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 140P

A sluice gate with free outflow is used to control the discharge rate of water through a channel Determine the flow rate per unit width when the gate is raised to yield a gap of 50 cm and the upstream flow depth is measured to be 2.8 m. Also determine the flow depth and the velocity downstream.

Expert Solution & Answer
Check Mark
To determine

The flow rate per unit width.

The flow depth.

The velocity for downstream.

Answer to Problem 140P

The flow rate per unit width is 2.068m3/s.

The flow depth is 0.293m.

The velocity for downstream is 7.06m/s.

Explanation of Solution

Given Information:

The height for the sluice gate is 50cm and the flow depth for upstream is 2.8m.

Write the expression for the flow depth.

  Dr=y1a...... (I)

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

Write the expression for the discharge rate.

  Q˙=Cdba2gy1...... (II)

Here, the discharge coefficient is Cd, the breadth of the gate is b and acceleration due to gravity is g.

Write the expression for the cross- sectional area at upstream.

  Ac1=by1...... (III)

Write the expression for the cross -sectional area at downstream.

  Ac2=by2...... (IV)

Write the expression for the upstream velocity of the fluid.

  V1=Q˙Ac1...... (V)

Write the expression for the downstream velocity of the fluid.

  V2=Q˙Ac2...... (VI)

Write the expression for the upstream specific energy.

  Ec1=y1+V122g...... (VII)

Write the expression for the downstream specific energy.

  Ec2=y2+V222g...... (VIII)

Calculation:

Substitute 2.8m for y1 and 50cm for a in Equation (I).

  Dr=2.8m50cm=2.8m50cm( 1m 100cm )=5.6

Refer to Figure 13.44 "Discharge coefficient for the drowned and free discharge from underflow gates" to obtain the coefficient of the discharge as 0.56 at depth ratio R=5.6.

Substitute 0.56 for Cd, 1m for b, 0.5m for a, 9.81m/s2 for g and 2.8m for y1 in Equation (II).

  Q˙=(0.56)(1m)(0.5m)2( 9.81m/ s 2 )( 2.8m)=(0.28m2)2( 9.81m/ s 2 )( 2.8m)=(0.28m2)(7.38m/s)=2.068m3/s

Substitute 2.8m for y1 and 1m for b in Equation (III).

  Ac1=(2.8m)(1m)=2.8m2

Substitute 2.8m2 for Ac1 and 2.068m3/s for Q˙ in Equation (V).

  V1=2.068 m 3/s2.8m2=0.738m/s

Substitute 1m for b in Equation (IV).

  Ac2=(y2)(1m)=y2m...... (IX)

Substitute y2m for Ac2 and 2.068m3/s for Q˙ in Equation (VI).

  V2=2.068 m 3/sy2m=2.068 m 2/sy2

Substitute 2.8m for y1, 0.738m/s for V1 and 9.81m/s2 for g in Equation (VII).

  Ec1=2.8m+( ( 0.738m/s ) 2 2( 9.81m/ s 2 ))=2.8m+0.0277m=2.827m

Since, the specific energy at downstream is equal the specific stream at upward stream hence Ec2=2.827m.

Substitute 2.827m for Ec2, 2.068m2/sy2 for V2 and 9.81m/s2 for g in Equation (VIII).

  2.827m=y2+( ( 2.068m/s y 2 ) 2 2( 9.81m/ s 2 ))2.827m=( 19.62m/ s 2 )y2+( ( 4.276 m 2 / s 2 y 2 2 ))( 19.62m/ s 2 )y232.827y22+0.218=0y2=0.293m

Substitute 0.293m for y2 in Equation (IX).

  Ac2=(0.293m)m=0.293m2

Substitute 0.293m2 for Ac2 and 2.068m3/s for Q˙ in Equation (VI).

  V2=2.068 m 3/s0.293m27.06m/s

Conclusion:

The flow rate per unit width is 2.068m3/s.

The flow depth is 0.293m.

The velocity for downstream is 7.06m/s.

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Chapter 13 Solutions

Fluid Mechanics: Fundamentals and Applications

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