Hot panels and plugs

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We found a couple of lighting panels today that service strickly electronic lighting. Circuit breakers {single pole 30a } hot to touch or some warm. After further inspection of the panels we found the following L-1 - 128A
L-2 - 125A L-3 - 118A , the neutral was some where around 60a . The circuits are using 30a single pole cb and pulling up to 22 amps a circuit. Each lighting section uses a 30a twist lock to a 120v plug to supply power, the plugs are burning up at a fast pace , both male and female plugs. There are 20 ballast in each section ,1 t5 bulb for each ballast.
Anyone have a reason the why the plugs are burning up?
Panels are three phase 120/208 with a 200a breaker.
 

SparkBox

Member
Hot Panel

Hot Panel

I did not see what size of conductors are feeding those 30 A twistlocks.
Could it be an undersized SO cable?
Also, since you're dealing with a 3 phase panel, and these lights are 120V, are every three circuits sharing one neutral?
Maybe they are out of phase. Other than that, just check to see if twistlock connections are secure.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Often the terminations inside the cord connectors fail. In many cases is seems that the terminations in the cord connectors do seem to work well with the fine stranded conductors found in SO cords even though that is their intended purpose. As far as the breakers being hot, you have to measure the temperature. The UL standard permits a 50?C rise over a 40?C ambient for the internal contacts. That can result in the exterior surface temperature feel "hot" to the touch.
 
Not sure of so cord size, all circuits have their own neutral conductor . The temp on some of the breakers are 170 -180 f . Used fluke amp . and temp. testers. They installed 3/0 to the 200a panel.
 

wireguru

Senior Member
Often the terminations inside the cord connectors fail. In many cases is seems that the terminations in the cord connectors do seem to work well with the fine stranded conductors found in SO cords even though that is their intended purpose. As far as the breakers being hot, you have to measure the temperature. The UL standard permits a 50?C rise over a 40?C ambient for the internal contacts. That can result in the exterior surface temperature feel "hot" to the touch.

What often happens with cord connectors, is the person installing them uses a drill to tighten the screws on the terminals, and strips the threads. Then the pressure plate isnt tight against the conductor and it heats leading to failure. Another common failure mode is the conductor is inserted too far into the terminal hole and the edge of the conductor insulation catches in the pressure plate when tightening. I guess what I am saying, is most of the plug failures I have seen resulted from improper installation.

And use good plugs, only hubbell, leviton, or marinco second to the first two. Stay away from the P+S and cooper ones....junk.
 

gar

Senior Member
091121-1707 EST

To convert from C to F multiply the C value by 9/5, the ratio of F to C, and add 32 for the temp F at 0 C.

From Don's post 40 + 50 = 90. So 90*9/5 = 810/5 = 162, and 32 + 162 = 194 F.

I measured the voltage drop on a standard 120 V 3 prong 15 A plug and socket with a 12 A load. The result were 22 MV and 33 MV, 0.26 W and 0.40 W. This would get warm, but not unreasonably. The respective resistances are 0.0018 ohms, and 0.0028 ohms.

Thru a Leviton GFCI 93 and 98 MV, total power 2.3 W. A 2 W molded wire wound resistor is about 0.25" dia and 0.6" long. With 2 W dissipated in this resistor you would not want to touch it. The surface area of the elements inside the GFCI are probably somewhat comparable to the said resistor. Thus, internally you can expect to get some substantial temperatures.

On a couple Hubbell receptacles (sockets) I measured 14 and 32 MV, and the other 16 and 25 MV.

I did a little searching for contact resistance specification, None found. However, I found this at Hubbell
Temperature Rise Max 30?C temperature rise at full rated current after 100 cycles of
overload at 150% of rated DC current.
This is at page 4 of
http://www.hubbell-wiring.com/Press/PDFS/H5225R09.pdf

.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
What often happens with cord connectors, is the person installing them uses a drill to tighten the screws on the terminals, and strips the threads. Then the pressure plate isnt tight against the conductor and it heats leading to failure. Another common failure mode is the conductor is inserted too far into the terminal hole and the edge of the conductor insulation catches in the pressure plate when tightening. I guess what I am saying, is most of the plug failures I have seen resulted from improper installation.

And use good plugs, only hubbell, leviton, or marinco second to the first two. Stay away from the P+S and cooper ones....junk.
I have seen it happen on all of the brands and started seeing it long before it was common to use a drill to tighten screws. Trying to clamp the fine strands of a flexible cable between two flat plates is just poor design and, in my opinion, that is the main cause of the failure.
 

__dan

Senior Member
High harmonic high frequency components in the load

High harmonic high frequency components in the load

In addition to the comments above:

I wired some electric resistance duct heaters, about 10 kw each if I recall, that modulated using SCR relays. I oversized everything, circuit feed and bolt on breakers, short run. I was surprised how warm everything ran, even the 30 ft of #8 cu in emt ran warm on a nameplate 28 amp load.

The SCR's were chopping the sinewave badly and the result was the load was drawing higher frequency non-sinewave noise. Losses due to inductive and capacitive circuit elements are very frequency dependent. Losses go up a lot as the frequency of the current draw waveform goes up.

