Kajian Saluran Irigasi Tersier di Desa Durian Lingga Daerah Irigasi Namu Sira Sira Kecamatan Sei Bingai Kabupaten Langkat

  Lampiran 1. Flowchart Penelitian Mulai

  Pengukuran Debit Saluran

  Menghitung Efisiensi Saluran

  Pengukuran Kehilangan Air Pada

  Saluran Menghitung Kecepatan

  Rata-Rata dan Kecepatan Kritis

  Merancang Saluran Data

  Analisis Data Selesai Lampiran 2. Perhitungan bulk density, particle density dan porositas.

  Volume Bulk Particle BTKO Porositas Saluran Total Density Density (gr) ^{3 3 3 (%)} (cm ) (gr/cm ) (gr/cm )

  I (dasar) 155,22 192,325 0,81 2,72

  70 I (tepi) 142,72 192,325 0,74 2,59

  71 II(dasar) 188,24 192,325 0,98 2,69

  64 II (luar) 108,17 192,325 0,56 2,40

  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 = 155,22 gr

  Ms

  B d =

  Vt 155,22

  3

  

3

  = gr/cm = 0,81 gr/cm

  192,325

  Tepi Saluran Ms = 142,72 gr

  Ms

  B d =

  Vt 142,72

  3

  3

  = gr/cm = 0,74 gr/cm

  192,325

  Kerapatan Parikel (particle Density) Dasar Saluran

  Berat Tanah = 155,22 gr

  Volume Tanah = 125 ml Volume Air = 350 ml Volume Air Tanah = 407 ml

  berat tanah

  P d =

  ( volume tanah−volume pori)

  Volume Ruang Pori = (volume air + volume tanah)- volume air tanah Volume Ruang Pori = (350ml+125ml) – 407ml

  = 68 ml

  155,22

  P d =

  ( 125−68)

  3

  = 2,72 gr/cm Tepi Saluran

  Berat Tanah = 142,72 gr Volume Tanah = 88 ml Volume Air = 350 ml Volume Air Tanah = 405 ml

  berat tanah

  P d =

  ( volume tanah−volume pori)

  Volume Ruang Pori = (volume air + volume tanah)- volume air tanah Volume Ruang Pori = (350ml+88ml) – 405ml

  = 33 ml

  142,72

  P d =

  88− 33

  = 2,59 gr/cm Porositas Dasar Saluran

  Bd

  Porositas = (1- ) x 100%

  Pd

  = (1-

  = 0,98 gr/cm

  3 Kerapatan Parikel (particle Density)

  = 0,56 gr/cm

  3

  gr/cm

  108,17 192,325

  =

  Ms Vt

  =

  d

  Ms = 108,17 gr B

  

3

Tepi Saluran

  3

  0,81 2,72

  gr/cm

  188,24 192,325

  =

  Ms Vt

  Ms = 188,24 gr B d =

  Saluran 2 Kerapatan Massa (Bulk Density) Dasar Saluran

  ) x 100% = 71%

  0,74 2,59

  ) x 100% = (1-

  Bd Pd

  Tepi Saluran Porositas = (1-

  ) x 100% = 70%

  Dasar Saluran Berat Tanah = 188,24 gr Volume Tanah = 150 ml Volume Air = 350 ml Volume Air Tanah = 420 ml berat tanah

  P d =

  ( volume tanah−volume pori)

  Volume Ruang Pori = (volume air + volume tanah)- volume air tanah Volume Ruang Pori = (350ml+150ml) – 420ml

  = 80 ml

  188,24

  P d =

  ( 150−80)

  3

  = 2,69 gr/cm Tepi Saluran

  Berat Tanah = 108,17 gr Volume Tanah = 93,3 m 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 = (350ml+93,3ml) – 395ml

  = 48,3 ml

  108,17

  P d =

  93,3− 48,3

  = 2,40 gr/cm Porositas Dasar Saluran

  Bd

  Porositas = (1- ) x 100%

  Pd 0,98

  = (1- ) x 100%

  2,69

  = 64% Tepi Saluran

  Bd

  Porositas = (1- ) x 100%

  Pd 0,56

  = (1- ) x 100%

  2,40

  = 77% Lampiran 3. Perhitungan debit pada saluran satu dan dua ₃ Debit (m /det)

  Saluran Hulu Hilir Tersier I 7,62 4,81 Tersier II 11,24 7,47

  Saluran 1 (hulu) a.

