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Posted
I've been told there's a formula for calculating energy savings on a motor after balancing. Anyone aware of this? Or provide direction?
 
Posts: 8 | Location: Midwest | Registered: 06 February 2009Reply With QuoteReport This Post
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It is something that seems to have been discussed quite a bit.

If someone has a formula, I'd love to see it.

But in my opinion the energy savings is generally pretty darned small and difficult to quantify. The big savings will generally come from improved reliability (reduced failure and maintenance costs)
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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I have not seen any either...

it is funny how people chase pennies and are loosing dollars....this is in the same category...

I do not believe that it would be measurable using standard test equipment. And to be honest chasing this would be the last thing on my to do list.


Kris, CMRP
 
Posts: 48 | Location: USA, GA | Registered: 05 March 2005Reply With QuoteReport This Post
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You can calculate force with masse and acceleration. So you can find the power dissipated in vibration before and after balancing. Maybe it is to be considered that adding balancing weights increases masse and inertia and will influence the energy saving.
 
Posts: 22 | Location: Belgium | Registered: 17 December 2008Reply With QuoteReport This Post
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I remember that when learning the basics of the balancing years back we had a little induction motor with double shaft. There was a disc on each side of the motor and you could attach balancing weights to them very easily.
The motor ran noticeably slower when the unbalance was large. “Of course, the motor runs slower”, the teacher explained, “the vibration uses energy”. I do believe it. But, when I try to calculate the energy, I don’t even know where to start.
Obviously, I don’t know the formula either, but it does not mean we can’t come up with one.
jank
 
Posts: 222 | Location: alberta, canada | Registered: 04 September 2005Reply With QuoteReport This Post
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Here is a link which branches out to several other discussion on this topic:
http://maintenanceforums.com/e...?r=20010627#20010627

I don't think you are going to extract any useful information from vibration. A forced response of a mass-spring system without damping can have lots of velocity, lots of energy changing between potential and kinetic energy but no damping and no energy dissipation. Damping does dissipate energy, but you genereally have a hard time getting a handle on it, and if you do (as I tried in one of the threads using impact ringdown characteristics), you will find the assoicated energy is pretty darned small. There is another effect that the loading on bearings can be increased due to misalignment or unbalance which in turn will increase the friction on the bearings... I think that is a bigger factor in energy efficiency than the system damping. But good luck figuring it out from vibration. As I have mentioned in another thread, you woudl probably have better luck judging energy losses with an infrared camera than with a vibraiton meter.

Jan - noticeably slower? I struggle to understand that. As you know for many motors 1% slower would mean motor full load is drawn. Spinning an unbalnace around does not consume energy except to the extent there is damping which generates heat and loading which heats the bearing. Did you see some sizeable fraction of the motor horsepower rating show up in heat?

This message has been edited. Last edited by: electricpete,
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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I actually did a test many years ago that showed a decrease in amps just by reducing the unbalance. I balanced a 100 hp motor that was ~5 mils out to roughly less than half a mil and the before and after amp readings decreased by one to two amps. If not mistaken, it indicated the same results at all three legs. I also found the formula in my dungeon of archives that illustrated savings after just one amp reduction. I do appreciate all your responses.
 
Posts: 8 | Location: Midwest | Registered: 06 February 2009Reply With QuoteReport This Post
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I don't dispute what anyone has seen with their own eyes.

Anything is possible and it is certainly possible that I am completely off base in my assessment of the power associated with unbalance. I have provided the linked threads which provide a lot of the basis for my opinion. I would like to brainstorm other possible influence (other than change in power draw) that could cause speed to decrease or current to increase during balancing :

  • Change in voltage during the test would affect the current and speed.
  • Change in temperature of the rotor could affect the speed... although would tend to make the machine slow down as rotor heats up... doesn't likely explain anything since the balanced condition generally occurs at the end of the evolution when the rotor is hottest.
  • change in grid frequency (not likely in the US but I understand common in some other places
  • The motor as we know draws inductive magnetizing current (associated with reactive power) and resistive load current (associated with real power). Is it possible that changes in unbalance can dynamically affect the airgap and therefore affect the magnetizing current fundamental or harmonics?
  • any others?


