002 Hidrogeologi Darcy Law

7/21/2019 002 Hidrogeologi Darcy Law

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Hidrogeologi dan Hidrotermal

Pertemuan ke-2

Hukum Darcy

Nazli Ismail, Ph.D.

Program Studi S1 Teknik Pertambangan FT Unsyiah

Groundwater

Groundwater flows slowly through the voidsbetween grains or the cracks in solid rock.

Basic principles of GW Flow

Porosity and effective porosityTotal porosity : the part of rock that's void space

n T = V v /V T = (V T V s )/V T V v : void volume,V T : total volumeVs: solid volume

void ratio

e = Vv/Vs Primary poro si ty : interstitial porosity (original in therock)Secondary poros i ty : fracture or solution porosity

Primary porosity range from 26% to 47% (usingdifferent arrangements and packing of ideal spheres).

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Rongga dalam berbagai batuan Porosity of Sediments and Rocks

Depending on grain size, particle shape andarrangement, diagenetic features, actual values ofporosity can range from 0 or ~ 0 to more than 60%.In general, for sedimentary rocks, the smaller theparticle size, the higher the porosity.Total poro si ty amount of pore space (does notrequire pore connection)Effective poro si ty : percentage of interconnectedpore space available for groundwater flow.Effective porosity can be one order of magnitudesmaller than total porosity (difference greatest in

fractured rocks). 6

M easur ement of porosity

In the lab, porosity is measured by taking asample of known volume (V),sample is dried in an oven at 105 oC until itreaches a constant weight (expelling moisture).Dried sample is then submerged in a knownvolume of water and allowed to remain in asealed chamber until saturatedVolume of voids is equal to original water volumeminus volume in the chamber after saturatedsample is removed. Result is effective porosity.

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M easur ement of porosity

Total porosity is found from:

bulk density is mass of sample after drieddivided by original sample volume, particledensity is oven-dried mass divided byvolume of solid determined from water-displacement test.In most rocks and soils, particle density isabout 2.65g/cc (2650kg/m^3).

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]/ (1[100 sbn

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Examples

n = 47.65 % n = 25.95%

(a) Cubic packing of spheres(b) Rhombohedral packing

Aliran Air TanahMuch of our knowledge depends on field andlaboratory observations. Here, for example, is anexperiment to measure head loss in an aquifer.

Darcys Law

Henri Darcy established empirically that the energylost h in water flowing through a permeableformation is proportional to the length of the sedimentcolumn L.The constant of proportionality K is called thehydraulic conductivity . The Darcy Velocity V D:

V D = K (h/L)

and since Q = V D A ( where A = total area)

Q = KA (dh/dL)


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Darcys Experiment

2. Head difference doesnt change with inclination of thesand filter

3. Again, Darcy related reduced flow rate to head loss andlength of column through a constant of proportionality K,V = Q/A = -K dh / dL

1. Velocities small, V ~ 0, so:

Piezometers before andafter sand. Pipe is full, so

flow rate is constant

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Darcy

s Data (One set of 10 experiments)L 0.58 m

diam. 0.35 m

n 0.38

A 0.096211 m 2

CalcExperiment Duration Q dp Ratio K K

No. (min) L/min (m) V/dp (m/min) cm/s1 25 3.6 1.11 3.25 0.019552 3.26E-02

2 20 7.65 2.36 3.24 0.019541 3.26E-02

3 15 12 4 3 0.018085 3.01E-02

4 18 14.28 4.9 2.91 0.017568 2.93E-02

5 17 15.2 5.02 3.03 0.018253 3.04E-02

6 17 21.8 7.63 2.86 0.017224 2.87E-02

7 11 23.41 8.13 2.88 0.017359 2.89E-02

8 15 24.5 8.58 2.85 0.017214 2.87E-02

9 13 27.8 9.86 2.82 0.016997 2.83E-02

10 10 29.4 10.89 2.7 0.016275 2.71E-02

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Plotted it. Note the strong coefficient of determination R 2 .Darcys allows an estimate of:

The velocity or flow rate moving within the aquifer The average time of travel from the head of theaquifer to a point located downstreamVery important for prediction of contaminant plumearrival

Confined Aquifer


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Darcy & Seepage Velocity

Darcy velocity V D is a fictitious velocity since itassumes that flow occurs across the entirecross-section of the sediment sample.Flow actually takes place only throughinterconnected pore channels (voids), at theseepage velocity V S .

