About thermal conductivity and determining temperatures in a 1-D slab - Help please! 3

[KevinWitt]

I have some questions about applying thermal conductivity equations to a physical system in order to calculate expected temperatures. In brief, I'd like to be able to determine the temperature of something under an insulating layer. I've made up a somewhat contrived scenario below, but it has the basics I'm interested in.

In a room that is at 20C (initial temperature of everything), I have an 1cm thick aluminum ironing board that has a surface area of 0.2 square meters (10cm x 20cm).
The ironing board has the following material/thermal properties:
- Density = 2700 Kg/ cubic meter
- Thermal Conductivity = 237 W/m-K
- Specific Heat = 205 J/Kg-K

On top surface of the ironing board there is an ironing pad made of foamed insulation that is 1mm thick (0.001 meters) that covers the top surface of the ironing board perfectly (0.2 square meters.) The ironing pad has the following material/thermal properties:
- Density = 32 Kg/ cubic meter
- Thermal Conductivity = 0.026 W/m-K
- Specific Heat = 1400 J/Kg-K

Onto the ironing pad I place a clothing iron that (which coincidentally has the same surface area of 10x20cm as the ironing board) for "t" seconds. This clothing iron has a surface temperature of 100C and supplies a continuous amount of power into the system of 600W. (This is not the electrical rating of the iron – it is the power source for this example: the iron provides 600 J/sec of power.) Assume this is a closed system with no edge effects, no radiative or other energy losses, and that all the heat from the iron is conducted perfectly to the top of the insulating pad.

After "t" seconds…
- What is temperature of the aluminum portion of the ironing board?
- What is the temperature at the bottom of the ironing pad (where it meets the table)?
Let's say that "t" is going to be on the order of a few seconds, 10 for example. Ultimately I will also look at the heat losses of the aluminum to the ambient air and try to determine the final, steady-state temperatures, but first-things-first!



I have been trying to get at these answers using conservation of energy and the 1-D version of Fourier's Law for Thermal Conduction I get myself confused over what "heat transfer" (q in Fourier's Law) means physically. Here's an example.

Let's say that the 600W iron is laid on ironing system (pad and board as described above) for 10s. In that time a total of 6000J of heat energy are added to the system. If we again assume a closed system, where does the energy go? If the amount going into the table can be determined, I can answer my question about table temperature. I'll work this out backwards in an attempt to show my thinking which will hopefully allow someone to point out where I am going wrong.

Let us assume that after 10s, the aluminum table reaches a maximum temperature of 52.27C. The heat transer associated with the ironing pad is:
q = (k/s)A(T2-T1)
= (0.026/0.001)*0.2*(100-52.27)
= 248.2 Watts
If the top of the pad is at 100C, and the bottom is at 52.27, and the 1-D temperature gradient is linear, the the average pad temperature will be (100-52.27)/2 = 76.1C. Given the heat capacity of the pad is 1400 J/Kg-K, the energy needed to raise the pad to this temperature is 503J. The iron puts out 600 J/s, so it takes about 0.84s to get the pad up to temperature. That leaves 9.16s of heat transfer to the table. (I am trying to simplify this a little by assuming the energy from the iron goes to heat the pad first. We can neglect the energy used to raise the pad temperature if this confuses matters as 503 is small compared to 6000.)

The application of Fourier's Law says there are 248.2W of heat transfer. So does that mean there are 9.16s of power applied to the table? That would be 2274 Joules provided to heat the aluminum. (248.2*9.16=2274). Or, is 248.2W the amount of power shielded/insulated by the padd, and power given to the table is 351.8W (600-248.2=351.8). In the former case the ironing table temperature would increase by 20.5C, whereas in the latter it would increase by 29C.

I've also tried looking at the extremes to answer this for myself. Suppose the temperature of the pad (T2) remained at 20C. This means the pad is a perfect insulator (no heat from the iron gets to the table). The 1D heat transfer in this situation is puts the heat transfer q at 416W which would be the maximum value possible in this example and unlikely given a 600W heater at 100C. (maybe this is some sort of background value?) At the other extreme, a lower pad temperature of 99C gives a heat transfer q=5.2W for a very poor insulator. This leads me conclude that good insulators have high heat transfer numbers (and poor insulators have low values) suggesting that the amount of heat actually transferred through the insulating layer (the pad in this case) is the not Fourier's heat transfer value (q), but the total power applied less (q), which in my example above would the case where the table rises by 29C. "Heat Transfer" then would be more akin to energy dissipation by the insulating layer than the amount of energy actually transferred. Or, have I gotten this all wrong? Help!

 

[IRstuff]

Is this for school? Student posting is forbidden.

I suggest you crack open a heat transfer text. Your concluding paragraph shows a total complete lack of understanding of even the term "insulator," which prevents heat transfer, so any notion that there is high power is completely wrong. This site is for working professionals to ask work related questions. You need a complete tutorial; I suggest re-reading the Wikipedia articles and then possibly look at Lienhard's text, which is free:

https://ahtt.mit.edu/

TTFN (ta ta for now)
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[Compositepro]

I stopped at the second paragraph. 10cm x 20cm is not 0.2 m2.

