jaringan aliran -...
TRANSCRIPT
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