Heat Treatment Oven Mathematics

Heat Treatment Oven Mathematics

In this topic, I will try to put together the calculation of the heating element for a tempering furnace as I have done for my own furnace. I do not take any responsibility if you build a furnace according to these instructions. If you have no knowledge about electrical matters, first consult an electrician. Hopefully, you are all old enough to think for yourselves. People tend to make mistakes, just like I do – I may have calculated slightly incorrectly or mixed something up while writing this, so please correct me. If you have any questions, feel free to ask. I will try to answer as best I can.


1. Furnace size and required power.
2. Wire in general: how much wire is needed and what diameter is reasonable.
3. Surface load of the heating coil.
4. Diameter of the heating coil + stretch factor.
5. Connecting multiple heating coils.
6. And if you have read and understood all this text (calculator).

1. Furnace size and required power

1.1 Power:

We use 230 V, and the maximum allowed current for one socket is 16 A. Let’s calculate the allowed power: P = U x I.
230 V x 16 A = 3680 W.

So, the most powerful furnace that can be used from one socket at 230 V is 3680 W. It is advisable to use about 10% less power to avoid the fuse tripping too easily when the furnace is in operation. This results in about 3312 W, so up to 3300 W is good, safe, and reasonable to use from one socket.

1.2 Furnace size:

What is the maximum internal size of the furnace that can be used with this power? This depends on several factors, including the insulation value of the furnace and the desired heating rate. It should be obvious that you cannot heat a 200-liter drum-sized furnace with a 1000 W element for any heat treatment. If someone happens to be a specialized engineer in this field, they can calculate the size and power requirements exactly, considering insulation value, heat loss, desired max temperature, heating rate, convection, radiation, conduction, mass of objects, and thermal conductivity. Since I cannot do this myself, I rely on information obtained from ceramic furnaces and commercially available furnaces. We generally do not consider the heating rate in our calculations. From the broad internet information, I have found that the heating power for the furnace’s internal surface area is 0.92 W/cm² to 1.3 W/cm². When looking at commercial furnaces, they usually provide internal surface area and power. Calculating according to this, we get results of 0.6 W/cm² to 1 W/cm². Some examples below:

Evenheat furnaces:

• KF18 = 0.668 W/cm²
• KF22.5 = 0.639 W/cm²

Paragon furnaces:

• PMT10 = 0.676 W/cm²
• KM24D = 1.08 W/cm²

I do not know the insulation values of these commercial furnaces, but I think they use insulation bricks and possibly ceramic wool. I have used both. To simplify calculations and ensure safety, it is reasonable to use a value of 1 W/cm² for the internal surface area of the furnace.


The internal chamber size of my self-built furnace is 34 cm x 12 cm x 10.5 cm. The reason for these dimensions is simply that I had such materials available. The total surface area of the chamber = 1826.5 cm², equivalent to 1826 W. As elements age over time, their diameter decreases due to oxidation. As the elements age, their diameter decreases due to oxidation, resistance increases, and furnace power decreases. To compensate for this and speed up heating, I tend to overbuild a bit. It would be reasonable to increase the power obtained from such calculations by about 10%, so 2010 W.

2. Wire in general: how much wire and what diameter

2.1 Wire

What resistance wire to use?

The two most useful sources of information I have found come from the manufacturer itself: the Kanthal Handbook or the Kanthal Furnace Mini Handbook. These are generally freely downloadable from the web and contain numerous valuable data and graphs needed later for element design. In short, I recommend using Kanthal A1 resistance wire, as it is used in ceramic furnaces and is widely available in various dimensions, making a long-lasting heating element. I have previously used Nicrothal80, whose main advantage is that it is much cheaper, but compared to Kanthal A1 wire, its lifespan at desired temperatures is 300% shorter than that of Kanthal A1 wire. This should help justify paying the higher price for Kanthal A1. You can use any other wire classes mentioned, such as APM, D, AF, Ncr80/20, whatever you can easily and cheaply obtain. Some are better, like APM, some worse, like Nichrome. Your heating elements won’t last as long if made from cheaper wires, but they will work well if properly designed. For calculations, you need the required wire specifications, which are generally provided in the previously mentioned books.

