×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

No load current on a motor

No load current on a motor

No load current on a motor

(OP)
I have a 480V 1800rpm 200hp motor with a full load of 233A and a P.f. of 84.5.

We have recently started these motors unloaded, and I noticed a no-load current of 90A.  This seems pretty high to me.  I would expect something like 50-60A such as I have seen on other similar motors.  I am trying to track down the datasheet but in the mean time am wondering if this sounds high?

Is there a rule of thumb for estimating no-load current on a motor?

RE: No load current on a motor

Rule of thumb is... no rules of thumb, or at least no particularly accurate ones. Doesn't sound outrageously high - don't forget that this is almost all reactive current at 90° to the voltage so it isn't doing any useful work. It is the magnetising current for the iron in the motor, plus a bit of in-phase current to provide the power consumed as windage and I2R losses.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: No load current on a motor

That's good information.  Write it down somewhere so you can find it later.

Scotty's correct.  Our motor shop used to do a no-load test on motors before they went out the door. but at "no load" the small increase in current caused by for instance, a bad bearing is going to be lost in the larger component of the magnetizing current.

Still, having a no-load current record has strengthened my position a few times in the age-old "It's not my process equipment, it's YOUR electric motor" arguments.

old field guy

RE: No load current on a motor

With a no load current of 233A power factor of 84.5 I would expect a no load current of over 120 Amps.
With a no load current of 90 Amps and a full load current of 233 Amps I would expect a power factor of over 0.9
Given that other motors are showing low current, have you checked your meter lately?
How is the voltage at the motor, loaded and unloaded?

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: No load current on a motor

From my experience, I can say that no-load current for 4-pole motor 200 HP, 480 V, 60 Hz should be approximately 60- 65 A, if the motor  is intended for permanent work.
 I must say that there are no strict rules and that it varies with different manufacturers, usually at low power motors.
The reason for the increased value of the no-load currents can be:
1 -  motor is provided by the manufacturer for a short overload and therefore must have increased  value of no-load current  or
2 - if the motor was already on the repair, the cause may be some changes in the windings ( turns, pitch, connections etc...) or improperly motor-heating in the process of removing the old windings.
Zlatkodo
 

RE: No load current on a motor

I have seen no-load cuurents as high as 80% of FLA's in ny repair shop. It doesn't seem that high in your case. However, if you are comparing the identical (similar doesn't cut it) motors from the same OEM, then it's a cause for concern. Is this an original winding or a rewind ?

Muthu
www.edison.co.in

RE: No load current on a motor

First approximation: We consider simple model neglecting leakage reactances. Then we have a magnetizing branch in parallel with a load branch.

The current in the load branch is PF*FLA=197A
The current in the magnetizing branch is (1-PF^2)*FLA=66.6A
Under this model, no-load current would be 66.6A.

What happens if you refined the model to include leakage reactance....  then the leakage reactance would consume vars at full load but not as much at no-load, so I think the no-load current would be even lower than predicted above.

Something sounds out of whack to me.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

If we use FLA as a starting point, then magnetizing current is FLA*sqrt(1-PF^2), which is 233*.535 = 125 A.

Then, a 480 V grid usually means a 460 V motor. That could change things, but not much (a few percents). Your 90 A no load seems OK, a little low, but not to worry about. As Bill says; checked your meters lately?

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: No load current on a motor

I was just coming back to add that sqrt which I left out in earlier calc. Gunnar beat me to it.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Another thing I had mentioned before - this approach is based on a model neglecting leakage reactances.  i.e. it assumes the vars remain the same at full load and no load, but there are actually some vars present at full load that are not present or reduced at no-load (rotor leakage reactance is present at full load and elimianted at no-load... stator leakage reactance VARS are reduced going from full load to no-load).

Using typical values of leakage reactance maybe 0.08 - 2 pu on both rotor and stator, one could estimate no-load current.   I suspect it won't bring the estimate down enogh to explain this reading, but might partial explain it (along with voltage... what was your voltage?). I might do that calculation later if I have a chance... gotta run now.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

(OP)
The motor in question is a brand new origonal motor.  