Long way of saying that's what I suspect is contributing to your problem. I get the impression the single lamp ballasts are very "cheap", and the internal ballast power supplies are drawing a lot of high harmonic high frequency noise from the line, also showing up as high neutral current.

Something does not match up in the numbers:
1. 22 amps per circuit (1 section) / 20 ballast per section = 1.1 amps per ballast.
2. 1.1 amp * 120 volts = 132 watts per ballast
3. The T-5 lamp is not drawing more than 8 watts per running ft of lamp. 132 wats per ballast = 19 ft of lamp length, not a single lamp per ballast.

First I would count the total running lamp length and multiply by 8 watts per ft of lamp. This should get you a number pretty close to what the load should be. Compare this to what the load is actually drawing.

If the load numbers are drawing what is necessary and the problem is caused by cheap ballast drawing high frequency noise from the line, one suggestion would be to try to split the load with additional circuits, going from 22 amp per to 11 amps per circuit. Maybe eliminating the twistlok plugs also. Sounds like you are stuck with the ballasts.
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
math brain fart here, help

math brain fart here, help

091121-1707 EST

To convert from C to F multiply the C value by 9/5, the ratio of F to C, and add 32 for the temp F at 0 C.

From Don's post 40 + 50 = 90. So 90*9/5 = 810/5 = 162, and 32 + 162 = 194 F.

http://www.hubbell-wiring.com/Press/PDFS/H5225R09.pdf
I understand and agree with your math. If I convert both temperatures to F prior to adding I get.

40C = 40*9/5=72+32= 104F
50C = 50*9/5=90+32= 122F

104+122 = 226F

226-194 = 32.

OK, shouldn't you be able to add 2 temperatures together in C and the equivalent temeratures in F and come up with the same answer?
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
I understand and agree with your math. If I convert both temperatures to F prior to adding I get.

40C = 40*9/5=72+32= 104F
50C = 50*9/5=90+32= 122F

104+122 = 226F

226-194 = 32.

OK, shouldn't you be able to add 2 temperatures together in C and the equivalent temeratures in F and come up with the same answer?
I should think a little more before I post. The equivalent temperature rise of 50C in F would be 90, not 122 since the +32 is only to change the base reference to the temperature that water becomes completely solid (and that had already been done).

Another DUH on me. :mad:
 
Yes,the twist lock prongs on the plugs are burned up showing a dark color.
The ballast are for t 5 uv bulbs about 5 maybe 6 ' long also there is some kind of solid state card or supply of some type in each module. There are 20 bulbs in each module. I am not formilar with the exack setup , I plan on taking a
look tommorow and get a better understanding of system. I believe the older ballast were magnetic and now are using electonic .
 

dbuckley

Senior Member
I wired some electric resistance duct heaters, about 10 kw each if I recall, that modulated using SCR relays. ... The SCR's were chopping the sinewave badly and the result was the load was drawing higher frequency non-sinewave noise.
Not a good design - for heating loads, burst power control is vastly better than phase shift control for exacly the reasons outlined. We chop sinewaves for lamps 'cos thats the most cost effective way of dimming a lamp, but for heaters, chopping waveforms is unnecessary and horrible.
 

neutral

Senior Member
Location
Missouri
Not a good design - for heating loads, burst power control is vastly better than phase shift control for exacly the reasons outlined. We chop sinewaves for lamps 'cos thats the most cost effective way of dimming a lamp, but for heaters, chopping waveforms is unnecessary and horrible.
I disagree that burst power control is vastly better than phase shift control, if you want to control heat to within 1 or 2 degrees of set point a chopped wave or phase angle firing is by far the best way to do it. The key is to make sure the firing occurs at the exact same time on each half cycle. At set point you only apply the necessary power required to keep the temperature where you want it. Most if not all industrial heating applications such as ovens, furnaces, and other heat forming tools are electronically controlled using scr?s or triac?s using mini computers. We were using 480 volt 3 phase controllers with Ignitron tubes and electronic controllers in the 70?s. Using contactors and applying full power is the cheap way to control heaters, but the least expensive.
 
If harmarnics is my problem than changing the transformer to a k-rated transformer should help along with larger conductors for each circuit. On vfds filters are used for harmonics is there anything for filtering circuits?
 

glene77is

Senior Member
Location
Memphis, TN
Something seems wrong with this,
(A)=128a ,(B)=125a, and (C)=118a and the
neutral has 60 amps?
Ben,

I calculate 8.89 Amps Neutral current,
at power factor = 1.

So, 60 amps indicates excessive power factor / harmonics.

Guess I could calc the power factor, but OP has problems!
:)
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
If harmarnics is my problem than changing the transformer to a k-rated transformer should help along with larger conductors for each circuit. On vfds filters are used for harmonics is there anything for filtering circuits?
Changing the transformer to a "K" rated one does nothing to reduce the harmonics in the circuits....it only provides a transformer that will not be damaged by the harmonics.
 
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