# Battery Powered Heater

1. Jan 14, 2012

### Big Uli

Hello every one

I would like some help regarding building a battery powered heater.

Let me describe the problem first
I shoot air rifles as a hobby and am restricted to 12foot pound of muzzle energy here in the UK
One of my rifles that I prefer to shoot is CO2 powered.
Unfortunately this results in a loss of power during this colder time of the year.
There is an option to wrap one of those battery powered heated socks around the barrel and CO2 tube but that's not an ideal solution to me.
I would prefer something neater.

If possible I would like to wrap some kind of cable around the 18 inches of the CO2 tube (24 inches long including the valve and hammer assembly) and then insulate it with a suitable material like foam carpet underlay or pipe insulation and adapt my custom stock to take the lot

The big question is what kind of cable and how many volts are needed to heat the reservoir to a constant 26 degree centre grade. I believe this is 80f

This is the rifle fitted with my DIY custom stock
The reservoir is the 1" pipe below the barrel.
The stock can be adapted within reason

2. Jan 14, 2012

### metiman

Sorry. I wrote a long post but got the dreaded:
So now I'll write a very short one instead. P=I^2*R. P=V^2/R. If you want 50 watts from a 12 volt battery: P=144/R and R=144/P. So R=144/50=2.88Ω. You would need a 2.88 Ω conductor to give you 50 watts. This also results in 4.16 amps of current. So watch your battery discharge curves. Any conductor material that can give you 2.88 Ω should work, but higher resistance materials like the traditional nichrome would use shorter and thicker wire. For instance you could use 20" of 24 gauge nichrome to give you about 50 watts. Unfortunately that wire might reach temps of over 1000 degrees F in still air. Not sure how hot it will be pressed against the rifle though. Other possible materials are stainless steel (half the resistance of nichrome), galvanized steel, aluminum and even thin copper wire. You could use about 44 feet of AWG 28 magnet wire, although keep in mind that the insulating varnish on the wire might burn off. The longer the wire the less the temperature per unit length should be. So just using copper or aluminum wire might not be such a bad choice.

3. Jan 14, 2012

### jim hardy

might see what's inside an old electric blanket or heatiing pad.
those heating wires might be self regulating, ie resistance goes up with temperature to limit how hot they get.
electric blankets have internal thermal switches that cut off on high temperature - that's the little 'lumps' you feel.

experiment.

4. Jan 14, 2012

### dlgoff

You could purchase some 12 Volt DC Self-Regulating Heat Cable (3.5 or 5 watt/ft). Just cut to length and use their 12 and 24 Volt Heating Cable Termination Kit.

I've never used their products but have used lots of bulk heat tape for various applications. I find the safety of the self regulating construction a big plus.

http://en.wikipedia.org/wiki/Trace_heating

5. Jan 15, 2012

### metiman

How is that heat tape "self-regulating"? If you mean that the conductor has a positive temperature coefficient so that the dissipation decreases as its temperature rises, well so does pretty much every non-ceramic conductor in the world including copper, aluminum, and iron. There are some ceramic materials with negative temperature coefficients so that as the temperature rises the resistance actually goes down, but I don't think you'd want to use those.

It looks like that heat tape uses a conductive polymer like my electric blanket used. The wikipedia article implies that this conductive polymer has a high PTC, but then so does nichrome. Without numbers it's impossible to judge whether that fancy heat tape will work any better than nichrome in terms of 'safety'.

According to this table (scroll down to middle of page) the temperature coefficients for copper/aluminum, iron, and nichrome are .0039, .005, and .0004 respectively. So the resistivity of nichrome increases 9.75 times more per unit temperature than copper or aluminum and 12.5 times more than iron. I don't know where conductive plastic fits in here. It's not a metal. So I wouldn't place any bets.

So let ρ = resistivity, α = temperature coefficient, and ΔT = the change in temperature from 20° C to whatever temperature you want to calculate for.
ρcopper at 20° C = 1.68 x 10-8 Ωm
ρnichrome at 20° C = 1.1 x 10-6 Ωm

@ 350° C:
ρcopper-350 = ρ20° + (α x ΔT x ρ20°) = 1.68 x 10-8 + (.0039 x 330 x 1.68 x 10-8) = 3.842 x 10-8 Ωm
ρnichrome-350 = ρ20° + (α x ΔT x ρ20°) = 1.1 x 10-6 + (.0004 x 330 x 1.1 x 10-6) = 1.245 x 10-6 Ωm

Now let's consider 2.88 Ω (@ 20° C) of each material. Let ρ = resistivity, L = length of conductor, A = cross-sectional area of conductor, and R = resistance.
For 20° C:
$$R = \frac{ρ_{20}L}{A} \hspace{4 mm} L = \frac{RA}{ρ_{20}}$$
So for a given target resistance in a given material if you increase the length you also have to increase the gauge of wire. Let's use 24 gauge for nichrome. 24 AWG wire has a diameter of .0201" or .51 mm and an area of 2.043 x 10-7 square meters. For copper let's use 30 AWG wire which has a diameter of .010" (10 mils) or .254 mm and an area of 5.067 x 10-8 square meters.
$$L_{nichrome}=\frac{(2.88)(2.043)(10^{-7})}{(1.1)(10^{-6})} = 0.535 \hspace{1cm} L_{copper}=\frac{(2.88)(5.067)(10^{-8})}{(1.68)(10^{-8})} = 8.69$$
So the 24 AWG nichrome wire would be .535 meters or 1.75 feet long and the 30 AWG copper wire would be 8.69 meters or 28.5 feet long. For a shorter length of copper you'd have to go to an even finer wire, which for this application might not be a great idea. I'd at least go up to aluminum to get a shorter length with a more durable wire thickness.