  Cara Tampung

  Waktu (t) Volume (v) Debit (Q) (det) (l) (l/det) 1 7,3 7,3 b.

  Sekat Ukur Thompson h = 12,5 cm

  ⁄ 5 2

  = 0,0138 ℎ

  5 2 ⁄

  = 0,0138 (12,5) = 7,62 l/det

  Saluran 1 (hilir) a.

  Cara Tampung

  Waktu (t) Volume (v) Debit (Q) (det) (l) (l/det) 1,5 6,63 4,42 b.

  Sekat ukur Thompson h = 10,4 cm

  Q =0,0138 h

  5 2 ⁄

  = 0,0138 (12,4)

  5 2 ⁄

  Sekat Ukur Thompson h = 12,4 cm Q =0,0138 h

  Debit (Q) (l/det) 1 7,1 7,1 b.

  Waktu (t) (det) Volume (v) (l)

  Cara Tampung

  = 11,24 l/det Saluran 2 (hilir) a.

  5 2 ⁄

  =0,0138 (14,6)

  Sekat ukur Thompson h = 14,6 cm Q =0,0138 h

  5 2 ⁄

  Debit (Q) (l/det) 1 10,63 10,63 b.

  Waktu (t) (det) Volume (v) (l)

  Cara Tampung

  Saluran 2 (hulu) a.

  2 = 6,21 l/det

  7,62+4,81

  = 4,81 l/det Q � =

  5 2 ⁄

  = 0,0138 (10,4)

  5 2 ⁄

  = 7,47 l/det 11,24+7,47

  � = Q = 9,36 l/det

  2 Lampiran 4. Ukuran saluran tersier Saluran 1

  ( 30,5cm+29,9cm+29,6cm)

  Kedalaman =

  3

  = 30 cm = 0,3 m

  ( 72cm+67,5cm+70,6cm)

  Lebar =

  3

  = 70,03 cm = 0,7 m Saluran 2

  ( 26,1cm+25,4cm+23,7cm)

  Kedalaman =

  3

  = 25,06cm = 0,25 m

  ( 82,6cm+78,5cm+79,1cm)

  Lebar =

  3

  = 80,07 cm = 0,8 m Lampiran 5. Perhitungan kehilangan air dari evapotranspirasi, perkolasi dan rembesan.

  Evapotranspirasi Perkolasi Rembesan Kehilangan Air No Lokasi (mm/hari) (mm/hari) (mm/hari) (mm/hari)

  1 Saluran 1 2,44 28 2907,16 2937,6

  Kehilangan air Saluran 1 Penurunan debit = (7,62-4,81) l/det

  = 2,81 l/det

• 3

  3

  = 2,81 x 10 m /det A = panjang x lebar

  = 120 m x 0,7 m

  2

  = 84 m

• _-3

  

3

2,81×10 m

�

det

  Maka jumlah air yang hilang = ₂

  84 m

• -3

  = 0,034 x 10 m/det = 2.937,6 mm/hari Saluran 2 Penurunan debit = (11,24-7,47) l/det

  = 3,77 l/det

• 3

  3

  = 3,77 x 10 m /det A = panjang x lebar

  = 90 m x 0,8 m

  2

  = 72 m _{-3 3}

  3,77×10 m

�

det

  Maka jumlah air yang hilang = ₂

  72 m

• 3

  = 0,05 x 10 m/det = 4320 mm/hari Jika jarak pengukuran sama (90 m) maka kehilangan air pada saluran satu yaitu:

  2937,6 mm/hari

  Jumlah air yang hilang = � � ×90 m

  120 m

  =2203,2 mm/hari = 1,607 l/det Sehingga debit hilir = 7,62 – 1,607 l/det = 6,01 l/det

