Transport Neutron

Komputasi dan Simulasi Transport Neutron Coaching Neutronik 2008 Computational Division PPIN – BATAN

last edited : august 6th 2007

Pendahuluan 

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Masalah utama dalam fisika reaktor nuklir adalah penentuan distribusi neutron dalam teras reaktor. Distribusi neutron menentukan laju terjadinya berbagai reaksi nuklir dalam teras reaktor. Dengan memahami keadaan populasi neutron maka stabilitas dari reaksi fisi berantai dapat diprediksi dengan baik.

Proses transport neutron.. 

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Untuk menentukan distribusi neutron dalam teras reaktor kita harus memahami dengan baik proses transport neutron. Yaitu proses yang terjadi selama neutron bergerak dalam teras reaktor, yang melibatkan berbagai interaksi neutron dengan inti penyusun teras reaktor berupa tumbukan hingga akhirnya neutron hilang karena diserap atau keluar dari teras reaktor.

Proses difusi.. 



Kebanyakan studi neutronik teras reaktor memperlakukan gerak neutron sebagai proses difusi. Dimana diasumsikan bahwa neutron cendrung untuk berdifusi dari daerah dengan densitas neutron tinggi ke daerah dengan densitas neutron lebih rendah, seperti difusi panas dari daerah bertemperatur tinggi ke temperatur rendah.

Keterbatasan ‘difusi’.. 

Namun,berbeda dengan penanganan difusi pada konduksi panas dan gas yang yang memberikan simulasi yang akurat, pendekatan difusi terhadap transport neutron memiliki validitas yang terbatas.

Diffusion’s limitation..(cont’d) 

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The reason for this failure is easily understood when it is noted that in most diffusion process the diffusing particles are characterized by very frequent collisions that give rise to very irregular, almost random, zigzag trajectories. However, the cross-section for neutron-nuclear collisions is quite small (about 10-24 cm2). Hence neutron tend to stream relatively large distances between interactions.

Diffusion’s limitation..(cont’d) 

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The mean free path (mfp) characterizing fast neutrons is typically on the order of centimeters. And the dimensions characterizing changes in reactor core composition are usually comparable to a neutron mfp. (noted that a reactor fuel pin is typically about 1 cm in diameter). Hence, it is required a more accurate description of neutron transport that takes into account the relatively long neutron mfp and neutron streaming.

Diffusion’s limitation..(cont’d) 

In practical problem, neutron diffusion theory is invalid near the boundary of a reactor, or near a highly absorbing material such as a fuel rod or control element.

More accurate ??? 



Such a description has been borrowed from the kinetic theory of rarefied gases (which are also characterized by long mfp). The fundamental equation describing dilute gases was first proposed more than one century ago by Boltzmann, and even today the Boltzmann equation remain the principal tool of the gas dynamicist.

Neutron transport equation  

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Its counter part for the neutron “gas” called ‘neutron transport equation’. It is far simpler than Boltzmann equation. Where it is linear equation while the boltzmann equation is non linear equation. Neutron transport equation is much simpler to derive, requiring only the concept of neutron conservation plus a bit of vector calculus, and easier to understand than the neutron diffusion equation. It is far more fundamental and exact description of the neutron population in reactor, indeed, it is the fundamental cornerstone on which all of the various approximate methods used in nuclear reactor analysis are based.

Transport problem.. 

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But, neutron transport theory has come to be associated with a hideous plethora of impenetrable mathematics, unwieldy formulas, and the expenditure of enourmos amounts of money on computer number-crunching. It is usually very dificult to solve the transport equation for any but simplest modeled problems

But.. 

