Diy heat treat oven help

[JTknives]

I know in the home building business you use r19 fir 2x6 walls and you put r10 or r13 2x4 walls. If you stuff r19 into a 2x4 wall thinking you gaining more insulation your wrong. You will actually be less then the r12 that’s not compressed. Air is a very good insulator but not when it’s allowed to move. The insulation traps it and prevents convention.

I would think ceramic wool would be the same. The fibers them selves have a higher thermal conductivity then air so the closer the fibers are together the faster heat can jump from fiber to fiber. Yes the wool blankets work real good if your wanting to add more insulation to something allready. At the mill it’s used on the out side shell of each combustion chamber in the boiler. The inside is fire brick on a steel structure and that steel has the wool on the outside. It’s weld in place with the spot welded spikes. It works good for this type of application because there is not much room and air flows through there to the under air grates.

I just was hoping someone had actual numbers on thermal conductivity of compressed wool.

 

[Cushing H.]

I just was hoping someone had actual numbers on thermal conductivity of compressed wool.

JT - I think t he best you are going to come with is the reference from Morgan that Natlek gave ( https://www.morganthermalceramics.c...al-ceramics-product-data-book-e-version_2.pdf ) . They report thermal conductivity as a function of density,from 64 to 128 kg/m^3. at 600C, they say it goes from 0.18 to 0.14 W/m/K (not an incredibly huge difference, and decreasing with compression.... which I do not understrand, but have to accept their data). In either case, the same document gives the thermal conductivity of firebrick as 0.18 W/m/K .... (which is basically the same as they give for fiber). My takeaway from that is that you would not need to worry a huge amount about compressing the fiber .... and that it would be pretty darned equivalent to using thicker bricks.....

 

[Cushing H.]

Of course, the question still remains: Does any of this really make that big of a difference for a 2.5" inch to 4.5" thick heat treating oven? How about after it's been soaking at around 2000F for 6 hours? What if the bricks are mortared vs not?

Drew - let me try to answer that for you. Do a thought experiment: a fact is that as you make insulation thicker, its insulating properties increase (typically linearly). The thought experiment is this: assume the insulating firebrick is something like 100 miles thick (yes I mean that) - so there is essentially no heat loss at all from the oven. if the chamber is perfectly sealed, you could put a 15W nightlight on the inside, and (given enough time) it would get pretty hot in there. The caveat as that it would take a LONG time for that little light to heat up all the material of the bricks immdiately around the chamber.

So .... yes, in terms of achieving hotter temperatures, going from 2.5 to 4.5 inches thick would make a difference to the inside temperature (like maybe (dont hold me literally to that) coming close to doubling the temperature above ambient you can reach) .... but at the cost of taking more time to get to that temperature. The answer to your question about what about after things have been soaking for 6 hours is a little more confusing. After things settle down and the oven is fully "soaked" - the temperature through the brick from inside to outside decreases, roughly speaking, in a linear fashion. But ... The thinner the brick, the hotter the outside temperature. Also, the hotter the outside temperature, the more heat is lost from the oven to the room around it. So .... if you have thin insulation, the outside gets hot .... and the heating element inside has a harder time pumping enough heat into the chamber to make up for the heat lost from the outside surface of the oven. With thin insulation, that little nigh light would not stand a chance, but increasing the wattage of the element, say by going from 120V to 220V will likely be enough to make up for the heat lost from the outside surface - even though if you dont change the insulation thickness the outside would still get hot and continue to loose that heat to the outside room.

I dont know .... maybe that is too confusing to follow. The basic set of trade offs then is that more insulation lets you hit higher temperatures, but it will take you longer to hit that higher temperature (and longer for the outside of the oven to come up to its highest temperature). You can compensate for that by using a higher wattage heating element, which will allow you to both heat faster .... and to reach a higher final temperature. The trade off there is that if you go to higher temperatures in the chamber without increasing the insulation, the outside of your oven will get hotter (maybe much hotter).

