1 / 14

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.

alanna
Télécharger la présentation

PERHIT DEBIT MAX

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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=…….?

More Related