The neutron flux in a reactor is 1.0x10 17 neutrons/(sm 2 ) in the moderator and

Question

The neutron flux in a reactor is 1.0x1017 neutrons/(sm2) in the moderator and cooling water.
a. If the reactor uses ordinary water as coolant and moderator how many deuterium nuclei are produced by neutron capture of 1H in one year per Kg of water?
b. If the reactor uses heavy water (D2O), how much tritium (3H) is produced in one year per Kg of heavy water? To account for the decay of tritium during the production assume that all produced tritium has been decaying for half a year.
c. Tritium is radioactive with a half-life of 12.3 years. The heavy water from part b) is now removed from the reactor. What is the activity of one Kg of the reactor water?
d. After which time has the activity fallen to 1/5th of the original activity?

Explanation / Answer

H-2 when abosrbs a neutron D-2 is produce and when D-2 abosrbs a neutron T-3 is produced

fast neutron flux moves through the moderator

1 mole of H2O = 18 gms contains 6.02e+24 molecules and every H2O molecule has 2 atoms of H-2

number H-2 atoms per Kg of H2O   = 6.02e+23 * (1000/18)*2   = 669e+23

abosrption cross section of Hydrogen for fast neutrons = 3.92e-5 b = 3.92 e-29 cm-2

Volume of 1 kg of water = 1000 cc , assume as cube

the pathlength = 10 cm

Flux of neutron = 1.0e17 neutrons/s/m2

                           = 1.0e13 neutrons /s/cm2

Neutrons abosrbed = 1.0e13 ( 1- exp(-669e+23 * 3.92e-29*10) )

                                 = 2.59 e+11 neutrons absorbed per kg of H2O

D-2 produced = (2.59e+11/6.02e+23 )*2 = 0.86e-12 gms per Kg of moderator /s

per year     = 0.86e-12 *365*24*3600 = 2.71 e-5 gms

D-2 fast neutron abosrption cross section = 5.34e-6 b

1 mole of D2O = 20 gm

1kg of D20 contains = 2*1000/20 *6.02e+23 = 1.0e+25 atoms of D2

we assume volume of 1 kg of D2O = 1000 cc aprox. for simplicity

path length = 10 cm ( as a cube of 1000cc)

neutrons abosrbed   = 1.0e+13 (1-exp( -1.0e+25 * 5.34e-30*10) )

                                  = 5.34e+9 atoms of T-3 /s

                                  = 3*5.34e+9 /6.02e+23 = 2.66 e-14 gms /s

we assume all the T-3 produced is decayed in half-year , we account ofr only hlaf-year produced

                    = 2.66e-14 * 365*12*3600 = 4.19e-7 gms /kg of D2O per yr

c) number atoms of T-3 per kg of T-3

                               = 5.34e+9

decay const = 0.693/12.3Yr   = 2.2e-8 /s

activity of T-3 in the reactor water = 5.34e+9*2.2e-8

                                                        = 117.48 Bq/s

d) T1/2 = 12.8 Yrs

A = Ao exp(-0.693 t /T1/2)

Ao /A = 5 = exp(0.693t/12.8 yr)

t = 28.57 Yrs , the activity will fall to 1/5 th

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