Can you be more specific about what you are trying to do.
If you are trying to relate the unbalance on the rotor (ounce-inches) to vibration velocity measurements on the bearing housing (inches per sec or mm/sec), you won't have much luck unless you can insert trial weights to calculate influence coefficients. The dynamic stiffness of the machine (transfer function which relates force to vibration) is generally unknown (unless you want to do some heavy-duty calculations to estimate it).
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Is this a small single stage impeller for an automobile water pump or a larger multistage (downhole?) pump that might be used in industry?
I am not really sure what you are trying to do but..if you need an estimate of the worst case unbalance forces that might be generated by the rotating assembly, you could try getting hold of a copy of ISO1940 (Balance Quality of Rotating Rigid Bodies) and assign say a Grade of (say) G16 to the rotating assembly and then follow the instructions (you need to know mass of rotating assembly and its speed) to calculate what this means in terms of allowable residual unbaklance (g-mm or oz-ins) for the assembly on a workshop balancing machine. Thats the mass*radius bit of the equation Force = mass*radius*(angular speed)^2.
You might want to check with the manufacturer of the pump to see what Grade they use when balancing these things and then add an allowance due to impeller wear from cavitation. They might even be able to give you a example of out-of-balance for used assemblies that have been sent in for balancing.
We get our pumps/compressors balanced in the workshop to better than G1.0. My suggestion of G16 takes account of material loss during operation (wear/cavitation/sea water erosion etc) - depends on what you are trying to do.
1) The grade is G6.3 according to ISO 1940
2) The max speed is 650 RPM
3) with 1) and 2) I can enter to the graph and compute the Maximum permissible residual unbalance eper
4) U is eper times the weight of the pump
5) Solve My"+Ky=UQ2 Sin(Qt)
M and K are matrices for a multi dof structure.
Note there are a number of masses to consider. Balance grade is related to rotor mass. In your equation of motion, there may be other stationary masses accelerating as a result of the force.
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Basic vibration stuff, can be found in the books of:
Hartog, Thomson, Rao and many others under the chapter Harmonically excited vibration.
In this case you have to model the (concrete) base as spring, the forcing agent = meW2, m is the mass of the rotor and e is the (off set) of the center of gravity to the center of true rotation
me = must compare to the desired Grade quality
Your biggest headache is to model the flexibility K of the concrete, or just put the pump on comercial springs.
I loved to do this calculations in school, but on the practical side, it is very difficult to measure the response if you do not have good equipment.
In practice if your calcualtions are wrong, you will not see the base move, but the concrete will be crumble and deteriorate when time goes by
Surely this is a simple equation where the LHS (containing the restraining forces) must be in equilibrium with the RHS (describing the exciting forces)
So...
I need m and r.: this is the g-mm or oz-in from ISO 1940
4) U is eper times the weight of the pump - this is incorrect, U is eper times the mass of the rotating element
In your equation of motion, there may be other stationary masses accelerating as a result of the force. These would be the M referred to in the following equation 5) Solve My"+Ky=UQ2 Sin(Qt)
K is the stiffness (and D not included would be the damping)
TPL – I'm not sure why you quoted those comments. Your comments are identical to mine in that we both identified the thought that the rotor mass only is related to balance grade, and a different mass is associated with “M” in the equation. The one additional point I’ve made is that it is worth considering whether a single degree of freedom model will be representative. It would be the case if rotor and bearings are stiff in comparision to the support while support has little mass in comparison to the pump.
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"4) U is eper times the weight of the pump" is wrong. End of
Your comments are identical to mine in that we both identified the thought that the rotor mass only is related to balance grade, and a different mass is associated with “M” in the equation.
6.3 g-mm
if the rotor mass is 5 kg, that means that the corrections must influence the center of gravity that it is e 0.00126 mm (1.26 micron) from its true center of rotation.
The exiting force caused by the imbalance would be
F = m * e * w2
The exiting force will be in Newton (sorry for the imperial guys
warning: w is the rotational speed in rad/sec don't use 600 RPM !!!
I’ll be amongst the first to admit that ISO1940 isn’t light reading, but let’s go back a bit on this one.
A) A balance grade of G6.3 does NOT mean that the allowable residual unbalance is 6.3 g-mm
B) Using the numbers in the example just given, for a rotor of mass 5kg rotating at 600rpm, to be balanced to G6.3 in accordance with IS01940:
Specific unbalance eper = G {60/(2*pi*Speed)}*1000
= 6.3{(60/(2*3.142*N)}*1000
= 100 g-mm/kg
Therefore, allowable G6.3 residual unbalance of this rotor:
= 100 *5 = 500g-mm
= 500*10^-6 kg-m
C) The force generated by this unbalance, F = m*r*?^2
where: m = effective mass of unbalance (not the rotor mass)
r = effective radius of point mass of unbalance
Since the value of m*r has already been calculated (500*10^-6kg-m) there is no need to try and identify the individual values of either m or r
So Force = 500*10-6 *(2*3.1412*{600/60})^2
= 1.97 Newtons
This is the force transmitted to the casing: vibration that would be measured on the casing (resulting from this force) is harder to determine, requiring a knowledge of the system (bearing housing/body foundation) stiffness and damping parameters.
If it helps, try thinking about it this way:
Imagine a perfectly balanced circular disk (unachievable in practice, but OK for this discussion) of total mass M – when it spins, F=0 at all speeds (m= 0, r = 0).
Add one 10g mass at a distance of 100mm from the mid-point of the disk. Now add an identical 10g mass at a distance of 100mm from its mid-point, exactly opposite the first 10g mass – in this situation, the disk remains perfectly balanced. Removing one of the 10g masses, now results in the disc being unbalanced by an amount of 10g * 100mm = 1000g-mm = m*r. The force generated by this unbalance is independent of the total rotor mass M.
For completeness, remembering that these definitions are purely academic, and can’t easily be used to calculate/predict vibration measured on a bearing casing:
The balance grade value G, is the velocity (in mm/sec) of an unrestrained rotor i.e in free space – this has no relationship to the vibration velocity measured on a bearing casing (unless you are able to carry out the research to establish the relationship of the casing stiffness and damping to unbalance force)
The balance grade value G can also be taken as being the equivalent of the eccentricity or displacement of mass centre (in microns) of a rotor running at a speed of 9500rpm.
That is a good explanation of the info required to convert balance grade to e_permissible, unbalance (gm-mm) and force (Newtons). I checked the numbers and came up with same results under the assumption rotor mass is 5kg. Also I would like to reiterate TPL’s caution that everything following “remembering these definitions are purely academic...” is not related to vibration that will be measured in the field.
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Thanks TPL
A rotor of 5 kg, with an offset of 1.26 mm, rotating at 600 RPM will generate a force of 895 Newton or 200 lbf on bearing structure. I agree that the ISO 1940 is not the most enlightning lecture.
With the balancing we shift the excentricity closer to zero.
Imagine if they didn't use precision lathes, or calipers, just plain manufacturing like the old black smith did
svanels - it looks like you started with 0.00126mm from your previous post which was not correct to begin with. Then multipliled by 1000 for some reason.
As related by TPL, 6.3 does not stand for 6.3gm-mm. In a very abstract mathematical sense it stands for 6.3 mm/sec (but don't expect to measure this value).
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