the following content is provided under a Creative Commons license your support will help MIT OpenCourseWare continue to offer high quality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu well it's promises we're gonna cover no new material today we've just hit the end of part 2 of the course which I think you'll agree with me was probably the most technically challenging part who would disagree I wonder I didn't think so so I wanted a bit of review and help you guys out with problem set seven since I've got a couple of fun problems I think they're fun because they're actually fairly realistic like the last one will it blend the ap1000 edition well you'll actually analyze if we've given the actual specification sheet of an ap1000 reactor which is a modern reactor that's being built now can you determine its k effective using the to group the two energy group approximation and just to show you guys that this isn't crazy I've got the ap1000 spec sheet up here so who got a chance to take a look at pset seven good that's a lot of hands alright I I highly recommend everyone look at it ahead of time because it is the doozy and it will be the last doozy of this course so part three of the course is lighter because most your other courses are gonna go nuts after Thanksgiving as far as I know right yeah this one's not so doing my best to equalize your total course load this semester by making this course crazy now and lighten up when final season comes so I've got the spec sheet for the ap1000 right here that actually goes over a description of what the core is like the materials that are used how many fuel rods there are how many assemblies there are basically what the core is made of and what I want to do is jump to the end where they actually talk about the analytical techniques used in the core design what I want to show you is for nuclear design of the core to get axial power distributions look at that to group diffusion Theory the same stuff that we just learned right here for axial power distribution control rod Worth's one-dimensional to group diffusion theory these are problems that you can solve with the stuff that we've done over the past week and a half they have a little more complexity and that they keep all the spatial variance in there so probably done with computers however when they make the same equations that you do so Westinghouse is making the same assumptions that we made in this course they have a lot more complexity in that there you know we just had cross sections or macroscopic cross sections as a function of energy but you may also think about it as a function of position and a function of temperature since as we started alluding to on what was today it's on Wednesday know Tuesday what day is it now it's Thursday thank you so definitely on Tuesday we started talking about how cross sections change with temperature and so really if we want to go crazy on that neutron transport equation these cross sections will be functions of energy temperature and position and so that's how this reactor would have actually been designed but you're going to do a simpler approximation and take all of the information about the core and an ap1000 blend it so homogenized it figure out what the average atomic fractions of all the different things in it are and calculate is K effective so I think is a pretty cool problem to do I haven't seen it done in the courses here but I want to see how you how it turns out and you might find it's surprisingly different from one because we make a lot a lot of simplifications that actually matter quite a bit but to help you parse this spec sheet well actually why don't I show you the problem first we'll go through a few of the different things that I've simplified the problem to make it not so tedious but I also want to make sure you understand what to do so the simple the simple statement is calculate K effective of the ap1000 using two group diffusion theory which means you've got a criticality condition from the two group approximation and I'd like to go over what that is right now let's so let's write out if we had a two group so there are two Energy Group equations for gains and losses of neutrons in the fast and thermal group what would it look like what are the gain terms yep there's gonna be some average new times Sigma fission fast times flux fast plus Sigma fission thermal times flux Thermal any other sources of neutrons into the fast group I don't think so what about sinks how do neutrons leave the fast group yep by absorption how else yep scattering from the fast to the thermal group how else leakage so there'll be some diffusion constant fast times some fast geometric buckling squared and yeah I think that's it no that needs a Phi as well fast okay cool what about the thermal group what are the sources of thermal neutrons yep scattering from the fast group so this same term right here fast a thermal times Phi fast and what are the losses leakage and absorption they look pretty familiar thermal by thermal plus the thermal BG squared by thermal and that's C T okay the hard part in this problem is going to be doing these averages this is the part that we haven't explicitly done on the board and I want to show you yeah this one top equation that should be a I'll make the apps really curly curve and those are straight T's these are curly FS okay great thank you so the hard part is going to be doing cross section averages we've just kind of written them as hey they're average cross sections in an average cross section would look something like the integral from a minimum to a maximum of the cross section as a function of energy times the flux over the same integral without the cross section this is where I want to point out something that I want you to remember for the rest of this course and the rest of your life you don't have to do things analytically if you don't want to unless it's explicitly stated that you have to so part of this pset is to drill in the idea that it's the future we have computers and you can do numerical integration with data remember on one of the second problem sets I showed you guys the web plot digitizer how you can extract information from a printed graph well a lot of times what you already have is that data and you'll have to then integrate that numerically in something