Winding a Kanthal A1 Element

Winding Your Own Elements

If you cannot find a ready-made Kanthal element to meet your needs, you can easily wind your own. This operation is 50% calculation and 50% execution.


Kanthal A1 is a high temperature heating wire. It contains iron, chromium and aluminum and can handle temperatures up to 1400°C (2550°F). Kanthal has a known amount of resistance per foot, generally labelled as Ohms/ft or Ohms/m (in metric).  Tables are available online, but the place where you buy your Kanthal wire will be able to tell you exactly what the resistance per foot is. I bought some Kanthal A1 from Pottery Supply House in Oakville Ontario, Canada. They sell all gauges by the pound. I ordered 18 AWG, of which a pound was about 255 feet of wire. The resistance per foot is 0.5369 Ohms. In the USA I have found similar Kanthal A1 18 AWG on eBay.

Ohms: Pottery Supply House kindly provides the data I needed to make the resistance of my element exactly right. The element I need is 19 Ohms.

Diameter: The next thing to look at is the diameter. I need my new element to be about 1/4" in diameter.

Length: Lastly I need the length of the element to be about 22" long. I am not super concerned about the exact length as it's going to get stretched to 75". The thing here is to go for shorter and stretching to length.

This is the calculation part...

Ohms required = 19
How many ohms per foot = 0.5369
Feet of wire required = 19/0.5369

35.38 ft.

Now that seems like a LOT OF WIRE to fit into the walls of a small oven! But the wire needs to be coiled into an element.

Coil calculation is roughly this:
pi = 3.14
Preferred outside diameter of coil = 0.25"

Each coil will use 3.14 * 0.25 = 0.785" of wire. (This is an estimation.)

But how long is the whole element going to be?

For this we need to know the thickness of the wire. The wire I have says 18 AWG is actually sold as 1.02 mm or about 0.0402".

Number of coils per inch is 1 / 0.0402 =  24.9
Number of coils per foot is 12 x 24.9 =  298

Number of feet of wire per foot of element 298 x 0.785 =  234 / 12 = 19.5
(Thanks for the correction here David!)

For 18 AWG wrapped into 1/4" coils, it takes almost 20 feet of wire to make a single foot of element!

So the length of the whole coiled up element is 35.38 / 19.5 = 1.81 ft (or 21.75 inches)

As I mentioned before, I plan on stretching this to become 75" so it will fit into the existing grooves in the fire bricks.

My Winder

 I used a power drill and clamped the handle in the vise. I set the trigger stop to make the drill go about 2 to 3 revolutions per second.
3/16" rod will give me very close to a 1/4" element (outside diameter). I drilled a 1/16" hole about an inch from the end and clamped it into the chuck.
 After cutting 35 1/2" feet off the big spool, I fed the wire into the small hole about 3" and started the drill. A quick bar clamp holds the trigger in. If you have a helper they can start the drill and stop it if you make a boo-boo.
To get through the firebrick, I need to leave about 3" uncoiled and the tail end. Once the wire is coiled. Shut the drill off and let the coil relax. It is a bit of a spring. Feed the end back through the small hole and slip the element off.

A quick check on the ohmmeter reveals we are right on track.

When installing the element, I know that I need it to go from anout 22" and stretch to around 75". This means that each coil of wire will be spaced apart from the next by about two diameters  of the wire. I did this stretching in small sections and stapled as I went.

The Leads

Fold back the wire on itself to pass through the fire brick. When you fold back the lead you create 1/2 of the resistance in that area and  the lead acts more like a wire than a heating element.

If you want to design your own oven, we can take this process as part of the whole.

General Heat Treating Oven Design Guide

Design Considerations

Step 1: Determine how big to make the INSIDE of the oven.
Your biggest knife will give you an idea, but smaller will take less power to heat. If you know the dimensions of the Insulated Fire Bricks (IFBs) this will allow you to make good size choices based on using full bricks.

Step 2: Determine the cubic feet of the inside of the oven.
Take the inside dimensions and convert them to decimal feet, so 6 inches becomes 0.5 foot and 9 inches is 0.75 foot.

A 6" x 6" x 24" oven is 0.5 cubic foot. This can be expressed as: 0.5 x 0.5 x 2 = 0.5

A 9" x 6" x 18" oven is  0.75 x 0.5 x 1.5 = 0.5625 cubic feet.

Remember that this volume will have to be heated and more volume takes more heat, so make the volume as small as practical. As a knifemaker, you'll likely never need a chamber that is 8" high, but you may need one that is 18" deep.

