1. Darcy’s Law: Q = KA(–dh/dl)
GEOTECNIA CI44A
Una generalización de la ecuación de Darcy para 2D y 3D
Darcy’s Law: Q = KA(–dh/dl)
1-dimensional

2. Balance de masa en un elemento diferencial
GEOTECNIA CI44A
Balance de masa en un elemento diferencial
Units

3. Shell Balance
Rate of mass accumulation = (flux x area)in – (flux x area)out

4. Shell Balance Group mass input and output at each face
Write differences as deltas

5. Shell Balance Multiply each term by 1 (e.g., across x face, Dx/Dx)
To yield:

6. Darcy’s Law Assume density does not vary too much spatially:
Substitute Darcy’s Law q = –KDh/Dl to yield:

7. Mass Accumulation – Short Form
Express DM/Dt in terms of specific storage
Ss = [m3 storage m–3 aquifer h–1]
Express in terms of mass
Take time derivative

8. S ~ SY for unconfined aquifers
Q: Where does water come from? (Part B)
A: In an unconfined aquifer, the answer is pretty straight forward. As the WT moves up and down, pores in the earth fill and drain.
How much water you would get depends on how much water can freely drain from the aquifer. (Do you recognize this phrase?)
The volume of water you would get out of a cubic meter of earth would be controlled by its Specific Yield (SY). (As you will later see, this is not 100% true, but is an excellent approximation)
We call this number the Storage Coefficient (or Storativity) (S)
(units - volume/volume = unitless!)
S ~ SY for unconfined aquifers
Storativity - the volume of water an aquifer releases from or takes
into storage per unit surface area per unit change in head

9. = volume of aquifer affected * Storage Coefficient
To calculate the volume released (or stored) by a change in head (dh) over an area A
Volume water released
= volume of aquifer affected * Storage Coefficient A
V = S (dh) A dh
Example:
If the water table rises by 3.0 feet in an unconfined aquifer with a Specific Yield of 20% over an area of 2.5 acres, how much additional water is stored in the aquifer?
V = S (dh) A S = SY
= 0.20 * 3.0 * (2.5*43560)
= ft3

10. In a confined aquifer, the answer is not quite so straight forward.
As pressure in the aquifer rises and falls (and therefore water levels in wells penetrating the confined aquifer rise and fall), the pores in the confined aquifer remain saturated. So where is the water coming from in this case?
Elastic storage - water is squeezed (expelled) from storage due to compressibility of the soil matrix.
Compressibility of water – water expands with a decrease in pressure
To fully answer this question, we must understand something about stresses in the earth, and the response of the earth to stresses.

11. sTot = se + P sT Effective Stress
Consider the stress applied across a plane in a fluid saturated porous media.
Some of the stress is accommodated by the fluid pressure
Some of the stress is accommodated by the grains
The stress felt by the grains themselves  Effective Stress sT
Fluid Pressure P
Effective Stress se
sTot = se + P
Total Stress = Effective Stress + Pore Water Pressure

12. sTot = P + se
If we change one of these total stresses, the change must be accommodated by changes in the other stresses
dsTot = dP + dse OR
dsTot – dP = dse

13. dsTot – dP = dse
In many situations, dsTot doesn't change very much in comparison to the change in pressure, so dsTot = 0 and
–dP = dse
As P decreases, se increases.
As se increases, the stress on the soil matrix increases, and the aquifer skeleton may consolidate or compact
shifting of mineral grains
mineral grain breakage
Such compaction causes release of water.

14. dV V In a 1- D case (where b = aquifer thickness, in the direction
Porous medium
Compressibility T V dV a s e
In a 1- D case (where b = aquifer thickness, in the direction
of compression

15. Water Release In terms of a pressure change
Therefore, water release due to a change in pressure
Some Typical Compressibilities
Clay 10 - 6
to 10 - 8 m 2 /N
Sand 10 - 7
to 10 - 9 m 2 /N
Gravel 10 - 8
to 10 - 10 m 2 /N
Sound Rock 10 - 9
to 10 - 11 m 2 /N

16. Compressibility of water
Compressibility: change in volume per change in stress
Compressibility of water w V dV b P
for water, b = 4.6x10–10 m2/N
Here we are talking about the volume of water, not total volume

17. compaction of the aquifer expansion of the water
In a confined aquifer, the water that is pumped out of a well results from lowered hydrostatic pressure around the well (and consequently results in an increase in effective stress in the aquifer around the well). This causes:
compaction of the aquifer
expansion of the water
Potentiometric
Surface Drops dh db b
A) Aquifer compresses
B) Water expands

