PRACTICAL HYDRAULICS TRAPEZOID CHANNEL OCEAN UNIVERSITY - PRAKTIKUM HIDROLIKA SALURAN TRAPESIUM UNIVERSITAS SAMUDRA ( ENG/IND )
![](https://steemitimages.com/DQmUsJV3yzGuDUuh4ty98cAikx8PVBVBqLYmTQWpSBVtRCH/IMG-20171019-WA0105.jpg)
[ENG]
Hallo Steemian ... Good Afternoon
On this occasion I will share the results of my post reply entitled Practical Hydraulics Trapezoid Channels in Ocean University of Langsa. On this task I as Chairman of the Group's guiding my friends to figure 8 the shape of the channel. One of them wrote will I share in this post is a form of Trapezoid Channel.
![](https://steemitimages.com/640x0/https://steemitimages.com/DQmX8FddLgBruwYpWZnne8KPb4Xa98PwHtVXJ7rTYTjG6Po/IMG-20171019-WA0106.jpg)
On this channel I served to calculate cross-sectional area, measure the height of the flow channel, channel slope, flow velocities, roving, wet fingers hydraulics and discharge on the trapezoid channel. To find out the speed of the flow, we measured him with 5 balls with different colors. We count him out using the Stopwach. After the fifth ball of his time to come by recently summed up and divided 5 to get results or average value.
![](https://steemitimages.com/DQmQtpwK3TuEb1hkmsAZvvZnPZwW91mqjvTo2nPqL4DVW6Y/IMG-20171019-WA0107.jpg)
Cross-sectional Area:
A = (B + Zh) h
= (0.80 + 0.30 × 0.30) 0.30
= 〖 〗 267m ^ 0, 2
Flow Rate:
V = s/t
= 4.1/〗 〖 5,937 = 0,691 ^ (m/s)
Discharge Flow:
Q = A x V
= 0,267 × 0,691
= 0.1845 m ^ 3/s
The Circumference Is Wet:
P = B + 2 h √ (1 + z ^ 2)
= 0.80 + 2 × 0.30 √ (1 + 〗 〖 0.30 ^ 2)
= 1.462 m
Hydraulic RADIUS:
R = (B + zh) h/(B + 2 h √ (1 + z ^ 2))
= (0.80 + 0.30 × 0.30) 0.30/(0.80 + 2 × 0.30 √ (1 + 〗 〖 0.30 ^ 2))
= 0,267/1,462
= 0.182 m
The Slope Of The Basic Channels:
I ^ (1/(2 = (V × n)/r ^ (2/3)))
= (0,691 × 0,013)/〖 〗 0,182 ^ (2/3) = (0,00899/0,3212) ^ 2
I = 0,0007816
![](https://steemitimages.com/640x0/https://steemitimages.com/DQmPiLofC4ZozkKtVFsvZJLXG8RJ6MkjqLJ1YcuR9J6pAEJ/persamaan%202.jpg)
Such was the reply can I share today.. May be beneficial for me and all readers ... Hopefully you liked him.
![](https://steemitimages.com/DQmUsJV3yzGuDUuh4ty98cAikx8PVBVBqLYmTQWpSBVtRCH/IMG-20171019-WA0105.jpg)
[IND]
Hallo Steemian.. Selamat Sore
Pada kesempatan ini saya akan membagikan hasil postingan saya yg berjudul Praktikum Hidrolika Saluran Trapesium di Fakultas Teknik Universitas Samudra Langsa. Pada tugas ini saya sebagai Ketua kelompok membimbing teman-teman saya untuk mencari 8 bentuk saluran. Salah satunya yg akan saya bagikan di posting ini adalah bentuk Saluran Trapesium.
Pada saluran ini saya bertugas untuk menghitung luas penampang, mengukur tinggi aliran saluran, kemiringan saluran, kecepatan aliran, keliling basah,jari-jari hidrolika dan debit pada saluran trapesium tersebut. Untuk mengetahui kecepatan aliran, kami mengukur nya dengan 5 buah bola dengan warna yg berbeda. Kami menghitung nya dgn menggunakan Stopwach. Setelah kelima bola didapat waktu nya baru dijumlahkan dan dibagi 5 untuk mendapat hasil atau nilai rata-rata nya.
Luas Penampang :
A=(B+Zh)h
=(0,80+0,30×0,30)0,30
=〖0,267m〗^2
Kecepatan Aliran :
V=s/t
=4,1/5,937=〖0,691〗^(m⁄s)
Debit Aliran :
Q=A x V
=0,267×0,691
=0,1845 m^3/s
Keliling Basah :
P =B+2h√(1+z^2 )
=0,80+2×0,30√(1+〖0,30〗^2 )
=1,462 m
Jari-jari hidrolika :
R=(B+zh)h/(B+2h√(1+z^2 ))
=(0,80+0,30×0,30)0,30/(0,80+2×0,30√(1+〖0,30〗^2 ))
=0,267/1,462
=0,182 m
Kemiringan Dasar Saluran :
I^(1⁄(2=(V×n)/R^(2⁄3) ))
=(0,691×0,013)/〖0,182〗^(2⁄3) =(0,00899/0,3212)^2
I=0,0007816
Demikian lah yg dapat saya bagikan hari ini.. Semoga dapat bermanfaat bagi saya dan pembaca sekalian.. Semoga anda menyukai nya.