So what I gather from this, is that a good aproximation for estimating no-load current (which is bascially all magnetizing current)is to calculate the magnetizing at full load current with the full load power factor, and use this vaule as the magnetising current for the full range of the motor ingoring any leakage reactance which will be minimal.

Is measuing the no-load current a good tool in trying to decide weather weather a high current situation is a result of a load issue or a motor issue?  Or will bearing problems and other problems with a motor not have a signifigant enough impact and to show an increased current in such cases as mentioned above?

The one thing that I dont understand is that I have a similar 200hp motor with an almost identical effeciency, and p.f. and this motor datasheet shows a no-load current of only 60A.  I am waiting to get my hands on the datasheet for the motor that I am referencing to see what it states, but in the mean time I am puzzled as to what would cause such a difference?

RE: No load current on a motor

The magnetizing component of the current is constant from no-load to full load. If it is a brand new motor, what does the test certificate say ?

Muthu
www.edison.co.in

RE: No load current on a motor

Here is an example of a hypothetical 200 HP motor to illustrate the difficulty of trying to check for bad bearings by monitoring no oad current.
Suppose the magnetizing current is 100 amps.
Suppose the windage and bearing load is about  8 or 9 HP. The real current component of this load will be about 10 amps.
This 10 amps is acting at right angles to the magnetizing current. Pythagoras tells us that the resultant current will be the root of the sum of the squares, or 100.5 Amps. Most of this load will be windage but suppose that a bad bearing doubles the load. That implies about 8 or 9 HP being dissipated in a bad bearing. That is about 6 kW being wasted in a dragging bearing. A catastrophic ailure may occur in minutes. This is a far far worse case of bearing failure than the early stages of failure we want to detect to avoid an unplanned outage.
But what will this do to our no load current?
We will still have 100 Amps of magnetizing current but our real component of current has risen from 10 Amps to 20 Amps.
The sum of the squares gives us 102 amps. 6 kW disipated in a bad bearing has caused an increase in the line current from 100.5 Amps to 102 Amps. A rise of only 1.5% in current. And a bearing would most likely fail long before the drag started wasting 6 kW.
Forget current. Check bearing health with vibration trending and temperature trending.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: No load current on a motor

Absolutely right. But, there is a simple way to check for active current, use a meter that shows kW instead of A.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: No load current on a motor

I agree looking at no-load current is not a good way to find a problem with a bearing.  

Quote:

So what I gather from this, is that a good aproximation for estimating no-load current (which is bascially all magnetizing current)is to calculate the magnetizing at full load current with the full load power factor, and use this vaule as the magnetising current for the full range of the motor ingoring any leakage reactance which will be minimal.
I agree it is reasonably good.  We also know the direction of the error (in calculating no-load current from full load current/power factor)...when we neglect leakage reactance the result will be higher than the actual no-load current predicted by the full equivalent circuit because there are vars "consumed" in the leakage reactance elements at full load that are reduced (stator leakage reactance) or eliminated (rotor leakage reactance) at no load.  I will try to do solme example calc sometime this weekend to check how significant is the difference.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Good point, Gunnar.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: No load current on a motor

There really is no good rule of thumb - you could estimate about 30% of FLA for a ballpark figure but motors can vary all over the place. I have seen small motors draw around 80% of FLA at no load and large motors draw <20% of FLA at no load.

The best test is to load the motor and ensure that temperatures and currents are within the motor ratings at full load.

 

RE: No load current on a motor

I used a little bit of brute force approach.  It is probably not the most intuitive approach, but gives a quantitative answer whose validity you are welcome to judge for yourself.

You gave us FLA, HP, FLPF.  From that data we can also calculate FL Eff =0.9455.

