Kajian Saluran Irigasi Tersier di Desa Namu Ukur Utara Daerah Irigasi Namu Sira-Sira Kecamatan Sei Bingei Kabupaten Langkat

  Lampiran 1. Flowchart Penelitian Mulai

  Pengukuran Debit Saluran

  Menghitung Efisiensi Saluran

  Pengukuran Kehilangan Air Pada

  Saluran Merancang Saluran

  Menghitung Kecepatan Aliran Rata-Rata dan

  Kecepatan Aliran Kritis Data

  Analisis Data Selesai Lampiran 2. Perhitungan Bulk Density, Particle Density dan Porositas

  

Volume Volume Bulk Particle

BTKO Porositas Saluran Total Partikel Density Density

  (gr) ^{3 3 3 3 (%)}

(cm ) (cm ) (gr/cm ) (gr/cm )

I (dasar) 177,73 192,325 150 0,92 2,96

  69 I (tepi) 135,10 192,325 108 0,70 2,70

  74 II (dasar) 170,76 192,325 146 0,89 2,94

  70 II (tepi) 120,95 192,325 108 0,63 2,69

  77 Dimana:

  BTKO = Berat tanah kering oven (massa tanah kering)

  1

  

2

Volume total = volume ring sample = t

  πd

  4

  1

  2

  = (3,14)(7 cm) (5 cm)

  4

  1

  3

  = �769,3 cm �

  4

  3

  = 192,325 cm Saluran 1 Kerapatan Massa (Bulk Density) Dasar Saluran

  Ms = 177,73 gr

  Ms

  B =

  d Vt 177,73

  3

  3

  = gr/cm = 0,92 gr/cm

  192,325

  Tepi Saluran Ms = 135,10 gr

  Ms

  B =

  d Vt 135,10

  3

  

3

  = gr/cm = 0,70 gr/cm

  192,325 Kerapatan Partikel (particle Density) Dalam Saluran

  Berat Tanah = 177,73 gr Volume Tanah = 150 ml Volume Air = 350 ml Volume Air Tanah = 410 ml

  berat tanah

  P d =

  (volume tanah - volume pori)

  Volume Ruang Pori = (volume air + volume tanah) - volume air tanah Volume Ruang Pori = (350 ml + 150 ml) – 410 ml

  = 90 ml

  177,73

  3 P = gr/cm d (150 - 90)

  3

  = 2,96 gr/cm Tepi Saluran

  Berat Tanah = 135,10 gr Volume Tanah = 108 ml Volume Air = 350 ml Volume Air Tanah = 400 ml

  berat tanah

  P =

  d (volume tanah - volume pori)

  Volume Ruang Pori = (volume air + volume tanah) - volume air tanah Volume Ruang Pori = (350 ml + 108 ml) – 400ml

  = 58 ml

  135,10

  108 - 58

  3

  = 2,70 gr/cm Porositas Dasar Saluran

  Bd

  Porositas = (1- ) x 100 %

  Pd 0,92

  = (1- ) x 100 %

  2,96

  = 69 % Tepi Saluran

  Bd

  Porositas = (1- ) x 100 %

  Pd 0,70

  ) x 100 % = (1-

  2,70

  = 74 % Saluran 2 Kerapatan Massa (Bulk Density) Dalam Saluran

  Ms = 170,76 gr

  Ms

  B d =

  Vt 170,76

  3

  3

  = gr/cm = 0,89 gr/cm

  192,325

  Tepi Saluran Ms = 128,95 gr

  Ms

  B =

  d Vt 128,95

  3

  3

  = gr/cm = 0,67 gr/cm

  192,325

  Kerapatan Partikel (particle Density) Dasar Saluran

  Berat Tanah = 170,76 gr

  Volume Tanah = 146 ml Volume Air = 350 ml Volume Air Tanah = 408 ml

  berat tanah

  P d =

  (volume tanah - volume pori)

  Volume Ruang Pori = (volume air + volume tanah) - volume air tanah Volume Ruang Pori = (350 ml + 146 ml) – 408ml