Step 3: Determine how many watts of heat you are going to put inside.
Steps 1 & 2 will point this out. Aim for 5000+ Watts per cubic foot. 3000 Watts or more for a 1/2 cubic foot oven is recommended. More power will heat up faster.

I did a survey a few dozen 'knife maker' units from both Paragon and Evenheat that puts the Watts per cubic foot in a broad range from 6,000 to 10,000. Of course the 10,000 W/cu.ft. unit is going to get there much faster. Note some of Paragon's Xpress models are over 10,000 W/cu.ft.

Step 4: Determine Voltage and Current requirements.
Larger ovens will need 240V supply. If you need more than 2400 Watts, you'll want to go to a 240 Volt supply. The most we normally get from 120 V circuits is 20A which is 2400 Watts. If your oven is over 0.5 cubic opt for 240 V, either 15A or 20A. Power is voltage times current. A quick way to work this out is Watts wanted, say 3500 W divided by 240 V = 14.6 Amps.

Step 5: Determine how the elements will be wired.
Elements in parallel will allow you to use smaller gauge heating wire.

Some basic arrangements are as follows:

3000 Watts @ 240 V
18 AWG Kanthal A1

2 parallel runs
38 Ohms per run
19 total Ohms
0.54 Ohms per foot
6.32 Amperes per run
12.6 total Amperes
3031.6 Watts
70.8 feet per run

2200 Watts @  120 V
16 AWG Kanthal A1

2 parallel runs
13 Ohms per run
6.5 total Ohms
0.34 Ohms per foot
9.2 Amperes per run
18.5 total Amperes
2215.4 Watts
38.5 feet per run

Step 6: Determine how the elements are going to be arranged.
For example, connections to the elements are made at the rear of the oven or at the side. Now estimate how long the elements are going to be.

From here you should be able to build the element from bulk Kanthal.

Other Design Notes:

In some designs it makes sense to have several elements in parallel. One helpful formula is the parallel resistor formula.

Rtotal =  1/(1/R1)+(1/R2)+(1/R3)...

For my 120 Volt oven, I made three elements. I measured each one with the ohmmeter and recorded the values.

R1 = 19.6
R2 = 19.4
R3 = 19.5

Plugging in the numbers I got:

1/(1/19.6)+(1/19.4)+(1/19.5) = 6.499 Ohms

120 Volts / 6.499 Ohms =  18.46 Amperes
120 Volts * 18.46 Amperes = 2215 Watts

This works well for me on a dedicated 120V 20A circuit.

Thermal Limit

There is a thermal limit at which the interior temperature cannot increase any further with a given amount of power. This is due to losses. Essentially what this means is you need more and more power to make the same increase in temperature. Your target temperature is going to be around 1100°C or slightly less. Kanthal A1 is rated for 1400°C and is ideal for a knife maker's needs.  


I have found the chore of computing coils for various voltages and power to be a bit taxing. I'd like to help everyone who asks, but I do have limited time. I made a spreadsheet that will assist in choosing wire gauge, length and an element path for your size of oven, voltage etc. 

It comes with no warranties, but it's free! 

As always, your comments and suggestions help make things better.

Good luck,



Hesham Abdo said...


DanCom said...

You are welcome Hesham. Thank you for visiting!

Greg Brister said...

How did you know what size wire to buy?

DanCom said...

I chose the wire size for heft and for the ohms per foot rating. For durability, choose either 18 AWG or 16 AWG. The seller of wire specifies the ohms per foot.


Unknown said...

How long does your heat treating oven take to warm up to your forge temperature?

David Dameron said...

Thanks, Dan. I found a small correction in your calculations. Length/ turn = .785 inches or .785/12 = 0.0654 foot/turn. This times 162turns/foot = 10.59 feet wire per foot of coiled element.

DanCom said...

Thanks David. Correction noted. :-)


DanCom said...

The oven is not intended for forging, rather it is for heat treatment. A gas forge is much more efficient for forging. I have published a curve on the
HT2100 Heat Treatment Oven page. Roughly 65 minutes to get to 1000°C (1832°F). This will depend on the load as well, heating lots of steel naturally takes longer.

Unknown said...

Thanks for posting your owen. I'm from Czech Republic and we have 220v el. supply here(actually more like 230v). I would like to ask you if any of these would work with your owen design and how would you configure these elements
Thanks Zdenek

DanCom said...

