compare the settling times of a spherical dust particle that is
25 microns (25x10^-6 m) in diameter from 1 km height in
the atmosphere of Earth. For this task, assume a constant velocity,
the upward and downward force acting on the particle, in
balance.
- Show Your Work And Include The Correct Units Calculate And Compare The Settling Times Of A Spherical Dust Particle That 1 (19.63 KiB) Viewed 17 times
was
4.625 x 10^10
Expanded Form:
46250000000
but I'm not sure what the decimal place is.
Please look at the medium sand and coarse silt examples I
provided above to help solve this with the correct decimal place
for my problem.
Multiplying and dividing out all the nonvariable terms, this expression reduces to Sg r² × 2.96 × 10 m/s Incorporating grain radius yields a settling speed of about 18.5 m/s for 500 μm grains, medium sand. For 63 µm grains, coarse silt, the settling speed is about 0.29 m/s. Hence, it would take sand grains only (1000 m/18.5 m/s) or 54 s to settle from a 1-km elevation, whereas it would take silt grains (1000 m/0.29 m/s) more than 3400 s or nearly an hour to settle the same distance. =
Fluid resistance (upward force) = Force due to gravity (downward force) 6πrμεν = 4/3πr3(pp - pi)g r is the particle radius, 12.5 pm u is the fluid viscosity (air, in this case), 1. 85 x 10-5 [kg /(m*s)] v is the particle velocity, ? Pp is the particle density, 2,700 kg/m³ pf is the fluid density, and 1.225 kg/m³ 8 is the acceleration of gravity. 9. 8 m/s² Rearrange equation 1 to solve for the velocity v. The resulting expression is known as Stokes' Law. Plug and chug the following values for Earth and Mars to find the particle velocity in each case, and then determine how long it takes to fall the proscribed height. Be mindful of units and unit conversions! Poised viscosity particle density air density gravity [kg/m³] 2700 [m/s²] [kg/(m*s)] [kg/m³] 1.225 Earth value 9.8 1.85x10-5 2700 3.71 0.01225 Mars value 1.23x10-5 Here, (1).