Unexpected Igniterman Igniters Results

Well, even though I = V/R, remember that P, the power (in Watts) dissipated by the resistance is given by P = V * I (since I first encountered Ohm's law from an electronics, rather than a physics, standpoint, I tend to use V for voltage, rather than E for Electromotive force. If you're more comfortable with E, simply substitute it for V in all of my equations).

Doing a bit of substituting, you get:

P = V * (V/R)

or 2 V P = ----- R

So, with your 9.6 V battery and 500 Ohm resistor, you dissipate 0.184 Watts, which is 184 mW. Just over 1/6 of a Watt doesn't sound like much, to me. However, what's missing is a description of how *small* the mass of conductive primer is.

One Watt is equal to one Joule of energy being transfered per second. So, 184 mW = 184 mJ/s. If the mass of the conductive pyrogen is small enough, it will not be able to disipate energy (as radiant heat) fast enough to keep up with the continuous influx of energy (at 184 mJ per second), and it will retain some fraction of that energy, and its temperature will increase.

My theory is twofold:

First, the *reason* that the resistance is so high is that the resistive bridge formed by the conductive primer is physically small compared to normal Igniterman igniters. Whether this is due to over-thinning the primer, under-thinning it, or by not dipping as much

Second, even though the power dissipated by the resistance is very low, the energy is transfered into a much smaller mass than normal. It takes less energy to cause the same temperature rise in a smaller mass, and I believe that the effect may be stronger than strict proportionality, since the surface area of an object is proportional to the square of the individual dimensions, and the amount of heat an object can dissipate by radiation as heat is roughly proportional to the surface area.

I realize that there's a lot of handwaving in the above explanation. It's a guess based upon a fairly intuitive grasp of the physics involved, but it's a guess, nonetheless. Bear in mind that, if you had told me that the igniters *didn't* work, I probably could have come up with a reasonable explanation for that, as well, involving the copper in the wires serving as a constant size heat sink for a smaller heat source....

- Rick "Armchair expert" Dickinson

Without delving into it any farther than speculation and handwaving, the question that pops into my mind is how much breakdown occurs prior to reaching ignition point, i.e. do thermal degradation, local pyrolysis cause a corresponding decrease in resistance resulting in a runaway situation ending up in ignition once the power dissipated is sufficient to raise the material to autoignition temperature. IOW, I wonder if the resistance remains relatively constant until ignition, or whether you have a situation of exponential rise in power dissipation over time due to breakdown. It woudl be interesting to record current over time for the latent period between power application and ignition.

Mike D.

Thanks, Rick and Mike! I mostly understood what you guys said. Now, if you could make an anology back to PV=mRT or Bernoulli, then simle fluids guys like me could understand it. :-)

BTW, Robert B and have been swapping some emails today about this. Rob thinks that the NC lacquer that I used may have impregnated the conductive primer, substantially increasing the resistance. Rob also, and deservedly, slapped my wrist for using an NC sealer, because his primer/pyrogen isn't designed to be sealed with NC lacquer. And Rob is VERY surprised that anything over 50 ohms ignited.

Mike, you want me to send you a 100-ohm igniter to run your little ignition test? :-) My DVM has RS232 pickoffs, but I doubt very seriously that the sampe rate is high enough to provide any useful data. And, my O-scope musta fallen outta my back pocket on the way home from the store. :-)

Just kidding about the shipping-an-igniter thing. Guess that the DOT wouldn't look too favorably upon that...

Chuck

That would probably be an issue if only ONE wire were heating, but with a SCB pyrogen several to many "wires" are simultaneously making contact and heating at varying rates.

What DO you use for a sealer?

Bob Kaplow NAR # 18L TRA # "Impeach the TRA BoD" >>> To reply, remove the TRABoD!

The sealer provided with the IM kit, when using the IM conductive primer and pyrogen. I have no idea what Robert uses to make the sealer. Trade secret, I guess... :-)

cp

So, experiment with it. Dip a few, and measure their resistance without the sealer. Then dip in the fingernail polish, and measure again. See if it affects the resistance. I've been using clear fingernail polish for years. It works, and it's cheap.

James