How do you read contactor nameplate? 3

[CtrlAltDel]

I was called to try and figure out why a contactor in an elevator panel keeps burning out. This was the second time it happened. Although I suspect the contactor was under-rated I want to make sure. I got this on the partially burned nameplate of the sprecher+schuh brand contactor:

AC-1...
AC-3
V 415 500 610
kW 25 31 32

600Vac, 72A

120V, 60Hz
110V, 50Hz

Can you please enlighten me on the conventions used so I can interpret contactor nameplate ratings correctly, especially what AC-1 and AC-3 means?

What is the relation of the 600Vac, 72A to the V and Kw rating under "AC-3?"

i=32kw/(1.732x610)
i=30.2A

Thanks!

 

[jraef]

AC-1... this is the rating for switching non-inductive (i.e. resistive) loads. This probably went with the 600VAC 72A meaning that it is rated for switching loads up to 72A @ up to 600VAC as long as they are not motors.

AC-3... this is the rating for switching inductive loads, i.e. motors. So at 415V, it is rated to switch a motor of 25kW max. At 500V, it is rated for 31kW max. At 610V, it is 32kW max.

V 415 500 610
kW 25 31 32

600Vac, 72A... as mentioned above, this is the maximum non-inductive current rating.

120V, 60Hz... these ratins are probably the coil voltage
110V, 50Hz

What is your load? Do those last numbers you posted mean that you have a 32kW motor? If it is at any voltage lower than 610, that contactor is undersized. Even then, the general rule on sizing contactors for elevator applications is to oversize them because the duty cycle is higher than normal.

"Venditori de oleum-vipera non vigere excordis populi"

 

[CtrlAltDel]

jaeref,

thanks for the very quick reply. it's all useful! I played with the numbers and tried to see what Amps I would get for a 610V line and 32kW (32kVA assumed) load: the figures came from the "AC-3". The initial figures and the AC codes reveal some clues already. I will return to site tomorrow to get more details. The load here is an elevator motor and the subject contactor was situated between the output side of an inverter and the motor terminals.

Thanks again, man!

 

[Skogsgurra]

CtrlAltDel,

You say that "The load here is an elevator motor and the subject contactor was situated between the output side of an inverter and the motor terminals"

That is not very common and it is a potential death trap for contactors.

I do not think that the contactor operates when the motor is running at full speed but opens when the motor has come to a stand-still. The motor then has close to full current to keep the elevator and also to keep the motor magnetized. That means that the contactor is opening with a very low-frequency current into a mostly inductive load.

The contacts are designed to open a 50 or 60 Hz current - not one that is close to DC. Make sure that the stopping sequence is something like this:

1 elevator is level with floor and speed is zero.

2 apply brakes

3 reduce current to zero

4 open contactor (I do not see why there is a contactor in the first place. A design flaw? Or is the inverter used with several elevators?).

 

[jraef]

I totally agree with skogsgurra. That is a bad application for a contactor, and in fact is also not too good for the VFD if the contactor should open under full load.

I should have also mentioned that it is highly unlikely that you will have 610V. I have never heard of that voltage level in building systems, but then again I am not an expert in worldwide power systems. What country are you in anyway?

"Venditori de oleum-vipera non vigere excordis populi"

 

[CtrlAltDel]

Hi Jraef, skogsgurra,

I am looking now at the diagram of the elevator controller. Its a vvvf system (I believe that would be variable voltage variable frequency.) I see that the contactor is located as mentioned before. I cannot post the diagram but here's as much as I can describe from line side to load side in series:

460V, 3-phase, 60Hz power supply
Fuse
Reactors
Rectifier-Inverter Section
.... Rectifier
.... Capacitor bank
.... Silicone Controlled Rectifiers
Contactors
AC Motor

There are other components like regenerative resistors, velocity encoders, fans, etc.

I used 610V only as an estimate. 460V and its correspponding kW is not given in the contactor nameplate. However if I use the given 500V paired 31kW (closer to 460V) I would get:

i=31kW/(1.732*500V)
i= 35.8A

(i=34.8A for 415V and 25kW)

I think this would be a conservative estimate of the rated amps of the contactor, don't you agree? BAsed on the meaning of AC-3, then this means that the contactor may not conduct more than 35.8A continuously for inductive loads, such as the motor in question.

