No load current on a motor 1

[rockman7892]

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?

 

[ScottyUK]

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.


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

 

[oldfieldguy]

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

 

[waross]

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

 

[zlatkodo]

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

 

[edison123]

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

http://www.edison.co.in

 

[electricpete]

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.

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[Skogsgurra]

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

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

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

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[electricpete]

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.

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[rockman7892]

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?

 

[edison123]

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

http://www.edison.co.in

 

[waross]

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

 

[Skogsgurra]

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

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

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

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.

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[waross]

Good point, Gunnar.

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

 

[LionelHutz]

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.

 

[Skogsgurra]

The Ossanna diagram is very illustrative. There is one at

http://www.ew.e-technik.tu-darmstadt.de/cms/fileadmin/pdf/EMA/Ossanna_circle.pdf

It is an old reference. But it may be useful for newcomers.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

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.


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[electricpete]

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"

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