Laporan Tugas Akhir Mata Kuliah Irigasi

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DISUSUN SEBAGAI PERSYARATAN MENGIKUTI UJIAN AKHIR SEMESTER DAN

MEMENUHI TUGAS MATA KULIAH IRIGASI

DISUSUN OLEH :

MUHAMMAD MISBAHUL NUR MUTAQIN C.111.19.0189

DOSEN PENGAMPU : Ir. MOCH. SEDIONO , BIE,ME DOSEN PEMBIMBING : KUKUH WISNUAJI W., ST.MT.

FAKULTAS TEKNIK JURUSAN TEKNIK SIPIL UNIVERSITAS SEMARANG

TAHUN 2020

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LEMBAR ASISTENSI

TUGAS : IRIGASI NAMA : M MISBAHUL N M.

ASSISTENSI : KUKUH WISNUAJI W., ST.MT. NIM : C.111.19.0189

NO TANGGAL KETERANGAN PARAF

HIMPUNAN MAHASISWA JURUSAN TEKNIK SIPIL

UNIVERSITAS SEMARANG

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1 UUD INDRASENA 5.9 1.1 1 2.1 1.9 3 0.65 100 125 150 175 200 225 250 2 BHAGAZT TRI JUNAIDI 5.8 1.2 1.2 2.2 1.8 3.1 0.65 250 225 200 175 150 125 100 3 ANANDA BAGUS WIRAYUDHA 5.7 1.3 1.3 2.3 1.7 3.2 0.65 125 150 175 200 225 250 100 4 TRIARTO FARIZ AROZY 5.6 1.4 1.4 2.4 1.6 3.3 0.65 150 100 150 175 200 225 125 5 ABBAS ABDURRAHMAN 5.5 1.5 1.5 2.5 1.5 3.4 0.65 175 125 125 150 175 200 150 6 BONDAN ARDI WIBOWO 5.4 1.6 1.6 2.6 1.6 3.5 0.65 200 150 100 125 150 175 175 7 MUSTAFA DANI BASKARA 5.3 1.7 1.7 2.7 1.7 3.4 0.65 225 175 125 100 125 150 200 8 ANDHIKA LUTFI ALVIN

DAMARA

5.2 1.8 1.8 2.8 1.8 3.3 0.65 250 200 150 125 100 125 225 9 BAGAS PRASETYA 5.1 1.9 1.9 2.9 1.9 3.2 0.65 125 225 175 150 125 100 250 10 CHANTARA PRAMUDYA

PUTRA

5 2 2 3 2 3.1 0.65 150 250 200 175 150 125 225 11 RINI SRIWULANDARI 4.9 2.1 2.1 2.9 1.9 3 0.65 175 100 225 200 175 150 200 12 GUNAWAN WIBISONO 4.8 2.2 2.2 2.8 1.8 3.1 0.65 200 125 250 225 200 175 175 13 ABDULLAH HAFIDH

SHOLIHIN

4.7 2.3 2.3 2.7 1.7 3.2 0.65 225 150 225 250 225 200 150 14 VICKY ARDIANSYAH 4.6 2.4 2.4 2.6 1.6 3.3 0.65 250 175 200 225 250 225 150 15 WAFIQ SYAHRUL FA'IZ 4.5 2.5 2.5 2.5 1.5 3.4 0.65 125 200 175 200 225 250 125 16 WANDA DWI NUR CAHYANI 4.4 2.6 2.6 2.4 1.4 3.5 0.65 150 225 150 175 200 225 100 17 ARISNA ALDI RIYANTO 4.3 2.7 2.7 2.3 1.5 3.4 0.65 175 250 125 150 175 200 125 18 PRAMUDYA WISNU

WARDANA

4.2 2.8 2.8 2.2 1.6 3.3 0.65 200 100 100 125 150 175 150 19 FATHINIYA SABDOWATI 4.1 2.9 2.9 2.1 1.7 3.2 0.65 225 125 125 100 125 150 175 20 MUHAMMAD MISBAHUL

NUR MUTAQIN

4 3 3 2.9 1.8 3.1 0.65 250 150 150 125 100 100 200

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T E K N I K S I P I L U N I V E R S I T A S S E M A R A N G | TUGAS IRIGASI 2021

