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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You are asked to design a unit to condense ammonia. The required condensation rate is 0.09kg/s. Saturated ammonia at 30 o C is passed over a vertical plate (10 cm high and 25 cm wide).The properties of ammonia at the saturation temperature of 30°C are hfg = 1144 ́10^3 J/kg andrv = 9.055 kg/m 3 . Use the properties of liquid ammonia at the film temperature of 20°C (Ts =10 o C):Pr = 1.463 rho_l= 610.2 kf/m^3 liquid viscosity= 1.519*10^-4 kg/ ms kinematic viscosity= 2.489*10^-7 m^2/s Cpl= 4745 J/kg C kl=0.4927 W/m CCalculate the surface temperature required to achieve the desired condensation rate of 0.09 kg/s( should be 688 degrees C) a) Show that if you use a bigger vertical plate (2.5 m-wide and 0.8 m-height), the requiredsurface temperature would be now 20 o C. You may use all the properties given as an initialguess. No need to iterate to correct for Tf. b) What if you still want to use small plates because of the space constrains? One way to getaround this problem is to use small…

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