Re. the mortar - there are those that maintain that using mortar restrains the expansion of the bricks so that they will break sooner. There are many who build ovens with no mortar for exactly that reason. Just more food for thought related to higher wattage heating elements: earlier in the summer when I visited my local supplier for firebricks - they cautioned me in very, very strong terms that heating an oven too fast will greatly increase the chance of cracking the bricks (they also recommended not using mortar). Especially when the bricks are new, residual moisture and other chemicals need to be burned off slowly. IIRC, they recommended an initial heating rate of 100 degrees F per hour (not a typo) during the first several heating cycles. And they still recommended a heating rate that is pretty slow during all later cycles. So ... another downside of a higher wattage heating element might be earlier cracking and breakage of the insulating bricks.

Anyway - not sure if all that helped give you some insight .... but hopefully it gave some....

 

[Natlek]

Looks like they're similar enough, depending on grade? Either way, some of the ceramic wools are a little bit closer to firebrick than I initially thought, though the brick still looks to have a slightly lower conductivity.

Of course, the question still remains: Does any of this really make that big of a difference for a 2.5" inch to 4.5" thick heat treating oven? How about after it's been soaking at around 2000F for 6 hours? What if the bricks are mortared vs not?

Mortared of course , much less energy loss .And sealed door of course !

 

[Natlek]

Drew - let me try to answer that for you. Do a thought experiment: a fact is that as you make insulation thicker, its insulating properties increase (typically linearly). The thought experiment is this: assume the insulating firebrick is something like 100 miles thick (yes I mean that) - so there is essentially no heat loss at all from the oven. if the chamber is perfectly sealed, you could put a 15W nightlight on the inside, and (given enough time) it would get pretty hot in there. The caveat as that it would take a LONG time for that little light to heat up all the material of the bricks immdiately around the chamber.

So .... yes, in terms of achieving hotter temperatures, going from 2.5 to 4.5 inches thick would make a difference to the inside temperature (like maybe (dont hold me literally to that) coming close to doubling the temperature above ambient you can reach) .... but at the cost of taking more time to get to that temperature. The answer to your question about what about after things have been soaking for 6 hours is a little more confusing. After things settle down and the oven is fully "soaked" - the temperature through the brick from inside to outside decreases, roughly speaking, in a linear fashion. But ... The thinner the brick, the hotter the outside temperature. Also, the hotter the outside temperature, the more heat is lost from the oven to the room around it. So .... if you have thin insulation, the outside gets hot .... and the heating element inside has a harder time pumping enough heat into the chamber to make up for the heat lost from the outside surface of the oven. With thin insulation, that little nigh light would not stand a chance, but increasing the wattage of the element, say by going from 120V to 220V will likely be enough to make up for the heat lost from the outside surface - even though if you dont change the insulation thickness the outside would still get hot and continue to loose that heat to the outside room.

I dont know .... maybe that is too confusing to follow. The basic set of trade offs then is that more insulation lets you hit higher temperatures, but it will take you longer to hit that higher temperature (and longer for the outside of the oven to come up to its highest temperature). You can compensate for that by using a higher wattage heating element, which will allow you to both heat faster .... and to reach a higher final temperature. The trade off there is that if you go to higher temperatures in the chamber without increasing the insulation, the outside of your oven will get hotter (maybe much hotter).

Re. the mortar - there are those that maintain that using mortar restrains the expansion of the bricks so that they will break sooner. There are many who build ovens with no mortar for exactly that reason. Just more food for thought related to higher wattage heating elements: earlier in the summer when I visited my local supplier for firebricks - they cautioned me in very, very strong terms that heating an oven too fast will greatly increase the chance of cracking the bricks (they also recommended not using mortar). Especially when the bricks are new, residual moisture and other chemicals need to be burned off slowly. IIRC, they recommended an initial heating rate of 100 degrees F per hour (not a typo) during the first several heating cycles. And they still recommended a heating rate that is pretty slow during all later cycles. So ... another downside of a higher wattage heating element might be earlier cracking and breakage of the insulating bricks.

