Sword Heat Treat Oven Design Review

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EDLK

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Jan 5, 2024
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Sword tempering oven design review.
I believe I have my design criteria outlined and would greatly appreciate review, comments, or corrections to the attributes before I complete material procurement.
Any input would be greatly appreciated

Chamber size 7" Wide x 5" High x 44" Deep
.58 x .41 x 3.66 = .87 CF volume

6000 watts power. 240V Single Phase
6000W / 240V= 25 Amps
240V / 25 A = 9.6 OHM

2 Coils in Series = 9.6 OHM x 2 = 19.2 OHM
2 Coils in Series =9.6 OHM /2 = 4.8 OHM
( I am unclear on which of these two options is more desirable for life span of the coils)

16 GU Wire Coiled at .375" = Coil Length of 3.5 ft (Stetched coil Length = 7.0 FT) X 2 Coils
14 ft of stretched coil fits nicely on the 2 side walls
(Thanks to Dan Comeau for his most excellent coil calculator spread sheet)

So a few things I still have not figured out how to calculate?
1. Based on the .87 CF of volume, how do I calculate the resulting Wattage for Oven based on my parameters?
I am shooting for Max temp to be 2100 F
2. What is the recommended ampacity of the protective fusing the following?
a. Line Voltage and Main Power Switch
b. line Voltage and SSR 2
c. Is there any additional protective fusing recommended.

Once again, thanks to anyone who can offer a review and commentary.
 
Welcome EDLK, Fill out your profile so we know a bit about you and where you live.

2 Coils in Series = 9.6 OHM x 2 = 19.2 OHM
2 Coils in Series =9.6 OHM /2 = 4.8 OHM
I believe the first one is Parallel, not series. Two 19.2 coils in parallel equals 9.6 Ohms.
the second is correct - two 4.8 Ohm coils in series equals 9.6 ohms.

You don't show your location, but perhaps the language translation has made a change you didn't intend. The oven to harden a blade is referred to as a Heat Treatment Oven. Tempering is often used incorrectly. Tempering is a lower temperatutevtreatment after hardening.
 
Stacy
Thanks for the correction , just a typo
Where on the site do i fill out my profile
Do you have any additional in put on the coil calculations
And in this case is series or parallel the better option ?
 
Look at the top right of the screen and click your name and avatar
Click Account Details
Add details in the spaces provided along the right side of the page
Click SAVE

Youb can change things anytime you want.
 
Series vs Parallel is a choice of design. Length of coils is often the determining factor. Sometimes they are done in combination. One top coil and two side coils with the sides parallel and both in series with the top. (for example only)

Take a look at the Paragon KM45T It is pretty much what you want to build, with a slightly less wide chamber. It has top and side elements and consumes 4500 watts, 20Amps
The top coil is .8 ohms and the two side coils are 5.6 ohms.
Your 6000W coils would work for your 5X7" chamber size.

The Paragon KM45PRO is a huge 300-pound commercial beast at 11500Watts ... 49 amps at 240VAC! It has then same internal volume as your build, but almost twice the power draw. It has 7 coils in a muffle type installation.

As far as fusing:
The breaker at the power panel should be at least 120% of the total draw of the oven - for your build you need a 30amp circuit with the proper gauge wires to the outlet. It is what protects the coils and SSRs.
The fuse on the oven usually only protects the electronics - The controller. 1 amp is normal.
 
Found the profile and filled it out, thanks

I checked out you ovens you referenced.
I have 50 amp circuit I can utilize for the oven and would like to build in a much wattage is a possible, and practical.
I am fortunate to have a solar array on the shop so controlling power consumption is not a design criteria.

With my chamber dimensions being .87 CF, and I have 50 amps overcurrent protection, using you 1.2 derate, I can utilize 40 amps of current.
Should I be sizing my coils to fit the length of available channel to evenly spread the heat? If so, 2 rows on the sides and I have apron 14.5 ft of channel.

I would like to max out the 40 amp draw and put as much wattage in the oven as possible. Still struggling with working back and forth with math on the coils
I have 16 Gauge Kathal A1 wire

I see that many of the commercial ovens have top coils, How critical is that?
 