Using the attached spreadsheet (requires analysis tookpak installed), I fitted the following equivalent circuit parameters to your data:
R_NL    =    40.27231006    ohms    Resistor simulate portion of No-Load losses - connected direct in parallel with the source
R_1    =    0.011574312    ohms    Stator Resistance
X_1    =    0.179466053    ohms    Stator Reactance
R_2    =    0.01157434    ohms    Rotor Resistance refd to stat - Should be ROUGHLY  FullLoadSlip* 3*VLN^2/FullLoadWatts
X_2    =    0.177699823    ohms    Rotor reactance refd to stator
X_M    =    3.66270788    ohms    magnetizing reactance

By the way I assumed a slip of 0.01.... it is not a critical assumption because R2 will vary to make up for any errors in estimating slip to keep R2/s nearly correct.... there are two other resistances in the model that can float to keep efficiency correct without imposing any constraint on R2.

The solution varies the equivalent circuit parameters to find a "weighted" best match to the target specifications.  The power factor, efficiency, and full load current were weighted most heavily.

Since there is are more parameters available than input data, there is some flexibility in the solution. Using very low weighting factors I weakly imposed three thumbrules:
X2/X1 = 1.0
R2/R1 = 1.0
X1/XM = 0.05

The results comparing calculated performance of the model vs target performance specification gives outstanding agreemetn:


FullLoadAmps    =    232.9999894    Amps    Target=    233    FractError=    -4.54736E-08    Weight=    1
FullLoadEff    =    0.945497223    none    Target=    0.945530445    FractError=    -3.51364E-05    Weight=    10
FullLoadPF    =    0.84489616    none    Target=    0.845    FractError=    -0.000122887    Weight=    10
FullLoadPower    =    148298.9084    watts    Target=    149200    FractError=    -0.006039488    Weight=    0
X2overX1    =    0.990158416    none    Target=    1    FractError=    -0.009841584    Weight=    0.01
R2overR1    =    1.000002482    none    Target=    1    FractError=    2.48166E-06    Weight=    0.01
X1overXm    =    0.048998189    none    Target=    0.05    FractError=    -0.020036217    Weight=    0.01

I will tell you from experience, R2/R1 is not very critical due to presence of R_NL.

X2/X1 is not critical for reasons that can be seen by examining the simplifying assumptions used in the Ossanna diagram.  The derivation of the Ossanna diagram starts by moving Xm to the line side of X1.  How can we do it!!??!.  The answer can be roughly seen by ignoring R1 and looking at the thevinin equivalent source of the circuit consisting of the voltage source, the leakage reactances and the magnetizing reactance.  In realistic situations we see that thevinin equivalent is practically unchanged.  What makes it happen?  Simply that Xm>> X1 and Xm>>X2.  If you work through the thevinin equivalent circuit with any of the leakage impedances moved to either side of the equivalent circuit, and then simplify the results using Xm>>X2, you will always end up with the same thevinin circuit which has
Voc ~ Vs
Isc ~ Vs/(X1+X2)
Rth = Voc/Isc

Since X1 and X2 appear only as a sum and not individually, it doesn't matter what their ratio is... only what their sum is.... we could lump all the total leakage reactance in either one (X1 or X2) and the thevinin circuit characteristics would not change much (based on the fact that Xm is so much larger than either one).  

This observation helps simplify the vector diagram for the Ossanna diagram, but also tells us the assumtion X2/X1 is not critical.

X1/Xm is more of a critical assumption because it gets directly at the subject we have been talking about, and it has a dramatic effect on the computed no-load current in this problem.  I noticed if you put the target at 0.1 or 0.2,  the program still wants to drag it down to 0.05.   If you put the target down below 0.5 the program drags the solution toward  non-credible results.

I think the ratio 0.05 is a credible ratio based on looking at similar size motors and certainly the design presented above is a very plausible equivalent circuit model of the motor from no load to full load (there may be others).  

Using the above equivalent circuit parameters, we can quickly calculate the no-load current as
NLA = VLN / (Xm + X1) = 277 / (3.66270788 + 0.179466053) = 69.1 Amps
if you add in the R_NL portion, it is 69.5 amps... not an important detail.

The bottom line, 70A would be my best prediction from the nameplate data you gave, and with a slightly different specified X1/XL it can easily change to 60A. I don't see any reason to suspect your measurement.