Anyway - not sure if all that helped give you some insight .... but hopefully it gave some....

Cushing my friend ,almost everything you wrote is wrong..............late I will explain you why .In the meantime I ask you why vacuum formed/which mean they are compressed/ ceramic board will heat oven faster then insulating brick and will cool oven much faster then INS. bricks?

 

[Natlek]

Drew - let me try to answer that for you. Do a thought experiment: a fact is that as you make insulation thicker, its insulating properties increase (typically linearly). The thought experiment is this: assume the insulating firebrick is something like 100 miles thick (yes I mean that) - so there is essentially no heat loss at all from the oven. if the chamber is perfectly sealed, you could put a 15W nightlight on the inside, and (given enough time) it would get pretty hot in there. The caveat as that it would take a LONG time for that little light to heat up all the material of the bricks immdiately around the chamber.

There would be heat lost , we cant stop that with INS.bricks ..........we can slow down heat transfer but we can not stop it 100% .
Thermal equilibrium !
It depend how big chamber you make .If it is knife furnace dimension , forget that you will get even 5 F more then ambient .All heat will be soaked in bricks from 15W source. As much small conductivity bricks have you need decent heat power to heat chamber ........we need more W then energy bricks can conduct to surface if we want to increase temperature ....all this can be calculated precisely but why we should waste time?

 

[Cushing H.]

Natlek - the steady state equations for heat loss through a wall are linear:

• heat loss through wall = (thermal conductivity) * area * 1/wall thickness * (change in temperature across wall)
• heat loss at outer surface = (compound heat loss coefficient) * (temperature of wall - temperature of surrounding room)

Ignoring transients during heat up, everything I said follows from those. During transients, if you have more wall material (more thermal mass) to heat up - its going to take longer, unless you increase the power of the heating element and pump more heat in faster.

Everything I said follows from what I just wrote above .... not quantitatively .... but as general guidance. If you have problems with it .... take it up with the engineering handbooks.

I am going to bow out of this thread - as this useless back and forth is just going to confuse people.

 

[Cushing H.]

There would be heat lost , we can stop that with INS.bricks ..........we can slow down heat transfer but we can not stop it 100%

Its called an adiabatic boundary condition. Seldom achieved in reality, but useful and commonly used to achieve increased insight. Like I said, it was posed as a "thought experiment".

Like I said, I am done. Drew - I apologize that this got drawn into a rabbit hole.

 

[Drew Riley]

Its called an adiabatic boundary condition. Seldom achieved in reality, but useful and commonly used to achieve increased insight. Like I said, it was posed as a "thought experiment".

Like I said, I am done. Drew - I apologize that this got drawn into a rabbit hole.

No need to apologize. It was/is relevant to the conversation. I guess I might be thinking about thermal conductivity/thermodynamics incorrectly, but to expand on your thought experiment some, but in the other direction:

Let's just say I have an oven that has only 1 inch thick walls in all directions, and is perfectly sealed, with a 1000F degree element heating the inner chamber. Will this not eventually reach 1000F on the outside as well? Or does the insulation limit the ultimate transfer of heat? I mean, I get that outside ambient temperature will absorb some of this heat as well (or does it?), but given enough time, how hot will the outside get?

 

[JTknives]

No need to apologize. It was/is relevant to the conversation. I guess I might be thinking about thermal conductivity/thermodynamics incorrectly, but to expand on your thought experiment some, but in the other direction:

Let's just say I have an oven that has only 1 inch thick walls in all directions, and is perfectly sealed, with a 1000F degree element heating the inner chamber. Will this not eventually reach 1000F on the outside as well? Or does the insulation limit the ultimate transfer of heat? I mean, I get that outside ambient temperature will absorb some of this heat as well (or does it?), but given enough time, how hot will the outside get?

you would have to know the thermal conductivity of the shell and the outside atmosphere. But I would also think the thermal conductivity/transfer would change as the surface got hotter. The hot shell is going to create an updraft with will carry heat with it But I’m no expert. But I can tell you my shell hits about a max of 250° no mater the temp inside and the length of time at that temp.