If at some point in the future, someone has the time for a "research project", it would be fascinating to document the circumstances when observed speed or current changed, ideally including the following:
  • actual speed (balanced and unbalanced).
  • voltage level (balanced and unbalanced - record to ensure it doesn't chagne)
  • fundamental current on all three phases (when balanced and unbalanced).
  • power factor (balanced and unbalanced)
  • current THD (balanced and unbalanced).
  • in lieu of measurements of current, power factor, current THD, could simply record real power if that's what the analyser measures
  • amount of balance change in inch-ounce
  • general machine description.
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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The vibration does seem to consume some energy. Attached is a ppt file with a simple experiment. The speed, watts and vibration were measured on a small 1/3 hp motor. The voltage was kept constant. The speed of the motor changed from 3553 RPM to 3539 RPM when the vibration increased from 0.085 in/sec to 1.5 in/sec. Also, the watts increased from 47.38 W to 61.95 W. The input voltage to the motor was held constant to about 40 Volts.
This experiment does not really explain anything. More it challenges the readers to come up with an explanation.
jank

PowerpointENERGY_AND_VIBRATION.ppt (1,311 KB, 26 downloads)
 
Posts: 222 | Location: alberta, canada | Registered: 04 September 2005Reply With QuoteReport This Post
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When balancing large, 900 RPM FD fans at a mill, I have seen a 1 to 2 RPM difference change from before to after. These are 30K# fans, with no adjustment made to dampers during the balancing. It was fairly common on all of them (8 fans). I can't say I have seen it on the faster fans, but maybe it wasn't as noticable.
On the big ones, I would go from 896 to 897 or 898.

D
 
Posts: 1045 | Location: Marietta, Oh | Registered: 15 April 2004Reply With QuoteReport This Post
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That is interesting data. Especially Jan's powerpoint which includes everything I asked for to rule out other influences.

Just starting to think about it, here are some thoughts:

The motor in question is a little unusual in that rated slip is so high for motors in general (3600-3450rpm)/3600 ~ 4% (but typical for very small 2-pole motors). When we operate at 1/3 voltage, the effect is magnified... torque speed curve reduces by (1/3)^2. Rated load would correspond to a slip speed of 9 * 150rpm = 1350 rpm (assuming we remained in the linear part of the curve). The conclusion is that this particular motor at this reduced voltage gives a very high rpm change for a very small fractional change in mechanical load.

Even so, the change in speed is more than I would have thought and the change in current is significant and confirmed by change in watts (not attributable to reactive power or change in voltage).

Just thinking through some numbers. I looked in catalogue at similar motor (1/3 hp 115vac 3600 rpm) and saw the motor weighs around 20 pounds. So I guess the rotor weighs around 7 pounds. Adding an unbalance of 1.15 inch-ounce at 3600 rpm creates an unbalance force of about 27 lbf or almost 4 times as much as the rotor weight. That is pretty extreme (!). But let's see if it explains the increase in watts from 47 to 63 watts.

As a first guess, if we increase bearing radial load by a factor of 4, we expect bearing frection losses to increase by a factor of 4. This is based on the NTN bearing catalogue gives formula for bearing friction losses in a form that suggests losses are directly proportional to bearing radial load. Although there is probably some difference between friction due to bearing static load (weight) and rotating load (unbalance) the catalogue doesn't make any distinction.

Let's say a fraction x of initial 47 watts total loss was due to bearing losses. When we add the unbalance, that fraction x increases by a factor of 4 and results in 63 watts total loss. Solve for x (fraction of initial no-load losses due to bearing loss).
63 = 47*(1-x) +4 * 47 * x = 47 * (1+3x)
63 = 47 + 141 x
16 = 141 x
x = 11%.

So we had a motor at no-load and 11% of the losses (before we added the weight) were due to bearing. Let's say no-load losses are half of full load losses, that means 5.5% of total full-load losses are attributable to bearing (remainder are windage losses, core losses, I^2*R losses etc). That does seem plausible. Although the reduced voltage complicates things a little.

