 2 pi into r cube minus r square r dr 0 to r by pi r square. So this becomes c pi by 2 by pi r square into r4 by 4 minus r4 by 2. So this is minus cr square by 8 alright. So that means you can substitute that expression for V average in the expression for Vz and once you substitute that you get Vz by V average is equal to 2 into 1 minus small r square by capital R square. So this is the velocity profile. Again the parabolic velocity profile is evident from this analysis. What we will do, we will find out either the dimensionless wall shear stress or the dimensionless pressure drop. Let us find out the dimensionless pressure drop. So you can write V average is equal to minus cr square by 8 and c is equal to what? 1 by mu dp dz that is 1 by mu delta p by l. Now so you can write in place of delta p we are basically working out the same steps that we worked out in the previous class just the coordinate description is a little bit different the geometry is a little bit different. So this is minus hf rho g where hf is the head loss due to friction. So V average is equal to hf rho g by mu l into r square by 8. So hf is equal to 8 mu vl by rho g r square and you can write it in terms of the flow rate by writing V average is equal to q by pi r square. So 8 mu ql by rho g pi r to the power 4. Many times in engineering convention we express the result in terms of the diameter of the pipe rather than the radius. So if you substitute r equal to d by 2 then we get hf is equal to 128 mu ql by rho g pi d to the power 4. You can find out the friction factor f from here the Darcy's friction factor just in the same manner in which we calculated in the previous class for the rectangular channel and you will get f equal to 64 by Reynolds number. So what we have done we have summarized the description for fully developed flow for parallel plate channel and for circular geometry. So let us summarize the important findings as we get back to the slides. So far we have discussed about the fully developed flow and fully developed laminar flow. So this diagram I have purposefully put in this context. I mean you can recall that this is nothing but the Moody's diagram what you see here and we will try to understand that which regime is important for us in microfluidic applications. In microfluidic applications we are so in the x axis there is Reynolds number. In the y axis on the left side there is friction factor either f or cf whatever you want to plot they are related and the right hand side it is either the roughness by the diameter or the diameter by roughness in this particular case it is diameter by roughness that is plotted. So either of the diameter by roughness or the roughness by diameter. Then basically we are operating in microfluidics in the low Reynolds number regime and in the low Reynolds number regime you have laminar flow. So this curve cf so this figure is for a circular geometry. So cf f was 64 by re so cf is f by 4 so 16 by re that is this. So in the log plot it is a straight line okay. So this is the regime in which we are operating and you can see that here the friction factor appears to be independent of the surface roughness. As you go to the high Reynolds number regime you can see that for different values of the relative roughness you are getting different values of the friction factor. So the friction factor is depending on the roughness but for fully developed laminar flow the roughness is apparently having no role to play because all the perturbations induced to the flow because of roughness are attenuated. Perturbations do not amplify in a laminar flow. So that makes it independent of the roughness but still the analysis does not take into account the length scales of the roughness. So the length scales of the roughness themselves can be important as compared to the system length scale when you are talking about microfluidic systems. So it is worth investigating that what is the role of roughness for laminar flow through I mean micro channels of whatever geometry that is the first remark that we have put. The second remark is that you see hf is proportional to muql by rho g pi dh to the power 4 hydraulic diameter to the power 4 in this case the hydraulic diameter is the diameter itself. So that means that the head loss for a given flow rate is inversely proportional to the fourth power of the diameter. So if you want to have a particular flow rate then if you reduce the diameter from 1 meter to 1 micron see 10 to the power 6 to the power 4 that is 10 to the power 24 times the head loss will increase. So you will require tremendous pumping power to drive the flow. So you cannot really achieve that high a flow rate in a pressure driven microfluidic system which you can achieve in a macroscopic system you have to compromise with the q because to achieve the same q you require to have a huge driving pumping power therefore you compromise on q but sometimes it is a blessing in disguise not in all of applications you require a large q. For example you are administering a medicine or a drug to a patient in a very minute and precision quantity. So there you do not require a high flow rate there in fact a precise delivery of the drug is what is important. So high flow rate may appear to be an important and useful thing but it depends on the context where other factors may be more important than high flow rate. So these are the 2 important observations that we have made from our analysis so far. Now whatever we have discussed so far is something which we normally covered in the undergraduate first level course of fluid mechanics but we want to go beyond that in this particular course. So now what we will see is that like we may still consider steady flow but instead of a parallel plate channel it can be a channel where the width is not infinitely large as compared to the depth. So that means the width and depth are comparable so how do you get a solution for such a problem. So we will try to address that I would partially work out this problem and then leave the remaining part of the problem as a homework to you. So we will work it out partially. So we will first make a schematic to understand that what we are talking about. So let us say that you have a channel of cross section like this. Let us say this is y axis, this is z axis and perpendicular to the plane of the board is the x axis and the pressure gradient is applied along the x direction. So there is a dp dx. Let us say that this is a by 2, this is a by 2, this is b by 2, this is b by 2 and we are interested to find out the velocity, the wall shear stress, the dimensionless wall shear stress or the dimensionless pressure drop, the standard parameters which we are interested. Now let us write the governing equation first before going for a solution and to write the governing equation I mean we can do a similar analysis like the parallel plate channel. Still we are assuming that it is fully developed. Let us write the x momentum equation. Let us write the x momentum equation because it is fully developed. The left hand side is 0, steady and fully developed. The right hand side, so let us write the viscous term. Now there will be 2 terms, one for variation along y, another for variation along z. For the parallel plate channel case this term was not there. But for the channel where the width and I mean these dimensions height and width are comparable these 2 terms may be of the same order. So they have to be each has to be written – dp dx. So clearly this does not have such a simple solution as that of the parallel plate channel case. The parallel plate channel case for a fully developed flow reduced to a ODE but here it still is a PDE. Now when this problem first came to engineers, see engineers traditionally have always tried to come up with a working solution and the working solution may be good or bad but it is some solution and sometimes it is important to get a solution rather than getting a very complex solution or no solution. So with that spirit, we will see that how to get an approximate solution to this problem. How does the approximate solution compare with the exact solution? We will work out the exact solution for this problem also and then I will give you a homework that you plot the approximate solution, you plot the exact solution and compare the error between the 2 solutions for different ratios of A by B. Possibly for the parallel plate channel case, you will find that that error is not very high but as the aspect ratio will become close to unity that is A by A is to B will be close to 1 is to 1 that error may be more and more significant. We will see why? This is not, this is x momentum equation, left hand side is, left hand side is the acceleration term, it is fully developed flow, del u del t plus u del u del x plus v del u del y plus w del u del z. What is del u del z? What is w del u del z? What is w? The flow is taking place in the yz plane, right. So w, so this is basically, so you have the flow is in the yz plane means what? The variation of the velocity profile is because of variation in z and y. So if you consider that you will see, I mean you have only one velocity profile, you have only one velocity component that is u and the other velocity components will be trivially 0 by applying boundary conditions by using the continuity equation, we have not worked out that details. So v and w are 0, okay. So in conjunction