1

22 WIDYAROSSA NUR FAJRINA 5.9 1.2 1.2 2.2 1.8 3 0.65 250 200 200 175 150 125 100 23 ABDUL HAMID 5.6 1.3 1.3 2.3 1.7 2.2 0.65 125 150 225 200 225 250 100 24 CHAIRISA PRAHASTI

ISTIFARANI

5.7 1.4 1.4 2.4 1.6 3 0.65 150 100 150 125 200 225 125 25 ISMAIL PANJI WIRANATA 5.4 1.5 1.5 2.5 1.5 4 0.65 175 125 125 150 150 200 150 26 MISBAKHUL KHOIRU

ROHMAN

5.5 1.6 1.6 2.6 1.6 2.5 0.65 200 150 100 125 150 225 175 27 MARSHANDA ALIFIA WARDA 5.2 1.7 1.7 2.7 1.7 3 0.65 225 175 125 100 125 150 250 28 DINGINE ISNAINI MAHMUDA 5.3 1.8 1.8 2.8 1.8 3 0.65 250 200 150 125 100 250 225 29 BUNGA MAY RIYADI 5.0 1.9 1.9 2.9 1.9 2.2 0.65 125 225 175 150 250 100 250 30 THARIESSA SEKAR AYU

WIRADIKA

5.1 2 2 3 2 2.1 0.65 150 250 200 150 150 125 225 31 FEBRI NUR MAJID 4.8 2.1 2.1 2.9 1.9 2.5 0.65 175 100 250 200 175 150 200 32 JEIHAN SALAHUDIN

MOHAMMAD

4.9 2.2 2.2 2.8 1.8 3 0.65 200 250 250 225 200 175 175 33 BINTAJAMARIA NAJMI

FATTAYA

4.6 2.3 2.3 2.7 1.7 2.2 0.65 125 150 225 250 225 200 150 34 ULUPY NOVA MALTA 4.7 2.4 2.4 2.6 1.6 2.3 0.65 250 125 200 225 250 225 150 35 JAZAAL AUFA 4.4 2.5 2.5 2.5 1.5 2.4 0.65 125 200 125 200 225 250 125 36 ADI MA'RUF NUGROHO 4.5 2.6 2.6 2.4 1.4 2.5 0.65 150 225 150 125 200 225 100 37 TEGAR ANDIKA SAPUTRO 4.2 2.7 2.7 2.3 1.5 2.4 0.65 175 250 125 150 125 200 125 38 RYAN CHANDRA RAMADHAN 4.3 2.8 2.8 2.2 1.6 2.3 0.65 200 100 100 125 150 125 150 39 MUHAMMAD ARIF

ANDRIANTO

4.0 2.9 2.9 2.1 1.7 2.2 0.65 225 125 125 100 125 150 125 40 MUHAMAD SARIFUDIN 4.1 3 3 2.9 1.8 2.1 0.65 250 150 150 125 100 150 200 41 MUHAMMAD IQBAL 5.7 1.1 1 2.1 1.9 3 0.65 120 125 150 175 200 225 250

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2

42 NANDHIKA JIHAN SAGARA 5.6 1.2 1.2 2.2 1.8 3.1 0.65 250 220 200 175 150 125 100 43 KANDIDA APRILIA REYAAN 5.5 1.3 1.3 2.3 1.7 3.2 0.65 125 150 250 200 225 250 100

SOAL : TUGAS BESAR IIGASI

Nama : ... NIM : ...

