GEOS 4430 Lecture Notes: Darcy’s Law

Dr. T. Brikowski

Fall 2013

0file:darcy law.tex,v (1.24), printed October 15, 2013

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Introduction

I Motivation: hydrology generally driven by the need foraccurate predictions or answers, i.e. quantitative analysis

I this requires mathematical description of fluid andcontaminant movement in the subsurface

I for now we’ll examine water movement, beginning with anempirical (qualitative) relation: Darcy’s Law

2

Darcy’s Observations

I Henry Darcy, studying public water supply development inDijon, France, 1856. Trying to improve sand filters for waterpurification

I apparatus measured water level at both ends and discharge(rate of flow L3

T ) through a vertical column filled with sand(Fig. 1)

3

Darcy Experimental Apparatus

L

h

h1

A

Cross−sectional area

Q

2

Figure 1: Simplified view of Darcy’s experimental apparatus. The centralshaded area is a square sand-filled tube, in contact with water reservoirson either side with water levels h1 and h2.

4

Empirical Darcy Law

I Darcy found the following relationship

Q ∝ A(h1 − h2)

L

or

Q = KAh1 − h2

L(1)

where K is a proportionality constant termed hydraulicconductivity L

T

5

Head as Energy

I Head as a measure of energy (1) was used without muchfurther consideration until the 1940’s when M. King Hubbertexplored the quantitative meaning of head h, and ultimatelyderived what he termed force potential Φ = g · h. Hedemonstrated that head is a measure of the mechanicalenergy of a packet of fluid

I mechanical energy of a unit mass of fluid taken to be the sumof kinetic energy, gravitational potential energy and fluidpressure energy (work).

I Energy content found by computing work to get to currentstate from a standard state (e.g. z = v = p = 0)

I energy units are kg m2

sec2 = Nt ·m = joule

6

Head Energy Components

I Kinetic Energy or “velocity work”. Energy required toaccelerate fluid packet from velocity v1 from velocity v2.

Ek = m

∫ v2

v1

vdv =1

2mv2 (2a)

Remember that the integral sign is really just a fancy “sum”,saying “add up all the tiny increments of change betweenpoints 1 and 2”

I Gravitational work. Energy required to raise fluid packet fromelevation z1 to elevation z2.

W =

∫ z2

z1

dz = mgz (2b)

7

Head Energy Components (cont.)

I Pressure work. Energy required to raise fluid packet pressure(i.e. squeeze fluid) from P1 to P2.

P =

∫ P2

P1

VdP = m

∫ P2

P1

V

m= m

∫ P2

P1

1

ρ

=1

ρ

∫ P2

P1

dP =P

ρ(2c)

where (2c) assumes a unit mass of incompressible fluid

8

Final Expression for Head

I the sum of (2a)–(2c) is the total mechanical energy for theunit mass (i.e. m = 1)

Etm =v2

2+ gz +

P

ρ(3)

I in a real setting, energy is lost when flow occurs Fetter (Sec.4.2, 2001), so Etm = constant implies no flow. No flow(Q = 0) in (1) implies constant head (h1 = h2).

I Following the analysis of Hubbert (1940) define head h suchthat h · g is the total energy of a fluid packet. Then

hg =v2

2+ gz +

P

ρ(4a)

h =v2

2g+ z +

P

ρg(4b)

= z +P

ρg(4c)

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Final Expression for Head (cont.)

where (4c) assumes v is small (true for flow in porous media).

I head is a combination of elevation (gravitational potential)head and pressure head (where P = ρghp, and hp is thepressure head, or pressure from the column of water above thepoint being considered)

I for a given head gradient ( ∆h∆x ) discharge flux Q is

independent of path (Fig. 4.5, Fetter, 2001).

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Head for Variable-Density Fluids

I (4c) indicates that head also depends on fluid density. This isonly an issue for saline or hot fluids. For constant density (4c)can be written as

h = z + hp

where hp is the height of water above the point of interest(eq. 4.11, Fetter, 2001).

I This implies that head is constant vs. z (for every meterchange in z there is an equal but opposite change in hp).