with that if you apply the fully developed flow consideration the left hand side will be 0, okay. So basically the flow variation is in the yz plane, the axial component of velocity varies in the transverse plane, that is what is the physical scenario, that is the physical concept that we can borrow from the, see why we start with the parallel plate channel? We could have started with this only, the parallel plate channel case gives us a lot of physical insight and we are just borrowing that physical insight directly in this example. Now you will see that if you write y momentum and z momentum equation you will have del p del y equal to 0 and del p del z equal to 0 because those velocity components themselves are 0. So that means p is not a function of y and z, so p is a function of x only, that is what I am saying that it is important, see if you want me to spend more time on deriving this equation by starting from the scratch I can do that. But at this level of maturity I expect that once I have covered the parallel plate channel you should be able to follow how this equation comes, it is just an additional dimensionality that has been added but physically the concept is very similar, okay. So again to repeat where from it follows that del p del x equal to dp dx if you refer to the y momentum and z momentum equations those velocity components are 0 that will give rise to del p del y equal to 0 and del p del z equal to 0 that means p is not a function of y and z. So p is a function of x only that means the del p del x will become dp dx. Now let us come back to the approximate solution, see let us say that I am a person who is not so conversant in advanced mathematics, still I want to solve this problem and want to design an engineering system out of that, so how can I get rid of that, see one of the clues that I can get is that possibly the velocity profile had it been a parallel plate channel would have varied quadratically with y that is what is the solution that we have seen for a parallel plate channel, had it been a parallel plate channel with variation along z that would have also quadratically varied with z. Now there is a combined variation with respect to y and z, so you really do not expect that from a fundamental mathematical perspective it will vary simultaneously quadratically with y and z but as an approximate solution we can consider that that is not from a very rigorous theoretical basis but from an approximate basis we can take a solution of the form assume u is equal to c into y square minus s square by 4 into z square minus b square by 4, so what is the motivation of taking this form, the motivation of taking this form is the quadratic variation of velocity with respect to y in the parallel plate case, had it been a parallel plate channel with variation along z it would have been this. Now this form we are taking quadratic but why this form, this form we are taking to satisfy the no slip boundary condition at y equal to plus minus a by 2 and z is equal to plus minus b by 2, see always keep one thing in mind many times you do CFD solution many times you do approximate analytical solution, many times you do detailed rigorous analytical solution whatever you do the first and foremost thing that we always check as a check to the solution that the solution satisfies your boundary conditions, even wherever we do a CFD problem we see that whatever contours or vectors whatever we are getting those are satisfying the boundary conditions that is the first check that we always make, so here the motivation of taking this form is 2 fold one is to keep a form that we get for the parallel plate channel, so because this form we get for the parallel plate channel possibly if this channel was such that it is a parallel plate channel that is either of a by b or b by a very large or very small then it would be quite accurate, the second motivation is that this will automatically satisfy the boundary conditions at the 4 boundaries, so once we do that this is heuristic remember this is not correct but this is something which is approximate and we will see that under certain limiting conditions it is not that bad, so the velocity profile may be specified completely if you know what is the value of c, so the question is how do we find out the value of c, no, no, no this is not imaginary wall this is the channel is like this, so these are walls these are these are all walls this is not I mean these are not imaginary walls this is a rectangular channel, problem is these are certain sort of mental blocks that we always have like from the second year undergraduate level we have understood that channel means parallel plate channel, so when we are putting a boundary I would say like if you tell to a person who has not learnt fluid mechanics this is something which is more obvious than a parallel plate channel, why should I suddenly assume a parallel plate channel I should better assume that it is bounded by the walls rather than infinite in one direction, so just do not keep any mental block, assume I mean you have all the boundaries not infinitely large in one direction, so this is what is u now to get c what we will do is we will use the integral form of this equation, so what we will do is we will substitute mu okay, so first let us work out the derivatives del u del y is equal to 2cy into z square-b square by 4 this is del 2u del y2, why we are calculating this we will require that to be substituted in the momentum equation, so you will see a very interesting thing let us complete this del 2u del z2 what will be that 2c into y square-s square by 4, if you add these 2 right, so that will become 2c into z square-y square-s square by 4-b square by 4 right hand side will be equal to dp dx and there is a mu adjusting factor, now there is no guarantee that these 2 will be equal right, so we can clearly see that this equation although or this expression although satisfies the boundary condition does not satisfy the governing equation right, so it is not a correct solution, it may be an approximate solution but it is not a correct solution but you may accept a level of approximation by accepting that it will not satisfy the governing equation in its differential form but maybe it will satisfy the governing equation when it is integrated over the cross section, so that is what we will do now, so what we will do is we will just write it a little bit like this 0 is equal to this as 1 by mu dp dx just a little bit readjustment, so this will be 2c into z square sorry y square-z square-s square by 4-b square by 4 is equal to 1 by mu dp dx. Let us integrate this over the cross section of the channel, so dy dz y is equal to what? y is equal to-a by 2 to a by 2 z is equal to-b by 2 to b by 2 okay, this also we integrate dy dz, we can safely keep dp dx out of the integral because p or dp dx varies with x only, it does not vary with y and z, I am not going to enter into the trivial algebra and waste time in this, you can yourself complete this to get the value of c, this is just a simple polynomial integration and at this level it is very straightforward, so once you calculate this it will be c into some number a right, the number a depends on the integral of this y square-z square all these things is equal to 1 by mu dp dx into ab, so c is equal to 1 by mu ab by adp dx, so that completes the description of the approximate velocity profile. Once you get the approximate velocity profile, you can find out the wall shear stress, how do you find out the wall shear stress, so wall shear stress depends on question now is which wall right, so you can calculate the wall shear stress here or I mean here and here it will be equivalent or here and here, when we are talking about parallel blade channel, when we say wall we know that right just one either of the parallel walls, now there are 4 walls, so either of this or of this and then you can non-dimensionalize that to get a friction coefficient, I am not working out that because that is just trivial algebra, now that solution if you work out, you will see that it will satisfy the boundary condition without satisfying the equation, so there is a level of approximation associated with it, question is what is the level of approximation, how accurate it is, is it good for some cases, is it okay for some cases or is it poor for all cases, to assess that we have to go for the exact analytical solution for this problem which we will attempt now, so this is the equation that we are solving, when you first look into a partial differential equation which is the linear partial differential equation, the non-linear terms have been dropped because of the fully developed flow condition, what is the technique which we are always tempted to use, that is method of separation of variables, what prohibits us to apply the method of separation of variables here directly, that is the inhomogeneity of the equation, the right hand side is not 0, right, so had we have the case when the right hand side is 0, it would have been easy and straight forward for us to apply the method of separation of variables, to convert this equation to that form, what we do is that we separate this equation into a collection of 2 problems, what are the 2 problems, problem 1 is a parallel plate channel problem, what is the parallel plate channel problem and maybe let us consider because of symmetry only one fourth of the