  = 88 ml

  170,76

  3 P = gr/cm d (146 - 88)

  3

  = 2,94 gr/cm Tepi Saluran

  Berat Tanah = 128,95 gr Volume Tanah = 105 ml Volume Air = 350 ml Volume Air Tanah = 395 ml

  berat tanah

  P d =

  (volume tanah - volume pori)

  Volume Ruang Pori = (volume air + volume tanah) - volume air tanah Volume Ruang Pori = (350 ml + 105 ml) – 395 ml

  = 60 ml

  128,95

  105 - 60

  3

  = 2,87 gr/cm Porositas Dasar Saluran

  Bd

  ) x 100 % - Porositas = (1

  Pd

  = (1 -

  Waktu (t) (det) Volume (V) (l)

  5 2 ⁄

  = 0,0138 (12,9)

  5 2 ⁄

  Q = 0,0138 h

  l/det = 7,95 l/det b. Sekat Ukur Thompson h = 12,9 cm

  1

  7,95

  =

  V t

  Q =

  Debit (Q) (l/det) 1,0 7,95 7,95

  Cara Tampung

  0,89 2,94

  Saluran I (Hulu) a.

  Hulu Hilir Saluran I 45 8,25 6,05 Saluran II 35 9,76 7,32 Saluran I 35 8,25 6,92

  Lokasi Jarak Pengukuran (m) Debit (l/det)

  Lampiran 3. Perhitungan Debit Pada Saluran 1 dan Saluran 2

  ) x 100 % = 77 %

  0,67 2,87

  ) x 100 % = (1-

  Bd Pd

  = (1-

  Tepi Saluran Porositas

  ) x 100 % = 70 %

  = 8,25 l/det = 8,25 l/det Saluran I (Hilir) a.

  Cara Tampung

  � = 8,25 + 6,05

  9,85 1,1

  =

  V t

  Q =

  Debit (Q) (l/det) 1,1 9,85 8,95

  Waktu (t) (det) Volume (V) (l)

  Cara Tampung

  Saluran II (Hulu) a.

  2 = 7,15 l/det

  = 6,05 l/det Q

  Waktu (t) (det) Volume (V) (l)

  5 2 ⁄

  = 0,0138 (11,4)

  5 2 ⁄

  Q = 0,0138 h

  l/det = 5,69 l/det b. Sekat ukur Thompson h = 11,4 cm

  7,4 1,3

  =

  V t

  Q =

  Debit (Q) (l/det) 1,3 7,4 5,69

  l/det = 8,95 l/det b. Sekat ukur Thompson h = 13,8 cm

  5 2 ⁄

  Q = 0,0138 h

  5 2 ⁄

  = 0,0138 (13,8) = 9,76 l/det

  Saluran II (hilir) a.

  Cara Tampung

  Waktu (t) Volume (V) Debit (Q) (det) (l) (l/det) 1,1 7650 6,95

  V Q = t 7650

  = l/det

  1,1

  = 6,95 l/det b. Sekat Ukur Thompson h = 12,3 cm

  5 2 ⁄

  Q = 0,0138 h

  0,0138 (12,3) = = 7,32 l/det

  5 2 ⁄

  9,76 + 7,32 Q � = = 8,54 l/det

  2 Lampiran 4. Ukuran Saluran Tersier Saluran I

  (19,5 cm + 16 cm + 15,5 cm)

  Kedalaman =

  3

  = 17 cm = 0,17 m

  (75 cm + 62 cm + 82cm)

  Lebar =

  3

  = 73 cm = 0,73 m Saluran II

  (24 cm+ 21,5 cm + 25 cm)

  Kedalaman =

  3

  = 23,5 cm = 0,235 m = 0,24 m

  (67 cm + 8 1 cm + 75 cm)

  Lebar =

  3

  = 74,3 cm = 0,74 m Lampiran 5. Perhitungan Kehilangan Air

  Jarak Kehilangan Evapotranspirasi Perkolasi Rembesan

  Lokasi Pengukuran Air No (mm/hari) (mm/hari) (mm/hari)

  (m) (mm/hari)

  1 Saluran I 45 2,38 61,7 5724,72 5788,8

  2 Saluran II 35 2,38 71,7 8064,8 8138,88 Saluran I 35 2,38 55 4438,32 4502,4

  Kehilangan air Saluran I (pada jarak pengukuran 45 m)

