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PotSherder

Kiln lid "caulk"?

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Hello everyone!  New to the forum and not at all sure this is the right group to post to, but here goes:

I'm building a one-off electric kiln, the goal of which is to get about 1 cubic foot of interior, reach at least cone 6,  but still run on household 120VAC.  

I took this on not only because of my personal need, but also as an engineering challenge since it seems to be in the "can't be done" category.

The idea was to use *lots* of insulation, since in theory with enough insulation you could reach any cone if you wanted to wait long enough, even with a wimpy heat source, as long as the element temperature is above the cone temperature (say, 1200 C for cone 6).

So far, I can only get to about 1150 C (about cone 3).    My calculations could be off, of course, but they've been pretty good on prior projects.   One possible problem is that the calculations are based on zero leakage, and the most likely culprit there is the lid seal.  I know this isn't as good as I'd like it to be, but fixing it is going to take a lot of work due to the design.  I'd like to be certain that lid leakage really *is* the problem before tackling that,  so my idea is to run a test where the lid is "caulked" to get a perfect seal.  Due to the nature of the extra lid/body insulation, I can't apply this caulk around the edge after the lid is closed; I will need to run a bead around the top edge of the kiln body, and have it squish into place when the lid comes down onto it.

So the question is what I can use that will be squishy enough to easily extrude and fill irregularities a the lid/body interface, yet not stick the lid and body together, so I can still open the lid after the test.  Since it will take a couple of minutes to run the bead around the edge, the caulk has to resist drying out from having its moisture sucked out due to contact with the surface (IFB with kiln wash over it), so it can still squish properly when the lid comes down.  However, my sense it that the goopier I make the caulk, to keep it squishy long enough, the worse it will stick after being fired.

I'm thinking of something that will pretty much turn to crumbly 'sand' after firing, like maybe flint or grog or alumina or ?  mixed up with an organic binder that will make it squishy to apply, but burn out later.  Library paste?  Flour and water?  (I will of course run tests on IFB scraps before risking the kiln getting stuck shut.)

Many thanks for any help, including where to ask if this isn't the right place!

 

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Yeah, I am not fully sold on this. Ordinary test kilns are about 0.6 cu ft with standard leaky construction and easily reach Cone six in five hours when following a schedule regulating their rate of climb. This seems easily achievable actually.

At 2000  F degrees there is very little atmosphere inside the kiln, most of the heating is by radiation. You have confirmed that you cannot achieve this with an actual test or is this only something calculated?

BTW in industry as we look for lower infiltration and better seals we simply keyway one surface into the other so it can expand and move yet stops radiation leaks cold. After that there are high temperature seals. Caulk is a mess.

The kiln below easily makes Cone six (actually it will go to cone 10) on roughly 2000 watts and is pretty leaky, nothing super special really.

 

FC11EA90-47BE-41CA-9F4C-46B28F9AAFF4.jpeg

Edited by Bill Kielb

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I have an Olympic doll kiln that runs off 120 VAC  (standard garage circuit) that I use for "test of concept".  I have fired to cone 10  cold to cold in less than 6 hours.   The difference between cone 3 and cone 10 is less than one hour.  There is more mass in the kiln  and furniture than in the ware.    Since I am firing to resolve issues of  concept  concerns rather than final effects, the differences between this electric fast fired kiln and my routine firing in a large gas kiln are not a problem.  

To fill the gaps between the lid and the lid  support, I would use either a fiber blanket or the clay-sand-aluminia hydrate-sawdust wadding often used to seal the door of a wood kiln.  

LT
 

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1 hour ago, Magnolia Mud Research said:

I have an Olympic doll kiln that runs off 120 VAC  (standard garage circuit) that I use for "test of concept".  I have fired to cone 10  cold to cold in less than 6 hours.   The difference between cone 3 and cone 10 is less than one hour.  There is more mass in the kiln  and furniture than in the ware.    Since I am firing to resolve issues of  concept  concerns rather than final effects, the differences between this electric fast fired kiln and my routine firing in a large gas kiln are not a problem.  

To fill the gaps between the lid and the lid  support, I would use either a fiber blanket or the clay-sand-aluminia hydrate-sawdust wadding often used to seal the door of a wood kiln.  