I when ρ isn’t constant, “fresh-water” head (the head for anequivalent-mass fresh water column, hf =

ρpρfhp) should be

used for computing gradients, or in flow models (Pottorffet al., 1987)

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Applicability of Darcy’s Law

I Darcy’s Law makes some assumptions, which can limit itsapplicability

I Assumption: kinetic energy can be ignored. Limitation:laminar flow is required (i.e. Reynold’s number low, R ≤ 10,viscous forces dominate)

I Assumption: average properties control discharge.Limitation: (1) applicable only on macroscopic scale (forareas ≥ 5 to 10 times the average pore cross section, Fig. 2)

I Assumption: fluid properties are constant. Limitation: (1)applicable only for constant temperature or salinity settings, orif head h is converted to fresh-water equivalent hf

12

Scale Dependence of Rock Properties

φ, K, k

Average

Homogeneous Medium

Po

ros

ity

, P

erm

ea

bil

ity

, o

r C

on

du

cti

vit

y

Distance

Heterogeneous M

ediumMicroscopic Macroscopic

Figure 2: Variation of rock properties with scale. Note heterogeneousmedia have average properties that vary positively (shown) or negatively(not shown) with distance. See also Bear (Fig. 1.3.2, 1972).

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True Fluid Velocity

I (1) gives total discharge through the cross-sectional area A

I specific discharge (q traditionally, v in the text) is the flux perunit area q = Q

A

I pore or seepage or average linear velocity (v traditionally, Vx

in the text) is the velocity at which water actually moveswithin the pores v = K

φdhdl

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Hydraulic conductivity (K )

I the proportionality constant in (1) depends on the propertiesof the fluid, a fact that wasn’t recognized until Hubbert(1956)

I Hubbert rederived Darcy’s Law from physical principles,obtaining a form similar to the Navier-Stokes equation1. Fromthis he found the following form for K :

K =kρg

µ(5)

where ρ is the density of the fluid and µ is its dynamicviscosity ( M

LT). k is the intrinsic permeability of the porousmedium Fetter (written as Ki eqn. 3-19, 2001).

I (5) demonstrates that hydraulic conductivity depends on theproperties of both the fluid and the rock

1http://en.wikipedia.org/wiki/Navier-stokes15

Permeability (k)

I k (units L2) depends on the size and connectivity of the porespace, and can be expressed as k = Cd2, where C is thetortuosity of the medium (depends on grain size distribution,packing, etc.; “unmeasurable”), and d is the mean graindiameter (a proxy for the mean pore diameter). See alsoFetter (sec. 3.4.3, 2001).

I standard units are the darcy, (1 darcy = 9.87x10−9cm2),which is defined as the permeability that produces unit fluxgiven unit viscosity, head gradient, and cross-sectional area(Sec. 3.4.2, Fetter, 2001)

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Measuring Rock Hydraulic Properties

I Porosity is measured in the lab with a porosimeterI sample is dried and weighedI then resaturate sample with water (or mercury), very difficult

to do completelyI measure volume or mass of liquid absorbed by sampleI yields effective porosity

I Permeability/Hydraulic Conductivity are measured in the labusing permeameters (Fig. 3), which apply Darcy’s Law todetermine K

I Constant Head permeameter best used for high conductivity,indurated samples (rocks)

I Falling Head permeameter best used for low-permeabilitysamples or soils

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Permeameters

(a) Constant HeadPermeameter: K = VL

Ath

(b) Falling Head Permeameter:

K =d2t

d2c

Lt

ln(hoh

)Figure 3: Constant and Falling-Head Permeameters. a) best for consolidated, high-permeability samples; b)best for unconsolidated and low-permeability samples. t is the duration of the experiment, V = Qt is the totalvolume discharged, and A is the cross-sectional area of the apparatus. After Fetter (Fig. 3.16-3.17, 2001), see alsoTodd and Mays (2005, p. 95-96).

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Applications of Darcy’s Law

I find magnitude and direction of discharge through aquifer

I solve for head at heterogeneity boundary in composite(two-material) aquifer

I find total flux through aquitard in a multi-layer system

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The Flow Equation (Continuity Eqn.)

