
#1
Jan709, 04:20 PM

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 F_{c} (N) = 4 m pi^{2} n^{2} r / 60^{2} 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 H_{2}0. 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, F_{c} (N) = 4 m pi^{2} n^{2} r / 60^{2} 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. 



#2
Jan709, 04:46 PM

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.




#3
Jan709, 05:14 PM

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. 



#4
Jan709, 06:40 PM

P: 619

Spinning Fluidmass Question.
Sure, for a consulting fee. Is that what you had in mind, or were you looking for a freeby?




#5
Jan809, 12:27 PM

P: 10

@Dr.D:
Firstly, if you're soliciting business: Now, is there anyone else who might be able to work with me here to elucidate the particulars of this problem? Thank you. 



#6
Jan809, 03:38 PM

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.




#7
Jan809, 03:58 PM

P: 10

@Dr.D:
You assume and allege much, sir. I have reported this incident to admin for their evaluation. Thank you. 



#8
Jan809, 05:04 PM

Mentor
P: 39,685

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:




#9
Jan809, 06:37 PM

Sci Advisor
HW Helper
P: 2,110

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. 



#10
Jan909, 12:37 PM

P: 10

@nvn:
Thanks for your help. I do have one question stemming from what you mentioned at the outset: Can you help me with this particular aspect of the problem? I do appreciate your assistance. 



#11
Jan909, 05:15 PM

Sci Advisor
HW Helper
P: 2,110

1000_Words: The equation I posted assumes the fluid top surface is approximately horizontal, and does not account for any significant slope at the fluid surface, if any. A completely filled, closed cylinder is especially consistent with this assumption.
The pressure on the cylinder wall varies along the wall height, as in any tank, if parameter "g" is greater than zero. If this cylinder is in a zerogravity environment, set g equal to zero. The equation assumes the cylinder axial centerline is vertical. 



#12
Jan909, 05:43 PM

P: 10

@nvn:
Of course. Due to the relatively small amount of pressure due to g (in this case), I had mentally just factored it out; hence my reference to a parabolic profile. Thanks for clarifying. I've "spread out" your math; and will work it soon. It looks really straightforward; but I'll come back with the final result for our example just for the record (and to see if I bumbled through it all properly). Cheers  



#13
Jan1209, 03:53 PM

P: 10

@nvn:
Hope you had a good weekend. I'd like to start by "spreading out" your original post to make things easy for us and all who may come this way in the future; and work down from there. To begin: 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, F_{c}, exerted on the cylinder wall vertical slice by the fluid is F_{c} (N) = s * h * [(po – (0.5 * rho * g * h)) + (0.5 * rho * (r * omega)^{2})], where (instant solutions for a .0254 meter segment of a .6 meter diameter X .0254 meter high spinning cylinder in blue) s = wall vertical slice horizontal arc length width = theta * r = 2 * r * x, .0254076 meters b = wall vertical slice horizontal chord width, .0254 meters r = cylinder inner surface radius, .3 meters h = total fluid depth, .0254 meters rho = fluid density (1000 kg/m^{3} for water), 1000 g = gravitational force; 9.80665 m/s^{2}, 9.80665 omega = angular velocity (rad/s) = pi * n/30, 314 x = asin(0.5 * b/r), 0.0423460 n = revolutions per minute (rpm), 3000 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. The pressure on the cylinder wall varies along the wall height, as in any tank, if parameter "g" is greater than zero. If this cylinder is in a zerogravity environment, set g equal to zero. The equation assumes the cylinder axial centerline is vertical. Now, setting the relatively small influences of g and po to 0 (for this instance), we have F_{c} (N) = .0254076 * .0254 * (0.5 * 1000 * (.3 * 314)^{2}) F_{c} (N) = .0254076 * .0254 * (0.5 * 1000 * 8873.64) F_{c} (N) = .0254076 * .0254 * 4436820 F_{c} (N) = 2863.315 Interestingly, this result is in fairly good agreement with that of the equation in the original post (which, for the benefit of some who have visited this thread, is nothing more than an algebraic reformulation of Newton's familiar F_{c} = m v^{2}/r) F_{c} (N) = 4 m pi^{2} n^{2} r / 60^{2}, where (instant solutions for a .0254 meter segment of a .6 meter diameter X .0254 meter high spinning cylinder in blue) h = total fluid depth, .0254 meters b = vertical slice wall segment width, .0254 meters r = cylinder inner surface radius, .3 meters rho = fluid density (1000 kg/m^{3} for water), 1000 n = revolutions per minute (rpm), 3000 and m = the spinning watermass segment's weight. .0967734 kg Note: m may be readily computed from [pi * r^{2} * h / ((pi * d) / b)] * rho. F_{c} (N) = 4 * 0.0967734 * 9.8596 * 9000000 * .3 / 3600 F_{c} (N) = 10304787 / 3600 F_{c} (N) = 2862.441 I may be missing something here, but all of this doesn't seem to be making an adjustment for the distributed mass along the horizontal slice's length: It seems to be giving us a value which simply assumes that all of m is placed directly at the distal spinning end r. Is this true? If so, how would we correct these results to reflect the actual mass distribution along the horizontal slice length (thereby adjusting F_{c} to accurately represent the local restraining force required of the outer wall)? Thanks again. 



#14
Jan1309, 08:01 AM

Sci Advisor
HW Helper
P: 2,110

1000_Words: Good point. Those two answers should not be the same. There is a mistake in my formula in post 9. Change the second 0.5 in post 9 to 0.333333. The revised formula is as follows, where all parameters are as previously described in post 9.
Fc = s*h*[po  0.5*rho*g*h + (1/3)*rho*(r*omega)^2] 



#15
Jan1309, 02:03 PM

P: 10

@nvn:
Thanks for that adjustment. To encapsulate, it looks as if the correction for mass distribution must reflect the center of mass for the section which is being evaluated here. In other words, the result from either approach may be corrected by multiplying what would be F_{c} by 2/3. This adjustment factors in the center of mass for the triangularshaped slice; which is located at 2/3 of the distance from the vertex (the radial origin) to the midpoint of the opposite side (center of the vertical slice wall segment). Correct? If so (out of academic curiosity), would a further (very) minor correction be in order to factor in the semicircular aspect of the cylindrical shell (defined by the wall vertical slice horizontal chord width)? 1000 words; 20 questions . . . Thanks again. 



#16
Jan1309, 08:50 PM

Sci Advisor
HW Helper
P: 2,110

1000_Words: The formula in post 14 already includes the cylinder arc. A description of the formula is mentioned in the first two sentences of post 9.




#17
Jan1409, 01:07 PM

P: 10

@nvn:
I think I get the point  Thanks again for your input. I hope this thread is edifying to the community... 


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