 

[KevinWitt]

IRstuff,
You have raised an important point which I would like to address first. I hereby swear and affirm that my post was not for school, nor am I a student. I am a working professional, and the topic I am asking about is ultimately work-related. The question I have posed is simplified and reduced(compared to the 3-D topic I am engaging) so as to help me understand the basic principles before tackling the real issue. Therefore, please rest assured that my posting violates no rules of this site and I thank you for checking on this first. I also wish to thank you heartily for your prompt reply. "Unhelpful" is not my first language and it has taken me some time to translate your words correctly. I believe I have distilled the essence of your words into my mother tongue: "Look it up."

This is surely sage advice and I offer you my most humble thanks. I'm sure it would not have occurred to me to actually research this topic prior to posting a question outside of my field of expertise. I feel privileged that someone who "can do anything" and "is an expert" was so gracious as to not imply I was cheating, or was unable to understand the purpose of this forum, took some their valuable time and deigned to provide such expert assistance. Surely a lesser guide might have corrected my misunderstanding directly, stated the correct approach explicitly, or possibly even shown me how to solve the problem in their reply posting thus depriving me of the valuable opportunity for self-learning. I see now that I am not worthy of such teaching and that your missive is in my best interests long term. I am lucky to have heard from you in this way, and for that you have my undying gratitude.

Compositepro: To you I owe a heartfelt and deep apology. I feel terrible that I have wasted your time and offended your delicate sensibilities by permitting a typographical error to creep into my posted request for help on solving an engineering problem. I can only imagine the agony you suffered having found that I had equated 10cm x 20cm with 0.1m2 thus forcing you to stop reading. How wicked and cruel of me to type 10cm when the value should have been 100cm for the length. Rest assured I am fully chastened. I am humbled by your ability to rise above your suffering and offer such germane and profound commentary on my flawed solicitation for assistance. Had I remained unaware of your plight, I might have continued in my wicked ways. It was selfish to ask a question with an error, and unreasonable to expect an answer to the actual question I asked. I hope you will forgive me.

It is so nice to know that my first foray into this professional community has resulted in such kind, gracious, and helpful assistance. So many times public forums harbor small souls who take the opportunity of someone's ignorance to make insulting remarks, proclaim unfounded judgments, or provide self-congratulatory or smug commentary that feeds their own need to feel superior. Today, with the kind assistance of those who have helped me answer my question, I am wiser and thankful. To you I raise the one-finger salute and wish you well in your endeavors.

 

[danschwind]

KevinWitt,

I'm confused on your statement regarding the 600W the clothing iron is supplying to your system. Before stating that, you said that the clothing iron has a constant surface temperature of 100C...don't you agree that the amount of heat the clothing iron is supplying to your system depends on the temperature gradient between said iron and the system? By your logic, if you supply a constant 600W to your system, after X seconds you will surpass the temperature of the clothing iron, which makes no sense.

What you have here is a transient problem (as you require the change in temperature x time) where one of the boundary conditions is a 100C constant source. You can't have a 100C constant temperature plus 600W constant heat input in the same boundary.

I suggest you to get the book that IRstuff kindly linked in his post (btw, thanks, I didn't have this one too!) and take a look at the chapters that start at page 203 and page 63. After this, you will probably want to look at natural convection, as it will be the main mechanism of heat loss of the ironing board.

 

[IRstuff]

Again, this site is not for tutorials about heat transfer, so as I pointed out, you have a fundamental misunderstanding, even at a very general level of what an thermal insulator does. Consider any ice chest or any other thermally insulated container, like a Thermos bottle. Clearly, the inside surface of the vacuum space is not at room temperature, but at the temperature of the stuff that's in the Thermos. Likewise, the outer surface is at the temperature of the ambient air. That's what distinguishes a thermal insulator; it can have the maximum possible temperature delta, but transfer little or no heat.

The equations you are using are all steady state equations, and cannot be directly used to solve your problem. Further, heat is measured in joules, and therefore, the time rate of heat transferred is joules/s = watts. You've not even applied any concept of thermal mass, i.e., specific heat, which is critical to knowing how much energy is required to change the temperature of a substance. Only then, can you being to possibly discretize the equations available to solve the time behavior of any thermal system.

All responders here are working professionals, for the most part, donating their time and effort to help fellow professionals with complex problems. What's possibly disrespectful is to treat these people as an on-demand Wikipedia, so rather than solving a complex problem, they wind up trying to answer questions whose answers can be found on Wikipedia. Rather than simply asking "I'm confused about Fourier's Law," you simply plowed ahead in a TLDR posting that is basically wrong at almost every statement, and wasting their time trying to figure how to solve your problem.

So, getting back to Fourier's Law and my first paragraph, it describes the rate of heat transferred across a substance as a function of the heat transfer coefficient and the temperature drop. Insulators will transfer very little heat per unit time, i.e., low watts, while thermal conductors, like copper, will transfer lots of heat per unit time, i.e., high watts. For a source with high temperature, but low thermal mass, a thermal conductor will effectively attempt to equalize the temperature on its two interfaces. That's why any typical iron has a huge chunk of steel as its thermal interface, so that the typical stuff that the iron touches, like cloth, with low thermal mass, will quickly get heated, but the iron will lose little heat and maintain a relatively constant temperature.