  Evapotranspirasi

  Saluran 1 Kc Rumput = 0,85

  o

  Tempertur (t) = 26,9 C Lama Penyinaran Matahari (P) = 3,9

  K×P(45,7 t+813) U=

  100 K=K ×K

  t c

  K =0,0311 t+0,240

  t

  = 0,0311 (26,9) + 0,240 = 1,077

  = 1,077 × 0,85 = 0,92

  0,92×3,9(45,7 (26,9)+813) U =

  100

  7327,88

  =

  100

  = 73,28 mm/bulan = 2,44 mm/hari Saluran 2 Kc Rumput = 0,85 Kc Kelapa Sawit = 1,2

  o

  Tempertur (t) = 26,9 C Lama Penyinaran Matahari (P) = 3,9

  K×P(45,7 t+813) U=

  100 K=K ×K

  t c

  K =0,0311 t+0,240

  t

  = 0,0311 (26,9) + 0,240 = 1,077

  = 1,077 × 1,03

  = 1,11 1,11×3,9(45,7 (26,9)+813)

  U = 100

  8842,25

  =

  100

  = 88,41 mm/bulan = 2,95 mm/hari

  Perkolasi

  Saluran 1

  Ulangan Penurunan air (mm)

  I

  30 II

  38 III

  16 Rata-rata

  28 Silinder 1

  h

  1 = 30 cm

  h = 27 cm

  2 h 1 −h

  2 P 1 = mm/hari t 1 −t

  2 ( 30cm−27cm)

  P =

  1 1 hari

  P

  1 = 3 cm/hari

  = 30 mm/hari Silinder 2 h = 33 cm

  1

  h

  2 = 29,2 cm h −h

  1

  2 t 1 −t

  2 ( 33 cm−29,2cm)

  P

  2 = 1 hari

  P

  2 = 3,8 cm/hari

  = 38 mm/hari Silinder 3 h

  1 = 29cm

  h

  2 = 27,4 cm h 1 −h

  2 P 3 = mm/hari t 1 −t

  2 ( 29 cm−27,4 cm)

  P

  3 = 1 hari

  P

  3 = 1,6 cm/hari = 16 mm/hari

  Perkolasi Rata-rata =

  3 ( 30+38+16)mm/hari

  =

  3

  = 28 mm/hari Saluran 2

  Ulangan Penurunan air (mm)

  I

  19 II

  32 III

  28 Rata-rata 26,3

  Silinder 1 h

  1 = 15,2 cm

  h = 13,3 cm

  2 h 1 −h

  2 P 1 = mm/hari t 1 −t

  2 ( 15,2 cm−13,3 cm)

  P =

  1 1 hari

  P = 1,9 cm/hari

  1

  = 19 mm/hari Silinder 2 h = 23,8 cm

  1

  h

  2 = 20,6 cm h 1 −h

  2 P 2 = mm/hari t 1 −t

  2 ( 23,8 cm−20,6 cm)

  P =

  2 1 hari

  P = 3,2 cm/hari

  2

  = 32 mm/hari Silinder 3 h = 23,2 cm

  1

  h

  2 = 20,4 cm h −h

  1

  2 P 3 = mm/hari t −t

  1

  2 ( 23,2 cm−20,4 cm)

  P

  3 = 1 hari

  P

  3 = 2,8 cm/hari = 28 mm/hari

  Perkolasi Rata-rata =

  3 ( 19+28+32)mm/hari

  =

  3

  = 26,3 mm/hari

  Rembesan

  Saluran 1 Rembesan = Kehilangan Air – (Evapotranspirasi+Perkolasi)

  = 2.937,6 mm/hari – (2,44 mm/hari + 28 mm/hari) = 2907,16 mm/hari

  Saluran 2 Rembesan = Kehilangan Air – (Evapotranspirasi+Perkolasi)

  = 4320 mm/hari – (2,95 mm/hari + 114,7 mm/hari) = 4202,35 mm/hari Lampiran 6. Perhitungan efisiensi saluran Saluran 1

