Element Design --- Wire Size and Pitch --- Part 3

This has been a very interesting time as I am trying to understand what makes a great element and, of course, get great elements in my kilns. The whole thing has been quite time consuming, but I feel like if I understand how to do this, I can maximize the life of my elements and the usability of my kilns. 

Here are my thoughts how to design elements for Papa Bear, which is a 29 inch diameter L and L kiln with 2 sections. There are 3 elements in each of the sections and 2 elements in the floor, so there are 8 elements in all. Since we want to have 60 amps, which will give us 14,400 watts, we divide the kiln into 3 different sections of 20 amps each. Each of these sections will be wired in parallel to each other. 

But within each section the elements will be wired in series. So the top section will have 3 elements wired in series, the bottom section will have 3 elements wired in series, and the bottom of the kiln will have 2 elements wired in series. That makes 8 elements in all. 

Let's look at the top section. We have 240 volts, we want 20 amps, so we need 12 ohms. (240/20 = 12) Since we have 3 elements in series each element needs to be 4 ohms. 

Now there are a lot of ways to get an element of 4 ohms depending on the diameter and type of wire, but I have decided that I am going to use either 11 gauge or 12 gauge Kanthal A1 wire. 11 gauge is thicker (.091 inches) than 12 gauge (.081 inches) and I would rather use 11 gauge than 12 gauge because it will last longer. 11 gauge wire also has less resistance than 12 gauge wire --- .106 ohms/foot as opposed to .134 ohms/foot. It's easier for the electrons to travel through a thick wire. 

However I have been told that 11 gauge wire is very difficult to work with because of the strength needed to wind the element, and also to install the wound element in the kiln. I don't have experience with this, but I do like the idea of using 11 gauge. 

Let's calculate the length of the wire needed for 4 ohms. 11 gauge is 4/.106 = 37.74 feet of wire and 12 gauge is 4/.134 = 29.85 feet of wire. 

Now we have to wind these wires on a mandrel and there are a couple of things we want to keep in mind, these courtesy of Kanthal. http://www.kanthal.com/C12570A7004E2D46/062CC3B124D69A8EC1256988002A3D76/D1D355F37C940491C12572B9003FD970/$file/1-A-5B-3%20UK%20resistance%20alloys.pdf?OpenElement

First of all we want the diameter of the element itself to be between 5 and 6 times the diameter of the element wire. This is so the heat can get out of the back of the element, and also so the element does not fall over when it is hot -- if the diameter is too big, the element won't be able to stand up when it is soft. 

Secondly, we would like to be able to stretch that element so that we have close to 1.25 element diameters between each pair of adjacent coils. We want to be able to take a piece of wire that the element is made of, insert it in the coils, and be able to wiggle it around a little. This is so the heat from the back of the element can make it out. Kanthal defines the "pitch" of the element to be the distance between the top of one coil and the top of the next coil and they specify that the pitch should be between 2 and 3. Our pitch above was 2.25. We don't want the elements to be stretched too much so we want to keep the pitch below 3 for an ideal element. 

When Papa Bear arrived the elements were of probably 17 or 18 gauge (They measure .040 inches) and the pitch was approaching zero. Just an unstretched element laying in the holders. I was only able to get a few cone 10 firings from them and now they will not even go to bisque. 

Next -- the winding of the elements. 

S Type Thermocouples --- for Alex

If you look very carefully inside the photo of Baby Bear you will see the S Type thermocouple sticking out into the kiln. The S Type thermocouple is much smaller than the K Type thermocouple, but the thermocouple wires, which are Platinum/10% Rhodium and Platinum, are totally enclosed. The wires are protected from the atmosphere of the kiln. 

The thermocouple does not in any way come into contact with the elements. 

I believe they will last indefinitely if you do not break them. They are easy to break because they are small and delicate. I broke one of mine and my friend Terry Fallon fixed it. 

The caustic nature of the crystalline atmosphere does not touch anything but the porcelain sheath. 

I believe Omega is one company that makes them and Cleveland Electric is another. And most kiln manufacturers probably stock them too. 

Element Design --- Some Basic Kiln Parameters for Potters

There are a couple of features that crystalline potters should have with their kilns and I would like to look at these as a little digression from the calculations of the elements. I would say that every potter with an electric kiln should at least consider these features. 

The first has to do with thermocouples. A crystalline potter needs an S Type thermocouple. There are other even more expensive thermocouples, but we are discussing the difference between the K Type thermocouples that normally come on kilns, and the S Type that is an upgrade if you order it when you order the kiln. The S Type thermocouple has several advantages. It will measure temperature accurately through thousands of firings. It does not wear out. However you can not just retrofit your kiln with one of these babies because you need to change a chip in the Bartlett Controller ($10.00), and you need different lead wires. It's cost is around $150.00 compared to a K Type at around $25.00. 

