I( x) t e. is the total mean free path in the medium, [cm] tis the total cross section in the medium, [cm ] A M

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2 t I ( x) I e x x t Ie (1) whee: 1 t is the total mean fee path in the medium, [cm] N t t -1 tis the total coss section in the medium, [cm ] A M 3 is the density of the medium [gm/cm ] v 3 N= is the nuclea density of the medium [nuclei/cm ] 4 A v is Avogado's numbe =.61 [nuclei/mole] M is the atomic weight [amu] -4 t is the micoscopic coss section, [1 ban =1 cm ] The attenuation facto in a thin shield of thickness x is: I( x) t e x I Taking the natual logaithm of both sides we get: I( x) tx ln I If the desied attenuation and the macoscopic coss section in the medium used as shield ae known, then we can estimate the needed thickness x fom: 1 I( x) x ln () I t If on the othe hand, the thickness of the mateial is known, and the attenuation is measue, then the value of the micoscopic coss section at a given neuton enegy can be estimated fom: 1 I( x) t ln Nx I (3) This appoach applies only to thin shields. Othewise the concepts of build-up factos o emoval coss sections can be used in thick shield and boad beam situations.

3 TIME EPENENT IFFUION EQUATION The continuity equation fo the neuton density n [neutons / cm 3 ] is: n a divj t (4) whee: J is the neuton cuent J J x y Jz div J=.J=,is the divegence opeato x y z is the souce stength is the neuton flux [neutons/(cm.sec)] -1 a is the macoscopic absoption coss section [cm ] Notice that the divegence opeato acts on the components of a vecto such as the cuent J and geneates a scala quantity. The neuton cuent is descibed by Fick s law as: J gad (5) whee: is the diffusion coefficient [cm] ˆ gad = = ˆ i j kˆ x y z iˆ, ˆj, kˆ ae the catesian unit vectos in the x, y and z diections It must be obseved hee that taking the gadient of a scala function such as the flux, geneates a vecto as the cuent J. Combining Eqns. 4 and 5 we obtain the diffusion equation fo the neuton flux: 1 ( x, y, z, t) a( x, y, z, t) div (gad ( x, y, z, t) v t ( x, y, z, t).( ( x, y, z, t)) (6) a whee: nv, n. v

4 TEAY TATE IFFUION EQUATION The steady state diffusion equation can be obtained by equating the patial time deivative to zeo:.( ( x, y, z)) ( x, y, z) (7) If we futhe assume that we have a unifom and homogeneous medium, the diffusion coefficient can be consideed as a constant: (. ( x, y, z)) ( x, y, z) Now the aplacian opeato is: a a. div gad = x x y y z z, x y z esulting in: (,, ) (,, ) (8) x y z a x y z ividing into the diffusion coefficient we get: a ( x, y, z) ( x, y, z) efining the diffusion aea as: [cm ] (9) a and the diffusion length as: [cm] (1) we can wite the steady state diffusion equation as: a 1 ( x, y, z) ( x, y, z) (11)

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7 and applying it to Eqn. 16, we get: d dv du ( uv) u v (17) dx dx dx d d ( w. ) w( ) dw d d d Multiplying both sides by, we get: d dw., d d w Taking the deivative with espect to and applying the chain ule of diffeentiation to the second tem on the ight: d d d dw w. d d d d dw d w dw 1. d d d dw d ividing by we get an expession fo the aplacian opeato in tem of the vaiable w: 1 1 d d d d d d w (18) ubstituting Eqn. 18 and 16 into Eqn. 15, we get: 1d w 1 d w Fo all values of not equal to zeo, this equation educes to a simple fom fo which we can eadily find a solution: dw d 1 w (19)

8 This equation is analogous to the equation of the simple hamonic oscillato with the time vaiable eplaced by the spatial adial vaiable, and a positive sign on the ight hand side, and allows an exponential solution with two constants of integation as: w() Ae Be () Using Eqn. 16, we can now get the solution fo the flux as; w() e e () A B (1) The second tem leads to a solution involving an infinite value fo the flux, which is not pactically feasible since the flux must be finite in magnitude, implying that the constant B =, and: e () A () To detemine the second constant of integation A, knowing that the solution does not apply at the oigin =, we suound the souce by a sphee of adius and integate the cuent J() ove the suface. As the adius of the sphee shinks to zeo, the cuent integated ove the suface tends to the stength of the souce : J (3) lim 4 ( ) Applying Fick s law fom Eqn. 5 to the flux solution in Eqn., using the chain law of diffeentiation, we get: d() d e J ( ) ( ) A d d d A. e A e ( ) e d 1 1 A e e (4) ubstituting fom Eqn. 4 into Eqn. 3 we get:

9 1 1 lim 4 J ( ) lim 4. A e e lim 4 A e e,. Taking the limit yields zeo fo the fist tem and unity and unity fo the second tem on the ight hand side, esulting in the value of the integation constant A: 4 A.1 A 4 (5) ubstituting A in the expession fo the flux we finally get: e () (6) 4 It is of inteest to compae this expession fo the flux in a diffusing medium compaed with the one fo the flux in a vacuum: one of the adial position factos in the denominato has been eplaced by a decaying exponential in the numeato including the diffusion length, and by the diffusion coefficient in the denominato estoing the appopiate units to the expession fo the flux. The solution is not valid at the oigin =, since was cancelled on both sides of an equation duing the deivation, which makes the solution valid eveywhee except at the oigin. PROCEURE TO ETIMATE THE FUX FROM A POINT OURCE IN A IFFUING MEIUM The following pocedue allows the display the esults fom Eqn. 6 equiing an input of the souce stength, diffusion coefficient, and diffusion length fo diffeent eato media.! Neuton flux fom a point souce in an infinite nonmultiplying! eato medium.! phi()=*exp(-/)/4*pi**! Pogam saves output to file:output1! This output file can be expoted to a plotting outine! M. Ragheb, Univ. of Illinois at Ubana-Champaign! pogam flux eal :: Pi = ! souce stength eal :: = 1.E+1! diffusion coefficient (O) eal :: =.6

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11 TWO ENERGY GROUP NEUTRON MOERATION Fo a and themal enegy goup of neutons the eating popeties ae diffeent as shown in Table 1. Table 1. Fast and themal neuton ages and diffusion aeas fo diffeent eatos. Modeato Fast goup age [cm ] Themal goup diffusion length [cm ] HO O 131 3x1 4 Be 1 48 Gaphite 368 3,5 One can wite two diffusion equations, one fo the goup, and one fo the themal goup. Fast Themal : : th th ath th (7) The slowing down density q is a loss tem in the goup equation, but a gain tem in the themal goup equation coupling the two goups: q (8) ividing the and themal goup equations by thei espective diffusion coefficients: ath th th th th (9) efining the neuton age as: (3) and the themal goup diffusion aea as:

12 th (31) ath we can ewite the equations as: 1 1 th th th (3) In spheical geomety the goup flux has a solution simila to that of Eqn. 6, yielding: e () (33) 4 ubstituting the value fo the flux in the second equation, one can solve the inhomogeneous equation: 1 e th th (34) th 4 to yield the themal goup flux as: th() e e 4 th( ) (35) FINITE MEIA IFFUION We conside a plana unit cell in a fission eacto suounded with a eato with a constant neuton souce. Ou goal is to estimate the fluxes in the two diffeent media of eato and. uch calculations ae impotant fo the estimation of the utilization facto f in the fou facto fomula. We define a souce of themal neutons in the eato egion alone: ( x) a x x b (36) In the egion, the neuton diffusion equation in a one dimensional catesian coodinate system is:

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14 ( a a) ( a a) J ( a) C sinh E cosh E E The cosh solution is etained, and the sinh solution is dopped (E = ), yielding: ( x a) ( x) Ccosh (4) In the eato egion, the diffusion equation with the souce is: 1 ( x) ( x) (41) The paticula solution is: p a whee : a 1 (4) The complementay solution is: c ( b x) ( b x) ( x) Acosh Bsinh (43) Adding the complementay and paticula solutions, we get: ( x) ( x) ( x) c p ( b x) ( b x) (44) Acosh Bsinh a Estimating the cuent in the eato egion yields: ( b x) ( b x) J ( x) A sinh B cosh (45) Applying the bounday condition:

15 ( b b) ( b b) J ( b) A sinh B cosh B B (46) Again, the cosh solution is etained, and the sinh solution dopped (B=), yielding: ( b x) ( x) cosh A (47) a Futhe applying the two inteface bounday conditions of the continuity fo the flux and the cuent at x =, yields the two equations in the two unknowns A and C: Flux continuity: a b Ccosh Acosh a (48) Cuent continuity: J J a b C sinh A sinh (49) Witing the two linea equations in the two unknowns A and C: b a Acosh Ccosh b a A sinh C sinh a (5) o: cosh b cosh A a b a B sinh sinh a (51) The solution fo the two constants A and C ae:

16 a sinh a A a b b a sinh cosh sinh cosh b sinh a C a b b a sinh cosh sinh cosh Multiplying the expessions fo A and C by, simplifies them to: A C a a a sinh a b b a sinh cosh sinh cosh sinh a b b a sinh cosh sinh cosh b (5) Thus the solutions fo the fluxes in the and in the eato egions ae: ( x a) ( x) Ccosh a b ( x a) sinh.cosh a b b a sinh cosh sinh cosh a x (53) ( b x) ( x) Acosh a a ( b x) sinh cosh 1 a a b b a sinh cosh sinh cosh x b. (54)

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20 They possess an oscillatoy o hamonic solution which can be witten in eithe one of the altenative foms: x( t) Acos( t) Bsin( t) ' jwt ' jwt x() t Ae B e (7) whee: j 1 since: o: jt e cos( t) jsin( t) jt e cos( t) jsin( t) e cos( t) e sin( t) jt jt e e j jt jt EXPONENTIA GROWTH AN ECAY OUTION (8) (9) Wheneve the ight hand side of Eqn. 6 has a positive athe than a negative sign: x t (1) ( ) x( t) The solution is a negative and positive exponential decay and gowth given by: x t A t B t '' '' ( ) cosh( ) sinh( ) ''' wt ''' wt x() t A e B e (11) whee cosh(x) and sinh(x) ae the hypebolic cosine and sine functions, with: t e cosh( t) sinh( t) t e cosh( t) sinh( t) (1) o:

21 e cosh( t) e sinh( t) t t e e t t (13) The exponential fom of the solution is usually adopted in the case of infinite media, wheeas the hypebolic sine and cosine fom is used in the case of finite media, facilitating the detemination of the constants of integation in eithe case. EXERCIE 1. Calculate the thickness of a shield made out of: a) Wate. b) Gaphite. that would attenuate a beam of neutons by a facto of: a) One million times (1-6 ). b) One billion times (1-9 ).. Plot the neuton fluxes away fom the oigin fo a point neuton souce of stength =1 1 [n/sec] in an infinite medium of the following eatos: a) HO: diffusion coefficient =.164 cm, diffusion length =.73 cm. b) O: diffusion coefficient =.6 cm, diffusion length = 116. cm. 3. Compae the fluxes geneated in the pevious poblem to that in a vacuum. 4. Estimate the flux and cuent in vacuum at the midpoint between two souces of stength each and sepaated by a distance of 1 cm. Hint: The cuent is a vecto and adds up vectoially, wheeas the flux is a scala. 5. Estimate the flux and cuent in a diffusing medium with diffusion length =.73 cm and diffusion coefficient =.164 cm (HO) at the midpoint between two souces of stength each and sepaated by a distance of 1 cm. 6. Estimate the flux and cuent in vacuum at the cente of: a) A squae, b) A cube, c) An equilateal tiangle with a souce at each cone and a side length of 1 cm. 7. Estimate the flux and cuent in a diffusing medium with diffusion length =.73 cm and diffusion coefficient =.164 cm (HO) in the fom of: a) A squae, b) A cube, c) An equilateal tiangle with a souce at each cone and a side length of 1 cm. 8. Though diect substitution pove that the diffeent geneal foms given fo the solution of the imple Hamonic Oscillato do indeed satisfy the undelying diffeential equation. 9. Compae the magnitude of the neuton flux geneated by a souce of stength =1 1 [neutons / second], at a distance of 1 cm fom the souce in spheical geomety, 1. In a vacuum.

22 . In a diffusing medium with diffusion coefficient =1 cm, and macoscopic absoption coss section equal to.1 cm Pove that the themal goup flux fo a two goup slowing down of a neuton souce in a eating medium is given by: th() e e 4 th( )