domain, u1 is a solution which is a parallel plate channel solution, what are the boundary conditions at y is equal to 0, du1 dy equal to 0, at y is equal to a by 2 u1 equal to 0, okay, this is problem 1, problem 1 is the parallel plate solution, see this is one important philosophical way of looking into education, we always try to build up from what we know to solve something what we do not know, so here we see that we have already worked out a problem of parallel plate channel, we are trying to build up a solution of this more complicated problem by starting with that as a basis, so that is problem 1, but there is a problem 2 which is this, problem 1 already takes care of the pressure gradient, so problem 2 does not have to bother about the pressure gradient, so this makes the governing equation at least homogeneous and this is a very standard partial differential equation in mathematics or mathematical physics, all of you know that this is called as the Laplace equation, okay, so Laplace equation in 2 dimensions, what are the boundary conditions at y is equal to 0 del u2 del y equal to 0, right, let me use this side because otherwise the board will not be visible, at y equal to 0 del u2 del y is equal to 0, then at y is equal to a by 2 u2 is equal to 0, right, y at y equal to a by 2 u2 equal to 0, see at y equal to a by 2 u has to be 0 and we want to use u as a superposition of u1 u2, this superposition we can use because of the linearity of the governing differential equation, because the governing differential equation is linear, when u equal to u1 is a solution and u equal to u2 is a solution, u equal to u1 plus u2 is also a solution, so that we can use, so u1 plus u2 is the solution which is 0 because we have already taken u1 equal to 0, u2 also has to be taken as 0, so that u1 plus u2 equal to 0, at z is equal to 0, what is the boundary condition, del u2 del z equal to 0, this is just because of symmetry, centre line symmetry, at z is equal to b by 2, what is u2, what is this, u2 is not 0, the reason is that at z is equal to b by 2, here we have no control over u1, so u1 is not 0 and u1 is a function of y, u1 as a function of y is solved from this problem, so we can say that u2 is equal to minus u1, right, u2 is equal to minus u1 which is a function of y, how are these problems coupled, this u1 as a function of y you will get from problem 1, okay and u is equal to u1 plus u2 is your solution, problem 1 we have already worked out, so I will not waste any time by working it out again, we will see how to work out the problem 2 and then the superposition you can get the solution, we see we can start with the solution in the form of method of separation of the variables because the governing equation is homogeneous, not only that 3 out of the 4 boundary conditions they are also homogeneous, does not matter one of the remaining one is inhomogeneous, still it is okay, so separation of variables u2 we can take as fy into gz, so del 2 u2 del y2 is equal to f double dash del 2 u2 del z2 is equal to f double dash del 2 u2 del z2 is equal to f double dash f g double dash okay, so this is the governing equation that needs to be satisfied, so if you want to satisfy this governing equation f double dash g plus f g double dash is equal to 0, so you have f double dash by f is equal to minus g double dash by g, this is a function of y only, this is a function of z only that means each is equal to a constant, question is and a very important question whether it is a positive constant or a negative constant okay, so we have to figure that out whether it is a positive constant or a negative constant, now here we can show and this I leave on you as an exercise that this is indeed a negative constant, you can arrive at this conclusion in 2 ways, one way is that you use a method of contradiction, you say that no it is a positive constant plus lambda square, you will see that that form will not be able to satisfy all the boundary conditions, otherwise from an intuitive view point if you see if you take this as minus lambda square then g will be coming in the form of sin and cos right or not, f will come in the form of sin and cos and g will be coming in the form of exponential and if you take it the other way then here you can see that f is coming in the form of exponential and g in the form of sin and cos, so in one case it is f in the form of exponential and g in the form of sin and cos, in one case g in the form of exponential and f in the form of sin and cos, now you have to figure out that out of these 2 which one is physically acceptable by considering the boundary conditions that u2 is 0 at both these boundaries but u1 is not 0 at both these boundaries or u2 is not 0 at sorry u2 is 0 at these 2 boundaries but not 0 at these 2 boundaries I am sorry I just may try to mention the other boundaries, so u2 is 0 at these 2 boundaries but it u2 is not 0 at both of these boundaries, so that gives the source of thought which will dictate that which one will be exponential and which will be sin and cos but even if you are not very confident with that and you start with the plus lambda square it will make no difference except that you will come up with a solution that will not satisfy the boundary condition. We will take up this issue in the next class for the time being thank you very much.


two days of I know the last two days you had a lot of information exchange from the experts and today we have we are going to talk about the implementation of a nuclear power program and a long-term commitment that's the overall objective of today and we have two experts from IAEA my colleagues from IAEA or here Ms. Amparo she's working for the the spend fuel management in the nuclear fuel cycle and material section she's an expert on the base management and the nuclear fuel cycle areas and she she would be giving you two presentations on these two topics and she has a doctorate in the chemistry and she also worked in the Spanish nuclear laboratory for over two decades and has a lot of experience in the waste characterization and other things maybe she would introduce herself in more detail as she goes through her presentation also today's chair who is going to chair the the entire session Mr. Sean Dunlop she works for the nuclear infrastructure development section she is I mean he is from the United States and he works for the IAEA and he has previously worked in the US Department of Energy and he does a lot of coordination with the IAEA member state countries particularly on embarking countries to develop nuclear power infrastructure in their countries so you will be having very interesting sessions with Sean in the afternoon the morning sessions will be from Amparo so I will hand over to Sean to take over these today's proceedings Sean please thanks good morning everybody I look forward to getting to know you a little bit better over the next couple days and without further ado I think I'll hand things over to Amparo. So good morning I'm Amparo Guanzales Espartero I'm the technical lead of spent fuel management in the nuclear fuel second material section in the INC and I joined the INC in 2015 and I was working previously for 24 years in the Spanish nuclear lab national lab who which supports and Resa who is the organization in charge of the radioactive waste management in Spain and also our regulatory body so as should I okay so as Ashok said I was asked to give a presentation another view of the nuclear fuel cycle and then second lecture I'm going to give is about radioactive waste management and the commissioning so my this presentation I will try to give you an overview of the main stages of the nuclear fuel cycle and we can talk about since we are going to talk about radioactive waste after the related waste produced in every stage so I think that we can give we can let the questions for the the end of the presentation so because sometimes I'm going to address some issues in the course of the lecture so the nuclear fuel cycle are the main is the industrial stages involved in the nuclear electricity and electricity production from nuclear power and it starts by the exploration of uranium mining and milling and ends with the eventual disposal of the waste produced the nuclear fuel cycle terminology there are the stages covering mining milling the conversion of the uranium to put it in a right form to be used as fuel the enrichment for different types of reactors of uranium and the fuel fabrication are called the front end of the fuel cycle and then after the electricity production the fuel has to be discharged from the core after more less three years in the core of the reactor and has to be a search or repossessed to recycling uranium and plutonium but producing and high level waste stream that has to be managed at the as the same with the same criteria as the spent fuel and has to be disposed in the deep geological repository and these stages are called the back end of the field cycle there are different options for the fuel cycles and so far there are two options commercially available one is the fuel cycle which consider the spent fuel as a waste so after a period of storage has to be disposed in the geological repository and the second option is the so-called closed fuel cycle or one through cycle and is the mono recycling of plutonium so so far the spent fuel after a period of storage is reprocessed to recover