A = total areaAv voids

Darcy & Seepage Velocities

From the Continuity Eqn. Q = constantPipe running full means Inputs = Outputs

Q = A V D = A V V sWhere:

Q = flow rate A = total cross-sectional area of material AV = area of voidsV s = seepage velocityV D = Darcy velocity

Since A > AV , and Q = constant, V s > V DPinch hose , r educe a rea , wate r goes fas te r

Darcy & Seepage Velocity:Porosity

Q = A V D = AV V s ,therefore V S = V D ( A/AV )Multiplying both sides by the length of themedium ( L) divided by itself, L / L = 1

V S = V D ( AL / A V L ) = V D ( Vol T / Vol V ) we getvolumes

Where: Vol T = total volumeVol V = void volume

By definition, Vo l v / Vo l T = n , the sedimentporosity

So the actual velocity: V S = V D / n

Turbulence and Reynolds Number

The path a water molecule takes is called astreamline.In laminar flow , streamlines do not cross,and the viscous forces due to hydrogenbonds are important.In turbulent flow acceleration and large

scale motion away from a smooth path isimportant(this is the familiar inertial force F = ma) andstreamlines cross .


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Streamline. laminar flow ,turbulent flow

Turbulence and Reynolds Number

We could take the ratio of inertial to viscousforces.When this number is large, inertial forces aremore important, and flows are turbulent.This ratio is known as the Reynolds number Re:

Viscosity

Viscosity is a fluids resistance to flow.Dynamic viscosity , units Pas = Ns/m 2, orkg/(ms) is determined experimentally.

If a fluid with a viscosity of one Pas is placedbetween two plates, and one plate is pushedsideways with a shear stress of one Pascal, itmoves a distance equal to the thickness of thelayer between the plates in one second.

Kinematic viscosity , is the dynamicviscosity divided by the density.

The SI unit of is m 2/s. Recall the ratio of Kinetic/Potential Energy (KE/PE) is the FroudeNumber (Fr) Fr = V / sqrt( g L)

Reynolds: Inertial/Viscous forces


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Limitations of Darcys Equation

1. For Reynolds Number, Re, > 10 or where the flowis turbulent, as in the immediate vicinity of pumpedwells.

2. Where water flows through extremely fine-grainedmaterials (colloidal clay)Q = KA (dh/dL)q = Ky (dh/dL)

Darcys Law works

for 1.0 < Re < 10

Example 1

A confined aquifer has a source of recharge.

K for the aquifer is 50 m/day, and porosity n is 0.2.

The piezometric head in two wells 1000 m apart is 55m and 50 m respectively, from a common datum.

The average thickness of the aquifer is 30 m, and theaverage width of the aquifer is 5 km = 5000m.

Q = KA (dh/dL)The hydraulic conductivity K is a velocity, length / timeand n = Vol voids / Vol total

A piezometer is a small-diameter observation well used to measure the piezometric head ofgroundwater in aquifers.Piezometrichead is measured as a water surface elevation, expressed in units of length.

Compute:

(a) the rate of flow through the aquifer (b) the average time of travel from the head of the

aquifer to a point 4 km downstream

Q = KA (dh/dL)

Solution

Cross-Sectional area= 30(5000) = 1 .5 x 10 5 m2

Hydraulic gradient dh/dL= (55-50)/1000 = 5 x 10 -3

Find Rate of Flow for K = 50 m/dayQ = (50 m/day) (1.5 x 10 5 m 2) ( 5 x 10 -3)

Q = 37,500 m 3/day

Darcy Velocity : V = Q/A

= (37,500m 3/day) / (1.5 x 10 5 m 2)

= 0.25m/day

Q = KA (dh/dL )


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Seepage Velocity:Vs = VD/n = (0.25) / (0.2) = 1.25 m/day (about4.1 ft/day)Time to travel 4 km downstream:T = (4000m) / (1.25m/day) =3200 days or8.77 yearsThis example shows that water moves veryslowly underground.

Lesson: Groundwater moves very slowly

Example 2

Confining LayerAquifer

30 ft

A channel runs almost parallel to a river, and they are 2000 ftapart.

The water level in the river is at an elevation of 120 ft . Thechannel is at an elevation of 110ft.

A pervious formation averaging 30 ft thick and with hydraulicconductivity K of 0.25 ft/hr joins them.

Determine the flow rate Q of seepage from the river to thechannel.

Example 2: Confined Aquifer

Consider 1-ft (i.e. unit) lengths of the river andsmall channel. Q = KA [(h 1 h2 ) / L]

Where: A = (30 x 1) = 30 ft 2K = (0.25 ft/hr) (24 hr/day) = 6 ft/day

Therefore,Q = [6ft/day (30ft 2) (120 110ft)] / 2000ftQ = 0.9 ft 3/day for each 1-foot length

002 Hidrogeologi Darcy Law

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