# channel design

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```					                                   Maximum rainfall for 1-5 consecutive days.
140

120

100
Precipitatoin (mm/day)

80

60

40

20

0
1            2              3            4       5
Time (Day)
Pump Design.
A drainage coefficient of 40cm was selected for this design.

Determining the gradient of the line using the points (0,40) and (4,125) “

Gradient = (y2 – y1) / (x2 – x1)

= (125 – 40) / (4)

= 21.25 mm

i.e. the pump must be able to remove 21.25mm of water from the area being drained per day.

Area being considered = 4 sq mile = 10.4 km2.

Volume to be removed per day = (21.25/ 1000)m x (10.4 x 106) m2

= 221000 m3

Design discharge for pump     = 221000m3 / (24*60*60) sec

= 2.56m3 / s

The pump must have a discharge of 2.6 m3/s in order to satisfy the maximum expected rainfall.
Channel Design.

Assuming a Manning’s n = 0.03 which is the accepted value for “weedy” earthen channels.

The area under consideration will be taken as 2 miles x 2 miles.

Primary Drains.

There will be 3 primary drains: a central drained designed to carry 1.5 m3/s and two perimeter
drains designed to carry 0.5m3/s each. The total drainage capacity will be 2.5m3/s.

The change in level from one end of the drain to the other will be taken as 1m.

Maximum channel velocity will be taken as 1.3 m/s to prevent erosion of channel banks.

Channel bed slope = 1m / 3.22 km       = 3.11 x 10 -4

Assuming a trapezoidal channel with 2m bed width, water depth of 1.1m and 1:1 side slopes:

Area of channel = 2(1.1) + 0.5(2)(1.1)(1.1)

= 3.41 m2

Wetted Perimeter = 2 + 2 √(1.12+1.12)

= 5.1m

R=A/P

= (3.41) / (5.1)

= 0.67m

Using Manning’s formula:

V = R2/3 S1/2 / n

V = (0.67) 2/3 (3.11 x 10 -4)1/2 / 0.03

V = 0.45 m/s

Discharge             Q = 0.45 x 3.41

Q = 1.53 m3 /s

The channel design is sufficient.
Perimeter Drains.

Perimeter drains are to carry 0.5 m3/s.

Assume a trapezoidal channel with 1m bed width, 1 m water depth and 1:1 side slopes.

Area    = 1(1) + 2(0.5) (1)(1)

= 2 m2

Wetted Perimeter       = 1 + 2 √2

= 3.83m

R       = 2 / 3.83

= 0.52m

Using Manning’s formula:

V = R2/3 S1/2 / n

V = (0.52) 2/3 (3.11 x 10 -4)1/2 / 0.03

V = 0.38 m/s

Discharge              Q = 0.38 x 2

Q = 0.76 m3 /s

The channel design is sufficient.
Secondary Drains.

Secondary drains will run perpendicular to the primary drains.

They will be located at 1km intervals on both sides of the central drain.

Each will be responsible for draining an area of 1km x 1.6km.

Using the previously found drainage requirement of 21.25mm/day:

Design Discharge       = (21.25/1000) * 1000 * 1600 /(24*3600)

= 0.4 m3/s

Assuming a trapezoidal channel with bed width 1m, water depth 0.8m and 1:1 side slopes:

Area    = 1(0.8) + (0.5*2)(0.8*0.8)

= 1.44m2

Wetted Perimeter       = (1) + 2√(0.82 + 0.82)

= 3.26m

R       = 1.44/3.26

= 0.44m

Using Manning’s formula:

V = R2/3 S1/2 / n

V = (0.44) 2/3 (3.11 x 10 -4)1/2 / 0.03

V = 0.34 m/s

Discharge              Q = 0.34 x 1.44

Q = 0.49 m3 /s

The channel design is sufficient.

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 views: 8 posted: 11/16/2010 language: English pages: 5