I'll let the HT oven build folks chime in on the rest of this build thread.
There are a couple folks who have considerable experience at it. Sadly, Dan can't help anymore. He passed away last spring.
His site still has info that is a must read.

What I will add is:
You choose the gauge and length to fit your three primary needs -1) Desired wattage, 2) Total length of the channels, 3) Durability of the coils.
12-16 gauge is the norm for high output ovens. You would need 6 ohm total resistance to get 40A at 240V = 9600W. Not sure why you want a wattage density over 11,000 w/sq.ft. Your first set of parameters at 25A was a better choice for a homebuilt oven, IMHO.

You will need BIG SSRs with BIG heat sinks. I suggest a fan across the fins. I usually recommend about twice the draw, so 75amp SSRs would be good.

I would suggest 3" thick K26 walls with kaowool around them.

I would suggest putting the control unit (PID or whatever) in a separate enclosure not attached to the oven.
 
Yes, Read through Dan's site, used his calculator to start understanding the coil construction
Thanks for the solid input.
If I go with the first set of parameters at 25 A, any idea what I could expect for the time required to come up to full temp?
I was going one the assumption the increased wattage would allow it to come up faster.
No need to grossly overbuild, just wanting to be sure I get it right on the first pass.
Thanks again Stacy
 
WIth some additional reading and some great insights from Stacy, here is where I am with the design parameters

Oven cavity will be 6.0" wide x 44" Long x 4.5" high
Converter to Square feet .50 x 3.66 x 4.5 = .68 CF
I would like to carry 4.5 Watts per cu in
.68 CF - 1175.04 Cubic Inches
1175 CI x 4.5 W = 5287.50 W minimum

Power source is 240 V Single Phase
Proposed design Wattage 6000 W / 240 V = 25 A
240 V / 25 A = 9.6 OHM

Using 16 Gauge Kathal A-1 the stretched coil length for a single coil would be only 3.5 ft (according to the coil calculator from D. Comeau)
I am looking to fill appox. 14 ft of stretch coil channel in the oven.

So this is the point where I am still scratching my head.

How do I modify the coil design to get the proper length of coil (14 Ft) , and hit my target wattage in the oven of 5500-6000 watts?
 
  • Figure the needed amperage to get 6000W.
  • Watt = Amp X Volt
  • Watt = Volt² / Ohm
  • Watt = Amp² X Ohm

Then use E=IR to get the needed resistance.

Use that number and a coil calculator to figure out the length and gauge of the Kanthall.
 
What is the E in the equation ?

Trying to back into it using 16 gauge wire because i have some
 
I'm not meaning to talk down to you, but if you do not know Ohm's law (E=IR) you need someone to help you with this build.

Here are some old threads on the calculations and making decisions.

HT Oven wiring schematic:

IIRC, a company called Kruger Pottery offers help with coils. They will col them for you or sell straight Kanthal.
 
Ok So I spent some time with all the math for this. I have just 2 more question


Design output 6000 Watts
Circuit Voltage 240 Volts
Ampacity 25 A

240 V x 25 A = 6000 W
240 V / 25 A = 9.6 Ohms

I am using 14 Guage Kathal A-1
.2123 Ohms Per Ft
Diameter .064"

9.6 Ohms / .2123 Ohms per ft = 45.21 ft of wire per coil
45.21 ft = 542.52 inches

Utilizing a 3/8 rod to wrap the coil with a circumference of 1.18"
542.52 inches of wire / 1.18 per wrap = 459 coil wrap
459 coil wraps x .064 (diameter of wire) = 29.42 inches un- stretched

3 x coil stretch = 3 x 29.42 = 88.26" which matches my layout

Final questions
1. If I wire them in series I would have 19.2 Ohm of Resistance
2. If I wire them in parallel I would have 4.8 Ohms Resistance

Do I have that correct?
which way should I run them , parallel or Series?

Thanks to anyone who can help
 
As for your parallel and series numbers -Yes that is correct.

BUT, you don't want 19.2 or 4.8 Ohms. You want 9.6 ohms. You would use half that length of coils in series or twice in parallel to get 9.6 ohms.

I think your question is about cutting the 88"coil into two sections, and that answer is series. 2X44"coil=9.6 ohms.
 
