MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
(OP)
HOW CAN FIND THE MOMENT OF INERTIA ON A NEW CENTRUFUGAL PUMP THAT I AM DESIGNING?
I NEED TO SIZE THE MOTOR CORRECTLY BECAUSE IS GOING TO BE A HIGH DOLLAR AMOUNT MOTOR.
I NEED TO SIZE THE MOTOR CORRECTLY BECAUSE IS GOING TO BE A HIGH DOLLAR AMOUNT MOTOR.





RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Moments of inertia should be explained in your kinematics textbook and/or in your machine design textbook. The Machinery's Handbook explains it too.
Is inertia really the primary load from your pump?
You are working on a project that requires some engineering training.
JHG
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Are you sure you have to design it? Centrifugal pumps are pretty commonplace. Take a look on Google, you may be able to find what you need.
As for finding the mass moment of inertia... as drawoh says, kinematics textbooks, Handbook of Mechanical Engineering, Machinery's Handbook, and Mark's should all have pretty good explanations.
V
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
It's posts like these that make me cringe whenever I need to buy new equipment.
I2I
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
If you're concerned about sizing the motor correctly due to cost, why do you feel it is not important to have accurate numbers from your calculations?
Another thing, your first question talks about the moment of inertia on a pump you are designing, but the statement quoted above is concerned with the cost of the motor.
Patricia Lougheed
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
If you want to include the bearings, use only the inner rings, don't forget the lock nuts and/or keys
If you succeed doing this by hand, compare the results with your 3D Cad, you could evaluate/quantify Electricpete's answer, which is also correct.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Ain't that the truth, just seen some shafts fail for just that reason! My favourite arrangement, two double row self aligning bearings per shaft.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
See for example here:
http://www.omel.com.br/pdf/32.pdf
In the above case, it is a relatively short shaft. The back-to-back bearing acts like one bearing as far as radial stiffness goes. Also vertical pumps with long shafts may have more than 2 radial bearings.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
However, if someone said to me 3 bearings on 1 shaft without clarification I would assume 1 each end and 1 in the middle - which equates to problems.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Here is my take:
J = Polar mass moment of inertia about axis of cylinder or disk
J = Sum (m * r^2)
J = Int (rho * r^2 dV)
Shell volume element of width dr is: has volume dV = L * 2*Pi*r * dr
J = Int rho * r^2 dV = Int rho * L * 2*Pi*r *r^2* dr
J = Int rho * L * 2*Pi*r ^3* dr
J = rho * L * 2*Pi * [r^4/4]
J = rho L Pi r^4 / 2 [<= in terms of r]
J = rho * L *Pi * [d^4/32] [<= in terms of d]
If you have outer an inner radius of your part:
J = rho L Pi (Router^4-Rinner) / 2 [<= in terms of r]
Plug in consistent units and you will get polar inertia "J" or "W*k^2".
Beware of the oddball notation "G*D^2" which is based on diameter, rather than radius. It differs by a factor of 4 from W*k^2.
Example:
Assume solid steel disk has OD = 6", Length = 1.5"
Radius =6/2*inch = 3 inch
rho:=500*lbm/ft^3; # steel
J= rho* Length* pi* Radius^4 / 2 *(ft/12/inch)^5
J = .3834951971*lbm*ft^2
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Check out thread404-218303: Inertia units conversion
Pounds mass is not good practise.
JHG
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
From first principles we can calculate it as
J = Sum m * r^2 = Int rho dv
Note the m in m* r^2 stands for MASS.
The m in lbm stands for MASS
If you object to lbm as a unit of mass, then I guess that's a whole 'nother discussion.
Personally, I saw a whole lot of "lb" in the other thread. That is what I would object to. lbm is not ambiguous. (hint: m stands for MASS).
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Let's say the motor is a general purpose NEMA frame motor with ratings 2hp, 1200 rpm.
Where can we find the max inertia (excluding motor inertia) that can be accelerated by a general purpose NEMA frame motor?
NEMA MG-1 (2003) table 12-7 provides the answer: The value is 30 and the units are given as "lb-ft^2"
One can see here from the context that lb is used as a unit of mass. Like many others, I choose to add "m" to the end (lbm) to make it clear what kind of lb is being talked about without having to analyse the context.
I would like for those in this thread who imply lbm is inappropriate or those in the linked thread who imply that use of lbm represents some misunderstanding to work this same problem above in your preferred units and compare to the NEMA limit.
