Kajian Koefisien Rembesan Pada Saluran Irigasi Tersier Di Desa Kuala Simeme Kecamatan Namorambe Kabupaten Deli Serdang

  Lampiran 1. Flow Chart Penelitian

  Mulai Menentukan lokasi pengukuran Menghitung luas penampang saluran

  Parameter :

  1. Bulk Density

  2. Particle Density

  3. Tekstur Mengukur nilai parameter

  Tanah

  4. Struktur Tanah

  5. Bahan Organik

  6. Evapotransp Melakukan pengolahan data irasi _8.

  7. Perkolasi rembesan Selesai

  Lampiran 5. Perhitungan kerapatan massa, kerapatan partikel dan porositas BTKO Volume Volume Bulk Particle Porositas

  Saluran (gr) total partikel Density Density

  3

  3

  3

  3

  (cm ) (cm ) (g/cm ) (g/cm ) Tepi kanan saluran 1 252,61 192,33 105 1,32 2,19 39,72 Tepi kiri saluran 1 228,72 192,33 100 1,19 2,12 43,86 Dalam saluran 1 212,67 192,33 100 1,11 1,69 34,31 Tepi kanan saluran 2 260,02 192,33 115 1,13 2,17 37,78 Tepi kiri saluran 2 217,23 192,33 100 1,35 2,28 50,43 Dalam saluran 2 228,10 192,33 100 1,19 2,16 44,90 BTKO = Berat tanah kering oven (massa tanah kering) Volume total = volume sing sample

  1

2 Volume Total = t

  4

  1

  2

  = (3,14)(7 cm) (5 cm)

  4

  1

  3

  = (769,3 cm )

  4

  3

  = 192,33 cm Saluran 1 Kerapatan Massa (Bulk Density) Tepi Kanan Saluran Ms = 252,61 g

  Ms = ρ b Vt

  3 252,61 g/cm

  =

  3 192,33 g/cm

  3

  1,32 g/cm

  =

  Tepi Kiri Saluran Ms = 228,72 g

  Ms = ρ b Vt

  3 228,72g/cm

  =

  3 192,33 g/cm

  3 = 1,19 g/cm

  Dalam Saluran Ms = 212,22 g

  Ms = ρ b Vt

  3 212,22 g/cm

  =

  3 192,33 g/cm

  3

  1,11 g/cm

  =

  Kerapatan Partikel Tepi Kanan Saluran Berat Tanah = 228,72 g Volume Tanah = 250 ml Volume air = 200 ml Volume air tanah = 340 ml

  berat tanah = ρ s (volume tanah

  −volume pori )

  Volume Ruang Pori = (volume air + volume tanah) – volume air tanah Volume Ruang Pori = (200 ml+ 250ml) – 340 ml

  = 110 ml ρ

  −volume pori )

  3 Dalam Saluran

  = 2,12 g/cm

  3 200 ml −100 ml

  = 212,67 g/cm

  berat tanah (volume tanah −volume pori )

  =

  s

  = 100 ml ρ

  Volume Ruang Pori = (volume air + volume tanah) – volume air tanah Volume Ruang Pori = (200 ml + 200 ml) – 300 ml

  = berat tanah (volume tanah

  s

  ρ s

  Berat Tanah = 212,67 g Volume Tanah = 200 ml Volume air = 200 ml Volume air tanah = 300 ml

  3 Tepi Kiri Saluran

  2,19 g/cm

  =

  3 200 ml −110 ml

  = 228,72 g/cm

  berat tanah (volume tanah −volume pori )

  =

  Berat Tanah = 252,61 g Volume Tanah = 350 ml

  Volume air = 200 ml Volume air tanah = 350 ml

  berat tanah = ρ s (volume tanah

  −volume pori )

  Volume Ruang Pori = (volume air + volume tanah) – volume air tanah Volume Ruang Pori = (200 ml + 350 ml) – 350 ml

  = 200 ml

  berat tanah

  = ρ

  s (volume tanah −volume pori )

