NOT A CIVIL ENGINEER in any way that could be interpreted

BUT the problem I had when trying to get this right was difficulty with the fact that while the stone has only an R1, there is no accommodation for what solar guys refer to as re-radiation.

But while I had adobe (mud brick) walls, which should only have an R1, in fact the building held temperature better than R1 walls should have, and the re-radiation effect "flattened" the swings in temp. that the R1 calculations indicated should be there.

Good luck!

...in fact the building held temperature better than R1 walls should have, and the re-radiation effect "flattened" the swings in temp. that the R1 calculations indicated should be there.

That comment reveals an ambiguity in the question. If the air outside stays at 18F and if other sources of heat like the sun or interior inhabitants stay constant, then the building will eventually reach some equilibrium temperature somewhat above 18F (because the sun is heating the building more than it is heating the air.)

But neither the air temperature or the sun are constant. The sun rises, travels in an arc across the sky, and sets. So the temperature inside will swing up and down around an average temperature daily, even if the air temperature is constant. Those swings might be large or small: if the building is massive enough, it might not swing at all. A cave could be considered a really really massive building, and temperatures deep inside a cave hardly change.

But it is impossible to tell how big those swings will be, or what will be the average value, from the facts given. If the building has a long side facing south, its temperature will swing up more during the day, and also probably be higher on average. If the building is subjected to strong wind all the time, then it will be cooler on average, because more heat will be carried away by direct conduction. Etc.

A cave could be considered a really really massive building, and temperatures deep inside a cave hardly change.

Agreed, but the "top level" calculations I was working with (since it isn't my field and was trying to teach myself out of an Audel's Guide to HVAC) only give you the option of Rock/Masonry wall = R1. (And as an aside didn't actually let me calculate ?insolation? either...seriously flawed) so south rooms and north rooms came out the same.

Which I think may be part of the problem this poster is running into.

Which led to calculations that showed triple digit BTUs per room on a windy winter night. And we just didn't have experiences that mirrored that. So really glad you came by!

How DO we calculate the heat needed for our "cave" f'rinstance?

GA from me.

Everything I would have written, but with none of the "Cuss" words that I may have added....

Hello karlc,

Please allow me to point out that, the OP said 18 ͦͦ Celsius.''''''

Take care

bb

the exterior temperature it is minus 18 grade Celsius

there isn't any heat source inside, and there won't be one.

after I will calculate the interior temperature (when outside are minus 18 grade Celsius) I will calculate the thermal resistence, that named R1, for an isolated wall because I will need to be inside around 2 celsius degree (when outside are minus 18 grade Celsius).

I somehow have the feeling that you have not understood many of the (good) posts already received....

No matter how much insulation you use, without a source of heat inside, the room will cool down till it is the same temperature as outside.

The insulation will slow this process down, but it canot stop it!!!

i understand the most, maybe all the posts. the problem it was that i've tried to find a way for that calculation. and i've thought that there it is a thing about heating transfer which i don't know and it can help me.

i wanted to talk with a proffesor or with somebody wich really knows thermodinamics.

thank you for answer me.

Believe me, you have all the information already. Without a heat source, the room will get colder and colder till it is at the outside temperature.

Insulation slows things down, nothing more nothing less.....

To me1234, some of the the replies you have received are what physics professors (or people who "really know thermodinamics" [sic]) would say. Others, though they don't exactly answer your question, are what experienced engineers and builders would say. Sorry to say it, but you don't understand enough physics to even ask an answerable question. The others have pointed out the flaw in your question, repeatedly.

To the others, there is a possible heat source in the house, and that is a human being, which gives off about 100 watts of heat, IIRC. You could assume there is one human being in the house for maybe 14 hours a day.

Well put, but unless that "human body" is there naked, you will not get much heat from it in the room, it will be too well insulated clothed!!

(sigh) As I and others have pointed out repeatedly, insulation doesn't affect what the temperature will be ultimately, it only affects how long it will take to reach that temperature. True, if the occupant is absent for half the time, there might not be enough time for the occupant to provide much heat. Or the occupant might provide a significant amount of heat, even when clothed. You would have to the calculations to find out!

I will kept for the calculations, the variables constant.

So how can I calculate the room temperature ? Because I don't know...

See my last post.

The temperature will continue falling (with no heat source) till it reaches the outside temperature. The rate will slow the nearer it gets to the outside temperature, but it will not stop, unless you have a heat source....

I found the posts by nick name extremely helpful. To try to further clarify some nomenclature, I will add the following. k = thermal conductivity [W/(m·K)]; C = thermal conductance [W/(m²·K)], sometimes called heat transfer coefficient (h); R = thermal insulance (m²·K/W) = 1/C. By definition, C = k/L, where L = length of material in direction of heat transfer (in this case, wall thickness). Therefore, substituting and/or rearranging, we have k = L·C, and R = L/k.