2.2 Wire length

How much wire of a specific diameter is needed to obtain a specific power element? You can calculate the required wire length using two methods: a more general but slightly simpler, mostly sufficient method, and a more accurate method that provides precise results.

Method A

Now that you have figured out how powerful your furnace needs to be, the next step is to connect it to the wire length. Each resistance wire has a specified resistance based on its diameter and wire specifications. Resistance is usually given in ohms per meter, or sometimes ohms per foot, but we will use the metric system for calculations. You can convert all calculations to inches/feet and use different voltages, it doesn’t change anything.

What we now know is:

• We need 2010 W
• Supplied voltage is 230 V

What we want to know is how many ohms the heating element’s resistance should be.
If you still remember some formulas from school physics lessons, such as R = U² / P, which gives, when you put in the corresponding numbers, 230² / 2010 = 26.32 ohms. If not, it might be wise to seek the help of someone more knowledgeable in electricity.
Assuming we have 1 mm diameter Kanthal A1 wire. This is not the right diameter because the lifespan of such a heating element may be very short. Why so will be clear in the next section. If you look up 1 mm Kanthal A1 wire data in the Kanthal Handbook, the resistance is given as 1.85 ohms per meter. Generally, the wire seller also provides the resistance per meter of the wire. And the calculation is then how many meters are needed to get a resistance of 26.32 ohms.
26.32 / 1.85 = 14.23 meters, which means that 14.23 meters of 1 mm Kanthal A1 wire is needed to make a 2010 W heating element.

Method B

The second method takes into account that the wire gradually changes its resistance as a specific temperature is reached. This provides a more accurate calculation. Having looked at the Kanthal A1 wire data in the Kanthal Mini Handbook, you can find a table with resistances and correction values at different temperatures. Kanthal A1 resistance is 1.45 ohms/mm². The correction factor is 1.04, as the maximum desired temperature would be 1100°C, so the total resistance is 1.45 x 1.04 = 1.508 ohms/mm².

Additionally, we need data:

• Resistance (see 2.1) = 26.32 ohms
• Cross-sectional area of the wire diameter. With 1 mm diameter wire, we get = 0.785 mm²

Wire length = (Ohms x wire cross-sectional area) / Resistance
(26.32 x 0.785) / 1.508 = ? m

As seen, there is a difference between the two calculation methods of ? cm. This equals a difference of ? watts. Since this power selection is variable, it is not a particular problem which to choose, but it is more reasonable to use the second option.

Surface Load

One of the most important factors affecting the expected lifespan of heating elements is the surface load. This term simply refers to how the total power is distributed across the surface of the element. If you apply too much power to a very thin wire, it will heat up quickly, oxidize, and eventually melt due to severe overheating. Consequently, the heating element will fail, necessitating a replacement in the oven. To prevent this, we must choose an appropriate surface load. Surface load is measured in watts per unit area of the wire surface.

It’s essential to note that almost all graphs and data in the Kanthal handbook are based on a wire thickness of 3 mm. Usually, thinner wires are used in knife-hardening ovens, but it’s generally assumed that thinner wires will last as long as 3 mm wires. Ideally, we would use 3 mm wire everywhere, but this is often not feasible. The Kanthal oven manual states on page 7 that at 1100 °C, the surface load of spiral elements should not exceed 3 W/cm², which is our typical operating condition. Freely radiating heating elements (non-zigzag without sharp bends) can be used with a higher surface load (5 W/cm²), but their disadvantage is that they take up more space and are much more challenging to install.

Note the small text at the bottom of the graph; it is crucial because it tells us that for on/off control (which a PID controller does), we should use about 20% less than the given value: spiral elements = 2.4 W/cm². If using thyristor control, you can opt for 3 W/cm². If you remember the underlined section, you should use an even smaller value if using wire less than 3 mm thick. This recommendation comes from the manufacturer, but it’s always a matter of trial and error. Always consider options like Evenheat or Paragon.

