Download - Tugas Hidroulika 3
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Nama : Doni Hidayat
NIM : H1A110082
MK : Hidroulika
Soal
15. Saluran trapesium dengan lebar dasar 15m dan kemiringan tebing 1:1 mengalirkan debit
100m3/d. Apabila koefisien Manning n = 0,02 hitung kedalaman kritis dan kemiringan
kritis dari aliran tersebut.
Penyelesaian :
Diketahui :
1:1
B
B = 15 m
Q = 100 m3/d
N = 0,02
Ditanya
a.yc = ?
b. Ic = ?
Dijawab :
![Page 2: Tugas Hidroulika 3](https://reader035.vdokumen.com/reader035/viewer/2022081210/55cf9dc1550346d033af0bdc/html5/thumbnails/2.jpg)
a. yc = π23
(π΅+2ππ¦π )
π(π΅+ππ¦π )3
yc = 10023
(15+2ππ¦π )
9.81(15+ππ¦π )3
yc =10,0641476 x (15+2π¦π )3
(15+π¦π )3
Asumsi
yc = 1
yc =10,0641476 x (15+2π₯1)3
(15+1)3
yc = 1.61736
yc = 0.9
yc =10,0641476 x (15+(2π₯0,9))3
(15+0,9)3
yc = 1.621124
yc = 0,8
yc =10,0641476 x (15+(2π₯0,8))3
(15+0,8)3
yc = 1.624885
yc = 0,7
yc =10,0641476 x (15+(2π₯0,7))3
(15+0,7)3
yc = 1.628641
yc = 0,6
yc =10,0641476 x (15+(2π₯0,6))3
(15+0,6)3
yc = 1.632391
![Page 3: Tugas Hidroulika 3](https://reader035.vdokumen.com/reader035/viewer/2022081210/55cf9dc1550346d033af0bdc/html5/thumbnails/3.jpg)
yc = 0,5
yc =10,0641476 x (15+(2π₯0,5))3
(15+0,5)3
yc = 1.636133
Jadi kedalaman kritis adalah yc = 0,5 m
b. Ic = ππ 2
π¦π 1/3
Ic = 9,81 π₯ 0,022
0,51/3
Ic = 0.004944
Jadi kemiringan kritis dari aliran tersebut adalah 0.004944 m