An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle Module 1 Segment 6. The Nuclear Fuel Cycle Calculations

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1 (Slide 1) Now we are going to learn how to do some simple problems dealing with the amount of uranium we have in a mass of uranium ore we take from a mine. Some of you out there may understand this all very well but some of you may be seeing this for the first time or haven t seen it for many years. Good or bad, this course will often get into scientific details. Every effort has been made to keep the modules as interesting and understandable as possible for the layperson. For students with more interest in specific information, references are given which can be used as starting points for further reading. (Slide 2) First, a simple problem that looks at how much uranium we get out of a metric ton of uranium ore. A metric ton (or sometimes called a tonne) is 1,000 kilograms of material. So here we want to calculate the mass of U-235 per metric ton (t) of uranium ore assuming the total uranium is 1 wt% of the ore. (Note: U has 0.711wt % U-235) (Slide 3) So please follow along. If you have a metric ton of ore weighing 1,000 kilograms, and the total amount of uranium is 1 weight % of the ore, then clearly we have 1000 kilograms times 0.01 kg of uranium per tonne of ore. or 10 kg. Now the U-235 is only weight % of uranium, so times 10 kg is kg or 71.1 grams of U-235. How typical is uranium ore having only 1 weight % of uranium? Well, we saw in a previous segment that high-grade ore had 2 weight % uranium and low-grade ore had 0.1%. So our 1 weight % is somewhere between highgrade and low-grade ore. (You can do this calculation with the Excel spread sheet entitled Segment 1.5 calculation for Slide 3.xlsx ) (Slide 4) To set us up for more difficult calculations, we need to introduce the concept of a mole and Avogadro s number. Our mole is not going to be a skin blemish present at birth, or a Mexican sauce, or a small mammal that lives underground. Our mole is a unit of measurement that tells us the number of atoms or molecules of a substance or element equal to x (Avogadro's number). For example: a mole of uranium has x uranium atoms that s 6 and some change followed by 23 zeros.. I haven t counted these atoms personally so you ll just have to push your I believe button. Avogadro was Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di Cerreto, Count of Quaregna and Cerreto (lived from August 1776 to July 1856). He is most noted for his contributions to molecular theory, including what is known as Avogadro's law. In tribute to him, the number of elementary entities (atoms, molecules, ions or other particles) in 1 mole of a substance, , is known as the Avogadro constant or Avogadro s number. Avogadro's number is simply how many molecules of a substance it takes in order to have a mole of the substance and it is equal to around x (Slide 5) The number of moles in a mass of material, call it m, is given by the mass of material in grams divided by the molecular mass M (gm/mole) where the molecular mass sometimes called the molecular weight is the standard atomic weight of an element as found in the standard periodic table that we will discuss later. For now, just push the I believe button again and consider that the molecular weight of uranium is 238 gms/mole and the molecular weight of oxygen is 16 gms/mole.

2 (Slide 6) Now let s ask what is the molecular mass of uranium oxide, U 3 O 8? Well, the chemist would combine 3 moles of uranium with 8 moles of oxygen to get one mole of U3O8. This mole of U3O8 has Avogadro s number (6.022 x ) molecules of U3O8 and it weighs 842 grams which we got by combining the molecular weight of 3 moles of uranium and 8 moles of oxygen. (Slide 7) Now let s do our next example problem. Uranium when mined is often in the form of Uranium Oxide, U 3 O 8. How many kgs of Uranium ore is in 100 kg of U 3 O 8? We just created the molecular mass of U 3 O 8. We took 3 parts of uranium at (3 moles)(238 gms/mole)= 714 grams plus 8 parts of oxygen at (8 moles)(16 gms/mole)= 128 grams to get a total of =842 gms. So the weight fraction of U in the U 3 O 8 is (714 gms/842 gms)= So if we have 100 kg of U 3 O 8 we have (0.848)(100 kg)=84.8 kg U. Again, you can work this out on the Excel spread sheet entitled Segment 1.5 calculation for Slide 6 & 7. xlsx (Slide 8) Now we are ready for a heavier problem. Presume the total weight of UO 2 fuel in a pressurized water reactor (PWR) is 150 metric tons where a metric ton is 1,000 kilograms). First off, we have to find out how many metric tons of just uranium are in 150 metric tons of UO 2? Here we have to recognize that not all of the UO 2 is U. So let s repeat the drill we just did with the U3O8. I give some choices for answers. Questions like this on the quizzes will be multiple choice questions. a)150mt, b)170 MT, c)750 MT, or d)150,000 MT (Slide 9) My answer is d, 150,000 metric tons. How did I get that? (Slide 10) Let s begin by taking one part of U and two parts of O to find that UO2 has 270 gms/mole and out of the 150 MT of UO2 we have the mass of uranium divided by the mass of UO2 to get the fraction as the part that is U. Or 132 MT of U in 150 MT of UO2. (Slide 11) As implied in the problem statement it s kind of involved. Let s look at the solution. So the MT of U in the 150 MT of UO 2 fuel is (0.8815)(150 MT) = 132 MT U of which according to the problem statement is 4 wt% is U-235. So the amount of U-235 in the core is (0.04)( MT) = 5.29 MT U-235. (Slide 12) How much natural uranium must be mined to get 5.29 MT U-235? Natural uranium consists of wt% U-235; hence, to get 5.29 MT of U-235 we must mine enough ore to give (5.29 kg/ MT U-235 per kg of U natural ) = MT U natural. If the U natural content of the ore is 0.5 wt%, we must mine MT/0.005 = 148,776 MT of ore or roughly 150,000 metric tons. So you can see that we had to mine a lot of ore to get the total weight of UO2 fuel. Again, you can work this out on the Excel spread sheet entitled Segment 1.5 calculation for