I will measure v and A today at the site. Should I get a current value more than 35.8A at the contactor input_output would it be logical to say that the contactor was under-rated for its specific use (AC-3?) and hence caused the early degradation and eventual burn out of the contactor? If so would it be wise to suggest to replace the contactor based on the actual current load, perhaps over-size it to the next higher rating for good measure?

You said that the contactor was in a precarious condition in the first place due to its unique location in the circuit. Also that the contactor was designed to operate on 60Hz/50Hz and not low frequency which it will certainly conduct. I wonder why it was designed this way.

I am in Manila, the Philippines. I work for a local engineering outfit and the customer in question is a 47-storey office building. I'm sure you'll understand if I defer to the elevator manufacturer and designer, which I will not name. This particular job is actually free for the customer, part of the benefit they get from our company.

Thanks guys. I learned a lot from you!

 

[Skogsgurra]

A 47 storey building! Wow! Not a single such building here in Sweden. I would be very concerned about the contactor in such a high-rise building. I would be in a five floor building too, for that matter.

When you go there. Do observe when the contactor opens. And keep an eye on your current clamp. By the way, you need a DC/AC clamp to measure those low frequencies. An AC clamp will not work. And you possibly also need a clamp that can take the interference from the PWM voltage. If available, bring a recorder or a portable scope to record the current.

If there is any substantial current when the contactor opens then you will probably also see an arc inside the contactor (do not remoce arc guards to see better! I have seen that being done - not good).

If current is zero, then the designer did it right (the 1-2-3-4 sequence above). If you have current and if you see an arc, you should contact the manufacturer and have him redesign the circuit and/or its control program. Best thing would be to remove the contactor altogether.

 

[jraef]

On thing I feel you should avoid doing is interpreting the switching rating of the contactor. The inductive ratings are different from the strictly thermal current ratings for a reason. Switching an inductive load has complex consequences on the life of the contact material beyond the heat produced by the current itself. Use the kW raitng as a guide. If it s rated for 31kW @ 500V, and you are feeding it with 480V, the kW rating will decrease by the ratio of the voltages. 480/500 = .96, so the 31kW x .96 = 29.8kW @ 480V, which equates to roughly 40HP. If your motor is more than 40HP, that contactor is too small.

Even if it is a 40HP motor, IMHO that would still be too small of a contactor based upon my experience. As Skogsgurra mentioned, switching at low frequency is more like switching DC, and if you looked up that contactor for a DC rating you would see it is much lower than 40HP because it draws more of an arc when the contacts open. Harmonics in the load side of the VFD also put additional thermal stress on contactors, and I always recommend oversizing the contactors by at least 1 size just for that reason.

The only possibility of success here is if the contactor is there strictly as a safety isolation device, meaning that it is NEVER intended to open or close under load. This actually may be the case because the elevator designer may have known better than to open the contactor on the load side of the VFD because of the risk of damaging the transistors. If that is the case, then only the thermal current rating matters, in your case, the 72A. If it is continually burning out, there may be a problem with the control system and it may be opening and closing that contactor when it shouldn't be.

It is also possible, being an elevator, that the contactor is tied to the Emergency Stop circuit, and someone using the elevator is pushing the E-Stop button frequently instead of just in an emergency. Have them install a hidden video camera and look for someone engaging in "elevator trysts"!

"Venditori de oleum-vipera non vigere excordis populi"

 

[Marke]

Hello CtrlAltDel

You have been given a mountain of good information and advice here, but I would like to add my two cents worth as well.

When the contactor fails, what is the mode of failure? This will give us a clue as to the cause.

Is the coil burnt? are the contacts welded?

The AC3 rating is based on the contacts carying an over load current for a short period or time. The total rating of the contacts is thermal. The AC1 rating is known as a resistive rating and is the maximum continuous current that the contacts should carry with no over loads at the rated temperature.
AC3 ratings allow for an overload current. The contacts will heat more due to the increase in current during the overload (i2r) and so the continuous current is reduced to ensure that the maximum temperature of the contact material is not overheated. The frequency of overload and the duration of the overload is a very important part of the AC3 rating. If there is a very frequent overload, the AC3 rating should be further reduced.

With the contactor in the position that you indicate, I would not expect it to operate frequently, or to experience overload currents. I would suspect the voltage on the coil is dropping and is reducing contact pressure. This will cause the coil to fail and can also cause the contacts to be burnt.

