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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Textbook Question
Chapter 3, Problem 81P

From Prob. 3-80, knowing that the density of the suspension depends on liquid depth and changes linearly from 800 kg m to 900 kg/m3 in the vertical direction, determine the resultant force acting on the gate ABC, and its line of action.

Expert Solution & Answer
Check Mark
To determine

Resultant force acting on gate and its line of action.

Answer to Problem 81P

Resultant force acting perpendicular to the gate is 144.129455kN.

Explanation of Solution

Given:

Density of liquid varies from 800kg/m3 to 900kg/m3 and the gate is parabolic.

Draw the diagram for the cross-section of the gate.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 81P , additional homework tip  1

Figure (1)

Draw the diagram for the variation of density with variation of depth.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 81P , additional homework tip  2

Figure (2)

Write the expression for the curve.

  x=y2   ...... (I)

Here, horizontal axis is denoted by the x and vertical axis is denoted by the y.

Write the expression for the density as a function of depth.

  ρ=(ρ2ρ1h2h1)h+ρ1   ...... (II)

Here, density is ρ2 at height of h2 and density is ρ1 at height of h1.

The density at a particular height is ρ.

Write the expression for the area of the elemental strip.

  dA=2xdy   ...... (III)

Here, area is dA.

Substitute y2 for x in Equation (III).

  dA=2ydy   ...... (IV)

Write the expression for the resultant force.

  dF=ρghdA   ...... (V)

Here, the resultant force is F.

Write the expression for the vertical depth of the centre of gravity of gate from the free surface.

  h=5ysin60°   ...... (VI)

Here, the vertical depth of the centre of gravity of gate from the free surface is h.

Calculation:

Substitute 800kg/m3 for ρ1, 900kg/m3 for ρ2 in Equation (II).

  ρ=( 900 kg/ m 3 800 kg/ m 3 5m0m)h+800kg/m3=( 1005kg/ m 2)h+800kg/m3=(20kg/ m 4)h+800kg/m3

Substitute (20kg/m4)h+800kg/m3 for ρ and Equation (IV) in Equation (V).

  dF=[(20kg/m4)h+800kg/m3]gh(2xdy)

Substitute Equation (VI) in above Equation.

  dF=[(20 kg/ m 4 )(5mysin60°)+800kg/m3]g(5mysin60°)(2xdy)=[900kg/m3(20 kg/ m 4 )y]g(5mysin60°)2xdy=[4500 kg/ m 2 ( 900 kg/ m 3 )ysin60°( 100 kg/ m 3 )ysin60°+20 kg/ m 4 ( y 2 ( sin60 ) 2 )]g2ydy=[[ ( 4500 kg/ m 2 ) y 1000 kg/ m 3 y 3/2 sin60° +( 20 kg/ m 4 ) y 5/2 sin 2 60°]g2]

Integrate both sides of the above Equation.

  F=[2g03[ ( 4500 kg/ m 2 ) y 1000 kg/ m 3 y 3/2 sin60° +( 20 kg/ m 4 ) y 5/2 sin 2 60° ]]=[2g[( 4500 kg/ m 2 ) 2 3 y 3/2 1000 kg/ m 3 ( 5 2 y 5/2 )sin60°+( 20 kg/ m 4 )( 7 2 y 7/2 ) sin 260°]]=2g[{3000 kg/ m 2 ×( 3 3/2 m)}{200 3 kg/ m 3 × 3 5/2 m}+{ 30 kg/ m 4 7×( 3 7/2 )m}]

Substitute 9.81m/s2 for g in above Equation.

  F=1.414×9.81m/s2[(15588.45725400+200.423022)kg]=144129.455kgm/s2=144129.455N

  F=144129.455N( 1kN 1000N)=144.129455kN

Conclusion:

Resultant force acting on gate is 144.129455kN

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Chapter 3 Solutions

Fluid Mechanics: Fundamentals and Applications

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