3 Slide 8 to 12. xlsx (Slide 13) Now we are in a position to calculate number of atoms or atom density using Avogadro s number. Note the formula on this slide. We take the density of a material times one over the molecular weight of the material times Avogadro s number. Notice how the units cancel out from all these terms to leave us with atoms per unit volume. Instead of using the density of the material let s say we are given a mass of the material. We use essentially the same formula except we use the mass of the material rather than the density. The molar mass of an atom (or molecule) is what the mass would be if you had x of those atoms. It is the total mass of all the protons, neutrons, and electrons within the atom/molecule. It is expressed in grams per mole (g/mol). The molar mass is the atomic mass and is the number just under the element's letter on the periodic table. For example: Uranium molar mass = (this number comes directly off the Chart of the Nuclides). This means 1 mol of Uranium = grams Thus, the molar mass of U is g/mol. (Slide 14) Let s use this formula to determine the number of atoms of uranium in a sample. Presume that a volume of pure U-235 is the size of a Tootsie Pop. How many atoms of U-235 will there be in the Tootsie Pop? (We will use this result in a future problem). Here are the steps that you need to go through: Estimate the volume of the Tootsie Pop using the formula for the volume of a sphere and get the density of uranium metal off the web as 19.1 gms/cm3. Calculate the volume with the formula Volume = 4 3 πr 3 where r is the radius of the sample. (You d be surprised how many undergraduate students calculate the volume as πr 2 (This of course is the formula for the area of a circle.) Again, you can work this out on the Excel spread sheet entitled Segment 1.5 calculation for Slide 14 to 16. xlsx (Slide 15) Now, having the volume, (presuming a radius of a sphere approximating the Tootsie Pop) we calculate the mass assuming the density of pure uranium metal and we get that the Tootsie pop s worth of U-235 metal is 220 grams or about a half a pound. (Slide 16) Now we have the information we need to calculate the number of atoms in the uranium mass with the size of a Tootsie Pop. We get 5.63 x atoms. We ll use this information later as we contemplate the amount of energy we can get from this amount of uranium (Slide 17) Let s do one more problem and that is to calculate the cost of nuclear fuel. (Slide 18) Fuel cost is a minor cost factor for nuclear power. Increasing the price of uranium would have little effect on the overall cost of nuclear power. Doubling the cost of natural uranium would increase the total cost of nuclear generated electricity by about

