For a channel to be in regime condition it should not have silting and scouring action. To obtain this silt entering channel must be carried away through channel, by channel section. The artificial channel to be in a regime it must have a fixed slope and fixed section. So, refore to achieve this flow in channel should uniform flow.

Given Q = 80 cumecs, S = 1/3000, m = 1.3, n = 0.021
Critical velocity (V_o) = 0.55my^0.64 (y = depth of channel)
= 0.55 x 1.3 x (2)^0.64
= 1.114 m/sec
Area (A) = Q/V_o = 80 / 1.114 = 71.8 m²
Now assume side slope as (1/2): 1(1/2H: V)
Now, A = y (b + y x 1/2)
71.8 = 2(b + 2 x 1/2)
71.8 = 2(b + 1)
b = 34.9 m
Perimeter (P) = b + 2 x ?5/2y
= 34.9 + 4.47
= 39.37 m
Now R = A/P = 71.8 / 39.37 = 1.82 m
V = ((1/n + (23 + 0.00155/s)) / (1 + (23 + 0.00155/s)n/\(\sqrt{RR}\)))\(\sqrt{RSRS}\)
V = 0.78 m/sec
Now calculating for critical velocity,
V_o = 0.55 x 1.3 x y^0.64
= 0.715 x y^0.64
Now assume y = 2.5 m
V_o = 0.715 x (2.5)^0.64 = 1.28 m/sec
A = 80 / 1.28 = 62.5 m²
62.5 = 2.5 (b + 1/2(2.5))
b = 23.75 m
P = 23.75 + 2 x ?5/2(2.5) = 29.35 m
R = A/P = 2.13
?R = 1.46
V = 75.27/1.4 x (1.46/54.8) = 1.432 m/sec
V > V₀
Now assume y = 3.2 m
V_o = 0.715 x (3.2)^0.64 = 1.51 m/sec
A = 80 / 1.5 = 53.33 m²
53.33 = 3.2 (b + 1.6)
b= 15.1 m
P = 15.1 + ?5(3.2)
P = 22.2 m
R = A/P = 53.33/22.2 = 2.4 m
?R = 1.56
V = (75.27/1.4) x (1.56/54.8)
= 1.53 m/sec
Therefore V = V_o
Hence use depth equal to 3.2 m and base width 15.1 m with slope 1/2:1.

The unit tractive force causing bed load to move is reduced by ripples in ratio of
S₀?= S₀ (n?/n) ^3/2 = 3.2 (0.0108/0.0222)^3/2 = 1.08 N/m².
The effective unit tractive force = S₀? ? S_c = 1.08 ? 0.5 = 0.58 N/m².

Shield was first person who designed non-scouring channels by providing semi-oretical analysis. He defined that bed particles need a drag force greater than or equal to resistance offered by particle for its movement.

Berms provide a path for inspection. Excessive loss of water due to seepage is checked. It is so because berms are formed of fine silt which makes berms fairly impervious.

The critical tractive stress is given by (d < 6 mm) ?
T_c = 0.155 + [0.409 d₂/(1 + 0.177 d₂)^0.5]
= 0.155 + [0.409 x 22/(1 + 0.177 x 22)^0.5] = 1.40 N/m².
The average shear stress = Y_w. RS = 9.81 x 0.8 x 1/3000 = 2.616 N/m²
Ratio = 1.40/2.616 = 0.53.

The main purpose of banks is to retain water in canal itself. This retaining action is always needed, in cases of low availability of water or during period of floods. The or purposes of banks include a means of communication and inspection paths.

Generally in nature, any two moving bodies oppose each or due to internal presence of friction between m. Here as sediment load moves along bed of channel, n internal friction is developed between m depending upon soil. Therefore to overcome this friction a minimum value of shear stress is needed. This minimum value is called critical shear stress (?_c).

Due to unusual high velocity in flooded streams, re is heavy sediment transportation and deposition downstream. The increased suspended load may be visible giving stream a muddy color.

Sometimes bed load is estimated to be in between 3 to 25% of total suspended load, depending upon reflexes of sediment to forces from sectors like physical, chemical, and biological. In between this percentage we commonly take bed load is 10% of total suspended load.