I had used this type of element on my oven build. They lasted about 1 year of firing. The heavier Kanthal A1 is a better choice. However, if you wish to use the pre-wound coil, you need to find a suitable resistance for your volume of interior space. A smaller oven may be okay with a single element at 3000 Watts. Two 3000W elements in parallel will give you lots of heat, but will suck about 26 A.

Zdeněk Herzán said...

Thanks for your reply, I really appreciate it.
I have another question will this work if I use with thermocouple from second link below_
Thank you

DanCom said...

The second thermocouple (1250°C) is suitable for a heat treatment kiln/oven. It is what I use, standard Type K thermocouple. You can simply set the controller to it's 1200°C range and connect the 1250° thermocouple. Works great!


Zdeněk Herzán said...

Thank you.

Derek said...

Think this kit from ebay would work?

Seems to have most of what you'd need, minus the element wire and some minor connectors, wire, etc.

DanCom said...

That should work. It has everything you need to run the oven.

Derek said...

Thanks, I'm confident in building the enclosure and running the coils, etc. I just have ZERO experience reading an electrical schematic so I'm not sure where to wire everything.

Crow077 said...

The formula for surface current load for Kanthal a-1 is between 14 and 32 watts per sq inch of wire surface area, The higher the temperature you intend to operate at the lower the current load capacity. To get the correct wire gauge you need to know your intended voltage and amperage (up to you, but you will need the correct ohms per foot to reach your intended amperage) The formula for surface area of a cylinder is as follows (I'm not including the top and bottom of the wire as it is not necessary for our case) (2 x pi) x radius x height. example 16AWG wire has a diameter of .0508 first divide .0508/2 = .0254 radius. Then multiply {(2 x pi) 3.14 x 2=6.28} .0254 x 6.28=0.159512. Now you must figure out how many feet of wire you need to reach your desired ohms. For this refer to the Other Design Notes: referenced above. For easy reference I'll use the 2200 watt @ 120V design of 2 element runs of 38.5 ft each. First we need to convert the total length of both wires to inches to attain the total surface area of both runs, which is as follows 38.5 x 2 = 77. 77ft converted to inches, 77 x 12 = 924 inches total length. Now we take our earlier calculation of .0254 x 6.28 = .159512 and multiply this by our 924" to get our total surface area in square inches. .0254x6.28x924= 147.389088 sq in.
Now that we have our total surface area we can determine our surface loading for our wire. Kanthal's furnace-mini-handbook pg 9 figure 4 (a. Kanthal a-1 spiral wound in grooves) shows for our desired temperature of approximately 1830F our maximum recommended surface load is 19 watts per sq in. So now we take our total of 147.389088 sq in and multiply by 19. 147.389088x19= 2800.392672 Watts surface loading capacity.
When deciding on wire gauge it's best to stay 10-20% below the max surface load in order to have the longest usable life of the element.


Michael Clerc said...

Hi Dan, I'm following this with tremendous interest and have begun the process of putting the parts and pieces together to build an oven like yours, thank you for going to all this trouble and effort to post all this information.

You mentioned earlier that it took somewhere around an hour to fully heat up, what would one have to do reduce that time by half, say to 30 minutes?


DanCom said...

Hi Mike,

Increasing the Watts will speed up the time to temperature. Or one could make a smaller oven (smaller interior dimensions) and reduce loading and speed up the temperature rise.

In my HT-2100 build, I was limited to the 120 V supply. If you are going to use 240 V supply you can get a lot more (2400+) Watts into your oven space.


Michael Clerc said...

Thanks for the reply Dan. To increase the voltage from 110vac to 220vac, since all of the components listed are rated at a level to handle the increased voltage, what additional items would need to be changed? Obviously the connections to the heater coils since you have two 110vac inputs instead of one, do you hook both "hot" ends to each end of the coil or is there another method?

Thanks again,

DanCom said...

Hi Mike,

For North American 240 VAC operation you'll need to use two SSRs, one for each hot. This is the only way to fully isolate the elements when the door is open. (One SSR would always leave one end of the coil connected to a hot). Also, you'll need to provide two fuses/breaker for overcurrent protection. Your controller will likely work at 240 V as they are usually universal 86 to 265 V rated. You'll need to source some 240 V lamps. Most of the other things, switches, wire etc. are already 250 V or 300 V rated.

Michael Clerc said...

Thanks a bunch Dan, that's sort of what I figured. Would you run both SSR's from the same two connections on the PID with identical switches and thermal cutout?

