rembesan danflow net
TRANSCRIPT
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HIDROLIKA TANAH
PERMEABILITASREMBESAN/SEEPAGEJARINGAN ALIRAN
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PERMEABILITAS
PENGERTIAN :KECEPATAN ATAU KEMAMPUAN AIR/CAIRAN MELALUI SUATU MEDIA BERPORI
SATUAN :m/s, cm/s
TUJUAN :Mengevaluasi jumlah rembesan (seepage) yang melaluibendungan/tanggulMengevaluasi gaya angkat atau gaya rembesan di bawahstruktur hidrolik untuk keperluan analisa stabilitasMengontrol kecepatan rembesanMengetahui laju penurunan konsolidasi (akan dibahas padatopik ke-7)
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PENENTUAN KOEFISIEN PERMEABILITASLABORATORIUM
TINGGI KONSTAN (CONSTANT HEAD)TINGGI JATUH (FALLING HEAD)
LAPANGANAKIFER BEBAS (UNCONFINED AQUIFER)AKIFER TERKEKANG (CONFINED AQUIFER)TINGGI AIR TIDAK TETAP
PERMEABILITAS
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PERMEABILITAS
TINGGI TETAP (CONSTANT HEAD)LEBIH SESUAI UNTUK TANAH BERPASIR, PASIR ATAU KERIKIL YANG MEMPUNYAI ANGKA PORI YANG BESAR
PERSAMAAN DASAR :
A.h.tQ.Lk
.tLhk. A. A.(k.i).t A.v.t Q
=
⎟⎠⎞
⎜⎝⎛===
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PERMEABILITAS
TINGGI JATUH (FALLING HEAD)LEBIH EKONOMIS UNTUK PENGUJIAN JANGKA PANJANGPERSAMAAN DASAR :
2
1
keluarmasuk
keluar
masuk
hhln
t.AL.ak
Lh.k.A
dtdha-qq
Lh.k.Ai.k.Av.Aq
dtdha-q
dtdh-v
=
=→=
===
=
=
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PERMEABILITAS
AKIFER BEBAS (UNCONFINED AQUIFER)
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PERMEABILITAS
AKIFER BEBAS (UNCONFINED AQUIFER)
( )
( )21
22
1
2
1
2
21
22
hh.rrln.Q
k
rrln
hh..kh.r..2.drdh.kq
−π=
−π=π=
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PERMEABILITAS
AKIFER TERKEKANG (CONFINED AQUIFER)
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PERMEABILITAS
AKIFER TERKEKANG (CONFINED AQUIFER)
( )21o
2
1
hh.h..2rrlog.Q.3,2
k−π
=
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PERMEABILITAS
LUBANG BOR DENGAN TINGGI AIR BERUBAH
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −⎟
⎠⎞
⎜⎝⎛ +
∆∆
=
Ly2.
rL20
ty.
yr.40
k
2r
∆yL
y
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PERMEABILITAS
RANGE NILAI-NILAI k
KORELASI EMPIRIS 210D.Ck = Cm/s
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PERMEABILITAS
KOEFISIEN PERMEABILITAS PADA TANAH BERLAPIS
Koefisien permeabilitas vertikal (kv’) ekivalenKoefisien permeabilitas horisontal (kh’) ekivalen
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PERMEABILITAS
Koefisien permeabilitas vertikal (kv’) ekivalenDasar Perhitungan
qmasuk = qkeluar
v konstan
n
nn
2
22
1
11 H
h.k...Hh.k
Hh.ki'.kvv =====
vh
kH.....
vh
kH;
vh
kH;
vh
kH n
n
n3
3
32
2
21
1
1 ====
n
n
3
3
2
2
1
1n321
kH...
kH
kH
kH
vh...