uranium and plutonium producing mox fuel that it's burned in reactors but only one time so once is the charge from the reactor is stored the mox fuel is stored for further use and so far is stored to be used in fast reactors so France is the main country producing irradiated mox fuel and in this case is produced a high level waste stream that has its vitrified are vitrified and then has to be stored also in the deep geological repository so there are two main fuel cycle options so if we start step-by-step let's talk by mining and milling so it depends on where the ore body is there are three options for to get uranium from the earth one is the open pit underground or in sea to leech and also there is another option that consists to recover uranium from natural rocks as the fertilizers and it's called by-product recovery so when the ore body is close to the surface the open pit is the mining option and in this case it's produced a huge amount or barb and rock and because the pit has to be big enough more bigger than the ore because it's mine in a slope to prevent collapse so as you can see there is a lot of surface disturbance in this type of mining so the rock produced the barb and rock has to be put underwater to prevent the natural radiation because you know the uranium is everywhere and the radon the emissions of radon are very common and to prevent these emissions the barb and rock it's put underwater and these tails are used after the mining lifetime and at the time of the commissioning and this is the example of one of Spanish mine in Andujar so it was during the remediation and the commissioning so the barb and rock was used to fill the pit and then this is the final situation of the mining and it was remediated the surface the second option is the underground and it's used when the ore body is underground and deep under the surface of the earth so in this case it's needed to build tunnels and the disturbance in the super in the surface is much less so much a smaller waste rock is produced with a less environmental impact and in this case ventilation is required for workers to be protected against the airborne radiation and the third option is in situ leech so in this case the ore body remains where it is so this is the picture of the surface and so the ore body has to be permeable to solution because the uranium is dissolved use leech using acid or alkaline solutions and then the liquid is pumped up to be recovered in this case this leech solution the uranium is recovered from the leech solution by solving solving destruction or ion exchange and and then is purified and obtain the uranium oxide so in this case the most important thing is to prevent not contaminate the ground water away from the ore body situation and if it's done well has less environmental impact. Uranium also can be recovered from phosphate rocks and so far 20,000 tons of uranium has already been obtained from phosphate rocks in this case there are two benefits one in one hand we recover uranium but in the other hand we reduce the amount of normed waste in the in the phosphate productions I don't know if you heard about it but when you when phosphates are produced it creates a stockpile of natural its norm is natural occurrence radioactive material and it's a big problem because they are natural this uranium is in the air but it has to be controlled under the radioactive requirements so if we recover uranium from these deposits it's a benefit from these areas that produce phosphates for fertilizers this is the map showing the distribution of resources so far of uranium so there are 30 countries that represent approximately the 96% of the total uranium resources and the main sources are in Australia and second in Kazakhstan and in the Russian Federation and Canada so once uranium is explored the sources and also mine close to every mine there is a milling where the rock obtained is crushed and ground and uranium is leaching using acid solution because the content of uranium in the rock is very low it's around 0.1 percent so once we get the solution uranium is obtained well first of all we have to separate the carbon rock from the liquid by filtration and after the solution is purified and concentrate getting U308 that is the way that uranium is packed and dispatched and this is the so-called yellow cake so the yellow cake can be yellow it's yellow but when it's casting it's very dark so this is the way that uranium is dispatched and contains the 80 contains around 85% of uranium and is packed in 220 liters drums and it ships to the conversion plant why because in this form uranium is not suitable to be used in the reactors so in the conversion plant the U308 is converted to uranium dioxide and in this case can be used by the reactors that use natural uranium that's they can do reactors of the heavy water pressurized reactors and in the case of those reactors like what the reactors we need to use uranium enriched uranium yellow cake is transformed in exaphrurite uranium which is the only gaseous compound of uranium because so far all enrichment process work in gaseous phase so let's go to the enrichment step so why we have to reach because the natural uranium 5 concentration in uranium the isotope composition is 0.71% of uranium 5 and most of the reactors like what the reactors work with 4% 5% of uranium 5 so the concentration of uranium 5 in the uranium has to be increased and how we can do that because it's very difficult to work with isotopes of the same element they have the same chemical properties so there are so far some several enrichment processors demonstrated but only two are commercially available both as close to different of mass between the isotope 5 and 8 and there are large commercial enrichment plants in countries such as France, Germany, Netherlands, UK, United States and Russia Federation and China is so it's now it's currently increasing their capacity as their own demand is increasing so during the enrichment we will see that we will get two streams one stream is the so-called low and rich uranium which will be four or five percent enriched and the other streams is called the plated uranium or tails so it's the uranium stream containing less than 0.7% of uranium 5 so the two options is the gaseous diffusion it was the first method used and in this case the exafluoride of uranium is forced to go through membrane to a high pressure and then physically the lighter atoms pass through the pores of the membrane meanwhile the heaviest one remain in the lower part so this process is repeated repeated many many times so to get around 4% of uranium 5 it has to be repeated around 1400 times so this process is very very energetic consumes a lot of energy and the energy that it's consumed during the process is measured in these separate working units we will see what does it mean and it consumes around 2,400 kilowatts per hour per zoom so they are all facilities dedicated to gaseous diffusion are going to be dismantling and stop when they reach their lifetime the lifetime and now the more used method is the centrifugal process that consists in tubes vacuum tubes with a rotor that speeds a very high speeds around 50,000 to 70,000 rpm and in this case the heaviest atoms are moved to the surface to the walls of the cylinder and the lighter remains in the middle so repeating the process several times we get the separation and the constant day of the to increase the concentration of uranium 5 in the final product and in this case you can see this process consumes much less energy so 50 kilowatts per hour per zoom so now is the the most used process and has to be done in cascade that means that there are many many tubes in Sirius and then if we repeat the process several times we get at the end the rich uranium and the depleted stream so what means soon the separative work unit is a very complex unit that indicates the energy input that it's need relative to the amount of uranium processed the degree of enrichment and the degree of depletion in the tails and can be referred to a kilogram soon or simply soon and for example if this is the evolution of soon depending on the enrichment and it depends also on the depletion of the tails and and the enrichment so we can play if it works for example if we refer okay so we can refer to one ton of product and we can put here how much we would like so if we as the fetus I is the natural uranium and we would like to get 4% for example of enrichment and we would like to have a tail of 0.3 or let's see 0.2 of the plate of uranium 5 in the depleted stream if we calculate then we get four thousand four thousand four hundred twenty-four soon but per kilograms of per ton of uranium enriched so we have a ton of uranium at what said at 4% with a tail of uranium depleted uranium we said 0.4 but for the same tone of uranium to 35 4% and we need to feed the system with seven tons of natural uranium 7.5 but what's happened if we would like to have a more depleted tail and if we keep the 4% we keep everything and then we will consume more energy so six kilo soon for the same uranium ton with a more depleted tail so we will consume 6.38 tons of uranium so we will need less feed less amount of uranium not natural uranium but we consume more energy so we can play during the enrichment with the tails and with the energy consume consumption so let me check if I can go back so once we have the