This is just one approach to the problem of estimating no-load currents from nameplate data.  It may seem like  overkill, but it's easy once you get the hang of the spreadsheet (that can be used for many many different tasks besides this).  A little bit of vector thinking will probably  give more insight into the behavior along the lines of the diagram linked by Gunnar.
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Correction in bold:
"X1/Xm is more of a critical assumption because it gets directly at the subject we have been talking about, and it has a dramatic effect on the computed no-load current in this problem.  I noticed if you put the target at 0.1 or 0.2,  the program still wants to drag it down to 0.05.   If you put the target down below 0.05 the program drags the solution toward  non-credible results"

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

If I put the ratio X1/Xm to 0.043, the no-load current changes to 90 amps and the agreement is even far better than above (I noticed I had 2% fractional error on the X1/XM).  I will post those results.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Here is the new model:
R_NL    =    41.19167808    ohms    Resistor simulate portion of No-Load losses - connected direct in parallel with the source
R_1    =    0.012288321    ohms    Stator Resistance
X_1    =    0.122340455    ohms    Stator Reactance
R_2    =    0.012288333    ohms    Rotor Resistance refd to stat - Should be ROUGHLY  FullLoadSlip* 3*VLN^2/FullLoadWatts
X_2    =    0.122339986    ohms    Rotor reactance refd to stator
X_M    =    2.845122833    ohms    magnetizing reactance

Here is the new performance which matches all targets to within fractional error of 5E-6 or better.
FullLoadAmps    =    233.0000919    Amps    Target=    233    FractError=    3.9432E-07    weight=    1
FullLoadEff    =    0.945530443    none    Target=    0.945530445    FractError=    -2.14925E-09    weight=    10
FullLoadPF    =    0.844999947    none    Target=    0.845    FractError=    -6.24187E-08    weight=    10
X2overX1    =    0.999996168    none    Target=    1    FractError=    -3.83158E-06    weight=    0.01
R2overR1    =    1.00000095    none    Target=    1    FractError=    9.50268E-07    weight=    0.01
X1overXm    =    0.043000061    none    Target=    0.043    FractError=    1.41176E-06    weight=    0.1

The high level of agreement to the X2/X1 and R2/R1 thumbrules is always easily attained over a variety of scenario's.... again not critical.... somehwat artificial.    The high level of agreement X1/XM is tougher to attain (higher error) and the error was dramatically reduced  when we decreased X2/X1 down from 0.05 to  X1/XM=0.043.

With X1/XM=0.043 what do we get for?  89.8A NLA  (including R_NL) which is exactly what was measured.    It is very good agreement on all parameters...certainly makes the model look more exact than it really is.   I don't think the model is that good... maybe just more lucky than good.  I'll admit I watched the no-load amps number as I nudged the X1/XM and stopped when it got close to 90..... I haven't carefully evaluated whether this is the exact best match or not I'm not sure but it is certainly better than the previoiusly-posted one with X1/XM= 0.05 because it moved the error associated with X1/XM way down.
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Clarification in bold:

Quote (electricpete):

By the way I assumed a full-load slip of 0.01.... it is not a critical assumption because...

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

I feel bad because I "cheated" on the last scenario and stopped varying X1/XM when I got a value of NLA that matched what I was looking for.  

So I repeated the scenario posted directly above varying the target X1/XM.  (The other 5 constraints remained constant:  FLA, FLPF, FLEFF, R2/R1, X2/X1).

The resulting
TX1/Xm SWSFE    No-LoadAmps    
0.04    1.24083E-13    94.41606374
0.041    1.01816E-13    92.97093277
0.042    1.14655E-13    91.42529287
0.043    5.49642E-13    89.75549295
0.044    1.72738E-12    87.9269906
where
TX1/XM is "target" X1/XM.   
SWSFE is sum of weighted squared fractional errors.  It describes the degree of "fit" of the model to the specified parameters.

The best fit occurs where the minimum SWSFE occurs which is at target X1/XM=0.41. That corresponds to a calculated no-load amps of 93.

In summary: 93A is the best estimate of no-load amps from this model (and using these thumbrules about target ratio R1/R2=1 and X1/X2=1 with assumed slip = 0.1).  

