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Author Topic: How do you calculate the total heat energy absorbed by a solid heated on 1 side?  (Read 4053 times)

briligg

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How do you calculate the total heat energy absorbed by a solid heated on 1 side?
« on: 28/08/2012 22:08:15 »
I'm trying to calculate how much heat energy will be absorbed by the thermal mass in a (hopefully) passive solar workshop. The floor and walls are essentially the same material, rammed clay soil stabilized with lime - that's the plan, anyhow. Let's say all thicknesses are 40cm for now - with a formula for decent calculations, we'll set about deciding what dimensions would be best given budget, climate, etc. If the proper calculation is very complicated, is there a way to come up with a decent estimate?
Probably there is a great page on this somewhere, i just don't know how to properly phrase the search.
Thx.

briligg

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Re: How do you calculate the total heat energy absorbed by a solid heated on 1 side?
« Reply #1 on: 28/08/2012 22:11:38 »
Maybe i should mention that the structure is intended to be insulated on the outside, and heated by a mostly transparent south wall letting sun in. So, i think you can mostly not worry about shifting sun changing the heat load on the building exterior, and think basically about the changing air temperature inside.

briligg

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Re: How do you calculate the total heat energy absorbed by a solid heated on 1 side?
« Reply #2 on: 28/08/2012 22:49:21 »
Could this be handled by an area under the curve approach? Material thickness on the x axis, watts on the y axis?
I can assume for now that area and temperature difference are 1 m2 and 1 K, and plug that stuff in later.
« Last Edit: 28/08/2012 22:53:32 by briligg »

briligg

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Re: How do you calculate the total heat energy absorbed by a solid heated on 1 side?
« Reply #3 on: 29/08/2012 16:39:30 »
Hm, this doesn't raise much reaction, i see. I think it is a matter for integral calculus, which i failed in high school but is the proper term for my vague 'area under the curve' memory.
Just in case this pops up in a google search in some future for someone who is in the same boat as me, my problem is that the material i'm considering has a thermal conductance of 0.8 W/mK, so if there is warm air on the interior face 10 K warmer than the rammed earth, 8 Wh would pass through a meter thickness of this material in an hour. The formula is H=kA(deltaT/x), where k is thermal conductance, A is area, delta T is temperature difference, and x is material thickness. The answer is expressed in watts, and since watt-hours are simply a quantity of watts sustained for an hour, adding in time is easy.
But notice that the lower x is, the higher H is. So, presumably, if a 10 K temperature difference is applied to one face of a cubic meter of rammed earth, 8 Wh would penetrate to the other side after an hour, but also, 16 Wh would penetrate 50 cm, 32 Wh would penetrate 25 cm, and so on. So how much energy would be absorbed altogether?
There has got to be an engineering handbook somewhere with a specific answer, at least for common materials like concrete. Maybe i need to go look for a layman's engineering forum.

William McCormick

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Re: How do you calculate the total heat energy absorbed by a solid heated on 1 side?
« Reply #4 on: 30/08/2012 04:17:47 »
I'm trying to calculate how much heat energy will be absorbed by the thermal mass in a (hopefully) passive solar workshop. The floor and walls are essentially the same material, rammed clay soil stabilized with lime - that's the plan, anyhow. Let's say all thicknesses are 40cm for now - with a formula for decent calculations, we'll set about deciding what dimensions would be best given budget, climate, etc. If the proper calculation is very complicated, is there a way to come up with a decent estimate?
Probably there is a great page on this somewhere, i just don't know how to properly phrase the search.
Thx.

I used to manufacture three phase heating equipment. And if I wanted to know what sized heaters I wanted to use, I first had to decide the material and weight of the material I wanted to heat. Then when I found out how many BTU's it would require, to heat that material a set number of degrees in an hour, I would size the heaters appropriately, to heat the material to the desired temperature in the time period that I wanted to get the job done in.

Sometimes if you don't have enough surface area to thickness or volume of material, it takes a long time to heat up the substance you want to heat. Because you do not want to put a red hot surface up against the material, to heat it up in the time you would like. It might burn or set fire to the substance.

From a quick estimate, I would say you will need about 15 kilowatt hours worth of power, to raise 1500 pounds of clay ceramic material, 70 degrees Fahrenheit. Because you are going to have a lot of losses. No matter how you do it.

So for that first initial heating 15 kwh is probably what you will need to raise it 70 degrees in one hour. If you want to raise it 70 degrees in two hours, it would require 7.5 kilowatt hours, for two hours. You get the idea. It will probably take days or weeks to actually raise it 70 degrees.

Your watts per square foot necessary, even at 90 watts per square foot in direct noon sunlight, will require 150 square feet to heat 1,500 pounds of material, that would be my approximation. Once you get the the ceramic or clay, heated, it is not so bad, it is just the losses. But you have  tons and tons of material to heat.

In Austria, they pile Hay and manure many feet thick, up against three foot thick stone walls, and that wall stays very warm all winter long. It rots all winter long making heat all winter.

I had a neighbor that used to pile his lawn clippings up a few feet high behind the shrubs of his house, or actually he used to have me do it for him, especially when hey went away for the summer. And that certainly did keep the foundation warm. If you exposed the very old grass it would actually smoke in the summer time.

Sincerely,

William McCormick

briligg

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Re: How do you calculate the total heat energy absorbed by a solid heated on 1 side?
« Reply #5 on: 30/08/2012 22:37:11 »
Wow, that's neat about the Austrian technique and the power of clippings and rot.

I got data on the specific heat of the material we want to use, its thermal conductivity, and on the light energy that falls on each m2 on an average day of each month in our building location. I have put together a spreadsheet with that data, and the data on heat loss through the building envelope, based on a rough plan, and it seems that we can reasonably hope to extend the dates when it would be a comfortable temperature inside to between about mid-March to the end of November. I think we could also keep it above freezing most of the rest of the year. The building would be done in summer so it would start out warm. Also, we plan to install insulated shutters that would cover all the south windows (the only windows) whenever there isn't enough sun to justify them being open, and the whole foundation will be just as well insulated as the exterior walls - something i think people sometimes neglect in projects like this.

After a while i started to worry, though, that the thick floor and walls would not suck up all the heat energy coming in the windows on sunny days fast enough to avoid losing a big portion of it back out the windows - that the air would heat up more and pass the heat right back to the windows. I have read that greenhouses heat up surprisingly quickly on sunny days even in very cold weather, but of course the heat isn't held, and shortly after sundown they are as cold as ever. The situation we are contemplating is very different, but the problem of temperature fluctuation has some similar dynamics. Thus the struggle with how quickly the heat gets absorbed - not how much will penetrate a given thickness, or how much that raises the (average) temperature of the mass being heated, but how much heat gets into it, total, each hour, when subjected to the heat source of sun directly on it, or air that is warmer than it is. The energy needed to heat it adequately does indeed come in the windows, according to my data. But will it stay there? There's the rub.

After scrounging data some more, and doing a very rough graph of the heat absorption curve and a simplistic calculation of the area under it, it seems that it will, handily. But i'm still going over that.