3/29/04 Nuclear Energy Economics andPolicy Analysis1The Economics of the NuclearFuel CycleMarch 29, 20043/29/04 Nuclear Energy Economics andPolicy Analysis23/29/04 Nuclear Energy Economics andPolicy Analysis43/29/04 Nuclear Energy Economics andPolicy Analysis63/29/04 Nuclear Energy Economics andPolicy Analysis9Enrichment Plant Material BalanceTails: W, xwFeed: F, xFProduct: P, xPMaterial balance on U: F = P + WMaterial balance on U-235: FxF = PxP + WxWF = Pxp- xWxF- xWÈÎ͢˚˙and for xP= 0.03, xF=0.00711, xW= 0.002, F/P = 5.53/29/04 Nuclear Energy Economics andPolicy Analysis103/29/04 Nuclear Energy Economics andPolicy Analysis133/29/04 Nuclear Energy Economics andPolicy Analysis14Source: US Energy Information Agency (http://eia.doe.gov/cneaf/nuclear/page/nuc_reactors/pwr.html)3/29/04 Nuclear Energy Economics andPolicy Analysis153/29/04 Nuclear Energy Economics andPolicy Analysis16Source: US Energy Information Agency (http://eia.doe.gov/cneaf/nuclear/page/nuc_reactors/pwr.html)3/29/04 Nuclear Energy Economics andPolicy Analysis17PWR Batch Refueling Scheme(Batch fraction = 1/3)Cycle No. Batch #I 1 2 3 Startup coreII 2 3 4 (new)III 3 4 5 (new)IV 4 5 6 (new)VDischargedafter 1 cycleBd~Bc*Dischargedafter 2cyclesBd~2 Bc*Dischargedafter 3cyclesBd~3 Bc**Useful approximation. In practice, needto track Bd with computer codes3/29/04 Nuclear Energy Economics andPolicy Analysis18If the batch fraction is 1/n, under steady state conditions each batch remains in the core for n cycles.Similarly, at steady state the energy produced by all n batches in the core during one cycle is equal to theenergy produced by one batch during its total residence time in the core (i.e., n cycles.).Thus the total electrical energy produced by a given batch during its in-core lifetime at steady state is:Eb(kwh(e)/batch) = 8766 (hrs/yr) x CF x K (kwe) x Tc(yr)where:CF = cycle average capacity factor (including refueling downtime)K = plant rating (kwe)Tc= cycle length (yrs) including downtimeWe can also write that the energy produced per batch is:Eb(kwhr(e)/batch) = Bd(MWD(th)/MTHM) x 24 (hr/day) x 1000 (kw/MW) x h x P (MT)where:Bd= discharge burnup of the fuel (MWD(th)/MT of heavy metal)h = thermodynamic efficiency (Mwe/MW(th))P = batch fuel inventory (MT of heavy metal)Also,Tb=nTcBd = n BcAnd P, the fuel inventory per batch = Total core inventory/nEnergy from a fuel batch3/29/04 Nuclear Energy Economics andPolicy Analysis193/29/04 Nuclear Energy Economics andPolicy Analysis20Material balance on the front end of the cycleAssume:K = 1000 MWeTc= 1.5 yearsCF = 90%h = 33%n=3In-core fuel inventory = 89.4 MTHMHence we can calculate:• Energy output per batch, Eb,(at steady state)• Mass of fuel (as heavy metal), P, per batch• Discharge burnup of spent fuel, Bd(MWDth/MTHM)3/29/04 Nuclear Energy Economics andPolicy Analysis21xp= 0.41201 + 0.11508 •n + 12n•BdÊËÁˆ¯˜+ 0.00023937 •n + 12n•BdÊËÁˆ¯˜2Source: Zhiwen Xu, NED DoctoralDissertation, 2003where n = number of batches(valid for xp< 20%)Approximate Correlation: Initial enrichment (xP)vs.discharge burnup (Bd)Image removed due to copyright considerations.3/29/04 Nuclear Energy Economics andPolicy Analysis22Material balance on the front end of the PWR fuel cycle(Basis: 1 steady state batch; 1000 MWe PWR; thermal efficiency = 33%; 90% capacity factor; 18-monthrefueling cycle; batch fraction = 1/3; discharge burnup = 50,000 MWD(th)/MT))FABRICATIONCONVERSIONCONVERSIONMINE & MILL1000 MWePWRCap. fact.= 90%Therm. eff.=33%Discharge burnup =50,000 MWD/MTHM29.8 MTHM of LEU4.51% U-2351% losses29.8/0.99 = 30.1MTHM of U30.1 x (270/238)= 34.15 MT UO20.5% lossesUF634.15.995¥352270= 44.74 MT UF6Product, PXP=4.51%Tails, WXW=0.3%458.3 - 44.74 =413.6 MT UF6Feed, FXF=0.711%F =4.51- 0.3.711 -0.3¥ 44.74= 458.3 MT nat. UF60.5% lossesYellowcake(U308)458.3.995¥8423352= 367.3MT U3O8≡ 367.3 ¥2388423= 311.4MT UNat.UF6Molecular WeightsU3O8842UF6352UO2270U 238Mill tailingsUranium oreSpent