EAS 44600 Groundwater Hydrology
Lecture 9: Flow Nets
Dr. Pengfei Zhang
Specific Discharge and Average Linear Velocity
Recall Darcy’s law:
h −h 
 dh 
Q = − KA a b  = − KA 
 L 
 dl 
(9-1)
It states that the flow (discharge) will be proportional to the hydraulic gradient (dh/dl), the
hydraulic conductivity (K), and the cross-sectional area of flow (A). The specific discharge
(Darcian velocity), q, of groundwater is equal to Q/A:
 dh 
q = −K  
 dl 
(9-2)
The specific discharge is an apparent velocity only, since flow of water does not occur through
the entire cross section (Figure 9-1A). Remember that flow occurs only in the void spaces
between grains (Figure 9-1A). Hence, the specific discharge given above underestimates the
actual velocity of water moving through the void spaces.
Flow
No
flow
A
B
Figure 9-1. Cross section of an R.E.V. (A) and average linear velocity (B).
To find the actual average linear velocity, v, at which the water is moving, one divides the
specific discharge by the porosity (n) to account for the actual open space available for flow:
v=
q
K  dh 
=−  
n
n  dl 
(9-3)
in terms of units:
Q L3w 1
q= =
A T L2REV
(9-4)
9-1
Q 1 L3w 1 L3REV LREV
=
=
v=
A n T L2REV L3w
T
(9-5)
Notice that in the above units for porosity it is assumed that all voids are filled with water
( L3w / L3REV versus L3voids / L3REV ), in other words, we are talking about groundwater, not the
unsaturated zone.
The units are useful because they help us to distinguish between specific discharge (L3w/L2REVT)
and the average linear velocity (LREV/T). The specific discharge concerns the volume of water
that passes by a particular cross section of the R.E.V. in a given amount of time, whereas the
average linear velocity concerns the length of R.E.V. that the average molecule of water travels
in a given amount of time. The specific discharge is less straightforward to understand
physically, and the units show that.
The average linear velocity is an average because the water molecules follow various tortuous
paths through the porous medium (Figure 9-1B). Some water molecules travel through paths of
greater tortuosity, resulting in lower linear velocities, whereas other water molecules follow
more straightforward paths, resulting in higher linear velocities. The value obtained above is the
average linear velocity, representing the average path of water through the medium.
The average linear velocity is extremely important. It is the value used to determine how long it
would take a contaminant to move from point A to point B in an aquifer. Since the average
linear velocity is determined from dh/dl, K, and n, one can understand why characterization of
these aquifer parameters is required in order to understand whether contaminants released at
point A could reach point B in a given amount of time.
Flow Nets
A simple visual method of capturing the major elements of a groundwater flow system is to use
flow nets. A flow net is a visual solution of a groundwater flow system that uses available
hydrologic data (e.g., hydraulic heads from piezometers, and places of relatively constant
hydraulic head such as lakes) along with fundamental principles of groundwater flow to
construct a qualitative model of the flow system.
A flow net is comprised of equipotential lines and flow lines. Equipotential lines are lines of
equal hydraulic head. Flow lines are imaginary lines that trace the path of a groundwater
molecule as it flows through an aquifer.
Since the movement of groundwater is from higher hydraulic head to lower hydraulic head, flow
lines cross equipotential lines. In an isotropic aquifer, the flow lines will cross the equipotential
lines at right angles (Figure 9-2A). This means that the groundwater flow is in the direction of
steepest hydraulic gradient. In an anisotropic aquifer, the flow lines may cross the equipotential
lines at angles other than 90 degrees (Figure 9-2B), since the hydraulic conductivity may be
highest in a direction different from the steepest hydraulic gradient.
9-2
Figure 9-2. Relationship of flow lines to equipotential field. A. Isotropic aquifer. B.
Anisotropic aquifer (Fetter).
A flow net is most useful in isotropic media. Flow net construction is based on the following
assumptions:
1.
2.
3.
4.
5.
6.
The aquifer is homogenous and isotropic.
The aquifer is fully saturated.
There is no change in the potential field over time.
The soil and water are incompressible.
