An engineer in collaboration with the NRCS and a watershed c

An engineer, in collaboration with the NRCS and a watershed conservancy district, has designed a flood alleviation scheme for a rural area. It is proposed to dredge the channel of a constructed ditch to achieve the design objectives to transmit the 1:20-year flood of 600 ft³/sec. As a recently graduated river engineer/scientist, you think that dredging may be unnecessary, and that the design objectives could be achieved by mowing vegetation on the sides of the channel, killing unwanted brush and weeds, and removing fallen trees from the river. You are provided the following information regarding the original channel;

                  Trapezoidal channel with 2:1 (horizontal to vertical) sidewalls

                  40 ft top width

                  5 ft maximum depth

                  0.002 bed slope

                  Current maximum discharge capacity of 300 ft³/sec

                  Banks are tree lined with overgrown shrubs

                  Gravel bed with fallen logs

                  Approximately 30 ft³, per linear foot of ditch, of sediment deposited in the ditch

a) Calculate the current roughness of the channel. Estimate the reduction in roughness needed to meet the objectives. Estimate the increase in conveyance capacity associated with only removing sediment and based on conducting all the proposed activities. Outline the long-term effects of dredging and straightening on channel stability. How can any adverse effects be minimized?

b) Develop a new two-stage channel design that has a narrow, 10-ft wide by 2-ft deep channel at the bottom of the ditch and then flat ‘benches’ on each side that have a total bench width at least two times the width of the narrow channel.   Model the flow in the channel using appropriate software.

Solution

the roughness is the resistance of the flood flows in the channel and it can be determined as: of the by manning\'s roughness formula that is:

V= (1.486/n) R^2/3S^1/2

where V = mean velocity of flow

R= hydraulic radius

S = bed slope

n= roughness coefficient

according to the roughness coefficient the surface roughness can be known from the standards.

so current roughness coefficient = (1.486/V) R^2/3S^1/2

V = discharge / area

area of trapazoid= (a+b)/2 * d = (40+20)/2*5 = 150 square ft.

dscharge= 300 cubic ft per sec

V = (300 cubic ft/sec )/ 150 square ft.= 2 ft/sec

now R= wetted area/ wetted perimeter

wetted area = full area when running full i.e 150 sq. ft.

wetted perimeter = 20+(under root 25+100) i.e 20+11.2 = 31.2ft.

R = 150/31.2 = 4.80

slope= 0.002

now n = (1.486/2)*4.8^0.6*0.002^0.5

= 0.743*2.86*0.044

= 0.0934

this indicates a very large amount of vegetation

the reduction in roughnes needed would be 0.08 as 0.01 wold be appropriate roghness coefficient for the canal.

conveyance can be calculated as K = (1.486/n)*A*R^2/3

current conveyance= (1.486/0.09)*150*2.86 = 7083 ft.

new conveyance or impoved conveyance = (1.486/0.01)*150*2.86 = 63749 ft.

that is how removing the unwanted vegetation long term effects cn be minimized.