Help designing kanthal coiled heating element

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
Hey buds, I'm really struggling to figure this out. I'm trying to make my own home made fiberglass jacketed heat tape. I need it 8' long(a lil longer is ok, say like 3meters if you wanted to stick to a meter scale) And was planning on doing kanthal wrapped around a 8mm rod to coil(8' of the 8mm coil to be stuffed into 1/2" fiberglass jackets) I'm on 120v AC but have a plethora of DC voltage supplies too, including a 24v 14amp, stacks of PC powersupplies converted into powersupplies, and better or smaller. About what gauge and how much kanthal wire am I gonna need to make the coil?

I THINK from my understanding I'm gonna need about 28-32w/inch to make this?(The tape is gonna be made into a 1L heating mantle so it needs to be able to reach at least 700c)

I've been trying to use the guidance here:

https://wiretron.com/wp-content/uploads/2017/04/NiCrTechTips.pdf
And from kanthal.com

And chatgpt says:(tho now that I'm copy pasting it I realize I asked the AI kinda weird.. so it's prob deff wrong?)

ME:

If I have a wire that is 1.8 Ohm/m how much wire would I need making a spiraled heating coil that would be 2.5m long and produce at least 5kW of heating

ChatGPT:

To calculate the length of wire needed to create a spiraled heating coil that is 2.5 meters long and produces at least 5 kW of heating, you can use the following steps:


  1. First, determine the resistance of the wire required to produce 5 kW of heating. The formula for power (P) in a resistive element is:
    P = V^2 / R
    where P is the power in watts, V is the voltage, and R is the resistance.
    Since you want to produce at least 5 kW (5000 watts) of heating, and you have a wire with a resistance of 1.8 Ohms per meter, you can rearrange the formula to solve for resistance:
    R = V^2 / P
    In this case, V is the voltage across the coil (which we'll assume to be constant), and P is 5000 watts.
    R = V^2 / 5000
  2. Now, let's calculate the resistance needed for 5 kW of heating. Assuming a constant voltage, you can calculate the resistance required:
    R = V^2 / 5000
    To achieve the desired heating power, you'll need a specific resistance value. However, for this calculation, we'll use a voltage of 120 volts (a common household voltage). Keep in mind that the actual voltage might be different depending on your specific application, so you should use the appropriate voltage value for your setup.
    R = (120^2) / 5000 R = 14.4 Ohms
  3. With the desired resistance of 14.4 Ohms, you can now calculate the length of wire required. You have a wire with a resistance of 1.8 Ohms per meter, so you can use the formula:
    Length (in meters) = R / (Resistance per meter)
    Length = 14.4 Ohms / 1.8 Ohms/meter Length = 8 meters

So, you would need approximately 8 meters of wire with a resistance of 1.8 Ohms per meter to create a spiraled heating coil that is 2.5 meters long and produces at least 5 kW of heating with a voltage of 120 volts.

But it seems to me the calculations in the guides demand I already know the answer to my question I'm trying to figure out? lol.. Maybe I'm just not getting it?

If someone could show their work with answer to I'd appreciate it, may help me understand better. TY so much ya'll. Feel dumb rn. I can do basic ohms law stuff but adding the coils into the mix my brain just won't do it lol.
 

MisterBill2

Joined Jan 23, 2018
17,791
First, more information is required because there is some maximum amount of current that the wire is rated for, as well as some power level that should not be exceeded.


Then the question of temperature. 700° C is quite hot. What are the specifications of this wire?
To make a spiral wrap you will need to know the inside diameter of the spiral, which will be the outside diameter of what you will be wrapping it on. And that material will need to not be damaged by the 700° C heat. And it will need to be a good insulator as well.

But you do not know the maximum current allowed through the wire and so it might easily be less than 20.833 amps.
So you need more information about the wire.
 
