An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
(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 0.711 weight % of uranium, so 0.00711 times 10 kg is 0.0711 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 6.022 x 1023 (Avogadro's number). For example: a mole of
uranium has 6.022 x 1023 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, 6.022×1023, 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 6.022 x 1023.
(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.
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
(Slide 6) Now let’s ask “what is the molecular mass of uranium oxide, U3O8 ?
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 1023) 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, U3O8. How many kgs of Uranium ore is in 100 kg of U3O8?
We just created the molecular mass of U3O8. 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 714+128=842 gms.
So the weight fraction of U in the U3O8 is (714 gms/842 gms)=0.848. So if we have 100
kg of U3O8 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 UO2
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 UO2? Here we have to recognize that not all of the UO2 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 0.881 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 UO2 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)(132.23 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 0.711 wt% U-235; hence, to get 5.29 MT of U-235 we must
mine enough ore to give (5.29 kg/0.00711 MT U-235 per kg of Unatural) = 743.88 MT
Unatural.
If the Unatural content of the ore is 0.5 wt%, we must mine 743.88 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
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
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 6.022 x
1023 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 = 238.0289 (this number comes directly off the Chart of
the Nuclides). This means 1 mol of Uranium = 238.0289 grams Thus, the molar mass of
U is 238.0289 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.
4
Calculate the volume with the formula Volume = π r 3 where r is the radius of the
3
sample. (You’d be surprised how many undergraduate students calculate the volume as
π r 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 1023 atoms. We’ll use this
information later as we contemplate the amount of energy we can get from this amount of
uranium -235.
(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
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
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 U235. 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
http://www.uxc.com/review/uxc_Prices.aspx
(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 $137.50 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
0.711 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.
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
(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 0.711 %. 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
Mf p−t
3.0 − 0.2
=
=
= 5.479
M p f − t 0.711− 0.2
It says that the mass of feed, Mf, over the mass of product, Mp, 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 5.479
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
χ
where the Greek
each quantity of enrichment. For example, V ( χ ) = (2 χ − 1)ln
1− χ
letter χ denotes the enrichment level we wish to evaluate. Let’s consider the evaluation
0.03
of χ for the product enrichment, p = 3.0. We find that V (0.03) = (2 * 0.03 − 1)ln
1− 0.03
evaluates to 3.2675. By similar calculations, we find that
V (0.002) = 6.1878 and V (0.0071) = 4.8689 .
Putting all these numbers into Equation 17-5 of Knief we get
SWU = V ( p) + V (t)(F − 1) − V ( f )F
SWU = 3.2675 + 6.1878 *(5.479 − 1) − 4.8689 * 5.479 = 4.306
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 5.479 kg of natural uranium feed as we calculated and 4.306
kgSWU to get 1 kg of product (uranium enriched to 3%)
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
(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 5.479 kg of natural uranium feed material and 4.306
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
(4th line on this slide) of $174.65 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 $956.91.
Now to get that 5.479 kg of natural UF6 into 1 kg of uranium enriched to 3% we have to
use 4.306 kg of SWU at a cost of $155 per kg SWU – so 4.306 times $155 gives us
$667.43 for the cost of enrichment.
Total cost is now $1,624.34 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 $2,084.34 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:
http://www.wise-uranium.org/nfcc.html
(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
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
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)(0.00711 kg U-235/kg U) = 0.85 kg U-235
2. By use of atomic weights of Uranium (238.029 gms/mole) and Oxygen (15.9994
gms/mole) what is the weight fraction of U in U3O8?:
a. 0.15
b. 0.38
c. 0.85
d. 1.0
Answer: c
U3O8 has 3*U+8*O=3(238)+8(16)=842
U3 has 3*U=3(238)=714 so the fraction of U in U3O8 = 714/842=0.848
3. 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, 4.964 kg U-238
b) 4.964 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
Answer: a
(1,000 kg)(0.005)=5 kg uranium
of the 5 kg, 0.00711 is U-235, or (5)(0.00711)=0.0356 kg of 3.56 gm; the rest is
essentially all U-238 or 5-0.0356 = 4.964 kg
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
4. Presume the total weight of UO2 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
0.711 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 UO2.