  3 252,61 g/cm

  = 350 ml −200 ml

  3 = 1,69 g/cm

  Porositas Tepi kanan saluran 1

  ρ _b

  Porositas = �1 − � x100%

  ρ _s

  3 1,32 g/cm

  = �1 −

  3 � x100% g/cm

  2,19

  = 39,72 % Tepi kiri saluran 1

  ρ _b

  Porositas = �1 − � x100%

  ρ _s

  3 g/cm

  1,19

  = �1 −

  3 � x100% 2,12 g/cm

  = 43,86 % Dalam saluran 1

  ρ _b

  Porositas = �1 − � x100%

  ρ _s

  3 1,11 g/cm

  = �1 −

  3 � x100% 1,69 g/cm

  = 34,31 % Saluran 2 Kerapatan Massa (Bulk Density) Tepi Kanan Saluran Ms = 217,23 g

  Ms = ρ b Vt

  3 217,23g/cm

  =

  3 192,33 g/cm

  3

  1,13 g/cm

  =

  Tepi Kiri Saluran Ms = 260,02 g

  Ms = ρ b Vt

  3 260,02 g/cm

  =

  3 192,33 g/cm

  3

  1,35 g/cm

  =

  Dalam Saluran Ms = 228,10 g

  Ms = ρ b Vt

  3 228,10 g/cm

  =

  3 192,33 g/cm

  3 = 1,19 g/cm

  Kerapatan Partikel Tepi Kanan Saluran Berat Tanah = 217.23 g Volume Tanah = 250 ml Volume air = 200 ml Volume air tanah = 300 ml

  berat tanah = ρ s (volu me tanah

  −volume pori )

  Volume Ruang Pori = (volume air + volume tanah) – volume air tanah Volume Ruang Pori = (200 ml+ 250ml) – 300 ml

  = 150 ml

  berat tanah

  = ρ

  s (volume tanah −volume pori )

  3 217,23 g/cm

  = 250 ml −150 ml

  3

  2,17 g/cm

  =

  Tepi Kiri Saluran Berat Tanah = 228,10 g Volume Tanah = 250 ml

  Volume air = 200 ml Volume air tanah = 300 ml

  berat tanah = ρ s (volume tanah

  −volume pori )

  Volume Ruang Pori = (volume air + volume tanah) – volume air tanah Volume Ruang Pori = (200 ml + 250 ml) – 300 ml

  = 150 ml

  berat tanah

  = ρ

  s (volume tanah −volume pori )

  3 228,10 g/cm

  = 250 ml −150 ml

  3

  2,28 g/cm

  =

  Dalam Saluran Berat Tanah = 260,02 g Volume Tanah = 250 ml Volume air = 200 ml Volume air tanah = 320 ml

  berat tanah = ρ s (volume tanah

  −volume pori )

  Volume Ruang Pori = (volume air + volume tanah) – volume air tanah Volume Ruang Pori = (200 ml + 250 ml) – 320 ml

  = 150 ml

  berat tanah

  = ρ

  s (volume tanah −volume pori )

  = 260,02 g/cm

  3 250 ml −130 ml

3 Porositas

  Tepi kanan saluran 2 Porositas =

  3 2,28 g/cm 3 � x100%

  � x100% = 44,90 %

  3

  3 2,16 g/cm

  1,19 g/cm

  �1 −

  � x100% =

  ρ _b ρ _s

  �1 −

  = 50,43 % Dalam saluran 2 Porositas =

  1,13 g/cm

  �1 −

  �1 −

  = 2,16 g/cm

  ρ _b ρ _s

  �1 −

  = 37,78 % Tepi kiri saluran 2 Porositas =

  3 2,17 g/cm 3 � x100%

  1,35 g/cm

  �1 −

  � x100% =

  ρ _b ρ _s

  � x100% = Lampiran 6. Perhitungan debit Perhitungan debit pada saluran satu dan dua

  Lokasi Saluran 1 (l/det) Saluran 2 (l/det) Hulu 2,34 2,19 Hilir 2,13 1,79

  Saluran 1 Hulu a.

