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1. Water flow in saturated soil D A Cameron
Introduction to Civil and Mining Engineering
3. Head of Water Pressure head = height to which water rises to in a standpipe above the point
4. Confined Aquifer A water bearing layer, overlain and underlain by far less permeable soils.
5. Steady flow in soils – Laminar flow Assumptions to theory:
Uniform soil, homogeneous and isotropic
Continuous soil media
Small seepage flow (non turbulent flow)
Darcy’s Law of 1850
6. Darcy’s 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)
7. Hydraulic Gradient, i
8. Hydraulic Conductivity Coefficient of permeability or just permeability
SATURATED soil permeability!
Hazen’s formula, for clean and almost uniform sands:
9. TYPICAL PERMEABILITIES Clean gravels > 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
10. Measuring Permeability [A] Laboratory
Constant head test
Falling head test
Borehole infiltration tests
11. 1. Constant head permeameter
12. Constant head test Suitable for clean sands and fine gravels
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?
13. 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
15. 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
16. 3. Field testing – drawdown test
17. 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)
18. Solution Axi-symmetric problem
By integration of Darcy’s Law,
19. 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.
21. 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
23. Flow Lines – shortest paths for water to exit
24. Equipotentials Are lines of equal total head
Can be derived from boundary conditions
and flow lines
25. The Flow Net
26. Flownet Basics Water flow follows paths of maximum hydraulic gradient, imax
flow lines and equipotentials must cross at 90°, since:
imax = [?(?h) / bmin]
29. 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 Darcy’s Law
30. Flow Net Calculations BUT a = b
AND total flow for Nf “flow channels”,
per unit width is :
31. Example: if k = 10-7 m/sec, what would be the flow per day over a 50 m length of wall?
32. Calculations Nf = 3 or 4
Nd = 9 or 10?
?h = 35 m?
k = 10-7 m/sec Answer: 6.72 m3
35. Finite Difference spreadsheet solution
Author: Mahes Rajakaruna
emailed to students today
39. Other numerical approaches – FESEEP
40. FESEEP Output (University of Sydney)
41. Critical hydraulic gradient The value of i for which the effective stress in the saturated system becomes ZERO!
no stress to hold granular soils together
?soil may flow ?
“boiling” or “piping” = EROSION!
44. Critical hydraulic gradient Fs = ?h?wA
= (h1 – h2) ?wA
Fw = (?sat - ?w)AL
Equating the 2 forces
i = ??/ ?w as before
45. Likelihood of Erosion
46. Minimising the risk of erosion 1. Add more weight at exit points
47. Lengthen flow path?
48. Summary Heads in soil
Coefficient of permeability
Measurement of permeability
Seepage from flownets
Piping, boiling or erosion
Critical hydraulic gradient