 

[Cushing H.]

to expand on your thought experiment some, but in the other direction:

I would also think the thermal conductivity/transfer would change as the surface got hotter.

Oh my .... I can see where this is going .... but Drew, if you are really interested, I will need to explain this in sequence, and will need pictures along the way. This is the realm of differential equations in heat transfer .... but the basic ideas can be got at (at least qualitatively) without the math. Also - JT brings up a really good point about the updraft, but that is incredibly complicated to predict accurately .... but at this level I think we can give a nod to it, and use that to understand how it will affect things like the outside and inside temp of the oven, without getting really accurate about it. It is too late tonight for me to start in on that .... but if you are really interested I can certainly try to explain (please give an affirmative answer? There are others here who will just say .... "oh .... cush is just spouting useless theory again.......".

I'll start though with your posed 1000 degree element. good for your for working to pose questions like that ..... but unfortunately with this stuff, you need to think at the lowest levels not about temperatures, but about rate of energy into and out of a system. If you thought only temperatures .... then with your same posed situation above, if your just had that 1000 degree element sitting out in your backyard and applied the same logic .... you would end up predicting that the single element would heat the entire atmosphere of the planet to 1000 degrees. Instead - you need to think about that element as putting energy into the oven at a certain rate - 15watts, 100W, 1200W, 2500W, etc. Then you need to think about how fast that energy moves through the insulating walls of the oven. THEN you need to think (like JT is talking about) how fast the energy that reaches the outside of the oven gets transferred to the room around it (which essentially acts as a heat sink). All of those interact - so that what is going on at the surface of the oven in terms of losing heat to the outside room definitely affects (to a lesser or greater degree, depending on the insulation) the temperature of the inside of the oven. This is literally why the subject is called "Heat transfer" .... not "temperature transfer".

Like I said - to reasonably explain this will take pictures. Let me know if your want that? If it makes more sense to start a new thread, we can do so.....

 

[Drew Riley]

Oh my .... I can see where this is going .... but Drew, if you are really interested, I will need to explain this in sequence, and will need pictures along the way. This is the realm of differential equations in heat transfer .... but the basic ideas can be got at (at least qualitatively) without the math. Also - JT brings up a really good point about the updraft, but that is incredibly complicated to predict accurately .... but at this level I think we can give a nod to it, and use that to understand how it will affect things like the outside and inside temp of the oven, without getting really accurate about it. It is too late tonight for me to start in on that .... but if you are really interested I can certainly try to explain (please give an affirmative answer? There are others here who will just say .... "oh .... cush is just spouting useless theory again.......".

I'll start though with your posed 1000 degree element. good for your for working to pose questions like that ..... but unfortunately with this stuff, you need to think at the lowest levels not about temperatures, but about rate of energy into and out of a system. If you thought only temperatures .... then with your same posed situation above, if your just had that 1000 degree element sitting out in your backyard and applied the same logic .... you would end up predicting that the single element would heat the entire atmosphere of the planet to 1000 degrees. Instead - you need to think about that element as putting energy into the oven at a certain rate - 15watts, 100W, 1200W, 2500W, etc. Then you need to think about how fast that energy moves through the insulating walls of the oven. THEN you need to think (like JT is talking about) how fast the energy that reaches the outside of the oven gets transferred to the room around it (which essentially acts as a heat sink). All of those interact - so that what is going on at the surface of the oven in terms of losing heat to the outside room definitely affects (to a lesser or greater degree, depending on the insulation) the temperature of the inside of the oven. This is literally why the subject is called "Heat transfer" .... not "temperature transfer".

Like I said - to reasonably explain this will take pictures. Let me know if your want that? If it makes more sense to start a new thread, we can do so.....

