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Water flow in saturated soil

- D A Cameron
- Introduction to Civil and Mining Engineering

SEEPAGE water pressures

- Water flows from points of high to low
- TOTAL head
- WATER HEADS
- head of water x ?w water pressure
- Total head elevation head pressure head
- i.e h (or hT) he hp
- Kinetic head is ignored in soils

Head of Water

- Pressure head height to which water rises to in

a standpipe above the point

No loss of head, h, in this soil mass, so no

flow - Steady State

Water table level

hp

h

he

Arbitrary datum

Element of soil within soil mass

Confined Aquifer

- A water bearing layer, overlain and underlain by

far less permeable soils.

standpipe

Water level in aquifer

Clay, silt - no free water

Sand aquifer

Clay, silt

Steady flow in soils Laminar flow

- Assumptions to theory
- Uniform soil, homogeneous and isotropic
- Continuous soil media
- Small seepage flow (non turbulent flow)
- Darcys Law of 1850

Darcys Law

- q kiA
- where q rate of flow (m3/s)
- i hydraulic gradient
- A area normal to flow direction (m2)
- k coefficient of permeability (m/s)

Hydraulic Gradient, i

?h

Area of flow, A

Flow rate, q

Length of flow, l

Hydraulic Conductivity

- Coefficient of permeability or just permeability
- SATURATED soil permeability!
- Hazens formula, for clean and almost uniform

sands

TYPICAL PERMEABILITIES

- Clean gravels gt 10-1

m/s - Clean sands, sand-gravel 10-4 to 10-2 m/s
- Fine sands, silts 10-7 to 10-4

m/s - Intact clays, clay-silts 10-10 to 10-7

m/s

Measuring Permeability

- A Laboratory
- Constant head test
- Falling head test
- Other
- B Field
- Pumping tests
- Borehole infiltration tests

A Laboratory How good is the sample? B

Field Need to know soil profile (incl. WT) and

boundary conditions

1. Constant head permeameter

Water tank - moveable

overflow

?ht

A

hpC

hpB

q

B

?he

C

D

soil

Constant head test

- Suitable for clean sands and fine gravels
- EXAMPLE
- If the sample area is 4500 mm2,
- the vertical distance between the 2 standpipe

points is 100 mm, - ?h is 75 mm
- Outflow is 1 litre every minute
- What is the coefficient of permeability?

Solution

- 1000 cm3/min
- OR q 16.7 cm3/sec 16.7x10-6 m3/sec
- i 75/100 0.75
- k q/(iA)
- (16.7x10-6)/(0.75x4500x10-6) m/sec
- k 5 x 10-3 m/sec
- Typically a clean sand or gravel permeability

2. Falling head permeameter

Falling head test

- Suited to low permeability materials
- silts and clays
- Soil sample length, L, and area, A
- Flow in the tube flow in the soil

3. Field testing drawdown test

Pumping well

q

r2

Water table

r1

h2

h1

Impermeable boundary

Drawdown -phreatic line

Drawdown test

- Needs
- a well-defined water table
- and confining boundary
- Must be able to
- pull down water table
- and create flow
- (phreatic line uppermost flow line)

Solution

- Axi-symmetric problem
- By integration of Darcys Law,

TUTORIAL PROBLEMS

- A canal and a river run parallel, an average of

60 m apart. The elevation of water in the canal

is 200 m and the river 193 m. A stratum of sand

intersects both the river and canal below the

water levels. - The sand is 1.5 m thick and is sandwiched between

strata of impervious clay. - Compute the seepage loss from the canal in m3/s

per km length of the canal, given the

permeability of the sand is 0.65 mm/s.

THE PROBLEM

Sand seam

RL 200 m

RL 193 m

canal

river

60 m

SOLUTION

- q kiA
- k 0.65 mm/s 0.65 x 10-3 m/s
- ?h 7 m
- q 0.65 x 10-3 x 0.117 x 1.5 m2/m length
- q 0.114 x 10-3 m3/sec /m length
- q 0.114 m3/sec/km length

THE PROBLEM

Hydraulic gradient, i 0.117

RL 200 m

RL 193 m

?h 7 m

l 60 m

Flow Lines shortest paths for water to exit

Phreatic surface

Equipotential line-

?h

hp1

Stream tube

hp2

he1

Dl

he2

1

Head reference line

Equipotentials

- Are lines of equal total head
- Can be derived from boundary conditions
- and flow lines

