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I am a complete newbie to anything electrical/electronics. I am doing a project where I need to heat a 1 km long (Nichrome/Kanthal) wire to up to 110 degree Celcius. The only power supply available is 12V from car cigarette lighter. Can someone explain what is the best way to attain this using the available resources please.

Thanks for your replies.

Thanks for your replies. I am intending to weave the wire onto a fabric which can easily withstand over 200 C. The wire will be in a continuous forms and the purpose of the wire is to heat all of the fabric. Imagine a blanket that is covering the back seat. something similar to heated car seats, but much hotter. It will only be used inside the car.

I could possibly connect the wires in parallel, but is not preferred due to the work involved. The thickness of the wire can be anywhere from 0.3 to 0.6 mm. Any thicker or thinner is not practical to lay into fabric. The 1 km of wire will be laid with in a 2-3 square meter of fabric.

By the way, this is to heat the whole car (atleast passenger area) so that the kids can sit in a warm car, rather than a cold car.

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    \$\begingroup\$ What's the diameter of the wire? It is a single piece or it can be cut in different segments? If you describe what problem you are trying to solve you could get better answers (i.e. what do you need such a long wire for?). The temperature the wire will reach will also depend on the environment where it is placed. \$\endgroup\$ – Lorenzo Donati May 10 '15 at 11:29
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    \$\begingroup\$ You will need 1 km of good-quality thermal isolation. \$\endgroup\$ – Wouter van Ooijen May 10 '15 at 12:10
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    \$\begingroup\$ @Wouter van Ooijen : Strange but true. Below a certain diameter, insulating a wire actually increases heat loss, because the insulator conducts heat (admittedly poorly) to a much larger outer surface which transfers heat much more efficiently to air. Now if you can find an insulator that actually conducts heat worse than air, you might be on to something. But consider that most good insulators work by ... trapping a layer of air. \$\endgroup\$ – Brian Drummond May 10 '15 at 13:01
  • \$\begingroup\$ 'Trapping' is the important point. But who says that the isolation must be air-trapping? Put the wire in a vacuum cyclinder with reflecting inner surface... \$\endgroup\$ – Wouter van Ooijen May 10 '15 at 14:10
  • \$\begingroup\$ Assuming you could get the 1km wire to 110 Celcius what could you practically do with the heat energy? The moment you tried to draw on it the wire would cool very rapidly. \$\endgroup\$ – JIm Dearden May 10 '15 at 15:44
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First, compute the power you need.

This will consist of radiated and conducted heat loss.

The radiated power can be computed from Stefan's Law. Assume an emissivity of 1.0 to be on the conservative side; you will also need the wire diameter in order to compute the radiative area.

This is likely to be relatively low. If you can keep the wire in a high vacuum, this will be the total power you need from the battery.

Conducted heat loss requires knowledge of the material surrounding the wire, such as air, water, insulating material, etc. Assuming still air, use Newton's Law. You need to plug in a "heat transfer coefficient" ... assuming the wire is horizontal, here's how to calculate it. For the value of k the thermal conductivity of air, use k = 0.024 W/m/K. You will need the wire diameter D, you can assume a Prandtl Number of 0.8 and a Rayleigh number of about 1E6 as a starting point, and you must make reasonable assumptions about the air temperature.

If the air is moving, cooling will increase and things get much more complex and less accurately calculable as calculating the heat transfer coefficient gets more difficult.

If this computation is too much, bypass it by experimenting with a metre of wire and a variable power supply.

This is likely to give a much higher power requirement than radiative cooling at the temperature given.

Add these figures and you now have a power requirement. Given 12V, you can calculate the required current, and thus the life of a car battery.

You can also calculate the resistance you need as a load. Compare that with the resistance for 1km of your chosen diameter of Nichrome wire and you will see that you cannot power a continuous 1km length from only 12V. Which means you need multiple short lengths of wire connected in parallel, or a much higher voltage than 12V.

