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

Reply to
edannemiller2002
Loading thread data ...

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!

Reply to
cr500r

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

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.

Reply to
MassiveProng

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 :-)

"MassiveProng" wrote in message news: snipped-for-privacy@4ax.com...

Reply to
cr500r

On Thu, 08 Mar 2007 01:29:29 GMT, "cr500r" Gave us:

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.

Reply to
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

Reply to
edannemiller2002

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!

"MassiveProng" wrote in message news: snipped-for-privacy@4ax.com...

Reply to
cr500r

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.

Reply to
cr500r

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

Reply to
ehsjr

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"

Reply to
contrex
[T (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.

Reply to
contrex

Just another made up word, not really that surprising... kind of like "SixPack-talk"

BTW I didn't use the word, the orig>

Reply to
cr500r

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

Reply to
edannemiller2002

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

Reply to
edannemiller2002

The Neher McGrath Paper published by IEEE in 1957 has information on materails other than copper and aluminum used as conductors.

Reply to
Gerald C Newton

OK, let's make sure I've got it right:

lamp operates at around 1500W 260V, with the current around 4.2A (steady state) Transient of 15A at startup. Thickness 0.000254 meters stainless steel 304, a length of .012 meters area of 836e-006 meters squeared. ( is this the cross sectional area, or the surface area of one side, or what?)

resistivity of 7.2e-007 ohms per meter (is this right?)

So it's not a wire, its a piece of thin (0.254mm) stainless sheet metal 12cm long and how wide?

Reply to
cr500r

I think I figured it out from equations I found here:

formatting link
If I'm guessing at your numbers right, then the below should be true:

cross area 0.000836 m^2

thickness 0.000254 m length 0.012 m

resistivity 7.20E-07 R = r*L/A 1.03E-05 Ohms of track

where

R is the end-to-end track resistance in Ohms r is the resistivity of the track material in Ohm Metres L is the track length in metres A is the track cross sectional area in square metres

I^2 R = 2.33E-03

2.33 milliwatts will be dissipated at 15A

I think you're safe if the above is true, by an order of magnitude

Also I found this:

formatting link

6.897E-07 ohm-m for 304 steel, even better!
Reply to
cr500r

The .836e-006m^2 is the surface area of the metal. The resistivity value is 7.2e-007 ohm-m at 20 degrees celcius. The length is actually

12 mm (I wrote the wrong unit before) and width is 5.3mm. It is a very thin, small piece of stainless sheet metal, which is why I think i'm having so much trouble. It is much easier to figure out the amount of current a wire can carry, mainly due to the fact there is tons of information out there about wire ampacity. Once again I appreciate all your help on this topic.

Thanks,

-Erik

Reply to
edannemiller2002

OK, updated with the new numbers:

width 0.0053 m thickness 0.000254 m length 0.012 m cross area=width*thickness 1.3462E-06 m^2 resistivity 7.20E-07 R = r*L/A 6.42E-03 Ohms of track

where

R is the end-to-end track resistance in Ohms r is the resistivity of the track material in Ohm Metres L is the track length in metres A is the track cross sectional area in square metres

I^2 R = 1.44E+00

1.44 watts will be dissipated at 15A

1.13E-01

0.113 Watts will be dissipated at 4.2A

A little higher, but still not bad.

Reply to
cr500r

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.