  Cara Tampung Ulangan Waktu (t) Volume (v) Debit (Q)

  (det) (l) (l/det) I 1,99 6,43 3,232

  II 2,22 6,52 2,937

  III 1,72 6,40 3,721

Q1+Q2+Q3

  Q =

  rata-rata

  3 3,232 l/ det + 2,937 l/ det + 3,721 l/det

  =

  3

  = 3,296 l/det b. Sekat Ukur Thompson

  H = 7,8 cm

  5/2

  Q = 0,0138 H

  5/2

  = 0,0138 (7,8 cm) =2,34 l/det Hilir a.

  Cara Tampung Ulangan Waktu (t) Volume (v) Debit (Q)

  (det) (l) (l/det) I 2,41 5,87 2,436

  II 2,08 5,82 2,798

  III 2,42 5,92 2,446

Q1+Q2+Q3

  Q =

  rata-rata

  3 2,436 l/ det + 2,798 l/ det + 2,446 l/det

  =

  3

  = 2,56 l/det b. Sekat Ukur Thompson

  H = 7,5 cm

  5/2

  Q = 0,0138 H

  5/2

  = 0,0138 (7,5 cm) =2,13 l/det

  Saluran 2 Hulu a.

  Cara Tampung Ulangan Waktu (t) Volume (v) Debit (Q)

  (det) (l) (l/det) I 2,23 5,18 2,322

  II 2,67 5,62 2,104

  III 3,20 6,28 1,962

Q1+Q2+Q3

  Q =

  rata-rata

  3 2,322 l/ det + 2,104 l/ det + 1,96 l/det

  =

  3

  = 2,129 l/det b.

  Sekat Ukur Thompson H = 7,6 cm

  5/2

  Q = 0,0138 H

  5/2

  = 0,0138 (7,6 cm) = 2,19 l/det

  Hilir a.

  Cara Tampung Ulangan Waktu (t) Volume (v) Debit (Q)

  (det) (l) (l/det) I 2,75 4,90 1,810

  II 1,69 3,19 1,887

  III 2,54 4,20 1,653

Q1+Q2+Q3

  Q =

  rata-rata

  3 1,810 l/ det + 1,887 l/ det + 1,653 l/det

  =

  3

  = 1,783 l/det b. Sekat Ukur Thompson

  H = 7 cm

  5/2

  Q = 0,0138 H

  5/2

  = 0,0138 (7 cm) = 1,79 l/det Lampiran 7. Ukuran saluran tersier Saluran 1

  16,1 cm +18,5 cm +16,2 cm

  Kedalaman =

  3

  = 16,93 cm = 0,17 m

  133 cm +135 cm +132 cm

  Lebar =

  3

  = 133,34 cm = 1,334 m

  Saluran 2

  16,1 cm +16,3 cm +15,2 cm

  Kedalaman =

  3

  = 15,86 cm = 0,16 m

  124 cm +129 cm +125 cm

  Lebar =

  3

  = 126 cm = 1,26 m Lampiran 8. Kehilangan Air Perhitungan Kehilangan Air Lokasi Kehilangan air (l/det)

  Saluran 1 0,219 Saluran 2 0,408

  Kehilangan Air Saluran 1 Kehilangan Air = Qhulu – Qhilir

  = (2,344 – 2,125) l/det = 0,219 l/det = 0,219 x 10

• 3

  m

  3

  /det Kehilangan Air Saluran 2 Kehilangan Air = Qhulu – Qhilir

  = (2,197 – 1,789) l/det = 0,408 l/det = 0,408 x 10

• 3

  m

  3

  /det Lampiran 9. Perhitungan Evapotranspirasi Saluran 1dan saluran2 Kc Rumput = 0,85 Temperatur (t) = 26,35 C Lama Penyinaran Matahari (P) = 4,79 Kt = 0,031 lt + 0,240

  = 0,031 l(26,35

  C) + 0,240 = 1,06

  K = Kt x Kc K = 1,06 x 0,85

  = 0,901 U =

  KP (45,7+813) 100

  =

  0,901 x 4,79 (45,7(26,35) + 813) 100 =

  87, 057 mm/bln = 2,90 mm/hari Lampiran 10. Perhitungan Perkolasi Saluran 1 Ulangan Perkolasi(mm/hari)

