Sta Kalibaku ng (mm/thn ) CH Wilayah (X) (mm/th n) 138, ,00 176, ,33 181,00 188, , , , ,00 135,66 133,00

Transkripsi

1 Tahu Margas ari (mm/th Dukuh Warigi (mm/th Kalibaku g (mm/th CH Wilayah (X (mm/th 38,66 36, 6,66 44,33 3 8, 88,66,66 6,66 5,66 39, 35,66 33, 4,66 34, , Jumlah 235,6 RataRata (X 53, std. Deviasi (Sx 9,5 Skews (Cs,58 Kurtosis (Ck 2,24 KT 25 2,45 KT5 2,594 94, R25 (mm 3 /det 24,82 R5 5 (mm 3 /det 25.5 R 5 (mm 3 /det (XX 2 (XX 3 (XX 4 226, , ,95 8 8,943 44, ,8 9 94, , , , , 232, ,8 4 2,95 522, ,3 2684, , , , , ,29 53,54 45,33 9,269 28,28 85,98 26,4 325, ,95 45, , , , ,3 585,3 8884,3 4 4,2 658,3 263, , ,22 66, , ,9 5953, , ,

2 Perhituga Curah Huja Recaa. Model Gumbel Perhituga dar Deviasi (Sx Sx= ( X X ² i= Sx= 5435,48 5 = 9,5 Perhituga Koefisie Skewwss (Cs Cs= ( ( 2 (X X ³ Sx ³ 5 Cs= (5 (5 29,5³ 5382,4 =,58 Perhituga Koefesie Kurtosis (Ck Ck= ( X X 4 ( ( 2( 3Sx 4 5² Ck= (5 (5 2(5 3Sx ,= 4 2,24 Tabel 3.2 Parameter Peetu Jeis Sebara Jeis Sebara Syarat Normal Cs = Ck = 3 Log Normal Cs =,63 Ck = 3 Log Pearso III Cs Ck =,5Cs + 3

3 Ck = 3,83 Gumbel Cs <,396 Ck < 5,42 Dari table diatas diperoleh ilai Cs<,396 (koef. Skewess da ilai Ck<5,42 (Koef. Kurtosis, maka termasuk kedalam jeis sebara Gumbel. Perhituga Curah Huja Recaa (R25 da R5 KT= 6 π {,52+l [ l ( T ]} T KT 25= 6 KT 5= 6 KT = 6 π {,52+l [ l ( π {,52+l [ l ( π {,52+l [ l ( 25 ]} = 2,45 5 ]} = 2,594 ]} = 3,36 R= X+ KT. Sx R 25 =53,+2,45(9,5 = 94, mm 3 /det R 5 =53,+2,594 (9,5 = 24,825 mm 3 /det R =53,+3,36(9,5 = 25,55 mm 3 /det 2. Metode Log Perso III Tahu Margasari (mm/th Dukuh Warigi (mm/th Kalibaku g (mm/th CH Wilayah (X (mm/th log X (log Xlog X² (log Xlog X³ ,6 2,42,228,56E5

4 , 2,34,2494, ,6 2,24,454, ,333 2,59,584,48E , 2,258,555, ,6 2,26,8539, ,6 2,224,84 6,865E ,6 2,29,632,5893E ,6 2,8 2,982E5,6289E , 2,43,633 6,625E ,6 2,32,265, , 2,24,35553, ,6 2,5,35 3,3496E , 2,2,38, , 2,248,4596,26822 Jumlah 235,66 32,526,432,996 Ratarata 2,83484 std. Deviasi (Sx, Koefisie Asimetri (Gx,49699 KT 25,98822 KT 5 2,38853 R25 93,534 R5 23,33 Logaritma Ratarata log X= Σ log x = 32,5 5 =2,83 Perhituga dar Deviasi (Sx X log X log 2 i= Sx= Sx=,432 5 =,543623

