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

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

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

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

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

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

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

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

= 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) = 65340 ft3

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.

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

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

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.

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

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

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

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

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

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!

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?

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)

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)

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)

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)

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!)

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

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 0.00001 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 10-6 to 10-8 m2/N Sand 10-7 to 10-9 m2/N Gravel 10-8 to 10-10 m2/N Sound Rock 10-9 to 10-11 m2/N Water b = 4.6 x 10-10 m2/N

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 10-8 10-10 10-11 Vol. Water Due To: Aquifer Compaction (arg) 9.8x10-4 m3 9.8x10-5 9.8x10-7 9.8x10-8 Water Expansion (bnrg) 1.8x10-6 m3 1.4x10-6 4.5x10-7 2.2x10-7 (b = 4.6x10-10 m2/N) Compaction:Expansion 544 70 2 0.45!!!! Typical numbers  rwater = 1000 kg/m3 g = 9.8 m/sec2

How Does One Recognize Different Aquifer Types? Property Confined Unconfined Stratigraphy Closed to atmosphere Open to atmosphere Storativity 0.001 to 0.0000001 0.1 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

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

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

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

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

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

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

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

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

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 : 

Mass Accumulation Term Substitute simplify

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

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

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

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

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

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

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

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

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

Unconfined/Confined Comparison 1-D Equations

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)

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