1. Transient Water Balance Eqn.
Inflow = Outflow +  S
Recharge
Discharge

2. OUT – IN =
x y z
= - V/ t
= change in storage
Ss = V / (x y z h)
V = Ss h (x y z) t

3. OUT – IN =
General governing equation for transient, heterogeneous, and anisotropic conditions

4. div (K grad h) = Ss (h t) – W
Law of Mass Balance + Darcy’s Law =
Governing Equation for Groundwater Flow
div q = - Ss (h t) +W (Law of Mass Balance)
q = - K grad h (Darcy’s Law)
div (K grad h) = Ss (h t) – W

5. 2D profile 2D confined 2D unconfined
Storage coefficient is either storativity (S) or specific yield (Sy).
S = Ss b (& Tx = Kx b; Ty= Ky b)
Ss is specific storage.

6. End Transient problems require multiple arrays of head values n+3 Time
planes t
n+2 t
n+1 t
Initial conditions

7. 2D confined
1D, transient, homogeneous, isotropic,
confined, no sink/source term

8. Reservoir Problem 1D, transient
Confined Aquifer
t > 0
1D, transient

9. (represents static steady state)
BC:
h (0, t) = 16 m; t > 0
h (L, t) = 11 m; t > 0
datum x
L = 100 m
t = 0
IC: h (x, 0) = 16 m; 0 ≤ x ≤ L
(represents static steady state)
Modeling “rule”: Initial conditions should represent a
steady state configuration of heads.

10. At t = tss the system reaches a new steady state:
datum x
L = 100 m
At t = tss the system reaches
a new steady state:
h(x) = ((h2 –h1)/ L) x + h1
[analytical solution: equation 4.12 (W&A)]
BC:
h(L) = h2
h(0) = h1
GE:

11. Governing Eqn. for transient Reservoir Problem
1D, transient, homogeneous, isotropic,
confined, no sink/source term
i-1,j
i+1,j
i,j
i+1/2
i-1/2

12. Explicit Approximation

13. Explicit Solution Eqn. 4.11 (W&A)
Everything on the RHS of the equation is known.
Solve explicitly for ; no iteration is needed.

14. Explicit approximations are unstable unless
small time steps are used.
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  equal to 1.
For the 1D governing equation used in the reservoir
problem, the stability criterion is: or

15. For the reservoir problem:
T = 0.02 m2/min
S = 0.002
x = 10 m

16. Explicit Solution
Spreadsheet

17. t = 5 min

18. Transient Water Balance Eqn.
Inflow = Outflow +Storage
 Storage = V2 – V1
Recharge t2
Discharge
V2 > V1

19. Transient Water Balance Eqn.
Inflow = Outflow +Storage
 Storage = V2 – V1
Recharge t1
Discharge t2
V2 < V1
Inflow -Storage = Outflow
Inflow + abs(Storage) = Outflow

20. Water Balance- version 1
Storage = V(t2)- V(t1)
IN > OUT then water going into storage
and Storage is +
OUT > IN then water is coming out of storage
and Storage is –
IN – OUT= Storage
MODFLOW Convention: Water coming out of storage goes into the aquifer (IN column).
Water going into storage comes out
of the aquifer (OUT column).
Total IN
Total OUT
Total IN = Total Out
Flow in
ABS(Flow out)
ABS(Storage)
Storage

21. Water Balance- version 2
Storage = V(t1)- V(t2)
IN > OUT then water going into storage
and Storage is -
OUT > IN then water is coming out of storage
and Storage is +
IN - OUT = Storage
Convention: Water coming out of storage goes into the aquifer (IN column).
Water going into storage comes out
of the aquifer (OUT column).
Total IN(+)
Total OUT(-)
Flow in
Storage
Flow out
Total IN = ABS(Total Out)

22. Reservoir Problem Water BalanceIN
OUT
t > 0
Storage
IN + change in storage = OUT
(where here the change in
storage term is intrinsically positive) + -
Flow in
Flow out
Convention: Water coming out of storage goes into the aquifer (+ column).
Storage
Total IN = ABS(Total Out)

23. Units + -
Flow in
Flow out
Storage
L3/T
(m3/min)

24. OUT – IN =
x y z
= - V/ t
= change in storage
Ss = V / (x y z h)
V = Ss h (x y z) t

25. Storage Term for the Water Balance
3D problem:
V = Ss h (x y z) t t
V = S h (x y) where S = Ss b t
2D problem:
In 1D Reservoir Problem,  y is taken to be equal to 1.

26. Water Balance
% Error = ABS(Total IN- Total Out) / Min(Total IN:Total Out)

27. Stablity criterion for explicit solution:
For the reservoir problem:
T = 0.02 m2/min
S = 0.002
x = 10 m
Judging by the water balance error,
t = 5 min may be too large.