Problem 5 (15%). The bed of a corrugated channel, with Manning's coefficient n = 0.019 s/m^1/3 d

1. First, we need to find the flow rate in the stream without the divider. We can use the Mannin

Problem 5 (15%). The bed of a corrugated channel, with...

Problem 5 (15%). The bed of a corrugated channel, with Manning's coefficient n = 0.019 s/m^1/3 drops 25 m every 1500 m. The channel has a semi-circular cross section with total width of W = 10 m (i.e. radius of the semi-circle is 5 m). 1. Given that the stream has a water depth of h = 2.5 m, find the flow rate in the stream. [10%] 2. A concrete divider with the same Manning's coefficient was installed at the middle of the stream. Given that the total flow rate in the stream stays the same, would the water depth in the stream change as a result of the divider? If so, would it increase or decrease? Would the stream velocity change? If so, would it increase or decrease? You may assume the thickness of the divider to be negligible. [5%] Figure 5: (a) Original channel cross section. (b) Cross section of channel with divider installed. Figure not to scale.

Transcript

00:01 In this question we have been given the bed of a corrugated channel with manning's coefficient is given to be 0 .019 second per cubic meter drops 25 meter every 1500 meter.

00:19 Okay, so this drop is 25 meter per 1500 meter.

00:28 The channel has a semicircular cross -section with total width.

00:31 So width is given to be 10 meter.

00:33 We need to find out what is the given that the stream of the water has a depth of 2 .5 meter.

00:41 We have to find the flow rate in the stream.

00:45 So flow rate is what we need to calculate.

00:50 So let us see how we are going to do this.

00:55 So for this let us use a formula beds drop 25 meter in every 1500 meter.

01:02 So i will now calculate here the bed slope.

01:06 Okay, so bed slope will be nothing but 25 over 1500 which is 1 over 60 and the radius r it is equals to 5 meter that is the radius of the semicircle.

01:22 Okay, now to find out the flow rate let us first draw circle semicircle rather and this is the center.

01:30 So i will drop one line from center to the curve c.

01:36 Okay, and then we can draw these two lines as well.

01:40 Let us join these two to form a triangle.

01:43 This angle i'm calling it as theta, this angle i'm calling it as theta and these sides are the ratio sorry r and here this value is equal to h.

01:53 So this will be r minus h.

01:57 So from this what i can say cosine of theta if i apply that will be r minus h upon r base upon hypotenuse.

02:06 Now area of circular arc i will write here the area of circular arc with angle 2 theta at the center it is given by pi r square over 2 times pi sorry this is pi here.

02:38 Okay, and this will be 2 times of theta.

02:41 So this comes out to be r square of theta.

02:44 Now we will find out what is the value area of upper triangle.

02:48 So the area of the upper triangle it is given by 2 times half of this which is given by r times r minus h sin of theta.

03:08 So this is equals to r times r minus h sin of theta.

03:15 Now what we can do hence for the area we get a is equals to r square theta r square theta which is equals to and minus with r times of r minus h sin theta that is the area we subtracted from the area of the upper triangle.

03:46 So this is what we are getting.

03:47 So i would say that my area will be r square theta minus r times r minus h sin of theta...

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