vh
vh
vh
++++=++++
LH...HHH n321 =++++
Lh'.kvv =
n
n
3
3
2
2
1
1k
H...kH
kH
kH
L'kv++++
=
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PERMEABILITAS
i'.kh.Lv.Aq ratarata == −
i.H.k...i.H.ki.H.ki'.kh.L nn2211 +++=
Koefisien permeabilitas horisontal (kh’) ekivalen
LH.k...H.kH.k'kh nn2211 +++
=
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CONTOH SOAL 1
Pertanyaan :Berapa Koefisien Permeabilitas Pasir dalam ft/min
q = 1 ft3/hr
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CONTOH SOAL 1
21t hhh ∆+∆=∆
1
1
1 Lh.k
Aq ∆
=
SECTION 1
k.AL.qh1
11 =∆
2
2
2 Lh.k
Aq ∆
= k.AL.qh2
22 =∆
k.AL.q
k.AL.qh
2
2
1
1t +=∆
SECTION 2
TOTAL
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CONTOH SOAL 1
k.10600.1
k.20400.120 +=
k = 4 ft/hour = 6,67x10-2 ft/min
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CONTOH SOAL 2
q
Pertanyaan :- Hitung h- Hitung q dalam cc/sec
Bagian 1Bagian 2
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CONTOH SOAL 2
Bagian 1
1111 A.i.kq =
Bagian 2
2222 A.i.kq =
25.40
h50.02,0q1−
= 25.40
5h.007,0q2−
=
21 qq =)5h.(007,0)h50.(02,0 −=−
h = 38,33 cm
Penentuan Tinggi h
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CONTOH SOAL 2
1111 A.i.kq = 2222 A.i.kq =
25.40
33,3850.02,0q −=
Penentuan debit air
atau
q = 0,15 cm/s
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REMBESAN
PENGERTIANBANYAKNYA JUMLAH AIR/CAIRAN YANG MASUK ATAU KELUAR PADA SUATU MEDIA/MASSA TANAH TERTENTU
TUJUANMENGETAHUI PENGARUH REMBESAN TERHADAP KESTABILAN BANGUNAN/BENDUNGANMEMPERKIRAKAN KECEPATAN ALIRAN DAN JUMLAH AIR PADA PEKERJAAN PEMOMPAAN/DEWATERING
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REMBESAN
PERSAMAAN ALIRAN AIRDASAR
HUKUM DARCY
HUKUM BERNOULLI
PERSAMAAN KONTINUITAS
i.kv =Lhi =
tankonsA.vA.vq 2211 ===
2w
222
1w
121 z.g
g.p
g2vz.g
g.p
g2v
+ρ
+=+ρ
+ = energi konstan
ni.k'v =
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REMBESAN
dxdyvdxdzvdydzvq zyxmasuk ++=Volume air yang masuk per satuan waktu :
Volume air yang keluar per satuan waktu :
dxdydzzvvdxdzdy
yv
vdydzdxxvvq z
zy
yx
xkeluar ⎟⎠⎞
⎜⎝⎛
∂∂
++⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂++⎟
⎠⎞
⎜⎝⎛
∂∂
+=
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REMBESAN
dtdVdxdydz
zv
yv
xv zyx =⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
+∂
∂+
∂∂
−
qmasuk = qkeluar
te
eo11
tW1
zv
yv
xv w
w
zyx
∂∂
+=
∂∂
γ=⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
+∂
∂+
∂∂
−
PERSAMAAN KONTINUITAS
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REMBESAN
0te=
∂∂
0zv
yv
xv zyx =⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
+∂
∂+
∂∂
KONDISI STEADY STATE :
KECEPATAN ALIRAN AIR :
zhkikv
yhkikv
xhkikv
zzzz
yyyy
xxxx
∂∂
−==
∂∂
−==
∂∂
−==
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REMBESAN
0zhk
yhk
xhk 2
2
z2
2
y2
2
x =∂∂
+∂∂
+∂∂
0zhk
zyhk
yxhk
x zyx =⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
0zh
yh
xhh 2
2
2
2
2
22 =
∂∂
+∂∂
+∂∂
=∇
TANAH HOMOGENk Konstan terhadap x,y,z
TANAH ISOTROPIkx = ky = kz = k
PERSAMAAN LAPLACE
0zh
xhh 2
2
2
22 =
∂∂
+∂∂
=∇DUA DIMENSI
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REMBESAN
SOLUSI SEEPAGECLOSED FORM SOLUTIONMODEL SOLUTIONSAPPROXIMATE SOLUTIONS
NUMERICAL SOLUTIONSGRAPHICAL SOLUTIONS FLOW NET
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JARINGAN ALIRAN / FLOW NET
PENGERTIANGabungan dari dua kelompok garisyang saling tegak lurus yaitu :
Garis Aliran (Flow Line)kumpulan titik atau garis yang menyatakanarah aliranGaris Ekipotensial (Equipotential Line)tempat kedudukan titik yang mempunyaitinggi tekanan air (head) total yang sama.