So I guess my initial thoughts are that as a rough order of magnitude, we expect the reported increase in loading (47 watts to 63 watts) if we assume bearing losses account for 5% of full-load losses and bearing losses increase according to load... which quadruples from 7 pounds (rotor weight) to 28 pounds (imbalance weight).

What is the bearing.. 6203? Sealed? I'll try using the bearing manufacturer's catalogue formula's to see if I can predict expected watts for the bearing.

** Late entry - my calculation of losses as fraction of full load losses did not properly account for reduced voltage operation. I will have to re-do that calculation later.

I'm still thinking it over. Interested in any other thoughts and theories.

This message has been edited. Last edited by: electricpete,
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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In one of the links I posted an analysis of energy dissipated by damping. Here is a repeat of same analysis, adjusted to match the 1/3 hp motor example. The bottom line for me is that I don't think the losses associated with damped vibration are significant. (The majority of any increased losses due to adding the unbalance are probably associated with bearing loading effects discussed above)

The analysis assumes single degree of freedom model. With generous assumption for damping factor (0.5), the losses for 20 pound mass vibrating at 1.8 ips is less than 0.5 watts

F = m X'' + c X' + k X
[Let multiplication by s correspond to differentiation (corresponds to assumption of solution exp(s*t)]
F = m s^2*X + c*s* X + k X
H(s) = X/F
H(s) = (1/m) / [s^2 + s*(c/m) + (k/m) ]
= (1/m) / [s^2 + s* 2*sigma + sqrt(sigma^2+wd^2)
= (1/m) / [s^2 + s* 2*sigma + w0^2]
= (1/m) / [s^2 + s* 2*zeta*w0 + w0^2]
where
sigma = c/(2m)
w0 = sqrt(k/m)
wd = sqrt(w0^2-sigma^2)
damping factor = zeta = sigma/w0 = cos(theta)
where theta is the angle between the real axis and the pole in the complex plane.

The response to an impact (bump test) might look something similar to:
x(t) ~ X0* exp(-sigma*t)*sin(wd*t)

Define ratio K such that the waveform drops factor of K between peaks (time difference 2Pi/wd)
exp(-sigma*2*Pi/wd) = K
-sigma*2*Pi/wd = ln (K) = logdec
let "logdec" = ln(K)
sigma = -logdec*wd / (2Pi).
Zeta = (sigma)/sqrt(sigma^2+wd^2)
= -logdec*wd / [(2Pi) *sqrt( (logdec*wd / (2Pi))^2+wd^2)]
= -logdec*wd / [wd * sqrt(logdec^2 + (2PI)^2)]
= -logdec / sqrt(logdec^2 + (2PI)^2)
For lightly damped systems logdec^2 << 2PI^2
Zeta ~ -logdec / (2Pi)

Now look at the attached bump test of a machine.
The ratio K ~ 0.95
LogDec~0.055
zeta = Log Dec / (2*Pi) = 0.0087
I think 1% or less is typical for a damping factor. Maybe a little higher for sleeve bearings which have viscious dampnig from the oil. Let's be generous and assume a whopping 5% damping factor

Let's try out these calculations on a 1/3 hp 3600 rpm motor. Weight ~20 pounds=9kg.
ASSUME resonant frequency 4500cpm=75hz, zeta = 5%=0.05

c = 2*m*sigma = 2*m*w0*zeta
c = 2*m*w0*zeta
c = 2* 9 kg * (2*Pi*75/sec)* 0.05 =424 kg/sec

Vibration level is 1.8 ips pk/0 = 45.7 mm/sec pk/0 = 0.0457 m/sec

P = 0.5 * c*v^2
P = 0.5 * 424 kg/sec*(0.0457 m/sec)^2 = 0.44 Newton*meter/sec = 0.44 watts

This 0.44 watts is the amount of watts expected to be dissipated by the support damping assuming SDOF model with 5% damping factor (very conservative assumption which overestimates the damping). There was also an assumption of resonant frequency assumed 4500 cpm which could be checked/adjusted if actual bump test were performed. If resonant frequency is lower as I suspect, the predicted watts would be even lower. If resonant frequency is higher than the predicted watts needed to be adjusted upwards in proportion to the ratio of resonant frequency above 4500. (Ridiculously high 9000cpm resonant frequency still gives less than 1 watt under this model).