uranium enriched at the amount necessary to be used in the light water reactors we go to the fuel fabrication and the fuel fabrication is a very very specific process because you know the first barrier is the pellet itself so once we have uranium natural uranium or enriched uranium we have to make the pellets by the ceramic pellets by a process called sintering at high temperature around 900 degrees and this process has to meet a very tight quality assurance specific specifics so the size of pellets is one centimeter per more or less one centimeter and these pellets are placed in tubes that call rods that could be a stainless steel or circalloy this is an alloy made by circonium and the rods with containing the pellets has to be placed in a very robust structures that are caused bundles or fuel assemblies and these structures has to be physical very robust to be able to support the high operating temperatures in the core of the reactor and also the intense flux of neutron radiation so these structures this is a fuel assemblies has to be resistant against the chemical corrosion because the water is in the core of the reactor the high temperatures and also the static loads and the vibrations and the fluid and mechanical mechanical impacts not only in the core of the reactor but also during the storage and all the handling steps that the fuel assemblies has to undergo there are different types of fuel assemblies depending on the type of the reactor so the can do reactors of the pressurized heavy-water reactors works with these 36 rods assemblies in circaloy and the PWR is the biggest fuel assemblies with 17 but 70 versus 17 pins in circaloy the BWR 9 plus 9 is the smallest one and also the Russian fuel assemblies the Russian version of the PWR is the BBR is used 3312 rods in circaloy and the advanced gas reactors use the pins and the pins are covered by graphite and we will see when we will talk about radioactive waste and the commissioning that this graphite is an issue today because the irradiated graphite is very difficult to manage and we will see what's happening during the commissioning and also this is the assembly the fuel assembly for the fast reactor the sodium fast reactors sorry the number of rods yes this square is four meters tall and it's I don't know what is the size that's the 17 17 rods and 17 so once there we have the fuel assemblies placed in the core and the electricity production in the oh sorry in the reactor uranium 5 is fission and also uranium 8 is a fertile material so uranium 8 it doesn't go under fission but can capture a neutron and it became to be plutonium 39 and plutonium 39 is fissionable so plutonium 39 also contribute to the energy production in one third so once the fuel it's in the core stays in the core for around three years more or less it becomes to be it doesn't meet the safety requirements so far so because the neutron it became to be more absorbed neutrons and also the metallic parts can be brittle so it's decided to consider the spent fuel the fuel is spent after three years in the core and the composition of the fuel has changed it from 95% of uranium 8 to around 92 and from 5% of uranium 5 to 1% of uranium 5 remaining and also we have now a new element that is plutonium in a concentration about 1% we have fission products and around a compensation of 5% and also the so-called minor actinides that are the actinides that are not uranium and plutonium so neptunium amyretium curium californium beccarellium and they are called minor actinides because they are in a very lower concentration compared to uranium and plutonium so once the spent fuel is considered the fuel is considered spent it should be it has to be stored underwater due to its characteristics because it's very the heat production but it can heat and because the water serves as biological shielding so the pools are located in the reactor building and the fuel assemblies are placed in racks that has materials that absorb neutron to avoid criticality so depending on the strategy to manage to spend fuel the spent fuel can be stored in the pools for from the cats or months and the water not only provides shielding but also allows visual inspection of the fuel assemblies and the chemistry of the water has to be strictly controlled so for example the routine controls are to the measure of the pH the bottom levels because the born is used by new as a neutron absorber the conductivity because the conductivity give us the idea of ions present in the water and also the gross alpha beta and gamma activity because it indicates if there is release of productivity from the fuel assembly that means that there are pinholes in the roads and give an idea so a leakage of radioactivity so there are different options and techniques to store the spent fuel and goes from wet at reactor or away from reactor and also dry options once the spent fuel spent time in water we can there are options to move the spent fuel to dry storage so let's see these are the options to store spent fuel so it could be at reactor storage for a long period of time and then go to the final disposition or can be in a reactor storage facility in pool and then the interim storage usually in dry could be on site and then to go to the final disposition there are other options that after a period at reactor if they spent fuel it moves to an interim storage away from reactor sometimes hundred of kilometers far away before the final disposition or could be a store at reactor and then to be moved to an interim storage on site and then to be moved to an interim storage off site and then go to final disposition so these are the different options so far for to a storage spent fuel oh there is no definition of interim storage so far is considered 60 years around 60 and now it's considered longer because only few countries have facing in a short term the final disposition so many countries needs a period of around 100 years to have a final disposition ready so interim storage now the facilities are licensed for 60 years more or less so on site you mean on site or away on site on site it depends on first of the lifetime in principle is the lifetime of the nuclear power because it's on site is in the pool of the reactor and it's around 40 years and now is the licensed period of the lifetime of the nuclear power and we will see some examples of nuclear powers in the United States for example that after the nuclear power plant has been shut down then the pools has to be empt and the spent fuel was moved to an interim storage on site dry storage and we will see a picture of a power plant completely dismantled and the dry storage remain on site so it happened now in the United States and it's a it's a problem because at the end the country has different interim storage spread around the country so it's an issue in the world we will see at the end of the presentation what is the status so what is the current situation of a storage in the world so every year is around 7000 tons of heavy metal discharged from reactors that goes undergoes to a storage from the 447 nuclear power plants in operation in 30 countries so at the end of 2016 there were 270,000 tons of heavy metal in the storage and the majority are on site at reactors the 80% but what's happened now most of the nuclear power plants are reaching the end of the lifetime and most of the pools are full so the nuclear power plants needs to to find a solution and some countries have decided to start dry storage on site so they are so far 151 away from reactor storage facilities and this name is a little bit confusing because if we if we speak properly at reactor means in the building of the reactor that's what I mean at reactor so the pool is at reactor building and away from reactor also mean on site but not in the same building and also mean away from reactor far away so away from reactor means could be on site and could be outside the containment but could be on site and also a hundred kilometers far away so so the newest facilities that doesn't mean the pool and they're inside the containment building and so now so far we have 151 away from reactor facilities in 27 countries and 80% of these facilities are dry storage and most of them have been deployed for more than 25 years so some member states such as Canada United States have more than 30% of the total spent fuel inventory in dry storage so they are I think you have received a lot of information about safety standards and all the storage facilities and storage activities has to meet the requirements safety requirements and there are two specific safety guides one is safety guide the WSG 6.1 is a storage of radioactive waste and the other is the specific safety guide 15 as its main and it's referred to the storage of a spent nuclear fuel and the main the overall objective of safety is to protect people environment from the harmful effects of the radiation and the storage systems has to meet the safety functions so they have to give appropriate containment of the radioactive material and the first barrier is the pellet the second barrier is the fuel assemblies and then there is the also the canister and the casks to keep the radioactive under control so the system has to be critical safety against criticality so it can never occur that it lets in a configuration that can be critical it has to have effective heat decay heat removal and in the case of