However, Steve Lewicki of L and L Kilns recommends that you change the K Type thermocouple every time you change your elements. He says they are accurate for only a few firings and then they begin to drift. When I got my new L and L kilns, the thermocouples were protected from the atmosphere by a porcelain tube. After about 25 firings I took the thermocouple out of the tube and then tapped the tube on the table and there was a large pile of metal shards that had spalled off the thermocouple. No way that thermocouple was accurate.

So, bottom line is that S Type thermocouples cost way more when you get them, but they do not need replacement and they are accurate throughout their life. They are essential if you need to have any accuracy at all of your kiln's temperature. 

I must say that they are easy to break and you must be careful with them. I take mine out of the kiln when changing elements, something I learned by experience. 

Another thing that Kanthal recommends is using some kind of relay that is more sophisticated than just on and off. The idea is somewhat like that of a teenager driving a car with normal relays. The element is full on and then it is full off so it's like pressing the accelerator of a car full blast and then taking your foot off the gas full blast. This has the tendency to wear out the car, and it also has the tendency to wear out the element. 

My friend Terry Fallon is designing an incredible kiln and it has solid state relays. They cycle on for 200 milliseconds and then they are off for 200 milliseconds. This has the effect of not shocking the element and also of not making the element get much hotter than the kiln. These relays are more sensitive and they need to be mounted on a heat sink so they do not burn up, but Kanthal suggests that they will increase element life.

Element Design --- Resistance (Part 2)

So now we know that Papa Bear needs 67 amps, Mama Bear needs 55 amps, and Baby Bear needs 37.5 amps. From this we can calculate the resistance of the elements that we need to generate this many amps from the formula already mentioned, V = I x R. 

So, for 

Papa Bear,  240 = 67 x R and R = 3.58 Ohms

Mama Bear, 240 = 55 x R and R = 4.36 Ohms

Baby Bear,   240 = 37.5 x R and R = 6.4 Ohms

OK   let's just look at Papa Bear here for a minute. The resistance of 12 gauge Kanthal A1 wire is 0.134 Ohms/Foot so to get 3.58 Ohms we only need 3.58/.134 or 26.72 feet of wire. Hey that's a very elegant solution and it won't cost much either. Just 27 feet of wire and we can fire the kiln. But there is a major problem and we need to understand what it is and what we can do about it. Just because it works mathematically does not mean it will work in real life. 

Let's consider that our 27 feet of wire is a cylinder, where the diameter of cylinder is .081 inches -- that's the diameter of 12 gauge Kanthal A1 -- and the height is 27 feet. So we have a cylinder of diameter 81 thousandths of an inch and a height of 27 feet. The surface area of the cylinder is the circumference of the circle times the height of the cylinder. So we have .081 x 3.14 x 27 x 12 = 82.4 square inches of surface area. Note I had to multiply 27 times 12 so that everything is in inches. 

So we have 82.4 square inches of surface area and we need to generate 16,000 watts of power so that each square inch of element must generate 194.17 watts of power. (16000/82.4)

The online Kanthal manual 

(http://www.kanthal.com/C12570A7004E2D46/062CC3B124D69A8EC1256988002A3D76/D1D355F37C940491C12572B9003FD970/$file/1-A-5B-3%20UK%20resistance%20alloys.pdf?OpenElement)

would like for the watt loading figure of the elements to be around 15 instead of 194. That is, a good long lasting element will only have to generate around 15 watts per square inch of surface area. 

My guess is that our 27 foot element would last only milliseconds with a full 240V passing through it. 

Element Design --- The Three Bears (Part 1)

Here is a photo of my 3 electric kilns. They are all built by L and L and I have found the company wonderful to deal with, and open to learning the special needs of crystalline potters. 

I must confess that the kilns have names and they are Papa Bear, Mama Bear, and Baby Bear. It's kind of embarrassing to admit, but it does make identifying them much easier so the large one is Papa Bear, and so forth. Each kiln is 18 inches high and their diameters are 29, 24, and 18 inches respectively. The cubic footage of each of them is 6.7, 4.8,  and 2.7, also respectively.

So the first question we have to ask is, "How much power do we need to fire these kilns?" There is a wonderful book by Robert Fournier called "Electric Kiln Construction for Potters" and it is full of information like this. Actually he gives 2 different tables, "Power/Volume Relationship in British Commercial Kilns," and " Power/Volume Relationship in American Commercial Kilns." The British table suggests more kW (Kilowatts) than the American table, and since we want our elements to last, I decided to use the British Table. The British table allots each kiln almost half again as much power, and since I want to power my kilns so that the elements do not have to work so hard,  using the British table seems like the correct choice. 