  Penurunan debit = (8,25 – 6,05) l/det = 2,2 l/det

• 3

  3

  = 2,2 x 10 m /det A = panjang x lebar

  = 45 m x 0,73 m

  2

  = 32,85 m _-3

  2,2 × 10

�

m3 det

  Maka jumlah air yang hilang = ₂

  32,85 m

• -3

  = 0,067 x 10 m/det = 0,067 mm/det x 24 jam/hari x 3600 det/jam = 5788,8 mm/hari

  Saluran II (pada jarak pengukuran 35 m) Penurunan debit = (9,76 – 7,32) l/det

  = 2,44 l/det

• 3

  3

  = 2,44 x 10 m /det A = panjang x lebar

  = 35 m x 0,74 m

  2

  = 25,90 m _-3

  2,44 ×10

�

m3 det

  Maka jumlah air yang hilang = ²

  25,90 m

• 3

  = 0,0942 x 10 m/det = 0,0942 mm/det x 24 jam/hari x 3600 det/jam = 8138,88 mm/hari

  Jika jarak pengukuran sama (35 m) maka kehilangan air pada saluran dua yaitu:

  5788,8 mm/hari

  Jumlah air yang hilang = � � × 35 m

  45 m

  = 4502,4 mm/hari = 1,33 l/det

  Sehingga debit hilir = (8,25 – 1,33) l/det = 6,92 l/det Evapotranspirasi

  Saluran I dan II Kc Rumput = 0,85

  o

  Tempertur (t) = 26,86 C Lama Penyinaran Matahari (P) = 3,8

  K × P(45,7 t + 813) U =

  100 K = K

  t

  80 III

  1

  P

  (16 cm – 10,5 cm) 1 hari

  =

  1

  mm/hari P

  1 = h ₁ -h ₂ t ₁ -t ₂

  = 10,5 cm P

  2

  h

  1 = 16 cm

  Silinder 1 h

  50 Rata-rata 61,7

  55 II

  × K

  I

  Ulangan Penurunan air (mm)

  Saluran I

  Perkolasi

  = 71,34 mm/bulan = 2,38 mm/hari

  7133,59 100

  =

  0,92 × 3,8 (45,7 (26,86) + 813) 100

  1,08 K = 1,08 ×0,85 = 0,92 U =

  = 0,0311 t + 0,240 = 0,0311 (26,86) + 0,240 =

  t

  K

  c

  = 5,5 cm/hari = 55 mm/hari Silinder 2 h = 18 cm

  1

  h

  2 = 10 cm h ^{1 -h 2} P = mm/hari

  2 t ^{1 -t 2} (18 cm – 10 cm)

  P

  2 = 1 hari

  P

  2 = 8 cm/hari

  = 80 mm/hari Silinder 3 h

  1 = 16 cm

  h

  2 = 11 cm h -h _{1 2} P 3 = mm/hari t -t _{1 2} �16 cm - 11 cm�

  P

  3 = 1 hari

  P

  3 = 5 cm/hari = 50 mm/hari P1 + P2 + P3

  Perkolasi Rata-rata =

  3 (55 + 80 + 50) mm/hari

  =

  3

  = 61,7 mm/hari Saluran II

  Ulangan Penurunan air (mm)

  I

  60 II

  85 III

  70 Rata-rata 71,7

  Silinder 1 h

  1 = 18 cm

  h = 12 cm

  2 h -h _{1 2} P 1 = mm/hari t -t _{1 2}

• – 12 cm) (18 cm

  P

  1 = 1 hari

  P = 6 cm/hari

  1

  = 60 mm/hari Silinder 2 h

  1 = 20,5 cm

  h

  2 = 12 cm h -h _{1 2} P 2 = mm/hari t -t _{1 2} (20,5 cm – 12 cm)

  P

  2 = 1 hari

  P

  2 = 8,5 cm/hari

  = 85 mm/hari Silinder 3 h

  1 = 18,5 cm

  h = 11,5 cm

  2 h -h _{1 2} P 3 = mm/hari t -t _{1 2} (18,5 cm – 11,5 cm)