18. sTot
sTot se P
As P decreases, se increases (and sTot remains constant)
Grains deform
Water expands (slightly)

19. What would happen to the fluid pressure in the confined aquifer if we put an additional load on the aquifer?
(for example move a train on top of the aquifer?)
 sTot increases, so both se and P increase
What happens to fluid levels in wells that penetrate the confined aquifer?
As P increases, rgh must increase (P=rgh)
Since r and g are essentially constants (actually r can vary a very little bit), h must do most of the changing to compensate – in other words water levels rise!

20. As we pump from a confined aquifer, the aquifer compresses and the water expands. Which is more important?
In other words, how much water do we get from a volume of aquifer per unit drop in the potentiometric surface from
Aquifer Compaction?
Water Expansion?

21. A) How much of the water comes from compaction of the aquifer?
dVw = -dVT (Vw = volume of water, VT = total volume)
recall that a = [-dVT/VT] /dse = aquifer compressibility
(definition of aquifer compressibility from above)
= [-dVT/VT] /dse = [dVw/VT] /dse (substitution)
dVw = a VT dse (rearranging)
for a unit volume (VT = 1)
dVw = a dse (substitution)
(continued)

22. dVw = a dse (from above)
since dse = -dP (from discussion of effective stress)
and P = rgh (from discussion of potentiometric surface)
dse = -d(rgh) (substitution)
dVw = a (-rgdh) (substitution)
For a unit drop in potentiometric surface (-dh = 1)
dVw = a rg
(a) (component of water from aquifer compression)

23. B) How much of the water comes from expansion of the water?
recall that b = [-dVw/Vw] /dP = water compressibility
dVw = -bVwdP (rearranging)
also recall that for saturated material
Vw = nVT (definition of porosity)
And dP = rgdh
(from discussion of potentiometric surface)
dVw = -b( nVT )(rgdh) (substitution)
For a unit volume (VT =1) and a unit drop in potentiometric surface (dh = -1)
 dVw = b(n)(rg)
dVw = b nrg (b) (component of water from water expansion)

24. Specific Storage (SS) = rg(a + nb)
Combining (a) and (b)
Volume of H2O released/(unit volume)/(unit change potentiometric surface)
(for now we’ll call a change in the potentiometric surface “change in head” – we’ll examine this in more detail shortly)
Vol. H2O/unit volume/unit Dhead
= arg + b nrg = rg(a + nb)
Specific Storage (SS) = rg(a + nb)

25. S = SS * b Volume of H2O released/(unit volume)/(unit change in head)
= rg(a + nb) = SS = Specific Storage
Specific Storage (SS) units = L-1 (e.g. m-1, ft-1, cm-1, etc.)
{M/L3 * L/T2 * L2/(ML/T2) = 1/L}
To convert this to a Storage Coefficient (S) (aka Storativity)
multiply SS by the aquifer thickness (b)
S = SS * b
Storativity = volume of water released from or taken into storage per unit surface area per unit change in head (units – unitless!)

26. S = SS*b V = SS*dh*b*A V = Sdh A
Volume of water released per unit area per dh (S)
S = SS*b
To calculate the total volume of water released by a confined aquifer due to a change in head, you need: SS
the area over which the dh takes place
and
the aquifer thickness
Volume of water released per unit thickness per dh (SS)
V = SS*dh*b*A
V = Sdh A

27. Some Typical Compressibilities (a)
Let's put Storativity and Specific Storage into perspective. How big a number are they???
For unconfined aquifers, we said that the Storage Coefficients are approximately equal to their Specific Yields.
SPECIFIC YIELDS high low avg.
Medium Gravel 26% 13% 23%
Medium Sand 32% 15% 26%
Silt 19% 3% 18%
Clay 5% 0% 2%
For confined aquifers, Storage Coefficients are related to their Specific Storages and their thicknesses. The typical storage coefficients for confined aquifers are MUCH smaller than for unconfined aquifers (0.001 to not uncommon)
To calculate some, you could use the following data, along
with typical porosities and typical thicknesses
SS = rg(a + nb)
S = b SS
Some Typical Compressibilities (a)
Clay to 10-8 m2/N
Sand to 10-9 m2/N
Gravel to m2/N
Sound Rock 10-9 to m2/N
Water b = 4.6 x m2/N

