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PERHIT DEBIT MAX

PERHIT DEBIT MAX. Adhi Muhtadi, ST., SE., MSi. 7an perhit:. Utk perenc bang air: sal pematusan, gorong2, siphon, norm sungai, bendung, sal pengelak dsb Tdk memperhatikan besar rambatan banjir. Metode2 debit max:. Metode Rasional Metode Melchior Metode Weduwen Metode Haspers.

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PERHIT DEBIT MAX

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  1. PERHIT DEBIT MAX Adhi Muhtadi, ST., SE., MSi.

  2. 7an perhit: • Utk perenc bang air: sal pematusan, gorong2, siphon, norm sungai, bendung, sal pengelak dsb • Tdk memperhatikan besar rambatan banjir

  3. Metode2 debit max: • Metode Rasional • Metode Melchior • Metode Weduwen • Metode Haspers

  4. Metode Rasional • Qp = 0,278 . α .I . A • I = R24 / 24 . (24/t) 2/3 • t = Tc • Tc = L/V dan V = 72 . (H/L)0,6

  5. Koef aliran (α) di Jepang: • Bergunung & curam : 0,75 - 0,90 • Pegunungan tersier 0,70 - 0,80 • Sungai dgn tanah & hujan di bag atas dan bwhnya : 0,50-0,75 • Tanah dasar yg ditanami: 0,15-0,60 • Sawah waktu diairi: 0,70-0,80 • Sungai bergunung: 0,75-0,85 • Sungai dataran: 0,45 – 0,75

  6. Contoh soal Rasional: • A = 100 km2 ; L = 10 km ; H/L = 0,001; R24 = 140mm; Q max = ? (Lihat Sholeh, hal:136)

  7. Penyelesaian: • V = 72 (H/L)0,6 = 72. (0,001)0,6 = 1,141 km/jam • Tc = L/V = 10 / 1,141 = 8,8 jam • I = R24/24 . (24/t)2/3 = 140/24.(24/8,8)2/3 = = 11 mm/jam • Daerah bergunung, mk dari tabel 10.1 harga =0,8 • Q = 0,278. α. I. A = 0,278. 0,80. 11 . 100 = = 244 m3/dtk

  8. Metode Melchior • Telah digunakan di Indonesia sejak thn 1933 • Q =α . β. q. A • = 0,52 • Angka reduksi (β): perbandingan hujan rata2 daerah aliran dg hujan max yg tjd di daerah aliran tsb • F = [1970 / ( - 0,12)] – 3960 + 1720 β (10.3)

  9. Melchior (2) • Utk hujan luar Jakarta, bila luas elips = 300 km2, dan lama hujan 4 jam, maka besar hujan rata2 per etmal X mm = (X/200) x 83,8 mm. • Hujan max setempat: Tc = 1000 L / 60 V • V = 1,31 x (β . q . A . i2) 0,2

  10. ContohSoal Melchior: Diket: Tinggi hujan rencana = 108 mm/et mal Panjang sungai = 28 km Luas = 86,45 m2 Beda tinggi hulu dgn lok bendung = 450 m Koef aliran = 0,70 Luas ellips nF = 120 km2

  11. Penyelesaian: • Q =α .β . q . A • Koef aliran = α = 0,70 • Angka reduksi = R/Rmax = 108/200 • hujan max setempat = q= ….? dgn perkiraan tabel 11.3 (Sholeh, h.139) nF=180, mk q=5,25 nF=120, mk q=………? nF=144, mk q=4,75

  12. Untuk selisih nF=36, selisih q = 0,50, jd perkiraan nF 120 = 5,25- (1/3 x 0,50) = 5,083. • Luas = A = 86,45 km2 • Q = α .β . q . A = 0,70 . (108/200).5,083 . 86,45 = 166,10 m3/dtk

  13. ContohsoalMetodeWeduwen: Diket: • Waktu pengamatan = 40 thn • Hujan max ke 2 = 205 mm • Luas DAS = 24 km2 • Kemiringan sungai rata2 = 0,005 • Debit max dlm 100 thn=…….?

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