EDIT : The new information changes things A LOT. You'll achieve this temperature with much less power because the wires are in close proximity so they are heating each other. Model the heated surface as a plane when calculating radiation losses and heat transfer coefficients.

Parallel 2m wires are the way to go, and in a 3m^2 (2m*1.5m) blanket they will only be 3mm apart. Why parallel? Because with one single wire, a single break will stop the entire blanket, and be difficult to find and repair, while with parallel wires, a few breaks won't even be noticed. As well as allowing much safer voltages around your children!

You'll also have to revise that temperature downwards - if the wire's at 110C the surface of the blanket could be around 100C, causing burns.

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  • \$\begingroup\$ You could use a boost converter but otherwise +1 for good words. \$\endgroup\$ – Andy aka May 10 '15 at 13:28
  • \$\begingroup\$ Yes, a boost converter is one approach to that "much higher voltage", but it's not likely to be an off-the-shelf item. \$\endgroup\$ – Brian Drummond May 10 '15 at 13:30
  • \$\begingroup\$ What is the maximum Volt that could be converted using a booster? Do not mind building one as long as it is compact and can be plugged into the cigarette lighter. \$\endgroup\$ – fairmount May 11 '15 at 13:34
  • \$\begingroup\$ @fairmount you need to decide what nichrome wire thickness you are going to use for this project - that wire thickness determines the current and the wire resistance determines the boosted voltage. \$\endgroup\$ – Andy aka May 11 '15 at 13:41
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A whole kilometer of wire? In open air?

Based on How do i find nichrome temperature

I'm going to estimate 300mA as the required current, and if we call it 20 ohms per meter, that's 20kOhm wire. That implies 6000V (!) driving it, at 1.8kW. You are not going to get that from a car battery. Your stated objective is impossible with that constraint.

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  • \$\begingroup\$ Or 500 2m segments connected in parallel, at 150A. Pretty tough on the battery, but easier than starting an engine. \$\endgroup\$ – Brian Drummond May 10 '15 at 13:04
  • \$\begingroup\$ 1.8kW can be drawn under 12V at 187A assuming 0.8 convertion efficiency. That's not an impossible figure. \$\endgroup\$ – Mister Mystère May 10 '15 at 22:26
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    \$\begingroup\$ @BrianDrummond: It may be impossible to split the cable in parallel segments if the cable is not meant to come back regularly to the supply - maybe the OP wants a big loop, who knows? \$\endgroup\$ – Mister Mystère May 10 '15 at 22:29
  • \$\begingroup\$ OP hasn't said one way or the other, the nichrome wire could even be underwater for all we know. Just a driveby question leaving everybody guessing, which is why I tried to keep my answer general. \$\endgroup\$ – Brian Drummond May 10 '15 at 22:58
  • \$\begingroup\$ Hi, Yes, it is easier to make a big loop than making short parallel connections. But, then again, isn't maximum Amp from a cigarette lighter restricted to 10A OR can that be increased? \$\endgroup\$ – fairmount May 11 '15 at 13:33
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You could wrap the wire (with thin insulation) into a bobbin or in a ball, and house it in an evacuated chamber. With a bit of care you could keep that km of wire 85 degrees C hotter than a presumed 25 degrees C ambient with very little power consumption. Best results would probably be from using a very thin wire and boosting the battery voltage with an efficient DC-DC converter.

For additional power savings, cover the interior of the chamber with spacecraft type multilayer insulation.

As an alternative to the DC-DC converter with 100% efficiency, wrap the wire around a heater of optimal resistance.

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    \$\begingroup\$ Ingenious. Note that the chamber must be evacuated to a very high vacuum - well below 1 Pa ... the thermal conductivity of air doesn't really start to fall until the mean free path is comparable to the chamber size : pressures below 0.5 Torr (65 Pa). See en.wikipedia.org/wiki/Pirani_gauge \$\endgroup\$ – Brian Drummond May 10 '15 at 13:39

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