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Flow through Soils (ch7)PowerPoint Presentation

Flow through Soils (ch7)

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### Flow through Soils(ch7)

Energies

Kinetic E (velocity)

Strain E (fluid pressure)

Potential E (elevation)

Head:convert each form of energy into the equivalent potential energy and express it as the corresponding height.

units of LENGTH

Heads

hv = velocity head (KE)

hp = pressure head (SE)

he = elevation head (PE)

h = total head = hv + hp + he

(Bernoulli)

units of LENGTH

h1

2

h2

1

l

Head lossFluid flows from point of high total head to point of low total head

head loss = Dh = h1 – h2

Hydraulic gradient

Rate at which the total head changes along a length

Heads in soils

Since velocity is slow through soils, we neglect the velocity head. Thus,

Pore water pressures

uhydrostatic = uh = due to hydrostatic condition only

uexcess = ue = due to additional processes

Hydraulic conductivity

“permeability” [cm/s]

Hydraulic gradient

Darcy’s LawAssumptions:

flow is laminar

soil properties do not D with time

soil

kMeasure of a soil-fluid system’s resistance to flow

depends on

Void size

Fabric (structure)

Void continuity

Specific surface (drag)

Viscosity

Mass density

SOIL

101 – 102

gravel

ksands

10-3 – 100

10-8 – 10-3

silts

clays

10-10 – 10-6

Probably soil’s most varying parameter (largest numerical range)

sand

clay “seam”

1

sand

2

Lab testingk1 = 10-2

k2 = 10-6

k – precision is on the order of +/- 50% or more!

Report values to one decimal place.

Hazen’s Correlation

k a pore size

~ (pore diameter)2

(pore diameter) ~ D10

For loose clean sands with 0.1mm < D10 < 3mm and Cu < 5

k = [cm/sec]

C = Hazen’s coefficient = 0.8 – 1.2 (typical = 1)

D10 = [mm]

clay

el. = 165m

clay

sand

seam

256 m

3.2 m

ExampleGiven: ksand = 4x10-2 cm/sec

reservoir length (into board) = 1000 m

Compute seepage loss (Q) through the sand seam

Q = kiA = 0.0115 m3/sec = 41.5 m3/hr

SolutionQ = kiA

k = 4x10-2 cm/sec

i = Dh/L = (167.3m – 165m) / 256m = 0.009

A = (3.2 m) (1000 m) = 3200 m2

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