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

typo correction:
"..with assumed slip = 0.01"

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

another similar typo correction in bold:
"The best fit occurs where the minimum SWSFE occurs which is at target X1/XM=0.041"

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

The no load current is primarily magnetizing current and while there are no rules, there are trends.
Large high speed motors tend to have lower magnetizing currents than small or low speed motors.
Motors that are water cooled (submersible pumps) typically have a magnetizing current that is significantly higher than air cooled motors.
In addition to the good information in the previous posts, the magnetizing current is dependent on the flux in the iron.
If the motor is operated on a voltage or frequency other than it's design condition, there can be a dramatic change in the magnetizing current.
If these motors are designed and optimized for 440 volt 60Hz and operated on 480 volts 60Hz, then the magnetizing current would be considerably higher. The increase in magnetizing current with supply voltage is not linear!
Likewise operating at a lower frequency without reducing the voltage will increase the magnetizing current.

If you supply voltage is higher than the "test" or "design" voltage, expect a higher no load current due to a higher magnetizing current. Full load current will be lower because the load current will reduce with increasing voltage.

Best regards,
Mark

Mark Empson
L M Photonics Ltd

RE: No load current on a motor

One slightly strange thing is that specifying the targets does not give a unique excel solver solution (of the parameters R_NL, R_1, X_1 etc).  The solver solution depends not only on the targets, but also to a small extent upon the initial starting point (initial values of those parameters) before running solver.  So telling you the targets without telling you the resulting solved parameters is not completely  precise, and if you repeat the experiment I described above (varying X1/XL) based only on the targets, you may get slightly different results.  So the more precise way to communicate the results is to include the calculated parameters

Here is the description of the parameters and results of my final best-fit model that gives no-load amps of 93:

R_NL    =    4.13E+01
R_1    =    0.012408
X_1    =    0.112810
R_2    =    0.012408
X_2    =    0.112810
X_M    =    2.751469

The performance against specified requirements is as follows:
FullLoadAmps    =    232.9999965    Amps    target=    233    FE=    -1.507E-08    WT=    1    WSFE=    2.27106E-16
FullLoadEff    =    0.94553045    none    target=    0.945530445    FE=    4.61964E-09    WT=    10    WSFE=    2.13411E-16
FullLoadPF    =    0.845000064    none    target=    0.845    FE=    7.53792E-08    WT=    10    WSFE=    5.68203E-14
X2overX1    =    0.999998473    none    target=    1    FE=    -1.52744E-06    WT=    0.01    WSFE=    2.33307E-14
R2overR1    =    1.00000091    none    target=    1    FE=    9.09733E-07    WT=    0.01    WSFE=    8.27613E-15
X1overXm    =    0.040999985    none    target=    0.041    FE=    -3.59819E-07    WT=    0.1    WSFE=    1.2947E-14
SWSFE =     1.01815E-13
where
FE = fractional error
WT = weighting factor
WSFE = weighted squared fractional error
SWSFE = sum of weighted squared fractional error.
Speadsheet showing this solution is attached.

I think I'm done now unless anyone has questions.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Mark - all the factors you mentioned (speed, submersible, flux density) affect the full load power factor, correct?

The full load power factor was given in the original post.

To my way of thinking, coverting that full load power factor into a no-load estimate does not require knowledge of the speed of the machine.  I'm not sure how I would alter my analysis if you told me the speed.  (although I could do a better fit if more data was provided such as full load slip and half-load efficiency and power factor).

Am I missing something? How would the speed of the motor influence your conversion of the original post data into an estimate of no-load current?

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

I apologize Mark.  I can see now that your comments were very  responsive to this part of the original post: "Is there a rule of thumb for estimating no-load current on a motor?"

rockman, FWIW, here are my general estimates for ratio of NLA/FLA for large motors:
Poles NLA/FLA
2 0.20
4 0.25
6 0.30
8 0.32
22 0.50
This would not include submersibles.

 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

please see its bearings are fine or not. because if its bearings are giving more friction to rotor then your motor draws more current, another factor is that the windings are when old the draw more current because of winding losses.