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Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
First, more information is required because there is some maximum amount of current that the wire is rated for, as well as some power level that should not be exceeded.
And the bad news is that to produce 5000 watts of heating using 120 volts the current will be 41.66 amps. (Power= Volts X Amps), then AMPS=Power/ volts=5000/120= 41.666 amps.
Then the question of temperature. 700° Cis quite hot.
To make a spiral wrap you will need to know the inside diameter of the spiral, which will be the outside diameter of what you will be wrapping it on. And that material will need to not be damaged by the 700° C heat. And it will need to be a good insulator as well.
So it is quite possible that you will want to power this heater with 240 volts.t Then 5000watts/240 volts = 20.8333 amps. and that can be done on a standard 240 volt 30 amp service connection. Next, V=Amps x Ohms . > so R (in ohms)=Volts/ Amps =240/20.833 =11.52 ohms. That is the resistance used to develop 5000 watts of heat with 240 volts applied to the wire and 20.8 amps flowing through the wire.
But you do not know the maximum current allowed through the wire and so it might easily be less than 20.833 amps.
So you need more information about the wire.
Shoot.. People made it seem like the tape could heat a flask NP running 12v 6amps or so? I'm gonna have a thermo couple and pid controller?
They have tapes for sale that can do what I want running off less power.. heres a link to pretty much exactly what I'm trying to make. https://cdn.shopify.com/s/files/1/0283/5140/files/LABORATORY_PRICE_LIST.PDF?1476
part # part # AWH-051-080D
Seems you're having the same issue i am. I don't know what wire or gauge to use OR the length I'll need to make the 8' coil, so I'm just at a total loss. There has to be a way to do it, other people do it XD I'm just not smart enough to figure it out. I'm just trying to heat a 1l flask to 600c or so with the tape. The tape will be 8' with 8' of coiled kanthal inside the sleeves so it'll actually be way more then 8' of kanthal cause it's coiled (home made heating mantle/stirrer)
Heres some examples of what I'm shooting for.:
 

MisterBill2

Joined Jan 23, 2018
17,791
It is not a lack of smart, it IS a lack of information about the heating wire. What temperature is the wire good for? and what is the allowable current?
The only error I see so far is asking AI to help.

And the bad news is that to produce 5000 watts of heating using 120 volts the current will be 41.66 amps. (Power= Volts X Amps), then AMPS=Power/ volts=5000/120= 41.666 amps.

Next, V=I x R, so R=V/I=120V/41.666amps =2.88 ohms.
SO 2.88 ohms/ 1.8 ohms/M =1.6M
That is a very short bit of wire to be dissipating 5000 watts. So it is time to stop and consider reality. Like, do you really need 5000 watts?
That is a whole lot of power.
If you explain the goal and what it is intended to achieve, probably a few folks here will have all of the experience needed to be able to share the correct answers.
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
It is not a lack of smart, it IS a lack of information about the heating wire. What temperature is the wire good for? and what is the allowable current?
The only error I see so far is asking AI to help.

And the bad news is that to produce 5000 watts of heating using 120 volts the current will be 41.66 amps. (Power= Volts X Amps), then AMPS=Power/ volts=5000/120= 41.666 amps.

Next, V=I x R, so R=V/I=120V/41.666amps =2.88 ohms.
SO 2.88 ohms/ 1.8 ohms/M =1.6M
That is a very short bit of wire to be dissipating 5000 watts. So it is time to stop and consider reality. Like, do you really need 5000 watts?
That is a whole lot of power.
If you explain the goal and what it is intended to achieve, probably a few folks here will have all of the experience needed to be able to share the correct answers.
You know mate, I'm not even sure where that 5Kw of heating came from. The more I research, the more it seems 600-1000 watts would be plenty of heating, especially contained in a thermos or pot and fiberglass to heat a 1ml flask to 600c.
Kanthal wire has a extremely high temp resistance, with continuous operation of 1400c being acceptable and 1500c melting point, I won't be going half that hardly, and I wouldn't be coming near it's limit there. I think they say factor 10-30% for extremely high temps in kanthal but since it will be controlled with a temp prob and PID I'm not so concerned about that(seems fairly negligible to me in this scenario?) I should probably remake all my posts ;.; this has been a nightmare trying to figure this out. I just wanna know what gauge kanthal an how much to coil ;___:
 

MisterBill2

Joined Jan 23, 2018
17,791
OK, it should work to use the same process, just substitute 1000 for the 5000 watts. That will give you one fifth the current and five times the resistance and length, And those numbers will be more reasonable. Hopefully the formulas are clear enough showing what I was doing so that you can know the process.
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
OK, it should work to use the same process, just substitute 1000 for the 5000 watts. That will give you one fifth the current and five times the resistance and length, And those numbers will be more reasonable. Hopefully the formulas are clear enough showing what I was doing so that you can know the process.
Yes, but what makes this calculation hard(impossible for me) is that it is coiled, not just a straight shot of wire. The coil HAS to be AT LEAST 8' long[total coil length]. There is special formulas, but one I tried seems like it uses a mix of metric and USA and doesn't specify were an when, an it also seems like it needs the #s I'm trying to figure out and that are unknown to me to figure it out, an it's just impossible to me to decipher, and the guide from kanthal is even harder to decode. =/ The wire will be coiled around a 8mm rod(to shape) and the coil needs to be 8' in length, to fully fit a 1l flask. and be able to produce 800-1000watts of heating. I've been on every forum there is and cannot get a answer. Seems only the scientists at wiretron an kanthal know this stuff >.<
 