  I 5,66

  II 7

  III 6 Rata-rata 6,22

  Ulangan 1

• h

  1 (Perkolasi di titik 0 m)

  P

  1 = 20,5 cm

  h

  2 = 20 cm h h

  1

  2 −

  P

  1 = mm/hari t t

  1

  2 − (20,5 cm −20 cm )

  P

  1 = 1 hari

  P

  1 = 0,5 cm/hari

  = 5 mm/hari

• h

  2 (Perkolasi di titik 15 m)

  P

  1 = 21 cm

  h = 20,3 cm

  2 h h

  1

  2 −

  P = mm/hari

  2 t t

  1

  2 − (21 cm −20,3 cm )

  P

  2 = 1 hari

  P

  2 = 0,7 cm/hari

  = 7 mm/hari

• 3 (Perkolasi di titik 30 m)

  P h

  1 = 21 cm

  h

  2 = 20,5 cm h h

  1

  2 −

  P

  3 = mm/hari t t

  1

  2 − (21 cm −20,5 cm )

  P

  3 = 1 hari

  P = 0,5 cm/hari

  3

  = 5 mm/hari

  

(P1+P2+P3)

  Perkolasi Rata-rata Ulangan 1 =

  

3

(5+7+5) mm /hari =

  3 = 5,66 mm/hari

  Ulangan II (Perkolasi di titik 0 m) -

1 P

  h

  1 = 20 cm

  h

  2 = 19,3 cm h h

  1

  2 −

  P

  1 = mm/hari t t

  1

  2 − (20 cm −19,3 cm )

  P =

  1 1 hari

  P

  1 = 0,7 cm/hari

  = 7 mm/hari P

• 2 (Perkolasi di titik 15 m)

  h

  1 = 20,2 cm

  h = 19,6 cm

  2 h h

  1

  2 −

  P = mm/hari

  2 t t

  1

  2 − (20,2 cm −19,6 cm )

  P

  2 = 1 hari P

  2 = 0,6 cm/hari

  = 6 mm/hari

• h = 20,4 cm

  3 (Perkolasi di titik 30 m)

  P

  1

  h

  2 = 19,6 cm h h

  1

  2 −

  P

  3 = mm/hari t t

  1

  2 − (20,4 cm −19,6 cm )

  P

  3 = 1 hari

  P = 0,8 cm/hari

  3

  = 8 mm/hari

  

(P1+P2+P3)

  Perkolasi Rata-rata Ulangan II =

  

3

(7+6+8) mm /hari =

  3 = 7 mm/hari

  Ulangan III (Perkolasi di titik 0 m)

• 1

  P

  h

  1 = 19,3 cm

  h

  2 = 18,5 cm h h

  1

  2 −

  P

  1 = mm/hari t t

  1

  2 − (19,3 cm −18,5 cm )

  P =

  1 1 hari

  P

  1 = 0,8cm/hari

  = 8 mm/hari P

• 2 (Perkolasi di titik 15 m)

h

  1 = 19,5 cm

  h = 19 cm

  2 h h

  1

  2 −

  P = mm/hari

  2 t t

  1

  2 − (19,5 cm −19,cm )

  P

  2 = 1 hari

  P = 0,5 cm/hari

  2

  = 5 mm/hari P

• 3 (Perkolasi di titik 30 m)

  h = 19,8 cm

  1

  h

  2 = 19,2 cm h h

  1

  2 −

  P

  3 = mm/hari t t

  1

  2 − (19,8 cm −19,2 cm )

  P

  3 = 1 hari

  P = 0,5 cm/hari

  3

  = 5 mm/hari

  

(P1+P2+P3)

  Perkolasi Rata-rata Ulangan III =

  

3

(8+5+5) mm /hari =

  3

  6 mm/hari

  = (5,66+7+6 ) mm /hari

  Nilai rata-rata Perkolasi Saluran 1 =

  3

  = 6,22 mm/hari Saluran 2 Ulangan Perkolasi(mm/hari)

  I 13,67

  II 11

  III 8,66 Rata-rata 11,11

  Ulangan 1 P

• 1 (Perkolasi di titik 0 m)

  h

  1 = 22,5 cm

  h

  2 = 21,4 cm h h

  1

  2 −

  P

  1 = mm/hari t t

  1

  2 − (22,5 cm −21,4 cm )

