In: Civil Engineering
3. a) A trapezoidal open channel is being designed to convey a discharge of 11.0 m3/sec. The channel will be excavated into clay and will have the following properties: side slope of m = 1.5, the bottom slope of So = 0.001, Manning’s n = 0.022, and bottom width b = 1.0 m. For these conditions is there a risk of channel erosion?
b) The earth-works contractor that will carry out the excavation suggests that the side slope angle can be increased to 45o (i.e. m = 1). Comment on the impact that such a change would have on 1) the required depth of excavation, and 2) the risk of erosion.
c) Calculation of the normal depth using Manning’s equation does not consider freeboard.
I need help with this question. Please help me, Thank you!
Ans) Given,
Discharge - 11 m3/sec
Side slope(m) - 1.50
Bottom slope(S) - 0.001
Manning Roughness Coefficient(N) - 0.022
Bottom width(b) - 1 m
We know,
Q = (A/N) R2/3 S1/2
where, Q = Discharge in channel
R = Hydraulic depth
R = A/P
Area of trapezoidal channel(A) = bd + md2
Wetted perimeter (P) for trapezoidal channel = b + 2d
Therefore ,R = bd + md2 / ( b + 2d)
Putting values,
R = d + 1.5d2 / (1 + 3.60d)
Putting values in Manning Equation,
Q = [(d + 1.5d2 ) / 0.022] (d + 1.5d2/ 1+3.6d )2/3 S1/2
or 11 = [d +1.5d2 /0.022] (d + 1.5d2 / 1+3.6d)2/3 (0.001)1/2
7.65 = (d + 1.5d2)5/3 / (1 + 3.6d)2/3
On solving above equation by hit and trial method, we get depth(d) = 2 m
Therefore ,Area = 8 m2
Wetted perimeter = 8.2m
Hydraulic radius (R) = 0.975 m
We know, V = Q/A
V = 11/8
V =1.375 m/s
(a) Froude Number(Fr) = V/(gA/T)0.5
where V = Velocity of flow
g = acceleration due to gravity
T = top width of channel ( b+2md)
Fr = 1.375 /(9.81x8/7)0.5
Fr = 0.41
Since, Fr<1, flow is subcritical ,therefore , the probability of erosion is low and channel is safe from erosion
(b) If slope is increased , then
1) Depth of excavation will increase which lead to more cost
2) Since depth increases, area also increases and velocity deceases which will reduce Froude number and hence risk of erosion also decreases
(c) Normal Depth calculated above as d = 2 m