Problems 5 Find the head required over a trapezoidal weir fo

Problems 5: Find the head required over a trapezoidal weir for a discharge of 180 cfs. The length of the weir is 35 ft and the side slopes are 2.5H:1V. The discharge coefficient is 0.63.

Solution

Solution:-

Given

Q = 180 cfs

C_w = 0.63

L = 35 feet

H = head above crest

Slope = tan = 1/(2.5)

= 21.8 degree

We know that for trapezoidal weir

Q = C_w *( L + 0.8 H tan (/2) )^1.5

180 = 0.63 * ( 35 + 0.8 * H * tan( 21.8/2) )^1.5

H = 54.35 feet   Answer