I'll redraw the schematic to allow for the 220vac and send it to you for a sanity check if that would be okay with you.


Michael Clerc said...

Hi Dan, sorry to be such a pest, another question if I may. Does the heating element have to be in three pieces 35.4 inches each or can it be one piece that is 106.2 inches in length?


DanCom said...

Hi Mike,

The three elements that I used are in parallel. This creates a lower overall resistance and more Watts. If you were to use one long element, the resistance would be high and there wouldn't be enough Watts to effectively heat the chamber.

For a single element, the formula looks like this:

Voltage / (Length of wire x Ohms per foot) = Amperes
Voltage x Amperes = Watts

If you're using 16 AWG Kanthal @ 0.34 ohms per foot it looks like this.

240/(106 x 0.34) = 6.6 Amperes
240 x 6.6 = 1584 Watts.

Pop me an email at knives(at symbol)


Michael Clerc said...

Hi Dan,

Well, what a pleasant surprise I just got, you re-drew the schematic for 220vac, YOU ARE THE MAN!!!


DanCom said...

Yes, check over on the HT2100 build page. It's a little simpler than the HT2100, but will work about 3000 Watts at 240 Volts.



Jay Gonsalves said...

Dan, I have thoroughly enjoyed your heat treat oven build. I've enjoyed it so much that I've ordered the stuff to make one for my shop. A question, could I use a single heating element in my oven? It appears that the 16ga kanthal I have is rated at.324 ohm/ft. I'm not the sharpest knife in the drawer when it comes to electrical!
Thanks for an very well done diy project description.
Jay Gonsalves

Wilson Bailey said...

Thank for all the calculations these are so helpful! I'm building a HT oven with a PID controller and maybe I'm just missing something simple but what is it that regulates the Amp draw? What prevents the element from just pulling Amps until it burns itself up or trips a breaker?

D. Comeau said...

Hi Wilson,

The wire itself is a resistance wire. It only can draw a limited amount of current based in the wire's built-in resistance. The only way to make more heat (or current) is to apply more voltage and thus make more watts (heat).

Hope this helps,


Cheyenne Meyer said...

Nice article. One correction: 16AWG wire is ~0.0508" not 0.074". I'm guessing your estimate is including wire insulation which, as you already know, this wire does not have. Wikipedia has a page which includes a formula for converting AWG towire diameter as well as a table of many gauge sizes.

Blaine Taylor said...

Hello. Very good read here, but i have a question on designing a kanthal element. I'm building a furnace and would like to be able to melt copper, which is nearly 2000 F and basically exceeds a safe working temp for NiChrome wire. I've found information on as to the amount of current required for a given size of nichrome wire to operate at a given temperature, but i can't find this info on kanthal. The first chart of this pdf is what i'm referring to.

I'm working with 240v and i'm using standard fire brick arranged to give me inside dimensions of 6.5"x 6.5"x 4.5". I've successfully used 16 gauge nichrome pulling 22A to melt aluminum, but shortening the coil to pull more amps to get into the higher temps results in a meltdown of the coil, so i want to give kanthal a try, but i can't find this type of info on kanthal. Ohms per foot and watts does not translate in to temperature. Do you have a any info concerning current vs temp range of kanthal? Thanks.

D. Comeau said...

Hi Blaine,

Kanthal A1 is rated for 1400°C (2552°F) and so should be able to withstand the temperatures required to melt copper. In fact I regularly take my heat treat oven to 1080°C (1975°F) with Kanthal A1 running @2000 Watts.

This is Kanthal's handbook for the product.

You will need to get your watts up if the loading is significant. I don't know what the firebrick is and the size the crucible. I don't think pulling 22 A through some 12 gauge wire would be a problem.

The fastest way to melt copper is with a gas setup, but there may be reasons for going electric such as CO emissions.

Good luck!

Blaine Taylor said...

Thanks Dan for the quick reply. The link you provided doesn't contain any info on A1. I have seen their furnace handbook but i didn't see anything in it concerning temperature vs amperage. If you'll look at the first table of this link for nichrome wire and we used your example of 22 amps through 12 gauge but used nichrome wire, the temperature would be less than 1000f, and the chart shows that 12 gauge nichrome would require over 30 amps to get hot enough to melt aluminum, or that 15 amps through 24 gauge nichrome would melt it. I'm looking for this type of information on kanthal A1.

Thanks, Blaine.

