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 10, Problem 125P
To determine

The terminal velocity of the particle.

Expert Solution & Answer
Check Mark

Answer to Problem 125P

The terminal velocity of the particle is 0.19357m/s.

Explanation of Solution

Given information:

The density of air is 0.8588kg/m3, the temperature of air is 50°C the pressure of air is 55kPa, the density of the particle is 1240kg/m3, and the viscosity of air is 1.474×105kg/ms.

Write the expression for the force acting on the particle in the downward direction.

  Fdown=πD36ρPg  ...... (I)

Here, the mass of the particle is m the density of the particle is ρP the diameter is D and the acceleration due to gravity is g.

Write the expression for the drag force.

  FD=3πμVD  ...... (II)

Here, the viscosity is μ, the terminal velocity is V and the diameter is D.

Write the expression for the buoyancy force.

  FBuoyancy=πD36ρPg  ...... (III)

Here, the mass of the particle is m the density of the particle is ρP the diameter is D and the acceleration due to gravity is g.

Write the expression for the upward force.

  Fup=FD+FBuoyancy  ...... (IV)

Substitute 3πμVD for FD, and πD36ρPg for FBuoyancy in Equation (IV).

  Fup=3πμVD+πD36ρPg  ...... (V)

Write the expression for the force Equilibrium.

  Fup=Fdown  ...... (VI)

Calculation:

Substitute 65μm for D, 1240kg/m3 for ρP and 9.81m/s2 for g.

  Fdown=π ( 65μm )36(1240kg/ m 3)(9.81m/ s 2)=π ( 65μm( 1× 10 6 m 1μm ) )36(1240kg/ m 3)(9.81m/ s 2)=5.56774π×109kg/ms2=1.749109kg/ms2

Substitute 65μm for D, 0.8588kg/m3 for ρP, 9.81m/s2 for g and 1.475×105kg/ms for μ in Equation (VI).

  Fup=[3π( 1.474× 10 5 kg/ ms )V( 65μm)+π ( 65μm ) 3 6( 0.8588 kg/ m 3 )( 9.81m/ s 2 )]=[3π( 1.474× 10 5 kg/ ms )V( 65μm( 1× 10 6 m 1μm ))+π ( 65μm( 1× 10 6 m 1μm ) ) 3 6( 0.8588 kg/ m 3 )( 9.81m/ s 2 )]=9.029×109Vkgm/s2+3.856×1013πkgm/s2=9.029×109Vkgm/s2+1.2114×1012kgm/s2

Substitute 9.029×109Vkgm/s2+1.2114×1012kgm/s2 for Fup, 1.749109kg/ms2 for Fdown in Equation (VI).

  9.029×109Vkgm/s2+1.2114×1012kgm/s2=1.749×109kgm/s29.029×109Vkgm/s2=1.749×109kgm/s21.2114×1012kgm/s2V=1.749× 10 9kgm/ s 29.029× 10 9kg/sV=0.1937m/s

Conclusion:

The terminal velocity of the particle is 0.19357m/s.

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Chapter 10 Solutions

Fluid Mechanics: Fundamentals and Applications

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