LT
 

2000 watts seems plenty for a small kiln with nothing special done to it, there must be an error here. I think he likely needs to work on the conduction and radiation losses first

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Thanks to all that have responded.  To be clear, this caulk idea is for a one-shot test to see if it gives a higher peak temperature.  (Yes, the 1150 C was measured.  My calculations showed  around 1300 C.)  I am familiar with the 0.6 cu.ft. test kilns, but notice that's about the biggest to be found.  That indicates to me that maybe it's not as easy to go bigger as I (and some of you) assumed.  My kiln now has 4 layers of 1" 8# fiber packed into 3.5" between the IFB and the outer sheet-metal shell.  The IFB is edgewise so only 2.5" thick.  Heating elements are 4 silicon carbide rods custom ordered for this job.    What I'm trying to determine by my quick-and-dirty caulk idea is just where my losses are actually coming from.  If it's not the lid, then I need to enlarge the shell to handle more fiber... not so easy!   

I considered trying to use fiber for this test, but although it does have magical insulation properties, I'm not convinced I could get a good seal... which is a slightly different matter. Not to mention that it would require the "closed" position of the lid to be increased to allow for the fiber thickness, a cumbersome adjustment with my simple hinge system that I am hoping to avoid with caulk that just squishes out to fit.  

The sort of wadding used to seal wood kilns was my inspiration for the caulk idea, but that (as far as I know) is applied to the outside of the sealing joint, where it is easy to remove later.  Dunno how it would be when compressed between two surfaces and fired into place...  considering the surface area involved, it wouldn't take very much tensile strength to bond the lid to the body so well that I couldn't lift it without damage to something.

I guess I'll have to run some tests with various compositions of refractory materials plus organic burn-out binders.  I'll report back here.  (Might take a while.)

Thanks again!

 

 

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2 hours ago, PotSherder said:

Thanks to all that have responded.  To be clear, this caulk idea is for a one-shot test to see if it gives a higher peak temperature.  (Yes, the 1150 C was measured.  My calculations showed  around 1300 C.)  I am familiar with the 0.6 cu.ft. test kilns, but notice that's about the biggest to be found.  That indicates to me that maybe it's not as easy to go bigger as I (and some of you) assumed.  My kiln now has 4 layers of 1" 8# fiber packed into 3.5" between the IFB and the outer sheet-metal shell.  The IFB is edgewise so only 2.5" thick.  

I guess I'll have to run some tests with various compositions of refractory materials plus organic burn-out binders.  I'll report back here.  (Might take a while.)

Thanks again!

 

 

This is pretty interesting in that losses can be measured somewhat easily  especially with an infrared camera or other means so that would be something I would check right away just to see the where and how much. That aside what is more interesting is it appears that you  have a low mass empty kiln and cannot make temperature, yet intuitively there are similar enough kilns that appear to make temperature loaded.

So the new wrinkle, silicon carbide heating elements? Hmm, I wonder now the difference between regular old high surface area heating element (Kanthal and Nichrome) and relatively low surface area  and confined nature of silicon  carbide rods? Intriguing for sure. I have never designed with silicon carbide though.

Just some food for thought below. The regular old leaky 0.88 cu ft kiln below is 120v and gets to cone 4 (loaded). This is not super insulated by any means.  

Radiation is the primary source of heat in a kiln as convection diminishes significantly at the upper temperatures, so we are left with radiation and conduction. Are the silicon carbide rods truly representative of common ceramics kilns in use today? Just asking, interesting!

It will be fun to see what you determine.

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You shouldn't need to do any sort of seal on the lid if everything else is done correctly. With that much insulation, the issue has to be the elements. If you must seal the lid, just lay a layer of fiber blanket around the top of the kiln and set the lid down on it. You won't get any better than that.

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2 minutes ago, Bill Kielb said:

More should be better actually ........ I think 

I guess it depends on how they work?  I was thinking if they're wired in series less would be more, but if they're wired in parallel more would be more. 

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Almost as soon as I had sent my last post, I had a 'duh' head-smack moment.  As noted in my original post, "in theory, with enough insulation you could reach any cone if you wanted to wait long enough, even with a wimpy heat source, as long as the element temperature is above the cone temperature".  It (finally) dawned on me that using 4 elements (for more-even heating) meant that the 2000 watts was being emitted from twice the surface area of my old 2-element designs.  That *has* to mean a lower element temperature. (Or just consider that each element is getting only half the watts.)  I'll have to crack open my dusty old undergrad "Heat Transfer" book and root around for the proper equation, or do some Googling, to get a computed temperature value.  But it's pretty simple to just move a couple of cables to omit two elements.  I'm betting that will solve my problem without any lid caulk nonsense.

