Transcript Flow Nets - Rice University

Flow Nets
Philip B. Bedient
Civil & Environmental Engineering
Rice University
Flow Net Theory
1. Streamlines Y and Equip. lines  are .
2. Streamlines Y are parallel to no flow
boundaries.
3. Grids are curvilinear squares, where
diagonals cross at right angles.
4. Each stream tube carries the same flow.
2
Flow Net Theory
3
Flow Net in Isotropic Soil
Portion of a flow net is shown below
Y
F
4
Flow Net in Isotropic Soil
The equation for flow nets originates from
Darcy’s Law.
Flow Net solution is equivalent to solving the
governing equations of flow for a uniform
isotropic aquifer with well-defined boundary
conditions.
5
Flow Net in Isotropic Soil
Flow through a channel between
equipotential lines 1 and 2 per unit
width is:
∆q = K(dm x 1)(∆h1/dl)
m
F1
Dq
Dq
Dh1
dm
n
F2
F3
Dh2
dl
6
Flow Net in Isotropic Soil
Flow through equipotential lines 2 and 3 is:
∆q = K(dm x 1)(∆h2/dl)
The flow net has square grids, so the head
drop is the same in each potential drop:
∆h1 = ∆h2
If there are nd such drops, then:
∆h = (H/n)
where H is
the total head loss between the first and last
equipotential lines.
7
Flow Net in Isotropic Soil
Substitution yields:

∆q = K(dm x dl)(H/n)
This equation is for one flow channel. If there
are m such channels in the net, then total flow
per unit width is:

q = (m/n)K(dm/dl)H
8
Flow Net in Isotropic Soil
Since the flow net is drawn with squares, then
dm  dl, and:
q = (m/n)KH
[L2T-1]
where:





q = rate of flow or seepage per unit width
m= number of flow channels
n= number of equipotential drops
h = total head loss in flow system
K = hydraulic conductivity
9
Drawing Method:
1. Draw to a convenient scale the cross
sections of the structure, water elevations,
and aquifer profiles.
2. Establish boundary conditions and draw one
or two flow lines Y and equipotential lines F
near the boundaries.
10
Method:
3. Sketch intermediate flow lines and equipotential
lines by smooth curves adhering to right-angle
intersections and square grids. Where flow
direction is a straight line, flow lines are an equal
distance apart and parallel.
4. Continue sketching until a problem develops. Each
problem will indicate changes to be made in the
entire net. Successive trials will result in a
reasonably consistent flow net.
11
Method:
5.
In most cases, 5 to 10 flow lines are usually
sufficient. Depending on the no. of flow lines
selected, the number of equipotential lines
will automatically be fixed by geometry and
grid layout.
6. Equivalent to solving the governing
equations of GW flow in 2-dimensions.
12
Seepage Under Dams
Flow nets for
seepage through
earthen dams
Seepage under
concrete dams
Uses boundary
conditions (L & R)
Requires curvilinear
square grids for
solution
13
Two Layer Flow System with
Sand Below
Ku / Kl = 1 / 50
14
Two Layer Flow System with
Tight Silt Below
Flow nets for seepage from one side of a channel through two different
anisotropic two-layer systems. (a) Ku / Kl = 1/50. (b) Ku / Kl = 50. Source: Todd15&
Bear, 1961.
Effects of Boundary Condition
on Shape of Flow Nets
16
Radial Flow:
Contour map of the piezometric surface near Savannah, Georgia,
1957, showing closed contours resulting from heavy local
groundwater pumping (after USGS Water-Supply Paper 1611).
17
Flow Net in a Corner:
Streamlines Y
are at right
angles to
equipotential
F lines
18
Flow Nets: an example
A dam is constructed on a permeable stratum
underlain by an impermeable rock. A row of
sheet pile is installed at the upstream face. If
the permeable soil has a hydraulic
conductivity of 150 ft/day, determine the rate
of flow or seepage under the dam.
19
Flow Nets: an example
Posit ion:
A B C D
E
F G H
I
J
Distance
f rom
f ront t oe
( f t)
0
3
22
37 .5
50
62 .5
75
86
94
10 0
n
16 .5
9
8
7
6
5
4
3
2
1. 2
The flow net is drawn with: m = 5
n = 17
20
Flow Nets: the solution
Solve for the flow per unit width:
q = (m/n) K h
= (5/17)(150)(35)
= 1544 ft3/day per ft
21
Flow Nets: An Example
There is an earthen dam 13 meters across
and 7.5 meters high.The Impounded water is
6.2 meters deep, while the tailwater is 2.2
meters deep. The dam is 72 meters long. If
the hydraulic conductivity is 6.1 x 10-4
centimeter per second, what is the seepage
through the dam if n = 21
K=
6.1 x 10-4cm/sec
= 0.527 m/day
22
Flow Nets: the solution
From the flow net, the total head loss, H, is
6.2 -2.2 = 4.0 meters.
There are 6 flow channels (m) and 21 head
drops along each flow path (n):
Q=
(KmH/n) x dam length
=
(0.527 m/day x 6 x 4m / 21)
x
(dam length)
=
0.60 m3/day per m of dam
= 43.4 m3/day for the entire 72-meter
length of the dam
23