Now let's heat things up (to 350° C).
$$R_{copper350°}=\frac{ρ_{350}L}{A}=\frac{(3.842)(10^{-8})(8.69)}{(5.067)(10^{-8})} = 6.59 \hspace{1cm} R_{nichrome350°}=\frac{ρ_{350}L}{A}=\frac{(1.245)(10^{-6})(0.535)}{(2.043)(10^{-7})}=3.26$$
So, unless my calculations are off, the copper resistance more than doubles from 2.88 Ω to 6.59 Ω, and the nichrome resistance increases by only 13% from 2.88 Ω to 3.26 Ω. Presumably this has something to do with the fact that nichrome has a much larger cross sectional area per unit length for a given resistance than copper and so is affected more by the coefficient than copper. So which one is more 'self regulating'?
$$P_{copper350°}=\frac{V^2}{R}=\frac{144}{6.59}= 21.85 \hspace{1cm} P_{nichrome350°}=\frac{144}{3.26}=44.17$$
So at 350° C the current in the copper is down to 12/6.59 = 1.82 amps and is heating with only 21.85 watts while the current in the nichrome is down to 12/3.26 = 3.68 amps and heating with a much greater 44.17 watts. Well I would say the copper is much more self regulating but actually quite a bit less practical because it self regulates too much. The nichrome barely drops in heat output at all at 350°, while the heat output from the copper is down to less than half.

Last edited: Jan 15, 2012
6. Jan 15, 2012

### jim hardy

Last edited by a moderator: May 5, 2017
7. Jan 15, 2012

### dlgoff

One thing about conductive polymers is the resistivity vs temperature isn't linear. Hence better regulation than just using a wire.

8. Jan 15, 2012

### dlgoff

9. Jan 15, 2012

### metiman

Those are some interesting links, Jim. Only the copper alloy would be of use to the OP though. The other two links are greatly useful for my project however. Thanks. I notice that those copper alloys have some awfully low 'working temperatures'. Especially some of the ones with the lowest coefficients. I wonder if the alloy melts if you exceed working temps or if the low temperature coefficient is only within those ranges.

10. Jan 15, 2012

### metiman

I didn't realize that that heating tape used was open circuit in terms of the metal conductor. So the current flows between the two wires through the conductive plastic. I have to say that is pretty damn cool and gives that heating tape some unique properties. For one thing the length doesn't matter. It could be a mile long and it wouldn't affect the heat output. The resistance is fixed. All you can do is increase or decrease the voltage to affect output. Now I'm really curious as to how my electric blanket was constructed.

11. Jan 15, 2012

### Big Uli

Hello
First i woyuld like to thank for taking the time to reply
Secondly, please excuse my ignorance but I didn't understand any of this but the quoted text

That does make sense to me so basically I get some of this cable and the termination kit and connect it to something like this battery
http://www.ebay.co.uk/itm/NP1-2-12-...rElectronics_Batteries_SM&hash=item41589988b7
That in theory will heat my reservoir
How much cable would I need to get it to a 80f or 25c.

The reservoir will have a temperature monitor on it as that will add safety by means of monitoring the temperature
Not only does performance drop significantly in cold weather but in hot weather there is also the danger of to high a pressure
CO2 is running at 850PSI standard but can go as hight as 1200PSI at around 30C or 90f
A temperature of above those perimeters is not desired at all
Cool temperatures of course result in a lower pressure than the 850PSI and that is what causes my problem

12. Jan 15, 2012

### dlgoff

Monitoring the temperature is a good idea; as you can also use it to help determine if insulation and/or another run (/longer) cable is needed. I would get enough cable to play with and start with one run parallel to the barrel and tank. Check if it brings the temperature to a useable value, as your ambient environment would be the determining factor. I like using cable ties to attach the wire because they can be clipped and the cable removed easily. It might be that you need heat cable on either side of your rifle with two separate parallel runs instead of one longe run, so maybe get two termination kits (it depends on how you want to route the cable). I'm thinking you shouldn't have to wrap the cable but again that depends on your environment.

13. Jan 15, 2012

### dlgoff

This battery will provide 1.2 Amp-Hours. So if say you are using 1 foot of the 3.5 watt/ft cable, the current needed would be 3.5watt/12volts=~0.3amp. So your battery would last for 1.2amp-hr/0.3amp=4hours. Similarly for 2 feet, your battery would last 2hours and for 10 feet it would last about 4/10hours=24minutes.