I would suggest that you look at the control voltage applied to the relay coil. I have experienced many strange voltages used in elevators. Next time you replace the contact, try to get one with an electronic coil and a wide voltage range. This may help.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[CtrlAltDel]

Hi Guys,

Thanks for all the information. The more I read about them the clearer the picture gets. Here's what I gathered so far:

1. The rectifier-inverter is IGBT (not SCR as I mentioned before)
2. Contactor rating at AC-3: 500V, 30kW
3. Motor rating (elevator hoist motor itself) loadside of contactor: 460V, 40Hp, 57A, 60Hz, 1160rpm
4. Actual maximum current conducted by contactor as measured using Fluke 43: 41.7A rms starting currrent, at least once every minute approximately. Normal running current is 20-30A depending on elevator climbing direction. The contactor opens and closes as often as once every minute on average, more when the building is busy with occupants going up and down.
5. Maximum voltage at contactor terminals varies between 0 volts and 396 volts rms.

Condition of burnt contactor: by merely looking only as we are not allowed to dissect it, coil seem intact, contacts appeared with soot and melted portions visible.

Line voltage during measurement: 466Volts

Please see these pictures for a better view of the V and A rms strip chart and inrush measurement (Ams peak) on contactor terminals:

http://us.f2.yahoofs.com/bc/426454cf_4e8e/bc/My+Documents/STRIPLR.BMP?bfKlFZCBy5FvhkqT

http://us.f2.yahoofs.com/bc/426454cf_4e8e/bc/My+Documents/INRUSHLR.BMP?bfKlFZCBxVVXlIbX

I am already formulating my findings and recommendations but I will go to the site again to gather other pertinent data.

Your analyses are welcome! Thank you!

 

[jraef]

Hmmm... neither link appears to work for me.

I still think your contactor is undersized. By my calcs it is good for about 37HP at 460V. Add an occasional high duty cycle and the location in the power circuit and you have a recipe for failure. When, as you say, there is high traffic on the elevator, the contacts will heat up significantly and if not given enough time between starts to cool off, they will begin to vaporize more rapidly. Add to that the extra heating effect of being on the load side of a VFD and you can expect relatively short contact life.

"Venditori de oleum-vipera non vigere excordis populi"

 

[CtrlAltDel]

So sorry about that invalid link above. Please try this other link again:

http://rapidshare.de/files/1344746/contractorProblem.zip.html

This is a free web hosting site so you might have to wait 30 seconds before the file is available for download. It contains images of the failed contactor and the rms and peak charts as described earlier. To the best of my knowledge it is free from viruses.

Once again, thanks for all the information!

 

[jraef]

Nope, still doesn't work. All I get is an advert for Rapidshare webhosting.

"Venditori de oleum-vipera non vigere excordis populi"

 

[Marke]

Hi Jeff
Got to the bottom of that page and there is a download option. You follow that link and it does work.

CtrlAltDel
I suspect that it is just related to the number of operations that you are doing. There does not seem to be a serious overload, but any overload will increase the heating of the contacts. If the operation is very frequent, then the average of the current squared will be high.
I suggest that you work out the worst case average of the current squared to get an indication of the heating effect of the current. The graph excerpt does not look like a problem from a current magnitude point of view.
The photograph suggests that the centre phase has been the hottest which is what I would expect from a current heating or the contacts.
The one thing that does concern me is that this is apparantly switching on the output of an AC drive and that just does not stack up at all. If you said that it was switching on the output of a DC drive, then that would make perfect sense, and in that case the DC component of the current would cause the contacts to burn.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jraef]

Thanks Mark, I see it now. I thought that was just another add to get me to sign up.

I concurr on the number of operations as the prime suspect, it does not appear as though the current ever exceeded 41A, well within the range of that contactor. But if you look at that 2nd strip chart, it appears to show 14 or 15 operations within 7 minutes. Although not necessarilly a problem in and of itself, combined with being on the output of a VFD with high harmonic content, those contacts are getting hot. And the center contact would get hottest because it has hot contacts on each side and cannot dissipate as quickly as the other 2.

"Venditori de oleum-vipera non vigere excordis populi"

 

[CtrlAltDel]

Hi Fellows,

Thanks very much for letting me pick your brains. I am now better informed to support my recommendation to the customer. I believe the prudent suggestion would be to have them replace the contactors (next time it failed) with one of higher rating. This is justified owing to the reasons just discussed.

Thanks a lot! You've been very helpful!