Flow Nets - libvolume3.xyzlibvolume3.xyz/.../flownets/flownetspresentation1.pdf · sheet pile is...

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Flow Nets Philip B. Bedient Civil & Environmental Engineering Rice University

Transcript of Flow Nets - libvolume3.xyzlibvolume3.xyz/.../flownets/flownetspresentation1.pdf · sheet pile is...

Page 1: Flow Nets - libvolume3.xyzlibvolume3.xyz/.../flownets/flownetspresentation1.pdf · sheet pile is installed at the upstream face. If the permeable soil has a hydraulic conductivity

Flow Nets Philip B. Bedient

Civil & Environmental Engineering

Rice University

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Flow Net Theory

1. Streamlines Ψ and Equip. lines φ are ⊥⊥⊥⊥.

2. Streamlines Ψ 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

Φ

Ψ

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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 φ1 and φ2 per unit width is: ∆q = K(dm x 1)(∆h1/dl)

dm

∆∆∆∆h1

dl

Φ1

Φ3

∆∆∆∆q

Φ2

∆∆∆∆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 Ψ and equipotential lines Φ

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 Dams

Flow 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 Ψ

are at right

angles to

equipotential

Φ 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 example Posit ion: A B C D E F G H I JDist ance

from

front t oe

( f t)

0 3 22 37 .5 50 62 .5 7 5 8 6 9 4 10 0

n 1 6.5 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