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Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
OK, it should work to use the same process, just substitute 1000 for the 5000 watts. That will give you one fifth the current and five times the resistance and length, And those numbers will be more reasonable. Hopefully the formulas are clear enough showing what I was doing so that you can know the process.
Even the coil wrap calculators online are totally useless because they demand I know the permiteres I'm trying to figure out before it'll calculate. makes no sense. =/
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
OK, it should work to use the same process, just substitute 1000 for the 5000 watts. That will give you one fifth the current and five times the resistance and length, And those numbers will be more reasonable. Hopefully the formulas are clear enough showing what I was doing so that you can know the process.
  1. Calculate the coil's resistance:
Given the total resistance required is 18 Ohms and the resistance of the Kanthal wire is 0.972 Ohms per meter, you can find the length of wire needed for a straight wire with the same resistance:
Length of straight wire = Required resistance / Resistance per meter Length of straight wire = 18 Ohms / 0.972 Ohms/meter ≈ 18.52 meters
  1. Account for coiling:
The actual length of wire needed for a coiled wire will be longer than the straight wire due to the coiling. To determine the coiled length, you need to consider the coil's geometry. If you have a uniform helical coil with an 8mm diameter (4mm radius), you can calculate the length of wire using the coil's circumference formula:
Circumference of coil = 2 * π * radius of coil
Circumference = 2 * π * 0.004 meters ≈ 0.0251 meters
Now, calculate the number of coils required to reach the straight wire's length:
Number of coils = Length of straight wire / Circumference of coil Number of coils = 18.52 meters / 0.0251 meters ≈ 738.24 coils
Finally, find the total length of coiled wire:
Total length of coiled wire = Number of coils * Circumference of coil Total length of coiled wire ≈ 738.24 coils * 0.0251 meters/coil ≈ 18.54 meters
  1. Convert the length to feet:
Since you want the coil to be 8 feet long:
Total length of coiled wire ≈ 18.54 meters * 3.281 = 60.79 feet
So, you would need approximately 60.79 feet of Kanthal wire with a resistance of 0.972 Ohms per meter to create a spiraled heating coil that is 8 feet long and produces 800 watts of heat when operated at 120 volts.

Hows this look to you?
 

MisterBill2

Joined Jan 23, 2018
17,791
Is there more than one diameter of this Kanthal wire available ? A thinner diameter will have a higher resistance.
And I am puzzled because I get 14.4 ohms, based on 1000 watts and 120 volts. P=VxI, so I=P/V=1000/120=8.333 Amps
Then V=Amps x Ohms, so Ohms=V/Amps=120/8.333=14.405 ohms to draw 1000 watts at 120 volts . And that number seems reasonable.The resistance of 18 ohms must have come from using 800 watts.
I will need to dig into an old book of engineering formulas to find the length per turn of a helix, which is more than a circumference because it has distance along in addition to distance around. (it is only 5:24 AM, been back from work an hour, almost.)
 

shortbus

Joined Sep 30, 2009
10,044
Why do you think that coiling the wire makes more heat? If you can decide on the length of wire that makes the heat needed coiling won't make a difference. Your 1L flask is only so big so how are you going to wrap eight feet of wire, coiled or not, around it? You are talking about an Erlenmeyer flask aren't you?
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
Is there more than one diameter of this Kanthal wire available ? A thinner diameter will have a higher resistance.
And I am puzzled because I get 14.4 ohms, based on 1000 watts and 120 volts. P=VxI, so I=P/V=1000/120=8.333 Amps
Then V=Amps x Ohms, so Ohms=V/Amps=120/8.333=14.405 ohms to draw 1000 watts at 120 volts . And that number seems reasonable.The resistance of 18 ohms must have come from using 800 watts.
I will need to dig into an old book of engineering formulas to find the length per turn of a helix, which is more than a circumference because it has distance along in addition to distance around. (it is only 5:24 AM, been back from work an hour, almost.)
Yes I think my calculations were based off 800 watts. This is way way harder then I thought It was gonna be. Seeing if kanthal or one of these wire techs email me back lol
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
Why do you think that coiling the wire makes more heat? If you can decide on the length of wire that makes the heat needed coiling won't make a difference. Your 1L flask is only so big so how are you going to wrap eight feet of wire, coiled or not, around it? You are talking about an Erlenmeyer flask aren't you?
That's so funny we had the same thought. I tried looking into it and if I increase wire size, it just doesn't work, or it does but I'd need to wrap around the flask like 7000 times lol. and I'd like to have the flask fairly wrapped in it so it'll be stuffed into 1/2" fiberglass sleeving.
8mm coiled wire fills the sleeve a alot better then a single tiny strand, trying to avoid hot spots around the glass and thin wire would cause that I think? I'm open if you have different work arounds =]
 

