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 106EP
To determine

The flow rate per unit width.

The Froude number of downstream gate.

Expert Solution & Answer
Check Mark

Answer to Problem 106EP

The flow rate per unit width is 10.9ft3/s.

The Froude number of downstream gate is 3.7140.

Explanation of Solution

Given Information:

The flow depth of upstream weir is 5ft and the height of the gate opening is 1.1ft.

Write the expression for the depth ratio.

  R=y1a...... (I)

Here, the flow depth of upstream weir is y1 and the height of the gate opening is a.

Write the expression for the volume flow rate.

  Q˙=Cdab2gy1...... (II)

Here, the breath of the channel is b and the acceleration due to gravity is g.

Write the expression for the upstream specific energy.

  Ec1=y1+Q˙22g( b y 1 )2....... (III)

Write the expression for the downstream specific energy.

  Ec2=y2+Q˙22g( b y 2 )2....... (IV)

Write the expression for the downstream velocity.

  V2=Q˙by2...... (V)

Write the expression for the Froude number of downstream.

  Fr2=V2gy2...... (VI)

Calculation:

Substitute 5ft for y1 and 1.1ft for a in Equation (I).

  R=5ft1.1ft=4.55

Refer to Figure 13.44 "Discharge coefficient for the drowned and free discharge from underflow gates" to obtain the coefficient of the discharge as 0.55 at depth ratio R=4.55.

Substitute 1ft for b, 1.1ft for a, 32.2ft/s2 for g, 0.55 for Cd and 5ft for y1 in equation (II).

  Q˙=(0.55)(1.1ft)(1ft)2( 32.2 ft/ s 2 )( 5ft)=(0.55)(1.1ft)(1ft)( 322 ft 2 / s 2 )=0.605ft2(17.94ft/s)10.9ft3/s

Substitute 1ft for b, 32.2ft/s2 for g, 10.9ft3/s for Q˙ and 5ft for y1 in Equation (III).

  Ec1=5ft+ ( 10.9 ft 3 /s )22( 32.2 ft/ s 2 ) ( 1ft×5ft )2=5ft+( 118.81 ft 6 / s 2 )( 1610 ft 5 / s 2 )=5ft+0.07379ft=5.074ft

Since, the specific energy of the fluid remains constant hence Ec2=5.074ft.

Substitute 1ft for b, 32.2ft/s2 for g, 10.9ft3/s for Q˙ and 5.074ft for Ec2 in Equation (IV).

  5.074ft=y2+ ( 10.9 ft 3 /s )22( 32.2 ft/ s 2 ) ( 1ft× y 2 )25.074ft=y2+( 118.81 ft 6 / s 2 )( 64.4 ft 5 / s 2 )y22(( 326.76 ft 5 / s 2 )y22)=(( 64.4 ft 5 / s 2 )y23)+118.81ft6/s2y2=0.643ft

Substitute 0.643ft for y2, 1ft for b and 10.9ft3/s for Q˙ in Equation (V).

  V2=10.9 ft 3/s( 0.643ft×1ft)=10.9 ft 3/s( 0.643 ft 2 )=16.9ft/s

Substitute 32.2ft/s2 for g, 16.9ft/s for V2 and 0.643ft for y2 in Equation (VI).

  Fr2=16.9ft/s ( 32.2 ft/ s 2 )( 0.643 )=16.9ft/s ( 20.7046 ft 2 / s 2 )=16.94.5502=3.7140

Conclusion:

The flow rate per unit width is 10.9ft3/s.

The Froude number of downstream gate is 3.7140.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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