TTFN (ta ta for now)
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[KevinWitt]

Thanks IRStuff,
I now see what you are saying and why. I appreciate the longer post.

Yes, I am trying to take a transitory phenomenon and apply steady state equations to simplify the situation. I was hoping to take advantage of the conservation of energy to look at the system after a long enough period of time that equilibrium had been reached, but not so long that everything had come to the same temperature (as I added the boundary condition of no other heat losses such as convective or radiative losses.) My thinking is as follows:
-in 10 s the "iron" will provide 6000J of heat energy.
- some of that energy will heat the pad material (q=mcDT) to bring it up to it's average temperature given the temperature gradient stated. That will consume 503J. That leaves 5497J unaccounted for.
- Assume that the first 503 watts coming out of the iron go to heating up the pad and establishing the temperature gradient across the pad, then it takes 0.838s to heat the pad. That means are 9.162s left for the output of the iron to go other places.
- With a thermal conductivity of 0.026 W/m2-K, and the dimensions given (0.2m2 given the correct length of 100cm)and the temperature difference stated, the conductive heat transfer of the pad is 248.2W.
- If the pad is transfering 248.2 W for 9.162s, then during that time, 2274 Joules are available to heat the underlying aluminum. Given the state properties and dimensions, that means there it will experience a rise in temperature of 20.5C, which having started at 20C means the final temp of the table will be 40.5C This leaves 3223 Joules unaccounted for.

Perhaps this whole idea of typing to switch between energy and power in this way is faulty accounting and I am mixing things that should not be gathered together. I can buy that since my Thermodynamics classes were about 35 years ago. So yes, I need re-education. But whenever I look at heat transfer in 1-D, I keep seeing examples where one determines the conductive heat transfer through a slab based upon q=-k dT/dx and the answer says X-many watts of heat transfer given the change in temperature etc. The window or brick transfers 200W.

There is the crux of my issue: What does it mean to have 200W of heat transfer through the slab in question. Is that 200W "available" on the low temperature side of the slab to heat up the stuff that resides there?

Which brings me back to the original question I asked in my first post.

After "t" seconds, what is the temperature of the aluminum? Forget my ham-fisted attempt at an answer. Aside from using COMSOL or CFD-ACE and modelling the system, how can I get an answer?

 

[IRstuff]

That equation is still a steady state equation, since it has no dependence on time; it's a 1D derivative in distance, not time.

Again, I suggest re-visit the HT articles in Wikipedia. Fourier's Law says that the heat transferred is proportional to the temperature difference, and goes from a place of higher temperature to a place of lower temperature. Everything else is essentially an empirical fit to that relationship. Thermal conductivity is typically a measured parameter and not predicted from first principles; which is why there are tables of values, rather than physical properties that you plug and chug.

q = -k dT/dx basically has a proportionality constant k that is thickness dependent, hence k/dx is the actual proportionality relationship. The temperature difference and power (rate of heat transferred, since heat is in joules) therefore, are not independent. If the high temperature source cannot support the heat flow in steady state, then the temperature of the source will drop, along with dT.

So, start with your "pad" 0.026 W/m-K 1-mm thick and 0.1 m * 0.2 m area, which results in 0.52 W/K. So, for a 100C iron and 20C aluminum on the other side, the instantaneous heat flow at t=0 cannot be any higher than 41.6 W, and, can only decrease as time increases, assuming that the aluminum heats up, thereby reducing the temperature difference over time.

The total thermal mass of the aluminum is 0.1 m * 0.2 m * 205 J//kg-K * 2700 kg/m^3 = 110.7 J/K
Since you only have a maximum of 41.6W over 10 seconds, you cannot have more than 416 J transferred, which means the aluminum, in steady state, cannot get higher than 23.76 C, although the surface directly next to the pad could get much higher, since Fourier's Law applies to the aluminum itself, i.e., in order for the temperature to rise elsewhere in the aluminum, there must be a temperature difference in order for heat to flow. Luckily for you the Biot Number is small, so the temperature difference will be relatively small, since the thermal conduction is 474 W/K in the aluminum.

TTFN (ta ta for now)
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[btrueblood]

KW: "in 10 s the "iron" will provide 6000J of heat energy."

In reality, the clothes iron probably has a thermostat on it that switches the heater on/off to maintain a (somewhat) constant 100 C. So the boundary condition of 600W from the iron is a mis-statement of the physics, since the pad cannot transfer more than 40-some watts at the given temperature differential; the system does not absorb anywhere near 600W thermal energy, the heater is not the driving constraint, the 100C temperature and the thermal mass of the iron's surface plate controls the heat transfer.

You could model the iron as a thermal mass of its own, with a heat input to it from the heater that varies depending on its temperature...or just model the iron as a constant temperature, one or the other. You can't model both boundary conditions simultaneously, it would be equivalent to modelling a cantilever beam that allows deflection of the cantilevered end, i.e. physically impossible.