Question

The sub division indicated in figure is comprised of five drainage areas that drain to two ditches. the runoff from each drainage area is is in the figure. Ditch #1 conveys the runoff from Area 1 and 2, and Ditch 2 conveys the runoff from Area 1 to 5. Ditch 1 ( from points A to B) will have a triangular cross section with 2:1 side slopes, a length of 500m, a longitudinal slope of 0.5% and will be covered with grass (Manning's n =0.045). Ditch 2 from points B to C will have a trapezoidal cross section with 3:1 side slopes, a 5m wide bottom, length of 400m, a longitudinal slope of 0.3% and will be eart partially obstructed with debris or weed (Manning's n =0.050)
Determine the design height of Ditchs 1 and 2 to convey the flows from the five drainage areas

Triangular Ditch #1 -Detail Area 1 Area2 Q= 1 m2ls Q=Q.75 mls Area 3 Q=1.2m1s Trhangular Ditch #1 Trapezoldal Ditch #2-Detail Tra Di Area 5 Area 4 Q=1 m/s Q=1,45 m21s 5m Drainage areas and design details of Ditches #1 and #2.

Answer 1

Hi,

Please find the solution in attached images.

I have used manning's equation to calculate the depth of Ditches.

Free board for ditch 1 has taken as 0.30m and ditch 2 as 0.50m.

We will solve the problem by Manning's formula which is as below Discharge from an open channel Q (R2(s/2 )/n Where, Hydraulic Radius= Area of flow in channel (A)/Wetted perimeter of channel (p) R S Hydraulic Gradient of the channel Design of Ditch 1 4d Q Discharge of area 1 Discharge of area 2 0.75 +1 - 1.75m3/s S 0.5% = 0.005 dvs n 0.045 Let, the depth of flow in ditch 1 d Since the side slope of ditch 1 is 2:1 (H:V), the area of flow in channel shall be as below: Area A x 4d x d 2d d5+dvs- 2d15 Wetted perimeter p Hydraulic radius R = A/p 2d' /2ds d/15 Now put the above values in manning's equation Q (R2s12/n 1.75 (d/v5) x (0.005V2)/0.045 (d/Ns)2/ 1.75 x 0.045/0.0052 - 1.1137 d (1.1137)x v5 = 2.63 meter Allow a 0.30m free board, Total Depth of Ditch 1 2.93m

Design of Ditch 2 Q Discharge of area 1-5 0.75 + 1 + 1.2 +1.45+ 1 = 5,40m/s S 0.3%= 0.003 n 0.050 Now put the above values in manning's equation, Q (R2/s/2/n 5.40 (R)0.0031//0.05 R2/35.40 x0.05/ (0.003)12- 4.930 R (4.930)32= 10.946 3d 3d Now R A/p Sm Bottom width of ditch 5m Let, the depth of flow in ditch 2 d Since the side slope of ditch 1 is 3:1 (H:V), the area of flow in channel shall be as below x (5+5+3d +3d ) x d 3d2 5d Area A dv10 +5+dv10- 2dv10+5 Wetted perimeter p A/p (3d+5 d)/(2dv10+5) Hydraulic radius R Now put the above values in manning's equation, 10.946 (3d2+5d) / (2d10+5) 3d2 64.228d-54.73 0 d 22.23m Allow a 0.50m free board, Total Depth of Ditch 2 22.23m

In case of any query you can leave comment.

Warm Regards

Saras Jain