4 5 percent. If the cost of natural gas were doubled, the cost of gas-fired electricity would increase by about 60 percent. (Slide 19) Here is a figure created in 2006 by the Uranium Institute that shows the impact of fuel costs on the cost of generated electricity. Note that the base case shows that portion of the cost of electricity due to the cost of the fuel used to make the electricity. Here you see that a 50% increase in the cost of the fuel would make hardly any change in the cost of nuclear generated electricity while the cost of electricity would change quite dramatically for coal and natural gas. But keep in mind that the cost of fuel is not the only story. The cost of building a nuclear plant is many times the costs of building either a coal plant or a natural gas plant. (Slide 20) Let s estimate the cost of 1 kg of uranium enriched to 3 wt% in the isotope U Again, you can work this out on the Excel spread sheet entitled Segment 1.5 Fuel Cycle Cost Calculator.xlsx We are going to use the web to find out the current prices of various things in the nuclear fuel cycle. And the web address that we will access is (Slide 21) Here is what that web page looks like. It s updated weekly. Here you can get the current price of U3O8, the price of converting the U3O8 to uranium hexafluoride (UF6), and the cost of enrichment shown as a SWU price (remember we talked previously about a SWU (pronounced swoo ) or Separative Work Unit. (Slide 22) And here is a historical price chart for the cost of U3O8. You can see that the price maxed out at about $136 per pound of U3O8 and currently the price is down around $50 per pound. (Slide 23) So the last time I did this problem, the price of yellowcake (purified U3O8) was $62.50 per pound or $ per kilogram. Now we need to take into account the cost of converting the U3O8 to UF6 to use in the enrichment facility. (Slide 24) So we go again to our web link and find that the cost of conversion varies with time. (Slide 25) At the time I did the problem before, the conversion cost was $12.50 per kilogram of UF6. Now we must factor in the cost of enrichment. We will assume that the entering UF6 is composed of natural uranium which has a U-235 concentration of weight percent. Let s assume that we want product from the enrichment plant that is enriched to 3 weight percent U-235 and that we will continue to extract U-235 from the enrichment stream until the tails the waste if you like have only 0.2% enrichment. Now note that if uranium is cheap and enrichment costs are high, we might even settle for a higher tails say 0.3%. On the other hand, if uranium is expensive and enrichment costs are low, we might take even more U-235 out of the natural uranium and settle on 0.2% tails. It s an optimization problem that you would understand well if you were responsible for buying uranium fuel enriched in U-235 to use in a nuclear reactor.

5 (Slide 26) At this point we re going to finesse the business of calculating the amount of natural uranium feed material and the Separative Work Units we need to get uranium enriched to 3% from natural uranium enriched to %. This eye chart is from Knief s book and we ll skip right to the next slide that shows what we want to take off of this table. (Slide 27) Here is an enlargement that s a little bit easier to see and we also have here the equations that we could use to calculate the amount of feed you need to get a kg of product and the amount of separative work you need to enrich the uranium to the desired product. If we have time, we will use those formulas. Let s take the time right now. The first formula we want to learn how to use is Equation 17-4 from Knief Eq.17 4 M f = p t = M p f t = It says that the mass of feed, M f, over the mass of product, M p, is equal to the enrichment of the product, p, minus the enrichment of the tails from the enrichment process, t, in the numerator divided by the enrichment of the feed, f, minus the enrichment of the tails from the enrichment process, t, in the denominator. If you use 3.0 % for the enrichment of the product, 0.2 % for the enrichment of the tails (or the waste if you like), and 0.711% for the enrichment of the feed (natural uranium) you get the result that you need kgs of natural uranium feed material to get 1 kg of product enriched to 3% in U-235. The calculation of the amount of SWUs is a little more complex for those of you who don t regularly use the natural log of a number (but it s easy to look up on the web or your hand calculator or your computer). We begin by getting the Value function V of each quantity of enrichment. For example, V(χ) = (2χ 1)ln χ where the Greek 1 χ letter χ denotes the enrichment level we wish to evaluate. Let s consider the evaluation of χ for the product enrichment, p = 3.0. We find that V(0.03) = (2 *0.03 1)ln evaluates to By similar calculations, we find that V (0.002) = and V (0.0071) = Putting all these numbers into Equation 17-5 of Knief we get SWU = V(p) + V(t)(F 1) V( f )F SWU = *( ) *5.479 = We ll use these relationships later in other calculations. (Slide 28) And here is a further enlargement of the figure from Knief s text something we can almost read. We see that in order to get 1 kg of uranium enriched to 3% assuming 0.2% tails, we require kg of natural uranium feed as we calculated and kgswu to get 1 kg of product (uranium enriched to 3%)