shortbus

Joined Sep 30, 2009
10,044
8mm coiled wire fills the sleeve a alot better then a single tiny
You do understand if you coil the wire you need someway to keep the coils from touching, don't you? I the coils touch they will act like a short circuit and effectively make the wire shorter. You still haven't said what type of flask your using, just saying "flask" has little meaning to the problem.

Have you seen this place? They have lots of flask heating methods. If you think your going to spend less on a DIY method you might only be fooling yourself. https://www.fishersci.com/us/en/browse/90056015/heaters-and-heating-mantles?page=1
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
You do understand if you coil the wire you need someway to keep the coils from touching, don't you? I the coils touch they will act like a short circuit and effectively make the wire shorter. You still haven't said what type of flask your using, just saying "flask" has little meaning to the problem.

Have you seen this place? They have lots of flask heating methods. If you think your going to spend less on a DIY method you might only be fooling yourself. https://www.fishersci.com/us/en/browse/90056015/heaters-and-heating-mantles?page=1
1l rbf 5something inch's base diameter. And I'll check that out tho I think this maybe the best route for me?
 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
Yes I think my calculations were based off 800 watts. This is way way harder then I thought It was gonna be. Seeing if kanthal or one of these wire techs email me back lol
OH AND YES. There is many different gauges and resistances of kanthal wire. I'm looking at 16gauge as my best bet?
 

shortbus

Joined Sep 30, 2009
10,044
I don't know what your doing(top secret?) but most flask heating mantles don't go to your 700C/1292F temperature. I'm not even sure the flask would stand it. But if your determined to DIY one, have you seen this guy's YouTube channel? The Canadian Chemist is the name. In the description he gives a link to the high heat tapes, don't think you can even get the Kanthal for what he claims to have paid for the tapes. Your temperature is more closely tied to a kiln not a flask mantle.


 

Thread Starter

Space_Dandy

Joined Oct 29, 2023
18
I don't know what your doing(top secret?) but most flask heating mantles don't go to your 700C/1292F temperature. I'm not even sure the flask would stand it. But if your determined to DIY one, have you seen this guy's YouTube channel? The Canadian Chemist is the name. In the description he gives a link to the high heat tapes, don't think you can even get the Kanthal for what he claims to have paid for the tapes. Your temperature is more closely tied to a kiln not a flask mantle.


Yes my friend I have watched all the homemade heating mantle vids there.
I am making a high-bred from the ones I linked. I need stirring to. And it prob won't ever be run that high I just need more then the standards offered. Or even the "powerful" 450c ones. The cost of the tape is no longer cheap like all the vids online say. For 8feet of heat tape that can do what I need I have been quoted 112 THE CHEAPEST. before tax. Before shipping.
60' of 16 gauge kanthal is like 30 bucks and the heat sleeves are like.. cheap.
The homemade heat tape will outperform the 112dollar tape and for half the cost. If I can figure this stupid stuff out. Lol.
But yeah I need magnetic stirring and I don't like the idea of just coiled wire stuffed in loose fiberglass. The tape looks a lot safer an snugger to me. I do appreciate the help. I really wish I could pull of a way to just make it single strand wrap around. =/ I'm kinda considering a sandbath with two long heat probes going down the side and a temp monitoring prob down the center with the mag stirrer below that. May save me this headache.. so much trouble for a heating element.. lol. And my first goal is distillation of sulfuric acid so I need those temps. Even tho I thought the 450c mantles would be plenty because of the bp of sulfuric acid I guess according to most you need to go beyond that for sulfuric distillations. Either way I can't really afford a digital heat stirrer off the shelves. I double checked fisher to an they are way out of the question
Higher prices then almost all the other sites I live scoured x.x really appreciate your help mate
 
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