  P =

  3 1 hari

  P

  3 = 7 cm/hari = 70 mm/hari

• P2 + P3 P1

  Perkolasi Rata-rata =

  3 (60 + 85 + 70) mm/hari

  =

  3

  = 71,7 mm/hari Rembesan

  Saluran I (jarak pengukuran 45 m) Rembesan = Kehilangan air di saluran – (Evapotranspirasi + Perkolasi)

  Saluran II (pada jarak pengukuran 35 m) E

  Q _hilir Q _hulu

  =

  e

  Jika jarak pengukurannya 35 m maka efisiensi penyaluran pada saluran I yaitu: E

  ×100 % = 75 %

  7,32 9,76

  ×100 % =

  Q _hilir Q _hulu

  =

  e

  ×100 % = 73,33 %

  = 5788,8 mm/hari – (2,38 mm/hari + 61,7 mm/hari) = 5724,72 mm/hari

  6,05 8,25

  ×100 % =

  Q _hilir Q _hulu

  =

  e

  E

  Lampiran 6. Perhitungan Efisiensi Saluran Saluran I (pada jarak pengukuran 45 m)

  = 4502,4 mm/hari – (2,38 mm/hari + 61,7 mm/hari) = 4438,32 mm/hari

  Saluran I (jarak pengukuran 35 m) Rembesan = Kehilangan air di saluran – (Evapotranspirasi + Perkolasi)

  = 8138,88 mm/hari – (2,38 mm/hari + 71,7 mm/hari) = 8064,8 mm/hari

  Saluran II (jarak pengukuran 35 m) Rembesan = Kehilangan air di saluran – (Evapotranspirasi + Perkolasi)

  ×100 %

  6,92

  E = ×100 %

  e 8,25

  = 83,88 % Lampiran 7. Perhitungan Kemiringan Pada Saluran 1 dan Saluran 2 Saluran 1

  Jarak (m) Beda tinggi (cm) 0 – 5 10 5 – 10 3,4

  10 – 15 5,5 15 – 20 2,3 20 – 25

  12 25-30

  0,352 m

  Kemiringan = × 100 %

  30 m

  = 1,17 % Saluran 2

  Panjang (m) Beda tinggi (cm) 0 – 5 16 5-10 6,1

  10-15 7,1 15-20 1 20-25 15,5

  25-30 15,6 Jumlah 61,3 0,613 m

  Kemiringan = × 100 %

  30 m

  = 2,04 % Lampiran 8. Perhitungan Kecepatan Rata-Rata Saluran I

  Q

  V =

  A _{-3 3} 7,15 x 10 m /s

  =

  (0,17 m x 0,73 m)

  = 0,057 m/s

  Saluran II

  Q

  V =

  A ₃ −3 x 10 m /s 8,54

  =

  (0,24 m x 0,74 m)

  = 0,048 m/s Lampiran 9. Perhitungan Kecepatan Kritis Saluran I

  D = 0,17 m

  0,64

  V = 0,546 D

  0,64

  = 0,546 (0,17) = 0,18 m/s

  Saluran II D = 0,24 m

  0,64

  V = 0,546 D

  0,64

  = 0,546 (0,24) = 0,22 m/s

  Lampiran 10. Perhitungan Rancangan Saluran Saluran I

  , lebar dan dalam saluran di lapangan

• Jika V = V V = Vo = 0,18 B = 0,73 m D = 0,17 m

  1 2 3 1 2 ⁄ ⁄

  V = R S

  N

  

2 3

⁄

  1 B×D

  1 2 ⁄

  0,18 = (S) � �

  0,0225 B+2D

  2 3 ⁄

  0,73×0,17

  1 2 ⁄

  0,18 = 44,44 S � �

  0,73+2(0,17)

  

2 3

⁄

  0,1241

  1 2 ⁄

  0,18 = 44,44 × S � �

  1,07

  1 2 ⁄

  0,18 = 10,67 S

  1 2 ⁄

  S = 0,017 S = 0,0003

  , kemiringan 0,03 %, dalam saluran di lapangan

• Jika V = V V = Vo = 0,18 D = 0,18 m S = 0,0003

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,18 = (S)

  � � 0,0225 B+2D

  

