contaminant simulation in the soil and water fileanalytical solution •persamaan kontaminan untuk...

CONTAMINANT SIMULATION IN THE SOIL AND WATER

JEFRI FERLIANDE

Technology and Environmental Management

Contaminant transport equation in the soil can be

solved by partial differential equation:

𝐷𝑥

𝜕2𝐶

𝜕𝑥2− 𝑣𝑥

𝜕𝐶

𝜕𝑥=

𝜕𝐶

𝜕𝑡

If condition involves adsorption effects, it relates

with retardation factor, so the equation is:

𝐷𝑥

𝑅

𝜕2𝐶

𝜕𝑥2−

𝑣𝑥

𝑅

𝜕𝐶

𝜕𝑥=

𝜕𝐶

𝜕𝑡

Solutions for Contaminant Transport in the

Soil

•Analytical Solution

•Numerical Solution

• Explicit

• Implicit

• Implicit using Crank Nicolson Method

Important Things

•Deciding condition of contaminant, steady

or unsteady

•Deciding :

• Initial Condition

• Boundary Condition

Analytical Solution

• Sangat berguna sebagai prakiraan pertama atas

penyebaran kontaminan

• Bermanfaat untuk verifikasi model misalnya dari

solusi numerik

• Dapat diterapkan pada kondisi yang lebih heterogen

• Dapat mengakomodasi reaksi kontaminan antara

kontaminan dengan tanah atau transformasi

kontaminan dalam dalam air yang mengikuti reaksi

yang tidak sederhana

Numerical Solution

Analytical Solution

• Persamaan kontaminan untuk sumber kontinu pada 1-D

menggunakan transformasi Laplace (Bear,1961)

• Jika dipengaruhi oleh faktor Retardasi (Bear, 1972)

𝐶 𝑥,𝑡

𝐶0=

1

2𝑒𝑟𝑓𝑐

𝐿 − 𝑣𝑥𝑡

2 𝐷𝑥𝑡+ 𝑒𝑥𝑝

𝑣𝑥𝐿

𝐷𝑥𝑒𝑟𝑓𝑐

𝐿 + 𝑣𝑥𝑡

2 𝐷𝑥𝑡

𝐶(𝑥,𝑡) =𝐶0

2𝑒𝑟𝑓𝑐

𝑅𝑥 − 𝑣𝑥𝑡

2 𝑅𝐷𝑥𝑡+ 𝑒𝑥𝑝

𝑣𝑥𝑥

𝐷𝑥𝑒𝑟𝑓𝑐

𝑅𝑥 + 𝑣𝑥𝑡

2 𝑅𝐷𝑥𝑡

Numerical Methods using Finite Difference

Forward Difference Approximation:

𝜕𝐶

𝜕𝑡=

𝐶𝑖,𝑗𝑛+1 − 𝐶𝑖,𝑗

𝑛

∆𝑡

Backward Difference Approximation:

𝜕𝐶

𝜕𝑡=

𝐶𝑖,𝑗𝑛 − 𝐶𝑖,𝑗

𝑛−1

∆𝑡

Backward Difference Approximation:

𝜕𝐶

𝜕𝑡=

𝐶𝑖,𝑗𝑛+1 − 𝐶𝑖,𝑗

𝑛−1

2∆𝑡

Explicit Solution

𝐶𝑖𝑘+1 = 𝛼 𝐶𝑖−1

𝑘 + 1 − 2𝛼 + 𝛽 𝐶𝑖𝑘 + (𝛼 − 𝛽)𝐶𝑖+1

𝑘

𝐷𝑥

𝜕2𝐶

𝜕𝑥2− 𝑣𝑥

𝜕𝐶

𝜕𝑥=

𝜕𝐶

𝜕𝑡

𝐷𝑥

𝐶𝑖+1𝑘 − 2𝐶𝑖

𝑘 + 𝐶𝑖−1𝑘

∆𝑥2− 𝑣𝑥

𝐶𝑖+1𝑘 − 𝐶𝑖

𝑘

∆𝑥=

𝐶𝑖𝑘+1 − 𝐶𝑖

𝑘

∆𝑡=

∆𝐶

∆𝑡

𝛼 =𝐷𝑥∆𝑡

∆𝑥2 𝛽 =

𝑉𝑥∆𝑡

∆𝑥

Implicit Solution using =1

𝐷𝑥

𝜕2𝐶

𝜕𝑥2− 𝑣𝑥

𝜕𝐶

𝜕𝑥=

𝜕𝐶

𝜕𝑡

𝜕2𝐶

𝜕𝑥2 = 𝛼𝐶𝑖+1

𝑘+1 − 2𝐶𝑖𝑘+1 + 𝐶𝑖−1

𝑘+1

∆𝑥2 − (1 − 𝛼)𝐶𝑖+1

𝑘 − 2𝐶𝑖𝑘 + 𝐶𝑖−1

𝑘

∆𝑥2

𝜕2𝐶

𝜕𝑥2 =𝐶𝑖+1

𝑘+1 − 2𝐶𝑖𝑘+1 + 𝐶𝑖−1

𝑘+1

∆𝑥2

𝜕𝐶

𝜕𝑥= 𝛼

𝐶𝑖+1𝑘+1 − 𝐶𝑖

𝑘+1

∆𝑥− (1 − 𝛼)

