FLOW NETS
Bernoulli's Equation
water travels very slowly through
soil as opposed to channel flow
p v
hz 
γw 2g
2
Total Head, m
0
Elevation Head, m
Fluid Pressure Head, m
Velocity Head, m
FLOW NETS
Bernoulli's Equation
For Seepage through soil:
Pore Water Pressure, kPa
u
hz
γw
FLOW NETS
Total HeadHydraulic
Loss,
In terms
h in
Gradient
of
water
Bernouli:
seeping
(Slope),from
i: A to B:
L
h
hA
uA
γw
h
head loss
i  or
L
distance over whichhead loss occurs
A
uB
γw
hB
B
zA
zB
 datum
FLOW NETS
Say
The
The
we
path
water
The
constructed
ofenergy
would
the flow
seep
driving
a tank
would
from
the
in be
the
the
seepage,
curved
lab
leftlike
chamber,
as
h?this
shown.
one.
through the soil and into the right chamber.
h
FLOW NETS
If
Lines
Line
we ab
stretch
ca and
is the
cefd
theupstream
tank,
are the
weequipotential
boundaries
have a mainly
of
Line bd is the downstream equipotential
horizontal
boundarychannel
this
where
flow
the
forchannel
the
totalseepage
head isflow
h
boundary where the total head is 0
from the left chamber to the right
h
FLOW NETS
at
In
The
If
What
bd
ca
order
we
water
h divide
would
= to
h
0would
determine
the
the rise
total
seepage
to
the
head
the
journey
total
be
same
at
head
into
level
theand
half
equally
on pore
the
way
water
hydraulic
spaced
pressure
drops
grade
mark
at
inline
(at
head
anyfrom
points
point
then
each
in
x,
wethe
yget
of
ormass
these
z)?
a flow
ofpoints.
soil
net.we
Each
point the
has flow
equalchannel
potential
therefore
the
subdivide
intoand
smaller
channels
line through them is an “equipotential”.
half way mark
h=h
h = 0.5h
x
y
z
h=0
FLOW NETS
If we recompressed the tank the flow net
would look something like this:
Construction of Flow Nets
To construct a flow net, you must start
Downstream Equipotential
with a scale drawing of theBoundary
hydraulic
structure:
Upstream Equipotential Boundary
1. Draw Flow Channel Boundaries
2. Draw Equipotential Boundaries
Construction of Flow Nets
Not
all
The
elements
first
trial:
are
“square”
The
bottom
flow
channel
intersects
the
It may
take
several
iterations
to finally
impervious
layer flow net.
come up with
a satisfactory
Construction of Flow Nets
water
pressure,
zof
5.
Using
the
given
scale,
elevation
4. 6.
At The
pointpore
P, the
total
headthe
is u10/12ths
P = (hp – head,
p)the
w
3.
Number
equipotentials
as
shown:
1.
free
water
surface
ispoint,
datum.
To
2.Downstream
determine
Show =(3.33+5.2)x9.8
the
And
total
the
the
total
head,
final
head
version
h
driving
at
any
is:
seepage.
P
=
83.3
kPa
z
is
-5.2
m
head driving
the seepage
P
h = 4.5-0.5 = 4.0m
hP  4.0 
10
 3.33m
12
FLOW NETS
Here’s some useful relationships:
1. Each channel carries an equal flow: ∆q = k∆h
2. Each drop in head is equal to:
h
Δh 
Nd
where Nd is the number of partitions or drops in potential
3. The total flow carried:
where Nf is the number of flow channel partitions
q = Nf∆q
Nf
q  kh
4. Or, the total flow carried:
Nd
nd
5. And, the head at any point P:
hp 
h
Nd
where nd is equipotential number (0 at downstream FWS)