Transcript Document

Governing Eqn. for Reservoir Problem
 2h
x 2
S h

T t
1D, transient, homogeneous, isotropic,
confined, no sink/source term
•
•
Explicit solution
Implicit solution
 2h
x 2
S h

T t
hi 1  2hi  hi 1
(x) 2
n 1
S hi  hi

T
t
n
Explicit Approximation
hi 1n  2hi n  hi 1n
(x) 2
n 1
n
S hi  hi

T
t
hi 1  2hi  hi 1
n
n
n 1
S hi  hi

T
t
n
x 2
n
Explicit Solution
hin 1

Tt hi 1  2hi  hi 1

(
)
2
S
(x)
n
hin
n
n
Eqn. 4.11
(W&A)
Everything on the RHS of the equation is known.
Solve explicitly for hin1 ; no iteration is needed.
• Explicit approximations are unstable with large
time steps.
• We can derive the stability criterion by writing
the explicit approx. in a form that looks like the SOR
iteration formula and setting the terms in the
position occupied by omega equal to 1.
• For the 1D governing equation used in the reservoir
problem, the stability criterion is:
2tT <
 1
2
S (x)
or
2
S
(

x
)
<
t  0.5
T
 2h
x 2
S h

T t
hi 1  2hi  hi 1
x 2
n 1
S hi  hi

T
t
n
Implicit Approx.
hi 1
n 1
 2hi
n 1
(x) 2
 hi 1
n 1
n 1
S hi  hi

T
t
n
Implicit Solution
hi 1
n 1
 2hi
n 1
x 2
 hi 1
n 1
n 1
S hi  hi

T
t
n
Solve for hi n1 to produce the Gauss-Seidel
iteration formula.
(hi
n 1 m1
)

n1 m1
n1 m
n
function{( hi 1 ) , (hi 1 ) , hi }
Could also solve using SOR iteration.
(hi, j
n1 m1
)
 (hi, j
n1 m
)   [(hi, j
n1 m1
)
 (hi, j
n1 m
) ]
Gauss-Seidel value from
previous slide.
n+1
t
m+3
Iteration
planes
m+2
m+1
n
Water Balance
Storage = V(t2)- V(t1)
IN > OUT then Storage is +
OUT > IN then Storage is –
OUT - IN = - Storage
+
-
Flow in
Flow out
Storage
Storage
Convention: Water coming out of storage
goes into the aquifer (+ column).
Water going into storage comes out
of the aquifer (- column).
Water Balance
V = Ss h (x y z)
t
t
V = S h (x y)
t
t
In 1D Reservoir Problem,  y is taken to be equal to 1.
 2h
h1
x
datum
0
2
 0
h2
x
L = 100 m
At t = tss the system reaches
a new steady state:
h(x) = ((h2 –h1)/ L) x + h1
(Eqn. 4.12
W&A)