5 Perhituga Koefisie Asimetri (Gx X log X log Gx= ( ( 2 Sx 3 5 Gx= (5 (5 2,543623,432 3 =,49699 Utuk Nilai KT, dilakuka Iterpolasi sebagai berikut : koefisie k 25 5,4,88 2,26,6,939 2,359,4969 9, ,38 85 Maka, didapat hasil sebagai berikut : KT 25 =,9882 KT 5 = 2,3885 Logaritma Huja racaga dega kala tertetu dega rumus : log R=log X+KT. Sx log R 25 =53,+2,45(9,5 = 2, R 25 = 93,534 mm3 /det log R 5 =53,+2,594 (9,5 = 2,38995 R 25 = 23,33 mm3 /det

6 3. Metode Log Normal Tahu Margasa ri (mm/th Dukuh Warigi (mm/th Kalibaku g (mm/th CH Wilayah (X (mm/th 38, , , ,6666,6666 6,6666 5, , , Jumlah Ratarata 235,666 Std. Deviasi (Sx, Factor frekuesi (K25, Factor frekuesi (K5 2,5 log X 2,49 2 2, , , , , , , ,8 2 2, , , ,526 2,2 48 2, ,52 6 2, (log X log X²,2 28,249 4,26 5,355 53,3 5,3 8,45 96,43 2 (log Xlog X³,56E5,2454,2588 6,48E5,4852 2, ,865E 5,5893E5,45 4,58 4,55 5,85 39, 84,63 2 2,982E 5,63 3,6289E 6,625E 5,3269 2, ,3496E 5,95 4,2682 2,99 6

7 R25 R5 88, , Logaritma Rata rata log x log X= = 32,526 5 = 2,83484 Perhituga daar Deviasi (Sx X log X log 2 i= Sx= Sx=,432 5 =, Faktor Frekuesi Meghitug faktor frekuesi dega cara iterpolasi : KT5,84 KT,28 KT2,64 KT25,83 33 KT5 2,5 KT 2,33 Logaritma huja racaga dega kala tertetu dega rumus : log R=log X+KT. Sx log R 25 =2,83484+,8333 (, = 2,

8 R 25 = 88, mm 3 /det log R 5 =2, ,5(, = 2, R 25 = 9,25446 mm3 /det Dari Ketiga Metode Perhituga didapat ilai R25 da R5 seperti pada tabel di bawah ii : Metode Gumbel Log Perso III Log Normal R R25 R5 94,68 24, ,5 23, ,9493 9, Probabilitas Tahu CH Wilayah (X (mm/th 38,666 Tahu CH Wilayah (X (mm/th P P% 2 33,625 6, ,25 2,5 3 6,666 35,666,85 8,5 4 44, , , ,325 3, ,666 39,35 3,5

9 ,666 4,666 3,435 43,5 8 6,666 44, , ,666 5,666 9, , ,666,625 62,5 35,666,666,685 68, ,666, ,666 5,825 8, ,85 8, ,666,935 93,5 Tr = 25 Tr = 25 =,4 Tr =q=,4 p= q=,96=96 R 25 = 9,426 Tr = 5 Tr = 5 =,2 Tr =q=,2 p= q=,98=98 R 5 = 93,88

10 Karea yag memeuhi syarat adalah aalisi dega gumbel maka yag diambil adalah metode gambel Metode Gumbel R R25 94,6 835 R5 24, Perhituga Debit Recaa Data Perecaaa : Luas catchme area telah (A = 4 Km 2 Pajag dasar sugai (L = Km Skala yag diguaka = : Kemiriga ratarata (i o Kemiriga (i = o Kemiriga (i2 = =, =,35

11 o Kemiriga (i2 = o Kemiriga ratarata (i = =,42 95,82+,35+,42 =, Metode Hasper Metode hasper biasaya diguaka pada luas DPS < 3 km 2 dega persamaa rumus rumus sebagai berikut : Q 25 =α. β.q. A t=,. L,8.i,3 R= t.rt t+ q= R 3,6t +,2. A, = +,5. A, t+3,.,4t =+. A,5 β t Dimaa : Q = Debit Bajir Recaa (m3/det α = Koefesie Pegalira (Ru Off Coefficiet β = Koefesie Reduksi (Reductio Coefficiet q = Bayakya air yag megalir tiap km, m 3 /det/km 2 A = Luas DAS (Catchmet Area km 2 Q25 Perhituga debit recaa dega periode ulag T tahu megguaka metode Hasper : t=,(,8 (,233,3 =4,234 Jam R= 4,234 (94, =56,93 4,234+

