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 53P
To determine

The diameter of the third channel.

Expert Solution & Answer
Check Mark

Answer to Problem 53P

The diameter of the channel 3 is 2.33m.

Explanation of Solution

Given Information:

The figure below represents the cross -section of three channels.

  Fluid Mechanics: Fundamentals and Applications, Chapter 13, Problem 53P

  Figure-(1)

The diameter of channel 1 is 1.8m, the diameter of channel 2 is 1.8m and the slope of the draining system is 0.0025.

Write the expression for flow rate from channel 1 and channel 2.

  V˙=an1Ac1Rh123So12+an2Ac2Rh223So12...... (I)

Here, the manning's coefficient for channel 1 is n1, the manning's coefficient for channel 2 is n2, the area of cross section of channel 1 is Ac1, the hydraulic radius of channel 1 is Rh1 and the bottom slope is So, the area of cross section of channel 2 is Ac2, the hydraulic radius of channel 1 is Rh2.

Write the expression for flow rate through the channel 3.

  V˙=an3Ac3(R c3)23(So)12...... (II)

Here, the manning's coefficient for channel 3 is n3, the hydraulic radius for channel 3 is Rc3, the area of the channel 3 is Ac3.

Write the expression for the cross sectional flow area of channel1.

  Ac1=πD128...... (III)

Here, the radius of the channel 1 is D1.

Write the expression for the cross sectional flow area of channel 2.

  Ac2=πD228...... (IV)

Here, the radius of the channel 2 is D2.

Write the expression for the cross sectional flow area of channel 3.

  Ac3=πD328...... (V)

Here, the radius of the channel 3 is D3.

Write the expression to calculate the wetted perimeter of the channel 1.

  

  p1=πD12...... (VI)

Write the expression to calculate the wetted perimeter of the channel 2.

  p2=πD22...... (VII)

Write the expression to calculate the wetted perimeter of the channel 3.

  p3=πD32...... (VIII)

Write the expression to calculate the hydraulic radius of channel 1.

  Rh1=Ac1p1...... (IX)

Write the expression to calculate the hydraulic radius of channel 2.

  Rh2=Ac2p2...... (X)

Write the expression to calculate the hydraulic radius of channel 3.

  Rh3=Ac3p3...... (XI)

Write the expression to calculate the diameter of the channel 3.

  D3=2R3...... (XII)

Calculation:

Substitute 1.8m for D1 in Equation (III).

  Ac1=π ( 1.8m )28=10.17m28=1.272m2

Substitute 1.8m for D2 in Equation (IV).

  Ac2=π ( 1.8m )28=10.17m28=1.272m2

Substitute 1.8m for D3 in Equation (VI).

  p1=π( 1.8m2)=2.82m

Substitute 0.9m for R2 in Equation (VII).

  p1=π( 1.8m2)=2.82m

Substitute 1.272m2 for Ac1 and 2.82m for p1 in Equation (IX).

  Rh1=1.272m22.82m=0.45m

Substitute 1.272m2 for Ac2 and 2.82m for p2 in Equation (X).

  Rh2=1.272m22.82m=0.45m

Substitute πR322 for Ac3 and πR3 for p3 in Equation (XI).

  Rh3= π R 3 2 2πR3=R32

Refer to Table (13.1), "Mean values of the Manning Coefficient for water flow in open channels" to obtain the value of n1 as 0.012 and n2 as 0.012.

Substitute 1m13/s for a, 1.272m2 for Ac1, 1.272m2 for Ac2, 0.012 for n1, 0.012 for n2, 0.45m for Rh1

  0.45m for Rh2 and 0.0025 for So in Equation (I).

  V˙=[ 1 m 1 3 /s 0.012×1.272 m 2× ( 0.45m ) 2 3 ( 0.0025 ) 1 2 + 1 m 1 3 /s 0.012×1.272 m 2× ( 0.45m ) 2 3 ( 0.0025 ) 1 2 ]=2( 1 m 1 3 /s 0.012×1.272m2× ( 0.45m ) 2 3 ( 0.0025 ) 1 2 )=6.22m3/s

Substitute 6.22m3/s for V˙, 1m13/s for a, 0.012 for n, R32 for Rh3, πR322 for Ac3 and 0.0025 for So in Equation(II).

  6.22m3/s=1 m 1 3 /s0.012( π R 3 2 2)( R 3 2)23(0.0025)126.22m3/s=(83.33×0.5π× ( 0.5 ) 2 3 ×0.005)R383R3=1.166m

Substitute 1.166m for R3 in Equation (XII).

  D3=2×1.166m=2.33m

Conclusion:

The diameter of the channel 3 is 2.33m.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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