DIKETAHUI : (lihat dari data soal) contoh sbb :

Eto (mm/hr) Kc (mm/hr)

P (mm/hr)

Re (padi) (mm/hr)

Re (palawija) (mm/hr)

WLR (mm/hr)

E total (mm/hr)

Koefisien Tanaman (C)

5.6 1.2 2 2.1 1.5 3 0.65 1.00

A (ha)

B (ha)

C (ha)

D (ha)

E (ha)

F (ha) G

(ha)

Nilai K

100 125 150 200 175 180 210 70

Efisiensi Irigasi Kode Presentasi

Saluran Tersier (e T) 80 % 0.80

Saluran Sekunder (e S) 85 % 0.85

Saluran Primer (e P) 90 % 0.90

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3

T.1

T.2

S.1 T.3

P.1 P.2 S.2

S.4

T.4 T.7

S.3 T.6

T.5

SOAL : Rencanakan jaringan irigasi sesuai denah diatas meliputi perhitungan tentang :

1. Kebutuhan air tanaman.

2. Demensi saluran irigasi . Saluran tersier (T 1 – T 7) Saluran sekunder (S1 – S4)

Saluran primer (P1 – P2)

B

G

F

E

D C

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4

3. Demensi saluran pembuang.

Saluran pembuang tersier (T1 – T7) Saluran pembuang sekunder (S1 – S4) Saluran pembuang primer (P1 – P2)

4. Gambar pintu Romyn.

5. Pintu pengambilan (Romyn) Demensi pintu Romyn (T1 – T7) Demensi pintu Romyn (S1 – S4) Demensi pintu Romyn (P1 – P2)

Note :

- Asistensi melalui email : [email protected]

- di email dilengkapi Subjek : Tugas Irigasi ( Nama: ,Nim: ,Kelas: ) - setiap asistensi memakai fitur balas pesan email ( jangan membuat pesan email baru )

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4 3 3 2,9 1,8 3,1 0,65 1,00

A ( ha )

B ( ha )

C ( ha )

D ( ha )

E ( ha )

F ( ha )

G ( ha )

Nilai K Strickler

250 150 150 125 100 100 200 70,00

Efisiensi Irigasi Kode Persentase Saluran Tersier ( e T ) 80 % 0,80 Saluran Sekunder ( e S ) 85 % 0,85 Saluran Primer ( e P ) 90 % 0,90

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1 Peta Skema Saluran Air

T.1

T.2

S.1 T.3

P.1 P.2 S.2

S.4 T.4

T.7

S.3

T.5 T.6

A

250 ha

B 150 ha

G 200 ha

F 100 ha

E 100 ha

D 125 ha

C 150 ha

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2

Rencanakan jaringan Irigasi seperti gambar Lay Out serta hitunglah :

1. Kebutuhan Air Tanaman 2. Dimensi Saluran Pembawa

Dimensi Saluran Tersier ( T.1 – T.7 ) Dimensi Saluran Sekunder ( S.1 – S.4 ) Dimensi Saluran Primer ( P.1 – P.2 ) 3. Dimensi Saluran Pembuang

Dimensi Saluran Tersier ( T.1 – T.7 ) Dimensi Saluran Sekunder ( S.1 – S.4 ) Dimensi Saluran Primer ( P.1 – P.2 ) 4. Gambar Pintu Romyn

5. Pintu Pengambilan ( Romyn ) Dimensi Saluran Tersier ( T.1 – T.7 ) Dimensi Saluran Sekunder ( S.1 – S.4 ) Dimensi Saluran Primer ( P.1 – P.2 )

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3 1. Kebutuhan Air Tanaman

1. Etc = Kc x Eto

= 3 X 4

= 12 mm/hari

2. Menghitung NFR (Need Field Requirement)

NFR = Etc + P - Re (Padi) + WLR

= 12 + 3 - 2,9 + 3,1

= 15,2 mm/hari

= ^15,2

8,64

⁼ 1,759 liter/detik 3. Menghitung Kebutuhan Air Padi ( IR Padi )

IR (Padi) = ^𝑁𝐹𝑅 𝑒_{𝑡𝑜𝑡𝑎𝑙}

= ^15,2

= 0,6523,385 mm/hari

= ^23,385 8,64

= 2,707 liter/detik

4. Menghitung Kebutuhan Air Palawija ( IR Palawija ) IR Palawija = ( 𝐸𝑡𝑐−𝑅𝑒(𝑃𝑎𝑙))

𝑒_{𝑡𝑜𝑡𝑎𝑙}

= ( 12 − 1,8 )