D. Comeau said...

Hi Blaine,

I don't understand the author's concept with respect to temperature and current. It's Watts that equate to energy. A large current at 2 V is not a lot of Watts if you get what I mean. If you wanted to use the wire act as a fuse, then this table is helpful. You will know at what current the wire exceeds it thermal capacity and burns up. I think you found this out already. ;-)

But there are a few other factors to look at. For example heating a wire in a poorly insulated, say regular masonry brick, or leaky environment will never reach intended temperatures. This is because the loading will always suck heat away from the coil and not contain it. This is called "loading."

The PDF I was looking for was no longer on Kanthal's web site, but I found it at Hi-temp's site. Although they don't reference temperature per Amperes, for reasons I mentioned above, they do have a furnace wall loading chart. This is what is of interest to us. Units are in Watts (not Amperes). On page 6, the curve shown for (a) indicates around 2800 Watts per square foot of wall to reach 2110°F (1100°C). Take your interior wall dimensions and determine what the square footage is. In your case, you are looking at about 0.7 square feet.

2800 Watts per square foot * 0.7 sq. ft. = 1960 Watts. (If you want faster heating times, crank up the 2800 W/ft2 to 3000 or more.)

To get 1960 Watts from some Kanthal we need to know the voltage supply. So we take 1960 Watts and divide by the Voltage to get Amperes.

1960 W / 120 V = 16.33 Amperes.

With the desired Amperes we can find the Ohms.

Ohms is 120 V /16.33 A = 7.34 Ohms.

Use the resistance per foot of Kanthal A1 for a suitable gauge to determine the length of wire needed to make 7.34 Ohms.

For 240 V circuits (2 times the Voltage), use two times the Ohms 14.7 ohms.

If you are concerned about the total current through a single coil, split the current into two or more parallel coils will create the Watts you need without the high Amperes in any one coil that will cause your coil to be a fuse. :-)

I hope this helps.



Blaine Taylor said...

Another example of amps vs temp. I'm using a 16 gauge cord to power my 2400W furnace. This is a cord from a typical 1500W space heater. It doesn't even get warm to the touch at 2400W.

The starter draw on my V-8 truck is right around 200A. At 12V, that's 2400W also. If i tried using the 16 gauge cord as the starter cable on my truck, it would become really hot really quick and melt. Was this melting temperature a result of 2400W of power? Nope. It was the 200 amps trying to pass through it.

It's the amperage that that has more to do with what temperature range you'll be in that what the wattage does. The wattage just needs to be great enough to overcome the load of the furnace size and insulating factors. Any wattage over that point will just get things up to temp faster.

Sure, there are a lot of variables in play here, but i feel amps vs temp for a given wire size is the bases on where to start.

So my question still stands on what a "suitable wire size" is for kanthal A-1.

At this point i feel that trying to reach high temps with 120V and stay within the recommended 80% operating range of a 20A breaker, i'd need to go with 18 gauge or smaller.
I suppose i'll need to spend some time and money and do my own research.

Thanks again Dan.

D. Comeau said...

18 AWG Kanthal A-1. Shoot for around 7 ohms @ 120 V (~2000 W). A well sealed furnace with K firebrick at that small volume will get to copper melting temps.


Blaine Taylor said...

I wealth of info on current vs temperature.

Blaine Taylor said...

Thanks for all of your help Dan. I'm trying to design a 240v element for the same furnace dimensions from my earlier post of 4.5" x 6.5" x6.5", which i come up with a total of 1.4 sqft. I'm struggling with the kanthal handbook that you had linked to before. To quote you:

"On page 6, the curve shown for (a) indicates around 2800 Watts per square foot of wall to reach 2110°F (1100°C)."

Ok, this is where i'm stuck, well, for starters anyway. If that is how the chart on page 6 is to be interpreted, it defies common sense. To follow your example and referring to the chart, we'd need over 4600w per square foot of wall to reach only 1470F. ???

Common sense tells us that if you want hotter temps, you need more power, not the inverse.

Honestly i feel they have the wattage and square footage flip-flopped. This chart would make a lot more sense if the column on the right was square feet per kilowatt, not kilowatts per square foot. Maybe i'm just confused.

I'm also find the element surface load chart confusing also. If i understand it correctly, the surface load is simply the wattage in relation to the amount of surface area of the element wire. Once again, my logic has it that in order to see higher temps, i'd need more wattage across the wire, not less, as their chart indicates. Their recommended maximum surface load for 2000F is 19W/square inch, while 1470F is 32W. Less where you really need it, and less where you don't. Honestly though, i do think i understand why.

