30A to 50A Adapter Causes MH Fire

[Sky_Boss]

OK...perhaps a bit of a sidebar but lets accept that coiled power cables create heat through induction. That would tell me there is added resistance. True?

So...would it make sense that avoiding coiled cables would also reduce the electric bill?

[WDK450]

One thing I have noted with RV power cables/park pedestals is that many of the high current electrical connectors are heat corroded, especially in the pedestals. Many times you can feel these connections getting hot, or see that your plug prongs are discolored. Short of finding another space with better connections, I recommend using a high quality electrical contact restorer spray each time you plug in. I like the Caig Cramolyn DeOxideIt product, which can be purchased in small cans from Radio Shack, or more larger economical cans from Amazon ( item # B00006LVEU ).

[dennis45]

Originally Posted by WDK450

One thing I have noted with RV power cables/park pedestals is that many of the high current electrical connectors are heat corroded, especially in the pedestals. Many times you can feel these connections getting hot, or see that your plug prongs are discolored. Short of finding another space with better connections, I recommend using a high quality electrical contact restorer spray each time you plug in. I like the Caig Cramolyn DeOxideIt product, which can be purchased in small cans from Radio Shack, or more larger economical cans from Amazon ( item # B00006LVEU ).

Thank you for this information. Good stuff to carry in the Electrical Bay.

[FlyingDiver]

Originally Posted by Sky_Boss

OK...perhaps a bit of a sidebar but lets accept that coiled power cables create heat through induction. That would tell me there is added resistance. True?

So...would it make sense that avoiding coiled cables would also reduce the electric bill?

Yes, but the actual additional current due to the added resistance will be fairly small. Probably not enough to effect your bill. Think how much heat a single light bulb puts out, and how much that one light will effect your usage. I'm not going to attempt to do the math right now, but I'd put a WAG at maybe 40-50 watts at most. Enough to heat up the wires, but pretty minimal on your bill, compared to AC units, etc.

[jerichorick]

Thank you for your kindness, Adaycj.

I started answering this early this morning but DW got up so coffee for her was priority.

After reading your great input I realized that I was posting from years of listening to rumours. When applying a little of my knowledge (very little I guess) it hit me. A swing neutral can potentially cause intermittent high resistance and sudden power surges if it makes solid. HVAC units, inverters and electronics are not very pleased when this happens and it may even be fatal to some of the equipment. Brownout conditions are never good for our equipment.

The term 'neutral' is a bit misleading. If both L1 and L2 are pulling the exact same load the neutral will have zero current. L1 and L2 are 180 degs out of phase. (Just for the unknowing to understand.) But it only seems reasonable if a load on L1 is greater than that of L2 the neutral is no longer neutral. Enough techno babble.

All 50A services have ~240V from L1 to L2 and ~120V from L1 or L2 to neutral and also to ground if it is functioning correctly. So, in summary, if the neutral connection or the L1, L2 connections become corrupted for any reason and build parasitic resistance the some of the usable voltage is lost at the parasite resistance. A strange thing happens in electrical circuits. When voltage goes down for a given load the current goes up. In the most simple of terms, a devise that require 10 watts to operate will demand it even when the voltage drops. If the design of the circuit can not handle the increased current to maintain the demanded watts the device burns out.

I hope this post is not too confusing. Again I that Adaycj for provoking me to think. He may be regretting it though!:>))

[Lifemember]

If the N goes open for any reason the full 240V is applied to the RV service panel and expensive things then happen.

this is incorrect! if the neutral goes open all 120 volt systems will not work. their is always 220 at the RV service panel with a 30 or 50 amp system.

when you coil up the cord it becomes a magnet generating flux lines and creating heat. the more you draw the hotter it will get.

add to that a dog bone with another fifty feet of cord and you have created more voltage drop. the more cord used the greater the resistance. I x E = R

[krautkamper]

Quote:
Originally Posted by 96 Wideglide
I said the same thing on this forum several months ago, and was corrected by many, that this is standard procedure with 30A coahes connected to 50A pedestals .

Originally Posted by gemini5362

I do not care how standard it is. It is wrong and dangerous

This is potentially dangerous. Let me explain why: Yes, there is a 30A breaker in your coach, however what about the wire PRIOR to the breaker, and whatever it runs through; transfer switch, etc? If you plug a 30 A cable, likely 10 AWG into a 50 A circuit breaker, that breaker will not trip until the current reaches 50+ amps. Your wire can melt and possibly catch on fire prior to that. If there is a short circuit of any type prior to your main house breaker; broken or pinched wire, loose screw, etc, you are at risk since the wire may fail before the breaker trips. Unfortunately this may be a standard practice, as if you only have a 50A supply available and 30A cord, what can you do besides add a protection device or move to another pedestal or campground?

Interesting discussion on the inductance vs. heating of wire. I don't dispute it from a physics standpoint, however I think the issue of heat dissipation may play as much of a role in the cables heating more when coiled as does the increased inductance. That is why the NEC rates cable ampacity as a function of the quantity of cables in an enclosed space vs. free air. If the wires are coiled (especially tightly), there is nowhere for the generated heat to dissipate. Uncoiling a wire yields a much greater surface area available to dissipate heat.

