Groundwater 40x larger than lakes+rivers

Groundwater 40x larger than lakes+rivers

• Ocean Volume = 1.4x 109 km3

Oceanic Evaporation Rate = 3.2x105 km3/yr• Groundwater Volume = 8.4x106 km3

• River+Lake Volume = 2x105 km3

• Atmosphere Volume = 1.3x104 km3

Runoff Rate = 3.6x104 km3/yr

Surface rock typically contains several

percent water within it

Permeability & Porosity

Impermeable

Permeable

Aquifer & Aquitard

(Retards flow)

(Better flow)

Factoid: Along the South Carolina Coast, half the water entering the ocean is entering by subsurface groundwater flow [a rare measurement]

Water Table & Vadose Zone

Vadose Zone

Why a water table?

(Discuss animation)

Regional Water Table

Water Flow Paths

Travel Times in Regional Aquifers

Hydraulic Gradient

L

LHydraulic Gradient

Darcy’s Law

Given a hydraulic gradient Dh/L

V=QA=−KDhL

V ms ≡

m3 sm2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Darcy Velocity V is a flow rate per unit cross-sectional area

(has units of velocity)

Hydraulic conductivity also has units of velocity since the hydraulic gradient is dimensionless

True Fluid Velocity larger than Darcy velocity

Darcy velocity V = (volume fraction fluid ) x (fluid velocity v)

so v = V / = (true fluid velocity)/ (volume fraction fluid)

e.g., for 10% porosity, the true fluid velocity is 10 times larger than the Darcy velocity

Hydraulic Head, Deviatoric Pressure, & Pressure

Hydrostatic pressure Pw is the fluid pressure from the weight of the overlying column of water: Pw = wgz

Deviatoric pressure p is the pressure at a given point that is in excess or deficit of the hydrostatic pressure, i.e. P = Pw + p

Hydraulic head h is the elevation h of the water level in a hypothetical well drilled at that point.

The total pressure P = wgh at that point: h = P/(wg)

Likewise, the deviatoric pressure is similarly related to the change in hydraulic head: p = wgh or h = p/(wg)

Conservation of MassIngredients needed to solve for groundwater flow:

(1) Darcy’s Law (relates flow to hydraulic head h)

(2) Conservation of mass (i.e., balance between what goes in and out of a region)

(3) Put together, (1) and (2) lead to the equation

V =−K∇h (≡ −K Grad h)e.g. Vx=−Kdh

dx; Vz=−Kdhdz

∇•V =0 (≡ Del dot V =0)

e.g. dVxdx

+ dVzdz

=0

∇• −K∇h( ) = −K∇2h = −K ∂2h∂x2 + ∂2h

∂z2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= 0 (here assume

constant K)

Using Mass Conservation to solve for flow into a well

(1)Note that the problem is symmetric: Assume the flow Vr depends only on the distance r from the well.

(2) By conservation of mass, the total flow (=extraction rate Q from the well) across each circle of radius r must be constant (the same amount flows across each surface of radius r and thickness b, or else mass would not be conserved)

(3) Therefore Vr*b*2r = Q,

or Vr = Q/(2r b)

Q

r

b

Vr

Darcy’s Law - Flow proportional to a ‘Potential Function’

Ingredients needed to solve for groundwater flow…

(1) Darcy’s Law (relates flow to hydraulic head h)

V =−K∇h =−∇Φ (≡ −K Grad h)e.g. Vx=−Kdh

dx; Vz=−Kdhdz

∇•V =0 (≡ Del dot V =0)

e.g. dVxdx

+ dVzdz

=0∇• −K∇h( ) = −K∇2h = −K ∂2h

∂x2 + ∂2h∂z2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= 0

(here assume constant K)

Using a potential function simplifies the problem in that we only need to solve for one function h (or its generalization Φ) instead of two individual velocity components Vx and Vz. However, the equation we need to solve for the potential has higher order derivatives and is usually more complicated to solve.

Flow into or out of a well (2)Q

Vr*b*2r = Q,

or Vr = Q/(2r b)

Can integrate this to find the Darcy flow potential

e.g. Vr =−Kdhdr=−

dΦdr =− Q

2brΦ=− Q

2blnr + C

The linearity of Darcy’s Law allows flow potentials to be added (e.g. can easily combine simple solutions to make a more complicated one…)

Flow from one well to another

Φ=− Q2πb

lnr1 + Q2πb

lnr2

The linearity of Darcy’s Law allows flow potentials to be added …

production wellinjection well

-xo xo

b

-Q Q

r1 r2

Vx =∂Φ∂x

-Q-

2b= [(x+xo)2 + y2x+xo ](x-xo)2 + y2x-xo-

Vy=∂Φ∂y

-Q-

2b= [(x+xo)2 + y2 y ](x-xo)2 + y2 y

-

(5)

(6)

Φ =

−Q4b

ln x+ x0( )2+ y2⎛

⎝⎞⎠−ln x−x0( )

2+ y2⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

y

x

Downdraw of the water table & cone of depression around a well…

(Discuss animation)

Regional Groundwater Flow (1)

(1)To solve for regional groundwater flow, in general need to use both conservation of mass and Darcy’s Law

z

x

hill

river

groundwater table

h

1/2

Darcy Flow

h is the elevation of the water table above the x-axis

Is the wavelength of the cosine curve representing topography (=2/k, where k is the wavenumber)

Regional Groundwater Flow (2) z

x

hill

river

groundwater table

h

1/2

Darcy Flow

(1) Darcy’s Law

(2) Mass Conservation

(3) Put together to create a relation for h

V =−K∇h ; e.g. Vx=−Kdhdx; Vz=−Kdh

dz∇•V = dVx

dx +

dVzdz

=0

∇• −K∇h⎛

⎝⎜

⎞

⎠⎟ = −K∇2h = −K ∂2h

∂x2 + ∂2h∂z2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= 0

Regional Groundwater Flow (3) z

x

hill

river

groundwater table

h

1/2

Darcy Flow

For constant K, this can be simplified to LaPlace’s equation.

−K ∂2h∂x2 + ∂2h

∂z2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= 0→ ∂2h

∂x2 + ∂2h∂z2 = 0

To solve this equation, we again assume (i.e.,hope for) a simplified form for the solution, that h = X(x)Z(z)

(Separation of Variables — perhaps the most useful trick in PDEs)








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Groundwater 40x larger than lakes+rivers