Answered: A gravelly rectangular channel, 7 m…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

A gravelly rectangular channel, 7 m wide and 2 m deep, is
flowing at 75 m³/s on a slope of 0.013. (a) Is this on a
mild, critical, or steep curve? (b) Approximately how
many meters downstream will the gradually varied solution
reach the normal depth?

Expert Solution

Step 1

Given data:

The width of the rectangular channel is b = 7 m.
The depth of rectangular channel is y = 2 m.
The flowing rate is Q = 75 m³/s.
The slop of flow is s = 0.013.

Step 2

(a)

The expression for the hydraulic radius is given by,

$R = \frac{A}{b + 2 y} R = \frac{b y}{b + 2 y}$

Here, A is the area of the channel.

Substitute all the known values in above equation,

$R = \frac{7 m \times 2 m}{7 m + 2 \times 2 m} R = 1.272 m$

The expression for the critical slop is given by,

$s_{c} = \frac{n g A}{α^{2} b R^{(\frac{4}{3})}} s_{c} = \frac{n g b y}{α^{2} b R^{(\frac{4}{3})}}$

Here, n is the manning factor and its standard value 0.025 for gravelly rectangular channel, g is the acceleration due to gravity and its standard value is 9.81 m/s², and α is the correction factor and its standard value is 1.486 for gravelly rectangular channel.

Substitute all the known values in above equation,

$s_{c} = \frac{(0.025) (9 . . 81 m / s^{2}) (7 m \times 2 m)}{{(1.486)}^{2} \times (7 m) \times {(1.272 m)}^{(\frac{4}{3})}} s_{c} = 0.16$

Here, the critical slop is more than the original slop (s). Thus, the channel is on mild curve.

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