⁼ 15,692 mm/hari 0,65

= ^15,692 8,64

= 1,816 liter/detik

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4 A. SALURAN TERSIER

A.1

Saluran Tersier (T.1) 1. Menghitung Debit Air ( Q )

Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐴 𝑒 𝑇

= 1 𝑥 1,759 𝑥 250 0,80

= 549,769 liter/detik

= 0,550 𝑚³/𝑑𝑒𝑡𝑖𝑘 2. Menghitung Kecepatan Air ( V )

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,550 )^0,182

= 0,377 m/detik 3. Luas Penampang Basah ( F )

F = ^𝑄

𝑉

= ^0,550

0,377

= 1,460 𝑚² 4. Dimensi Lebar (b) dan Tinggi (h)

Berdasarkan table hubungan Q dan m apabila Q diantara 0 s/d 0.5 𝑚³/𝑑𝑡 maka m = b/h adalah 1 atau 1:1 sehingga b = h dan m = 1

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 1,460

ℎ² = ^1,460

2

h = 0,854 m

jadi b dan h = 0,854 m 5. Mencari Tinggi Jagaan ( W )

W = √0.5ℎ

= √0.5 𝑥 0,854

= 0,654 m

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5

6. Mencari Panjang Keliling Basah ( O )

O = b + 2h√1 + 𝑚²

= 0,854 + 2 x 0,854√1 + 1²

= 3,271 m 7. Mencari Radius Basah Efektif ( R )

R = ^𝐹

𝑂

= ^1,460

3,271

= 0,446 m

8. Menentukan Kemiringan Saluran ( I ) → dengan Rumus Strickler

V = K x 𝑅²³ x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,352

70 x 0,383 2 3

√𝑖 = 0,009214 𝑖 = 0,000085

= 0,0085 %

b

h

W Q = 0,550 m³ / dt

V = 0,377 m³ F = 1,460 m² b & h = 0,854 m O = 3,271 m R = 0,446 m I = 0,000085

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6

Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐵 𝑒 𝑇

= 1 𝑥 1,759 𝑥 150 0,80

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,330 )^0,182

= 0,343 m/detik

3. Luas Penampang Basah ( F )

F = ^𝑄

𝑉

= ^0,330

0,343

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 0,961

ℎ² = ^0,961

2

h = 0,693 m

W = √0.5ℎ

= √0.5 𝑥 0,693

= 0,589 m

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7

O = b + 2h√1 + 𝑚²

= 0,693 + 2 x 0,693√1 + 1²

R = ^𝐹

𝑂

= ^0,961

2,654

= 0,362 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,323

70 x 0,316 2 3

√𝑖 = 0,009651 𝑖 = 0,000093

= 0,0093 %

b

h

W Q = 0,330 m³ / dt

V = 0,343 m³ F = 0,961 m² b & h = 0,693 m O = 2,654 m R = 0,362 m I = 0,000093

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8

Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐶 𝑒 𝑇

= 1 𝑥 1,759 𝑥 150 0,80

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,330 )^0,182

F = ^𝑄

𝑉

= ^0,330

0,343

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 0,961

ℎ² = ^0,961

2

h = 0,693 m

W = √0.5ℎ

= √0.5 𝑥 0,693

= 0,589 m 6. Mencari Panjang Keliling Basah ( O )

O = b + 2h√1 + 𝑚²

= 0,693 + 2 x 0,693√1 + 1²

= 2,654 m

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9

7. Mencari Radius Basah Efektif ( R )

R = ^𝐹

𝑂

= ^0,961

2,654

= 0,362 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,367

70 x 0,419 2 3

√𝑖 = 0,009651 𝑖 = 0,000093

= 0,0093 %

b

h

W Q = 0,330 m³ / dt

V = 0,343 m³ F = 0,961 m² b & h = 0,693 m O = 2,654 m R = 0,362 m I = 0,000093

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Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐷 𝑒 𝑇

