Question

Suppose a gust of wind has carried a 44-μm-diameter dust particle to a height of 310 m. If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density of 2700 kg/m3, the viscosity of 25∘C air is 2.0×10−5 N⋅s/m2, and you can assume that the falling dust particle reaches terminal speed almost instantly.

Answer 1

Time=185.459 s in seconds

Time=3.091 min in minutes

Step-by-step explanation:

Given

density=2700 kg/
`m^(3)`

Diameter=44 μm

radius=diameter/2

radius=44/2=22 μm

Volume=(4/3)*π*
`r^(3)`

Volume=
`(4/3)*(3.14)*(22*10^(-6) )^(3)`

Volume=
`4.46*10^(-14)`
`m^(3)`

mass=Volume*density

mass=2700*4.46*
`10^(-14)`

mass=1.2042*
`10^(10)`

density of air=1.1839 kg/
`m^(3)`

Projected area=
`\pi *r^2`

Projected area=
`3.14*(22*10^(-6))^2`

Projected area=1.5197*
`10^(-9)`

`C_(d)` for sphere=0.47

Terminal velocity=v=
`\sqrt{(2*m*g)/A*C_(d) *density }`

where density is the air density.

Substitute values we will get

v=1.6715 m/s

Time=distance/velocity

height=distance=310 m

Time=310/1.6715

Time=185.459 s in seconds

Time=3.091 min in minutes

Answer 2

t=39.76 min

Step-by-step explanation:

Given that

D= 44 μm

R= 22 μm

h=310 m

`density = 2700\ kg/m^3`

`viscosity = 2.0* 10^(-5) Ns/m^2`

We know that drag force

F = 6πμRV

V=Terminal velocity

R=Radius

μ=Dynamic velocity

At the terminal condition

m g = 6πμRV -----------1

m=ρ x Volume

`Volume= (4)/(3)\pi R^3`

`Volume= (4)/(3)* \pi * (22* 10^(-6))^3\ m^3`

`Volume=4.4* 10^(-14)\ m^3`

`m=4.4* 10^(-14)* 2700\ kg`

`m=1.1* 10^(-10)\ kg`

Now by putting the value in equation 1

`1.1* 10^(-10)* 9.81=6* \pi * 2.0* 10^(-5)* 22* 10^(-6)* V`

`V=(1.1* 10^(-10)* 9.81)/(6* \pi * 2.0* 10^(-5)* 22* 10^(-6))\ m/s`

V= 0.13 m/s

So time taken by particle

t= h/V

t = 310/0.13

t=2384.6 sec

t=39.76 min