Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 13, Problem 50P
To determine

(a)

The dimension for the circular channel.

Expert Solution
Check Mark

Answer to Problem 50P

The diameter of the circular channel is 3.446m.

The depth of the circular channel is 1.732m.

Explanation of Solution

Given Information:

The water flow rate is 10m3/s and the bottom slope is 0.0015

Write the expression for the volume flow rate of the water.

  Q=anAcRh2/3So1/2  ...... (I)

Here, the manning coefficient is n, the constant is a, the hydraulic radius is Rh, the bottom slope is So and the cross sectional flow area is Ac.

Write the expression for the radius of the semicircle.

  Rh=AcP...... (II)

Here, the diameter of the semicircle is D.

Write the expression for the area of the circular channel.

  Ac=π8D2...... (III)

Here, the diameter of the circle is D.

Write the expression for the perimeter of the circular channel.

  P=π2D...... (IV)

Substitute π8D2 for Ac and π2D for P in Equation (II).

  Rh=π8D2π2D=D4...... (V)

Write the expression for the flow depth of the circular channel.

  y=D2...... (VI)

Calculation:

Refer to table 13-1 "Mean values of the manning coefficient n for water flow in open channels" to obtain the manning coefficient as 0.016.

Substitute D4 for Rh, 1m1/3/s for a, 0.016 for n, π8D2 for Ac, 0.0015 for So and 10m3/s for Q in Equation (I).

  10m3/s=( 1 m 1/3 /s )0.016(π8D2)( D 4)2/3(0.0015)1/210m3/s=62.5m1/3/s(0.3926D2)(0.3967D 2/3 )(0.03872)10m3/s=(0.369 m 1/3 /s)D8/3D=3.446m

Substitute 3.446m for D in Equation (VI).

  y=3.446m2=1.732m

Conclusion:

The diameter of the circular channel is 3.446m.

The depth of the circular channel is 1.732m.

To determine

(b)

The dimension for the rectangular channel.

Expert Solution
Check Mark

Answer to Problem 50P

The dimension for the rectangular channel are-

The bottom breadth of the rectangular channel is 3.124m.

The depth of the rectangular channel is 1.562m.

Explanation of Solution

Given Information:

The water flow rate is 10m3/s and the bottom slope is 0.0015

Write the expression for the area of the rectangular channel.

  Ac=yb...... (VII)

Here, the depth of the rectangular is y and the bottom breath of the rectangular channel is b.

Write the expression for the flow depth of the rectangular channel.

  y=b2...... (VIII)

Substitute b2 for y in Equation (VII).

  Ac=b(b2)=b22...... (IX)

Write the expression for the perimeter of the rectangular channel.

  P=b+2y...... (X)

Substitute b2 for y in Equation (X).

  P=b+2(b2)=2b...... (XI)

Substitute b22 for Ac and 2b for P in Equation (II).

  Rh= b 2 22b=b4...... (XII)

Calculation:

Substitute b4 for Rh, 1m1/3/s for a, 0.016 for n, b22 for Ac, 0.0015 for So and 10m3/s for Q in Equation (I).

  10m3/s=( 1 m 1/3 /s )0.016( b 2 2)( b 4)2/3(0.0015)1/210m3/s=62.5m1/3/s(0.5b2)(0.3967b 2/3 )(0.03872)20.864=b8/3b=3.124m

Substitute 3.124m for b in Equation (VIII).

  y=3.124m2=1.562m

Conclusion:

The bottom breath of the rectangular channel is 3.124m.

The depth of the rectangular channel is 1.562m.

To determine

(c)

The dimension for the trapezoidal channel.

Expert Solution
Check Mark

Answer to Problem 50P

The bottom breath of the trapezoidal channel is 1.902m.

The depth of the trapezoidal channel is 1.647m.

Explanation of Solution

Given Information:

The water flow rate is 10m3/s and the bottom slope is 0.0015

Write the expression for the area of the trapezoidal channel.

  Ac=y(b+ytanθ)...... (XIII)

Here, the trapezoidal angle is θ, the depth of the trapezoidal is y and the bottom breath of the trapezoidal channel is b.

Write the expression for the flow depth of the trapezoidal channel.

  y=32b...... (XIV)

Write the expression for the inclined height of the trapezoidal section.

  c=ysinθ...... (XV)

Write the expression for the perimeter of the trapezoidal channel.

  P=b+2c...... (XVI)

Calculation:

Substitute 60° for θ and 32b for y in Equation (XV).

  c= 3 2bsin60°= 3 2b 3 2=b

Substitute 60° for θ and 32b for y in Equation (XIII).

  Ac=32b(b+ 3 2 b tan60°)=32b(b+ 3 2 b 3 )=334b2

Substitute b for c in Equation (XVI).

  P=b+2b=3b

Substitute 3b for P and 334b2 for

  Ac in Equation (II).

  Rh= 3 3 4b23b=34b

Substitute 34b for Rh, 1m1/3/s for a, 0.016 for n, 334b2 for Ac, 0.0015 for So and 10m3/s for Q in Equation (I).

  10m3/s=( 1 m 1/3 /s )0.016( 3 3 4b2)( 3 4b)2/3(0.0015)1/210m3/s=62.5m1/3/s(1.299b2)(0.5723b 2/3 )(0.03872)5.55=b8/3b=1.902m

Substitute 1.902m for b in Equation (XIV).

  y=32(1.902m)=0.8660(1.902m)=1.647m

Conclusion:

The bottom breath of the trapezoidal channel is 1.902m.

The depth of the trapezoidal channel is 1.647m.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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