|
0 members (),
220
guests, and
12
robots. |
|
Key:
Admin,
Global Mod,
Mod
|
|
|
|
Joined: Nov 2002
Posts: 794 Likes: 3
OP
Member
|
In Wedensday's Star Ledger there was a story about a fatal house fire caused by a faulty outlet. http://www.nj.com/news/ledger/jersey/index.ssf?/base/news-4/1156310312169680.xml&coll=1 from the story: The culprit was an old electrical outlet in the first-floor living room, Rubbinaccio said. The outlet no longer held a tight connection with a floor lamp, and the resulting heat in the wall started the fatal fire, the prosecutor said. [This message has been edited by wa2ise (edited 08-23-2006).]
|
|
|
|
|
Joined: Jul 2004
Posts: 625
Member
|
I'd love to see some more information on this. I'm wondering if the outlet box was plastic or metal, and why it didn't contain the failure. And what changes might have been made to better contain the failure.
|
|
|
|
|
Joined: Jul 2002
Posts: 8,443 Likes: 3
Member
|
Let's look at it this way,
If you have a contact resistance of 1 ohm at a given point on a circuit, say a receptacle hot contact to it's corresponding pin contact, you have:
(Assuming a 120V supply voltage)
P= V(squared)/R
= 120(squared)= 14400/R
= 14400/1
P = 14400W
P is the power dissipated at that point in the circuit, usually as heat.
Now, 1 ohm is quite high as far as contact resistance is concerned and most receptacles give contact resistances in the order of milli-ohms.
I could be totally wrong with this calculation above too, but it does show how a poor connection anywhere (say at a junction box in a roof void) could pose a fire hazard.
|
|
|
|
|
Joined: Aug 2002
Posts: 1,691
Member
|
The outlet box could have gotten hot enough to light something flammable?
Or maybe the sparks that shot out the front (the plate and face of the socket are usually plastic anyway), ignited whatever was in front of the device?
|
|
|
|
|
Joined: Jul 2004
Posts: 625
Member
|
Trumpy, the formula you used only applies if the full 120V is across the 1 ohm. The formula you want in this case is P = I^2*R. To make the numbers round, suppose the lamp had a 120 watt light bulb (try to find that one on the shelf somewhere!). Then:
The load current is 120W/120V = 1 amp.
The power dissipated at the bad receptacle is (1 amp)^2 * 1 ohm = 1 watt.
I would certainly expect that a box, whether plastic or steel, could contain that.
I suspect that the resistance of the receptacle was higher than 1 ohm. Still, if you assume that the resistance of a light bulb is a constant (it isn't), and solve for the worst case at the box, you will find that the maximum power that the receptacle could be dissapating with our hypothetical 120 watt light bulb is 30 watts. Which is quite a bit of power in a box. And which brings me back to my question of the relative performance of plastic and steel boxes.
[This message has been edited by SolarPowered (edited 08-24-2006).]
|
|
|
|
|
Joined: Dec 2003
Posts: 751
Member
|
What is the greater likely hood to start a fire is the arcing fault. The spark from a small current can draw a high temperature spark, which can ignite any nearby easily ignited flammables. (dust bunnies, paper, even charred wood from this arcing fault over a period of months)
Why else do you think the arc-fault CB was required?
Earl
|
|
|
|
|
Joined: Dec 2005
Posts: 869 Likes: 4
Member
|
Yes Mike, the formula you used is ok but the application was wrong.
The contact resistance and heat generated from it is dependant upon the load flowing through the contact area.
The voltage is not immediately the crucial factor although it can help to sustain an arc across the contacts for a period of time which could start a fire.
Say at 10 Amps and 1 Ohm.
P=I²R
P=10²*1 = 100 Watts dissipated at the contact surface area.
In case the Neutral contacts are poor and of similar resistance an other 100 Watt may be dissipated here too.
Certainly not 14 kW.
The product of rotation, excitation and flux produces electricty.
|
|
|
|
Posts: 43
Joined: September 2013
|
|
|
|