The Flow Net

Flow lines - run parallel to impervious

boundaries and the phreatic surface. Phreatic

surface the top flow line Equipotential lines

- line of constant total head - the total head

loss between consecutive equipotentials is

constant 2 consecutive flow lines constitute

a flow tube

Flownet Basics

- Water flow follows paths of maximum hydraulic

gradient, imax - flow lines and equipotentials must cross at 90,

since - imax ?(?h) / bmin

Since ?q is the same, a/b will be constant for

all the squares along the flow tube

Flow ?q

?h

Flow Lines

?(?h)

M

b

a

square M

Equi- potential lines

Impervious boundary

Discharge in flow direction, ?q / flow tube

Equipotentials

h3

90º

l

h2

Flow lines

h1

Flow Net Calculations

- Nd equal potential drops along length of flow?

Then the head loss from one line to another is - ?(?h) ?h / Nd
- From Darcys Law

Flow Net Calculations

- BUT a b
- AND total flow for Nf flow channels,
- per unit width is

But only for squares!

Example if k 10-7 m/sec, what would be the

flow per day over a 50 m length of wall?

Calculations

- Answer 6.72 m3

- Nf 3 or 4
- Nd 9 or 10?
- ?h 35 m?
- k 10-7 m/sec

Example what is the hydraulic gradient in the

square C?

Calculations

Answer 0.14

- ?h / Nd 35/9
- 3.9 m head / drop
- Average length of flow is about 23 m

Finite Difference spreadsheet solution

- Author Mahes Rajakaruna
- emailed to students today

ROWCO

A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

P

Q

R

S

T

U

V

W

L

Soil level

1

100

104

2

100

104

3

100

104

Cell H5

4

100

104

5

100

104

6

100

104

7

100

104

8

100

104

Interior cell value (H4I5H6G5)/4

9

100

104

104

104

104

104

104

104

10

100

11

100

Impermeable boundary

12

100

13

100

14

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

100

Flow lines from finite difference program

(spreadsheet)

Equipotentials from finite difference program

(spreadsheet)

Other numerical approaches FESEEP

cutoff

Mesh of foundation soil

FESEEP Output (University of Sydney)

flownet

increasing

pore pressures

Critical hydraulic gradient

- The value of i for which the effective stress in

the saturated system becomes ZERO! - Consequences
- no stress to hold granular soils together
- ?soil may flow ?
- boiling or piping EROSION!

Seepage Condition upward flow of water

- ?satz total stress
- ?u due to seepage
- iABz(?w)
- (represents ?h ?hp)
- ?? ? - u
- (?satz - ?wz) - iz?w
- ?? ??z - iz?w

B

z

A

?? 0, when ??z - iz?w 0 OR i

??/ ?w

Critical Hydraulic Gradient

Free surface, end area , A

h2

L

Small cylindrical element of soil

h1

FW

Flow direction

Seepage force, Fs

Critical hydraulic gradient

- Fs ?h?wA
- (h1 h2) ?wA
- Fw (?sat - ?w)AL
- (??)AL
- Equating the 2 forces
- i ??/ ?w as before

Likelihood of Erosion

When the effective stress becomes zero, no stress

is carried by the soil grains Note when flow

is downwards, the effective stress is

increased! So the erosion problem and ensuing

instability is most likely for upward flow, i.e

water exit points through the foundations of dams

and cut-off walls

Minimising the risk of erosion

- 1. Add more weight at exit points

permeable concrete mats?

Lengthen flow path?

1. Deeper cut-offs 2. Horizontal barriers 3.

Impermeable blanket on exit surface

Summary

- Heads in soil
- Darcys Law
- Coefficient of permeability
- Measurement of permeability
- Flownets
- Flownet rules
- Seepage from flownets
- Piping, boiling or erosion
- Critical hydraulic gradient

- Exercises
- a) Draw a flow net for seepage under a vertical

sheet pile wall penetrating 10 m into a uniform

stratum of sand 20 m thick. - b) If the water level on one side of the wall is

11 m above the sand and on the other side 1.5 m

above the sand, compute the quantity of seepage

per unit width of wall. k 3 ? 10-5 m/s - What is the factor of safety against developing

the quick condition on the outflow side of the

wall? ?sat 21 kN/m3

(No Transcript)

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