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

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

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

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

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.

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

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

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

(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.

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:

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

Explicit Approximation

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

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

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

Explicit Solution Spreadsheet

t = 5 min

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

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

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

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)

Reservoir Problem Water Balance IN 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)

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

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

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.

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

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.