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

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However that is quite all right, since it is not the intent to attack the transport equation head on (for a while). Rather the job of the reactor analyst is to develop suitable (calculationally feasible and accurate) approximation to it. Usually, only by comparing these various approximation theories to the transport equation from which they originated can one really assess their range of validity. The effort in understanding the neutron transport equation will provide one with a much deeper and more thorough understanding of the approximate methods

Some Introductory Concept

Neutron Density and Flux 

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Start by defining the neutron density N(r,t) at any point in reactor core by N(r,t) d3r ≡ expected number of neutrons in d3r about r at a time t. It is a statistical theory in which only mean or average values are calculated. The neutron density N(r,t) is of interest because it allows us to calculate the rate at which nuclear reactions are occuring at any point in the reactor.

Neutron Density and Flux (cont’d) 

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Let us suppose that all the neutrons in the reactor have the same speed ν. The frequency with which a neutron will experience a given neutronnuclear reaction in terms of the macrocospic cross section characteizing that reaction Σ and the neutron speed v is vΣ = interaction frequency

Neutron Density and Flux (cont’d) 

Hence, the reaction-rate density F(r,t) at any point in the system is defined by multiplying the neutron density N(r,t) by the interaction frequency vΣ : F(r,t) d3r ≡ vΣ N(r,t) d3r ≡ expected rate at which interactions are occuring in d3r about r at a time t.

Neutron Density and Flux (cont’d) 

Example : Neutron density of N=108 cm-3 in a graphite medium where its total cross section Σt=0.385 cm-1, neutron speed 2.2x105 cm/sec. We would find a reaction rate density of 8.47x1012 reactions/cm3/sec. In this particular case, most of these reactions would consist of scattering collisions.

Neutron Density and Flux (cont’d) 

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These concept can easily be extended to the case in which the neutron density is different for various neutron energies E by defining : N(r,E,t) d3r dE ≡ expected number of neutrons in d3r about r, energies in dE about E, at time t. Also the reaction rate density F(r,t) d3r dE ≡ vΣ(E) N(r,E,t) d3r dE

Neutron Density and Flux (cont’d) 

 

The product vN(r,t) occurs very frequently in reactor theory, and therefore it is given a special name Φ(r,t) ≡ vN(r,t) ≡ neutron flux Its unit is [cm-2 sec-1] Noted that neutron fux is scalar quantity not as others definition of flux in electromagnetic or heat conduction.

Angular Densities and Currents 

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So far, we already use three variable to characterize the state of individual neutron; the neutron position (r), its energy (E), and the time (t) at which the neutron is observed. Yet, notice that to specify the state of the neutron, we must also give its direction of motion characterized by the unit vector Ω=v/|v|.

Angular Densities and Currents (cont’d) 

By introducing this new variable lets generalize the concept of density by defining the angular neutron density :

n(r,E,Ω,t) d3r dE dΩ = expected number of neutrons in d3r about r, energy dE about E, moving in direction Ω in solid angle dΩ at time t. 

This is the most general neutron density function we need to define since it happens that one can derive an essential exact equation, the neutron transport equation, for the angular neutron density n(e,E,Ω,t).

Angular Densities and Currents (cont’d) 

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Angular neutron flux φ(r,E,Ω,t) ≡ v n(e,E,Ω,t) Angular current density j(r,E,Ω,t) ≡ vΩn(e,E,Ω,t) ≡ Ωφ(r,E,Ω,t) Notice that since Ω is a unit vector, the angular flux is actually nothing more than the magnitude of the angular current density. |j|=|Ω|φ = φ

Angular Densities and Currents

(cont’d) 

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The angular current density has a useful physical interpretation. j(r,E,Ω,t) dA dE dΩ ≡ expected number of neutrons passing through an area dA per unit time with eergy E in dE, direction Ω in dΩ at time t. We can also define an angular interaction rate f(r,E,Ω,t) = v Σ(r,E) n(e,E,Ω,t) = Σ(r,E) φ(r,E,Ω,t)

Angular Densities and Currents

(cont’d) 

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All of the angle-dependent quantities can be related to the earlier definition by simply integrating over the angular variables. For neutron density : N(r,E,t) = ∫4πdΩ n(e,E,Ω,t)

further N(r,t)