  Q

  hilir

  W = ×100% Q

  hulu 4,81

  = × 100%

  7,62

  = 63,12% Jika jaraknya 90 m maka,

  Q

  hilir

  W = ×100% Q

  hulu

  6,01 W= ×100%

  7,62 =78,87% Saluran 2

  Q

  hilir

  W = ×100% Q

  hulu 7,47

  = × 100%

  11,24

  = 66,46% Lampiran 7. Perhitungan kemiringan pada saluran 1 dan 2 Saluran 1

  Jarak (m) Beda tinggi (cm) 0-5 6,5 5-10 2,8 10-15 14,9

  15-20 16,1 20-25 6,5 25-30 10,9 Jumlah 57,7

  0,577 m

  Kemiringan = ×100%

  30 m

  = 1,92 % Saluran 2

  Panjang (m) Beda tinggi (cm) 0-5 4,8 5-10 11,3 10-15 2,1

  15-20

  4 20-25 11 25-30 7,4

  Jumlah 40,6 0,406 m

  Kemiringan = ×100%

  30 m

  = 1,35 % Lampiran 8. Perhitungan kecepatan rata-rata (V) Saluran 1

  Q

  V =

  A −3

  3 6,21 / x 10 m s

  =

  ( 0,7m x 0,3m)

  = 0,03 m/s Saluran 2

  Q

  V =

  A −3

  3 9,36 / x 10 m s

  =

  ( 0,8m x 0,25m)

  = 0,047 m/s = 0,05 m/s Lampiran 9. Perhitungan kecepatan kritis (V ) Saluran 1 D = 0,3 m

  0,64

  V = 0,546 D

  0,64

  = 0,546 (0,3) = 0,25 m/s Saluran 2 D = 0,25 m

  0,64

  V = 0,546 D

  0,64

  = 0,546 (0,25) = 0,22 m/s

  Lampiran 10. Rancangan Saluran Saluran 1

  , lebar dan dalam saluran dilapangan

• Jika V = V V = Vo = 0,25 B = 0,7 D = 0,3

  1 2 3 1 2 ⁄ ⁄

  V= R S

  N 2 3 ⁄

  1 B×D

  1 2 ⁄

  0,25= (S) � �

  0,0225 B+2D

  2 3 ⁄

  0,7×0,3

  1 2 ⁄

  0,25=44,444 S � �

  0,7+2(0,3)

  2 3 ⁄

  0,21

  1 2 ⁄

  0,25=44,444 × S � �

  1,3

  1 2 ⁄

  0,25= 13,1 S

  1 2 ⁄

  S =0,019 S =0,00036

  , kemiringan 0,04 %, lebar saluran dilapangan

• Jika V = V V = Vo = 0,25 B = 0,7

  S = 0,0004

  1

  2 3 1 2 ⁄ ⁄

  V= R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,25= (S) � �

  0,0225 B+2D

  2 3 ⁄

  0,7D

  

1 2

⁄

  0,25=44,444 (0,0004) � �

  0,7+2D

  2 3 ⁄

  0,7D 0,25=0,889 ×

  � � 0,7+2D

  2 3 ⁄

  0,7D 0,281=

  � � 0,7+2D

  0,7D 0,149=

  0,7+2D 0,1043 + 0,298 D = 0,7 D 0,1043 = 0,402 D D = 0,259 m = 25,9 cm

  , kemiringan 0,04 %, B = 2D

• Jika V = V V = Vo = 0,25 S = 0,0004

  1

  2 3 1 2 ⁄ ⁄

  V= R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,25= (S)

  � � 0,0225 B+2D

  2 3 ⁄

  2D×D

  

1 2

⁄

  0,25=44,444 (0,0004) � �

  2D+2D

  2

  2 3 ⁄

  2D 0,25=0,89 ×

  � �

  4D

  2

  2 3 ⁄

  2D 0,281=

  � �

  4D

  2

  2D 0,149=

  4D

  2

  0,596 D = 2D 0,596 = 2D D = 0,298 m = 29,8 cm

  B = 2 x 0,298 m = 0,596 m = 59,6 cm , kemiringan 0,032 %, lebar saluran dilapangan

• Jika V = V V = Vo = 0,25 B = 0,7 S = 0,00032

  1

  2 3 1 2 ⁄ ⁄

  V= R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,25= (S) � �

  0,0225 B+2D

  2 3 ⁄

  0,7D

  1 2 ⁄

  0,25=44,444 (0,00032) � �

  0,7+2D

  2 3 ⁄

  0,7D 0,25=0,796 ×

  � � 0,7+2D

  2 3 ⁄

  0,7D 0,314=

  � � 0,7+2D

  0,7D 0,176=

  0,7+2D 0,1232 + 0,352 D = 0,7 D 0,1232 = 0,348 D D = 0,354 m = 35,4 cm

  , kemiringan 0,032 %, B = 2D

• Jika V = V V = Vo = 0,25 S = 0,00032

  1

  2 3 1 2 ⁄ ⁄

  V= R S N

  2 3 ⁄

  1 B×D

  1 2 ⁄

  0,25= (S)