Papa Bear, which is 6.7 cubic feet will require about 16kW, Mama Bear at 4.8 cubic feet will need 13kW, and Baby Bear, at 2.7 cubic feet gets 9kW. 

So now we need to know 2 simple formulas. 

The first is Watts = Volts x Amps or W = V x I where I is the current in Amperes, V is the voltage and W is the power to the kiln. 

The second is V = I x R where R is the resistance in Ohms. 

For my kilns the voltage is 240 so this will remain constant throughout the calculations. 

OK. 

Papa Bear needs 16kW of power at 240 volts so 16,000/240 = 66.67 amps. We may not be able to supply that much because of the wire leading from the breaker to the kiln. But we will be able to improve on the 38 amps that the kiln was originally designed for, and which proved to be inadequate. 

Mama Bear needs  13kW of power at 240 volts so 13,000/240 = 54.16 amps.

Baby Bear needs 9 kW of power at 240 volts so 9000/240 = 37.5 amps.

Tonight I lead the chanting in our Temple and I have to sign off for the moment. 

But below is a photo of the inside of Baby Bear, and the double sets of elements. For some reason the hairpin turns offend my sense of craftsmanship --- I suppose it is the fact that the hairpin was not planned, someone just took the element and stretched it  ---- and I am going to try to eliminate the hairpins. 

Element Design for Electric Kilns

I'm in the process of designing new elements for my 3 electric kilns. 

For potters who do crystalline glazes, normal elements may last from 15 to 40 firings, and this means that we buy lots of elements, and that we spend lots of time changing them. But if an element is designed correctly, it will last for many more firings, and this was proven to me by redesigning the elements in my smallest kiln -- so far they have lasted 136 firings to cone 10-11. 

This redesign was the inspiration of my friend Jesse Hull. Steve Lewicki of L and L kilns actually did the calculations and made the new elements. It was a redesign of the whole kiln, as we now have twice the element wire as before, and it is much thicker -- gauge AWG 12 to be exact. 

The proper information to design elements is available online but it is much more complicated than at first glance. It's not as easy as it looks and you have to figure out basic parameters and then balance variables so that the elements work well. It would seem to be easy but it is not. However I am going to figure it out. 

This information is really essential if you do crystalline glazes but most helpful if you are anyone who wants to maximize the life of your elements. 

Before we get to the process of calculation -- and I admit that I do not have this all crystallized in my mind -- the most important variable is what is called watt loading. This is defined as how many watts per square inch of surface area do the elements have to generate. If they have to generate a lot of watts, they will not last as long. Bottom line. 

I wish it were this easy. Just design the kiln with thick elements and all is well. But it is much more complicated than that. Thicker elements have less resistance per foot, so we now have to have a longer wire to have the same resistance as before, and that longer wire may not fit in our kiln.

I am going to try to figure it all out and post it here. Dull subject but of great value. 

Gas Kiln Firing

Today we unstacked the gas kiln which I thought was going to be a disaster. The firing was fine even though it was too hot in the bottom and I struggled to get it even while keeping the atmosphere steady. The lower half was about a cone too hot. 

If you look very carefully you can see that there is a test tile which is very near to the left uppermost bowl. Too near as it turns out. I was able to remove it and do some work on the bowl, but it will always have a little mark. 

Somehow I expect my pots to be perfect. I'm not perfect and MY body certainly has 64 years of dings and doinks and I accept (mostly) these. I have a chiropractor and a personal trainer to help with balance, flexibility, and strength because I realize that I need to address these issues of imperfection before they become debilitating. Why can't a pot have a little ding and that be OK?

At cone 012 that same bowl was also holding up the cone but I was able to move the cone with a metal rod. 

There were some refired oilspot bowls in the kiln and this was the nicest of them, at least to my eye. The stoneware bowls are part of a dinnerware set for my friends Dean and Betty. 

I also managed to drop one of the peephole hard bricks on the Oxy Probe and it broke the porcelain tubing but did not otherwise hurt the probe. It will have to be fixed before using it again. This was the result of the leather gloves having a hole in the finger and me using a tea towel to remove the brick, the tea towel catching on fire, and me not being aware enough to figure out what is happening. 

I've been dropping stuff in the past week. Hopefully this is not the beginning of something permanent. I have to learn to be more aware. 

It's time to take the burners apart and make sure everything is working correctly. The fans seemed like they could use some oil. I have 21 year old Dedell forced air burners and they need some attention, especially since this is spring, and mud daubers love to build nests in small orifices.