= ∫0∞ dE N(r,E,t) = ∫0∞ dE ∫4πdΩ n(e,E,Ω,t)

Angular Densities and Currents

(cont’d) 

For neutron flux Φ(r,E,t) = ∫4πdΩ φ(e,E,Ω,t)

and Φ(r,t) = ∫0∞ dE Φ(r,E,t) = ∫0∞ dE ∫4πdΩ φ(e,E,Ω,t)

Angular Densities and Currents

(cont’d) 

For neutron current J(r,E,t) = ∫4πdΩ j(e,E,Ω,t) J(r,E,t) is called neutron current density. Also, J(r,t) = ∫0∞ dE J(r,E,t) = ∫0∞ dE ∫4πdΩ j(e,E,Ω,t)

More about J(r,t) and Φ(r,t) 

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Notice that J(r,t) is actually what would be reffered to as the ‘flux’ in other fields of physics, since if we have a small area dA at a position r, then J(r,t).dA = net rate at which neutrons pass through a surface area dA. The unit of both J(r,t) and Φ(r,t) are identical [cm-2 .sec-1]. However, J is a vector quantity that characterize the net rate at which neutrons pass through a surface oriented in a given direction, whereas Φ simply characterize the totalrate at which neutron pass through a unit area, regardless of orientation.

More about J(r,t) and Φ(r,t) 

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Such an interpretation would suggest that J is a more convenient quantity for describing neutron leakage or flow , while Φ is more suitable for characterizing neutron reaction rates in which the total number of neutron interactions in a sample is of interest. Although the angular flux and current density are very simply related, we will find that there is no simple analogous relationship between J and Φ.

Persamaan Transport Neutron

Pendahuluan 

Persamaan yang menggambarkan kerapatan neutron angular pada sistem nuklir akan diturunkan dengan melakukan akuntansi terhadap proses-proses yang dapat memunculkan neutron dan menghilangkan neutron dari sembarang volume v dalam sistem.

Neutron pada volume v 

Untuk sembarang volume V, jumlah neutron dengan energi E dalam dE, dengan arah Ω dalam dΩ dalam V adalah 3

( ∫ n(r, E, Ωˆ , t)d r )dE ⋅ dΩˆ V

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Laju perubahannya terhadap waktu diberikan oleh kesetimbangan berikut ∂ ∂t

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(∫ n(r, E, Ωˆ , t )d r )dE ⋅ dΩˆ = muncul pada V – hilang dari V 3

V

Bila volume V diasumsikan tidak bergantung waktu, maka ∂ ∂t

(∫ n(r, E, Ωˆ , t )d r )dE ⋅ dΩˆ =  ∫ ∂∂nt d 3

V

V

3

u/ simpifikasi notasi variabel dihilangkan

 ˆ r dE ⋅ dΩ 

Mekanisme pada volume V 

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Mekanisme neutron ‘muncul’ : 1. sumber neutron dalam volume V 2. neutron yang terhambur dengan variabel akhir E,Ω dari sembarang E’, Ω’. (ruang energi dan arah) 3. neutron masuk volume V melalui permukaan S.(ruang spasial) Mekanisme neutron ‘hilang’ : 4. neutron bocor melalui permukaan S. 5. neutron dalam V (dengan variabel E, Ω) mengalami tumbukan sehingga variabelnya menjadi E’, Ω’.

Mekanisme Neutron Muncul 1.Sumber neutron pada V, dengan definisi sumber berikut

(

)

ˆ , t d 3 r ⋅ dE ⋅ dΩ ˆ s r, E, Ω Maka suku sumber dinyatakan sbb:

[∫ ( V

) ]

3 ˆ ˆ s r , E , Ω, t d r dE ⋅ dΩ

Mekanisme Neutron Muncul 2. Neutron muncul karena tumbukan dan terhambur ke ‘ruang’ V. Laju neutron terhambur dari suatu ruang (E,Ω) ke (E’, Ω’) adalah