(A refresher of the problem: solid steel disk with OD = 6", Length = 1.5", rho rho:=500*lbm/ft^3. Find out if it meets the NEMA-specified inertia limit for a 2 hp 1200 rpm motor... namely 30 lb-ft^2)
Then, please compare how many unit conversions were required to finish this task under your preferred unit system to how many I used in my analysis above.
Thanks in advance.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Let's say I have a very thin hollow cylinder. The radius is 1 ft and the mass is 10lbm. (or the weight on the equator of planet earth is 10 lbf, if you prefer).
What is the J?
J = M*R^2 = 10 lbm * 1 ft^2 = 10 lbm-ft^2.
You can compare it directly to the NEMA table which carries the same units and see it is less than their limit of 30. I didn't need any unit conversions to get there.
So you guys are suggesting the correct approach would have been to use slugs or something? I would like to understand.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Now what?
h
Ted
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
There is a system for calculating MOI based on English units and one based on metric units. This is due to the fact that pounds are a unit of weight (force), not mass. The gravitational constant must be accounted for, somehow.
My personal preference, as noted in the thread, above, is to systematically do mass unit calculations, and use m=w/g. You can use the force unit system in English, and the mass unit system in metric. You can refuse to recognize the existence of either English or metric units. A trained professional should be able to interpret the various handbooks and get all this right, especially if they do a unit balance.
Some of us are questioning the OP's level of training. We are, or at least I am, surprised that the impeller inertia is an important factor in a pump design. I do think that a person who understands the fluid density and viscosity and their effect on the impeller would have asked the original question. The pump seal will also use up motor torque.
JHG
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Darn! I hit the Submit button insead of the Preview button.
That should say...
I do not think that a person who understands the fluid density and viscosity and their effect on the impeller would have asked the original question.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
You can get to the MG standards for free. Just give them your email and sign in here:
http://www.nema.org/stds/mg1condensed.cfm#download
You'll get MG-1 2007 condensed version. Table 45 of that document on page 62/76 is the same as NEMA MG-1 Table 12-7.
Take a look at motor OEM sites. They show inertia in units LBF-FT^2 also. For example look here:
http:/
Now let's go back to fundamentals. Polar inertia J (as in Torque = J * d^2 theta/dt^2) is the sum of mass times radius squared. Choose a unit for mass and choose a unit for length (radius) and you're done.
In the SI system of units, mass is given in units of kg and distance is given in units of meters, so we would have polar mass moment of inertia in kg*m^2.
In the "American/British Mass (Scientific) System" of units, mass is in units of lbm and distance is in units of feet (Refernce: Applied Dimensional Analysis and Modeling by Thomas Szirtes, Pal Rozsa ISBN: 0123706203), so we would have polar mass moment of inertia in units of lbm*ft^2.
It is really that simple. I have never heard of "weight moment of inertia". I'm sure that some people who use English units are careless about mixing up weight and mass as suggested by the US motors link above. But I cannot be held responsible to defend every misuse of units commited by someone who happens to use English units. Any attempt to introduce weight into discussion of this quantity is in my view completely irrelevant since polar mass moment of inertia depends on mass, not weight (that's what the "mass" stands for). To prove it, send your pump up to the moon and try to accelerate it. Acceleration time will be the same (for the same motor torque profile), even though weight as changed.
I assume the same people who are annoyed by lbm-ft^2 would have no problem with kg-m^2. Now why should that be?
I am not telling anyone which units to choose. Choose what you want as long as you apply it correctly and get the right answer. But for someone to suggest that my choise of lbm as a unit of mass is incorrect (when those are in fact the units captured in NEMA/ANSI standard documents) is a little bit wacky imo.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
drawoh - I think we are in agreement that any units applied correctly give the correct answer. I assume that means you retract your comment about lbm.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
"They show inertia in units LBF-FT^2"
should have been
"They show inertia in units LB-FT^2 "
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
The original poster said he wanted to know pump inertia as an input to motor selection. When you size a motor to match a load, you need to know two distinct characteristics of the load: the load torque-speed curve (which is where the seals and viscosity part comes in) and the rotating inertia. Both are needed. If the inertia is too high, it can drive you toward a larger motor for successful starting, regardless of the level of the load torque speed curve. Picture a tiny motor trying to start a huge flywheel on frictionless magnetic bearings. The motor will be so slow coming up to speed that it will overheat (or trip off-line if properly protected).
