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?
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
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RE: No load current on a motor
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 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
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
Muthu
www.edison.co.in
RE: No load current on a motor
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.
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RE: No load current on a motor
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
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RE: No load current on a motor
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RE: No load current on a motor
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.
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RE: No load current on a 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
Muthu
www.edison.co.in
RE: No load current on a motor
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
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: No load current on a motor
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.
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RE: No load current on a motor
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: No load current on a motor
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
It is an old reference. But it may be useful for newcomers.
Gunnar Englund
www.gke.org
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RE: No load current on a motor
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.
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RE: No load current on a motor
"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"
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RE: No load current on a motor
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RE: No load current on a motor
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.
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RE: No load current on a motor
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RE: No load current on a motor
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).
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RE: No load current on a motor
"..with assumed slip = 0.01"
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RE: No load current on a motor
"The best fit occurs where the minimum SWSFE occurs which is at target X1/XM=0.041"
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RE: No load current on a motor
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
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.
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RE: No load current on a motor
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?
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RE: 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.
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RE: No load current on a motor
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.
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 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.
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RE: No load current on a motor
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RE: No load current on a motor
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RE: No load current on a motor
I am not so sure if I would buy an induction motor from Mr. KIM or Mr. LEE
Gunnar Englund
www.gke.org
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RE: No load current on a motor
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If we learn from our mistakes I'm getting a great education!
RE: No load current on a 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
----------------------------------
If we learn from our mistakes I'm getting a great education!
RE: No load current on a motor
htt
Here are some more 200hp, 1800rpm, 460 vac motors.
http
FL PF = 0.883, NLA = 53.5
http
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
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RE: No load current on a motor
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).
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RE: No load current on a motor
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
RE: No load current on a motor
Gunnar Englund
www.gke.org
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RE: No load current on a motor
RE: No load current on a motor
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.
ht
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.
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RE: No load current on a motor
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RE: No load current on a motor
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