Ampacity question/problem

I am trying to calculate the ampacity through a piece of .01"(. 254e-003m) metal. The metal is stainless steel 304, it has a
resistivity of 7.2e-007 ohm-m, a length of .012m, and an area of . 836e-006m^2. There are tons of references to ampacity ratings of copper and aluminum wires, but no resources I could find that tells you how they calculate the current carrying capacity of a particular wire. I just want to know the amount of current this piece of metal can carry without melting. The melting point of this particular type of metal is 1400-1455 degrees celcius. Any help or references would be helpful.
Thanks in advance,
-Erik
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From ohms law we know that I=SQR(P/R) meaning the Current = the square root of ( Power / Resistance ) if you calculate how much power that length of wire can radiate in its environment (based on its surface area, surrounding air etc.) without going above 1400 C, (BTW the resistance will change with temperature making this a bit harder) Take that power (in watts) divide it by the resistance (in ohms) and take the square root... and you have the current in amps!

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On Wed, 07 Mar 2007 21:28:09 GMT, "cr500r"

Ampacity ratings are not about melting points of materials. They are about preventing temperatures which can cause flashpoint values for surrounding media.
So the MAX temp you want to reach is well below 450F.
Gaineth thyself a clue.
His solution is to epoxy a temp probe end to the center of a test section of his medias and feed more and more amperage to it until a SAFE desired maximum settled in temperature is reached.
THEN, one only REALLY wants to feed such a conductor a huge percentage LESS than that value in a good, proper design.
IF he wishes to use it as some fusing element, the rep[eatabiltiy from one to another will likely be pretty bad, and it will pose hazards if not contained in a glass of flame resistant fiber tube.
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Your right, but I was just answering the question he asked, assuming nothing about it. My guess is it's from a textbook problem in some class, and has about as much to do with the real world as does your attitude :-)
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On Thu, 08 Mar 2007 01:29:29 GMT, "cr500r"
You're nothing but a top posting Usenet retard.
It is "You're", dipshit.
You should have paid attention is school, and you should bone up on the conventions of a forum BEFORE you invade it, interloper.
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On Mar 8, 3:04 am, MassiveProng

I did pay attention is school...most of the time. The piece of metal will be welded to the base of a lamp (1500W lamp). I just want to be able to calculate the amount of current that can pass through the material before the temperature rise becomes too great. Sorry for asking a question
-Erik
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Erik, He was just faming me, don't worry about him. So what voltage lamp? Is it running off a ballast like a HID lamp?
I really am trying to help here.
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The highest wattage lamp is 1500W, and yes they are HID lamps that run off a ballast. The piece of metal is welded to the lamp lead wire on the arc tube. It sits in a detent that is molded into the glass, where the threads are for screwing on the base. Once the base is screwed on there is a portion of the metal that folds over the outside of the base and is secured with a high frequency resistance spot welder, to prevent the base from coming loose when the lamp is taken out of a socket. I was using the equation you had stated before to find the actual resistance of the piece of metal, but i'm still having trouble relating that to the current carrying capacity(steady state and transient) that it can handle. I know that the current capacity needs to be 7A steady state and 15A max(transient), so maybe i'm going about this wrong and just need to figure out the amount of heat that is transfered at those current ratings. Any help is greatly appreciated. I am a programmer and physics is not my strongest point.
Thanks, -Erik
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snipped-for-privacy@yahoo.com wrote:

If you are welding this to the base of a lamp, then ampacity is a mute point. You cannot use exposed metal as a conductor and be safe. You cannot allow the design to cause inductive heating in that plate and be safe. You need to use mechanical and electrical construction that will keep any electrical current away from exposed metal on that lamp.
MP gave a great description of ampacity. Many could learn from it.
Ed
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On 8 Mar 2007 05:22:43 -0800, snipped-for-privacy@yahoo.com Gave us:

I was talking to the other person that top posted, not you.
You didn't do anything incorrectly, until you forgot how to trace a thread or even read the fact that I quoted HIM in my response to HIM.
It's ok though dude. If you are worried, back up the steel with a copper strap along it.
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Connecting two different metals and exposing them to air will cause corrosion, adding heat will accelerate the corrosion.
BTW I'm going to keep top posting just to piss off MassiveProng, he doesn't add anything to this group except insults.
message ... If you are worried, back up the steel with a

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On Sat, 10 Mar 2007 21:28:40 GMT, "cr500r"

Yeah, right. Aren't you the twit that thinks that ampacity refers to a melting point?
Were not all your pathetic calculations based on a melting point?
The two are NOT related... at all!
You refer to FUSING point. An entirely different realm of electrical engineering and physics.
Get a clue, dipshit. And YES, you deserve any name I throw at you for your retarded top posting baby bullshit, especially when you even declared that you were doing it deliberately.
I do not expect you, however, to be mature enough to see how adolescent you are.
You are proof that numerical age does not an adult make, little boy.
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No, Erik started off with the melting point, then he realised it wasn't part of the solution. My calculations proved that the stainless steel he had in mind would work.
First you think those calculations are overkill, and now they are pathetic. I guess that means you don't understand them

Further proving you don't understand, Erik understood, why can't you?

I must admit it's a little fun to see you rant and rave, but I'm bored now.
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On Sat, 10 Mar 2007 22:12:52 GMT, "cr500r"

You have not proven anything... ever. Not in this group or anywhere else. The only thing you have proven is that you ignore standards and conventions due to being a lazy TOFU fucktard.
That alone should have gotten you ignored by all from the get go.
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It never ceases to amaze me how these self appointed moderators get off on correcting grammar and spelling. I'm surprised he didn't use the word "thus" in his stereotypical flame of a post. I wonder, does he walk around correcting grammar in real life? A great way to win friends and influence people!
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Well, here's another surprise for you. People outside the USA (They do exist) are going to wonder what on earth "ampacity" means, until they realise it's SixPack-talk for "current carrying capacity"
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http://en.wikipedia.org/wiki/Ampacity
wrote:

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[Top-posted Wikipedia link]
http://en.wikipedia.org/wiki/Ampacity
Yup, as the Wikipedia page says:
(I quote)
"The examples and perspective in this article or section may not represent a worldwide view of the subject. Please improve this article or discuss the issue on the talk page."
"Categories: Limited geographic scope | Electricity"
On the talk page:
"If the word "ampacity" is only used in the US, the article should state this. Otherwise, the article should be edited to be less US- centric (is there an international standard about ampacity?) Jushi 11:27, 30 November 2006 (UTC)"
Like I said. SixPack-talk.
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Just another made up word, not really that surprising... kind of like "SixPack-talk"
BTW I didn't use the word, the original poster did.
wrote:

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And the lamp operates at around 260V, with the current around 4.2A. I don't know if i mentioned that before.
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