[∫ υ '⋅Σ ( E ' → E, Ωˆ ' → Ωˆ ) ⋅ n(r, E, Ωˆ , t )d r ]dE ⋅ dΩˆ 3

V

s

Karena harus diperhitungkan neutron dari semua ‘ruang’ lain maka

( (

) (

∞  3  ∫V d r ∫4dπ Ωˆ ' ∫ dE ' υ '⋅Σ s E ' → E , Ωˆ ' → Ωˆ ⋅ n r , E , Ωˆ , t 0 

))

  dE ⋅ dΩˆ 

Mekanisme Neutron ‘hilang’ 5. Neutron yang terhambur ke ‘ruang’ lain dari V. Laju neutron mengalami interaksi adalah

(

)

(

f t r , E , Ωˆ , t = υ ⋅ Σ t ( r , E ) ⋅ n r , E , Ωˆ , t Maka neutron yang terhambur ke ‘ruang’ V dinyatakan sebagai berikut

)

[∫ υ ⋅ Σ ( r, E ) ⋅ n(r, E, Ωˆ , t )d r ]dE ⋅ dΩˆ 3

V

t

Mekanisme ‘hilang’+’muncul’ (3+4) Bocor kedalam dan keluar volume V digabung. Dengan konsep rapat arus angular j, maka laju pada E,Ω akan bocor dari permukaan dS adalah

(

)

(

)

ˆ , t ⋅ dS = υ ⋅ Ω ˆ ⋅ n r, E, Ω ˆ , t ⋅ dS j r, E, Ω Untuk seluruh permukaan, total bocor keluar dan masuk,

(

ˆ ⋅ n r, E, Ω ˆ ,t dS ⋅ υ ⋅ Ω ∫ S

Dari pers. Gauss berikut Didapat,

)

∫ dS ⋅A(r ) =

∫ ( d r∇ ⋅ A(r )) 3

[∫ dS ⋅υ ⋅ Ωˆ ⋅ n( r, E, Ωˆ , t )]dE ⋅ dΩˆ = ∫ d r(∇ ⋅ υ ⋅ Ωˆ ⋅ n( r, E, Ωˆ , t ))dE ⋅ dΩˆ ˆ ⋅ ∇ n( r , E , Ωˆ , t ) )dE ⋅ dΩˆ ( d r υ ⋅ Ω Atau ∫ 3

S

V

3

V

S

V

Total semua mekanisme 

Dengan mensubstitusi semuanya ke pers. Awal diperoleh :

∞  ∂n    ˆ ˆ ˆ ˆ ∫V d r  ∂t + υ ⋅ Ω ⋅ ∇n + υ ⋅ Σt ⋅ n − ∫0 dE ' ∫4π dΩ' v'⋅Σs ( E ' → E , Ω → Ω' ) ⋅ n  − s dE ⋅ dΩˆ = 0  

(

3

)

Karena volume V sembarang maka integran diatas harus nol. Maka didapat hubungan kesetimbangan berikut : ∞

(

)

∂n + υ ⋅ Ωˆ ⋅ ∇ n + υ ⋅ Σ t ⋅ n(r , E , Ωˆ , t ) = ∫ dE ' ∫ dΩˆ ' v'⋅Σ s ( E ' → E , Ωˆ → Ωˆ ' )n(r , E , Ωˆ , t ) + s(r , E , Ωˆ , t ) 4π ∂t 0

Formulasi Persamaan Transpor 

Dengan menggunakan notasi fluks angular maka persamaan transport biasa ditulis sbb: ∞

(

)

1 ∂ϕ ˆ + Ω ⋅ ∇ ϕ + Σ t (r , E ) ⋅ ϕ (r , E , Ωˆ , t ) = ∫ dE ' ∫ dΩˆ ' Σ s ( E ' → E , Ωˆ → Ωˆ ' )ϕ (r , E , Ωˆ , t ) + s(r , E , Ωˆ , t ) 4π υ ∂t 0 Dimana : -Syarat awal :

