Question

8. The discharge flowing in a trapezoidal canal is equal to 30m3/s and a velocity of.12m/s. (a) What would be the width of the canal for a most efficient section? (b) If the width of the canal is 1.5 of its depth, compute the width of the canal if the sides are sloping at an angle of 50 degrees. (c) If the sides are sloping at an angle of 54 degrees, compute the width of the base for a minimum seepage.

Answer 1



Que Given data @= 30mls 12 0.12 ms. Portal Given that Most ellicient channel. ly mag B ndy A= (Bimy ly P= B+2 Jmetiy P = f my + 5 Hmly for efficient Poringa Minisnom discuroge Maximum de dy Channed we get Bt any y slama 2 for Best Slop. de 2 m = STAMP O Tin we geli Die ma 3 O = 600 th = Colo

from Bt ayo aym (1+ 11-? ) 11 Bt24 2J1+y B: 27 1B= 1.1544 B

Discharges - Area = Areax Velocity 30 = 250m² 012 250 (Bt my ly (11 suy+0.Sittyly = 250 18 = 12 B = kishy = B = 13.86m

bol b) Given Br tisd B. l.sy Soo 0.839 on= colo Aroa - Bramy ly y my my Disclcenge Area xvelocity Axu 30 A 10.12 250m2 = А (Btmyly & 250 11.54 +0.9394)y = 250 y= 10.33m Ba Is. som from ☺ ona Cote Date) for Minimum Soepage Perimeter of Channel ii Minimum from eanation Blamy yst me - 07265 2 B+1.4539 = 1.2364 B = 1.0194

с Area > Зо о: 2 A = 250m² Area с (Bt myly 250 = ( 1-ot 99 +0-12 653 )8 (3= (-46ң