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Now I know why I prefer the SI system. Isn't the imperial fundamental unit of mass the slug?
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Pound-mass is a unit of mass in the "American/British Mass (Scientific) System"
Once again, my reference for the above statements is: "Applied Dimensional Analysis and Modeling" by Thomas Szirtes, Pal Rozsa ISBN: 0123706203).
I am not particularly interested in all the arcane historical twists of unit systems. The simple fact is the industry standard document and vendor documents I cited above give inertia is units lb-ft^2. It doesn't take too much thought to see the lb here is lbm.
If you prefer SI, I can certainly understand that and have no objection. I would never presume to correct anyone based on the fact that their personal preference for a unit system was different than mine.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
I think the following is a poor analogy for a centrifugal pump, although you are not particularly refering to a pump in this statement - it will probably be read that way by many who haven't followed the discussion.
" Picture a tiny motor trying to start a huge flywheel on frictionless magnetic bearings. The motor will be so slow coming up to speed that it will overheat (or trip off-line if properly protected). "
Your earlier statement is more to the point for a centifugal pump and to my mind fully answers the original poster - who by the way seems to have now vanished :-
"electricpete (Electrical) 13 Jun 08 17:51
For a single stage centrifugal pump, the pump rotating inertia is often relatively insigificant compared to the motor rotating inertia."
In my 30+ years in the pump industry, the only time I have needed pump inertia is to fill in the blank on a data sheet because a consultant thought it should be included - what they did with the info once given is anyone guess.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Both torque speed and load inertia are needed for motor starting calculation or validation.
For small and medium motors (up to approx 200 hp), it typically amounts to a validation that the inertia is less than the maximum inertia listed in the NEMA tables linked above. Typically would not be a challenge to achieve for a centrifugal pump. For example the example I gave above - 2hp 1200 rpm motor can start a load up to 30 lbm-ft&2 which is pretty darned large - a hollow cylinder of 1' radius (2' diameter) with 30 pounds mass on the circumference which is much bigger than any 1200 rpm 2 hp pump I'd (I'll leave it to others to calculate how many slugs that is).
But for large motors (perhaps 300-500hp and up, depending on speed), NEMA doesn't give us any table of maximum inertia. In that sense each large motor purchased to NEMA specs will be a custom engineered motor. The motor manufacturer will need to know the driven load torque-speed characteristics AND inertia to evaluate starting performance under worst case voltage conditions. He will develop a worst case starting profile (current vs time) and compare it to the motor damage curve (current vs time) and verify there is enough margin to provide protective relay function between those two curves.
And while typically the pump inertia is typically a very small part of that equation, it is not always so. I have in front of me a motor starting calculation for a vertical 3500 hp 324 rpm motor driving a single stage axial pump (power plant circ water pump). The pump impeller is huge like an airplane propeller. The motor inertia is 110,000 lbm-ft^2 and the pump inertia is 20,000 lbm-ft^2. So the pump inertia is more than 15% of the total inertia which is not insignificant. As a result of presence of the pump inertia, it takes the machine 15% longer to accelerate than it would without the pump inertia.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
"The pump impeller is huge like an airplane propeller."
should have been
"The pump impeller is huge like an cruiseship propeller."
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Thanks
Patricia Lougheed
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
It's years since I've heard that term!
One good thing about metrication was the clear division between FORCE (in Newtons) and MASS (in kilograms)
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
I
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
I must agree completly with your big pump scenario but these installations are rare for the normal day to day pump/motor selections. However,as youself,others and myself are aware, pump installations of this nature need to be engineered from the ground up and certainly entail more complex data than pump inertia which becomes a fairly minor problem in the overal scheme of things.
Like Patricia (vpl) - your effort in this is certainly worth more than 1 star - so another star from me for your effort and detail.
Just hope our budding new pump designer has taken note.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Your approach to the problem is correct, sum the moments of inertia of component elements about the rotation axis.
However, your conclusion is not correct. The values given in the NEMA table for Wk^2 are weight inertia, not mass inertia. W is the weight of the rotating component(s) in lbf. K is the radius of gyration in ft. This weight inertia is used so that easily measured units of weight and distance can be used to calculate moment of inertia.