= 1 𝑥 1,759 𝑥 125 0,80

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,275 )^0,182

F = ^𝑄

𝑉

= ^0,275

0,332

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 0,828

ℎ² = ^0,828

2

h = 0,643 m

W = √0.5ℎ

= √0.5 𝑥 0,643

O = b + 2h√1 + 𝑚²

= 0,643 + 2 x 0,643√1 + 1²

= 2,463 m

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11

R = ^𝐹

𝑂

= ^0,828

2,463

= 0,336 m

V = K x 𝑅²³ x √𝑖

√𝑖 = ^𝑉

K x 𝑅²³

√𝑖 = ^0,359

70 x 0,401 2 3

√𝑖 = 0,009812 𝑖 = 0,000096

= 0,0096 %

b

h

W Q = 0,275 m³ / dt

V = 0,332 m³ F = 0,828 m² b & h = 0,643 m O = 2,463 m R = 0,336 m I = 0,000096

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12

Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐹 𝑒 𝑇

= 1 𝑥 1,759 𝑥 100 0,80

= 0,220 𝑚³/𝑑𝑒𝑡𝑖𝑘

2. Menghitung Kecepatan Air ( V )

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,220 )^0,182

F = ^𝑄

𝑉

= ^0,220

0,319

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 0,690

ℎ² = ^0,690

2

h = 0,587 m

W = √0.5ℎ

= √0.5 𝑥 0,587

= 0,542 m

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13

O = b + 2h√1 + 𝑚²

= 0,587 + 2 x 0,587√1 + 1²

R = ^𝐹

𝑂

= ^0,690

2,248

= 0,307 m

V = K x 𝑅²³ x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,352

70 x 0,383 2 3

√𝑖 = 0,010012 𝑖 = 0,000100

= 0,0100 %

b

h

W Q = 0,220 m³ / dt

V = 0,319 m³ F = 0,690 m² b & h = 0,587 m O = 2,248 m R = 0,307 m I = 0,000100

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14

Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐸 𝑒 𝑇

= 1 𝑥 1,759 𝑥 100 0,80

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,220 )^0,182

F = ^𝑄

𝑉

= ^0,220

0,319

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 0,690

ℎ² = ^0,690

2

h = 0,587 m

W = √0.5ℎ

= √0.5 𝑥 0,587

O = b + 2h√1 + 𝑚²

= 0,587 + 2 x 0,587√1 + 1²

= 2,248 m

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15

R = ^𝐹

𝑂

= ^0,690

2,248

= 0,307 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,343

70 x 0,363 2 3

√𝑖 = 0,010012 𝑖 = 0,000100

= 0,0100 %

b

h

W Q = 0,220 m³ / dt

V = 0,319 m³ F = 0,690 m² b & h = 0,587 m O = 2,248 m R = 0,307 m I = 0,000100

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16

Q = 𝐶 𝑥 𝑁𝐹𝑅 𝑥 𝐺 𝑒 𝑇

= 1 𝑥 1,759 𝑥 200 0,80

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,440 )^0,182

F = ^𝑄

𝑉

= ^0,440

0,362

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 1,216

ℎ² = ^1,216

2

h = 0,780 m

W = √0.5ℎ

= √0.5 𝑥 0,780

O = b + 2h√1 + 𝑚²

= 0,780 + 2 x 0,780√1 + 1²

= 2,985 m

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17

R = ^𝐹

𝑂

= ^1,216

2,985

= 0,407 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,343

70 x 0,363 2 3

√𝑖 = 0,009403 𝑖 = 0,000088

= 0,0088 %

b

h

W Q = 0,440 m³ / dt

V = 0,362 m³ F = 1,216 m² b & h = 0,780 m O = 2,985 m R = 0,407 m I = 0,000088

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18

1. Menghitung Debit Air ( Q )

Q = 𝑄 𝑇1 + 𝑄 𝑇2 𝑒 𝑆

= 549,769 + 329,861 0,85

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 1,035 )^0,182

F = ^𝑄

𝑉

= ^1,035

0,423

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 2,449

ℎ² = ^2,449

2

h = 1,106 m

W = √0.5ℎ