Just FYI, the elements are driven by a massive old transformer with multiple taps, along with a control panel with an ammeter and big Coarse and Fine contactor knobs, all stripped from an ancient Burrell industrial test furnace circa 1940s vintage.   For those not familiar with silicon carbide elements, they don't like the full-on, full-off of the standard "infinity" controls typically used with wire elements.  You have to bring the power up more smoothly... hence the manual control knobs.  I watch the thermocouple readout and adjust as needed every hour... a colossal pain.  When I get everything working, the plan is to buy or build an electronic controller (essentially a giant lamp-dimmer) that can be automated.

So everyone who smelled something fishy about my whole project, your nose was right on!

Again, many thanks to all who responded.

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52 minutes ago, PotSherder said:

 For those not familiar with silicon carbide elements, they don't like the full-on, full-off of the standard "infinity" controls typically used with wire elements.  You have to bring the power up more smoothly... hence the manual control knobs.  I watch the thermocouple readout and adjust as needed every hour... a colossal pain.  When I get everything working, the plan is to buy or build an electronic controller (essentially a giant lamp-dimmer) that can be automated.

 

Sounds like a typical multi tap auto transformer, popular then and still popular now actually.  Just curious when you say smoothly I got the impression that this was zero to full power in seconds or minutes. Are you adjusting the input voltage (wattage)  on the hour as you fire?  

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19 hours ago, Bill Kielb said:

Sounds like a typical multi tap auto transformer, popular then and still popular now actually.  Just curious when you say smoothly I got the impression that this was zero to full power in seconds or minutes. Are you adjusting the input voltage (wattage)  on the hour as you fire?  

Yes, to control the rate of rise.  You are correct that the elements could go to full power more quickly without harm to the elements... it's the pots that I'd be concerned about in that case!  So I'm manually doing the firing curve for now.  What silicon carbide doesn't like is full power until a target temp, full off until below, full on, etc.  Electronic  controllers for silicon carbide are like lamp dimmers, in that they control the average power by only switching on for a fraction of the AC cycle.  My understanding is that for maximum element life it's even better to chop the cycle even finer, so that instead of getting small bursts once or twice per cycle you get multiple narrower pulses.  I probably won't go to that much trouble, since I expect the elements to outlast me anyway!

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So I misunderstood, you are currently testing this  fully loaded  (which means wares, shelves and furniture) with four silicon carbide rods and modulating the power by hand every hour or so by gradually turning up the voltage.  What  has been your peak energy input thusfar (Wattage) and what is your terminating rate of rise before you call it quits?

Might save you some money

As far as dimmers go they are zero crossing devices limiting their switching speed. In practice you  generally can only switch them on/off once every half cycle; however this has to be fast enough because right now they are naturally being operated  this way since they are AC devices. Maybe that can save you something exotic and just use a standard pid enabled controller and some solid state relays. It does not appear you will need to build your own sine wave to do this if their longevity seems adequate at turning off and on 120 times per second.

Food for thought

 As the kiln heats up the atmosphere or remaining air in the kiln becomes so thin it contains very little energy transfer ability. So under fully loaded conditions the rods need  to provide as much available line of sight surface area to your wares as possible since the bulk of the heat will be direct radiation. I don’t know your imbed philosophy with respect to the rods and your wares positionally but maximizing this unobstructed view is important. You may want to reevaluate your choice of elements as silicon carbide rods will likely become less and less ideal as the volume of the kiln increases.

I am curious what wattage and rate of rise you call it quits at though?

Edited by Bill Kielb

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6 hours ago, Bill Kielb said:

So I misunderstood, you are currently testing this  fully loaded  (which means wares, shelves and furniture) with four silicon carbide rods and modulating the power by hand every hour or so by gradually turning up the voltage.  What  has been your peak energy input thusfar (Wattage) and what is your terminating rate of rise before you call it quits?

