The updraft velocity of air motion.

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Answer 1

Question

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

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Students have asked these similar questions

A 0.1-mm-diameter dust particle whose density is 2.1 g/cm3 is observed to be suspended in the air at 1 atm and 25°C at a fixed point. Estimate the updraft velocity of air motion at that location. Assume Stokes law to be applicable. Is this a valid assumption?

A circular sign has a diameter of 52 cm and is subjected to normal winds up to 160 km/h at 10°C and 100 kPa. Determine the drag force acting on the sign. Also, determine the bending moment at the bottom of its pole, whose height from the ground to the bottom of the sign is 1.5 m. Disregard the drag on the pole.

A tennis ball with a mass of 57 g and a diameter of 6.4 cm is hit with an initial velocity of 105 km/h and a backspin of 4200 rpm. Determine if the ball falls or rises under the combined effect of gravity and lift due to spinning shortly after hitting. Assume air is at 1 atm and 25°C.

Answer 2

Question

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

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Answer 3

Question

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

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Answer 4

Question

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

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Answer 5

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

Answer 6

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

Answer 7

Chapter 11, Problem 64P

To determine

The updraft velocity of air motion.

Answer 8

Expert Solution & Answer

Answer to Problem 64P

The updraft velocity of air motion is 0.6177 m/s.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The diameter of the dust particle is 0.1 mm, the density id 2.1 g/cm3, the pressure of the air is 1 atm, and the temperature of the air is 25°C.

Write the expression for the weight of the body.

W=FD+FB …… (I)

Here, the drag force is FD and the buoyant force is FB.

Write the expression for weight of the body in terms of density.

W=ρDgV …… (II)

Here, the density of dust particle is ρD, volume of the dust particle is V, and the gravitational acceleration is g.

Write the expression for the drag force as per stokes law.

FD=3πμV′D …… (III)

Here, the velocity of the dust particle is V′, the viscosity of air is μ, and the diameter of dust particle is D.

Write the expression for the buoyant force.

FB=ρagV …… (IV)

Here, the density of air is ρa.

Write the expression for the volume of the dust particle.

V=πD36 …… (V)

Calculation:

Substitute 0.1 mm for D in Equation (V).

V=π( 0.1 mm)36=π( 0.1 mm( 1 m 1000 mm ))36=π( 0.0001 m)36=5.23×10−13 m3

Substitute 9.81 m/s2 for g, 2.1 g/cm3 for ρd and 5.23×10−13 m3 for V in Equation (II).

W=(2.1 g/cm3)(9.81 m/s2)(5.23×10−13 m3)=(2.1 g/cm3( 10 3 kg/ m 3 1 g/ cm 3 ))(9.81 m/s2)(5.23×10−13 m3)=(2.1×103 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=1.077×10−8 kgm/s2

W=1.077×10−8 kgm/s2(1 N1 kgm/ s 2)=1.077×10−8 N

Refer to Table A-9, “Properties of air” to obtain the values of ρa as 1.184 kg/m3 and μ as 1.849×10−5 kg/m⋅s at temperature 25°C.

Substitute 9.81 m/s2 for g, 1.184 kg/m3 for ρa and 5.23×10−13 m3 for V in Equation (IV).

FB=(1.184 kg/m3)(9.81 m/s2)(5.23×10−13 m3)=6.074×10−12 kgm/s2(1 N1 kgm/ s 2)=6.074×10−12 N

Substitute 6.074×10−12 N for FB, 1.077×10−8 N for W, 3πμV′D for FD in Equation (I)

1.077×10−8 N=3πμV′D+6.074×10−12 N3πμV′D=(1.077×10−8 N)−(6.074×10−12 N)V′=1.0764×10−8 N3πμD …… (VI)

Substitute 1.849×10−5 kg/m⋅s for μ and 0.1 mm for D in Equation (VI).

V′=1.0764×10−8 N3π(1.849× 10 −5 kg/m⋅s)(0.1 mm)=1.0764×10−8 N( 1 kgm/ s 2 N)3π(1.849× 10 −5 kg/m⋅s)(0.1 mm( 1 m 1000 mm ))=1.0764×10−8 kgm/s23π(1.849× 10 −5 kg/m⋅s)(0.0001 m)=0.6177 m/s

Conclusion:

The updraft velocity of air motion is 0.6177 m/s.

Answer 12

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The updraft velocity of air motion.

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The updraft velocity of air motion.

Answer to Problem 64P

Explanation of Solution

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Chapter 11 Solutions