# Spinning Fluidmass Question.

by 1000_Words
Tags: fluidmass, spinning
 P: 10 Good day. This is a reformulation of another post which I made with the physics guys at the Forum ;) OK. Here's the scoop: Through some helpful input from another source, I have found that the following handy equation Fc (N) = 4 m pi2 n2 r / 602 can be used to determine the restraining force (N) required in the following example: We have a square tube 1" in diameter X 2' long spinning at 3,000 RPM (n); filled with H20. Metric: The spinning radius is .3 meters (r); the watermass weight (for the .3 meter radial segment) is .1944 kg (m). On the basis of the preceding equation used to figure centripetal force vs. RPM, we get a figure of about 5,750 N as the result. However, this basic arithmetic assumes that all of the mass in question is concentrated at the distal end of r; giving nothing to the fact that the mass is uniformally distributed along r from the center of rotation outward. To solve for this, I have been informed that one merely needs to adjust for the distributed mass within the tube by dividing the result (5,750 N) by 2 to obtain the actual distal restraining force (2,875 N) required at the outer end of the spun water column at n RPM. Assuming that all of this is correct (yes?), I now can get to my question. How would we use our original equation, Fc (N) = 4 m pi2 n2 r / 602 to solve for a homogeneous mass of water centrifugally slung against the outer wall of a rotating cylindrical housing (think spinning barrel; not tube) at n RPM? Could this setting for the math conceivably involve evaluating for a triangular section? As a quick working example along these lines, let's take the case of a flat, closed cylinder 1" high by 2' in diameter, filled with water; and spinning at 3,000 RPM. In this case, could someone take a one square inch (outer wall footprint) "pie slice" triangular segment of the homogeneous fluidmass in question and solve for that one element using the cited equation? Would this accurately represent the local restraining force required of the outer wall? If so, how would we correct for the distributed mass in this situation? Sorry if this has been a bit prolix -- Thanks for any help.
 P: 619 Rather than depend on an equation for which you are not sure just how it applies (and evidently did not derive it for yourself in the first place), let me suggest that you work whatever problem you are truly interested in from first principles. By this I mean that I suggest that you apply F = ma to the actual fluid mass involved for the case you are interested in, where you develop the proper expression for the mass of the fluid, the location of the center of mass, and the acceleration of the center of mass. This will give you an expression for the net force acting on the fluid and you will have a much better understand of just how you got there if you go this way. This is much more satisfactory in the long run than simply trying to adapt a formula that you found somewhere.
 P: 10 @Dr.D: Thank you for your response. However, I am not a student: I am simply looking to apply a quick, handy form of one of Newton's equations to a problem which I have run across. I do not have time to reinvent the wheel with this particular matter; and am hoping that someone with greater knowledge of the subject can presently assist me. For the record, I am as certain as I can be about the integrity of this particular equation; but I always call upon as many resources (people) as I can to verify my data. So, in consideration of your superior knowledge in this matter, can you provide me with the data which I need to finish this tool and move on to the next challenge which is present to me? Thank you again.
 P: 619 Spinning Fluidmass Question. Sure, for a consulting fee. Is that what you had in mind, or were you looking for a freeby?
P: 10
@Dr.D:
 Sure, for a consulting fee. Is that what you had in mind, or were you looking for a freeby?
I see a couple of possibilities in these remarks which the TOS could bear upon.

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Good day to you, sir.

Now, is there anyone else who might be able to work with me here to elucidate the particulars of this problem?

Thank you.
 P: 619 You really should not get so high and mighty when you come shopping for professional services, or should I say begging, among the students. If you want professional services, you should be prepared to pay for them. Would you go to the medical school hoping for free advice? Would you go to the law school expecting free legal work? Trust me, I would not touch your work under any circumstance.
 P: 10 @Dr.D: You assume and allege much, sir. I have reported this incident to admin for their evaluation. Thank you.
Mentor
P: 41,041
I've seen the post report. However, to a large extent, I agree with Dr.D. Unless someone here understands the equation that you posted, and understands how it applies to your experiment, it probably makes no sense to try to answer the questions in your original post (OP). The integration that Dr.D suggests is the most straightforward way to solve the problem that you pose:

 Quote by Dr.D Rather than depend on an equation for which you are not sure just how it applies (and evidently did not derive it for yourself in the first place), let me suggest that you work whatever problem you are truly interested in from first principles. By this I mean that I suggest that you apply F = ma to the actual fluid mass involved for the case you are interested in, where you develop the proper expression for the mass of the fluid, the location of the center of mass, and the acceleration of the center of mass. This will give you an expression for the net force acting on the fluid and you will have a much better understand of just how you got there if you go this way. This is much more satisfactory in the long run than simply trying to adapt a formula that you found somewhere.
Relying on canned equations without knowing how they are derived is generally not a good thing in engineeing.
 Sci Advisor HW Helper P: 2,124 1000_Words: The pressure on the cylinder wall will vary along the wall height. If you take a vertical slice of that wall, where the slice has a horizontal arc length width of s, then the total centrifugal force, Fc, exerted on the cylinder wall vertical slice by the fluid is Fc = s*h*[po - 0.5*rho*g*h + 0.5*rho*(r*omega)^2], where s = wall vertical slice horizontal arc length width = theta*r = 2*r*asin(0.5*b/r), b = wall vertical slice horizontal chord width, r = cylinder inner surface radius, h = total fluid depth, rho = fluid density (1000 kg/m^3 for water), g = 9.80665 m/s^2, omega = angular velocity (rad/s) = pi*n/30, n = revolutions per minute (rpm), and po = fluid pressure at origin (origin is located at the centerpoint of the cylinder bottom surface). Note that if the air pressure above the fluid is atmospheric pressure, then po = rho*g*h. If you compute wall arc length s from chord width b, be sure the angle given by asin, in the above formula, is in units of radians, not degrees.
P: 10
@nvn:

I do have one question stemming from what you mentioned at the outset:
 The pressure on the cylinder wall will vary along the wall height.
I'm not sure how this would apply to a completely filled and completely enclosed spinning cylinder; as there would apparently be no way for the entrained (incompressible) fluidmass to form a parabolic profile.

Can you help me with this particular aspect of the problem?