Tahu Margas ari (mm/th Dukuh Warigi (mm/th Kalibaku g (mm/th 35 5 3 2 3 28 43 3 22 9 29 4 3 42 6 5 65 253 25 6 22 25 39 64 55 84 8 8 63 4 9 29 46 36 5 24 2 53 2 2 6 8 6 3 29 29 4 25 52 25 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, 5 8 68 83, 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 4 226,33 5 33,68 3 526,95 8 8,943 44,68 3 22,8 9 94,5 8 345,8 9 5555,68 3 296,6 3 824, 232,5 84 42,8 4 2,95 522, 2 9839,3 2684,2 39 33,92 554553,4 5 493, 284 393,49 6 63,29 53,54 45,33 9,269 28,28 85,98 26,4 325,6 2 428,95 45,6 9 388,52 8 542,3 2 5435, 48 383,3 585,3 8884,3 4 4,2 658,3 263,2 49 5382, 4 46836,22 66,6 46 83998,2 8 244,9 5953,9 3 294, 6 332653,
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 332653,= 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
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 35 5 3 38,6 2,42,228,56E5
2 3 28 43 36, 2,34,2494,2454 3 22 9 29 6,6 2,24,454,25886 4 3 42 6 44,333 2,59,584,48E5 5 65 253 25 8, 2,258,555,48522 6 22 25 39 88,6 2,26,8539,8426 64 55 84,6 2,224,84 6,865E5 8 8 63 4 6,6 2,29,632,5893E5 9 29 46 5,6 2,8 2,982E5,6289E 36 5 24 39, 2,43,633 6,625E5 2 53 2 35,6 2,32,265,32692 2 6 8 6 33, 2,24,35553,2994 3 29 29 4,6 2,5,35 3,3496E5 4 25 52 25 34, 2,2,38,954 5 8 68 83, 2,248,4596,26822 Jumlah 235,66 32,526,432,996 Ratarata 2,83484 std. Deviasi (Sx,543623 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
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 9,98 82 2,38 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 = 2,8244253 R 25 = 93,534 mm3 /det log R 5 =53,+2,594 (9,5 = 2,38995 R 25 = 23,33 mm3 /det
3. Metode Log Normal Tahu Margasa ri (mm/th Dukuh Warigi (mm/th Kalibaku g (mm/th 35 5 3 CH Wilayah (X (mm/th 38,6666 2 3 28 43 36 3 22 9 29 4 3 42 6 6,6666 44,3333 33 5 65 253 25 8 6 22 25 39 64 55 84 8 8 63 4 9 29 46 88,6666,6666 6,6666 5,6666 36 5 24 39 2 53 2 35,6666 2 6 8 6 33 3 29 29 4,6666 4 25 52 25 34 5 8 68 83 Jumlah Ratarata 235,666 Std. Deviasi (Sx,5436 23 Factor frekuesi (K25,8333 33 Factor frekuesi (K5 2,5 log X 2,49 2 2,3353 89 2,245 46 2,5936 66 2,25 86 2,2569 52 2,22444 2,2862 5 2,8 2 2,43 48 2,324 32 2,2385 6 2,526 2,2 48 2,249 33 32,52 6 2,834 84 (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 2,842 6 6,865E 5,5893E5,45 4,58 4,55 5,85 39, 84,63 2 2,982E 5,63 3,6289E 6,625E 5,3269 2,299 4 3,3496E 5,95 4,2682 2,99 6
R25 R5 88,9493 86 9,254 46 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 =,543623 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 (,543623 = 2,26345483
R 25 = 88,949386 mm 3 /det log R 5 =2,83484+2,5(,543623 = 2,29498922 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 24,8246 35 93,5 23,3 34 3 88,9493 9,254 86 46 4. Probabilitas Tahu CH Wilayah (X (mm/th 38,666 Tahu CH Wilayah (X (mm/th P P% 2 33,625 6,25 2 36 4 34,25 2,5 3 6,666 35,666,85 8,5 4 44,333 2 36 33,25 25 38,666 5 8,325 3,25 6 88,666 39,35 3,5
,666 4,666 3,435 43,5 8 6,666 44,333 4 33,5 5 9 5,666 5,666 9,5625 56,25 39 8 6,666,625 62,5 35,666,666,685 68,5 2 33 3 6,666,5 5 3 4,666 5,825 8,25 4 34 5 8,85 8,5 5 6 88,666,935 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
Karea yag memeuhi syarat adalah aalisi dega gumbel maka yag diambil adalah metode gambel Metode Gumbel R R25 94,6 835 R5 24,824 6 3. 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 = 5554 55 =,82 46 55 554 =,35
o Kemiriga (i2 = o Kemiriga ratarata (i = 554 5522 =,42 95,82+,35+,42 =,233 3 3.2. 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 2 +5 2 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+
q= 56,93 m3 =,3369 3,6(4,234 det /km2 α=,5 β 4,234+3,.,4(4,234 =+. 4,5 4,234 2 +5 2 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 4,234 2 +5 2 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 =,48322 3,6(4,234 det /km2 α=,5 β 4,234+3,.,4(4,234 =+. 4,5 4,234 2 +5 2
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
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
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