Last edited:
EDLK said:
Yes, Read through Dan's site, used his calculator to start understanding the coil construction
Thanks for the solid input.
If I go with the first set of parameters at 25 A, any idea what I could expect for the time required to come up to full temp?
I was going one the assumption the increased wattage would allow it to come up faster.
No need to grossly overbuild, just wanting to be sure I get it right on the first pass.
Thanks again Stacy
Click to expand...
Maybe I missed that in your post ..................you need to calculate the load on the wire which must not exceed 3W/cm2 ? That affect the service life of the wire , not just thickness of wire .
Calculating the load on the wire is done in such a way as to obtain the total area of the wire by which we divide our desired power.
To get the area of the wire, the formula is 2∏ x r x length_of the wire
Other rule is that the diameter of the rod on which coil is wound must be at least 5-6 times greater than the thickness of the wire . .. ..
 
Natlek said:
Maybe I missed that in your post ..................you need to calculate the load on the wire which must not exceed 3W/cm2 ? That affect the service life of the wire , not just thickness of wire .
Calculating the load on the wire is done in such a way as to obtain the total area of the wire by which we divide our desired power.
To get the area of the wire, the formula is 2∏ x r x length_of the wire
Other rule is that the diameter of the rod on which coil is wound must be at least 5-6 times greater than the thickness of the wire . .. ..
Click to expand...
 
To create the length of coil to match the size of my oven channels, I need to determine the optimum stretch ration
I have been doing alot of reading on posts and kiln sites. The consensus seem to be 2-3 times the un-stretched length
when I adjust my coils for 19.2 ohms to get them to 9.6 ohms in parallel, I am coming up with 1.75 x un-streched to stretched.
Is that an adequate stetch or should I look at increasing the radius of the coil to shorten unstreched coil and increase the stretch ratio?
 
EDLK said:
To create the length of coil to match the size of my oven channels, I need to determine the optimum stretch ration
I have been doing alot of reading on posts and kiln sites. The consensus seem to be 2-3 times the un-stretched length
when I adjust my coils for 19.2 ohms to get them to 9.6 ohms in parallel, I am coming up with 1.75 x un-streched to stretched.
Is that an adequate stetch or should I look at increasing the radius of the coil to shorten unstreched coil and increase the stretch ratio?
Click to expand...
I don t know from where that come ? I know that distance between spiral must be 2-3 times greater than the thickness of the wire ?
Edit...........which is maybe same thing ?
 
Here is my math on the loading of the wire
Not sure if it matters in the equation, however I converter to metric

14 Guage wire diameter .16 cm
Length of wire. 2717 cm
Circumference 3.14 x .16 = .5026
Surface area = 2717 cm x .5026 = 1365 cm2
Surface loading = 3000 / 1365 = 2.19 w/cm2
(under 3w/cm2)

Back to the coil length option
Assuming the coil has no gaps when unstretched, I think we are saying the same thing,

Is two times the radius minimum? will 1.75 radius be problematic with overheating

I have 89 inches of channel per coil in the oven

If I use 7/16 rod to wrap the coil my OD is
.4375"+ .064 + .064 = .565"
circumference of 7/16 rod is 1.37
Length of wire for 19.2 Ohm 19.0hm / .2123 Ohm/ft = 90 ft wire
90 ft x 12= 1080 inches of wire
1080/ 1.37 = 788 coil wraps
788 x .064 = 50.5 inches

89 inches of channel / 50.5 unstreched coil = 1.72
Stretch length allows 1.72 x the radius of the wire


If I use 1/2 rod to wrap the coil my OD
.5"+ .064 + .064 = .628"
circumference of 1/2 rod is 1.57
Length of wire for 19.2 Ohm 19.0hm / .2123 Ohm/ft = 90 ft wire
90 ft x 12= 1080 inches of wire
1080/ 1.57 = 687 coil wraps
687 x .064 = 43.9 inches

89 inches of channel / 43.9 unstreched coil = 2.02
Stretch length allows 2.02 x the radius of the wire

Based on getting 2 times the radius of the wire in the stretch
the 1/2 ID seems to be the way to go
With the OD of .628 my channel would need to be a little over 5/8
I can't find any information on the amount of space is optimal in the channel for the coil?
Are there any standard on this ?
 
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