as simple as Excel or as complicated as MATLAB or something worse whatever tool you choose to use I'll be using excel because it's kind of the lowest common denominator and so to show you guys you don't need to know any fancy software to actually solve these problems so let's go through getting some of this data right now let's say that the cutoff between fast and thermal so if we were doing a fast cross section the cutoff would be at one Evie and our max would be let's say 10 MeV which would be the top of the fission birth spectrum or the Chi spectrum one UV that would be 10 MeV so all that's left is you need tabulated values for the cross sections and the fluxes and then you can perform this numerical integral I've given you tabulated values for the fluxes wherever that is yep use the attached ap1000 tabulated Neutron flux profile which opens that's awesome so I've given you the approximate Neutron flux in neutrons per centimeter squared per second as a function of neutron energy and so you can tell for a low Neutron energy there aren't any ultra cold neutrons in this problem though there are in the problem right before because remember the we'll see now is when we'll see but if we scale down we get down to the thermal regions where it's in the EVM levels you start to see pretty significant Neutron fluxes on the rail of 10 to the 14th neutrons for centimeter squared per second so that value of 10 to the 14th for flux that we've been using in all our previous problem sets it's because that's what we actually get this flux spectrum and the picture of it that we have here was taken from the MIT reactor because it's representative of a pressurized water reactor the only difference here is this is the spectrum from the fast flux trap you don't usually see that fast to thermal ratio in a thermal reactor but it's the closest spectrum that I was even easily able to get my hands on and it's not that unrepresentative and the reaction rates for things aren't gonna be that different because most of the reaction rates the cross-sections down here are in the like thousands of barns level and the cross-sections here are in the one barn level it's pretty much any fast cross-section for anything about a barn that's a good rule of thumb so it's not going to change total reaction rates that much and that's why I'm not worried about making a taking the fast flux spectrum from the MIT reactor and pretending like it's the ap1000 it's not horribly that far off and it's at the right order of magnitude which is important so that data I give you let's talk about how to get this data the macroscopic cross-sections so if you remember a macroscopic cross-section is a microscopic cross-section times a number density and that's for one single isotope if you have a mix of isotopes then your total averaged cross-section for all different types of atoms is going to be a sum and make it very different over all of your isotopic possible ores all your possible isotopes of the atom fraction of that isotope times the number density of that isotope times the microscopic cross-section of that reaction for that isotope I'm sorry the atom fraction is included in the number density let's just simplify this a little bit the total number density of that isotope that'll be in atoms per cubic centimeter times the cross-section for that particular isotope which is in centimeters squared which leads you to a one over centimeter macroscopic cross section and so let's say we were summing up something like stainless steel which happened to be iron 18 chrome 10 nickel not only would you have to then get the number densities of iron chrome and nickel but you have to look at which isotopes there are so if we want it to get the macroscopic cross section for stainless steel we'd have to split this into the stable isotopes of iron the stable isotopes of chrome and the stable isotopes of nickel and then say it's this number density times of cross-section Plus this number density times a cross-section and so on and so on and so on the easy way to do get those number densities if you take the number density of your stainless steel times the atom fraction of that isotope that should give you the number density of that isotope this makes sense to everybody does anyone not know how to get a number density of a material from its basic chemical properties okay so a number density is in atoms per cubic centimeter and usually we would have something like its density which would be grams per cubic centimeter so if we take density in grams per centimeter cubed and x Avogadro's number and divide by I'll just put a / symbol here the molar mass of a god rose number is given in atoms per mole and divided by molar mass units are given in what is it grams per mole so that's going to be like moles per gram the grams cancel the moles cancel and you get atoms per cubic centimeter so to get a number density you can take the density of the material you know for stainless steel it's like eight grams per cubic centimeter times Avogadro's number say 6 times 10 to the 23rd this would be like 8 grams per centimeter cubed or so time or divided by the molar mass or the average molar mass of stainless steel I'm guessing that's around 56 grams per mole and then you'll get a number density and typical solid number densities tend to range from like 10 to the 26 to 10 to the 28 it'll be atoms per meter cube so it's gonna be around like 21 to 23 atoms per centimeter cubed so if you end up with something way outside those bounds you've probably got some sort of unit or power error so that'll help you check your math to make sure you get the number densities right if you get the number densities right and you know the atom fraction and you have the number density of each isotope in the number of atoms of that isotope cubic centimeter and then you multiply by your cross-section your microscopic cross-section and you get your macroscopic cross-section yeah yep atom fraction is a fraction between zero and one of what proportion of the atoms in your material are that isotope so again if your atom fractions are outside the bounds of zero to one that's not physically significant if you're number densities are around to really far from those bounds then they're probably not right unless you're talking about a gas or a neutron star but solid matter