The EXPLICIT SOLUTION was given as equation 9 on the finite difference main page:



to ensure stability:



so use 0.07 days and note that you can solve for head at any node by:



for the first time step:

volume of flow into the left boundary for time step is:

VOL_left = Q dt = Ti dx DT

VOL = 0.06 ft²/d (h₁-h₂)/y dx DT

= 0.06 ft²/d * (8.2-8.2)ft/3ft * 1 ft * 0.07d = 0 ft³

volume of flow out of the right boundary is:

VOL_right = Q DT = Ti dx DT

VOL = 0.06 ft²/d (h₄-h₅)/dy 1 ft 0.07d

= 0.06 ft²/d (6.06-3.6)ft/3ft 1 ft 0.07d = 3.4x10^-3 ft³

the change in storage all occurred in cell 4:

Change in S₄ = (h₄⁰-h₄^0.07)*S*dx*dy

= (8.2ft - 6.06ft)*0.001*(3ft²) = 6.4x10^-3 ft³

Mass Balance

Mass Balance = (Inflow - Outflow) / ((Inflow+Outflow) / 2)

water coming out of storage is seen as an inflow

(6.4x10^-3 ft³ - 3.4x10^-3 ft³) / 4.93^-3 ft³ = 0.61

does not look good!

h₁ through h₅ are: 8.2 8.2 8.2 6.06 3.6

VOL_left = 0

VOL_right =3.2x10^-3 ft³

Change in Storage (cells 3 and 4) =3.4x10^-3 ft³

Mass Balance = 0.06

the mass balance is looking better

using the heads from time = 0.14 to solve for heads at time = 0.21:



and so on ...

Notice that given the character of the explicit equations the pressure front can only move one cell at a time, consequently large time steps lead to fairly large inaccuracies in the solution

You may wish to explore the following exercise related to the material you just studied now, in which case follow the steps below. Otherwise continue the discussion for this unit by going "BACK" and exploring the implicit option, then take up the exercises later when you reach the exercise section on the main page for this unit.

For the same parameters you were using above:

1. Calculate h₄ @ 0.07 day increments to 0.7 days using the explicit approach

2. Repeat #1 @ 0.14 day increments to 0.7 days using the explicit approach

3. Graph your results as head vs time

4. Explain what you see