RE: No load current on a motor

Marke wrote:
"The no load current is primarily magnetizing current and while there are no rules, there are trends.
Large high speed motors tend to have lower magnetizing currents than small or low speed motors.
In addition to the good information in the previous posts, the magnetizing current is dependent on the flux in the iron."

I think its true. If we are talking about two similar large motors that are properly connected to the proper voltage and frequency, and have different no-load currents, we can ask: why?
One motor  can have a no-load current 30% of the nominal current and the other 60%.
That's because the manufacturer wants to make a motor with greater power than the iron core allows.  The reason is commercial in nature.                                                                                                                 

How can this be achieved?
Manufacturer enters the calculation with the maximum allowed (sometimes even higher) values for the air-gap flux density, which can cause excessive flux density in teeth and/or  in  back-iron.
Then  tries to compensate this  by increasing the insulation class, with increased ventilation,( high efficiency) etc, but sometimes these attempts are not entirely successful.
If we make the calculation of turns / coil depending on the dimensions of stator-iron for this motor with common values for the air-gap flux density, I am sure that the result would be higher (less power) , than the number of turn / coil which we have already in the this motor. This means that the manufacturer tried to get more power than we would normally expect. Often, such a motor is not designed for continuous operation at full load.
 
That's why I think it is desirable to keep the no-load current in normal limits.

 A rough but satisfactory estimate, for large motors gave Electricpete:
"here are my general estimates for ratio of NLA/FLA for large motors:
Poles NLA/FLA
2 0.20
4 0.25
6 0.30
8 0.32
22 0.50"
Zlatkodo
 

RE: No load current on a motor

I just realized, a few years ago we changed out motor with the exact same horsepower rating (200), voltage rating (460) and speed (1800 rpm) as the one in the original post.

I went and dug up that documentation and attached it.

Page 5 is motor summary data
Page 7 includes current, pf, efficiency curves
page 8 is nameplate photo.  By the way, ours in an ODP motor.

Page 5 and 7 allow comparision to your motor  This motor of ours has FLA = 235A,   FL PF = 0.84.... so it continues to look very similar to the motor in the original post (which had FLA=233A, PF = 0.845).

But Page 7 also plots current all the way down to zero load... from which we can read a no-load current of around 60A!  (I have to admit I was surprised it was that low).

A small summary of results of review so far:
* considering constant vars (neglecting leakage reactance), we predicted no-load current of 233*SQRT(1-0.845^2)=125A
* my model-fitting excercize with some rather arbitrary thumbrules came up with no load amps of 93A.  The ratio of leakage to magnetizing reactance was not too far from similar size example motors shown in Krause' motor analysis book. But NLA changes pretty dramatically with small changes in that ratio.
* a very similar motor in the attachment had no-load current of 60A
* the thumbrule 25% FLA would predict 58A.

Some conclusions we might draw:
* This shows that neglecting leakage reactance (assuming constant vars from full load to no load) gives a high estimate of NLA... up to a factor of 2 (125/60) too high.
* It makes me wonder a little more how realistic my 93A was... either it was unrealistically high, or else there can there be substantial variability in NLA (93/60>150%)  among motors that have the same FLA and same full load power factor.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

By the way, for me it is an academic excerize to try to figure out the no load current.  I would NEVER be inclined to condemn a motor based on NLA magnitude unless I knew for a fact it was way out of line.... which we don't... the only valid comparision would be manufacturer data or previous meausrement of same/identical motor at same voltage.  Also, there are certainly not many faults that would cause the no-load current to change (in absence of abnromal voltage) without affecting other parameters or tripping the motor.  And there are more relevant parameters to look at (current balance under load, off-line testing etc).

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

If you do find the motor is drawing unusual no-load amps compared to OEM data, then you might check the connections (in addition to voltage).  But again, I am not used to investigating no-load current magnitude.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

I thought it could be interesting to compare Pete's data with a RL data sheet. But you have to be careful when doing so. Have a look at this sheet http://motor.hyosung.co.kr/lsp/LSP5447.pdf

I am not so sure if I would buy an induction motor from Mr. KIM or Mr. LEE  smile

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: No load current on a motor

Very flat torque curve, efficiency is down and PF is low. Slightly unusual for a standard motor.
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: No load current on a motor

And NLA is 0 A. Also very unusual for an induction motor.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: No load current on a motor

At zero slip there's zero torque and zero magnetising current... must be a special purpose motor. wink
  

----------------------------------
  
If we learn from our mistakes I'm getting a great education!
 