  P

  1 = 1 hari

  P = 1,1 cm/hari

  1

  = 11 mm/hari P

• 2 (Perkolasi di titik 15 m)

  h

  1 = 22 cm

  h

  2 = 20,5 cm h h

  1

  2 −

  P

  2 = mm/hari t t

  1

  2 − (22 cm −20,5 cm )

  P

  2 = 1 hari

  P

  2 = 1,5 cm/hari

  = 15mm/hari (Perkolasi di titik 30 m) -

3 P

  h

  1 = 22 cm

  h

  2 = 20,5 cm h h

  1

  2 −

  P

  3 = mm/hari t t

  1

  2 − (22 cm −20,5 cm )

  P

  3 = 1 hari

  P = 1,5 cm/hari

  3

  = 15 mm/hari

  

(P1+P2+P3)

  Perkolasi Rata-rata Ulangan 1 =

  

3

(11+15+15) mm /hari =

  3 = 13,67 mm/hari

  Ulangan II (Perkolasi di titik 0 m) -

1 P

  h

  1 = 21 cm

  h

  2 = 20,5 cm h h

  1

  2 −

  P

  1 = mm/hari t t

  1

  2 − (21 cm −20,5 cm )

  P =

  1 1 hari

  P

  1 = 0,7 cm/hari

  = 7 mm/hari P

• 2 (Perkolasi di titik 15 m)

  h

  1 = 20,5 cm

  h = 19 cm

  2 h h

  1

  2 −

  P = mm/hari

  2 t t

  1

  2 − (20,5 cm −19 cm )

  P

  2 = 1 hari P

  2 = 1,5 cm/hari

  = 15 mm/hari

• h = 20,5 cm

  3 (Perkolasi di titik 30 m)

  P

  1

  h

  2 = 19,3 cm h h

  1

  2 −

  P

  3 = mm/hari t t

  1

  2 − (20,5 cm −19,3 cm )

  P

  3 = 1 hari

  P = 1,2 cm/hari

  3

  = 12 mm/hari

  

(P1+P2+P3)

  Perkolasi Rata-rata Ulangan II =

  

3

(7+15+12) mm /hari =

  3 = 11 mm/hari

  Ulangan III (Perkolasi di titik 0 m)

• 1

  P

  h

  1 = 20,5 cm

  h

  2 = 19,3 cm h h

  1

  2 −

  P

  1 = mm/hari t t

  1

  2 − (20,5 cm −19,3 cm )

  P =

  1 1 hari

  P

  1 = 1 cm/hari

  = 10 mm/hari P

• 2 (Perkolasi di titik 15 m)

h

  1 = 19 cm

  h = 18,1 cm

  2 h h

  1

  2 −

  P = mm/hari

  2 t t

  1

  2 − (19 cm −18,1 cm )

  P

  2 = 1 hari

  P = 0,9 cm/hari

  2

  = 9 mm/hari P

• 3 (Perkolasi di titik 30 m)

  h = 19,2 cm

  1

  h

  2 = 18,5 cm h h

  1

  2 −

  P

  3 = mm/hari t t

  1

  2 − (19,2 cm −18,5 cm )

  P

  3 = 1 hari

  P = 0,7 cm/hari

  3

  = 7 mm/hari

  

(P1+P2+P3)

  Perkolasi Rata-rata Ulangan III =

  

3

(10+9+7) mm /hari =

  3

  8,66 mm/hari

  = (13,67+11+68) mm /hari

  Nilai rata-rata Perkolasi Saluran 2 =

  3

  = 11,11 mm/hari Lampiran 11. Perhitungan Koefisien Rembesan No Lokasi Koefisien Rembesan (mm/hari)

