Determining locked rotor current

[Jraef]

When first energized, it takes a cycle or so for the electromagnet, i.e. coil and core, to BECOME an electromagnet via induction and mutual induction. So in that first cycle, the ONLY thing impeding the flow of current is the resistance of the conductors. That "instantaneous" current draw to establish the magnetic field is extremely high amplitude, but of extremely short duration. THAT is the current that is often higher in an Energy Efficient design, because one way they save energy is to keep resistance/impedance as low as possible. This is why the NEC changed to add 1700% as a last resort setting for magnetic trips on circuit breakers, once you have shown that lower settings (i.e. 1300%) couldn't work.

Motor STARTING current is the maximum sustained current the motor will draw when accelerating from a stop to about 80-90% rated speed. That is what will be per the Locked Rotor Code, but is generally estimated at from 500-700% of FLA and generalized at 600%. It's also the amount of current the motor will draw with the rotor unable to turn, hence the term "Locked Rotor Current". Motor Power Factor is so low at that time that most of that current is reactive, so even though you are at 600% current, you are at only 150-160% shaft torque. Peak TORQUE takes place at that 80-90% speed point, called Break Down Torque, and is at around 200-220% of FLT, but by then most of the current is active, i.e. doing work, so it drops. From that point on, FLA % = FLT% (roughly).

 

[Ingenieur]

can't read name plate clearly, assume
440/3, 22 kw, 38.3 A, pf 0.83

total real P = sqrt3 x 38.3 x 440 x 0.83 = 24220 w
delivered shaft P = 22000 w
R losses = 2220 w
P = 2220 = i^2 x R or R = 2220/38.3^2 = 1.51 Ohm
eff = 22000/24220 = 0.91, losses 9%

you could do shaft P = 22000 = sqrt3 x 38.3 x 440 x 0.83 x eff
eff = 0.91
total P = 22000/0.91 = 24220
losses = 24220 - 22000 = 2220 w

lr i = v/R = 440/1.51 = 291 A or 7.6 x fla

if my assumptions are wrong just plug in the correct values
next time please clean the plate

 

[electrofelon]

Thanks iggy. Yeah I took a WAG at 9x and gave the utility that. This morning I filled out one of those Internet "contact us" forms. I'm not holding my breath for a reply.

Lol: remember my rant a week or so about the cad drawing? They asked for cad drawing, I sent, they said send as pdf.....well met with the utility project manager today, he asked for a cad drawing :slaphead: this is going to be a fun one, and just a dinky 200a overhead.

 

[topgone]

Thanks iggy. Yeah I took a WAG at 9x and gave the utility that. This morning I filled out one of those Internet "contact us" forms. I'm not holding my breath for a reply.

Lol: remember my rant a week or so about the cad drawing? They asked for cad drawing, I sent, they said send as pdf.....well met with the utility project manager today, he asked for a cad drawing :slaphead: this is going to be a fun one, and just a dinky 200a overhead.

It could be late but here's the nearest I have found in the WEG catalog:

Output: 30 HP
Poles: 2
Frequency: 50 Hz
Torque: 53.0 lb.ft
Voltage: 400 V
Frame: 284TS
RPM: 2935
Full Load Amps: 40.4 A
Efficiency (100%): 91.3
Power factor (100%): 0.86
Insulation: F
Noise: ---
No load current: 11.0 A
Locked rotor current (II/In): 6.0​

The catalog number is 03036OT3E284TS.

 

[gadfly56]

It could be late but here's the nearest I have found in the WEG catalog:

Output: 30 HP
Poles: 2
Frequency: 50 Hz
Torque: 53.0 lb.ft
Voltage: 400 V
Frame: 284TS
RPM: 2935
Full Load Amps: 40.4 A
Efficiency (100%): 91.3
Power factor (100%): 0.86
Insulation: F
Noise: ---
No load current: 11.0 A
Locked rotor current (II/In): 6.0​

The catalog number is 03036OT3E284TS.

I salute your determination, but it appears that the IP rating doesn't match, if that makes a difference. IP of 21, 23 versus what appears to be 56. I suspect this was a custom one-off.

