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Well, I want to heat up 100 wires with 20cm of length each made of Kanthal (A1) with 20 awg gauge (0,8mm/0.03inch diameter and 0.817 ohms/in) to 130 ºC from room temperature of 30 ºC continuously. And let's say this wire heats up from 30 to 130 in half a second.

I'm listing all these numbers because I've made a question about induction heating before and people said they were useful for that question, I don't know if it is useful for this question about a resistive heater.

  • Specific heat of Khantal A1: 0.460 J/g-°C (at 20.0 °C) 0.110 BTU/lb-°F (at 68.0 °F).
  • Each wire weights 0.41 grams, and since there are 100 of them, everything weights 41 grams.
  • In order to heat up 1 gram of Kanthal A1 by 1ºC you need to provide 0.46 J, so assuming the room temperature is 30ºC and we heat one gram to 130 ºC, you would need to use 46 J to heat all of the 100 wires.
  • But since all the 100 wires weights 41 grams together, you would need 1886 joules in total.

Obviously, if one would supply the same energy continuously, the wire would simply heat beyond that value (130 ºC), but let's assume that this is heating 100 different wires every time that it takes to heat up to heat up 100 wires from 30 ºC to 130 ºC.

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    \$\begingroup\$ Do you keep these wires at that temperature, or do you immediately let them cool again? If you keep them hot for a period of time, the steady-state losses will dominate, which means that you'll simply have to measure the power consumption. \$\endgroup\$ Nov 24, 2022 at 18:39

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You want to heat 41 grams by 100 degrees C. So that is 0.460 J/g-°C \$\times\$ 41 g \$\times\$ 100°C = 1886 J. Yes, I calculate the same number that you did.

You asked about heating them up and not keeping them hot. To keep them hot you need to compensate for the rate at which they cool down. That is more difficult to calculate and you might find it easier to simply measure it. Not much heat is lost in half a second, so if you just need to heat them up once, then you can ignore this.

You asked how long the battery would last when doing this continuously (swapping in a new set of wires every time they get to 130°C). Note: the battery specification was changed since this answer was originally posted. Here is the updated battery information.

This battery will be way too small for the power requirement. 3886W / 12V is over 300A. From a 160Ah battery (how I originally understood the battery), that's pretty high but not impossible. But it's way too high for an 8Ah battery (much smaller) - it's basically a short circuit. If we calculate \$\frac{12V \times 8Ah \times 3600 \frac{h}{s}}{3886W}\$ we can expect the battery to run out after about 90 seconds. However, if we look at the datasheet of some random 12V 8Ah battery we can see that actually the capacity is quite a lot lower if you discharge the battery faster. When the battery is discharged in one hour, you get only 4.62Ah, only a bit more than half its full capacity - and you want to discharge it much much faster than that. A 90-second discharge is basically a short circuit and you won't get nearly the power you want as it will all be dissipated inside the battery's internal resistance. You really need a much bigger battery or a connection to the power grid.

There's also the question of voltage and current and resistance - is 12V enough to heat one of your wires to the needed temperature in half a second? If not then you may use a boost or buck converter, or a different battery, to change it accordingly. I assume you've already worked that out.

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    \$\begingroup\$ Of course, using a converter will affect the efficiency and you will end up losing some additional energy to things that aren't the wire. \$\endgroup\$
    – Hearth
    Nov 25, 2022 at 4:47
  • \$\begingroup\$ "I assume you've already worked that out.", no I didn't, I'm a mess and I don't know where I'm doing, lol \$\endgroup\$
    – Fulano
    Nov 25, 2022 at 11:14
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    \$\begingroup\$ @Fulano voltage equals current times resistance; power equals voltage times current; your wire has a certain resistance and if the voltage is too low you just won't get the current you need. Have to know the wire's resistance to estimate that (and it will be an estimation since the resistance changes based on temperature) \$\endgroup\$ Nov 25, 2022 at 12:02
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    \$\begingroup\$ @SpehroPefhang The question has been edited. Originally it said 8A 20h (i.e. 160Ah) and now it says 8Ah. I will edit the numbers in the answer. \$\endgroup\$ Nov 25, 2022 at 12:23
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    \$\begingroup\$ @Fulano I misunderstood the battery you had. Now it looks to be way too small. The amount of power you want would discharge the battery in 90 seconds, and batteries do not work well at such incredible discharge rates (you will not get most of the energy out of the battery). You need a much bigger battery or a power outlet, or the machine needs to work much slower. \$\endgroup\$ Nov 25, 2022 at 12:41

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