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 3, Problem 80P

An open settling tank shown in the figure contains a liquid suspension. Determine the resultant force acting on the gate and its line of action if the liquid density is 850 kg/m3. The gate is parabolic as sketched, looking straight at the gate.

Expert Solution & Answer
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To determine

The Resultant force acting on gate and its line of action.

Answer to Problem 80P

The Resultant force acting on gate is 140.524kN and its line of action of the hydrostatic force of liquid from the bottom is at 1.63m.

Explanation of Solution

Given:

Density of liquid is 850kg/m3, the gate is parabolic.

Draw the cross-sectional view of the gate.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 80P , additional homework tip  1

Figure (1)

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 vertical distance between the centre of the elemental axis and the x axis.

  y=y   ...... (II)

Here, the vertical distance between the centre of the elemental axis and the x axis is y.

Write the expression for the area of the elemental strip.

  dA=2xdy   ...... (III)

Here, area is dA.

Write the expression for the centre of gravity.

  CG=0 y maxydA0y maxdA   ...... (IV)

Here, centre of gravity is CG.

Substitute 3 for ymax, Equation (I), (II), and (III) in Equation (IV).

  CG=03y( 2xdy)032xdy   ...... (V)

Draw a view of gate and liquid.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 80P , additional homework tip  2

Figure (2)

Write the expression for the length of gate which is in contact with the liquid.

  l=zsin60°...... (VI)

Here, length of the gate is l and perpendicular height from free surface is z.

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

  hCG=ICG   ...... (VII)

Here, the mass moment of inertia is I and the centre of gravity of the gate from the free surface is hCG.

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

  h'=hCGsinβ   ...... (VIII)

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

Write the expression for pressure acing on the gate.

  Pa=ρgh'   ...... (IX)

Here, pressure acing on the gate is Pa.

Write the expression for the area of gate resisting the pressure on the gate.

  Agate=032xdy   ...... (X)

Here, the area of gate resisting the pressure on the gate is Agate.

Write the expression for the resultant force acting on the gate.

  FR=PaAgate   ...... (XI)

Here, the resultant force acting on the gate is FR.

Write the expression for the moment of inertia.

  I=03y2dA   ...... (XII)

Write the expression for the action of resultant hydrostatic force.

  yp=CG+ICGA   ...... (XIII)

Here, the action of resultant hydrostatic force is yp.

Write the expression for the centre of line of action from the free surface.

  hp=ypsin60°   ...... (XIV)

Here, the centre of line of action from the free surface is hp.

Write the expression for the centre of location of the force from the bottom.

  H=5mhp   ...... (XV)

Here, the centre of location of the force from the bottom is H.

Calculation:

Substitute y2 for x in Equation (V).

  CG= 0 3 y( 2 y 2 dy )032 y 2 dy=03( y y )dy03( y)dy=[( y 5 2 5/2 )( y 3/2 3/2 )]03

  =( 3 5y)03=95m=1.8m

Substitute 5m for z in Equation (VI).

  l=5msin60°=5m( 3 2 )=103m

Substitute 103m for l and 95m for CG in Equation (VII).

  hCG=( 10 3 m95m)=5.7735m1.8m=3.9735m

Substitute 3.9735m for hCG and 60° for β in Equation (VIII).

  h'=3.97m×sin60°=3.97m×0.866=3.44m

Substitute 850kg/m3 for ρ and 9.81m/s2 for g and 3.44m for h' in Equation (IX).

  Pa=850kg/m3×9.81m/s2×3.44m=838.5×3.44kg/ms2=28684.44N/m2

Substitute y2 for x in Equation (X).

  Agate=032 y 2 dy=203ydy=2[y 3 2 32]03=26m2

Substitute 28684.44N/m2 for Pa and 26m2 for Agate in Equation (XI).

  FR=28684.44N/m2×26m2=140524.483N( 1kN 1000N)=140.524N

Substitute y2 for x in Equation (XII).

  I=032( y 2 )y2dy=203y 5 2dy=2( y 7/2 7/2)03=18.896m4

Substitute 1.8m for CG, 18.896m4 for I and 26m2 for Agate in Equation (XIII).

  yp=1.8m+18.896m4( 1.8m×2 6 m 2 )=1.8m+2.1428m=3.9m

Draw the diagram for different parameter.

  Fluid Mechanics: Fundamentals and Applications, Chapter 3, Problem 80P , additional homework tip  3

Figure (3)

Substitute 3.9m for yp in Equation (XIV).

  hp=3.9m×sin60°=3.9m×0.866=3.37m

Substitute 3.37m for hp in Equation (XV).

  H=5m3.37m=1.63m

Conclusion:

Resultant force acting on gate is 140.524kN and its line of action of the hydrostatic force of liquid from the bottom is at 1.63m.

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Chapter 3 Solutions

Fluid Mechanics: Fundamentals and Applications

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