ˆ ,0) = ϕ (r , E , Ω ˆ , t) ϕ (r , E , Ω 0

- Syarat batas :

ˆ , t) = 0 ϕ (rs , E , Ω

Mis.Syarat batas vakum

Bila

ˆ • eˆ < 0 Ω s

u/ semua rs pada S

Persamaan Diffusi Satu Energi 

Dengan akuntansi yang sama, untuk asumsi satu energi diperoleh persamaan berikut  3  1 ∂φ d r − S + Σ φ + ∇ ⋅ J a ∫V υ ∂t =0  Sehingga

1 ∂φ = −∇ ⋅ J − Σ aφ + S υ ∂t

Dari pers. Diatas, untuk dapat diselesaikan lebih lanjut diperlukan hubungan antara J dan Φ. Ini diberikan oleh Hukum Fick’s berikut

J ( r , t ) ≅ − D(r )∇ ⋅ φ (r , t )

Konstanta diffusi

Persamaan Diffusi Satu Energi 

Setelah disubstitusikan kembali maka diperoleh 1 ∂φ − ∇ ⋅ D(r )∇φ (r , t ) + Σ a (r )φ (r , t ) = S (r , t ) υ ∂t

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Untuk D yang homogen : 1 ∂φ − D∇ 2φ (r , t ) + Σ a (r )φ (r , t ) = S (r , t ) υ ∂t

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Pers.Difusi Satu Grup

Lebih jauh, untuk masalah statis :

− D∇ 2φ (r ) + Σ a (r )φ (r ) = S (r ) Persamaan Helmholtz

Pers.Difusi : Kasus 1-D 

Untuk satu dimensi (mis.X) maka d 2φ − D 2 + Σ aφ ( x) = S (r ) dx

φ (0) = φ (a) = 0 

Diskritisasi ruang, operator diff. menjadi :

2 dφ 2d φ φ i + 1 ≡ φ ( xi + 1 ) = φ i + ∆ + ∆ 2 + ... dx i dx i

φ i −1 ≡ φ ( xi −1 )

2 dφ 2d φ = φi − ∆ +∆ − ... 2 dx i dx i

φi +1 − 2φi + φi −1 d 2φ ≅ 2 dx i ∆2

Pers.Difusi : Kasus 1-D 

Setelah substitusi diperoleh,  φi +1 − 2φi + φi −1  − D  + Σ aφi = S i 2 ∆   Dengan pengaturan variabel :

D D  2D  − 2 φi −1 +  2 + Σa φi − 2 φi +1 = S i ∆ ∆ ∆  Atau (untuk i=1,2,…,N-1)

ai ,i −1φi −1 + ai ,iφi + ai ,i +1φi +1 = S i

Aφ = S

A  matriks (n-1)x(N-1) Φ,S vektor kolom (N-1)

Bentuk lebih umum u/ 1-D 

Pers.diff umum 1-D pada geometri bidang datar : −



d dφ D( x) + Σ a ( x)φ ( x) = S ( x) dx dx

Cara memecahkan persamaan ini terbagi kedalam dua langkah : 1. menurunkan persamaaan beda (diskritisasi). 2. menyelesaikan persamaan beda dengan algoritma tertentu.

Diskritisasi 

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Metoda umum untuk memperoleh pers.beda (difference eq.) adalah dengan melakukan integrasi terhadap pers.diff pada sembarang mesh interval. Integrasi dari tiap suku pers.diff dilakukan terhadap mesh interval berikut : xi −

x i −1

∆i 2

xi +

xi

∆ i +1 2

x i +1

Integrasi tiap suku suku sumber dan penyerapan 

Suku sumber ∆ xi + i +1 2

xi



∆ ∆ i i+ 1  dx ⋅ S ( x ) ≅ S + i  ∫∆  2 2   − i 2

Suku penyerapan ∆ xi + i +1 2

∆i +1  ∆i dx ⋅ Σ ( x ) φ ( x ) ≅ Σ φ + ai i  ∫∆ a  2 2   x − i i

2

Integrasi tiap suku suku bocor 

Suku bocor : ∆ xi + i +1 2

∫dx ⋅

∆ xi − i 2

∆ xi + i +1 2

d dφ dφ D( x) ≅ D( x) dx dx dx xi −∆i 2

Suku ini memerlukan beberapa langkah detail berikut :

