00:01
So we know that the force is equaling the viscosity times the velocity times the cross -sectional area divided by l, the length between the plates.
00:11
We know that the volumetric flow rate, q, would be equalling the difference in pressures, p.
00:20
Sub 1 minus piece of 2 over r.
00:22
We know that we poise's equation this is simply saying that here r would be equal to 8 times and the viscosity multiplied by l the length divided by pi times the radius to the fourth power so capital r is simply represents the resistance and then our lowercase r represents the radius and so we can then say that q, poiseuio's law, prasayu's law for flow in a tube, for flow in a tube would be equalling the change in pressure, p sub 2 minus piece of 1 times pi r to the fourth power, divided by 8 times the viscosity times the length.
01:23
And so we can say that's then the volumetric flow rate, q, is directly proportional to, the change in pressure.
01:36
And so for part a, we can say that the volumetric flow rate would be equal in 100 centimeters cubed per second.
01:45
And here, the pressure difference increases by a factor of 1 .5.
01:50
So you simply have to multiply this by 1 .5 because the volumetric flow rate is, again, directly proportional to the pressure difference.
01:59
And so this is giving us 150 centimeters cubed per second for part b we can say then that the volumetric flow rate is equaling 100 cubic centimeters per second and here the the new fluid is three times of a greater viscosity now here we know that this volumetric flow rate is inversely proportional to one over the viscosity and so you simply have to multiply this if the viscosity is three times greater that means that you just simply have to multiply by one third and so this is giving us 33 .3 centimeters cubed per second now here for part c we know that the volumetric fluorate is inversely proportional to one over or rather inversely proportional to the length of the tube and so the volume metric fluorite here would be 100 centimeters cubed per second.
03:04
The length of the tube is now four times is has increased by a factor of four and so the new volumetric fluorate has decreased by a factor of four and this is going to be equalling 25 .0 centimeters cubed per second.
03:19
For part d we know that the radius is 0 .100 times the original again here the volumetric fluorate is directly proportional to the radius to the radius raised to the fourth power.
03:37
And so the volumetric flow rate q would be equaling 100 centimeters cubed per second.
03:44
Again, the radius is essentially 0 .1 .00 times the original radius...