tends to have approximately those number densities good question you'll determine those fractions from the ap1000 spec sheet so in the ap1000 spec sheet it tells you things like total weight of the fuel as uranium dioxide and so you can go from total weight of the fuel to let's say molar or atom fraction of the fuel if you know the weight of the fuel it's nice they give you the way to the fuel they gave you the way to the clad and what let's see which materials did we say you have to talk think about said you have to talk about four materials the coolant of the moderator water the fuel you owe to the clotting where you can assume piers or cone iam forget all the crazy Zircaloy because that's just busywork and structural materials assume pure iron and so on this spec sheet luckily they just tell you the mass of the fuel they tell you the mass of the clad I do not believe they give you the mass of the water but they do give you the volume of the core and so you can figure out if you've got a core and you subtract off the volume of the fuel and the weight and the cladding and the structural materials all you're left with this volume of the water in the core and that'll give you your total weight of water and once you have all the weights then you can go to atom fractions and then you've got all the information you need to get the macroscopic cross-sections is that unclear to anybody cool so we talked about how to get the ends let's show you how to get the Sigma's so there was a comment that came in that said please teach us how to use these databases we just kind of throw them around well I want to teach you how to use this database let's say we're going to get the cross-sections for oxygen in uranium dioxide that's nice it easy because there's only one stable isotope of oxygen you have to consider is oxygen 16 I highly recommend using the java version of Janus because it is a lot easier to use less clunky and more intuitive so if you don't have on your machine first of all it runs on everything like phones tablets Linux Mac Windows whatever and second of all it'll just make your life easier so a little time investment now will make the piece that take less so once you have Java it should just open cleanly and it may show up with nothing it may show up something depending on what you last looked at in this case it's looking at whatever last library I looked at so let's pretend we're starting over so sometimes you may just see this database as NEA that's the that's all the databases that come with Janis we expand this make sure that you go to incident Neutron data because these are cross-sections for Neutron reactions that we want to go for and there are a lot of databases in here and in the problem set specifically say to use the most recent en d.f or evaluated nuclear data file just to make sure that we're all using the same data set you'll notice that there are discrepancies between the data so different groups have measured things with different uncertainties and different values and so these cross-sections are necessarily fundamental constants of nature they're measurements of those constants with whatever uncertainty and error which are two separate things that could be in there so let's open up the most recent endif library and click on cross-sections and now whatever you see here the green squares are the elements that have tabulated cross-sections just to make sure that we only have to consider oxygen-16 let's go to the table of nuclides we'll never stop using this table it's like the most useful thing in this class and check to make sure that there's no other stable isotopes of oxygen that we have to worry about as long as the Internet's working did they immigrate to Korea website crashed to interesting well we'll let that load for a bit and let's start looking at oxygen-16 so if you double-click on the element of interest it'll usually take a little while because it's Java it's loading cross-sections and then you can pick the nuclear reaction of choice from here and unfortunately I can't easily make this bigger I will take a very quick detour and see if I can make these things a bit larger so you can see them but if not then whatever no no oh well so let's say you wanted to get the elastic scattering cross-section for oxygen the first the first letter before the comma is going to be the incident particle notice that sometimes it says N and sometimes it says Z and specifically means neutrons z means whatever incident particle you chose so we know that here Z means neutrons coming in and elastic that's elastic scattering so we can then click on cross-section if you want to see what it looks like you can check P for plot and this will give you a logarithmic plot of the scattering cross-section as a function of energy you can see that it's pretty boring for oxygen there aren't a lot of nucleons there aren't a lot of different energy levels there are not that many resonances or things going on so I wouldn't be that upset if you just approximated the fast scattering cross-section for argon as that value whatever it is but let's do this completely let's tabulate this so if you click on T you actually get a table of data and you can choose how much data you export by default you get like thousands upon thousands of entries which is just gonna make your life horrible so what you can do is either pick the original values starting at one evey and we know we only have to go up to ten MeV so you can pick your bounds and all of a sudden there is only a few thousands worth of data or you can interpolate them you can interpolate values either linearly or logarithmically I recommend logarithmic because this is such a large energy range and get maybe five values per decade so between one and ten MeV you only need five numbers that's still pretty intense oh I didn't uncheck the original values ah isn't that better so now there's only 20 or 30 entries it glosses over the residences it sure does but are they that important that's up to you guys to decide you can try doing one of these calculations with the original values and with the interpolated values and see just how different they really are in to group theory and hint is not very much so then you can actually export that data so either you can just copy paste it so I just highlighted copied start Excel and