• The fraction of the UO2 that is just U is (Molecular Wt of U)/(Molecular Wt UO2)
= (238)/(238 + 32) = 0.8815.
• So the MT of U in the UO2 fuel is (0.8815)(150 MT) = 132.23 MT U of which 4
wt% is U-235.
• So the amount of U-235 in the core is (0.04)(132.23 MT) = 5.29 MT U-235.
• How much natural uranium must be mined to get 5.29 MT U-235?
• Natural uranium consists of 0.711 wt% U-235; hence, to get 5.29 MT of U-235
we must mine enough ore to give (5.29 kg/0.00711 MT U-235 per kg of Unatural) =
743.88 MT Unatural.
• If the Unatural content of the ore is 0.5 wt%, we must mine 743.88 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
Answer: a
o (1000 kg U ore)(12 kg U/100 kg ore)(0.00711 kg U-235/kg U) = 0.85 kg
U-235
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
6. Calculate the mass of U-235 in 1 kg of uranium assuming the U-235 is enriched to 1
wt%.
a) 0.0711 kg U-235
b) 9.875 gm U-235
c) 0.711 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. 5.479
c. 7.436
d. 9.00
Answer: d
The first formula we want to learn how to use is Equation 17-4 from Knief
Eq.17 − 4 from Knief
Mf p−t
4.0 − 0.3
=
=
= 9.002
M p f − t 0.711− 0.3
8. How many SWU’s will be needed to get one kg of enriched product out of natural
uranium in the previous problem?
a. 4.110
b. 5.277
c. 8.609
d. 9.002
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 ) = (2 χ − 1)ln
p
t
f
(X)
0.04
0.003
0.0071
χ
1− χ
2*X-1
-0.9200
-0.9940
-0.9858
(X)/(1-X)
0.04116667
0.00300903
0.00716091
ln (X)/(1-X)
-3.1780538
-5.8051385
-4.9191176
V(X)
2.9238
5.7713
4.8689
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
SWU=2.9238+5.7713(9.002-1)-4.8689(9.002)=5.2765 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. 47.93 kg natural U feed and 38.315 kg SWU/kg EUP to get 1 kg of 20% EUP.
b. 47.93 kg natural U feed and 47.93 kg SWU/kg EUP to get 1 kg of 20% EUP.
c. 38.315 kg natural U feed and 38.315 kg SWU/kg EUP to get 1 kg of 20% EUP.
d. 38.315 kg natural U feed and 27.5 kg SWU/kg EUP to get 1 kg of 20% EUP.
Answer: a
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. 4.909 kg of 20% EUP and 21.74 kg SWU/kg EUP to get 1 kg of 97% EUP
b. 4.909 kg of 20% EUP and 27.14 kg SWU/kg EUP to get 1 kg of 97% EUP
c. 21.74 kg of 20% EUP and 21.74 kg SWU/kg EUP to get 1 kg of 97% EUP
d. 21.74 kg of 20% EUP and 21.14 kg SWU/kg EUP to get 1 kg of 97% EUP
Answer: a
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.
Answer: a
We want 4.909 kg of the 20% EUP to get 1 kg of 97% EUP
It costs 47.93 kg natural U and 38.315 SWU to get 1 kg of 20% EUP
So we’ve got to do this 4.909 times
4.909 x 47.93 kg U nat = 235.29 kg U nat required for 1 kg 97% EUP
4.909 x 38.315 SWU = 188.09 SWU spent to get the necessary 20% EUP
So the investment to get 4.909 kg of 20% EUP is 235.29 kg natural U and 188.09
SWU
To take that the rest of the way to get 1 kg 97% EUP, we need 21.74 more SWU for a
total of 21.74 + 188.09 = 209.83 SWU
So we’ve spent 188.09 SWU to get the 4.909 kg of 20% EUP and we only need 21.74
to get the rest of the way. (almost 90% of the way)
An Introduction to Nuclear Power and a Grand Tour of the Nuclear Fuel Cycle
Module 1 – Segment 6
The Nuclear Fuel Cycle – Calculations
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
Answer: a
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: http://en.wikipedia.org/wiki/Radioactive_waste
Source: http://www.iaea.org/Publications/Factsheets/English/manradwa.html