𝐶𝑖+1𝑘 − 𝐶𝑖

𝑘

∆𝑥

𝜕𝐶

𝜕𝑥=

𝐶𝑖+1𝑘+1 − 𝐶𝑖

𝑘+1

∆𝑥

𝜕𝐶

𝜕𝑡=

𝐶𝑖𝑘+1 − 𝐶𝑖

𝑘

∆𝑡

𝛼 =𝐷𝑥∆𝑡

∆𝑥2 𝛽 =

𝑉𝑥∆𝑡

∆𝑥

𝐶𝑖𝑘 = −𝛼 𝐶𝑖−1

𝑘+1 + 2𝛼 − 𝛽 + 1 𝐶𝑖𝑘+1 + (−𝛼 + 𝛽)𝐶𝑖+1

𝑘+1

2𝛼 − 𝛽 + 1 −𝛼 + 𝛽 0 0 0−𝛼 2𝛼 − 𝛽 + 1 −𝛼 + 𝛽 0 00 −𝛼 2𝛼 − 𝛽 + 1 −𝛼 + 𝛽 00 0 −𝛼 2𝛼 − 𝛽 + 1 −𝛼 + 𝛽0 0 0 −𝛼 2𝛼 − 𝛽 + 1

𝐶𝑖𝑘+1

𝐶𝑖+1𝑘+1

𝐶𝑖+2𝑘+1

𝐶𝑖+3𝑘+1

𝐶𝑖+4𝑘+1

=

𝐶𝑖𝑘 + 𝛼𝐶𝑖−1

𝑘

𝐶𝑖+1𝑘

𝐶𝑖+2𝑘+1

𝐶𝑖+3𝑘+1

𝐶𝑖+4𝑘+1

known unknown known

Matrix : 𝐀 ∙ 𝑩 = 𝑪 , 𝐦𝐚𝐤𝐚 𝐁 = 𝐢𝐧𝐯(𝐀)𝑪

A B C

Implicit Solution using =0.5

𝐷𝑥

𝜕2𝐶

𝜕𝑥2− 𝑣𝑥

𝜕𝐶

𝜕𝑥=

𝜕𝐶

𝜕𝑡

𝜕2𝐶

𝜕𝑥2 =𝐶𝑖+1

𝑘+1 − 2𝐶𝑖𝑘+1 + 𝐶𝑖−1

𝑘+1

2∆𝑥2 −𝐶𝑖+1

𝑘 − 2𝐶𝑖𝑘 + 𝐶𝑖−1

𝑘

2∆𝑥2

𝜕𝐶

𝜕𝑥= 𝛼

𝐶𝑖+1𝑘+1 − 𝐶𝑖

𝑘+1

∆𝑥− (1 − 𝛼)

𝐶𝑖+1𝑘 − 𝐶𝑖

𝑘

∆𝑥

𝜕𝐶

𝜕𝑥=

𝐶𝑖+1𝑘+1 − 𝐶𝑖

𝑘+1

2∆𝑥−

𝐶𝑖+1𝑘 − 𝐶𝑖

𝑘

2∆𝑥

𝜕2𝐶

𝜕𝑥2= 𝛼

𝐶𝑖+1𝑘+1 − 2𝐶𝑖

𝑘+1 + 𝐶𝑖−1𝑘+1

∆𝑥2− (1 − 𝛼)

𝐶𝑖+1𝑘 − 2𝐶𝑖

𝑘 + 𝐶𝑖−1𝑘

∆𝑥2

𝜕𝐶

𝜕𝑡=

𝐶𝑖𝑘+1 − 𝐶𝑖

𝑘

∆𝑡

𝛼 =𝐷𝑥∆𝑡

2∆𝑥2 𝛽 =𝑉𝑥∆𝑡

2∆𝑥

𝐶𝑖−1𝑗+1

−𝛼 + 𝐶𝑖𝑗+1

1 + 2𝛼 − 𝛽 + 𝐶𝑖+1𝑗+1

𝛼 − 𝛽 = 𝐶𝑖−1𝑗

𝛼 + 𝐶𝑖𝑗

1 − 2𝛼 + 𝛽 + 𝐶𝑖+1𝑗

(𝛼 − 𝛽)

1 + 2𝛼 − 𝛽 −𝛼 + 𝛽 0 0 0−𝛼 1 + 2𝛼 − 𝛽 −𝛼 + 𝛽 0 00 −𝛼 1 + 2𝛼 − 𝛽 −𝛼 + 𝛽 00 0 −𝛼 1 + 2𝛼 − 𝛽 −𝛼 + 𝛽0 0 0 −𝛼 1 + 𝜕

𝐶𝑖𝑘+1

𝐶𝑖+1𝑘+1

𝐶𝑖+2𝑘+1

𝐶𝑖+3𝑘+1

𝐶𝑖+4𝑘+1

=

1 − 2𝛼 + 𝛽 𝛼 − 𝛽 0 0 0𝛼 1 − 2𝛼 + 𝛽 𝛼 − 𝛽 0 00 𝛼 1 − 2𝛼 + 𝛽 𝛼 − 𝛽 00 0 𝛼 1 − 2𝛼 + 𝛽 𝛼 − 𝛽0 0 0 𝛼 1 − 𝛼

𝐶𝑖𝑘 − 2𝛼𝐶𝑖−1

𝑘

𝐶𝑖+1𝑘

𝐶𝑖+2𝑘

𝐶𝑖+3𝑘

𝐶𝑖+4𝑘

Matrix: 𝐀 ∙ 𝑩 = 𝑪 ∙ 𝑫 𝒎𝒂𝒌𝒂 𝑩 = 𝒊𝒏𝒗(𝑨) ∙ 𝑪 ∙ 𝑫

A B

C D

known unknown

known known

Compare between Analytical, Explicit, Implicit

(=1), and Implicit Crank Nicholson (=0.5)

KESIMPULAN

Hasil simulasi menunjukkan bahwa metode finite difference

menggunakan implisit Crank Nicolson memiliki hasil yang

lebih stabil dibandingkan dengan metode numerik secara

eksplisit dan implisit sederhana.

THANK YOU