Might save you some money

As far as dimmers go they are zero crossing devices limiting their switching speed. In practice you  generally can only switch them on/off once every half cycle; however this has to be fast enough because right now they are naturally being operated  this way since they are AC devices. Maybe that can save you something exotic and just use a standard pid enabled controller and some solid state relays. It does not appear you will need to build your own sine wave to do this if their longevity seems adequate at turning off and on 120 times per second.

Food for thought

 As the kiln heats up the atmosphere or remaining air in the kiln becomes so thin it contains very little energy transfer ability. So under fully loaded conditions the rods need  to provide as much available line of sight surface area to your wares as possible since the bulk of the heat will be direct radiation. I don’t know your imbed philosophy with respect to the rods and your wares positionally but maximizing this unobstructed view is important. You may want to reevaluate your choice of elements as silicon carbide rods will likely become less and less ideal as the volume of the kiln increases.

I am curious what wattage and rate of rise you call it quits at though?

No, I don't test with a full load, just a couple of test bars sitting on the floor.   I don't have the exact peak watts at hand, but it's around 2500 or so.  I called off the test at 1152 C, which was a rise of only 13 C from the prior hour (1139).  At turn-off the kiln had been at max for 4 hours , starting at 1039 C.   Amps were steady at 24.5 through the elements (all in series).  I don't have the tap voltage at hand, but it's somewhere just over 100 V.

 I'm an old hand at dimmers (electrical engineer since 1971), and actually have a homebuilt job on the output of the transformer.  I normally keep it all the way on to prevent RF noise.  I haven't played around with PID controllers, though I'm familiar with the theory.  If they have some sort of PWM output that could drive an SSR, that would be perfect.  But first I have to get the kiln working right!

The 4 rods run fore and aft, near the corners of the front face (not too close to the walls, as they don't like that!), with the intention of maximizing unobstructed view.  When I cut back to only two elements for my test, it will be the 2 at the bottom since that's easiest to re-wire.   If putting the same peak watts through only 2 elements allows me to go above 1200 C, then I can consider changing their placement at my leisure.  I chose silicon carbide not only because I was already familiar and had the transformer and stuff, but also because they are simpler to build into a rectangular kiln that uses ordinary rectangular IFB.   I was not interested in trying to carve wire element channels into brick faces.    Might have been simpler to buy normal electric kiln bricks (or an old kiln and rewire it), and build my own metal jacket big enough to hold the extra insulation.  But that would mean a lot of fancy sheet metal bending, compared to rectangular.

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Yes most controllers today can either be configured as PWM or  4-20 ma current  or ordered as such So building your own for reasonable dollars is very doable today. Simple pid control  with most having auto tune for the pid and usually contain enough memory for multi segment programming.

The one  below, is $46. 00, I believe. 

2500 watts at 100 volts wired four in series. Interesting, these things are on the order of 1 ohm each. Very interesting! Not sure I am sold on the silicon carbide to be truthful. Hope it eventually works out for you.

431C1222-02FE-4614-880D-86C769F5D8CD.png

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After a lot more thinking (always dangerous!) I no longer believe the number of elements has any bearing on the ultimate kiln temperature, except regarding temperature distribution.  It's all down to watts, as others suggested.  My claim that the max temperature couldn't exceed the element temperature was true, but spreading the watts over more elements shouldn't make any particular difference.  The reason is that the *total* power is what is heating the chamber, and the temperature of the chamber will keep increasing until the losses balance the power input.  The added elements may start off cooler at the early stages when convection is important, but as things heat up and shift toward radiant, the element can't get rid of excess heat until it is hotter than the walls... heat only flows from hotter to colder.  So it will keep heating up until its temperature balances the wall losses, no matter how low its individual wattage, since the total watts are heating the walls.  The formula I use is

Temperature = (Watts * wall thickness) / (wall area * thermal conductance)

I use meters and meters^2 for thickness and area, and W/(m * K) for conductance, and since we are talking about temperature change relative to ambient,  it's OK to use Celsius instead of Kelvin since the degrees are the same size.  The trick is dealing with the fact that some of the thickness is IFB and some is fiber, with different conductances.  I've used approximations instead of anything fancier, which may be one factor in my calculation coming up with 1300 C but actually getting 1150 from the kiln, though in the past it has been pretty close.