The flow is laminar (Darcy’s law is valid).
All boundary conditions are known.
There are basically three types of boundary conditions. The first type of boundary condition is a
“constant head boundary”, such as a lake surface. Since a constant head boundary is also an
equipotential line, flow lines intersect constant head boundaries at right angles. Another
boundary condition is a confining layer, e.g., clay above or bedrock below. This boundary
condition is known as a “no-flow boundary”, since flow is not allowed to cross the boundary.
Flow lines will run parallel to a no-flow boundary, and equipotential lines will cross the
boundary at right angles. For unconfined aquifers, there is a “water-table boundary” at the
surface. The water table is neither a flow line nor an equipotential line; it is merely a surface
where the head is known. If there is recharge or discharge across the water table, flow lines will
be at an oblique angle to the water table. If there is no recharge across the water table, flow lines
can be parallel to it.
Following RULES must be obeyed in drawing flow nets:
1. Flow lines and equipotential lines are always perpendicular and form curvilinear squares in a
homogeneous, isotropic medium.
2. Equipotential lines are always perpendicular to no-flow boundaries.
3. Equipotential lines are always parallel to constant head boundaries.
4. Flow lines can never intersect. In other words, flow lines constitute imaginary no-flow
boundaries in that there is no flow across a flow line.
9-3
The following hints might be helpful in construction of a flow net:
1. Ensure that the use of a flow net is valid (e.g., steady-state flow, third dimension is constant,
and boundary conditions are known or can be approximated well).
2. Prepare for a trial-and-error solution.
3. Ensure that the boundaries are drawn to scale.
4. Always identify the boundary conditions (e.g., equipotential or constant-head boundary; flow
line or no-flow boundary; known-head boundary, commonly the water table). Remember
RULES 2 and 3!
5. Ask yourself the following: Where is the groundwater entering the system? Where can it
leave? What path might it take between the two? Remember that what comes in must go
out.
6. Always look for any symmetry in the boundary conditions. This can reduce the size of the
region where the flow net must be drawn (i.e., make the problem easier).
7. Decide on the number of flow tubes you want to use (any number of tubes can be used).
8. Where flow tubes constrict, higher hydraulic gradients are needed to move the same quantity
of water through the flow tube. Therefore, the equipotential lines will be more closely
spaced.
The following example (Figure 9-3) gives the steps in construction of a flow net:
1. Set the boundary conditions. Notice that Figure 9-3 could be interpreted as either plan or
cross-sectional view.
2. Draw a trial set of equally spaced flow lines. The flow lines will run parallel to no-flow
boundaries, and will intersect equipotential lines at right angles.
3. Draw a trial set of equipotential lines orthogonal to the flow lines. Make squares!
Why should one make squares? Well, if we know the hydraulic conductivity, K, for the material,
we can calculate the discharge through the system. The area between two adjacent flow lines is
called a streamtube or a flowtube. If the flow lines are equally spaced, the discharge through
each streamtube is the same. In Figure 9-3, the flow through a particular “square” with unit
depth normal to the page is:
 dh 
dQ = K  dm
(9-6)
 ds 
where dh is the head drop over distance ds. This discharge is the same along the entire length of
the streamtube as long as flow lines are parallel. If dm = ds (i.e., squares are used):
For a system with m streamtubes:
dQ = Kdh
(9-7)
Q = mKdh
(9-8)
If the total head drop across the net is H, and there are n divisions of head in the flow net, then:
Q=
mKH
n
9-4
(9-9)
No flow boundary
Step 1
Constant
head
boundary
60m
50m
Constant
head
boundary
50m
Constant
head
boundary
50m
Constant
head
boundary
No flow boundary
No flow boundary
Step 2
Constant
head
boundary
60m
No flow boundary
No flow boundary
Step 3
Constant
head
boundary
ds
60m
dm
No flow boundary
Figure 9-3. Steps in making a flow net.
For Figure 9-3, m = 4, n = 7, and H = 10 meters. Therefore, Q = 5.7K. If K=10-5 m/s,
Q = 5.7×10-5 m3/s.
9-5