Tahap 1 Lengkung Debit.docx

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BAB I LENGKUNG DEBIT Data Soal Bentuk Penampang

: Trapesium

Material Dasar

: Kerikil

Lebar Dasar Sungai (b) : 25 m Kemiringan Sungai (I0) : 0,00002 Q100

: 318 m3/s

m

:

Gambar 1.1 Penampang Saluran Trapesium

1

Tabel 1.1 Perhitungan nilai d3 dengan metode coba - coba b1 25 25 25 25 25 25 25 25 25

d3 1 3 5 10 10,5 10,8 10,85 10,853 10,85351

A 26,000 84,000 150,000 350,000 372,750 386,640 388,973 389,113 389,136

P 27,828 33,485 39,142 53,284 54,698 55,547 55,688 55,697 55,698

R 0,934 2,509 3,832 6,569 6,815 6,961 6,985 6,986 6,986

C 49,437 58,283 62,548 68,425 68,846 69,089 69,129 69,132 69,132

v 0,214 0,413 0,548 0,784 0,804 0,815 0,817 0,817 0,817

Q 5,556 34,678 82,137 274,494 299,592 315,178 317,814 317,973 318,000

Lengkung Debit 12 10

10

10.853 10.85351 10.85 10.510.8

h (m)

8 6 5 4 3 2 1 0 0

50

100

150

200

250

Q (m3/s)

Grafik 1.1 Lengkung debit sungai

300

350

Contoh Perhitungan :  Nilai n (koefisien manning) n = 0,02 (bahan dasar sungai adalah kerikil, dengan asumsi kerikil yang digunakan adalah kerikil halus) Sumber : Pedoman Perencanaan Saluran Terbuka, Pusat Penelitian dan Pengembangan Pengairan Dep. PU, 1986.

 Nilai Luas Penampang Sungai (A) 𝐴 = 𝑑3 (𝑏1 + π‘š Γ— 𝑑3 ) = (10,85351)(25 + 1 Γ— 10,85351) = 389,136 π‘š2  Nilai Keliling Basah (P) 𝑃 = 𝑏1 + 2𝑑3 √1 + π‘š2 = 25 + 2(10,85351)√1 + 12 = 55,698 π‘š  Nilai R 𝑅=

𝐴 389,136 π‘š2 = = 6,986 π‘š 𝑃 55,698 π‘š

 Nilai v (Rumus Chezy) Sebelum menghitung v, konversi dulu koefisien manning ke koefisien chezy : 𝐢=

1 1 1 1 Γ— 𝑅6 = Γ— 6,9866 = 69,132 𝑛 0,02

𝑣 = πΆβˆšπ‘…πΌ = 69,132 Γ— √6,696 Γ— 0,00002 = 0,817 π‘š/𝑠  Nilai Debit (Q) 𝑄 = 𝐴 π‘₯ 𝑣 = (389,136)(0,817) = 318,000 π‘š3 /𝑠

Kesimpulan: Dengan menggunakan perhitungan Tabel 1.1 Perhitungan nilai d3 dengan metode coba – coba, maka dengan Q 100 = 318 m3/dt di dapat ukuran penampang saluran, yakni d3= 10,85351 m

b = 25 m

Gambar 1.2 Dimensi Penampang Saluran Trapesium

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