A rectangular channel is 8 m wide, has bed slope of 0.0025, and Manning roughness of 0.020.DetermineThe flow rate in m3secm3sec if the normal depth is 4.1 m for this channel.The critical depth in m for this channel.Whether the channel slope is mild or steep.The depth of flow in the channel at the point where x = 115 m is 2.0 m. Complete the direct step method table below and find the x-positions of the next two depth values shown below. Show the equations used and demonstrate the calculations for one row across the table.y (m)A (m2)P (m)R (m)V (m/s)E (m)SeSeSebarSebar∆E (m)∆x (m)∑x∑x(m)2.0?? 1152.2 2.4

The shape of open flow channel = Rectangular Width, B = 8 m Bed slope, S = 0.0025 Manning's roughness, n = 0.020 To determine Flow rate for normal depth, y = 4.1 m Critical depth, yC Nature of the slope, mild/steep For x= 115 m depth of flow, y = 2.0 m and to determine values of x for depths of 2.2 m and 2.4 m 1) Area, A = By =8×4.1 =32.8 m2 Wetted perimeter, P = B +2y = 8+2×4.1= 16.2 m Hydraulic depth, R = AP = 32.816.2= 2.02 m Now, from Manning's formula, the flow rate is given as Q = 1nAR23S12=10.02×32.8×2.0223×0.002512=131.033 m3sec Thus, for a depth of 4.1 m, the flow rate is 131.033m3sec. 2) For critical depth, yc yc = q2g13=131.0334.129.8113=4.704 m Thus, the critical depth, yc = 4.704 m 3) Nature of the slope. As the critical depth is greater than the normal depth, the slope is specified a steep slope.Specific energy, E = y +v22g = y +Q22gB2y2 Change in specific energy with change in distance, ∆E∆x=So-Sf Where So is bed slope and Sf is water surface slope. Sf = QnAR232 As the depths of flow is less than the normal depth, it is a case of S3 profile Depth, y Area, A Wetted perimeter R E Sf ∆E = Ei+1-Ei ∆x 2.0 8×2 = 16 8+2×2 =12 1.33 2 +131.03322×9.81×82×225.418 131.033×0.0216×1.332320.0183 - - 2.2 17.6 12.4 1.42 5.025 0.0144 5.025-5.418= -0.393 -0.3930.0025-0.0183=24.873 2.4 19.2 12.8 1.5 4.774 0.0108 -0.251 30.24 If the depth in the channel is 2.0 m, at a distance x = 115 m And as it is a S3 profile Thus, For y = 2.2 m, x = 115+24.873 = 139.873 m For y = 2.4 m, x = 139.873+30.24 = 170.113 mFlow rate for normal depth, y = 4.1 m, Q = 131.033 m3/sec Critical depth, yC = 4.704 m Nature of the slope = steep For y = 2.2 m, x = 139.873 m and for y = 2.4 m, x = 170.113 m

Answered: A rectangular channel is 8 m wide, has…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A rectangular channel is 8 m wide, has bed slope of 0.0025, and Manning roughness of 0.020.

Determine

The flow rate in m3secm3sec if the normal depth is 4.1 m for this channel.
The critical depth in m for this channel.
Whether the channel slope is mild or steep.
The depth of flow in the channel at the point where x = 115 m is 2.0 m. Complete the direct step method table below and find the x-positions of the next two depth values shown below. Show the equations used and demonstrate the calculations for one row across the table.

y (m)	A (m2)	∑x∑x(m)
2.0	??	115
2.2
2.4

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