dφ dx dφ dx

xi +

∆ i +1 2

xi −

∆i 2

≅

φ i +1 − φ i ∆ i +1

φi − φ−1i ≅ ∆i

xi +

∆i +1 2

xi

x i +1

xi −

x i −1

∆i 2

xi

Integrasi tiap suku suku bocor 

Untuk nilai D, ∆  1  D xi + i +1  = [ Di +1 + Di ] ≡ Di ,i +1 2  2  ∆  1  D xi − i  = [ Di −1 + Di ] ≡ Di ,i −1 2  2 



Sehingga total suku bocor, xi +

∆ i +1 2

Di ,i +1  Di ,i +1 Di ,i −1  d dφ Di ,i −1 ∫∆dx ⋅ dx D( x) dx ≅ ∆ i φi −1 −  ∆ i +1 + ∆ i φi + ∆ i +1 φi +1 i

xi −

2

Hasil integrasi 

Substitusi hasil integrasi terhadap pers.diffusi awal sbb:

ai ,i −1φi −1 + ai ,iφi + ai ,i +1φi +1 = S i Dimana koefisiennya adalah  Di + Di −1 a i ,i −1 = −  ∆i 

 1   ∆ +∆ i +1  i

 D + Di Di −1 + Di ai ,i = Σ a +  i +1 + ∆ ∆i i +1 

 Di +1 + Di ai ,i +1 = −  ∆i 

 1   ∆ i + ∆ i +1

 1  ∆ +∆ i +1  i

Diperoleh N -1 pers.beda tiga titik (three-point difference equations) untuk N+1 variabel tak diketahui yaitu Φ0,Φ1,…, ΦN.

Syarat batas 

Syarat batas umum dapat diberikan sbb:

a 0, 0φ0 + a 0,1φ1 = S 0

a N , N −1φN −1 + a N , N φN = S N

Solusi ‘pers.differensial 3-titik’ 

Persamaan terakhir yang kita dapatkan adalah

A ⋅φ = S Lebih eksplisitnya  a11   a11  0     0 

a11 a11 a11  0

0 a11 a11  0

 0  φ1   S1       0  φ2   S 2   0  φ3  =  S 3       ⋅           ⋅ ⋅ φN −1   S N −1 



Matrik tridiagonal dapat langsung dipecahkan dengan eliminasi Gaussian. Sehingga diperoleh matriks berikut : 1  0 0  ⋅ 0 

A1 1 0

0 A2 1

0 0 A3

0 0

⋅ 

 0

 φ1   α1       φ2   α2   φ  = α   3   3  ⋅           1 φN −1  αN −1 

dimana An =

a n , n +1 a n,n + a n,n −1 An−1

αn =

S n − a n ,n −1α n −1 a n ,n − a n ,n −1 An −1

A1 =

α1 =

a1, 2 a1,1 s1 a1,1



Maka, nilai fluks diperoleh dengan substitusi kembali, dan diperoleh :

φ N −1 = α n −1

φ N − 2 = − AN − 2φ N − 1 + α N − 2 = − AN − 2α N − 1 + α N − 2 dst,

Dekomposisi LU 

Secara formal yang telah dilakukan adalah dekomposisi LU berikut 0 0  a11  0  a 21 (a 22 − a 21 A1 ) A=  0 a32 (a33 − a32 A2 )  0   0  ⋅ ⋅ ⋅ 

0 0 0  ⋅

⋅  1  ⋅  0 ⋅  0  ⋅  ⋅ ⋅   0

A1 0 0   1 A2 0  0 1 A3   0 ⋅  ⋅ 0  0 1 

Sehingga penyelesaiannya sebagai berikut

A⋅φ = L ⋅U ⋅φ = S

Forward elimination

−1

U ⋅φ = L S = α Back substitution

A ⋅φ = U

−1

−1

⋅L ⋅S =U

−1

⋅α

Perhitungan Kritikalitas 

Sekarang kita beralih kepada perhitungan yang sangat penting, yaitu tingkat kritikalitas suatu sistem nuklir dengan mengetahui komposisi bahan dan geometrinya.