6 (Slide 29) Whew! Now we can go on with our calculation of fuel cost for a nuclear reactor. This slide shows the information that we have developed so far. On top of the cost of yellowcake and the cost of conversion, we recognize that to get 1 kg of uranium enriched to 3% in U-235 we need kg of natural uranium feed material and kgswu. Now we have to find the cost of a kg SWU. (Slide 30) So we go again to our web link to current costs and we see here that the SWU cost is around $155 per kgswu. (Slide 31) Moving on the next slide we combine our cost of yellowcake and conversion (4 th line on this slide) of $ per kg of natural uranium times the number of kg of feed we need (5.479) to get the cost of UF6 (natural uranium feed) as $ Now to get that kg of natural UF6 into 1 kg of uranium enriched to 3% we have to use kg of SWU at a cost of $155 per kg SWU so times $155 gives us $ for the cost of enrichment. Total cost is now $1, to get one kg of uranium enriched to 3% in U-235. (Slide 32) Finally, let s add in the cost of fabricating that one kg of uranium enriched to 3% in U-235 into fuel pellets that are formed into rod which are assembled into fuel assemblies. I ll just pull a ball-park number out of the air of $460 per kg. (Fuel manufacturers don t advertise those prices to college professors.) Final result: a total cost of $2, to get one kg of fuel material ready to put into a reactor. Of course, this reactor has to have many kg of fuel to make many megawatts of electrical energy. Maybe we ll extend this calculation later to figure the cost per kilowatt hour. (Slide 33) I think this is a good place to stop. We ve covered a lot of ground related to the nuclear fuel cycle. There s much more of course but we can t do everything in one short Coursera course. Now you can do this problem on the web by going to this URL: (Slide 34) Thank you for your attention. Here s the problem I want you to be able to do. Word Count = 3,049 Time estimate = 21 minutes

7 Implied Objectives: Get to the point where we can do the Tootsie Pop problem Get to the point where we can do the Uranium Mining Problem Get to the point where we can do the Iranian enrichment exercise Get to the point where we can do the fuel cost problem Question Bank for Segment 1.5 (self grading only) 1. What is the mass of U-235 in a metric ton (1,000 kg) of uranium ore assuming the total uranium is 12 wt% of the ore. a) 0.85 kg U-235 b) 1.78 gm U-235 c) 17.8 gm U-235 d) 85 gm U-235 Answer: Correct choice is (a) (1000 kg U ore)(12 kg U/100 kg ore)( kg U-235/kg U) = 0.85 kg U By use of atomic weights of Uranium ( gms/mole) and Oxygen ( gms/mole) what is the weight fraction of U in U 3 O 8?: a b c d. 1.0 Answer: c U 3 O 8 has 3*U+8*O=3(238)+8(16)=842 U 3 has 3*U=3(238)=714 so the fraction of U in U 3 O 8 = 714/842= The masses of U-235 and U-238 per metric ton (t, tonne) of uranium ore, assuming the total uranium is 0.5% of the ore is: a) 35.6 g U-235, kg U-238 b) g U-235, 35.6 kg U-238 c) 5 kg U-235, 95 kg U-238 d) 0.7 kg U-235, 99.3 kg U-238 (1,000 kg)(0.005)=5 kg uranium of the 5 kg, is U-235, or (5)( )= kg of 3.56 gm; the rest is essentially all U-238 or = kg

8 4. Presume the total weight of UO 2 fuel in a PWR is 150 metric tons (1 metric ton = 1,000 kg). If the enrichment of U is 4 wt% U-235, how many tons of uranium ore (with wt% U-235) must be mined if the total uranium in the ore is 0.5 wt%? (This is kind of involved, so think through the steps carefully) a) 150MT b) 170 MT c) 750 MT d) 150,000 MT Answer: d. about 148,776 MT of ore (or 150,000 metric tons ore) Consider 150 metric tons (MT) of UO 2. The fraction of the UO 2 that is just U is (Molecular Wt of U)/(Molecular Wt UO2) = (238)/( ) = So the MT of U in the UO 2 fuel is (0.8815)(150 MT) = MT U of which 4 wt% is U-235. So the amount of U-235 in the core is (0.04)( MT) = 5.29 MT U-235. How much natural uranium must be mined to get 5.29 MT U-235? Natural uranium consists of wt% U-235; hence, to get 5.29 MT of U-235 we must mine enough ore to give (5.29 kg/ MT U-235 per kg of U natural ) = MT U natural. If the U natural content of the ore is 0.5 wt%, we must mine MT/0.005 = 148,776 MT of ore. 5. What is the mass of U-235 in a metric ton (1,000 kg) of uranium ore assuming the total uranium is 12 wt% of the ore. a) 0.85 kg U-235 * b) 1.78 gm U-235 c) 17.8 gm U-235 d) 85 gm U-235 o (1000 kg U ore)(12 kg U/100 kg ore)( kg U-235/kg U) = 0.85 kg U-235