This message has been edited. Last edited by: electricpete,

PowerpointDampingFromLogDec.ppt (98 KB, 14 downloads)
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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I looked at two different ways of estimating bearing losses (one from FAG and one from NTN) and could not get any to come up with anything near even a 1 watt change in losses for 6203 at 3600 rpm when adding 26 lbf load to a single bearing (the reported change was 14 watts total and the 26 lbf may have been split among bearings). The two calculations that I looked at didn't consider grease vs oil, so may have been a little low if they assume oil vs grease. Also there is the matter of static load vs rotating load... seems like it could be a big difference, but not accounted for in these calcs provided by the bearing manufacturers. So I can't think of any way to confirm or disprove whether losses as high as 14 watts might come from increased bearin load.

It could be that the simplifying assumptions used in my calculation above based on damping factor were too simple and created significant error. (By the way I assumed the same damping constant applied at running speed as was calculated from resonance).

Either way it seems beyond my ability to understand or explain. In a world with unlimited resources, I would ask for thermographic images of motor in both conditions to try to guess where the extra heat is being generated (it is my "belief" that more energy dissipation requires more heat generation and I would expect to see it on the bearings... or perhaps it will show up somewhere else or perhaps both machines will have exactly the same thermal profile in which case my "belief" would be in serious question). It would also be interesting to see bump test to try to extract damping factor and w used in the very rough calculation above. But even with these there is no guarantee the picture would get clearer.

Maybe we should post a link in the vib forum asking those guys to weigh in regarding the explanation and theory for this documented increase in losses?
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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EDITED TO ADD: DISREGARD THIS MESSAGE - SEE MESSAGE 18 February 2009 06:38 AM INSTEAD.

I mentioned that the references I reviewed assume a static direction load. I think there probably is a big difference in friction from static load and rotating load due to "skidding".

As we know, higher static-direction load reduces skidding. Bearing OEM's publish minimum load to prevent skidding, not just for cylindrical roller bearings but also for ball bearings. At first glance, this seems a little counterintuitive since higher static-direction load toward the bottom direction in a horizontal machine will increase clearance at the top position. But yet we know that higher radial load decreases skidding because all the bearing manufacturers tell us so. The only way I can reconcile this is as follows: if we have only a very small radial load, then the load will be distributed primarily in the bottom 30 degrees of the bearing and the balls have to travel the other 330 degrees essentially unloaded which leads to skidding. If we add higher magnitude static direction radial load, then the balls may be loaded in the bottom 120 degrees of the bearing and have to travel only 240 degrees arc unloaded, which reduces skidding (less arc while unloaded).

Now apply the same logic to see what happens if I add a rotating load far higher than rotor weight (like 4 times as high in this example). There is a ball 180 degrees opposite the high spot which is ALWAYS unloaded (unloaded through the full 360 degrees of travel of that particular ball around the bearing). That ball and it's neighbors are surely more likely to skid by my intuition (especially for high speed and poor lubrication conditions). And that would dramatically increase friction.

So, it seems plausible to me that a high rotating load causes skidding which increases friction much higher than would be predicted for a comparable stationary load. This may explain why the estimates above do not predict the full magnitude of increase of bearing friction in presence of rotating load. If this is the case, we would expect to see those bearing housings show up hotter in a thermography scan in the unbalanced scenario. The skidding would probably show a raised noise floor in vibration also. We can't see it on a linear scale with 1.8 ips 1x magnitude.

Jan - Can you show a log scale of both vibrataion spectra?

This message has been edited. Last edited by: electricpete,
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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DISREGARD PREVIOUS MESSAGE ABOUT SKIDDING.

For a stationary load zone, a ball moves at FTF relative to the load zone.

For a rotating load zone rotating at shaft speed, a ball moves at a faster speed (1X - FTF) relative to the load zone and relativie to unloaded zone and therefore spends less time in the unloaded zone.

Based on this, there is no reason to suspect that skidding is more likely for rotating load zone (in fact we might suspect the opposite).
 
Posts: 4026 | Location: Texas Gulf Coast | Registered: 20 February 2005Reply With QuoteReport This Post
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