Hidrolika Air Tanah (2)

HIDROLIKA ALIRAN TANAH

AKUIFER TERKEKANG (confined Aquifer) Dasar Penurunan Rumus

Persamaan dasar untuk aliran air tanah pada akuifer terkekang dalam keadaan aliran tunak adalah persamaan Laplace, yang merupakan pengembangan dari hukum Darcy dan Kontinuitas. 2

    x

2

2



    y

2

2



   2

 z 



0

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Dengan anggapan bahwa pada umumnya aliran mendatar lebih berperan dibandingkan aliran tegak, maka persamaan menjadi : 2 2        x

2



 y

2

0

Persamaan di atas berlaku untuk akuifer terkekang, dengan anggapan bahwa tebal akuifer adalah konstan.

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Untuk aliran satu arah, hukum Darcy : Vx   K 

 

Vy = 0 ; Vz = 0

 x

Persamaan Kontinuitas : Q  V  x . D. B Q   K . D. B.

   x

Persamaan Laplace : 2

    x



2

, merupakan persamaan Umum

0

 2 



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Syarat batas x = 0,  



 H 1

maka

x = L,



 H 2

maka, H2 = A. L + B

 

H1 = B

 A



 2



 1

 L

Sehingga :  



 H  2

1

  H 

 L

. X     H 1

Debit yang mengalir : Q   B. D.Vx Vx   K 

   x

 H 2  H 1

disebut persamaan tinggi air tanah

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Contoh soal :

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Pernyelesaian : - Persamaan tinggi air tanah

 2    x 2   0     x    A     Ax   B Syarat Batas Syarat batas x = 0,  



 H 1



15m

maka B = 15

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Persamaan tinggi air tanah :   

0.05 x  15

Atau dipakai rumus langsung :  H 2  H 1    . X    H 1  L

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AKUIFER SETENGAH TERKEKANG ( Semi Confined Aquifer/ Leaky Artesian Aquifer) Dasar Penurunan Rumus : Pada akuifer Setengah terkekang : - adanya aliran kearah vertikal. Walaupun aliran kearah horisontal lebih

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Sehingga persamaan kontinuitas : 2

    x

2

2



    y

2

0

Atau dapat ditulis berdasarkan hukum Darcy :

  2   2     K  2  2  x. y.D  0  y      x

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Jumlah air yang melalui lapisan 1 persatuan waktu adalah adal ah :    H 1  K 1  x. y d 1 Dimana : H1 = tinggi muka air tanah pada lapisan di atas lapisan 1 (m) K1 = nilai kelulusan air lapisan 1 (m/hari)

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sehingga persamaan menjadi :

  2   2      H 1    H 2  K  2  2    x. y. D  K 1  x. y  K 2  x.y  0 d 1 d 2  y     x   2   2      H 1    H 2  KD 2  2    0

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Dibawah ini diberikan dua persoalan yang sering terjadi pada akuifer setengah

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Bila semua aliran hodrisontal tegak lurus sumbu Y, maka persamaan dasar

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Persamaan karakteristiknya adalah :

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Persamaan Tinggi Muka Air Tanah :

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1. Jika persoalan seperti gambar berikut :

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

Pada kondisi batas : Dalam kondisisepanjang -L ≤ x ≤ L ini tidak terjadi rembesan disebab kan tanggul kedap air, sehingga perhitungan dilakukan dengan prinsip aguifer terkekang. Persamaan umum: φ = A.x + B Pada x = -L

φ = H2 

H2 = A.(-L) + B

Pada x = L

φ = H3 

H3 = A. L + B H2 + H3 = 2B

 A 

 H 2  B  L



 H 2



  H   H   1

2

2

 L

3

B = ½ (H2 + H3)

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Persamaan menjadi:

kecepatan aliran:

1   H 2   H 3         x    H 2   H 3  2   2 L  

      H 2  H 3      v  k    x  x   2 L  

  H 2  H 3      2 L  

  H 2  H 3      2 L  

v  k .  

v  k .

debit yang lewat aquifer: q = V . D

  H 2   H 3      2 L  

q2 = k . D

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

φ2 = H4

Pada x =

Pada x = L

φ = H4

φ = H3





H4 –  H  H4 0

= A.e + B.e = A.e~ + 0

H3 –  H  H4 H3 –  H  H4

= A.eL/ λ + B.e L/ λ = B.e L/ λ

B

=

 H 3 e

 H4 φ –  H

diperoleh persamaan:

φ φ  

 

=



 A = 0

  H 4  L /  

 H 3 e

  H 4  L /  

-x/λ  e-x/λ

-x/λ = H4 + (H3  –  H  H4) e λ . e-x/λ X)/ λ = H3 + (H3  –  H  H4) e(L - X)/λ

1

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Debit aliran yang melalui aquifer selebar B pada x = L Q=vxDxB



q=

Q  B

  H 3   H 4   q3= k . D        Besarnya nilai H2 dan H3 dapat dihitung berdasarkan hukum kontinuitas : q1 = q2 = q3 dari persamaan q1 = q2 akan didapat:

  H 1  H 2   k . D =      

  H 2   H 3   k . D     2 L  

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: H1 = 10 m , H4 = 13 m , L = 15 m maka



 H 2



 H 2



,

 

= 200 m

2 15 10  200 H 3 200  2(15) 1,304  0,87 H 3

………………………….….…(1)

dari persamaan q1 = q3 akan didapat:

  H 1  H 2     H   H 4    = k . D 3             

k . D

(H1  –  H  H2)

= (H3  –  H  H4) H2 = H1 + H4  –  H  H3 H2 = 23  –  H  H3

Dari persamaan (1) dan (2):

……………………………(2)

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Nilai-nilai yang telah diketahui dari hasil hitungan, maka persamaan head dari masing-masing batasan adalah sebagai berikut: 

Pada kondisi batas φ = H1 + (H2 – H  – H1) e(L + X)/λ

φ = 10 + (11,4 – 10) –  10) e(15 + x) / 200

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

φ = H1 + (H2 –  H  H1) e(L + X)/λ

q=V.D



V    k 

   x



   x  



  H 2   H 1





11 4  10

1

e

 L  x  /  

  1

e

(15  X  ) / 200

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untuk x < -15, misal diambil x = - 40 (25 m dari sisi kiri tanggul) 7

5,6 10



5,6 10



q



q

 

q

 

7

e e

(15 40) / 200

( 0,125)

5,6 10 7 (0,8825) 

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Soal No. 2

Titik awal x = 0 pada saluran sebelah kanan. Untuk x = 0 q



2

k H2

 H12 

  L   N  x 

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Dengan menggunakan persamaan sebagai berikut: H

2

 N 

x

k 

2

 C1  C 2

Boundari condition;

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Persamaan head air tanah adalah:

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