jaringan aliran -...

JARINGAN ALIRAN

Aliran Air Lewat Bendungan

Lapisan

Garis Phreatic

Tanah di atas muka airLapisan Drainasi

Aliran air

∂∂

∂∂

2

2

2

20

h

x

h

z+ =

z

x

Penggambaran jaringan aliran

1. Equipotentials Lines / garis yang menghubungkan titik yang headnya sama ,h(x,z)

Equipotential (EP)

Garis Phreatic

2. Flow lines Lintasan air ( tegak lurus EPL )

Lanjutan penggambaran jaringan aliran

Flow line (FL)

Equipotential Line (EPL)

Sifat garis Equipotential

Flow line (FL)

Equipotential (EP)

h(x,z) = constant (1a)

∂∂

∂∂

h

xdx

h

zdz+ = 0Thus: (1b)

Equipotenial slopedz

dx

h x

h zEP

= − ∂∂∂∂ ∂∂∂∂

∂∂∂∂ ∂∂∂∂/

/(1c)

∆∆∆∆z∆∆∆∆x

Geometry

vzvx

Kinematics

Sifat Garis Aliran

Flow line (FL)

Equipotential (EP)

Dari geometrinya (2b)

Hukum Darcy

Sehingga (2c)

dx

dz

v

vFL

x

z

=

v kh

xx = −∂∂

dx

dz

h x

h zFL

=

∂ ∂∂ ∂

v kh

zz = −∂∂

Bentuk Orthogonal antara garis aliran dan garis equipotential

Flow line (FL)

Equipotential (EP)

dz

dx

h x

h zEP

= − ∂∂∂∂ ∂∂∂∂

∂∂∂∂ ∂∂∂∂/

/

dx

dz

h x

h zFL

=

∂ ∂∂ ∂

Pada equipotential

Pada garis aliran

Sehinggadx

dz

dx

dzFL EP

×

= − 1 (3)

T

Y

Z FL

vQ

yx= ∆∆∆∆

v kh

zt= ∆∆∆∆

(4a)

(4b)

Dari definisi aliran

Menurut hukum Darcy

Sifat Geometri jaringan aliran

hh+∆h

h+2∆h

EP

∆∆∆∆Q

Xy

z

t X

FL ∆∆∆∆∆∆∆∆Q

k h

yx

zt=

∆∆∆∆∆∆∆∆Q

k h

YX

ZT=

(4c)

(4d)

Gabungkan (4a)&(4b)

Atau

∆∆∆∆Q

Kesimpulan

yx

zt

YX

ZT= (5)

vQ

cd= ∆∆∆∆

v kh

ab= ∆∆∆∆

(6a)

(6b)

From the definition of flow

From Darcy’s law∆∆∆∆Q

D

B

Ch

Geometric properties of flow nets

FL

∆∆∆∆Q

EP( h )

∆∆∆∆∆∆∆∆Q

k h

cd

ab=

∆∆∆∆∆∆∆∆Q

k h

CD

AB=

(6c)

(6d)

Similarly

Combining (6a)&(6b)

Conclusioncd

abCDAB

=

ab

c

d Ah h+ ∆∆∆∆EP ( h + ∆∆∆∆h )

• When drawing flow nets by hand it is most convenient to draw them so that

• Each flow tube carries the same flow ∆Q

• The head drop between adjacent EPs, ∆h, is the

Geometric properties of flow nets

• The head drop between adjacent EPs, ∆h, is the same

• Then the flow net is comprised of “SQUARES”

Geometric properties of flow nets

Demonstration of ‘square’ rectangles with inscribed circles

Drawing Flow Nets

To calculate the flow and pore pressures in the ground a flow net must be drawn.

The flow net must be comprised of a family of orthogonal lines (preferably defining a square mesh) that also satisfy the boundary conditions.

WaterH-z

H

Common boundary conditionsa. Submerged soil boundary - Equipotential

hu

zw= +

Datum

z

H

(7)

hu

z

now

u H z

so

hH z

z H

w

w

w w

w

w

= +

= −

= − + =

γγγγ

γγγγ

γγγγγγγγ

( )

( )

Permeable Soil

Flow Linevn=0

vt

Common boundary conditionsb. Impermeable soil boundary - Flow Line

Impermeable Material

Common boundary conditionsc. Line of constant pore pressure - eg. phreatic surface

hu

zw

w

= +γγγγ

hu

zw

w

= +γγγγ

∆∆∆∆∆∆∆∆

∆∆∆∆

Head is given by

and thuswγγγγ

h z=∆∆∆∆ ∆∆∆∆

uw

=∆∆∆∆ 0now if pore pressure is constant

and hence (8)

Common boundary conditionsc. Line of constant pore pressure - eg. phreatic surface

Procedure for drawing flow nets

• Mark all boundary conditions

• Draw a coarse net which is consistent with the boundary conditions and which has orthogonal equipotentials and flow lines. (It is usually easier to visualise the pattern of flow so start by drawing the visualise the pattern of flow so start by drawing the flow lines).

• Modify the mesh so that it meets the conditions outlined above and so that rectangles between adjacent flow lines and equipotentials are square.

• Refine the flow net by repeating the previous step.

Value of head on equipotentials

Phreatic line

∆ hH

Number of potential drops= (9)

15 m

Datumh = 15m

h = 12m h = 9m h = 6mh = 3m

h = 0

Calculation of flowPhreatic line

15 m

h = 15m

h =12m h = 9m h = 6mh = 3m

h = 0

For a single Flow tube of width 1m: ∆Q = k ∆h (10a)

For k = 10-5 m/s and a width of 1m ∆Q = 10-5 x 3 m3/sec/m (10b)

For 5 such flow tubes Q = 5 x 10-5 x 3 m3/sec/m (10c)

For a 25m wide dam Q = 25 x 5 x 10-5 x 3 m3/sec (10d)

Q kH

NN

hf=Note that per metre width (10e)

Calculation of pore pressurePhreatic line

P5m

15 m

h = 15m h = 0

P5m

hu

zw

w

= +γγγγ

(11a)Pore pressure from

h = 12m h = 9m h = 6mh = 3m

Calculation of pore pressurePhreatic line

P5m

15 m

h = 15m h = 0

P5m

hu

zw

w

= +γγγγ

(11a)

uw w= − −[ ( )]12 5 γ (11b)

Pore pressure from

At P, using dam base as datum

h = 12m h = 9m h = 6mh = 3m

StrandedVessel

Water Supply

Example Calculating Pore Pressures

20 m

Soft SeaBottom

Well Point

ReactionPile10 m

Step 1: Choose a convenient datum. In this example the sea floor has been chosen

Then H1 = 40 mH2 = 1 m.

The increment of head, ∆h = 39/9 = 4.333 m

A B C D E

Step 2: Calculate the head at points along the base of the vessel. For convenience these are chosen to be where the EPs meet the convenience these are chosen to be where the EPs meet the vessel (B to E) and at the vessel centerline (A). Hence calculate the pore water pressures.

At B Head = H1 - 5 ∆h = H2 + 4 ∆h = 18.33 m

Pore pressure at B = 18.33 γw = 179.8 kPaww zhu γ)]([ −=

Step 3: Calculate the upthrust (Force/m) due to pore pressures

0

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10

Dis tance from ce ntre line (m )

Po

re W

ate

r P

res

su

re (

kP

a)

×

++×

++×

++×

+× 7.02

3.529.948.1

2

9.943.1375.2

2

3.1378.1795

2

8.1791.2012

= 3218 kN/m

Without pumping Upthrust = 20 × 1 × 9.81 = 196 kN/m

Upthrust due to Pumping = 3218 – 196 = 3022 kN/m

200

250

Po

re W

ate

r P

res

su

re (

kP

a)

Flow required, h

f

N

NHkQ = = 24 108.1

9

1439103 −− ×=××× m3/m/sec