Please correct my math if this is incorrect. To calculate the surface area of a wire, you should multiply the diameter by pi, then multiply that by it's length. 18ga for example:
.040" x pi = .125 square inches. Multiply this by 12" and it gives us 1.5 square inches of surface area per foot of 18ga wire.

So let's apply this to the 2000w 120v element configuration that you had recommend i try to get to copper melting temps, which i thought sounded just about right to me.

"18 AWG Kanthal A-1. Shoot for around 7 ohms @ 120 V (~2000 W)"

For a 7 ohm we'd need 12.9 feet of 18ga wire. At 1.5 square inches of surface area per foot, that gives us a total surface area of 19.4 square inches.

At 2000w, the element would have a surface load of 102w/sq in.

That's over 5 times their recommended maximum load of their chart. Maybe i suck at math. Maybe their charts do.

What gives?

Thanks. Blaine.

Bradley Howes said...

How did you come you with needing 19 ohms for your elements?

D. Comeau said...

Hi Bradley,

I intended the total watts to be around 2200. At 120 Volts that would require current of 2100/120 = 18.3 A. 120 V divided by 18.3 A = 6.55 Ohms. With 18 gauge wire I could use three elements in parallel to get close to this. Adding elements in parallel lowers the resistance. I used the formula Rtotal = 1/(1/R1)+(1/R2)+(1/R3) where R1, R2 and R3 are close to 19 ohms. This gives us close to 6.5 ohms.

Hope this helps.


James Vansandt said...

Dan, is there a practical guideline for the "recommended" maximum current to put through a run of 18 Ga or 16 Ga. I'm looking at sizing some elements. I know in conventional wiring 15A - on 14Ga Copper but since the goal here is to produce heat not limit heat.. It's a bit uncharted to me.. I can assume the real benefit wit 16 vs. 18 Ga would be longevity.. Any opinions along that line would be welcome..

D. Comeau said...

Hi James,

Yes it's kinda weird from a conventional copper wire perspective where we do not want to melt the insulation off. Resistance wire has no insulation and no current ratings. It does however have a melting temperature.

One piece of 16 gauge would reach melting point in an enclosed space while another would never reach melting point with a fan blowing on it. Even if the current was the same. So the environment has a role to play as well. Let's assume your oven will not have a fan inside your oven moving air around. LOL

Obviously a very short or low resistance piece of wire will get very hot, reach melting point and fail, much like a fuse.

I am pulling close to 6 A per leg of 18 AWG Kanthal A1 and the temps are closing in on 1100°C (2000°F). For a maximum I would have no problem with moving twice as much current through the same wire. However, I have never found a "maximum rated current" table for Kanthal. Some data on Nichrome wire suggest 23 A for 18 AWG and 32 A for 16 AWG.

So I have no solid answer, but I hope that the empirical data is helpful.


James Vansandt said...

Thank you sir, any thoughts on element longevity between 16 and 18 gauge ? since both are readily available I'll work out designs for both as an exercise..

D. Comeau said...

Hi James,

I can only comment from my experience with 18 gauge. The latest 18 gauge elements have been running in my oven for a couple of years, whereas the imported 22 gauge ones I got from eBay failed in less than a year. If we dared to assume, it makes sense that heavier wire will be more durable. I've worked with a few makers that went with 16 gauge and their ovens are working fine. IMHO 18 is fine, 16 is better. ;-)


James Vansandt said...

Thank you sir, some very nice write ups by the way.. After running the calcs.. to get the resistance for the target wattage 16ga was going to take up too much length for the design I have in mind.. So 18ga it will be.. the plan is to build the oven in two segments that can be coupled together for sword length blades or use a single segment for knife length projects .. should be interesting.. Thanks again

BernieThePie said...

