Mba admission in india
20
Admission in India 2015 By: admission.edhole.com
Transcript of Mba admission in india
321 δδδzyx
grad∂∂+
∂∂+
∂∂=
grad is a vector
q = - K grad h
321)( δδδz
h
y
h
x
hhgrad
∂∂+
∂∂+
∂∂=
Darcy’s law
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321 δδδ zyx qqqq ++=
q is a vector
t
hSW
z
q
y
q
x
qs
zyx
∂∂−=−
∂∂+
∂∂+
∂∂
Transient mass balance equation:
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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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Wt
hS
z
hK
zy
hK
yx
hK
xszyx −
∂∂=
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂
)()()(
Anisotropic medium: K at a point varies with directionKx ≠ Ky ≠ Kz
orKx = Ky ≠ Kz
Heterogeneous medium: K varies in space
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Homogeneous, isotropic Homogeneous, anisotropic
A A B B
Heterogeneous, isotropic Heterogeneous, anisotropic
AA
B
B
Characteristics of K in two dimensions
Kz
Kx
Kx ≠ Kz
••
Kx = Kz = K
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Homogeneous, isotropic Homogeneous, anisotropic
A A B B
Heterogeneous, isotropic Heterogeneous, anisotropic
AA
B
B
Characteristics of K in two dimensions
Kz
Kx
Kx ≠ Kz
••
Kx = Kz = K
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Wt
hS
z
hK
zy
hK
yx
hK
xszyx −
∂∂=
∂∂
∂∂+
∂∂
∂∂+
∂∂
∂∂
)()()(
2
2
)(x
h
x
h
x ∂∂=
∂∂
∂∂ 0
2
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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To t h Pr o b le m ( 2 D)
Impermeable Rock
Groundwater divide
Groundwater divide
2 D, s t e a d y s t a t e
02
2
2
2
=∂∂+
∂∂
z
h
x
h
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i, jj
i
The 5 point starin a regular grid
i+1, ji-1, j
i, j-1
i, j+1
∆x
∆y
∆x = ∆y
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04 ,1,1,,1,1 =−+++ −+−+ jijijijiji hhhhh
4
11,1,
1,1,11
,
+−+
+−++ +++
=m
jim
jimji
mjim
ji
hhhhh
Solution by iteration
Gauss-Seidel Iteration
4/)( 1,1,,1,1, −+−+ +++= jijijijiji hhhhh
Iteration indices
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Iterationplanes
initial guessesm
solution
m+2
m+1
m+3
41,1,,1,11
,
mji
mji
mji
mjim
ji
hhhhh −+−++ +++
=
4
11,1,
1,1,11
,
+−+
+−++ +++
=m
jim
jimji
mjim
ji
hhhhh
Gauss-Seidel Iterationadmission.edhole.com
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 examplewith specified headboundary conditions
(Assume an initialguess of h =10 forall interior nodes)
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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
Examplewith specified headboundary conditions
9.75
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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
Examplewith specified headboundary conditions
8.44
9.75
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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
Examplewith specified headboundary conditions
9.75
9.94
8.44
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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
Examplewith specified headboundary conditions
9.94
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i, jj
i
Gauss-Seidel Iteration
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C4=(C3+C5+B4+D4)/4admission.edhole.com
4
33243
ECDDD
+++=
Example of spreadsheet formula
Toth Problem
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