= √0.5 𝑥 0,956

O = b + 2h√1 + 𝑚²

= 1,106 + 2 x 1,106√1 + 1²

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19

R = ^𝐹

𝑂

= ^2,449

4,236

= 0,578 m

V = K x 𝑅²³ x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,423

70 x 0,578 2 3

√𝑖 = 0,008701 𝑖 = 0,000076

= 0,0076 %

b

h

W Q = 1,035 m³ / dt

V = 0,423 m³ F = 2,449 m² b & h = 1,106 m O = 4,236 m R = 0,578 m I = 0,000076

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20

Q = 𝑄 𝑇3 + 𝑄 𝑇4 𝑒 𝑆

= 329,861 + 274,884 0,85

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,711 )^0,182

F = ^𝑄

𝑉

= ^0,711

0,395

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 1,802

ℎ² = ^1,802

2

h = 0,949 m

W = √0.5ℎ

= √0.5 𝑥 0,949

O = b + 2h√1 + 𝑚²

= 0,949 + 2 x 0,949√1 + 1²

= 3,634 m

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21

R = ^𝐹

𝑂

= ^1,802

3,634

= 0,496 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ⁰³⁹⁵

70 x 0,496 2 3

√𝑖 = 0,009001 𝑖 = 0,000081

= 0,0081 %

b

h

W Q = 0,711 m³ / dt

V = 0,395 m³ F = 1,802 m² b & h = 0,949 m O = 3,634 m R = 0,496 m I = 0,000081

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22

Q = 𝑄 𝑇5 + 𝑄 𝑇6 𝑒 𝑆

= 219,907 + 219,907 0,85

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 0,517 )^0,182

F = ^𝑄

𝑉

= ^0,517

0,373

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 1,389

ℎ² = ^1,389

2

h = 0,833 m

W = √0.5ℎ

= √0.5 𝑥 0,833

O = b + 2h√1 + 𝑚²

= 0,833 + 2 x 0,833√1 + 1²

= 3,190 m

(31)

23

R = ^𝐹

𝑂

= ^1,389

3,190

= 0,435 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,373

70 x 0,435 2 3

√𝑖 = 0,009265 𝑖 = 0,000086

= 0,0086 %

b

h

W

Q = 0,517 m³ / dt V = 0,373 m³ F = 1,389 m² b & h = 0,833 m O = 3,190 m R = 0,435 m I = 0,000086

b

(32)

24

Q = 𝑄 𝑆3 + 𝑄 𝑇7 𝑒 𝑆

= 517,429 + 439,815 0,85

= 1.126,169 liter/detik

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 1,126 )^0,182

F = ^𝑄

𝑉

= ^1,126

0,429

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 2,624

ℎ² = ^2,624

2

h = 1,145 m

W = √0.5ℎ

= √0.5 𝑥 1,145

O = b + 2h√1 + 𝑚²

= 1,145 + 2 x 1,145√1 + 1²

= 4,385 m

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25

R = ^𝐹

𝑂

= ^2,624

4,385

= 0,598 m

V = K x 𝑅

2

3 x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,429

70 x 0,598 2 3

√𝑖 = 0,008634 𝑖 = 0,000075

= 0,0075 %

b

h

W Q = 1,126 m³ / dt

V = 0,429 m³ F = 2,624 m² b & h = 1,145 m O = 4,385 m R = 0,598 m I = 0,000075

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26

1. Menghitung Debit Air ( Q )

Q = 𝑄 𝑆2 + 𝑄 𝑆4 𝑒 𝑃

= 711,465 + 1.126,169 0,90

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 2,042 )^0,182

F = ^𝑄

𝑉

= ^2,042

0,478

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 4,269

ℎ² = ^4,269

2

h = 1,461 m

W = √0.5ℎ

= √0.5 𝑥 1,461

= 0,855 m

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27

O = b + 2h√1 + 𝑚²

= 1,461 + 2 x 1,461√1 + 1²

R = ^𝐹

𝑂

= ^4,269

5,593

= 0,763 m

V = K x 𝑅²³ x √𝑖

√𝑖 = ^𝑉

K x 𝑅 2 3

√𝑖 = ^0,478

70 x 0,763 2 3

√𝑖 = 0,008181 𝑖 = 0,000067

= 0,0067 %

b

h

W Q = 2,042 m³ / dt

V = 0,4788 m³ F = 4,269 m² b & h = 1,461 m O = 5,593 m R = 0,763 m I = 0,000067

(36)