 

[electrofelon]

I was curious what IP was. Google led me to the answer from this document, which is a nice summary/primer of NEMA and IEC motor data:

https://pdhonline.com/courses/e156/e156content.pdf

 

[Ingenieur]

what does the name plate say?
460?
38.3?
etc

 

[Jraef]

I salute your determination, but it appears that the IP rating doesn't match, if that makes a difference. IP of 21, 23 versus what appears to be 56. I suspect this was a custom one-off.

There are a couple of other anomalies on that nameplate that make me agree with you. One is that somehow, the mechanical kW rating (as in the IEC version of HP) is listed as the same at both 50 and 60Hz. That would not be the case, because a mechanical power rating is a factor of torque AND SPEED so if the torque is the same (480/60 vs 400/50), and the speed is 20% higher, the kW should be 20% higher too. Also, IEC motors with an S1 duty rating are typically NOT going to be designed to have a 1.15 Service Factor for NEMA ratings. I'm thinking maybe this is a 30kW IEC motor design that has be de-rated to 22kW (30HP) so that it can have that 1.15 SF at 480V, which would explain the earlier report of the higher than typical LRC, because the motor is actually larger than they show. By not showing it as 40HP, they don't have to worry about it being "abused" (by IEC standards) in the US when someone drives it into the SF.

 

[electrofelon]

Interesting. Thanks for the comments. Why do I always get the odd ducks :rant:

 

[Jraef]

Interesting. Thanks for the comments. Why do I always get the odd ducks :rant:

It builds character...

 

[Ingenieur]

what is the marking below v and f?
All jxxx ?
or 'delta' ll jxxx ?

what is the last v? 440?
and its associated power looks blurred or restamped to 26 or 27?

 

[electrofelon]

what is the marking below v and f?
All jxxx ?
or 'delta' ll jxxx ?

what is the last v? 440?
and its associated power looks blurred or restamped to 26 or 27?

Yeah not sure....it's in a real tough spot to see. Couldn't even look at it directly, could only take photo. I don't think its dirt and grime either, looks like poor stamping job and/or restamped, corossion, delaminating maybe

 

[kwired]

And I should have added, the PoCo will be interested in Locked Rotor Current because that can have an effect on voltage drop, not the instantaneous/inrush/magnetizing current that only lasts a few cycles.

Too add to what you said, if large enough motor POCO may want some sort of reduced voltage starting method to lessen the "inrush at starting" but motor can still reach a locked rotor condition, like say an auger that is plugged to the point the motor can't drive it at all.

 

[gar]

180301-2134 EST

Jraef:

I disagree with your description in post #21. The statement does not correspond to electrical circuit theory, and analysis.

When first energized, it takes a cycle or so for the electromagnet, i.e. coil and core, to BECOME an electromagnet via induction and mutual induction. So in that first cycle, the ONLY thing impeding the flow of current is the resistance of the conductors.

Start with basics. Instantaneously you can not change the current thru an inductor. Create a series circuit consisting of a battery, switch, resistor, and inductor. Assume zero initial current in all of the components of said series circuit at a time t = 0. Before t = 0 there is no current anywhere. There is a voltage drop across the battery, but none anywhere else.

Just after the switch closes, t = 0, there is no current flow because you can not instantaneously change the inductor current. Following t = 0 the current gradually rises to an asymptotic value of V/R.

A reasonable discussion with circuits and plots is ---
https://www.electronics-tutorials.ws/inductor/lr-circuits.html

I agree with your generalized curve of current for a ferromagnetic series circuit under some conditions. In motor plots I have recorded I have not seen in-rush current. Can you show me plots where an in-rush current is significant at the beginning of the motor starting current plot?

.

 

[Jraef]

gar,
This is a dance we have done before, so I’ll just recycle my response too.
http://forums.mikeholt.com/showthread.php?t=176400&page=3
And no, still no plots, I’m admittedly just accepting established tenets of the industry, as most of us do who have to move on to other more important things. I understand your objection, I can’t personally disprove it, but there are others who have published a different viewpoint that also fits my experience, without public challenge that I am aware of (other than yours). So I choose to accept that.