in it comes and right there's the data that you can start to use to do your numerical integration and I don't want to give away how to do the numerical integration though I kind of did symbolically I'd like you guys how to figure out how to do this numerical integration mathematically given that you can get the data now so is there any step here that's unclear to anyone yeah every single material yes you are sounds horrible isn't it that's why I made some simplifications so if you notice we are simplifying the cladding as Zr and we're simplifying stainless steel as pure iron I don't want you guys to just do tons and tons of repetitive stuff but I do want you to get roughly the right answer I want you to answer that question so how would you consider the isotopes of uranium in this question that's right it's not the natural values it's the enriched values so they will say in the spec sheet you wanna don't want to be on a list you don't have to look up and rich to uranium on Google you can look it up on Janus so if you go to Janus right are you looking for what the enrichment level is yeah oh that's in the ap1000 spec sheet where'd it go we search for enrichment fuel enrichment first cycle weight percent you can either average these or just pretend it's five because that's that's a pretty pretty typical enrichment level is 5% atomic fraction u-235 or if you want to get really technical you can average these noting how many fuel assemblies are in each of the regions but I don't really care if you do that you 238 yep so there's only two isotopes of uranium one of oxygen however many of iron and zirconium hint there's not that many and then there is h2o and there's only one hydrogen and one oxygen so it's not really that much busy work it's just enough for you to get to the right answer so I wouldn't even call it busy work because there's a point to it so in that way you can determine all of these cross sections we give you these fluxes these new values you can also get from Janus they're not labeled as new but they are labeled as Neutron multiplication factor and they're usually way down here and we probably need to find a fissile isotope for that so let's ditch oxygen for now go up to uranium okay so let's do you 235 it'll pull up the data and then near the bottom noubar total neutron production check it out for pretty much all energies until you get to the fast region it's the value we talked about two point four four and then in the fast region you suddenly can get more neutrons from fission why do you guys think that is well what sort of additional things happen when you're incident particle energy increases let's think back to binding energy and all the stuff at the beginning of the course there's more different kinds of fission products that can be made because you're increasing the Q value of this reaction by increasing the initial kinetic energy so there are more fission products that can be made and some of them let's say give off more neutrons than others so you'll be able to get your noubar total from this one near the bottom and everything else will be near the top they'll either be a fission cross-section an absorption cross-section a scattering cross-section how do you get your diffusion coefficients they're not tabulated but they're close yep exactly so let's render this nicely the diffusion constant is one over three times the total cross-section minus mu not scattering average average average and this average cosine is sorry about two-thirds times the atomic mass so you can get these diffusion coefficients from tabulated data the total cross section is found at the top better wise better known as well it's just as total so you can get total cross-sections here the only one that might be a little tricky for you to find from notation is absorption you're not going to see something labeled absorption in this database however you will see this gamma reaction or how else do you get it if you take the total - fission - scattering you're left with absorption if you don't count end-to-end reactions or really esoteric high-energy things so if you can't find it that's okay because you can calculate it because we know Sigma absorption is Sigma total minus Sigma scattering minus Sigma fission - others that we don't care about and since you're getting these anyway tabulating Sigma absorption should be trivial it's just an Excel subtraction yep yep inelastic scattering is not considered absorption so inelastic scattering means one neutron goes in one neutron comes out but at a very different energy level so I guess we would also say minus Sigma in elastic and inelastic does happen for just about every other element but the nice thing is those don't turn on until about one MeV so it's not going to matter much but you can quantify how much it matters and just check so if you can make a justified assumption to say here's the one of the calyx with and without inelastic scattering and if they don't differ by much then forget it as long as you show me your math so let's say we've got news we have cross-sections I give you fluxes we have DS you can calculate buckling from the geometry of the reactor which is given in the ap1000 spec sheet the only other trick will be what's Sigma scattering from fast to thermal so you'll have to figure out what is the cross section not just of scattering total but the cross section or the probability that a fast Neutron enters the thermal group and I don't want to give that away either but it's not terribly mathematical yep not quite so even if you only have one isotope for each material if you have more than one material you've got to average those cross sections because this criticality criterion is for the entire reactor and all the stuff in it that's why it's like in a blender so even if iron has one isotope and zirconium as one isotope and uranium has one isotope which wouldn't really be a reactor then you'd still have to take atom fractions of those to get the total criticality condition speaking of which I forgot to write where the K is and since everything else here will be tabulated you can solve for K effective so K effective is the only variable unknown in this whole equation yep that's right so yeah let's go to the rest of the problem set since you mentioned it for one of the other questions North Korean nuclear