2 3

⁄

  0,17B

  1 2 ⁄

  0,18 = 44,44 (0,0003) � �

  B+0,34

  2 3 ⁄

  0,17B 0,18 = 0,7688 ×

  � � 0,34+B

  2 3 ⁄

  0,17B 0,23 =

  � � 0,34+B 0,17B

  0,11 = 0,34+B

  0,0374 + 0,11 B = 0,17 B 0,0374 = 0,06 B

  B = 0,62 m = 62 cm

  , kemiringan 0,02 %, lebar saluran di lapangan

• Jika V = V V = Vo = 0,18 B = 0,73 m S = 0,0002

  1 2 3 1 2 ⁄ ⁄

  V = R S

  N 2 3 ⁄

  1 B×D 1 2 ⁄

  0,18 = (S) � �

  0,0225 B+2D 2 3 ⁄ 0,73D

  1 2 ⁄

  0,18 = 44,44 (0,0002) � �

  0,73+2D 2 3 ⁄ 0,73D

  0,18 = 0,6284 × � �

  0,73+2D 2 3 ⁄ 0,73D

  0,29 = � �

  0,73+2D 0,73D

  0,16 =

  0,73+2D

  0,1168 + 0,32 D = 0,73 D 0,1168 = 0,41 D D = 0,28 m = 28 cm

  , kemiringan 0,02 %, B = 2D

• Jika V = V V = Vo = 0,18 S = 0,0002

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  

2 3

⁄

  1 B×D

  1 2 ⁄

  0,18 = (S)

  � � 0,0225 B+2D

  

2 3

⁄

  2D×D

  1 2 ⁄

  0,18 = 44,44 (0,0002) � �

  2D+2D

  2

  

2 3

⁄

  2D 0,18 = 0,6284 ×

  � �

  4D _{2 ⁄} 2 3

  2D

  0,29 = � � ₂

  4D

  2D

  0,16 =

  4D

  2

  0,64 D = 2D 0,64 = 2D D = 0,32 m = 32 cm B = 2 x 0,32 m = 0,64 m = 64 cm

  , kemiringan 0,04 %, B = 2D

• Jika V = V V = Vo = 0,18 S = 0,0004

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,18 = (S) � �

  0,0225 B+2D

  2 3 ⁄

  2D×D

  1 2 ⁄

  0,18 = 44,44 (0,0004) � �

  2D+2D _{2 ⁄} 2 3

  

2D

  0,18 = 0,89 × � �

  

4D

2 3 ⁄

  2

  2D 0,20 =

  � �

  4D ₂

  2D

  0,089 =

  4D

  2

  0,36 D = 2D

  0,36 = 2D D = 0,18 m = 18 cm B = 2 x 0,18 m = 0,36 m = 36 cm

  , kemiringan 0,04 %, lebar saluran di lapangan

• Jika V = V V = Vo = 0,18 B = 0,73 m S = 0,0004

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,18 = (S)

  � � 0,0225 B+2D

  2 3 ⁄

  0,73D

  1 2 ⁄

  0,18 = 44,44 (0,0004) � �

  0,73+2D

  2 3 ⁄

  0,73D 0,18 = 0,89 ×

  � � 0,73+2D

  

2 3

⁄

0,693D

  0,20 = � �

  0,693+2D 0,73D

  0,089 =

  0,73+2D

  0,065 + 0,178 D = 0,73 D 0,065 = 0,552 D D = 0,12 m = 12 cm Saluran II

  , lebar dan dalam saluran di lapangan

• Jika V = V V = Vo = 0,22 B = 0,74 D = 0,24

  1 2 3 1 2 ⁄ ⁄

   V = R S N 2 3

  ⁄

  1 B×D

  1 2 ⁄

  0,22 = (S) � �

  0,0225 B+2D

  2 3 ⁄

  0,74×0,24

  1 2 ⁄

  0,22 = 44,44 S � �

  0,74+2(0,24)

  1 2 2 3 ⁄ ⁄

  0,22 = 44,44 × (0,15) S

  

1 2

⁄

  0,22 = 12,4432 S

  1 2 ⁄

  S = 0,018 S = 0,0003

  , kemiringan 0,03 %, dalam saluran di lapangan

• Jika V = V V = Vo = 0,22 D = 0,24 m S = 0,0003

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,22 = (S) � �

  0,0225 B+2D

  

