1.6 Fuel Cycle Coursera Calculations Presentation
Transcript of 1.6 Fuel Cycle Coursera Calculations Presentation
Nuclear Engineering Program
A Look at Nuclear Science and Technology
Larry Foulke
Module 1.6 Grand Tour of the Nuclear Fuel Cycle – Supplemental File of Calculations
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)
Mass and Energy Calculations
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)
Answer:
1000 kg of ore x (0.01 kg U/kg ore) = 10 kg U U-235 is 0.711 wt% of U so mass of U-235 = 0.00711 x 10
kg U = 0.0711 kg or 71.1 grams
Mass and Energy Calculations
See Segment 1.6 calculation for Slide 3. xlsx
• Let n represent the number of moles • A mole is the number of atoms or
molecules of a substance or element equal to 6.022 x 1023 (Avogadro's number, NAV).
For example: a mole of uranium is roughly 6.022 x 1023 uranium atoms.
Concept of a Mole
Image Source: See Note 1
• The number of moles, n, of a mass, m, of material is given by:
n moles = mass of material, m, in grams divided by the molecular mass M (gm/mole)
• M – molecular mass (weight) in gm/mole – For uranium, MU ≈ 238 gm/mole
– For oxygen, MO ≈ 16 gm/mole
– The molecular mass of an isotope is just its atomic weight
Concept of a Mole and Molecular Mass
nmoles= m (grams)
M grams / mole( )
• What is the molecular mass of MU3O8 ? • We combine 3 moles of uranium with 8 moles of
oxygen to get:
• So the molecular mass of U3O8 is 842 gm/mole
Concept of a Mole
3moles( ) U gm /mole( )+ 8moles( ) Ogm /mole( )= 3( ) 238( )+ 8( ) 16( ) = 842gm /mole
See Segment 1.6 Calculation for Slide 6.xlsx
Uranium when mined is often in the form of Uranium Oxide, U3O8. How many kg of Uranium is in 100 kg
of U3O8?
Answer: We just created the molecular mass of U3O8
We took 3 parts of uranium at (3 moles)(238 gms/mole)= 714 grams
We took 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
Mass and Energy Calculations
See Segment 1.6 Calculation for Slide 6.xlsx
• 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).
• Choices for answers
a) 150MT
b) 170 MT
c) 750 MT
d) 150,000 MT
Mass and Energy Calculations
See Segment 1.6 Calculation for Slides 8 – 12.xlsx
• 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)
• Answer: about 148,776 MT of ore (or 150,000 metric tons ore)
How did we get that?
Mass and Energy Calculations
See Segment 1.6 Calculation for Slides 8 – 12.xlsx
• What is the molecular mass of MUO2 ? • We combine one mole of uranium with two moles of
oxygen to get:
• So the molecular mass of UO2 is 270 gm/mole and the fraction of the mass that is U is (238/270)=0.881
• So there is (0.881)(150 MT)= 132 MT of U in the UO2
Concept of a Mole
1moles( )U gm / mole( )+ 2moles( )O gm / mole( )= 1( )238( )+ 2( )16( )= 270 gm / mole
See Segment 1.6 Calculation for Slides 8 – 12.xlsx
• As implied in the problem statement – kind of involved. Let’s look at the solution.
• Consider 132 metric tons (MT) of U in 150 MT of UO2.
• So the 132 MT U is enriched to 4 wt% in U-235.
• So the amount of U-235 in the core is • (0.04)(132.23 MT) = 5.29 MT U-235.
Mass and Energy Calculations
See Segment 1.6 Calculation for Slides 8 – 12.xlsx
• 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) = 743.87 MT Unatural.
• If the Unatural content of the ore is 0.5 wt%, we must mine 743.87 MT/0.005 = 148,773 MT of ore.
Mass and Energy Calculations
See Segment 1.6 Calculation for Slides 8 – 12.xlsx
Now we are in a position to calculate the number of atoms in a mass of material.
Here is a formula that is used often:
Or
Number of Atoms Or Number Density
N (atoms / volume) = ρ gm
cm3
⎛
⎝⎜⎞
⎠⎟x 1
Mmolegm
⎛
⎝⎜⎞
⎠⎟x N AV
atomsmole
⎛
⎝⎜⎞
⎠⎟
N (atoms) = m gm( )x 1
Mmolegm
⎛
⎝⎜⎞
⎠⎟x N AV
atomsmole
⎛
⎝⎜⎞
⎠⎟
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 (Volume =
• Calculate the mass assuming the density of pure uranium metal
• Calculate the number of atoms in the uranium mass
Mass and Energy Calculations
Volume = 43π r( )3 where r = radius
Get the density of uranium metal off the web as 19.1 gm/cm3
See Segment 1.6 Calculation for Slides 14 – 16.xlsx
• Estimate the volume of the Tootsie Pop (Volume =
Presume r = 2.8 cm/2 = 1.4 cm so
• Calculate the mass assuming the density of pure uranium metal
So the mass of a Tootsie Pop’s volume of pure uranium-235 is
Mass and Energy Calculations
Volume = 43π 1.4cm( )3 = 4
3(3.14) 2.744( )=11.5 cm3
Get the density of uranium metal off the web as 19.1 gm/cm3
mass = 19.1 gm
cm3
⎛
⎝⎜⎞
⎠⎟11.5cm3( )≅ 219.5gms U − 235
See Segment 1.6 Calculation for Slides 14 – 16.xlsx
Now the final step using our formula
• Calculate the number of atoms in the uranium mass
Mass and Energy Calculations
N (#) = 220 gmsU − 235
235gmsmole
x 6.022x1023( ) #mole
⎛
⎝⎜⎞
⎠⎟= 5.63x1023 atoms
See Segment 1.6 Calculation for Slides 14 – 16.xlsx
• Let’s do one more problem to calculate the cost of nuclear fuel.
Fuel Cost Impact
• Fuel cost is a minor 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 5 percent.
• If the cost of natural gas were doubled, the cost of gas-fired electricity would increase by about 60 percent.
Fuel Cost Impact
Image Source: See Note 2
Estimate the cost of 1 kg of 3 wt% enriched uranium.
Yellowcake (U3O8 cost) = Conversion cost = Enrichment SWU’s required=
See Segment 1.6 Fuel cycle cost calculator. xlsx
Estimate the cost of 1 kg of 3 wt% enriched uranium.
Image Source: See Note 3
http://www.uxc.com/review/uxc_PriceChart.aspx?chart=spot-u3o8-full
Image Source: See Note 4
Estimate the current cost of 1 kg of 3 wt% enriched uranium.
Yellowcake (U3O8 cost) = $62.50/lb U3O8= (2.2 lb/kg)($62.50) = $137.50/kg U3O8 Conversion cost = Enrichment SWU’s required=
See Segment 1.6 Fuel cycle cost calculator. xlsx
http://www.uxc.com/review/uxc_g_2yr-price.html
Image Source: See Note 4
Estimate the current cost of 1 kg of 3 wt% enriched uranium.
Yellowcake (U3O8 cost) = $62.50/lb U3O8= (2.2 lb/kg)($62.50) = $137.50/kg U3O8= Conversion cost = $12.50/kg U as UF6 Enrichment SWU’s and feed required = (See Knief, Table 17-3 and assume 0.2% tails)
=3( ) 238( )+ 8( ) 16( )⎡⎣ ⎤⎦kg U3O8
3( ) 238( )kg Ux$137.50kg U3O8 = $162.15 / kg U
See Segment 1.6 Fuel cycle cost calculator. xlsx
Enrichment Requirements (From Table 17-3 of Knief)
Image Source: See Note 5
Eq.17 − 4M f
M p
=p − tf − t
=3.0 − 0.20.711− 0.2
= 5.479
Eq.17 − 5SWU =V (p)+V (t)(F −1)−V ( f )F
Eq.17 − 6
V (χ ) = (2χ −1)ln χ1− χ
See Segment 1.6 Fuel cycle cost calculator. xlsx Image Source: See Note 5
3% enriched
0.2% tails Mf/Mp=5.479
SWU = 4.306 kgSWU/kg product
See Segment 1.6 Fuel cycle cost calculator. xlsx Image Source: See Note 5
Estimate the current cost of 1 kg of 3 wt% enriched uranium.