I Approach: use “control” volume, itemize flow in and out, andcontent of volume (sec. 4.7, Fetter, 2001), see Fig. 4

I Simplifications: assume steady-state, constant fluid properties

I Mass fluxI must describe movement of mass across the sides of the

control volume, i.e. determine massarea·time or m

t·l3I known fluid parameters: density ρ

(ml3

), velocity ~q

(lt

)(really

specific discharge, or velocity averaged over a cross-sectionalarea; see references such as Bear (1972); Schlichting (1979);Slattery (1972) for discussion of volume-averaging and REV’s)

I Then mass flux = ρ · ~q

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Representative Control Volume

q xq x x+

∆ ∆ ∆

∆ z

∆ y

∆ x

ρ ρ∆ x

(x,y,z)

(x+ x, y+ y, z+ z)

Figure 4: Mass fluxes for control volume in uniform flow field.

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Control Volume Mass Balance

I Content of control volume (storage)I Steady-state ⇒ no change in content with timeI then sum of in and outflows must be 0 (otherwise content

would change)

I Sum of flows for control volume in uniform flowI from above know that form will be:{

rate ofmass in

}−{

rate ofmass out

}= 0 (6)

I mass flux in{rate ofmass in

}= ρ~qx︸︷︷︸

mass flux per unit area

· ∆y∆z︸ ︷︷ ︸area of cube face

(7)

I mass flux outI depends on ~qout , let ~qout = ~qin + change

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Control Volume Mass Balance (cont.)

I express change as distance · (rate of change) and rate ofchange as ∆qx

∆xI to be most accurate, determine rate of change over a very

small distance, i.e. take limit as ∆x → 0I that’s a derivative

lim∆x→0

∆qx∆x

= dqdx

I since ~q can vary as a function of y and z as well, we write itas a partial derivative, holding other variables constant

d~qdx

∣∣∣y,z constant

=∂~q

∂x

I then in the form required for (6){rate ofmass out

}= ρ

(qx + ∆x

∂qx∂x

)∆y∆z

I net flux in x-direction: −ρ ∂qx∂x ∆x∆y∆z = −ρ ∂qx

∂x ∆V

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Continuity Equation

I sum of flows for general case (y and z sums are same form asfor x)

−ρ∆V

(∂qx∂x

+∂qy∂y

+∂qz∂z

)= 0(

∂qx∂x

+∂qy∂y

+∂qz∂z

)= 0

∇ · ~q = 0 (8)

I this is the divergence of the specific discharge, a measure ofthe fluids to diverge from or converge to the control volume

I assume this equation applies everywhere in problem domain,i.e. that fluid is everywhere, and fluid/rock properties andvariables are continuous

I then rock and fluid are overlapping continua

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Flow Equation

I Approach: governing equations (8) and (1) contain similarvariables, can we simplify?

I Combined equationI substituting Darcy’s Law (1) for ~q in (8):

∇ · ~q = ∇ · ~(−K∇h)

= ∇2 · h

=

(∂2h

∂x2+

∂2h

∂y2+

∂2h

∂z2

)= 0

I This is our governing equation for groundwater flowI Assumptions: many were made above

I steady-state: no time variationI continuum: fluid and rock properties and variables continuous

everywhere in problem domainI constant density: incompressible fluid, no compositional

change, no temperature changeI no viscous/inertial effects: low flow velocities

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References

Bear, J.: Dynamics of Fluids in Porous Media. Elsevier, New York, NY(1972)

Fetter, C.W.: Applied Hydrogeology. Prentice Hall, Upper Saddle River,NJ, 4th edn. (2001), http://vig.prenhall.com/catalog/academic/product/0,1144,0130882399,00.html

Hubbert, M.K.: The theory of ground-water motion. J. Geol. 48,785–944 (1940)

Hubbert, M.K.: Darcy’s law and the field equations of the flow ofunderground fluids. AIME Transact. 207, 222–239 (1956)

Pottorff, E.J., Erikson, S.J., Campana, M.E.: Hydrologic utility ofborehole temperatures in Areas 19 and 20, Pahute Mesa, Nevada TestSite. Report DRI-45060, DOE/NV/10384-19, Desert ResearchInstitute, Reno, NV (1987)

Schlichting, H.: Boundary-Layer Theory. McGraw-Hill, New York (1979)

Slattery, J.C.: Momentum, Energy, and Mass Transfer in Continua.McGraw-Hill, New York (1972)

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References (cont.)

Todd, D.K., Mays, L.W.: Groundwater Hydrology. John Wiley & Sons,Hoboken, NJ, 3rd edn. (2005), http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP000351.html

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