核工原理第四章作業參考解答
DESCRIPTION
nuclear engineeringTRANSCRIPT
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1
4.8
P=250 MWt (thermal ) w: (
) C=0.88
fission U Example 4.1 converter 235U
2.7 1021(/)
6.02 1023(/) 235(
) = 1.05P(
)
239Pu
1.05
0.001
365
239235
= 85.75 kg/year
4.11 238U 239Pu tDe=10 years specific power = 0.6 MW/kg(Pu)
= Gwm
m(t) = 0 = 20
=ln 2
= P/m= 0.6 MW/kg w= 1.067 (/ MW-Day) =
ln 2
=ln 2
() 1.05(
) ( ) 365(
) 0.001(
)
=
ln 210 1.05 0.6 365 0.001
= 0.301
4.18 3000 MW 3000 MW 365 Day = 3000 365 MWD 1.05 235U 1 MWD 1.23 235U
3000 365 MWD 1.23(
) = 1.35
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2
4.25 10000 kg 3 w/o UF6 MF = MP +MT xFMF = xP MP + xTMT
MF =
(a)
MF = 0.03 0.002
0.00711 0.002 10000 = 54795
(b) SWU = MPV(xP) + MTV(xT) - MFV(xF)
V(x) = (1 2x) ln 1
MT = MF - MP SWU = MP [V(xP) - V(xT)] - MF [V(xF) - V(xT)]
V() = V(0.03) = (1 2 0.03) ln 1 0.03
0.03 = 3.268
V() = V(0.002) = (1 2 0.002) ln
1 0.0020.002
= 6.188
V() = V(0.00711) = (1 2 0.00711) ln
1 0.007110.00711
= 4.869
SWU = 10000(3.268 6.188) 54795(4.869 6.188) = 43075 kg 205 USD 130.75 Cost = 43075 130.75 = 5632056 USD
4.30 (a) MP = 25 kg xP = 90 w/o xF = 20 w/o xT = 0.2 w/o
MF = 0.9 0.0020.2 0.002
25 = 113.38
(b)
V() = V(0.9) = (1 2 0.9) ln 1 0.9
0.9 = 1.758
V() = V(0.002) = (1 2 0.002) ln
1 0.0020.002
= 6.188
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3
V() = V(0.2) = (1 2 0.2) ln
1 0.20.2
= 0.832
SWU = 25(1.758 6.188) 113.38(0.832 6.188) = 496.51 kg (c) From natural uranium to 20 w/o: MP = 25 kg xP = 20 w/o xF = 0.72 w/o xT = 0.2 w/o
MF = 0.2 0.002
0.00711 0.002 25 = 968.69
V() = V(0.2) = (1 2 0.2) ln
1 0.20.2
= 0.832
V() = V(0.002) = (1 2 0.002) ln
1 0.0020.002
= 6.188
V() = V(0.00711) = (1 2 0.00711) ln
1 0.007110.00711
= 4.869
SWU = 25(0.832 6.188) 968.69(4.869 6.188) = 1143.8 kg SWU(natural to 20 w/o) > SWU( 20 w/o to 90 w/o)