9 6. Calculate the mass of U-235 in 1 kg of uranium assuming the U-235 is enriched to 1 wt%. a) kg U-235 b) gm U-235 c) kg U-235 d) 10 gm U-235 Answer: d 7. We wish to enrich natural uranium (0.711% U-235) to 4% in the isotope U-235. Presume the tails from the enrichment cascade will be 0.3% U-235. How many kilograms of natural uranium feed is needed to provide one kilogram of enriched product. Assume no losses in the enrichment process. a. 4.0 b c d Answer: d The first formula we want to learn how to use is Equation 17-4 from Knief Eq.17 4 from Knief M f = p t = M p f t = How many SWU s will be needed to get one kg of enriched product out of natural uranium in the previous problem? a b c d Answer: b The formula we must use is shown on Slide 27 of the Power Point for Segment 1.5. It is: Eq.17 5 from Knief SWU = V(p) + V(t)(F 1) V( f )F Eq.17 6 from Knief V(χ) = (2χ 1)ln χ 1 χ (X) 2*X-1 (X)/(1-X) ln (X)/(1-X) V(X) p t f

10 SWU= ( ) (9.002)= SWU 9. (a) We read that a nation has enriched natural uranium to 20% in U-235. How many kg of natural uranium feed did it take to get 1 kg of 20% EUP (Enriched Uranium Product)? How many kg SWU did it take them to get the 1 kg of 20% EUP? a kg natural U feed and kg SWU/kg EUP to get 1 kg of 20% EUP. b kg natural U feed and kg SWU/kg EUP to get 1 kg of 20% EUP. c kg natural U feed and kg SWU/kg EUP to get 1 kg of 20% EUP. d kg natural U feed and 27.5 kg SWU/kg EUP to get 1 kg of 20% EUP. 9. (b) Now that they have the 20% EUP, how many kg of this material will it take to get 1 kg of 97% EUP (weapons grade)? How many more kg SWU will it take them to get the 1 kg of 97% EUP? You may presume there is sufficient 20% EUP available. a kg of 20% EUP and kg SWU/kg EUP to get 1 kg of 97% EUP b kg of 20% EUP and kg SWU/kg EUP to get 1 kg of 97% EUP c kg of 20% EUP and kg SWU/kg EUP to get 1 kg of 97% EUP d kg of 20% EUP and kg SWU/kg EUP to get 1 kg of 97% EUP 9. (c) Are they at least half way there with enrichment work to the 1 kg of 97% EUP? a. Yes; they are almost 90% of the way there in terms of enrichment work. b. Yes, they are a little over 50% of the way there in terms of enrichment work. c. No, they are a little under 50% of the way there in terms of enrichment work. d. No, they are a little over 510% of the way there in terms of enrichment work. We want kg of the 20% EUP to get 1 kg of 97% EUP It costs kg natural U and SWU to get 1 kg of 20% EUP So we ve got to do this times x kg U nat = kg U nat required for 1 kg 97% EUP x SWU = SWU spent to get the necessary 20% EUP So the investment to get kg of 20% EUP is kg natural U and SWU To take that the rest of the way to get 1 kg 97% EUP, we need more SWU for a total of = SWU So we ve spent SWU to get the kg of 20% EUP and we only need to get the rest of the way. (almost 90% of the way)

11 Research Question 10: Part a: The generation of electricity from a typical 1000 MW(e) nuclear power station produces approximately tonnes of high level solid packed waste per year. a. 30 tonnes b. 12,000 tonnes c. 300,000 tonnes d. unknown Part b: By way of comparison a 1000 MW(e) coal plant produces some tonnes of ash alone per year, containing among other things radioactive material and heavy metals which end up in landfill sites and in the atmosphere. a. 30 tonnes b. 12,000 tonnes c. 300,000 tonnes d. unknown Answer: c Source: Source: