66. A nuclear power plant uses 87 tons of uranium in its core. The first core co
ID: 2269065 • Letter: 6
Question
66. A nuclear power plant uses 87 tons of uranium in its core. The first core consists of three equal batches with enrichments of 2.5%, 2.9%, and 3.1%. For subsequent cycles, the reactor uses batches with enrich- ments equal to 2.6% and 3.2% in alternate years. One-third of the core is discharged each year and is replaced by new fuel. Assume a 30-year lifetime and 0.2% tails during the life of the plant. Calculate: (a) The number of SWUs needed for the first core, the number of SWUs every year after the first cycle, and the total SUWs over the life of the plant. (b) The amount of U,Os and UF6 needed for the first core, every year after, and the total U,Os and UFs over the life of the plant. Give your result in metric tons of uranium, UF6, and UOs. core and the cost of the fuel per year for the life of the plant. U3Os: $65.00/kg Conversion: $7/kg; 0.5% loss SWU price: $105 Fabrication/transportation: $250/kg; 0.8% loss (c) Assume the following prices and calculate the cost of the firstExplanation / Answer
Ans :- In nuclear power plant 87 tons Uranium .so that as on problem we are use these equation of SWU of the first core of three equal batches so on the slide found in 30 years of the lifetime and 0.2% trail. that for NP = 2.5%, 2.6 % SWU need if NT = 2.9%, or 3.2 % SWU if NT = 3.1%. so we are again Recall the value of NF for natural uranium is 80 tons. so our ratio SWU-to-product, SWU/P.
Now we can arrange the given formula by dividing by rate feed is denoted as F.
F = P+T 1=P /F + T/F
FNF = PNP + TNT [Rightarrow] NF =( P/F) NP + (T/F) NT
Now be add 2nd and 4th equation of above result for P/F.
NF = P/F NP + (1-P/F) NT [Rightarrow] P/F = NF - NT/ NP- NT
So that is the 1st result. we are handling the reverse of the equation for P/F so gives the result for F/P.
F/P = 1+T/P = NP -NT /NF - NT
SWU/F is distributed equation for SWU by F to and use for P/F equation. now we derived to found the N term of the equation for SWU/F.
SWU/F = P/F V(NP) + T/F V(NT)- V(NP) = NF -NT/ NP - NT V(NP) + ( 1 - (NF - NT / NP - NT)) V(NT) - V(NF)
SWU/F = (NF - NT / NP - NT) V(NP) + (NP -NT- NF + NT /NP - NT) V(NT) =- NP - NT / NP - NTV(NF)
SWU/F = (NF - NT ) + (NP -NF) V (NT) - (NP - NT) V(NF) / NP - NT
WU/F = V (NP) + T/P V(NT) - F/P V(NP) = V(NP) + (1- NP-NT / NF - NT) V(NT) - NP - NT / NF- NT V(NF)
that the concentrations of nuclear enrichment must be explained as a fraction also not as %
v
v(n)
V9 (0.03) = [ 2(0.03)) -1] log (0.03 / 1-0.03) = 3.268
V( 0.0071) = [2(00071) - 1] log (0.0071 / 1-0.0071) = 4.780
V( 0.0025) = [2(0.0025) - 1] log (0.0025 / 1-0.0025) = 5.959
V( 0.005) = [2(0.0015) - 1] log (0.0015 / 1-0.0015) = 6.481
so that proper fraction
SWU / F = (NF -NT) V (NP) + (NP - NF) V (NT) - (NP- NT) v (NF)
= (0.0071 - 0.0025) (3.2) + (0.2 - 0.0071)( 5.989) - (0.03 -0.025) (4.78) / 0.03 -0.0025= 0.638
So that the above equation of four number of SWU needed in for first core 3.268, 4.780 ,5.959, 6.481 and SWU after the year of first cycle& total life time 0.638 %
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