00:01
Consider this diagram.
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We're told that in pure water, 20 degrees, enters the retention tank, rising to a level of two meters when it stops flowing in.
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And we're asked to find the shortest amount of time possible or required for all of the sediment.
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Volume of the spheres, given as four -thirds pi are cubed.
00:22
Okay.
00:26
And we're told that the density asked to determine the shortest time needed for all sediment particles with a diameter of 0 .05 millimeters or gross.
00:34
To settle to the bottom.
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We're given the density of the particles, and we're told how to calculate the volume of a sphere.
00:44
So our kinematic viscosity, let's figure this out, our density of air will be equal to 9, 9, 3, or 8.
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0 .3 kilograms per meter cubed at 20 degrees, and our kinematic viscosity will be equal to 10 to the minus 6.
01:14
Let's convert density.
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So we're given our density of 1 .6.
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That'll be 1 .6 times 10 to the third kilograms per meter cubed.
01:43
And let's think of our sum of forces.
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So we'll have fb to buoyancy and our drag and our weight.
02:07
Okay.
02:07
Our mass of our particle will be our density times our volume, and that will equal 1 ,600 kilograms per meter cubed times 4 thirds pi are cubed.
02:40
So that'll be 1 ,600.
02:42
I'll leave my units off this part, times 4 thirds pi, and then this will be 0 .025 meters.
02:53
Over 1 ,000 cubed.
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So my mass of my particle will equal 1 .04719 times 10 to the minus 10th kilograms.
03:08
Underline that for when i need it...