Question

a) An earthen irrigation channel is 10km long and it is suggested to have a capacity of 18m3 /s for a flow depth of 1.2m. The cross-section is trapezoidal with a bed width of 10m and side slopes of 4 horizontal to 1 vertical. The bed slope is 1:1,000. What value of n was used in the design? How does this compare with the literature?

b) The channel in a) has floodplains of 250m on the left bank with a scattered brush, and 750m of cultivated land without crops on the right bank. The channel overtops and floods at a depth of 1.5m. What is the flow depth for a flood of 250m3 /s?

Answer 1

Ans a) We know,

Q = (A/N) R^2/3 S^1/2

where. N = Manning roughness coefficient

A = cross sectional area of channel

R = Hydraulic radius

S = channel bed slope

Area of trapezoidal channel(A) = BD + mD²

Side slope ( m : 1) = 4 : 1 (given)

=> m = 4

=> A = 10(1.2) + 4(1.2)² = 17.76 m²

Wetted perimeter (P) = B + 2mD (1 + m²)^1/2

=> P = 10 + 2(4)(1.2) (17)^1/2

= 49.58 m

Hydraulic radius (R) = A/P = 17.76/49.58 = 0.358 m

Putting values,

18 = (17.86/N) (0.358)^2/3 (0.001)^1/2

=> 17.86 / N = 1129.36

=> N = 0.0158 or 0.016

  Literature value of Manning roughness coefficient for earthen channel = 0.018 .So , calculated value is approximately same as give in literature value.

Ans b) Manning roughness for scattered brush = 0.050

Manning roughness for cultivated land with no crops = 0.030

Average value = 0.04

We know , Q = (A/N) R^2/3 S^1/2

R = A/P = 10d + 4d²/ (10 + 33d)

Putting values,

250 = [(10d + 4d²)/0.04] [10d + 4d²/ (10 + 33d)]^2/3 (0.001)^1/2

=> 316.22 = (10d + 4d)^5/3 / (10 + 33d)

By solving above equation by hit and trial , d = 7.2 m