  � � 0,0225 B+2D

  2 3 ⁄

  2D×D

  1 2 ⁄

  0,25=44,444 (0,00032) � �

  2D+2D

  2

  2 3 ⁄

  2D 0,25=0,796 ×

  � �

  4D

  2

  2 3 ⁄

  2D 0,314=

  � �

  4D

  2

  2D 0,176=

  4D

  2

  0,704 D = 2D 0,704 = 2D

  D = 0,352 m = 35,2 cm B = 2 x 0,352 m = 0,704 m = 70,4 cm

  Saluran 2 , lebar dan dalam saluran dilapangan

• Jika V = V V = Vo = 0,22 B = 0,8 D = 0,25

  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 ⁄

  0,8×0,25

  1 2 ⁄

  0,22=44,444 S � �

  0,8+2(0,25)

  2 3 ⁄

  0,2

  0,22=44,444 × S

  1 2 ⁄

  � � 1,3

  1 2 ⁄

  0,22= 12,488 S

  1 2 ⁄

  S =0,018 S =0,00032

  , kemiringan 0,03 %, lebar saluran dilapangan

• Jika V = V V = Vo = 0,22 B = 0,8 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 ⁄

  0,8D

  

1 2

⁄

  0,22=44,444 (0,0003) � �

  0,8+2D

  2 3 ⁄

  0,8D 0,22=0,756 ×

  � � 0,8+2D

  2 3 ⁄

  0,8D 0,291=

  � � 0,8+2D

  0,8D 0,157=

  0,8+2D 0,126 + 0,314 D = 0,8 D 0,126 = 0,486 D D = 0,259 m = 25,9 cm

  , kemiringan 0,03 %, B = 2D

• Jika V = V V = Vo = 0,22 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 ⁄

  2D×D

  

1 2

⁄

  0,22=44,444 (0,0003) � �

  2D+2D

  2

  2 3 ⁄

  2D 0,22=0,756 ×

  � �

  4D

• Jika V = V , kemiringan 0,02 %, lebar saluran dilapangan

  1 2

⁄

  (S)

  1 2 ⁄

  0,22=44,444 �

  0,8D 0,8+2D

  �

  2 3 ⁄

  (0,0002)

  0,22=0,622 × �

  B+2D �

  0,8D 0,8+2D

  �

  2 3 ⁄

  0,354 = �

  0,8D 0,8+2D

  �

  2 3 ⁄

  0,211= 0,8D

  2 3 ⁄

  � B×D

  0,291= �

  2

  2D

  2

  4D �

  2 3 ⁄

  0,157=

  2D

  2

  4D 0,628 D = 2D

  0,628 = 2D D = 0,314 m = 31,4 cm

  1 0,0225

  B = 2 x 0,314 m = 0,628 m = 62,8 cm

  V=Vo = 0,22 B = 0,8 S = 0,0002

  V=

  1 N R

  2 3 ⁄

  S

  1 2 ⁄

  0,22=

  0,8+2D 0,169 + 0,422 D = 0,8 D

  0,169 = 0,378 D D = 0,447 m = 44,7 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 ⁄

  1 B×D

  1 2 ⁄

  0,22= (S) � �

  0,0225 B+2D

  2 3 ⁄

  2D×D

  

1 2

⁄

  0,22=44,444 (0,0002) � �

  2D+2D

  2

  2 3 ⁄

  2D 0,22=0,622 ×

  � �

  4D

  2

  2 3 ⁄

  2D 0,354=

  � �

  4D

  2

  2D 0,211=

  4D

  2

  0,844 D = 2D 0,844 = 2D D = 0,422 m = 42,2 cm B = 2 x 0,422 m = 0,844 m = 84,4 cm Lampiran 11. Gambar Gambar 1. Saluran Sekunder Gambar 2. Petak Tersier Gambar 3. Saluran Tersier 1

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