So, you can imagine why the previously described element may not last long. Let’s calculate the surface load, and everything should become clear.

First, we need the surface area of the used wire. A wire is nothing but a very long, very thin cylinder, so for 14.23 meters with a thickness of 1 mm, we get: 2 x pi x radius x length = 447.05 cm² surface area. Then, we divide the power by the surface area. For our 2010W 1mm element, we get:
2010 W / 447.05 cm² = 4.5 W/cm²

So far, our planned element is almost one and a half times the recommended surface load. Assume its lifespan might be somewhat shorter. All our calculations have been correct, but the heating element isn’t as durable because we chose too small a wire diameter. How do we determine the appropriate wire diameter? It’s simple to calculate the best option or use the calculator at the end of this discussion.

4. Heating Coil Diameter and Stretch Factor

Finally, we come to the actual element and its diameter. This is also a significant factor, but not as crucial as surface load. The core diameter of the spiral element must be at least 5–7 times the wire thickness. The biggest challenge is fitting the element into the oven. The element must also be minimally stretched between individual coils. The distance between individual coils should be 2-3 times the wire diameter. This prevents:

• Overheating,
• Short circuits between individual coils, and ensures longer usage.

Be particularly careful that the coils do not touch the inner curve, especially in the corners or turnbacks of the spiral. To prevent this, stretch the element a bit more at the turns. Therefore, it’s recommended not to choose too small a stretch factor, otherwise, the spiral turns may get too close over time and overheat in some sections.

Next, calculate how long the heating element is in its spiral and stretched state. You will find that the thicker the wire, the better the surface load, but this means a longer and thicker heating element. Eventually, your calculations will reach a point where the element size is such that you don’t have enough space in the oven to fit the element. Usually, elements are only placed on the sides of the chamber, so ensure enough space before making it. It would be nice if we could use thick wire in small ovens, but due to space constraints, this is generally not possible.

5. Connecting Multiple Elements

Using only one heating element in an oven is relatively tricky because the oven needs to be heated relatively evenly. Therefore, two heating elements are generally placed on the sides of the oven. They can then be connected in series or parallel. Each method has its advantages and disadvantages. Although the heating elements do not have to be the same length, to avoid confusion, the following example connects two equal heating elements.

To calculate, we need three values:

• Voltage – typically 230V in our case
• Resistance – we use the previously found resistance of 26.32 ohms
• Element surface – we use the previously calculated result of 447.05 cm²

5.1 Series Connection:

What happens if we take the previously calculated 1 mm wire element and connect two of them in series? Using the previously mentioned values, the changing values will be current and power. When resistors are connected in series, the rules are:
R_x = R_1 + R_2 + R_3…
R_x = 26.32 + 26.32 = 52.64 \Omega
Calculate the current:
I = \frac{U}{R_x} = 4.37 A
Calculate the power:
P = U \times I = 1005 W
Since we doubled the wire length, we doubled the surface area:
The surface load is now lower:
\frac{1005 W}{894.1 cm²} = 1.12 W/cm² instead of 4.5 W/cm².

By connecting two same elements in series, we have significantly improved the surface load but reduced the power by half.

5.2 Parallel Connection:

When resistors are connected in parallel, the rule is:
\frac{1}{R_x} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}…
For two parallel elements:
\frac{1}{R_x} = \frac{1}{26.32} + \frac{1}{26.32}
R_x = 13.16 \Omega
Calculate the current:
I = \frac{U}{R_x} = 17.48 A
Calculate the power:
P = U \times I = 4020 W
Surface load:
\frac{4020 W}{894.1 cm²} = 4.5 W/cm²

So, connecting elements in parallel maintains the surface load but increases power. Using three parallel elements would increase power further while maintaining the same surface load.

6. If You’ve Read and Understood This Text (Calculator)

This calculator will be gradually updated over time.