I feel bad putting you to too much trouble, but you're definitely helping already to clear up some of my questions (while generating a few more at the same time, haha). If you want to further explain with pictures, I would be interested in getting to know more about it, but I also understand that those types of explanations take time and effort, and you certainly don't owe me any of that. Also, I realize that we've somewhat derailed the original thread, so many a new thread would be appropriate? Then again, I suppose that all of this applies to "DIY ovens" as much as anything else does.

 

[Natlek]

The reason I ask is because I can’t imagine the wool not compressing when you put it between the shell and brick. The way I stack my brick I put it under compression with bolts and this allows me to span 2 bricks across the roof (18”). It just seams like some people look down on only using brick around here but if your not gaining anything from the wool then why not just use the thicker brick.

Why you don t use ceramic fiber board ? Easy to build oven and you don t need to use that much force for compression and you can do any dimension on chamber you want .Considering that you heat treat many steel daily , the width of the chamber is more important to you so you can HT more blanks at once ? With ceramic board you can make it wide as much you want ? Ceramic board will never crack , oven will heat much faster , will cool much faster and you will save some energy ? You HT different steel on different temperature , so fast heat and fast cooling of oven is good thing for you if time is money ?
You can even make it combined ? Floor ins.bricks ,rest ceramic board .Or you can use bricks for all wall except top one ? Corners bricks rest board .....So many combination there .............

 

[Cushing H.]

Ok Drew .... just to start this out I will be using variations of the following diagram. The two vertical black lines are meant to represent the inside and outside walls of an oven, with the adjacent inside and outside spaces marked. The horizontal red line is meant to represent the temperature at the locations varying from inside the oven, through the wall, and then into the room outside the oven. I will be ignoring variations in temperature by location inside the oven, and in the room around the oven - these will be viewed as (each) being all one temperature. When that horizontal red line moves up in any location, that is meant to represent a hotter temperature at that location. So .... in this diagram, the oven is off and dead cold, so all temperatures are at room temperature.

Make sense?

 

[Drew Riley]

Ok Drew .... just to start this out I will be using variations of the following diagram. The two vertical black lines are meant to represent the inside and outside walls of an oven, with the adjacent inside and outside spaces marked. The horizontal red line is meant to represent the temperature at the locations varying from inside the oven, through the wall, and then into the room outside the oven. I will be ignoring variations in temperature by location inside the oven, and in the room around the oven - these will be viewed as (each) being all one temperature. When that horizontal red line moves up in any location, that is meant to represent a hotter temperature at that location. So .... in this diagram, the oven is off and dead cold, so all temperatures are at room temperature.

Make sense?
View attachment 1482873

So far so good.

 

[Cushing H.]

Ok .... now imagine that you have a magic "perfectly insulating but infinitely thin" layer on the outside of the wall, and then you turn on a heater on the inside of the oven (does not matter if it is that 15W night light, or a 2400 W 240V coil. (note - for now I am ignoring the fact that it takes time for heat to travel through the thickness of the wall, and so the profile is non linear in reality - but I will treat the profile as linear. This is called a "steady state" or "pseudo steady state" assumption - it just makes things easier).

But the two points starting here, and for the rest of this are:

• the heat put into the inside of the oven can not just "dissappear" - something has to happen to it:

1. it either goes to heating up a material (insulation or air)
2. or it escapes to the outside world - which is treated as big enough that the escaping heat does not change the temperature of the outside world.

ok .... so after a short amount of time, you have a situation that looks like this (the temperature of the oven inside and wall goes up but stays uniform (and NO heat escapes to the outside world)

a little later it looks like this:

and later yet, it looks like this:

and so on .... until things burn or melt. Because no heat escapes to the outside world, ALL of the heat put into the inside of the oven goes to heating up the chamber and the walls. It does not matter if it is a 15W bulb - that will just take longer. (this is what I was trying to get at with yesterdays "100 mile thick" wall (which would be a REALLY good insulator). Does that make sense?

If so, then the other thing to take note of (though the reason is not clear yet) is that the rate of movement of heat through the wall is (for a given unchanging thermal conductivity) proportional to the SLOPE of the temperature curve through it. In this case, the slope of the temperature curve is zero - so there is no net heat movement through the wall to the outside world.

