Flow NetsPhilip B. Bedient

Civil & Environmental Engineering

Rice University

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Flow Net Theory1. Streamlines and Equip. lines are .2. Streamlines 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.

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Flow Net Theory

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Flow Net in Isotropic Soil

Portion of a flow net is shown below

Stream tube

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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.

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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)

dm

h1

dl

q

h2

q

n

m

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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.

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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

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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

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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 and equipotential lines near the boundaries.

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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.

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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.

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Seepage Under DamsFlow nets for seepage through earthen dams

Seepage under concrete dams

Uses boundary conditions (L & R)

Requires curvilinear square grids for solution

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Two Layer Flow System with Sand Below

Ku / Kl = 1 / 50

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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: Todd & Bear, 1961.

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Effects of Boundary Condition on Shape of Flow

Nets

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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).

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Flow Net in a Corner:

Streamlines are at right angles to equipotential lines

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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.

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Flow Nets: an examplePosition: A B C D E F G H I JDistancefromfront toe(ft)

0 3 22 37.5 50 62.5 75 86 94 100

n 16.5 9 8 7 6 5 4 3 2 1.2

The flow net is drawn with: m = 5 n = 17

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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

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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

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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