It`s a little bit like asking:

"My mains voltage is 230V AC and my cable is 10 m long. How much current does my toaster draw ?"

You have to assume an amount of energy from your radiator inside in order to answer that question. If you dont spend this amount of energy then you have only to wait long enough and your temperature inside will (with a rough simplification) always be the same as the temperature outside - because, where else should come the difference from ?

When you switch of your refrigerator the inside temp will also become the outside temp.

Regards Uwe

Its going to be cold, I should imagine that sleeping in 32°F or 22°F or whatever is not going to be much fun.

If the outside of the house is unattractive and you like the look of the stones from the inside, insulate the roof and the whole outside of the house. It will take longer to warm up, but at night when the fire goes down, you will find that the temp goed down real slow as the walls give up their stored heat. Thats how I like it.....

If you want the outside to remain as it is, insulate the inside, but from then on, your rooms will be slightly smaller and hanging stuff on the walls in some cases almost impossible - cupboards and pictures. But it will warm up quicker with a fire going..... also cool down quicker....

Take your choice.....

Isolating from the inside has to be planned well - an often observed error occurs when the dew point suddenly lies inside the wall and leads to condense moisture then.

Correct but there are special plastic sheets that need to be sealed together to prevent any breathed out moist air getting to the cold walls.....something that many (most?) forget to do....even professionals!!!

I sealed up a room some years ago and had to tell the professional to shutup and use the sheets as I was paying, not him!!!! In places we had 10" of fiberglass insulation (uneven walls.....)

We have had zero problems......our heating costs went down dramatically......but I only hang pictures on those walls, nothing more......

Recently we insulated the whole house on the outside, much easier and I feel much netter.....you live and learn.....

Agreed with both of you

I was looking at how/where to insulate and seal and ran into a study on masonry wall retrofits (I believe former factory spaces) that explicitly discussed considerations on moisture.

Traditional adobe and stucco "breathes" and I was considering a solution that didn't, so pondered carefully, and decided to use a barrier to keep the "wall" dry.

And after much consideration, like you Andy, I decided if ever (project currently on hold) I would rather insulate outside and add more stucco to finish.

Since the construction was circa 1970, while I then wanted to control interior moisture with an air-to-air heat exchanger, practically I would have to use a typical bath-to-exterior fan initially.

With all this longwindedness, I would end up storing heat, but in a dry wall.

But this house has areas where a pencil can be inserted around doors, and cheap (although double-pane) windows, so the significance of air-to-air vs outlet fan will be a while and a lot more work, before it gets "out of the grass."

I might have written your post, the cellar had more air movement than outside on windy days before, but not anymore!!!!

"But it will warm up quicker with a fire going..... also cool down quicker...."

If insulation is increased it works as well for heating as for cooling the same, then why the difference? It warms quicker because less energy is lost to outside, since less energy is lost per time unit then it will cool SLOWER. Or am I wrong?

The energy balance equation is: accumulated energy = generated energy - lost energy

Accumulated energy = M*cp* dT M= mass inside kg, cp =specific heat (mean value) in J/kg; T inside temperature °K or °C

Lost energy = (T-Tout)*A/R *dt A= heat transfer area (walls, windows, aso) R= averaged thermal resistance °K*m²/w dt= element of time

Generated energy= H( watt)*dt

For Tout= constant (approximation) dT= d(T-Tout)

Equation becomes: M*cp*d(T-Tout)= dt*(H-A/R*(T-Tout)) it is a differential equation which leads to an exponential solution with a time constant τ=(R*M*cp)/A

If H>0 then there is a heating and if H=0 the temperature will drop but in both cases the time constant is the same. Or am I wrong?

With outside insulation, the stone walls take a lot of heat at first and store it. When the fire goes out overnight, this heat, because the outside is well insulated, can only leach back into the room...which is really nice.

When insulating on the inside, you do not have any sort of storage block, so it warms up and cools down quicker, but of course still much better economically than without insulation!! Usually, unwanted drafts have also been cut down on to assist further.....

I did not make myself clear at all before......sorry.

It is ok my comment was only in order to see if there was an analysis or only a "say so".

Of course with out side insulation the internal mass increases and the time constant as well. So that the time to reach a higher temperature inside increases too since not only the air has to we warmed but also the walls.

Its really onlly a "say so" but anyone with minor knowledge of physics will "cotton on" quickly I feel.....

if I will isolate the inside there will be a problem: the condens.

Only if you do it wrongly!!

As with all jobs, there is a correct way and a large number of incorrect ways.....take your pick!!

is it because I lack any formal high and technical education ,that I wonder why no one is mentioning the the question for the relation between the walls area and the inner volume?

As much as I know,the square/cube rate playes an important role in any thermodynamic calculations.

Hello az native,

Thats a fair remark. But it sounds as if the building has been built already, and is just awaiting the stucco?

bb

Mentioned in #8

Hello nick,

sorry for the duplication.

Take care......

bb