  1 Tepi kanan saluran 1 3.775,68

  2 Tepi kiri saluran 1 10.368,8

  3 Tepi kanan saluran 2 34.621,51

  4 Tepi kiri saluran 2 9.562,13 Saluran 1 Perkolasi = 6,22 mm/hari x Luas saluran

  6,22 mm /hari

  2

  x 37,8 m =

  86,400

• 6

  3

  = 2,72 x 10 m /det

• 3

  = 2,72 x 10 l/det Evapotranspirasi = 2,90 mm/hari x Luas saluran

  2,90 mm /hari

  2

  = x 37,8 m

  86,400

• 6

  3

  = 1,27 x 10 m /det

• 3

  = 1,27 x 10 l/det Debit Rembesan = Kehilangan air – (Perkolasi + Evapotranspirasi)

• 3 3 -3 -3

  3

  = 0,219 x 10 m /det – (2,72 x 10 + 1,27 x 10 m /det)

• 3

  3

  = 0,215 x 10 m /det = 0,215 l/det q

  2 = Debit rembesan per satuan panjang saluran

  3 −3 0,215 x 10 m

  =

  30 m

• 6

  2

  = 7,2 x 10 m /det d kanan = 45 cm = 0,45 m d kiri = 124 cm

  = 1,24 m h = 38,5 cm

  1

  = 0,385 m Perhitungan koefisien rembesan

  q2 x 2d

  Tepi kanan =

  2 h1

  2 −6 7,2 x 10 m / det x 2 x 0,45m

  =

  2 (0,385 m )

• 4

  = 4,37 x 10 m/detik = 3.775,68 mm/hari

  q2 x 2d

  Tepi kiri =

  2 h1

  −6 7,2 x 10 / det x 2 x 1,24 m

  =

  

2

(0,385 m )

• 3

  = 1,20 x 10 m/detik = 10.368,8 mm/hari

  Saluran 2 Perkolasi = 11,11 mm/hari x Luas saluran

  11,11 mm /hari

  2

  x 40,02 m =

  86,400

• 6

  3

  = 5,15 x 10 m /det

• 3

  = 5,15 x 10 l/det Evapotranspirasi = 2,90 mm/hari x Luas saluran

  2,90 mm /hari

  2

  = x 40,02 m

  86,400

• 6

  3

  = 1,34 x 10 m /det

• 3

  = 1,34 x 10 l/det Debit Rembesan = Kehilangan air – (Perkolasi + Evapotranspirasi)

• 3 3 -3 -3

  3

  = 0,408 x 10 m /det – (5,15 x 10 + 1,34 x 10 m /det)

• 3

  3

  = 0,401 x 10 m /det = 0,401 l/det q

  2 = Debit rembesan per satuan panjang saluran

  3 −3 0,218 x 10 m

  =

  30 m

• 5

  2

  = 7,26 x 10 m /det d kanan = 105 cm = 1,05m d kiri = 29 cm = 0,29 m h

  1 = 26,5 cm

  = 0,265 m Perhitungan koefisien rembesan

  q2 x 2d

  Tepi kanan =

  2 h1

  −5 1,34 x 10 / det x 2 x 1,05m

  =

  

2

(0,265 m )

• 4

  = 4,0 x 10 m/detik = 34.621,51 mm/hari

  q2 x 2d

  Tepi kiri =

  2 h1

  −5 1,34 x 10 / det x 2 x 0,29 m

  =

  

2

(0,265 m )

• 4

  = 1,1 x 10 m/detik = 9.562,13 mm/hari Lampiran 12. Perhitungan efisiensi saluran No Lokasi Jarak pengukuran Efisiensi(%)

  1 Saluran 1 30 m 91,02

  2 Saluran 2 30 m 81,02 Saluran 1 Q hulu = 2,34 l/det Q hilir = 2,13 l/det

  Qhilir

  W = x 100%

  Qhulu 2,13 l/det

  = x 100%

  2,34 l/det

  W = 91,02% Saluran 2 Q hulu = 1,79 l/det Q hilir = 2,19 l/det

  Qhilir

  W = x 100%

  Qhulu 1,79 l/det

  = x 100%

  2,19 l/det

  W = 81,73%

  Lampiran 13. Dokumentasi Penelitian Pengukuran Debit Saluran 1 Pengukuran Debit Saluran 2

  Pengukuran Perkolasi Saluran 1 Pengukuran Perkolasi Saluran 2