[Libero]

WOW - sooo much chatter on this subject, some sound electrical info and some real crap and some in the middle with good background info.

The OP said the 50-30A connector was shorted and caused the fire. But why did it not blow the 30A breaker at the pedistal if there was a short circuit? The fire inspector should and likely did look at that aspect. Perhaps even though the guy was careless, dumb or both for doing what he did vis-a-vis making that connection inside his coach and not using a short cable to convert 50A to a 30A outlet and then leaving his 50A cable coiled up inside the coach, the circuit should have popped if there was a short circuit. The wire is burnt back from the connector. Just an Observation. He might have a claim against the campground for faulty breakers. and good luck with that. I would think there was a bad connection in the connector causing increased resistance, more voltage drop, hence more heat generated (not not more current drain necessarily) and that got so hot it melted the wire and started a fire. Just one dumb RVer putting himself, his family and others at risk. A good object lesson to others. Nothing is ever useless, it can always be used as a bad example.

Pushman - good reporting. It makes people think and may save some other electricially unsavvy RVer but fixing stupid is sometimes difficult.

But looking at all the buzz on this gave me a lot of chuckles and others going off on tangents about plugging a 30A coach into a 50A supply. Some valid stuff there but a bit of the subject of the post. For Example:

one poster said if the neutral wire is open then full 240V is applied to the panel. HUH - the full 240V is there all the time but not normally used as 240 in an RV. The fact that a Neutral is open does not change the fact that the 240V is at the panel??

others talk about "pulling current" HUH - current is pushed by voltage (EMF) and is the product of the voltage and resistance. Current is NEVER Pulled.

Others talk of the inductive load because of coiled wire on the 50A cord causing heat that caused the fire. Questionable in this case, albeit a tightly coiled cord will provide inductive reactance but because of the fire source being near at the plug and not in the coiled wire and of course, Pushman was there, saw the fire, took the pictures and said it was the plug. I believe the plug was the cause. The 50A cord has a lot of insulation seperating the wires, it is not a tightly wound coil and with the thick insulation on it, and while it might get warm, probably not hot enough to start a fire with less than 30A going through it. Speculation on my part because I have not tested it. It is easy to test. Check the voltage at the plug and at the transfer switch and see the voltage drop. The calculate the power loss, which will be dissapated in heat or measure the heat with IR thermometer to see if it is near the combustion point of the materials around it.

Here is a real zinger - "current flows in one direction in one wire and the oppoisite direction in the other direction" I almost fell off my chair laughing at that one. Really??? What a hoot.

Now this one really made me chuckle also - "the more cord the greater the resistance" - true, but then followed by this "I x E = R". is this the new Ohms Law.

Current is directly proportionally to the voltage and inversly proportional to the resistance. That formula does not cut it. I x E = Power

how about some who said an increase in resistance will cause an increase in current. See Ohms Law above. Really funny.

bottom line, good reporting, some interesting and worthwhile discussions and some a real hoot.

Back in the day just after I got my MEng in Aeronautical Engineering, I taught electronics theory and advanced calculus to young engineering students. Been about 39 years since I did that and in my 70s so the grey matter loses a lot, but the basics are still there, sometimes I forget what day it is now.

Good OP

[myredracer]

Originally Posted by steveclv

All I can do is point you to a Tutorial on AC Inductance and Inductive Reactance HERE

There will be a test when you have read it

Oh gawd... I used to know all that sort of stuff once upon a time.... Now it just makes me feel queazy and want to go lie down. Thanks for that!

[Lifemember]

but then followed by this "I x E = R". is this the new Ohms Law.

My fault typo. IXR =V it was over twenty years ago.

[Libero]

Originally Posted by Lifemember

My fault typo. IXR =V it was over twenty years ago.

Nice try, but current is directly proportional to the voltage and inversly proportional to the resistance.

This would make the formula for resistance as follows:

R= E/I

R= E2/P

R= P/I2

E2 = Voltage Squared

I2= Current Squared

P= Power in watts.

It is a real long time for me but since I taught and used this stuff for more years than I would like it to be, but some of it is carved in my stone head. It is easy to have a lapse in memory, I am certain you had this nailed down at one time. It provided me with a chuckle, and I had no doubt you knew OHMs law but just had the relationship mixed up a bit.

Only taught for 3 years at military college, the rest of my time I was doing real engineering work. Don't even ask me to do advanced calculus now. Use it or lose it is true.

[Cobra1]

You gotta love all of the info available on this site. Here is my contribution.






I believe the bottom line is that everyone needs to be aware of their adaptors serviceability and be aware that bad connections get hot under load and will cause problems.
Just another thing to add to the list of items to check every so often.

[Lifemember]

Ok you made me take the old books out ...lol

V = I x R (Voltage = Current multiplied by Resistance)
R = V / I (Resistance = Voltage divided by Current)
I = V / R (Current = Voltage Divided by Resistance)

22 years as an electrician. in my first life... but that seams like hmmm it is 20 years ago. dang time fly's use it or lose it!

[steveclv]

It's a bloody ac circuit!!! Not dc