28

Q = 𝑄 𝑆1 + 𝑄 𝑃1 𝑒 𝑃

= 1.034,858 + 2.041,816 0,90

V = 0,42 x ( 𝑄 )^0,182

= 0,42 x ( 3,419 )^0,182

F = ^𝑄

𝑉

= ^3,419

0,525

F = (

2 2h b b  

)* h

2ℎ² = F

2ℎ² = 6,508

ℎ² = ^6,508

2

h = 1,804 m

W = √0.5ℎ

= √0.5 𝑥 1,804

O = b + 2h√1 + 𝑚²

= 1,804 + 2 x 1,804√1 + 1²

= 6,906 m

(37)

29

R = ^𝐹

𝑂

= ^6,508

6,906

= 0,942 m

V = K x 𝑅²³ x √𝑖

√𝑖 = ^𝑉

K x 𝑅²³

√𝑖 = ^0,525

70 x 0,942 2 3

√𝑖 = 0,007807 𝑖 = 0,000061

= 0,006 %

b

h

W Q = 3,419 m³ / dt

V = 0,525 m³ F = 6,508 m² b & h = 1,804 m O = 6,906 m R = 0,942 m I = 0,000061

(38)

30 A. SALURAN TERSIER

A.1 Saluran Tersier ( T.1 ) 1. Menghitung Debit Air ( Q )

Dm = Modulus Pembuang = 1 liter/dt/ha

Qd = 1,62 x Dm x

𝐴

^0,92

= 1,62 x 1 x

250

^0,92

Vd = 0,42 x ( 𝑄𝑑 )^0,182

= 0,42 x ( 0,260 )^0,182

Fd = ^𝑄𝑑

𝑉𝑑

= ^0,2660

0,329

Fd = (

2 2h b b  

)* h

2ℎ² = Fd

2ℎ² = 0,792 ℎ² = ^0,792

2

h = 0,629 m

Wd = √0.5ℎ

= √0.5 𝑥 0,629

= 0,561 m

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31

Od = b + 2h√1 + 𝑚²

= 0,629 + 2 x 0,629√1 + 1²

Rd = ^𝐹𝑑

𝑂𝑑

= ^0,892

2,409

= 0,329 m

Vd = K x 𝑅𝑑²³ x √𝑖𝑑

√𝑖d = ^𝑉𝑑

K x 𝑅𝑑 2 3

√𝑖𝑑 = ^0,329

70 x 0,329²³

√𝑖𝑑 = 0,009860 𝑖d = 0,000097

= 0,0097 %

b

h

W Qd = 0,260 m³ / dt

Vd = 0,329 m³ Fd = 0,792 m² b & h = 0,629 m Od = 2,409 m Rd = 0,329 m Id = 0,000097

(40)

32

Qd = 1,62 x Dm x

𝐵

^0,92

= 1,62 x 1 x

150

^0,92

Vd = 0,42 x ( 𝑄𝑑 )^0,182

= 0,42 x ( 0,163 )^0,182

Fd = ^𝑄𝑑

𝑉𝑑

= ^0,163

0,302

Fd = (

2 2h b b  

)* h

2ℎ² = Fd

2ℎ² = 0,539 ℎ² = ^0,539

2

h = 0,519 m

Wd = √0.5ℎ

= √0.5 𝑥 0,519

Od = b + 2h√1 + 𝑚²

= 0,519 + 2 x 0,519√1 + 1²

= 1,988 m

(41)

33

Rd = ^𝐹𝑑

𝑂𝑑

= ^0,302

1,988

= 0,271 m

K x 𝑅𝑑 2 3

√𝑖𝑑 = ^0,302

70 x 0,271²³

√𝑖𝑑 = 0,010289 𝑖d = 0,000106

= 0,0106 %

b

h

W Qd = 0,163 m³ / dt

(42)