2 3

⁄

  0,24B

  1 2 ⁄

  0,22 = 44,44 (0,0003) � �

  B+0,48

  2 3 ⁄

  0,24B 0,22 = 0,7688 ×

  � � 0,48+B

  2 3 ⁄

  0,24B 0,29 =

  � � 0,48+B 0,24B

  0,16 = 0,48+B

  0,0768 + 0,16 B = 0,24 B 0,0768 = 0,08 B

  B = 0,96 m = 96 cm

  , kemiringan 0,02 %, lebar saluran di lapangan

• Jika V = V V = Vo = 0,22 B = 0,74 m S = 0,0002

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  1 B×D 1 2 ⁄

  2 3 ⁄

  0,22 = (S) � �

  0,0225 B+2D 2 3 ⁄

0,74D

  1 2 ⁄

  0,22 = 44,44 (0,0002) � �

  

0,74+2D

2 3 ⁄

  0,74D 0,22 = 0,6284 ×

  � � 0,74+2D

  

2 3

⁄

0,74D

  0,35 = � �

  0,74+2D 0,74D

  0,21 =

  0,74+2D

  0,1554 + 0,42 D = 0,74 D 0,1554 = 0,32 D

  D = 0,49 m = 49 cm

  , kemiringan 0,02 %, B = 2D

• Jika V = V V = Vo = 0,22 S = 0,0002

  1

  2 3 1 2 ⁄ ⁄

  V = R S N

  2 3 ⁄

  2 3 ⁄

  B = 2 x 0,42 m = 0,84 m = 84 cm

  V = Vo = 0,22 S = 0,0004 V =

  1 N R

  2 3 ⁄

  S

  1 2 ⁄

  0,22 =

  1 0,0225

  � B×D

  B+2D �

  (S)

  2

  1 2 ⁄

  0,22 = 44,44 �

  2D×D

  2D+2D �

  2 3 ⁄

  (0,0004)

  1 2 ⁄

  0,22 = 0,89 × �

  2D

  2

  4D �

  0,84 = 2D D = 0,42 m = 42 cm

  4D 0,84 D = 2D

  0,22 =

  1 2 ⁄

  1 0,0225

  � B×D

  B+2D �

  2 3 ⁄

  (S)

  1 2 ⁄

  0,22 = 44,44 �

  2D×D

  2D+2D �

  2 3 ⁄

  (0,0002)

  0,22 = 0,6284 × �

  2

  2D

  2

  4D �

  2 3 ⁄

  0,35 = �

  2D

  2

  4D �

  2 3 ⁄

  0,21 =

  2D

• Jika V = V , kemiringan 0,04 %, B = 2D

  2

  2 3 ⁄

  2D 0,25 =

  � �

  4D

  2

  2D 0,125 =

  4D

  2

  0,5 D = 2D 0,5 = 2D D = 0,25 m = 25 cm B = 2 x 0,25 m = 0,50 m = 25 cm

  , kemiringan 0,04 %, lebar saluran di lapangan

• Jika V = V V = Vo = 0,22 B = 0,74 m S = 0,0004

  1

  

2 3 1 2

⁄ ⁄

  V = R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,22 = (S) � �

  0,0225 B+2D

  2 3 ⁄

  0,74D

  1 2 ⁄

  0,22 = 44,44 (0,0004) � �

  0,74+2D

  2 3 ⁄

  0,74D 0,22 = 0,89 ×

  � � 0,74+2D

  2 3 ⁄

  0,74D =

  0,25 � �

  0,74+2D 0,74D

  0,125 = 0,74+2D

  0,0925 + 0,25 D = 0,74 D 0,0925 = 0,49 D D = 0,19 m = 19 cm Lampiran 11. Gambar Gambar 1. Saluran Sekunder Gambar 2. Bangunan Pembagi ke Saluran Tersier Gambar 3. Saluran Tersier 1

  Gambar 4. Saluran Tersier 2 Gambar 5. Pengukuran Perkolasi Gambar 6. Pengukuran Debit Saluran