Electroplated Impulse Heaters

The main purpose of the copper is to lower the contact resistance. This is very important with steel as there is always an oxide coat on steel that gives high resistance, and thus high heat at the contact point. With the copper layer, the contact area will stay the same temp as the rest of the heated steel since it has very low contact resistance. Since you don't have water on the contact, the copper will corrode and flake off very slowly, and since copper melts at 1980 F it will stay on the steel for a long time.

Thanks for the reply Guest!!! Would the copper plating also give that area better heat dissapation capabilities? Or is it that the lower resistance in that area causes less power to be to be "used" in that area if you look at the heater as 3 resistors in series and apply P=I2R? I ask because the ends of the heaters tend to run about 100F cooler than the unplated area of the heater.

Killerkaz,

You have the correct idea.

"the lower resistance in that area causes less power to be to be "used" in that area - P=I²R"

The copper plated low resistance contact areas don't produce heat and provide disspation of heat between the heat producing high resistance element and the connections you don't want to fry.

Jon

Though thin, the 0.001" Cu film can carry quite a bit of current. Since this film is parallell to the heater The power density in the plated section will be much lower. The lower I²Rin this zone will not allow the temperature to reach the melting point of copper, especially in impulse heating.

Also, some of the heat at the ends is being conducted away into the connecting wires.

The popular school experiment could explain this.

A cone made out of a tissue paper, filled with water, heated directly by a candle flame, would never burn the paper.

Like frying and egg in a brown paper bag.

Jon

I did a version of that in Bot Scouts. Boiling an egg in a paper cup.

Question for you:

Do you know how American indians boiled broth on a wooden bowl more than 5 centuries ago?

Yahlasit

Hot rocks.

Nothing like stone soup.

Jon

Hi,

I brazed the ends of similar heaters with AgCu alloy, simple, very effective, non-oxidising if heated in air.

RHABE

Thanks for all the replys. CR4 is such an awesome recource.

I will like to comment regarding the question you made in the following paragraphs:

"The company I work for does a lot of soft goods heat sealing. We use inconel strip heaters that are plated with .001 inches of copper on each end to keep the heat in an area between the the plated ends. I have wondered what prevents that minimal ammount of copper from just melting off as it is transmitting up wards of 100amps to a heater that is 4" wide by 8" long. If anyone here could explain it to me I'd appriciate it. Thanks for your time! "

Here is my answer:

The copper plated ends have a very little electrical resistance (R) because it is a very small strip,in other words it is a resistor of very insignificant value, therefore the energy (Heat) dissipated by it, is also very insignificant, which is expressed in the formula:

Power Dissipated P= V X I where V= voltage drop through a Resistor and I= Current flowing thru the resistor. Also V= R x I

Therefore P= R x I X I , and given that the R is insignificantly small value the end result is that power dissipated is very small. Say the R was 0.0001 ohms and as you say I=100 amps then P= 0.0001 ohms X 100 X 100 = 1 volt-amp or P = 1 watts

Because the resistance of the copper ends, as I said is insignificant, the voltage drop that occurs in those ends are also insignificant, because.

Remember that in a wire or strip, the Resistance R= Unit resistance X Length of wire or strip, thus if the copper ends are very short (L is insignificantly small), therefore R will be also insignificantly small as I said before.

Rafael L. Lama (BSME) P.E.

Former Engineering Mgr. of a High Volume Printed Wire Circuit Facility Owned by Digital Equipment Corporation in PR, and Engineering Manager for the two high volumes mfg plants DEC had in PR.

Rafael,

You have put the icing on the cake.

Jon

I will like to comment regarding the question you made in the following paragraphs:

"The company I work for does a lot of soft goods heat sealing. We use inconel strip heaters that are plated with .001 inches of copper on each end to keep the heat in an area between the the plated ends. I have wondered what prevents that minimal ammount of copper from just melting off as it is transmitting up wards of 100amps to a heater that is 4" wide by 8" long. If anyone here could explain it to me I'd appriciate it. Thanks for your time! "

Here is my answer:

The copper plated ends have a very little electrical resistance (R) because it is a very small strip,in other words it is a resistor of very insignificant value, therefore the energy (Heat) dissipated by it, is also very insignificant, which is expressed in the formula:

Power Dissipated P= V X I where V= voltage drop through a Resistor and I= Current flowing thru the resistor. Also V= R x I

Therefore P= R x I X I , and given that the R is insignificantly small value the end result is that power dissipated is very small. Say the R was 0.0001 ohms and as you say I=100 amps then P= 0.0001 ohms X 100 X 100 = 1 volt-amp or P = 1 watts

Because the resistance of the copper ends, as I said is insignificant, the voltage drop that occurs in those ends are also insignificant, because.

Remember that in a wire or strip, the Resistance R= Unit resistance X Length of wire or strip, thus if the copper ends are very short (L is insignificantly small), therefore R will be also insignificantly small as I said before.

Rafael L. Lama (BSME) P.E. rlama@caribe.net

Former Engineering Mgr. of a High Volume Printed Wire Circuit Facility Owned by Digital Equipment Corporation in PR, and Former Engineering Manager for the two high volumes mfg plants DEC had in PR.