34

Qd = 1,62 x Dm x

𝐶

^0,92

= 1,62 x 1 x

150

^0,92

Vd = 0,42 x ( 𝑄𝑑 )^0,182

= 0,42 x ( 0,163 )^0,182

Fd = ^𝑄𝑑

𝑉𝑑

= ^0,163

0,302

Fd = (

2 2h b b  

)* h

2ℎ² = Fd

2ℎ² = 0,539 ℎ² = ^0,539

2

h = 0,519 m

jadi b dan h = 0,519m 5. Mencari Tinggi Jagaan ( W )

Wd = √0.5ℎ

= √0.5 𝑥 0,519

Od = b + 2h√1 + 𝑚²

= 0,519 + 2 x 0,519√1 + 1²

= 1,988 m

(43)

35

Rd = ^𝐹𝑑

𝑂𝑑

= ^0,539

1,988

= 0,271 m

K x 𝑅𝑑²³

√𝑖𝑑 = ^0,302

70 x 0,271 2 3

√𝑖𝑑 = 0,010289 𝑖d = 0,000106

= 0,0106 %

b

h

W Qd = 0,163 m³ / dt

(44)

36

Qd = 1,62 x Dm x

𝐷

^0,92

= 1,62 x 1 x

125

^0,92

Vd = 0,42 x ( 𝑄𝑑 )^0,182

= 0,42 x ( 0,138 )^0,182

Fd = ^𝑄𝑑

𝑉𝑑

= ^0,138

0,293

Fd = (

2 2h b b  

)* h

2ℎ² = Fd

2ℎ² = 0,470 ℎ² = ^0,470

2

h = 0,485 m

Wd = √0.5ℎ

= √0.5 𝑥 0,485

Od = b + 2h√1 + 𝑚²

= 0,485 + 2 x 0,485√1 + 1²

= 1,856 m

(45)

37

Rd = ^𝐹𝑑

𝑂𝑑

= ^0,470

1,856

= 0,253 m

K x 𝑅𝑑 2 3

√𝑖𝑑 = ^0,293

70 x 0,253²³

√𝑖𝑑 = 0,010447 𝑖d = 0,000109

= 0,0109 %

b

h

W Qd = 0,138 m³ / dt

(46)

38

Qd = 1,62 x Dm x

𝐸

^0,92

= 1,62 x 1 x

100

^0,92

Vd = 0,42 x ( 𝑄𝑑 )^0,182

= 0,42 x ( 0,112 )^0,182

Fd = ^𝑄𝑑

𝑉𝑑

= ^0,112

0,282

Fd = (

2 2h b b  

)* h

2ℎ² = Fd

2ℎ² = 0,397 ℎ² = ^0,397

2

h = 0,446 m

Wd = √0.5ℎ

= √0.5 𝑥 0,446

Od = b + 2h√1 + 𝑚²

= 0,446 + 2 x 0,446√1 + 1²

= 1,707 m

(47)

39

Rd = ^𝐹𝑑

𝑂𝑑

= ^0,397

1,707

= 0,233 m

K x 𝑅𝑑 2 3

√𝑖𝑑 = ^0,282

70 x 0,233²³

√𝑖𝑑 = 0,010643 𝑖d = 0,000113

= 0,0113 %

b

h

W Qd = 0,112 m³ / dt

(48)

40

Qd = 1,62 x Dm x

𝐹

^0,92

= 1,62 x 1 x

100

^0,92

Vd = 0,42 x ( 𝑄𝑑 )^0,182

= 0,42 x ( 0,112 )^0,182

Fd = ^𝑄𝑑

𝑉𝑑

= ^0,112

0,282

Fd = (

2 2h b b  

)* h

2ℎ² = Fd

2ℎ² = 0,397 ℎ² = ^0,397

2

h = 0,446 m

Wd = √0.5ℎ

= √0.5 𝑥 0,446

Od = b + 2h√1 + 𝑚²

= 0,446 + 2 x 0,446√1 + 1²

= 1,707 m

(49)

41

Rd = ^𝐹𝑑

𝑂𝑑

= ^0,397

1,707

= 0,233 m

K x 𝑅𝑑 2 3

√𝑖𝑑 = ^0,282

70 x 0,233²³

√𝑖𝑑 = 0,010643 𝑖d = 0,000113

= 0,0113 %

b

h

W Qd = 0,112 m³ / dt

(50)

42

Qd = 1,62