 

[GoldDigger]

gar,
The DC example is not particularly helpful to understand the AC case:
0. In the AC case the steady state current will not be V/R.
1. The current at time t = 0 will indeed be zero, but that does not mean that the current at some point in the first cycle will not be greater than the peak of the steady state current amplitude (but still with locked rotor.)
2. The inductive component of the locked rotor impedance will be affected by a ferromagnetic core and subject to hysteresis related effects and dependency on the residual flux from the last run. The effect will not be as pronounced as with a transformer, but will be present.
3. It is not, IMHO, correct to assume that the starting and locked rotor current will be limited primarily by the resistance of the windings.

Sent from my XT1585 using Tapatalk

 

[gar]

180302-0842 EST

Jraef:

Mathematically and experimentally your description that initial current at t = 0 is the instantaneous applied voltage divided by R for a series circuit containing an inductor is plain wrong. It is wrong to be repeat this. On the other hand it is correct for a series RC circuit.

It is not correct that inductance is zero at t = 0.

In a series circuit containing an inductor the current just after t = 0 is the same as just before t = 0. In most cases the current is 0 at this time. Note: t = 0 is at whatever time you want to start your analysis.

If a coil of wire is mechanically stable, then its inductance is never less than its air core value whether the core is air or a ferromagnetic material. You can slightly reduce the apparent inductance of an air core coil with an external shorted turn, or some non-ferromagnetic conductive material. The prior history of a ferromagnetic core coil will determine its inductance at t = 0, and this inductance will likely change after t = 0, but not below its air core value.

The above applies for AC or DC circuits.

What happens after t = 0 is a function of the waveshape of the applied voltage and the characteristics of the core material, and any initial flux state of the core material.

Ferromagnetic materials can have a non-zero residual flux state at t = 0 in ordinary circuits, such as, transformers and DC magnets used for lifting or holding. In most AC induction motors I believe there is usually little residual flux and therefore it is unlikely to have a huge in-rush current. Note: by my definition in-rush and starting current are two different things. I want to see worst case experimental evidence of in-rush current for for an AC induction motor.

GoldDigger I will respond later.

.

 

[gar]

180302-1221 EST

Jraef:

At photo P6, see http://www.beta-a2.com/EE-photos.html , which I have shown various times, there is a peak inrush current of about 40 A. The DC coil resistance is about 1.6 ohms. The peak current thru a 1.6 ohm resistor from a 120 V RMS volts sine is about 106 A. Not close to my measured peak. Thus, even though inductance under core saturation conditions is less than under steady-state the inductance is still significant. Additionally you do not see a step rise in current.

With an RC circuit a step rise in current would occur.

With a purely resistive load, if voltage is applied at any time other than a zero crossing, then there is a step change in voltage or current to the value defined by the instantaneous values of voltage, or current, from ohms law and the value of resistance. The largest possible current is Vpeak/R.

Most power transformers are designed with minimum air gap. Thus, if one looks at the transformer unloaded the ratio of input impedances between saturation and non-saturation is quite high. As you increase air gap this ratio diminishes. Also with increased air gap the inductance goes down.

In many ways a motor is different than a transformer. A motor probably has a larger air gap. A motor is less likely to have much residual flux in the core after the motor stops. Thus, major saturation at startup is not likely.

.

 

[gar]

180204-1305 EST

Using my 120 V 60 Hz single phase motor, mechanically unloaded, as a test load, and randomly cycling many times I see no indication of any in-rush current.

I believe this is because after AC input power is removed we have energy stored in the mechanics of the system, this energy back drives the rotor and the motor becomes a generator, as that stored mechanical energy is dissipated the oscillating generated voltage gradually diminishes, and thus gradually demagnetizes any residual flux in the core material to near zero.

I have no direct core flux measurement capability, but motor induced voltage after turn off implies this.

Current and voltage information can be obtained with a scope.

I would like to see any evidence of high motor in-rush current on an induction motor at start up, and an explanation of what might be the cause.

.