Pers.Difusi 

Persamaan Difusi yang harus dipecahkan,misalnya (pada kasus sederhana) − D∇ φ (r ) + Σ a (r )φ (r ) = S (r ) 2



Pada perhitungan kritikalitas sumber hanya diberikan dari reaksi fisi

− D∇ 2φ (r ) + Σ a (r )φ (r ) = υ Σ f φ ( r )

Mencari Kesetimbangan 

Untuk menentukan komposisi agar diperoleh kesetimbangan maka diberikanlah koefisien k berikut 1 − D∇ 2φ (r ) + Σ a (r )φ (r ) = υΣ f φ ( r ) k



Cara lain … dengan menganggap v variabel, dimana keadaan kritis dicapai pada nilai v tertentu yaitu vC

− D∇ 2φ (r ) + Σ a (r )φ (r ) = υ C Σ f φ ( r )



Hubungannya… υ υC

k =

Perhitungan Kritikalitas 

Secara sederhana persamaan yang akan dipecahkan berbentuk 1 M ⋅φ = F ⋅φ k dengan

M ⋅ ≡ − D∇ 2 ⋅ +Σ a (r ) ⋅

F ⋅ = υΣ f (r ) ⋅

Operator ‘destruksi’

Operator ‘sumber’

Metoda iterasi 

Solusi dilakukan dengan metoda iterasi berikut, diawali dengan memberi sumber awal dan k tebakan.

S (r ) ≡ F ⋅ φ ≅ S ( 0) (r )

k ≅ k (0)

Lalu tentukan flux Φ(1) sbb :

M ⋅ φ (1) ≡ − D∇ 2 ⋅ φ (1) + Σ a (r ) ⋅ φ (1) =

1 (0) S (1) k

Dengan hasil diatas dapat kita hitung sumber dan k baru sbb

S

(1)

= F ⋅φ

(1)

= υΣ f φ

(1)

Bagan Algoritma Input geometri dan komposisi bahan Tebak sumber awal (S(0)) dan k(0) Iterasi dalam

Mφ ( n +1) =

1 k (n)

Fφ ( n +1)

S ( n +1) = F ⋅ φ ( n +1) k

( n +1)

∫d

≅

1 k

(n)

3

Iterasi

rS ( n +1) ( r )

∫d

3

k ( n ) − k ( n −1) S ( n ) − S ( n −1) < ε1 < ε2 k (n) S (n)

Yes

Keff

luar

rS ( n ) ( r )

No



  

Tim Java [Arya dan Sinta AW,Aniq) Tim Visual Basic [Elfrida,Utaja] Tim Fortran [Marsodi, Sangadji] Tim MATLAB [Mike,Entin,Wahyu,Dinan]

Eksplisit ‘Code programming’   

Input Solver Output

Input INPUT Geometri

Material

Panjang bahan

Bil.riil

L

Lebar partisi

Bil.riil

H

Pen.lintang absorpsi,fisi, konstanta difusi

Array 1-D Bil.riil pla[n], plf[n], D[n]

Neutron per fisi

Array 1-D bil.riil

D[n]

Array 2-D bil.riil

Fl[n,i]

Array Bil.riil

K[i]

Perhitungan Fluks awal

K-awal

Solver 



Iterasi dalam inloop( Iterasi luar



Untuk memantapkan pemahaman kita, mari kita simak penjelasan untuk hal yang sama dari pengembang MCNP F.Brown dari Los Alamos National Laboratory.