 

[Cushing H.]

Oh ... ok Drew. I did not see your “like” earlier. Will try to post more in a little while

 

[Cushing H.]

Ok ..... next little nibble. Keep in mind that energy (heat) always flows from areas of high temperature to areas of low temperature - the reasons for this follow from statistical mechanics (which we dont want to go in to here) .... but should make pretty much intuitive sense. So .... "heat" (or a roughly equivalent term "energy" - but I will stick with "heat") is NOT the same at temperature - the two being related by the "heat capacity" of a material, which I will mention, but not spend a lot of time on here.

Also, I lied - I do need to use an equation or two (simple one though). The heat transport equation that describes the movement of energy through an insulating wall, at any infinitessimally thin "slice" on the inside of the wall is:

Q = - k * dT/dx, where is the amount of heat energy moving through an area of unit area, k is the thermal conductivity of the material, and dT/dx is the temperature gradient (the slope of the temperature curve from inside of the oven to the outside wall). the negative sign is there so that the heat flux is a positive value (because the slope of the temperature is negative - remember, heat moving from areas of high temp to areas of low temp). remember, this equation holds independently at every location through the thickness of the wall. In words:

heat flux through a segment of wall of area "1", at any location or depth in the wall, = thermal conductivity multiplied by the local change in temperature versus distance towards the outside of the wall.

if that is understandable, then you might think that the slope of the temperature curve at different depths in the wall can be different. BUT, again we are considering "steady state" here .... and an important thing to keep in mind is that, to avoid energy being created or destroyed (which it can not), the heat flux at every location in the wall has the be exactly the same as every other location in the wall. so "Q" (the local heat flux), must be some fraction of the energy put in by the heating element on the inside of the oven. for convenience - I will assume all the energy put in by the heating element goes through a single unit area .... so I will just say that Q is the same as the power of the heating element.

Because the energy put in by the heating element does not change, and (we assume here) that the thermal conductivity does not change with wall depth, then at every "depth slice" through the wall, flux is constant, the heat capacity is constant, and so the temperature gradient must be constant at all locations through the wall (it is constant).

By the equation, Q = - k * dT/dx, if Q is constant, and K is constant, then dT/dx must also be constant.

Also, if you have a bigger heating element, then the heat flux through the wall must be bigger (must be the same as the input of the heating element), and if the thermal conductivity is constant, then the gradient of temperature through the wall must be bigger.

by the equation Q = - k * dT/dx, if Q is bigger, and if k is constant, then dT/dx must be larger.

If I have not lost you in the above, then we can look at that with a picture. Assume a situation where instead of perfectly insulating the outside wall of the oven, you find a way to absolutely pin the temperature there right at room temperature (say, blowing a really strong fan over the wall, or putting the oven in a water bath (dont try that for real), or by putting a cooling plate on the outside with room temperature water flowing through it....), then you have a situation like the below (it is important to remember here that the slopes of the temperature lines are defined by the power of the heating element, not by the fact that the temperature of the oven is higher (the inside temp is defined from the slopes and the starting known temperature on the outside):


All the above essentially explains why, all else being equal, when you use a higher power heating element, the inside temperature of the the oven must be higher than if you used a lower power heating element (here we are assuming that the heating element is going full-bore - 100% output all the time). Starting with the known temperature on the outside of the oven, the higher slope of the temperature curve(which is required when the heat flux is higher) must result in a higher temperature inside the oven.

This is still a pretty simplified and idealized view of the thing, but is meant to try to give intuition as to how the different parts of the system interact and affect each other. Let me know if/when you have digested this .... and when we have gotten there I can move on to more complicated situations - like when you are not explicitly controlling the outside temperature of the oven....

Oh ..... "extra credit" . From the above, if you have the same power heating element, walls with the same thermal conductivity, BUT the wall is twice as thick - can you see why the temperature of the inside of the oven must be still higher yet?

