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