pool the water is the coolant and in the case of dry is the natural convention convection and it also has to ensure that those for workers and also for the public remains under the limits and also has to ensure that the variability of the spent fuel to undergo to the next steps of the spent fuel management so we have to talk about to store the spent fuel and the water so the pool storage is a mature technology since it's been used from the beginning of the nuclear power generation and so they accumulated more than 50 years of operating experience and as we said the chemistry of the water has to be strictly controlled to ensure the minimum concentration of retinoglyce in the water to keep the clarity for inspection and to minimize the corrosion of the metal surfaces the roads and the fuel assemblies so the benefits of water is because the water gives is the coolant efficiency and shielding it's biological shielding and the main issues is the water always provoke corrosion so we have to keep the water and their strictly conditions to prevent the corrosion of the metal parts and this is the clap concept in Sweden because they decided to storage the internet storage of a spent fuel and the water and it's away from the reactor and instead of using dry storage they use water storage and sorry of course because the water if you put something in water is corrode so you have to it's day first of all it's day on inside water because the present of different ions can prevent corrosion chlorine for example and the ions we have in the normal water so first the whole pool is full by day on ice water to prevent the corrosion and also you can control it's very if you see under control still water or you have bacteria very easily so you cannot see and it's not clear the water so you cannot see you cannot expect visually what is the status of the fuel assemblies so the bacteria's and also the bacteria's corrodes everything and so in one hand you have to prevent corrosion is the most important thing and you have to also to prevent the levels of boron too for criticality controls for neutral absorbers yeah the fuel assemblies is directed you can see I think it's better seen in the other picture here you can see this is the fuel assemblies only puts it in how you call this in racks so it's not without any containment so this is a rack and the fuel assembly it's put like this directly in the water so that's why you have to prevent all corrosion and also is a way to control the red activity in the water so you cannot have a concentration of red activity in the water first for the workers and also because it means that there is some leakage in the fuel assemblies and you have to be careful so the corrosion rate oh I don't know by heart I can check I don't I don't have this information as I don't mind but they can check if there is any guidelines any limits I will tell you so there are a lot of projects there is yeah I don't know if there is any guidelines or I don't know I have to take so this is the dry storage options so you see we have casks metal casks we have also concrete canisters and bolts so the cucks and the canisters are modular and also can be monolithic and they are circular the casks in the cross section and the heat is the decay heat is removed by convection a natural convection and in the case of canisters they can be vertical or horizontal and the bolts are buildings that are also modular so can be added more and more a rise of cavities and can be above or below ground level and also the canister can be below above or below ground level but in dry storage the main say the main benefit of the dry storage is this is a passive system so that doesn't need any systems active systems so the system is controlled by itself and in the case of the water or the pools now there is a lot of the Fukushima there is a lot of work done and perform and trying to find solutions to make the pools more passive than they were before so the goal is to have the most passive systems for storage so they spend the dry storage options and technologies have been applied to the all different types of spent fuels and has the dry storage technologies have evolved since their concept conception and at the beginning they were all only designed for storage but today they are designed for storage and transportation and they are the so-called dual proposed casks and the idea is to design multiple systems and to be licensed for a storage transportation and also for disposal so the idea is not to once the spent fuel is placed in the casks and the casks license for all the stages is to avoid additional handling and additional to open the casks to handle the spent fuel from one cask to another casks so the idea is to simplify the process to avoid manipulate too much the fuel assemblies so this is an example of a casks design in Germany so you can see the fuel assemblies this is the basket for the fuel assemblies and there are many layers outside with neutron absorbers to prevent the dose and the spent fuel inside the casks around 300 Celsius degrees more or less and the decay heat is eliminated by natural convention and there are some layers and leads to close the casks some are bolted and also welded so this is the main issue now to try to find a way to license the cask for the three stages it's not that easy so this is the the main issue because then you can you don't need to retrieve the spent fuel in it from the casks use for a storage and transportation so if it's licensed for disposal too you can use it to be disposed so you don't need to to retrieve the spent fuel that is a very difficult but also is a sources of damage in the fuel assembly if you have to handle after 30 years of storage or 60 years of storage so you have the same casks licensed for disposal and it's the issue how to get the license how to ensure that a system that was licensed for storage 60 years ago is still valid to be disposed so now we for example in our section in my team we are starting working in transportation issues for casks after a long period of storage because some of the casks the dual proposed casks have been licensed for storage and transportation but you store you store the casks in the nuclear power plant and you transport but what's happened after 60 years of storage in the interim storage and you have to transport these casks to the disposal facility even to retrieve but what's happened after 60 years so now we are working in these type of issues and there is an international group in safety dedicated to dual proposed systems for the license to to find the requirements and all the aspects and parameters to be taken into account for the license process so so far it's when we consider so once we store the spent fuel we can dispose the eventually dispose of the spent fuel in the deep geological repository but for those countries which don't consider the spent fuel as waste they consider spent fuel as an asset they reprocess the spent fuel to get the uranium and plutonium for recycling so what is the status so we we were we talk about the storage but the total amount of spent fuel discharge globally is around 400,000 tons of heavy metal so far and one third of this amount it has been or is under reprocessing so far so 1000 tons of heavy metal is discharged every year 7,000 tons no sorry 10,000 tons are discharged 7,000 tons are stored and the remaining 3,000 tons so one third is repossessed the annual reprocessing capacity in the world so far is 4,600 tons of heavy metal per year although now it's not all is currently used now it's not operational so in which consists the reprocessed is to to get the plutonium and uranium to be used again as a mox fuel and it's produced a high level waste that means the the composition is mainly deficient products and the mine reactinates so so far France UK India and Russian Federation are reprocessing Japan now is it was four or five months ago the government clearly said that they will continue going for closing close cycle and the Rokashomura it's under licensing process and China has a pilot plan and also it's building or it's going to be a new reprocessing plan but they are facing some difficult build by a ribbon but they are facing some a stakeholder issues to in the site decided for to build a reprocessing plan so by the end of 2016 around 120 kilotons of heavy metal had been repossessed and the major reprocessing facilities are planned to be commissioned in China Japan and in the Russian Federation since UK is going to shut down a third plan and also they are going to open cycle so they are changing the fuel cycle option and they are moving UK from close cycle to open cycle so this is the main process for reprocessing is the Putex process is plutonium uranium restriction and after mechanical disassembly and the roads has to be shopped and the spent fuel is dissolved in nitric acid it is generated at a gas stream that is treated by and it's filtered and main of mainly iodine and the novel gases produced are trapped in particular trapping absorbers and the nitric acid once this is dissolved it's a stream that has to be treated so the acid solution undergoes through solvent extraction using tributyl phosphate and this is the molecule in aliphatic diluent as kerosene or dodecaene so and their liquid-liquid extraction uranium and plutonium are recovered and the rest of the fission products and monorectinites remain