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 129P
To determine

The flow rate through the channel.

The effective manning coefficient.

Expert Solution & Answer
Check Mark

Answer to Problem 129P

The flow rate through the channel is 37.2m3/s.

The effective manning coefficient is 0.0386.

Explanation of Solution

Given information:

The channel angle is 0.5°.

Write the expression for flow rate using manning's equation.

  V˙=anAcRh2/3tanα...... (I)

Here, dimensional constant is a, area is Ac, manning's coefficient is n, hydraulic radius is Rh,channel angle is α.

Write the expression for net volume flow rate.

  V˙=anAc1Rh12/3tanα+anAc2Rh22/3tanα...... (II)

Here, hydraulic radius for first channel is Rh1, hydraulic radius for second channel is Rh2, area of first channel is Ac1 and area of first channel is Ac2.

Write the expression for area of rectangular section.

  Ac=l×b...... (III)

Here, length is l and width is b.

Write the expression for total area.

  Ac=Ac1+Ac2...... (IV)

Write the expression for wetted perimeter.

  p=l+b+d...... (V)

Here, depth is d.

Write the expression for total wetted perimeter.

  p=p1+p2...... (VI)

Write the expression for hydraulic radius.

  Rh=Acp...... (VII)

Write the expression for equivalent hydraulic radius.

  Rh=Acp

Calculation:

Substitute 3m for l, 2m for b and Ac1 for A in Equation (III)

  Ac1=3m×2m=6m2

Substitute 10m for l, 1m for b and Ac2 for A in Equation (III)

  Ac2=10m×1m=10m2

Substitute 6m2 for Ac1 and 10m2 for Ac2 in Equation (IV)

  Ac=6m2+10m2=16m2

Substitute 3m for l, 2m for b, p1 for p and 1m for d in Equation (V).

  p1=3m+2m+1m=6m

Substitute 10m for l, 1m for b and p2 for p in Equation (V).

  p2=10m+1m=11m

Substitute 6m for p1 and 11m for p2 in Equation (VI).

  p=6m+11m=17m

Substitute 10m2 for Ac2, 11m for p2 and Rh2 for Rh in Equation (VII).

  Rh2=10m211m=0.909m

Substitute 6m2 for Ac1, 6m for p1 and Rh1 for Rh in Equation (VII).

  Rh1=6m26m=1m

Substitute 16m2 for Ac, 17m for p and Rh for Rh in Equation (VII).

  Rh1=16m217m=0.94m

Substitute 6m2 for Ac1, 10m2 for Ac2, 1m for Rh1, 0.909m for Rh2 1m1/3/s for a, 0.022 for n and 0.5° for α in Equation (II).

  V˙=antanα(A c1R h1 2/3 +A c2R h2 2/3 )=1 m 1/3 /s0.022tan0.5°(( 6 m 2 ) ( 1m ) 2/3 +( 10 m 2 )( 0.909m))=0.396675m1/3/s(( 6 m 2 ) ( 1m ) 2/3 +( 10 m 2 )( 0.909m))=37.2m3/s

Substitute 16m2 for Ac, 11m for p2, 0.941m for Rh1m1/3/s for a, n for n, 37.2m3/s for V˙ and 0.5° for α in Equation (I).

  37.2m3/s=1 m 1/3 /sn(16m2)(0.941m)2/3tan0.5°n=( 1 m 1/3 /s )( 16 m 2 ) ( 0.941m ) 2/3 tan0.5°37.2 m 3/sn=0.0386

Conclusion:

The flow rate through the channel is 37.2m3/s.

The effective manning coefficient is 0.0386.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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