Dust particles having an average diameter of 0.05 mm and an average density of 450 kg / m^3 ar

step 1 In this problem, we're asked to consider dust particles. An average diameter of 5 millimeters. S

Dust particles having an average diameter of 0.05 mm and an...

Dust particles having an average diameter of $0.05 \mathrm{~mm}$ and an average density of $450 \mathrm{~kg} / \mathrm{m}^{3}$ are stirred up by an airstream and vblown off the edge of the 600 -mm-high desk into a horizontal steady wind of $0.5 \mathrm{~m} / \mathrm{s}$. Determine the distance $d$ from the edge of the desk where most of them will strike the ground. The air is at a temperature of $20^{\circ} \mathrm{C} .$

Key Concepts

Terminal Velocity

Terminal velocity is the constant speed that a falling object reaches when the force of gravity is balanced by the drag force exerted by the fluid (air) through which it falls. For small particles, once they quickly accelerate to this speed, they continue to fall with a constant settling speed, which determines the time it takes to descend from a given height.

Stokes' Law

Stokes' law provides a relation for the drag force experienced by small, spherical particles moving at low Reynolds numbers in a viscous fluid. It allows one to compute the terminal settling velocity by equating the gravitational force minus buoyancy with the viscous drag force, using the particle’s diameter, its density, and the fluid's viscosity.

Projectile Motion in a Fluid

Projectile motion in a fluid considers the motion of an object that is subject to both gravitational forces and fluid resistance (drag). In scenarios where particles are carried by wind, the horizontal displacement from the initial point depends on the wind speed and the time taken to fall under gravity, which is influenced by the terminal velocity.

Wind Advection

Wind advection refers to the transport of particles by the horizontal wind flow. In this context, while the dust particles descend with their terminal velocity, the wind continuously moves them horizontally. The overall distance at which the particles hit the ground is determined by the product of the horizontal wind speed and the fall time.

Transcript

00:01 In this problem, we're asked to consider dust particles.

00:03 An average diameter of 5 millimeters.

00:05 So i'm going to switch colors here.

00:09 And do that my diameter will be 5.

00:17 Let's just do this.

00:25 05 times 10 to minus 3 meters.

00:27 Our density of our dust, we're told, is 450.

00:34 Kilograms per meter cubed.

00:37 Our speed of our dust will be 0 .5.

00:40 And our height is 0 .6 meters.

00:47 Okay.

00:48 And at 20 degrees c, we know that our density of air is 1 .202 kilograms per meter cubed.

01:02 And our dynamic viscosity is 18 .1 times 10 to the minus 6 newton seconds per square meter.

01:14 And we're finding the distance.

01:17 So we're finding the distance.

01:21 Distance of where the dust particles hit the ground, distance where they hit.

01:30 So let's think about this.

01:31 Here's my dust particle, and here will be my force of b, here will be my force of d, and here will be my w.

01:46 Okay.

01:49 So we've only got y forces.

01:50 We know that our force sum of y has to equal zero.

01:54 This will be w minus fd, minus fb, equals 0 and fd is the force of drag acting on the dust particle and fb is the buoyant force okay so i'm going to go to the next page here and this will equal the density of dust times g times the volume of the dust so this will be w equals 450 times 9 .81 times 4 pi and then 0 .05 times 10 to the minus 3 divided by 2 squared divided by 3.

03:09 Solving for w here we'll get 2 .8892835 times 10 to the minus 10th newton's.

03:25 Then my drag force can be calculated by 3 pi.

03:45 Okay, and this will be equal to 3 times pi times 18 .1 times 10 to the minus 6 times v -dust times 0 .05 times 10 to the minus 3...

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