in the high level waste stream so the main of the minor actinites are in trivalent state so that's why they are not extracted by TVP that only extract uranium in the only extract in exavalent and for balance state so once uranium plutonium are recovered from the organic solution the solvent is treated is clean and is reused again in order not to not produce a lot of organic spin solvent because the organic solutions are very difficult to manage as secondary waste so the TVP in kerosene is treated and cleaned to be reused several times so once uranium and plutonium are recovered they are split into streams in plutonium dioxide and the so-called repue uranium and they are stored for further use so plutonium is mixed with the depleted uranium or natural uranium to produce mox fuel and so far repue is stored to be used in the future TVP is tributyl phosphate it's a molecule it's a phosphate with three butyl changed so you mean reprocess the fuel so the mox fuel as far as I know the mox fuel is produced with mixing plutonium and oxide and natural uranium or depleted uranium if you use repue repue has different uranium composition because it remains 1% of uranium 5 in the repue and also there is other isotopes of uranium I think it's uranium 232 that is a very strong gamma emitter so the company isotope composition of repue is different to the natural uranium or the depleted uranium so I'm not engineering the core so but you have to have into consideration that the isotope composition of repue is different to reprises so it depends on the contract because for example some countries when you sign the contract with the reprises facility you can receive mox fuel you can receive only the high level waste so you have to receive the high level waste because any country can get waste from other countries so even if you send your spent fuel to reprises you will receive the high level waste I think you will depends on the contract but you cannot if you would if you don't want to receive the repue you don't need to receive the repue as a few you will receive mox you will receive mox fuel you will never see repue okay so you will receive okay so it depends on the contract you sign if your core allows you to use repue you can receive repue if your core allows you for example in the case of Spain we send some of our spent fuel from Bandejoz for example to France for reprises and we are not allowed to use mox in our reactors and we are not allowed to use repue as well so that we are going to receive is a high level waste but we don't have so far any storage to store high level waste in Spain so our waste remain in France waiting for our country to have a suitable storage facility but we only will receive the waste because we are obliged to receive it but we are not going to receive any additional fuel because we are not allowed in our nuclear power plants are not licensed to use mox fuel and repue and this is a contract that is pain sign so it depends on your the conditions of your license of your regulatory body and also that your reactor is not because sorry because if you cannot use plutonium you cannot use repue the facility the reprises facility can use it and can sell this material to another country so can be used for another country or using the cells sorry can be used as a fuel but the core should be changed a little bit at the design of the core and should change and the problem with repue is the composition of uranium after reprisesing the isotope composition change and now it needs additional safety requirements because there is one of the isotopes of uranium is 232 and it's a very strong gamma emitter so and you need a shield for the workers or to remote handling but can be used very rich again and can be used as a fuel of plutonium and repue because the repue is the 95% of a spent fuel is yeah so if you consider for example one road and the final composition of the spent fuel is around 92% of uranium and around 1% of plutonium and this is if we consider that the process is 100% efficient we will get at the end the plutonium is 1% of in the beginning and repue it's 92% because the bulk of the spent fuel is uranium that changed the composition at the beginning the composition was 95% to uranium 8 and 5% of uranium 5 but at the end repue change in the core the composition change from 95% of uranium 8 is now 92 because some of the uranium 8 has been consumed producing plutonium 9 and uranium 5 has been has been fission so and it's consumed from 5% from the beginning to around 1% at the end and during the fission process and the neutron capture process that change also changes in the composition of uranium and there are new isotopes produced but if we consider that we started with a pellet at the end we will have 1% of plutonium and 92% of repue so since in the enrichment process there's a lot of depleted uranium produced it's easier to use the depleted uranium or natural uranium to fabricate Mox fuel because you can handle this uranium without any shielding for the workers but if you use repue you need a facility a modified facility because the isotope composition of uranium has changed and there are strong ammeters I don't remember the name but not to run itself this this element will give a high gamma rate high gamma emission so even for yesterday we talked about refrigeration for 1, 2, 3, 2, by the way for thorium and like this for full cycle you can by the way it's very difficult to produce a gamma which will conclude but if you have online or online refueling or you can remove 2, 3, 2, you run it 2, 3, 2, before it scores to neutrons so it's mean online you can you have a problem with our challenge with the operation so uranium as it is for gamma emission the problem with that it scores to neutrons more neutrons and in this case you will have another element maybe you don't remember okay but I mean it's not the same because for the handling of this uranium the facility yeah for an operation issues yeah okay so I will give you after the talk reference because there is a very interesting technical report dedicated to repuel it's a nuclear energy series so and in this report there is a lot of the all options for to manage repuel and to use repuel so far so I will give you the reference after the talk the represents uranium as we talk it can be re-enriched again and reused in the reactor fuel and plutonium is used to produce mox fuel is mixed oxide fuel so to produce mox fuel plutonium and depleted uranium are mixed and pelletized are loaded in the roads to fabricate new fuels and for example in France most of the of the of their reactors are licensed to use mox fuel and certainly the fuel and the fuel assembly are identical to the oxide uranium fuel but in this case for the fuel fabrication it's necessary to make it under glove box due to the plutonium is has problems with inhalation as art is because it's a very strong alpha emitter so it should be handled in glove boxes yes sure yeah because in this case if you are very rich again because the enrichment is around 100 percent if you range rich again you don't need any other fission at all you have the uranium as is fission even so you will never receive the plutonium because you are not okay yeah it depends on your contract in the case yeah so you can yeah if you even you need a license to use reprocess uranium for example in our case we don't we cannot use it in spain but if you are allowed to use you will receive the reprocess uranium reenriched or you can reenrich and reuse again and the plutonium as i mentioned before it's mixed with the plated uranium because the the fissile material is plutonium and uranium the plated uranium act as a matrix and as a 13 material so as a summary we have talked about the open cycle one of the fuel cycle options is the open cycle so it's considered they use fuel as a waste the one through cycle or the close cycle is not fully closed because plutonium is used once as a mox fuel and then is stored or is considered waste and now for the future and for the sustainable sustainability of the nuclear energy there is a lot of effort applied to use and to develop as a commercial scale fast neutron reactors to fully close the cycle and to reuse all the time plutonium and to get the equilibrium in the plutonium so in the ideal world it will it won't be necessary to have more uranium and with plutonium we can sustainable and we have the stable inventory of plutonium so it's reduced in a huge amount the waste produced and the burden of the waste so this is the the R&D so far is dedicated to to have this fuel cycle option and there are countries as russia that has the only operating fast reactor in the world is the bn 800 that was connected to the grid in 2016-15 and we had last june the fast reactor conference there in diakaterin world with 800 attendees and with the all the community of fast reactor and associated fuel cycles so this slide it means the radio toxicity and so we can see this is a radio toxicity evolution of the open cycle and it's due to plutonium mainly the plutonium if we avoid plutonium in the high-level waste the radio toxicity decrease and it's due to minor actinides but in the case of fully use the fast reactors and the advanced fuel cycles and the fully use of uranium plutonium recycling plus the minor actinide recycling only fission products remain in the high-level waste and due to the half-lives of