Question

4 m3/s of water flows in a Rectangular channel with base width 3 m, base slope J= 0.0016 and roughness coefficient K= 80.
a) find the normal depths and critical depth of the current.
B) determine the current regime.

roughhness coefficient given 80

Answer 1

Ans A) We know,

Q = (A/N)

where, Q = Flow rate = 4 m3/s

N = Manning roughness coefficient = 1 / K = 1 / 80 0.013

A = Area = BD

R = Hydraulic depth = Area / Wetted Perimeter

J = Base Slope = 0.0016   

Given,base width (B) = 3 m

=> A = 3 D

Wetted perimeter = B + 2D = 3 + 2D

=> R = (3D / 3 + 2D)

Putting values,

=> 4 = (3D/ 0.013)

=> 4 = (230.77 D) x 0.04

=> 100 = (230.77 D)

=> 0.433 = D x () /

=> 0.433 = 2.08 /

=> 0.208 = /

On solving above equation ,we get D 0.705 m

Hence, normal depth ,D = 0.705 m

Now,

For rectangular channel, critical depth (Dc) = (/ g

where, q = Discharge per unit width = Q / B = 4 / 3 = 1.33 m2/s

=> Dc = ( / 9.81

=> Dc = 0.565 m

Hence, critical depth, Dc = 0.565 m

.

Ans B) We know,

Froude number (Fr) = V /

where, V = Flow velocity = Q / A = 4 / (3 x 0.705) = 1.89 m/s

R = Hydraulic depth = A / P = (3 x 0.705) / (3 + 2(0.705)) = 0.48 m

=> Fr = 1.89 /

=> Fr = 0.87 < 1

Since, Froude number is less than 1, current flow regime is sub critical