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 54P
To determine

The flow rate through the channel and the equivalent manning's coefficient.

Expert Solution & Answer
Check Mark

Answer to Problem 54P

The flow rate is 116.42m3/s and the equivalent manning's coefficient is 0.0217.

Explanation of Solution

Given Information:

The figure below shows the cross -section of the channel.

  Fluid Mechanics: Fundamentals and Applications, Chapter 13, Problem 54P

  Figure-(1)

The bottom slope of the channel is 0.002.

Write the expression for flow rate.

  V˙=an1Ac1R123So12+an2Ac2R223So12...... (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 R1 and the bottom slope is So, the area of cross section of channel 2 is Ac2, the hydraulic radius of channel 1 is R2.

Write the expression for area of channel 1.

  Ac1=Aa+Ab+Ac+Ad...... (II)

Here, the area of section a is Aa, the area of section b is Ab, the area of section c is Ac and the area of section d is Ad.

Write the expression to calculate the area of channel a.

  Aa=ba×ha...... (III)

Here, the breadth of section a is ba and the height of the section a is ha.

Write the expression to calculate the area of channel b.

  Ab=12(bb×hb)...... (IV)

Here, the breadth of section b is bb and the height of the section b is hb.

Write the expression to calculate the area of channel c.

  Ac=bc×hc...... (V)

Here, the breadth of section c is bc and the height of the section c is hc.

Write the expression to calculate the area of channel d.

  Ad=12(bd×hd)...... (VI)

Here, the breadth of section d is bd and the height of the section d is hd.

Write the expression to calculate the perimeter of the concrete channel.

  p1=2(ha+ ( h b )2+ ( h a )2)...... (VII)

Write the expression to calculate the perimeter of the light brush channel.

  p2=b+ha...... (VIII)

Here, the breadth of the light brush channel is b.

Write the expression to calculate the total perimeter.

  p=p1+p2...... (IX)

Write the expression to calculate the equivalent hydraulic radius.

  R=Ap...... (X)

Write the expression to calculate the area of light brush channel.

  Ac2=b×ha...... (XI)

Write the expression for total area.

  A=Ac1+Ac2...... (XII)

Write the expression for hydraulic radius for channel 1.

  R1=Ac1p1...... (XIII)

Write the expression for the hydraulic radius for channel 2.

  R2=Ac2p2...... (XIV)

Write the expression for equivalent flow rate.

  V˙=anA(R)23(S0)12...... (XV)

Here, the equivalent manning's coefficient is n.

Calculation:

Substitute 6m for ba and 1.5m for hb in Equation (III)

  Aa=6m×2m=12m2

Substitute 2m for bb and 1.5m for ha in Equation (IV)

  Ab=12(2m×1.5m)=1.5m2

Substitute 2m for bc and 1.5m for hc in Equation (V)

  Ac=2m×1.5m=3m2

Substitute 2m for bd and 1.5m for hd in Equation (VI)

  Ad=12(2m×1.5m)=1.5m2

Substitute 12m2 for Aa, 1.5m2 for Ab, 3m2 for Ac, 1.5m2 for Ad in Equation (II).

  Ac1=12m2+1.5m2+3m2+1.5m2=18m2

Substitute 10m for b and 2m for ha in Equation (XI)

  Ac2=10m×2m=20m2

Substitute 18m2 for Ac1 and 20m2 for Ac2 in Equation (XII)

  A=18m2+20m2=38m2

Substitute 2m for ha and 1.5m for hb in Equation (VII)

  p1=2(2m+ ( 2m ) 2 + ( 1.5m ) 2 )=2(2m+2.5m)=9m

Substitute 10m for b and 2m for ha in Equation (VIII)

  p2=10m+2m=12m

Substitute 9m for p1 and 12m for p2 in Equation (IX).

  p=9m+12m=21m

Substitute 21m for p and 38m2 for A in Equation (X).

  R=38m221m=1.81m

Substitute 18m2 for Ac1 and 9m for p1 in Equation (XIII).

  R1=18m29m2=2m

Substitute 20m2 for Ac2 and 12m for p2 in Equation (XIV).

  R2=20m212m=1.667m

Refer to the Table (13.1), "Mean values of Manning coefficient for water flow open channels" to obtain the value of n1

  0.014 and n2 as 0.050.

Substitute 0.014 for n1, 0.050 for n2, 1m13/s for a

  18m2 for Ac1, 20m2 for Ac2, 0.002 for So, 2m for R1, 1.667m in Equation (III)

  V˙=1 m 1 3 /s0.014×(18m2)(2m)23(0.002)12+1 m 1 3 /s0.050×(20m2)(1.667m)23(0.002)12=91.27m3/s+25.15m3/s=116.42m3/s

Substitute 1m13/s for a, 38m2 for A, 1.81m for R and 0.002 So.and 116.42m3/s for V˙ in Equation (XV).

  116.42m3/s=1 m 1 3 /sn(38m2)(1.81m)23(0.002)n=1 m 1 3 /s116.42 m 3/s(38m2)(1.81m)23(0.002)n=0.0217

Conclusion:

The flow rate is 116.42m3/s and the equivalent manning's coefficient is 0.0217.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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