CIV3248
Groundwater,seepage and
environmental engineering
•Workshop (4)
•Flownet sketching
•Keith H McKenry
CIV3248 Flow Net Workshop
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Intention:
•Mathematical insight.
•Practise sketching skills.
•Flow net compared with SEEPW
output.
•Interpretation of flow nets
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Mathematics.
• Steady state: conservation of mass,
continuity equation, stream function (),
streamlines.
• Energy balance: Potential function (),
total head lines, irrotational flow (flow
around a curve by distortion rather than
rotation).
• Laplaces Equation for  & .
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Streamlines
• Seepage velocity vector is tangent to
streamline.
• Streamlines cannot cross.
• Commence on inflow boundary, finish at
outflow boundary.
• Streamlines have constant stream
function value.
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Streamlines
Unconfined flow: 1 is
a boundary streamline
 stream function value
dQ = 2 - 1
1
Stream tube
dQ
2
3
Seepage
velocity
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Streamlines and stream
function (x,y),
• Two streamlines define a streamtube.
• Streamtube width is inversely
proportional to local seepage velocity.
• Difference in in stream function value
equals the discharge between two
streamlines defining the stream tube.
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Total head lines
• Total head = -  / k
• Lines of constant value of potential
function.
• Zero hydraulic gradient zero flow
• orthogonal to streamlines (cross at right
angles).
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Equipotential / Total head lines
F = equipotential value = - H*k m3/s/m
1
3
2
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Flownet definition…1
• Selection of streamlines and total head
lines.
• Constant difference in stream function
values between all streamlines.
• Constant difference in potential function
values between all total head lines.
• Same difference value for both stream
function and potential function.
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Flownet definition…2
F = 24
Streamlines
F = 27
F = 30
F = 33
dq = 3
Equipotential
lines
(constant Total
Head)
dq = 3
 = 11
=5
=8
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Flownet features
• Orthogonal net.
• Inscribed circles.
• Diagonals to curvilinear squares form
another orthogonal net.
• Ratio of number of streamtubes to
number of potential drops is constant for
a given flow boundary.
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Flownet Examples
• Based on explicit functions for total
head and stream function.
• Prior to the advent of cheap computers much
mathematical effort was expended on techniques to
allow explicit equations to fit a wide range of
boundary geometries.
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Basis
• For a given boundary geometry, only
ONE true flownet exists.
• Find the true flow net by “trial and
correct”.
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Strategy
• Identify the two streamlines and two total
total head lines that define the problem.
• For 4 or 5 STREAM TUBES draw the 3
or 4 internal streamlines.
• Draw TOTAL HEAD lines to attempt to
form curvilinear squares (a true flownet).
• Adjust the trial lines to get better
curvilinear squares.
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Confined Flow example
Boundary total head lines
Total head
difference
“Pervious”
Boundary streamlines
“Impermeable”
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Cross-section to scale: 10 mm = 1 m. W1
C

D

E 
F

B
 A
Datum RL 60m
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Cross-section to scale: 10 mm = 1 m. W2
C

D

E 
F

B
 A
Datum RL 60m
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Cross-section to scale: 10 mm = 1 m. W3
C

D

E 
F

B
 A
Datum RL 60m
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Cross-section to scale: 10 mm = 1 m. W4
C

D

E 
F

B
 A
Datum RL 60m
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Flownet Interpretataion
• Discharge (m3/sec/m)
• Pore pressure
• Hydraulic gradient.
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Discharge per unit width
(Q m3/sec/m)
Q = k.HL.(N /N)
• k - coefficient of permeability.
• HL - drop in Total Head across flow.
• N - number of stream tubes.
• N - number of potential drops.
• Note N might not be an integer.
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N = 3
Q = k.HL.(N /N) = 7 x 10-5 . 4 .(3/8)
= 10.5 x 10-5 m3/s/m
N = 8
HL = 4 m
0
8
0
1
7
1
K = 7 x 10-5 m/s
2
6
2
3
4
3
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5
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Pore Pressure “u” at a point
• Identify total head value of all constant
total head lines relative to datum.
• Find total head at point by interpolation
“H”.
• Identify elevation head of point “z”.
• u = (H - z) * w (w = unit weight of
water)
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u = (H - z) * w
uA =(94 – 80)*9.8 = 137 kPa
uBD =(94 – 75)*9.8 = 186 kPa
uB =(92.5 – 83)*9.8 = 93 kPa
RL95m
RL94m
RL92.5m
RL91m
H=91m
H=95m
H=91.5m
H=94.5m
RL83m
RL80m
RL75m
H=94m
B
A
H=92m
D
H=93m
H=92.5m
H=93.5m
Total head datum = RL 0m
RL72m
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u = (H - z) * w
uB =(91.75 – 83)*9.8 = 86 kPa
uA =(92.5 – 80)*9.8 = 123 kPa
RL93m
RL91.75m
RL92.5m
RL91m
H=91m
H=93m
H=91.25m
H=92.75m
RL83m
RL80m
B
A
H=91.5m
H=92.5m
H=92m
H=91.75m
H=92.25m
Total head datum = RL 0m
RL72m
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Hydraulic gradient at a point - “i”
• Interpolate streamline through the point.
• Determine difference in total head
between constant total head lines either
side of point “dH”.
• Measure real distance between these
constant total head lines along the
interpolated streamline “dL”
• i = dH /dL
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dH = 0.5m
(-)iC = dH /dL = 0.5 / 4.5 = 0.11
RL95m
RL91m
H=91m
H=95m
H=91.5m
H=94.5m
dL = 4.5m
C
H=92m
H=94m
H=93m
H=92.5m
H=93.5m
Total head datum = RL 0m
RL72m
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Upward seepage flow:
Critical hydraulic gradient at exit - Icrit
• Liquifaction of soil may occur if the exit
hydraulic gradient exceeds a critical value
such that effective stress equals zero.
• Icrit = (Gs - 1)/(e +1)
Gs - Specific gravity of soil particles
e - Void Ratio
• Icrit ~ 1.0
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Seepw solution
70
60
50
40
A
5.8120e-003
30
20
10
0
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
Feet
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Conclusion: flow net sketching
• Limited to isotropic soils, homogeneous
flow
• Quick, simple equipment,
• Analysis of multiple geometry for design.
• Check on computer packages.
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