28. Let’s go back to that question about the relative importance of
aquifer compaction
and
water expansion
and put some numbers on them now…
How much water do we get from a unit volume (1 m3) of aquifer per unit (1 m) drop in the potentiometric surface from:
rigid
clay sand rock rock
n = 40% 30% 10% 5%
a = 10-7 m2/N
Vol. Water Due To:
Aquifer Compaction (arg) 9.8x10-4 m3 9.8x x x10-8
Water Expansion (bnrg) 1.8x10-6 m3 1.4x x x10-7
(b = 4.6x10-10 m2/N)
Compaction:Expansion !!!!
Typical numbers 
rwater = 1000 kg/m3 g = 9.8 m/sec2

29. How Does One Recognize Different Aquifer Types?
Property Confined Unconfined
Stratigraphy Closed to atmosphere Open to atmosphere
Storativity to to 0.01
Water Level (Topographic) Does not reflect topography Reflects topography
Water Level (Pump) Pressure response (fast, wide) Drainage response (slow, local)
Water Level (ET) Does not respond diurnally to ET May respond to ET
Water Level (Barometric) Responds to barometric change Does not respond to barometric change
Water Level (Load) Responds to load Does not respond to load
Water Level (Tidal) Strong Tidal Response Weak to no tidal response

30. Equate Mass Accumulation and Net Flux
Express DM/Dt in terms of specific storage
Density and volume cancel

31. Final Forms
In the limit as D goes to zero,
K here is a tensor

32. Mass Accumulation Mass in differential element: Mass accumulation:
Chain rule:

33. Mass Accumulation = Net Mass Input
Note that the term cancels
Mass accumulation term:
Flux term:

34. Mass Accumulation Term
The flux term is written in terms of head
Relate Dn/Dt and Dr/Dt to changes in head ? ?

35. Dn/Dt Term Definition of matrix compressibility
Defined relative to a constant mass of solids
Replace VT with Vv + Vs

36. Dn/Dt Term
Divide through by Dt
Divide through by VT
= n
= 0
= 1

37. Dn/Dt Term Relate DP/Dt to changes in head Divide through by Dt
Recognize Dz/Dt = 0. Solve for DP/Dt, substitute

38. Dr/Dt Term  Definition of compressibility
Defined relative to a constant mass of fluid
Chain rule, then solve for Dr/Dt
Substitute Dh/Dt for DP/Dt : 

39. Mass Accumulation Term
Substitute
simplify

40. Groundwater Flow Equation
Substitute, cancel r and DxDyDz, let D go to zero

41. Homogeneous Isotropic
Hydraulic conductivity does not depend on position or direction
K here is a scalar

42. Homogeneous Isotropic
It applies to transient flow in 3-D
Homogeneous and isotropic conditions
Specify K, a, n, b, r
Solution h(x,y,z,t) describes the head at any position, at any point in time
Sometimes known as the diffusion equation
Multiply both sides by aquifer thickness b
Where and therefore

43. Steady State Conditions
Mass accumulation is equal to zero
The
LaPlace
Equation

44. Unconfined Aquifer Consider 1-D flow
Head and y-component of flow area are now the same ho hL
h(x) y L z x

45. Unconfined Aquifer - Flux
(flux x area)in – (flux x area)out = h o h L y z x

46. Unconfined Aquifer - Flux
Substitute Darcy’s Law q = –Kdh/dl to yield:
Assume density, Dz and Dx are constant

47. Unconfined Aquifer – Mass Accumulation
Mass in control volume:
Mass accumulation term (incompressible media):
Substitute
Cancel density, Dz and Dx, shrink D to zero

48. Unconfined Aquifer Non-linear PDE:
Homogeneous, 1D and 2-D (Boussinesq Eq):
Changes in h are small, treat as a constant (average)

49. Unconfined/Confined Comparison
1-D Equations

50. Modeling
We have the model: head (h) is described in the equations in terms of x, y, z, and t
Need to solve the differential equation!
If: boundary conditions can be described algebraically
If: the aquifer is homogeneous and isotropic
Then: we may be able to solve the mathematical model by integral calculus
E.g., separate variables and integrate
predict hydraulic head and flow
More typically, we solve numerically (using computer models)
OR solve graphically (using FLOW NETS)

51. Flow Nets Graphical solution to LaPlace’s Equation
Graphical depiction of equipotential lines and flow lines
Valid for 2-D, incompressible media and fluid, steady state conditions