weapons why is it that just putting together a super critical mass of plutonium does not constitute an effective bomb we're lucky for that too that making nuclear weapons is a lot harder than just getting nuclear material and this is a lot of the reason why there's a been you know 500 tonne yield or kiloton yields duds it's really really really really hard and I want you to think about what would you need to do to turn a super critical mass of nuclear material into an effective weapon and why is it that it's so difficult to do and actually really glad it's so difficult to do it's one of the reasons that a lot of folks don't have them but yeah if you only have one isotope like you do in this problem you don't have to do atom fractions you just take number density times microscopic cross-section from Janus and you get the macro cross-sections that's why it's one of the skill building problems where it's it shouldn't be that hard so you'll have a criticality condition you'll be able to get cross-sections and number densities and you'll be able to tell what that radius of the sphere is given the form of buckling for a sphere which would be in your reading yep that's being sent to you today so everyone oh so you guys all did power manipulations from the reactor I wanted to get him from Monday the guys were busy doing an in core experiment installed they've promised me the data today so you guys will be able to take a look at it and using the transient stuff that we talked about Tuesday explain why it doesn't look the linear feedback or intuitive yep so we'll also I'll be giving each of you that data today we talked about nuclear weapons there's top one I want you to do on your own because it's other repetitions of the intuitive criticality examples that we did on Tuesday the last one we haven't talked about is the ultra cold nuclear reactor or the will see problem so here I want you to actually get a criticality condition for the case where you can have ultra cold neutrons where your moderator is let's say liquid hydrogen really really cold way below the thermal level and you have to split your reactor into three energy groups fast thermal and ultra cooled and the trick to this here is somebody asked can you ever have ultra up scattering why yes if you have an ultra cold moderator but you've got hot fuel you can actually scatter up in energy where the surrounding atoms could be hotter than some of the neutrons hitting it and will impart energy to those neutrons and send it up the energy spectrum yep mmm well how do you get old for cold neutrons mm-hmm but some of them might scatter up yeah they can scatter about they can scatter down by hitting something cold and scatter back up by getting knocked by a hot item yeah so it's all gonna be in the formulation of the equations for this problem like 80 or 85 percent of the credit is did you formulate the three group equations correctly so I want you to think about what are the actual sources and sinks in each case what are the fractions of prompt and delayed neutrons where do they go and what terms matter where so it's doing this but for the case that we've given you here solving it is a lot of algebra and therefore not a lot of credit so is that clear to everybody figured this piece up with worth explaining and not just saying have fun you know I'll see you on Piazza Sunday night yeah that's why I check it yeah cool I've also got problem set four for everybody here I want to mention a couple of quick things please when you if you handwrite your psets which is fine please make sure to scan them legibly and to write legibly we can't give partial credit for things we can't read and so this will be a lesson to some of you guys so depending on who got what grade there are times when you may have written stuff for partial credit but we just honestly couldn't make it out so please do make sure that your submissions are legible and for things like these these are handwritten problem sets but they were scanned with either a scanner which does the correct contrast or writing them on one note which apparently works pretty well or apps like cam scanner there's an app on your phone you can get that scans pieces of paper and automatically contrast enhances them it knows the paper should be white and the writing should be black and it also does it in color so if you do color graphs it recognizes that there are multiple colors and will take care of that for you it only takes an additional minute but I can give you you know double the points and a problem set because we can read stuff so I'll give back pset for now we're working on five and six the big delay in grading there was I went to Russia and Russia doesn't have as much internet as we do at least not where I was and it was busy so working on those solutions now yeah yep oh okay yeah good question so problem one throwing quarters into the directly into the core like a wishing well it used to be that back in the day they would take folks on reactor tours to look down into the core because when the lid is off you can see it you can see the Cherenkov radiation the nice blue light you can actually see the fuel elements because the water is sufficient shielding for you and the distance is sufficient shielding to keep you away from the gammas not to mention the reactor is usually off when the lid is off so it's not that hot problem is you can't watch what everyone's doing all the time and someone he dropped a quarter into the reactor they were like oh it's a wishing well well it took something like six dive robots to go into the core and fish it out because each of those robots lasts you know 10 minutes before the intense radiation fries it and if you didn't find the corridor you got to take into out put down another one and these are like radiation hard you know narrow whatever diving robots this story was relayed to me through someone that related to them through whatever so it's been through the telephone chain but I dunno that's one of the big reasons you can't look down on the core anymore it's because folks abused the privileges of torus yeah yep interesting then you could use the web version which is just through the blah browser and no plugins required yeah cool any other questions about the pset that we didn't cover together does looking a little more doable you