Hi Dan,

I would like to sincerely thank you for all the info and experience that you have shared with us, really appreciate It :) I made my own furnace for heat treating blades (I mainly work with 80crv2 @ approx 840 degC) and used your blog for the most part of the build and It works perfect. I didnt know anything about Kanthal up till then and only just few days ago I stumbled accross some info that 1 mm (0.0403 in) 18 awg is a bit thin for 3 kw of power at the elements lenght of approx 9 meters and that should be more like 3 mm (0,118 in) thick regarding the power to surface ratio of the Kanthal element and thus shortened lifespan. 18 awg was pleasure to work with and would choose It again Its just that I dont know, yet, how does It look when Kanthal wire is gone; ok, obviously It doesnt work any more, but is it like on/off or It just starts to drop in performance and It is hard to notice It? I left "the ceiling" uncemented at first to provide myself with the service access to the Kanthal element but during past months I have competely forgotten about It and cemented the ceiling to prevent any heat losses so It will be quite an unpleasant experience replacing It, geeeee. Sorry for any weird english as It is not my first language :)

Thank you again, It really Is one awesome blog. Big thumbs up :)
All the best,


D. Comeau said...

Hi Bernard,
Thank you for commenting.

The recommendation was for two runs of 18 AWG Kanthal A1 in parallel, which will be fine at 3 kW. Being in parallel, the current in each leg is half of the total current. 3000 / 240 = 12. 5 amperes. 12.5 divided into two runs is 6.25 amperes per run. This also halves the power in each leg and the surface is doubled. Elements should last a long time, even at 1100­°C usage.

After replacing my first set of "store bought" elements, I inserted some small diameter (1/8" ~3mm) steel rods to hold the ceiling bricks together. This was done to accommodate future element change-outs. This was noted in the HT-2100 build page.

Again, thank you for pointing this out. Readers and myself always appreciate feedback from people who have actually made their own heat treatment ovens.

Best wishes,


BernieThePie said...

Hi Dan,

yes, my "grand design in one piece" was entirely my fault, of course :) I am not best friends with electicity in general so I apparently gladly skipped the concept of two runs of wire. But as there is always positive side of a negative situation, my oven will be a testing ground for Kanthal`s lifespan under not so ideal conditions. I have found that post about Kanthal`s calculations If it is of any interest to anyone

Thank you and best wishes,


Ukzz said...

1. Find out temperature in a given size room in perod of time? Room size 17"x25"
Element 1500w , 230v,600x5mm size, No of piece =2(i.e.,1500x2) , i need 200°c temperature so how much time it take to get that much temperature in a given room?

D. Comeau said...

Unfortunately there are too many unknown parameters. What are the thermal properties of the walls? Ambient temperature outside the room. Is there going to be thermal mass inside the room? etc. If the walls are if typical K23 fire brick, with little thermal mass at room temperature, 1500 W will reach 200°C in a very short time. I'd guess based on my oven less than 30 seconds.

Tod Livingston said...

This is my attempt to explain a method of choosing the Kanthal gauge and resistance based on the maximum kiln operating temperature. I am not a heating pro and the following info is my best guess and may not be completely accurate. Hopefully others will chime in with corrections as needed.
Basic physics dictates that as we increase the wattage, we increase the temperature. Here is the problem; The higher the maximum kiln temperature the lower the recommended wattage per square inch of Kanthal.(Per the manufacturer) I believe this inverse relationship exists to minimize the temperature differential between the coils and the load. This means we will need much more coil material to expend the same amount of wattage at higher temperatures.(To prevent damage to the Kanthal?) See page 7 of Kanthal Handbook. There is an associated maximum wattage per square inch for each max temp design. Example: for 1,470 deg F the max recommended wattage per square inch is 32 watts or less. for 2,370 deg F the max recommended wattage per square inch is about 15 watts or less. Crow077 posted on April 8, 2016 at 10:58 AM He shows how to calculate the recommended resistance for a kiln designed to operate at a maximum of 1,830 deg F. According to my calculations, For a 120V design with a desired maximum temperature of 2,375 deg F, using 14 watts a square inch, 16 ga diameter of .0508 in, .324 ohm per ft of Kanthal A1, the Wattage will be 1,095 (40.8 ft wire) requiring multiple parallel coils to reach the 3,500 to 7,200 watts a square foot as shown by Mr. Comeau. 120V design with a desired maximum temperature of 2,375 deg F, using 14 watts a square inch, 18 ga diameter of .0403 in, .515 ohm per ft of Kanthal A1, the Wattage will be 775.(36.5 ft of wire) For the same temp design at 240V 18 ga produces 2,185 Watts (81.5 ft wire) 16 ga produces 1,545 Watts (72.6 ft wire)

John ______ said...

Thanks very much for the excel file... helps A WHOLE LOT !!! I was close without it but I couldn't be sure, so the spread sheet gave all the information I needed. I'm needed to make a 400W element for an Arbe style wax injection machine I'm constructing (except mine will have PID temperature controller w/ thermistor feedback loop) for injecting molten wax into molds. to eventually do investment casting.

arda belati said...

