Re: centrifuge exercises

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

EUREKA? PROBABLY NOT.

This is perhaps oversimplified, but it is better than ignoring
water resistance. I calculated terminal velocity corresponding
to the constant artificial gravity and used it to calculated the
fall time. Terminal velocity for part (b) was calculated as for
part (a) except that g was replaced by r*w^2. In my problem
w=100*2Pi and r=0.095. Using the formula 9.28 (from S&F)
the "average terminal" velocity (for Al particle of radius 1e-6 m)
was found to be 0.014 m/s (instead of 3.7e-6 m/s in gravitational
field). If this were correct then the sedimentation time would
be 0.7 seconds (instead of 2700 s, as in part a)

At least different times for particles of different size are now
now accounted for. Thus:

r=1e-8 m --> time=7000 seconds
r=1e-7 m --> time=70 seconds
r=1e-6 m --> time=0.70 seconds -->(this is 70 rotations)

This is better than an hour ago. I will check if the "averaging
approach" is acceptable with a computer program (Euler
integration). But this will not tell me if Stoke's formula is
valid for v as large as 0.014 m/s (r=1e-6) and above.

I WILL NOT POST ANYTHING TILL THE PROGRAM
IS WRITTEN, UNLESS PROMPTED BY SOMEBODY.
RESPONDING TO MY OWN MESSAGES IS LIKE
TALKING TO MYSELF. WHY SHOULD THIS BE
DONE IN PUBLIC? SORRY.

Ludwik Kowalski wrote:

> I DID FORMULATE THE PROBLEM (SEE BELOW) BUT
> I CAN NOT SOLVE PART B WITHOUT IGNORING WATER
> RESISTANCE. IGNORING WATER RESISTANCE (STOKE)
> IS RIDICULOUS; IT LEADS TO A CONCLUSION THAT
> THE SEDIMENTATION TIME DOES NOT DEPEND ON
> SIZES OF PARTICLES. THE BASIS FUNCTION OF THE
> CENTRIFUGE IS NOT ACCOUNTED FOR.
>
> CAN IT BE ASSUMED THAT STOCKES FORMULA IS
> VALID UP TO VELOCITIES OF SEVERAL M/S  IN
> WATER? PROBABLY NOT. I NEED HELP AGAIN.
>
> IF THERE WERE NO WATER RESISTANCE THEN A
> PARTICLE OF ANY SIZE, AT REST WHEN R=9 CM,
> WOULD HIT THE BOTTOM OF THE TUBE AT
> V=27.3  M/S (WHEN W=100 RPS) AFTER ONLY
> 0.00073 S. THIS IS A SMALL FRACTION OF ONE
> ROTATION (LIKE A COIN SLIDING IN A GROOVE).
>
> John D wrote:
>
> > Earlier you had declined repeated suggestions to invoke
> > Stokes law of viscosity because it hadn't been covered
> > yet.  Since you now seem to have changed your mind
> > about that, we can begin to address the bio-lab centrifuge.
>
> I did use Stoke in part (a) but can I also use it in part (b).
>
> ************************************
> Problem for Chapter 9 (Serway and Faughn).
> ************************************
> Assume that the inner cavity of a glass tube is cylindrical.
> A muddy water is poured into the cavity to the height of
> 1 cm. To simplify assume that all mud particles are
> aluminum spheres whose radii are 0.001 mm
>
> a) The tube is positioned vertically at rest. How long will
> it take for all particles to settle? Assume the water
> temperature is 20 degrees C? (THE ANSWER IS 2700 s)
>
> b) An identical tube with the same amount of muddy water
> is placed into a centrifuge, as illustrated in Figure 9.46.
> The position is horizontal and the bottom of the container
> is 10 cm away from the axis of rotation. (Imagine that a
> cover is used to make sure the liquid is confined to the
> bottom of the tube.) How long will it take for all particles
> to settle if the tube is making 100 rotations per second?
> The problem is to be solved by ignoring gravity.
>
> Hint: (Use the average radial acceleration between r=9 cm
> and r=10 cm.)
>
> c) Why can the effect of gravity be neglected at w=100 rps?
>
> Answers:
> 2700 seconds for (a) and 0.7 seconds for (b). Earth gravity has
> a negligible effect when r*w^2 >> 9.8 m/s^2.