Yellowcake (U3O8 cost) = $62.50/lb U3O8= (2.2 lb/kg)($62.50) = $137.50/kg U3O8= Conversion cost = $12.50/kg U as UF6 Enrichment SWU’s and feed required = (See Knief, Table 17-3 and assume 0.2% tails) Natural U feed required = 5.479 kg SWU required = 4.306 kg SWU
=3( ) 238( )+ 8( ) 16( )⎡⎣ ⎤⎦kg U3O8
3( ) 238( )kg Ux$137.50kg U3O8 = $162.15 / kg U
See Segment 1.6 Fuel cycle cost calculator. xlsx
http://www.uxc.com/review/uxc_PriceChart.aspx?chart=spot-u3o8-full
Image Source: See Note 4
Estimate the current cost of 1 kg of 3 wt% enriched uranium. Yellowcake (U3O8 cost) = $62.50/lb U3O8= (2.2 lb/kg)($62.50) = $137.50/kg U3O8= Conversion cost = $12.50/kg U as UF6 Yellowcake + Conversion to UF6=$162.15+$12.50=$174.65/kgU Enrichment SWU’s and feed required = (See Knief, Table 17-3 and assume 0.2% tails) Natural U feed required = 5.479 kg SWU required = 4.306 kg SWU Cost of UF6 = (5.479 kg)($174.65/kg) = $956.91 Cost of enrichment = (4.306 kg SWU)($155) = $667.43 Total Cost = $956.91 + $667.43 = $1624.34
=3( ) 238( )+ 8( ) 16( )⎡⎣ ⎤⎦kg U3O8
3( ) 238( )kg Ux$137.50kg U3O8 = $162.15 / kg U
See Segment 1.6 Fuel cycle cost calculator. xlsx
Estimate the current cost of 1 kg of 3 wt% enriched uranium. Yellowcake (U3O8 cost) = $62.50/lb U3O8= (2.2 lb/kg)($62.50) = $137.50/kg U3O8 Conversion cost = $12.50/kg U as UF6
Yellowcake + Conversion to UF6=$162.15+$12.50=$174.65/kgU Enrichment SWU’s and feed required = (See Knief, Table 17-3 and assume 0.2% tails) Natural U feed required = 5.479 kg SWU required = 4.306 kg SWU Cost of UF6 = (5.479 kg)($174.65/kg) = $956.91/kg Cost of enrichment = (4.306 kg SWU)($155) = $667.43 Total Cost = $956.91 + $667.43 = $1,624.34 Plus cost of fuel fabrication of $460/kg for total cost of $1,624.34 + $460 = $2,084.34 for 1 kg of 3 wt% U-235
=3( ) 238( )+ 8( ) 16( )⎡⎣ ⎤⎦kg U3O8
3( ) 238( )kg Ux$137.50kg U3O8 = $162.15 / kg U
See Segment 1.6 Fuel cycle cost calculator. xlsx
Image Source: See Note 6
Image Source: See Note 6 Image Source: See Note 4
Presume that a volume of pure U-235 is the size of a Tootsie Pop. If the average annual energy consumption for a U.S. residential utility customer is approximately 11,496 kw-hrs*
per year, how many years would this amount of U-235 power the average home if all the U-235 atoms were fissioned and converted to electricity? Assume 180 MeV of recoverable
energy per fission and a conversion efficiency of 0.33 (thermal energy to electrical energy).
In this hypothetical case, what would be the approximate volume of
waste produced after all the U-235 is fissioned?
*In 2010, the average annual electricity consumption for a U.S. residential utility customer was 11,496 kWh, an average of 958 kilowatt hours (kWh) per month. See http://www.eia.gov/tools/faqs/faq.cfm?id=97&t=3
Mass and Energy Calculations
1. Public domain: http://en.wikipedia.org/wiki/File:Avogadro_Amedeo.jpg
2. Reprinted with permission from World Nuclear Organization. http://www.world-nuclear.org/info/Economic-Aspects/Economics-of-Nuclear-Power/#.UWh_Q0rSl4c
3. Reprinted with permission from Ux Consulting. http://www.uxc.com/review/uxc_prices.aspx
4. Reprinted with permission from Ux Consulting. http://www.uxc.com/review/uxc_PriceChart.aspx?chart=spot-conv-ful
Image Source Notes
5. Reprinted with permission from the American Nuclear Society. Nuclear Engineering – Theory and Technology of Commercial Nuclear Power by Ronald Allen Knief, 2nd Edition, American Nuclear Society. Copyright 2008 by the American Nuclear Society, La Grange Park, Illinois. Table 17-3.
6. WISE Uranium Project. Free for non-commercial use. http://www.wise-uranium.org/nfcc.html
Image Source Notes