I am trying to find out how many Amperes I will need to reach a certain temperature in a certain time (Milliseconds)

I need a thick wire to slow the melting of the wire. "Nichrome is the wire I'll be using" now in the American Gauge. I want to use 12 as its thick but not too thick and on Wikipedia, when I was checking how many amps I will need to reach a specific temperature using that gauge type it said I will need around 39.03A to reach around 1400F but in what time period will I reach that? Thicker wires are much slower to heat up but smaller wires are very weak.

P S I am very new to the American Wire Gauge so forgive me if I'm wrong.

I need a thicker, more sturdy wire that can put in a high PSI environment and still reach temperatures of 2-4 millisecond. The problems i see is thicker wire don't have as much resistance and a smaller mutil-twisted smaller wire. Is there away i can change that and still reach my desired temp of 1400F?

  • 1
    \$\begingroup\$ Given a certain power (which is set by e.g. specifying current and resistance) the time to reach a certain temperature depends a bit on the mass of the wire and, at that high temperature, heavily on the radiance characteristics of the surroundings. So, without specifying the surroundings you cannot calculate anything. \$\endgroup\$ – Janka Feb 21 '17 at 1:06
  • \$\begingroup\$ 39 Amp is correct for 12 gauge. \$\endgroup\$ – Misunderstood Feb 21 '17 at 2:59
  • \$\begingroup\$ yes. strange question. More current. \$\endgroup\$ – Marcus Müller Feb 21 '17 at 6:03
  • \$\begingroup\$ So if i increase the current if will get hotter, faster? Do i also need to increase voltage? Sorry i am a complete noob. \$\endgroup\$ – DeusIIXII Feb 21 '17 at 6:05
  • \$\begingroup\$ P=I²·R , apply Ohm's law, thicker wire has lower R. You forgot to tell us what you're heating with this. Thicker wire namely also has more mass, needing more energy to heat up in the first place. However, you seldom heat wires for the sake of having a hot wire, but for the sake of heating some other mass along with the wire. \$\endgroup\$ – Marcus Müller Feb 21 '17 at 6:07

James Clerk Maxwell popularized this method of dimensional analysis; in fact, before him, dimensions were a mishmash, unstandardized, and such analysis was impossible.

I'll give you silicon, which has specific-heat about 3X higher than tungsten.

At 2 picoJoules/(cubic_micron*degree Centigrade), suppose we want to short the output driver of a microcontroller? That driver is 100 micron * 100micron (its a powerful transistor), with 0.1 amp short-circuit ability. Assume 2.5 volts.

How long before the transistor reaches 1,025 degrees Centigrade? starts at 25C.

Our power is I*V = 0.1amp * 2.5 volts = 0.25 watts, or 250Billion picoJoules/second. The volume of silicon? Assume we'll only heat the top 100microns of the Integrated Circuit during our experiment (that depth has a thermal TimeConstant of 114 microseconds, and in that time "most" of the heat remains in that 100micron thickness. Our total volume is 100*100*100U or 10^6 cubic microns.

What is our rate-of-change-of-temperature? we want degrees/second as dimensions for our answer.

The only bit of info we have with seconds is the power: 4 seconds/joule

We want to cancel the "joules" so multiply 4seconds/joule by specific-heat of silicon

$$4 seconds/joule * 2 picoJoule/(cubicmicron * degree Cent) $$

Our rate of temperature rise is $$8 picoseconds/(cubicmicron *degree Cent)$$

And we have 1Million cubicmicrons of silicon. We need to cancel 'cubicmicron' in our answer, so multiply the answer by 10^6 cubicmicron, and we get

$$8 Million picoseconds/degree Cent$$ or $$8microseconds/degree Centigrade$$

We wanted 1,000degree Cent increase in temperature, thus 8,000 microseconds or 8 milliSeconds is the answer.

We initially assumed ALL THE HEAT would remain inside 100*100*100 micron cube. In 8 milliSeconds, heat will have moved outside the cube. A different method is needed for a correct answer.

And thank Maxwell, also the investigator of viscosity, for this method.

  • \$\begingroup\$ No idea if you are right, but I like the answer, you got my vote. Me, I'd apply 39 Amps, use a thermometer and and stop watch. \$\endgroup\$ – Misunderstood Feb 21 '17 at 4:51
  • \$\begingroup\$ In deep submicron (OK lets use 0.25micron), that output transistor would be 10micron by 10micron, containing 10--20 stripes each 10micron long. Point is, the mass to heat is now a 10 micron cube, thus volume is 1,000X smaller; this (modern) FET heats to 1,000deg Cent in only 8 microSeconds. Hence the need to NOT SHORT THE OUTPUT PINS of microncontrollers. \$\endgroup\$ – analogsystemsrf Feb 22 '17 at 13:59

You can reach any given temperature in nano-seconds if you ram in enough power(+).

Calculate the mass of the wire, length*area*specific_gravity. Look up the specific heat capacity of the wire material, Kaye and Laby online is my usual resource, and calculate the heat capacity of the wire. Note that the capacity will vary with temperature, so you'll need to use a mean figure, or do a numerical integration, if you want reasonable accuracy. Mean figure is probably OK for the accuracy needed. Multiply by the number of degrees you want to heat through, that gives you the heat in Joules you need to deliver during the heating phase. Now divide by your target heating time, that gives you your minimum power in watts for heating. This is a minimum figure, the adiabatic figure, as heat will be lost from the wire by radiation, and by conduction though the ends, during heating. To a first approximation, for a short heating time, this amount will be small. Pick a voltage or current that will deliver that power (for any given heater, once one is chosen, the other is automatically fixed).(++)

How is the wire temperature going to be regulated? If by temperature feedback, say from the wire resistance, then your controller will handle the transition from heating to staying warm OK. If by dead reckoning, simply using a defined voltage or current through the wire so that power in matches power out at the target temperature, then you may have a problem. It's likely the heating power will exceed the keep-hot power, if you want to heat fast. This means you'll need either a timer (to do it by dead reckoning), or a resistance measurement (OK if using steel for instance, more difficult if the wire is a low tempco resistance alloy wire) to switch from heating power to keep-hot power.

As a general rule, if you want to use a single power setting for both heating and keep-hot, as wire mass (heating time) goes by radius^2, and wire area (radiation loss, convection loss, keep-hot power) goes by radius^1, you'll get fastest heating finest wire you can use.

(+) Look up Sandia's Z-pinch machine, that makes wire quite hot, quite quickly.

(++) At this point, you're only one step away from computing the \$I^2t\$ of the wire, often specified for fuses and power semiconductors as a measure of what sort of pulse will get them to a given temperature (a max for survival, min for destruction).


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