RE: No load current on a motor

Reliance allows you to look up motor data by characteristics
http://www.reliance.com/cgi-bin/mtrquery.pl#modelchar

Here are some more 200hp, 1800rpm, 460 vac motors.

http://www.reliance.com/pdf/pdf/aced/W06432-A-A001.pdf
FL PF = 0.883, NLA = 53.5

http://www.reliance.com/pdf/pdf/aced/E08055-D-N003.pdf
FL PF = 0.873, NLA = 64.1A

http://www.reliance.com/pdf/pdf/aced/L1642A-V.pdf
(designed for variable speed)
FL PF = 0.867, NLA = 70.5
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

The last linked motor above has equivalent circuit parameters listed, which are in the same ballpark as the ones I had computed.  On the left are the values I posted 8 Jan 10 16:58, on the right are the values from http://www.reliance.com/pdf/pdf/aced/L1642A-V.pdf

R_1    =    0.011574312    ohms      0.0208
R_2    =    0.01157434    ohms      0.0154
X_1    =    0.179466053    ohms     0.148
X_2    =    0.177699823    ohms   0.153
X_M    =    3.66270788    ohms   3.75
R_NL    =    40.27231006    ohms   N/A

We see X1 ~ X2 and X1/XM ~ 0.05.    I think we all initially suspected based on the relatively small values of X1 and X2, that the vars consumed in X1 and X2 at full load would be relatively small.  But on the contrary,  in round numbers, the var consumed at full load by X1+X2 turned out to be approximately equal to the vars consumed in XM (since the reactive current is approx 120A at full load and approx 60A at no-load).

The reason is Q = 0.5*I^2 *X.    

While the sum of X1 and X2 is only about 10% of Xm,  the current through X1 and X2 at full load is on the order of 3x the current through Xm, so the square of current through the leakage reactances at full load is approx 3^2~10x the square of current through magnetizing reactance. One tenth the reactance times 10 times the [current-squared] gives approximately same Q consumed in leakage reactances at full load as is consumed by the magnetizing reactance (very round numbers... just for illustration of how the small leakage reactances can end up playing a large role).

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

To determine the magnetizing current for power factor correction, I calculate as above, but instead of using the full load figures, I use the half load figures. (these are usually given in the data sheet) This reduces the errors due to the leakage reactance.
One can use the full load figures and the half load figures to determine what the leakage reactance is and separate out the magnetizing current only.

I have the impression that the new higher efficiency motors tend to have a higher leakage reactance than the older motors, but could be wrong there.

Best regards,
Mark.

Mark Empson
L M Photonics Ltd

RE: No load current on a motor

For your question "Is there a rule of thumb for estimating no-load current on a motor?", Zlatkodo's genearl estimates are reasonable for "standard" enclosed motors. In case of open type motors as IP23, induction is often designed to be higher and thus the saturation of iron is a lot higher. In this case 90A could be roughly ok.

RE: No load current on a motor

Pete, you know that I sometimes find your math excercises a bit trying. But, this time, you have shown that there is more to induction motors than I thought. The fact that leakage reactance consumes a significant part of the full load vars is a valuable insight. Thanks for that!

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: No load current on a motor

Usualy the no load current can get to 30% of nominal current. So 90A seams a little high; First step is to check the bearing health. One cause that can get to a higher current is if you just put vaseline to you bearings; in this case at the beginning the current will be a little high but if you measure the current after 1 or 2 hours, you'll see that you current will decrease. Another cause can be old vaseline to bearing that have been strenghten in time, but if is a new motor we can eliminate this cause, and anyway this problem can rise you current just with 1 or 2 amps. And the last, but the moust probably cause is a bed bearing. You can have a ball bearing that have been damaged, or maybe the bearings was not install corect. You can check this by vibration measurements.
 

RE: No load current on a motor

Thanks Gunnar.   