Hi Dan,

I am new in knife making and would like to build this HT oven to learn further in the journey or knife making.
I tried to read through all the information but I just want to confirm if I understand correctly.

I live in Australia and we have 240V - 10Amp in our normal household.

using the excel spreadsheet calculation - to reach 2400Watt - I am planning to use 1 element only.
The chamber is about 0.648 cubic feet.

Will this allow me to reach 1100 deg C mark?

Thank you

D. Comeau said...

From the experience I gained with my build, I would try reducing the volume of your chamber to around 0.5 cu. ft. (4800 W/cu.ft.) if possible. Getting that last 100°C is going to take a lot of time. I am running 2100 W into 0.44 cu. ft. or 4700 W/cu.ft. I can hit 1070°C after about 90 minutes. (see HT-2100 Thermal Curve)

Be sure to seal everything nice and tight. The door needs to be fitted so the IFB faces are in contact and no gaps. Any leakage to the room will reduce the maximum temperature and slow the heating times.

Hope this helps.


Unknown said...

This is very helpful in planning my oven build! In reviewing your Kanthal A1 calculator spreadsheet, is the coil diameter the inside or outside diameter of the coil?

My oven is going to be 17.5" deep, 6.5" wide and 4" tall. I have the ability to run 240V, and was hoping for approximately a 3000W element for faster ramp up to temp. Running a single element in the oven in two rows along the walls, I would need about 80" in stretched length, so the 16ga wire in your spreadsheet looks to be about perfect for what I want, especially for longevity of the element.

D. Comeau said...

The coil diameter parameter on the spreadsheet is the intended OD of the coil. Subtract 2 times the diameter of the wire to get the form diameter for winding the element. I used some 3/16" steel rod as a form to make a 1/4" (OD) coil. It's close enough for practical use.

3kW will be good power into a space that size.

Let me know how it goes,


Unknown said...

Hi. How you determined that you needed the AWG 18 calibre wire?

D. Comeau said...

I chose 18 AWG resistance wire based on the ohms per linear foot rating of the wire. In some locations resistance wire is available in metric sizes and you may be able to get say 1 mm or 1.2 mm equivalents. The idea is to match the resistance to the intended wattage and still be within a suitable total current draw. If your supply is limited to a known current, (in my case 20 Amperes), plan around that constraint and choose resistance wire that gives you a suitable wattage. As the wire gauge increases, so does the ohms per foot rating and less wire is required to make your target resistance. Very thin wire is more prone to breakage and harder to work with. I will recommend using either 18 AWG or 16 AWG for DIY electric heat treatment oven elements. There is a spreadsheet that helps with the choosing the wire size, length and number of elements linked on this page.


Pete said...

I'm thinking of using you 240v setup for a brick pizza oven I'm building. About 1.8cu.ft and i wont need more than 700-800°F. I'm thinking about 3000w but on a single run. I would love to hear your input on this idea.

D. Comeau said...

Hi Pete,

I am not sure what the loading and losses would be like and if you would be able to overcome these with the wattage provided. The pizza ovens I am thinking of are usually open front and the losses to the room would be significant. Are you thinking of a closed oven or an open front oven?


D. Comeau said...

Hi Pete. Sorry I inadvertently deleted your reply. Is the inside of the oven going to be fire brick? What insulation will surround the interior, insulating brick, steel & fiberglass? My concern is loss of watts and this is hard to determine unless the materials and construction are known. A typical 30" range/oven at home will be somewhere in the 4 to 6 kW range. Although these are larger in volume they are insulated, including the door.


Pete said...

Inside 2" fire brick dome shaped(igloo) surrounded with 1" ceramic whool insulation + 1" aerated cement(also insulating properties) + mosaic tile.
Floor 1 1/2 pizza stone under that 1" insulating ceramic board under that 1" aerated cement.
I realize the biggest heat loss will be at the door.
So to be safer should increase the wattage?
How whould you do that? Making the loops larger to take a longer wire? Heating element will go around the perimeter about 72" long
Thanks for the reaply

D. Comeau said...

Hi Pete, I can't see a problem with 3 kW if you have 1" ceramic wool for insulation and can keep the heat in with the cast iron door until it gets up to temp. Once the mass is warmed up the system will become very stable.

Pete said...

Thats what I thought. Thanks for the input. Great instructions btw
Thank you