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
CARA PENGGAMBARAN FLOW NETPermukaan atas air baik di hulu maupun di hilirmerupakan garis ekipotensialGaris interface antara air dan tanah merupakangaris ekipotensialPerpotongan garis alir dan garis ekipotensialmembentuk sudut tegak lurusPermukaan suatu batas yang kedap air (impermeable) merupakan garis alirKotak yang dibentuk dari garis alir dan garisekipotensial membentuk bangun bujursangkar
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
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JARINGAN ALIRAN / FLOW NET
bh.k).1.a(i.k.Av.Aq ∆
===∆d
21
Nhhh −
=∆
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=∆∑=
d
21f N
hhba.k.Nqq 21 hhH −=∆
⎟⎠⎞
⎜⎝⎛∆=
baH.
NNkq
d
f H.NNkq
d
f ∆=a = b
h
h+∆h
∆q
a b
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CONTOH SOAL 3
4,50 m
8,60 m
A B C D
E
Datum
Sheet Piling
6,0 m
0,5 m
k = 1,5 x 10-6 m/s
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CONTOH SOAL 3
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CONTOH SOAL 3
Nd = 12
Nf = 4,3
∆H = 4,0 m
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CONTOH SOAL 3
H.NNkq
d
f ∆=
H.Nnh
d
dP ∆=
m.s/m10x15,200,4.12
3,4.10x5,1q 366 −− ==
m33,34.1210hP ==
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CONTOH SOAL 4
k = 2,5 x 10-5 m/s
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CONTOH SOAL 4
Nd = 15 Nf = 4,7
∆H = 4,0 m
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CONTOH SOAL 4
H.NNkq
d
f ∆=
m.s/m10x1,300,4.15
7,4.10x5,2q 355 −− ==
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GAYA REMBESAN/SEEPAGE FORCE
H
h2
L
h1
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GAYA REMBESAN/SEEPAGE FORCE
A).hh.(A.L.F 21wt −γ−γ=∑
GAYA TOTAL
GAYA BADAN (BODY FORCE)
volumeTotal_Gaya)F(force_Body =
L Berat tanah = γt.L.A
γw . h2 . A
γw . h1 . A
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GAYA REMBESAN/SEEPAGE FORCE
wbouyant
wtwt
21wt
.iF
)i1(L
LHF
A.LA).hh.(A.L.F
γ−γ=
+γ−γ=⎟⎠⎞
⎜⎝⎛ +
γ−γ=
−γ−γ=
w.i γSEEPAGE BODY FORCE (j)=
KONDISI KRITIS
e11Gi
0.i
s
w
bouyantc
wbouyant
+−
=γ
γ=
=γ−γ
γbouyant = γt - γw
H
h2
L
h1
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CONTOH SOAL 5
k = 1x10-3 cm/s
n = 0,67
Pertanyaan :
1. Debit
2. Kecepatan Pengaliran
3. Kecepatan Rembesan
4. Gaya Rembesan titik A
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CONTOH SOAL 5
Debit Pengaliran
Kecepatan Pengaliran
A.i.kq = 144
LHi ===
A10x1A.1.10x1q 55 −− ==
i.kv =
s/m10x11.10x1v 55 −− ==
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CONTOH SOAL 5
Kecepatan Rembesan
Gaya Rembesan
nv
ni.k'v ==
s/m10x5,167,0
10x1'v 55
−−
==
ws .iF γ=
2s m/kg10001000.1F ==