Good point Mark – the stuff we're discussing should be built into power factor correction calcs.   I googled and found this Siemens document about selecting power factor capacitors.

https://w3.energy.siemens.com/cms/us/US_Products/CustomerSupport/TechTopicsApplicationNotes/Documents/TechTopics20Rev0.pdf

Quote (Siemens):

The leakage reactance current is relatively small, so that the total reactive current is relatively constant (compared to the kW variation) over the range of motor no-load to motor full-load. For a range of medium voltage machines sampled, the ratio between full-load reactive current and no-load reactive current varied from 140-260% (depending on machine design, speed, and voltage).   For perspective, the ratio between full-load kW and no-load kW is of the order of 4000%!
 
Because the variation in reactive current is relatively low over the load range of the machine, a capacitor sized to compensate to a desired power factor level at full load, will maintain the power factor in the near vicinity of the desired level over the entire load range.  Typically, a capacitor sized to correct full-load power factor to 95% will maintain power factor in the 95-98% area over the full range from no-load to full-load.
 
....
Over-correction (and self-excitation):
It is important not to over-correct when sizing capacitors that are connected in parallel with the motor.  The motor requires reactive power (kVARs) to create the magnetic flux.  The power factor correction capacitor can supply the kVARs required by the motor when the motor is switched off.  At the instant that the motor is switched off, the motor and the driven load are at full speed.  When the motor is switched off, the motor and load inertia will continue to drive the motor.  If the magnetizing current required by the motor is available from the charged capacitor, the motor will operate as a voltage generator, and maintain the voltage on the motor.
In the preferred situation, the power factor correction capacitors are sized at or below 90% of the no-load kVAR requirement of the motor .  If the capacitors are too large, the motor can be subjected to self-exciation, which will result in excessive voltages applied to the capacitors and motor.  The capacitors are sized based on 90% of the no-load kVAR requirement because the manufacturing tolerance of the capacitors is –0%, +15%.

The 2nd sentence of the quote describes a conclusion very similar to ours:  the ratio of reactive current at full-load over reactive current at no-load is 1.4 – 2.6..... average is 2.  i.e. no-load reactive current is 50% of full load reactive current..... the same number we came up with as a ballpark. (although they mentioned medium voltage motors, we were working with 200hp low voltage motor  example)

However the way it is described in the first sentence is odd "the total reactive current is relatively constant... over the range of motor no-load to motor full-load".   They justify this statement by comparing it to the change in real power using a ratio comparision that is a little unnatural imo..... and disguises the point that there is substantial variation in reactive current (that factor of 2).

They suggest correcting FL power factor to 0.95 and imply this will keep PF from 0.95 to 0.98 from full load down to no-load.... seems like a dubious claim to me.   In attached spreadsheet I have used the simplifying  factor 50%  (no-load motor vars = 50% of motor full load vars) instead of the range.  By my calculations in attached spreadsheet, if I start with 94% full-load power factor and correct to 95% full load, then I'll end up at 99% PF at no-load.  If I start with 83% full-load power factor and correct to 95% full load power factor, then I end up with 100% no-load power factor.   If I start below 83% and correct to 95% at full load, then I end up overcorrected at no-load

Later in the section on over-correction, they suggest to use 90% no-load reactive vars... quite a different approach which results in far less correction than correcting to 95% at full load.  I assume this would apply to caps switched with the load because of the cited potential for overvoltage during coastdown after switch-off.
 

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Quote:

Later in the section on over-correction, they suggest to use 90% no-load reactive vars... quite a different approach which results in far less correction than correcting to 95% at full load.  I assume this would apply to caps switched with the load because of the cited potential for overvoltage during coastdown after switch-off.
I should clarify the 90% number is a fraction of no-load vars, the 95% number is a target PF at full-load.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: No load current on a motor

Quote:

Later in the section on over-correction, they suggest to use 90% no-load reactive vars... quite a different approach which results in far less correction than correcting to 95% at full load.  I assume this would apply to caps switched with the load because of the cited potential for overvoltage during coastdown after switch-off.
Not always far less... depends on initial power factor.   

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources