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 61P

Repeat Prob. 13-60 for a weedy excavated earth channel with n = 0.030.

Expert Solution & Answer
Check Mark
To determine

The depth.

Answer to Problem 61P

The depth is 1.073m.

Explanation of Solution

Given information:

The channel is made of unfinished concrete, the bottom angle of trapezoidal channel is 1°, the base width is 5m, the side surface slop is 1:1, the flow rate is 25m3/s, the angle is 45° and the manning coefficient is 0.030.

Below figure represent the cross- section of the trapezoidal channel.

Fluid Mechanics: Fundamentals and Applications, Chapter 13, Problem 61P

Figure-1

Write the expression of height of water in the channel as shown in Figure-1.

  h=xtanθ........ (I)

Here, the parallel distance of water from below corner is x and the angle is θ.

Substitute 45° for θ in Equation (I).

  h=xtan45°=x×1=x

Write the expression of surface width of the channel as shown in Figure-1.

  c=(5m)+2x........ (II)

Here, the parallel distance of water from below corner is x.

Substitute x for h in Equation (II).

  c=(5m)+2h

Write the expression of bottom slop.

  S0=tanα........ (III)

Here, the slop angle or bottom angle is α.

Substitute 1° for α in Equation (III).

  S0=tan1°=0.0175

Write the expression of area of cross -section of flow.

  Ac=(b+c2)h........ (IV)

Here, the width is b, the surface width of channel is c and height of the channel is h.

Substitute 5m for b and (5m)+2h for c in Equation (IV).

  Ac=( ( 5m )+( 5m )+2h2)h=( 10m+2h2)h={(5m)+h}h=(5m)h+h2

Write the expression of wetted perimeter of the channel.

  p=hsinθ+b+hsinθ........ (V)

Here, the height of the channel is h, the width of the channel is b and the angle is θ.

Substitute 5m for b and 45° for θ in Equation (V).

  p=hsin45°+(5m)+hsin45°=h0.707+(5m)+h0.707=2h0.707+(5m)=2.828h+(5m)

Write the expression of hydraulic radius of the channel.

  Rh=Acp........ (VI)

Here, the cross-sectional area of flow is Ac and the perimeter of the channel is p.

Substitute (5m)h+h2 for Ac and 2.828h+(5m) for p in Equation (VI).

  Rh=(5m)h+h22.828h+(5m)

Write the expression of flow rate.

  V˙=anAcRh2/3S01/2........ (VII)

Here, the constant is a, the manning coefficient is n, the area of cross -section of flow is Ac, the hydraulic radius is Rh and the slop is S0.

Substitute 25m3/s for V˙, 1m1/3/s for a, 0.030 for n, (5m)h+h2 for Ac, (5m)h+h22.828h+(5m) for Rh and 0.0175 for S0 in Equation (VII).

  25m3/s=( 1 m 1/3 /s )0.030×{(5m)h+h2}{ ( 5m )h+ h 2 2.828h+( 5m )}2/3(0.0175)1/225m3/s=(33.33 m 1/3 /s)×(0.1322)×{( 5m)h+ h 2} { ( 5m )h+ h 2 } 2/3 { 2.828h+( 5m )} 2/3 25m3/s=(4.4 m 1/3 /s)× { ( 5m )h+ h 2 } 5/3 { 2.828h+( 5m )} 2/3 25 m 3/s( 4.4 m 1/3 /s )= { ( 5m )h+ h 2 } 5/3 { 2.828h+( 5m )} 2/3

  (5.68m 2.66)= { ( 5m )h+ h 2 } 5/3 { 2.828h+( 5m )} 2/3 (5.68m 2.66)×{2.828h+( 5m)}2/3={( 5m)h+h2}5/3{( 5m)h+h2}5/3(5.68m 2.66)×{2.828h+( 5m)}2/3=0........ (VIII)

Apply hit and trial method in Equation (VIII).

Substitute 1m for h in Equation (VIII).

  {( 5m)×( 1m)+ ( 1m )2}5/3(5.68m 2.66)×{2.828( 1m)+( 5m)}2/3=0(6 m 2)5/3(5.68m 2.66)×(7.828m)2/3=0(36m 3.32)(44.46m 3.32)=08.46m3.320

Hence 1m for h does not satisfy the Equation.

Substitute 1.073m for h in Equation (VIII).

  {( 5m)×( 1.073m)+ ( 1.073m )2}5/3(5.68m 2.66)×{2.828×( 1.073m)+( 5m)}2/3=0(6.52 m 2)5/3(5.68m 2.66)×(4m)2/3=0(22.74m 3.32)(22.74m 3.32)=00=0

Hence 1.073m for h satisfies the equation.

Conclusion:

The depth is 1.073m.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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