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1. Admission in India 2015
By:
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2. 321 δδδ
zyx
grad
∂
∂
+
∂
∂
+
∂
∂
=
grad is a vector
q = - K grad h
321)( δδδ
z
h
y
h
x
h
hgrad
∂
∂
+
∂
∂
+
∂
∂
=
Darcy’s law
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3. 321 δδδ zyx qqqq ++=
q is a vector
t
h
SW
z
q
y
q
x
q
s
zyx
∂
∂
−=−
∂
∂
+
∂
∂
+
∂
∂
Transient mass balance equation:
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4. 0=
∂
∂
+
∂
∂
+
∂
∂
z
q
y
q
x
q zyx
The dot product of
grad and q is
div q = 0
grad • q = 0
Steady State Mass Balance Equation with W = 0
∇ • q = 0 (del notation)
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5. W
t
h
S
z
h
K
zy
h
K
yx
h
K
x
szyx −
∂
∂
=
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
)()()(
Anisotropic medium: K at a point varies with direction
Kx ≠ Ky ≠ Kz
or
Kx = Ky ≠ Kz
Heterogeneous medium: K varies in space
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6. Homogeneous, isotropic Homogeneous, anisotropic
A A
B B
Heterogeneous, isotropic Heterogeneous, anisotropic
A
A
B
B
Characteristics of K in two dimensions
Kz
Kx
Kx ≠ Kz
••
Kx = Kz = K
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7. Homogeneous, isotropic Homogeneous, anisotropic
A A
B B
Heterogeneous, isotropic Heterogeneous, anisotropic
A
A
B
B
Characteristics of K in two dimensions
Kz
Kx
Kx ≠ Kz
••
Kx = Kz = K
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8. W
t
h
S
z
h
K
zy
h
K
yx
h
K
x
szyx −
∂
∂
=
∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
)()()(
2
2
)(
x
h
x
h
x ∂
∂
=
∂
∂
∂
∂
02
2
2
2
2
2
=
∂
∂
+
∂
∂
+
∂
∂
z
h
y
h
x
h
2
h = 0
Laplace equation
homogeneous, isotropic porous medium: Kx = Ky = Kz = K
steady state conditions: 0=
∂
∂
t
h
W = 0
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9. Tot h Pr ob le m ( 2 D)
Impermeable Rock
Groundwater
divide
Groundwater
divide
2 D, st e ad y st at e
02
2
2
2
=
∂
∂
+
∂
∂
z
h
x
h
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10. i, j
j
i
The 5 point star
in a regular grid
i+1, ji-1, j
i, j-1
i, j+1
∆x
∆y
∆x = ∆y
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11. 04 ,1,1,,1,1 =−+++ −+−+ jijijijiji hhhhh
4
1
1,1,
1
,1,11
,
+
−+
+
−++ +++
=
m
ji
m
ji
m
ji
m
jim
ji
hhhh
h
Solution by iteration
Gauss-Seidel Iteration
4/)( 1,1,,1,1, −+−+ +++= jijijijiji hhhhh
Iteration indices
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12. Iteration
planes
initial guesses
m
solution
m+2
m+1
m+3
4
1,1,,1,11
,
m
ji
m
ji
m
ji
m
jim
ji
hhhh
h
−+−++ +++
=
4
1
1,1,
1
,1,11
,
+
−+
+
−++ +++
=
m
ji
m
ji
m
ji
m
jim
ji
hhhh
h
Gauss-Seidel Iterationadmission.edhole.com

13. j
i
10
10
10
10
10
10
9 8
9 8
7
7
6
6
5
5
1 2 3 4
1
2
3
4
5
6
2D example
with specified head
boundary conditions
(Assume an initial
guess of h =10 for
all interior nodes)
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14. j
i
10
10
10
10
10
10
9 8
9 8
7
7
6
6
5
5
1 2 3 4
1
2
3
4
5
6
Example
with specified head
boundary conditions
9.75
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15. j
i
10
10
10
10
10
10
9 8
9 8
7
7
6
6
5
5
1 2 3 4
1
2
3
4
5
6
Example
with specified head
boundary conditions
8.44
9.75
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16. j
i
10
10
10
10
10
10
9 8
9 8
7
7
6
6
5
5
1 2 3 4
1
2
3
4
5
6
Example
with specified head
boundary conditions
9.75
9.94
8.44
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17. j
i
10
10
10
10
10
10
9 8
9 8
7
7
6
6
5
5
1 2 3 4
1
2
3
4
5
6
Example
with specified head
boundary conditions
9.94
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18. i, j
j
i
Gauss-Seidel Iteration
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19. C4=(C3+C5+B4+D4)/4
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20. 4
3324
3
ECDD
D
+++
=
Example of spreadsheet formula
Toth Problem
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