12 q= 56,93 m3 =,3369 3,6(4,234 det /km2 α=,5 β 4,234+3,.,4(4,234 =+. 4,5 4, β=,8528 Q 25 =,5 x,8528 x,3369 x 4=89,36 m 3 /det Q5 t=,(,8 (,233,3 =4,234 Jam R= 4,234 (24,825 =65,534 4,234+ q= 65,534 m3 =,943 3,6(4,234 det /km2 α=,5 β 4,234+3,.,4(4,234 =+. 4,5 4, β=, 8528 Q 5 =,5 x,8528 x,943 x 4=89,36 m 3 /det Q t=,(,8 (,233,3 =4,234 Jam R= 4,234 (25,55 =4,654 4,234+ q= 4,654 m3 =, ,6(4,234 det /km2 α=,5 β 4,234+3,.,4(4,234 =+. 4,5 4,

13 β=,8528 Q =,5 x,8528 x,48322 x 4=99,6569 m 3 /det 3.3 Metode Nakayasu metode akayasu diguaka utruk meperkiraka debit bajir yag aka terjadi, metode ii dapat dilakuka dega aalisi Raifall (Ruoff Model dega persamaa rumus rumus yag diguaka berikut : A. R Qp= 3,6(,3. T p +T,3 tg=,2 L, utuk L<5km tg=,4 +,58 Lutuk L>5 km tr=lama hujaefektif yag besarya,5 Tp=tg +,8 tr AL,25,4 α= T,3 =α.tg Dimaa : Qp = Debit Pucak Bajir (m 3 /det R = Huja Satua (mm Tp = Teggag waktu(time log dari permulaa huja sampai pucak bajir (jam T,3 = Waktu yag diperluka oleh peurua debit, dari debit pucak sampai mejadi 3% dari debit pucak (jam A = Luas DAS (Catchmet Area km 2 Sedagka harga α mempuyai criteria sebagai berikut : a. Daerah pegalira biasa α=2

14 b. Bagia aik hidrograf yag lambat da bagia meuru yag cepat α=,5 c. Bagia aik hidrograf yag cepat da bagia meuru yaglambat α=3 Perhituga debit recaa dega periode ulag T tahu megguaka metode Nakayasu : Perhituga Qp25, Qp5 da Qp koef. pegalira=,5 R 25 = 94, m 3 /det R = R25. α = 94, m 3 /det x,5 = 98,9435 m3 /det tg=,2(, =,25 Tr = diambil ilai sebesar, Tp=,25+,8 (,=,685 jam 4 x,25,4 α= T,3 =2 x,25=2,25 4 x 98,9435 Qp25= 3,6(,3 x,685+2,25 =398,938 m3 /det koef. pegalira=,5 R 5 = 24,825 m 3 /det R = R5. α = 24,825 m 3 /det x,5 = 4,468 m3 /det

15 tg=,2(, =,25 Tr = diambil ilai sebesar, Tp=,25+,8 (,=,685 jam 4 x,25,4 α= T,3 =2 x,25=2,25 4 x 4,468 Qp5= 3,6(,3 x,685+2,25 =42,833 m3 /det koef. pegalira=,5 R = 25,55 m 3 /det R = R. α = 25,55 m 3 /det x,5 = 9,96 m3 /det tg=,2(, =,25 Tr = diambil ilai sebesar, Tp=,25+,8 (,=,685 jam 4 x,25,4 α= T,3 =2 x,25=2,25 4 x 9,96 Qp= 3,6(,3 x,685+2,25 =443,446 m3 /det