So it looks like I'm back where I started, with fixing lid leaks versus extra insulation.  Of course we *know* it has to be insulation, since that's the most work! <g>

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20 minutes ago, PotSherder said:

After a lot more thinking (always dangerous!) I no longer believe the number of elements has any bearing on the ultimate kiln temperature, except regarding temperature distribution.  It's all down to watts, as others suggested.  My claim that the max temperature couldn't exceed the element temperature was true, but spreading the watts over more elements shouldn't make any particular difference.  The reason is that the *total* power is what is heating the chamber, and the temperature of the chamber will keep increasing until the losses balance the power input.  The added elements may start off cooler at the early stages when convection is important, but as things heat up and shift toward radiant, the element can't get rid of excess heat until it is hotter than the walls... heat only flows from hotter to colder.  So it will keep heating up until its temperature balances the wall losses, no matter how low its individual wattage, since the total watts are heating the walls.  The formula I use is

Temperature = (Watts * wall thickness) / (wall area * thermal conductance)

I use meters and meters^2 for thickness and area, and W/(m * K) for conductance, and since we are talking about temperature change relative to ambient,  it's OK to use Celsius instead of Kelvin since the degrees are the same size.  The trick is dealing with the fact that some of the thickness is IFB and some is fiber, with different conductances.  I've used approximations instead of anything fancier, which may be one factor in my calculation coming up with 1300 C but actually getting 1150 from the kiln, though in the past it has been pretty close.

So it looks like I'm back where I started, with fixing lid leaks versus extra insulation.  Of course we *know* it has to be insulation, since that's the most work! <g>

Well it depends.  Since you're controlling input voltage, and your elements are wired in series, each is receiving a fraction of total available power, right?

So if you remove half of the elements, they will receive twice as much power at the same input, creating more heat.  The problem comes when you have reached the maximum input (120v @20 amps?) And it's cut in 4.  So yes there's more points of heating, it will be more evenly heated, but with each element putting out a quarter of the heat.  Now if you turn that down to two elements, you stop running into that false temperature ceiling because those two elements are able to run at a higher wattage. But the heating won't be as even.

At least that's the way I see your problem.  You're thinking only of total wattage and not wattage per element.

I could be entirely wrong because I'm not an electrician or engineer, and really have almost zero experience in either, but just from a pure math problem perspective I see an oversight

Edited by liambesaw

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What I was trying to say is that since only total watts matter for peak temperature, not watts per element, there is no reason to even try the 2-element test, and I can stick with the current 4-element design for even heating.  (And I have to look elsewhere to get the peak heat I need, either plugging the lid leak or adding more insulation.)

In the now-abandoned 2-element approach, I would have of course controlled the element voltage to have the same total power input.  If I hadn't, halving the load resistance at the same voltage would have doubled the total power (for a few milliseconds until the breaker tripped).  Power is voltage squared over resistance, so I'd have had to cut the voltage to about 70.7%.   Each element is about 1 ohm and the original voltage was about 100 V, yielding power of 10000 / 4 = 2500 watts.  With 70.7 V and 2 ohms it's about 5000 / 2, also 2500 watts..

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On 9/14/2019 at 8:17 AM, PotSherder said:

What I was trying to say is that since only total watts matter for peak temperature, not watts per element, there is no reason to even try the 2-element test, and I can stick with the current 4-element design for even heating.  (And I have to look elsewhere to get the peak heat I need, either plugging the lid leak or adding more insulation.)

In the now-abandoned 2-element approach, I would have of course controlled the element voltage to have the same total power input.  If I hadn't, halving the load resistance at the same voltage would have doubled the total power (for a few milliseconds until the breaker tripped).  Power is voltage squared over resistance, so I'd have had to cut the voltage to about 70.7%.   Each element is about 1 ohm and the original voltage was about 100 V, yielding power of 10000 / 4 = 2500 watts.  With 70.7 V and 2 ohms it's about 5000 / 2, also 2500 watts..

So now curious because that wattage seems like it should work, where are your losses? 

Any chance you have some infrared pics or temp measurements?

what does your measured R or U value end up to be top, side and bottom? I ask because L&L used to publish a table about their kilns and I always took it for granted that it likely was correct for simple design  and waste heat purposes. They all fall into the 3-4 watts per square inch but none are super insulated. A 1 cu ft kiln should be on the order of 750 sq. Inches (round) so 2100- 2800 watts seems doable.

 

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