the radionuclides of the fission products the radio toxicity get the radio toxicity of the uranium in around 100 years so 300 400 years so compared to million of years is an increased in the reduction an improvement in the reduction of the parten of the radio toxicity in the high-level waste and this is the future that is foreseen in nuclear energy technology so let me give you some examples of the so to apply these scenarios we need to have so uranium and plutonium recycling from the chemistry point of view is have been already proved with purex process but to get minor actinides is not that easy task because the minor actinides and some of the fission products have the same chemistry so there are three elements and it's not that easy to separate them so there are international efforts applied to this and there are some european projects and also american projects and we try now in the INC to coordinate all these r&d efforts in the world so we try to set a project with the european community with american japan korea china and we are working on that so for example this is the american process is urex plus and it's a modification of the purex process and the p disappear from purex and now it's urex because they work not to have plutonium separated for the proliferation issues and so we they separate uranium they separate plutonium with the rest of minor actinides basically with neptunium so and they conduct the process in steps so they separate uranium and also they separate those elements that produce heat decay and they are cesium and strontium and they separate plutonium with neptunium the minor actinides with the lanthanides that are the main fission products and they are very difficult to separate among them and finally there is a process dedicated to separate minor actinides from lanthanides this is the american approach the european approach is to separate once the bulk of uranium it's separated the plutonium is extracted with the rest of minor actinides and the process is called ganics it's the group actinide extraction there is another approach and it's as the american approach in steps so it's the so-called diamex process is called diamex because it's using a big licoramide and it's dedicated to separate minor actinides with lanthanides from the high-level ways as the american process and then once we have minor actinides with lanthanides the separation between minor actinides and lanthanides is done with by using the sanics process it's selective actinide extraction using a dedicated molecule that is selective for actinides and not for lanthanides and then there is another process it's called cesame to separate america and corium to produce the blankets for for transmutation so all this process has have been tested at a pilot plant scale but none of them have been already demonstrated at a commercial scale this is the weight the weight methods and also they are under studied the dry methods that consist in dissolving the spent fuel in a molten salt bath and it's called the electro refining process and using the difference in electro activity between the elements we can get a cathode if the bulk of uranium and we can have a cathode a liquid cathode with the rest of the minor actinides and plutonium and with some of uranium and this is the real liquid cathode after the separation and this is the real uranium deposited cathode with the uranium deposit and this process has been has been also tested at a pilot scale but it's not commercially available and in korea they are working hard and they built a pilot plant to test the process called in a cold test not using radioactive material but using cold simulators and to demonstrate the capability of this process to be at commercial scale so this is the future and in any case we need at the end deep geological repository to dispose of the spent fuel or the high level waste produced during the reprocessing activities so this is a picture of a deep geological repository and so far there are four countries in the development of the license step so france has made a lot of effort and to implement the industrialization of the industrial implementation of the deep geological repository for our high level waste and they have underground laboratory and it's called CGO in in the northeast part of france. Finland has admitted the license for the deep geological repository and they received in 2015 the positive feedback from the regulatory body expressed in a letter to the government and they recommend to start the building of the facility and their certain circumstances so they are now building the deep geological repository and also sweden has admitted the license it's in the licensing process and they don't have received so far the final approval by the regulatory body but they have some positive feedback and informal positive feedback and the unique ever submitted license was in united states in 2008 and the nrc and the staff the regulatory body staff shows their positive feedback from for the repository but they found at the end some issues with the rights the only rights for the land for the water that stopped the project and so it happened the change of administrative administration so obama administration decided to stop the project but not for the technical issues because the project was approved by the nrc but was political and societal issues so it shows that it's very important that stakeholder involvement from the beginning in this in all the nuclear development energy development and stages but specifically in the sites for the repository that has to last for a thousand years so the community has to be engaged and to be not to be asked at the end so they have to be engaged from the beginning in the project and i so i let you some sources of information so this is our technical areas you can find here our projects on the on storage i spent a few storage and also some interesting information and i will give you later the the reference of the repu and report and that's it thank you for your attention


TfNN, headline, news update. Welcome folks. We have the Dow Industries trading down 30. Nasdaq is off 20. S&P's up five and a half. Gold contract up $16.10, trading at $18.72 an ounce. We have Silver up 28 cents, $23.62 an ounce. Late sweet crude, up a buck, $32.93, 36 cents, a barrel, notes and bonds. Ten year, up two ticks, trading $125.26, the 30 year up two at $150.26, and $king dollar. $king dollars down 294 ticks, $95.693, Euro 113, yen 150 and a half, and the British pound $136 to one U.S. dollar. We'll get over and take a look at the S&P first folks. We're looking at it out here. We've got a bottom line, we've got a little bounce going. That being said, guess what? You're going to have even more contraction of volume. My take is that we are going to get up, though, to the top of this range. Inside the spy, it's $457. Right now you're at $446. We take a look at the NDX100, the three cues. Bottom line, even weaker. That's still down 36 cents, but that's going to be a lower price, rejection of price. Light volume, I suspect. This also, the one with the cues is going to get interesting, too. Because what has happened with the cues is, the last two highs out here, well, not the last two highs, but what I'm talking about the last couple of weeks, they were lower highs. Now, we didn't get lower lows. We have that high volume low that wants to get tested at 334. But I suspect somehow the cues this week, they want to get up into that level also. Gold, we take a look at the gold contract out here. What we have with gold. Gold reached 1881 yesterday. Right now you're at 1872. You need volume. We need more volume here. There's no doubt about it. 127,000 contracts. Now, when you read this, though, this is where it gets interesting, because it's an inside day. So you're like, okay, did you reject the low in light of volume today, or what did you do? So bottom line is that what we have got, no doubt, is that we've got two days of good volume pushing into that swing. We'll see if we can get a catalyst to get us over that 1882. But what you also have to remember is this. This was 100% move for move, meaning that you started out at 1882, you're back down, you go back up to that level. And most times when you do something like that, the bottom line is that you're going to have to build more cars in order to get to higher price. Stay right there, folks, we'll be coming right back.


This study compared the accuracy of manual muscle testing to detect strength differences among grade 4 and 5 subjects with shoulder flexor weakness at varying percentages of dynamometric mean. The study found that manual muscle testing had higher diagnostic accuracy than dynamometry at 10% strength deficit for both non-dominant and dominant shoulders. However, the study also noted that the positive predictive value, PPV, and specificity of manual muscle testing was better at 10% strength deficit in comparison with dynamometric measures, whereas negative predictive value, NPV, and sensitivity was better at 20% strength deficit. Additionally, the study found that there was a contrasting association between PPV and sensitivity, suggesting that manual muscle testing may be more accurate when used as a screening tool for shoulder flexor weakness. This article was authored by Sadia Purways, Muhammad Waqar Afzal, and Gulam Fadima. We are article.tv, links in the description below.