 

[Drew Riley]

Ok ..... next little nibble. Keep in mind that energy (heat) always flows from areas of high temperature to areas of low temperature - the reasons for this follow from statistical mechanics (which we dont want to go in to here) .... but should make pretty much intuitive sense. So .... "heat" (or a roughly equivalent term "energy" - but I will stick with "heat") is NOT the same at temperature - the two being related by the "heat capacity" of a material, which I will mention, but not spend a lot of time on here.

Also, I lied - I do need to use an equation or two (simple one though). The heat transport equation that describes the movement of energy through an insulating wall, at any infinitessimally thin "slice" on the inside of the wall is:

Q = - k * dT/dx, where is the amount of heat energy moving through an area of unit area, k is the thermal conductivity of the material, and dT/dx is the temperature gradient (the slope of the temperature curve from inside of the oven to the outside wall). the negative sign is there so that the heat flux is a positive value (because the slope of the temperature is negative - remember, heat moving from areas of high temp to areas of low temp). remember, this equation holds independently at every location through the thickness of the wall. In words:

heat flux through a segment of wall of area "1", at any location or depth in the wall, = thermal conductivity multiplied by the local change in temperature versus distance towards the outside of the wall.

if that is understandable, then you might think that the slope of the temperature curve at different depths in the wall can be different. BUT, again we are considering "steady state" here .... and an important thing to keep in mind is that, to avoid energy being created or destroyed (which it can not), the heat flux at every location in the wall has the be exactly the same as every other location in the wall. so "Q" (the local heat flux), must be some fraction of the energy put in by the heating element on the inside of the oven. for convenience - I will assume all the energy put in by the heating element goes through a single unit area .... so I will just say that Q is the same as the power of the heating element.

Because the energy put in by the heating element does not change, and (we assume here) that the thermal conductivity does not change with wall depth, then at every "depth slice" through the wall, flux is constant, the heat capacity is constant, and so the temperature gradient must be constant at all locations through the wall (it is constant).

By the equation, Q = - k * dT/dx, if Q is constant, and K is constant, then dT/dx must also be constant.

Also, if you have a bigger heating element, then the heat flux through the wall must be bigger (must be the same as the input of the heating element), and if the thermal conductivity is constant, then the gradient of temperature through the wall must be bigger.

by the equation Q = - k * dT/dx, if Q is bigger, and if k is constant, then dT/dx must be larger.

If I have not lost you in the above, then we can look at that with a picture. Assume a situation where instead of perfectly insulating the outside wall of the oven, you find a way to absolutely pin the temperature there right at room temperature (say, blowing a really strong fan over the wall, or putting the oven in a water bath (dont try that for real), or by putting a cooling plate on the outside with room temperature water flowing through it....), then you have a situation like the below (it is important to remember here that the slopes of the temperature lines are defined by the power of the heating element, not by the fact that the temperature of the oven is higher (the inside temp is defined from the slopes and the starting known temperature on the outside):
View attachment 1483535

All the above essentially explains why, all else being equal, when you use a higher power heating element, the inside temperature of the the oven must be higher than if you used a lower power heating element (here we are assuming that the heating element is going full-bore - 100% output all the time). Starting with the known temperature on the outside of the oven, the higher slope of the temperature curve(which is required when the heat flux is higher) must result in a higher temperature inside the oven.

This is still a pretty simplified and idealized view of the thing, but is meant to try to give intuition as to how the different parts of the system interact and affect each other. Let me know if/when you have digested this .... and when we have gotten there I can move on to more complicated situations - like when you are not explicitly controlling the outside temperature of the oven....

Oh ..... "extra credit" . From the above, if you have the same power heating element, walls with the same thermal conductivity, BUT the wall is twice as thick - can you see why the temperature of the inside of the oven must be still higher yet?

I had to read that a couple of times, but I think I’m getting there.

 

[Cushing H.]

I had to read that a couple of times, but I think I’m getting there.

Want to take a stab at the extra credit? If you can explain that, then you basically do get it...