Using your disk example: the disk weight is 12.27 lb and the radius of gyration is .0313 ft. Wk^2 = .384 lbf-ft^2
It is numerically the same as your lbm approach only because lbf and lbm are numerically equal. For consistant units when you use lbm you must also use the conversion constant gsubc = 32.17 lbm-ft/lbj-sec^2 to convert lbm to consistant mass units. lbf = (lbm/gsubc)*a. Here on earth at sea level a = 32.17 ft/sec^2. So lbf = lbm, numerically.
Mass inertia units are lbf-sec^2-ft. Weight inertia is calculated by multiplying mass inertia by gravitational acceleration g = 32.17 ft/sec^2. The weight inertia units are lbf-ft^2.
Radius of gyration is defined as k = sqrt(I/m). In the case of the disk, k = sqrt((mr^2/2m) or k^2 = r^2/2. For a cylinder, k^2 = (R^2-r^2)/2. To calculate Wk^2, use the weight(lbf) of the component and radius of gyration. Breaking an impeller into disks and cylinders, calculating and summing the Wk^2 for each will result in a figure to be compared to the NEMA chart.
My references are Mark's Standard Handbook for Engineers and Hartman's Dynamics of Machinery.
Ted
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
On 16 Jun 08 11:11, I described the problem and concluded J = .384*lbm*ft^2
You are saying the solution of the same problem should be instead Wk^2 = .384 lbf-ft^2
So, you are saying Polar mass moment of inertia J and this other mysterious quantity WK^2 are not the same thing, and the definitions of the relevant quantities are:
J = m*k^2
and
WK^2 = m*g * k^2 = Weight * k^2
It may be the case, and it would explain what the W in WK^2 stands for. But the quantity WK^2=m*g*k^2 would have absolutely no physical meaning in calculation of rotational acceleration, J=m*k^2 is the only sensible quantity to use for that purpose.
I don't doubt that there may have been bizarre historical reasons similar to your logic that lead to the strange name "WK^2". But I am still skeptical that WK^2 = m*g*k^2 is the current accepted definition of that term. People use this quantity to solve problems using first principles of physics like T = J d^2 theta / dt^2, and this process demands using sensible units lbm*ft^2. Can you please point me to the specific portion of Marke's Handbook of interest?
Regardless of the historical definition W*K^2, the calculation of J as presented above is correct. You can compare the result directly against the NEMA table values labled as "lb-ft^2". (which if interpretted as inertia must be lbm-ft^2, same as J). If you prefer to consider the quantity tabulated is represents m*g*k^2 in lbf-ft^2, then you can divide by g to get the desired quantity J = m*k^2 and then multiply by the unitless conversion g*lbm/lbf to get expected units lbm-ft^2 with the same numerical value.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Ted - I did find this in "Handbook of Mechanical Engineering Calculations" 2nd ed by Hicks, Tyler
So, you are right about the definition of WK^2.
But the practical aspect as described in the paragraph are the same as I mentioned above. We can get to the mass moment of inertia J (which is what we need), by using the same numerical value with units changed to lbm-ft^2 instead of lbf-2. It was obviously the intent of whoever concocted WK^2 that it be used this way.
Before the chorus of people telling me to switch to metric chimes in again, NEMA MG-1 is what we are given and what we have to work with. The sensible way to work with it IMO is to treat the values as J in units lbm-ft^2.
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
"units changed to lbm-ft^2 instead of lbf-2"
should be
"units changed to lbm-ft^2 instead of lbf-ft^2"
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
If it wasn't for bizzare or archaic what would we have to discuss.
Ted
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Who is Hicks to "decide" that weight moment of inertia is incorrect in favor of mass moment of inertia. I submit that each is equally valid and that mass moment of inertia is only a convenience for making dynamic calculations.
Moreover, why has this thread degenerated into a boring contest of units. I think the answer to the OP question has long ago been made and this tiring overlong discussion where people are overgenerously awarding "stars" for almost no important input should come to a conclusion.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
I have a question for you, though. What possible purpose would anyone have for calculating polar weight moment of inertia, other than as a means to calculate polar mass moment of inertia ?
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
FYI, whenever I have done a calculation of polar inertia ,I first sum the WEIGHT moment of inertia and then make the conversion to mass moment of inertia. I have never seen engineering tables of materials of density given in mass units. Have you? Would you do it otherwise?
I still think the subject is boring and anybody who continues to follow this should get a book on basic dynamics,get a day job or spend more time doing the work they are being paid to do.
BTW, I am self employed so this doesn't apply to me.
RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
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RE: MOMENT OF INERTIA ON A CENTRUFUGAL PUMP
Ted