I was wondering since I have checked but not found how to calculate inertia in materials?
The simple part is calculating the k-value for walls or slabs, but I haven't found how to calculate the time perspective and how it's affected since some materials like steel conduct heat quickly while wood and gypsum conduct slowly, and for foam and mineral wool it's hard to find how quickly heat is removed through it.
Another aspect is how to calculate storage in materials like concrete, leca, or earth since I can only find information on calculating it to maintain an even temperature, and then basement floors and walls or slabs on the ground involve storing heat in many different materials, each with different inertia from wooden floors to concrete to insulation and then into the drainage material, and then the soil, and after a few years, one has heated the various layers to different extents.
So I'm wondering how to calculate it with a HEAVY structure since I've only found a little about light structures and then you're supposed to calculate average temperatures over periods between 100-300 hours depending on whether you have a cast slab or a crawl space, but I can't find anything on how to calculate it.
Because it's not enough to calculate how much energy is stored in the structure as a garage in leca 10x5x3m with 20cm thick walls and a cast slab of about 10cm will weigh around 23 tons, and the k-value for just the walls will be 1 and 90 sqm, so you lose heat very quickly making it cold (it takes around a few days to lose heat so you're only a few degrees more than the outdoor temperature is, but in reality, it holds heat much more than what one can calculate as stored in the materials, and if it’s insulated externally I can’t find how that should be calculated.
So does anyone have some insights on how to calculate this and how to calculate periods for isoras or the different insulation blocks available?
(hope it wasn't too confusing what I'm asking for)
The simple part is calculating the k-value for walls or slabs, but I haven't found how to calculate the time perspective and how it's affected since some materials like steel conduct heat quickly while wood and gypsum conduct slowly, and for foam and mineral wool it's hard to find how quickly heat is removed through it.
Another aspect is how to calculate storage in materials like concrete, leca, or earth since I can only find information on calculating it to maintain an even temperature, and then basement floors and walls or slabs on the ground involve storing heat in many different materials, each with different inertia from wooden floors to concrete to insulation and then into the drainage material, and then the soil, and after a few years, one has heated the various layers to different extents.
So I'm wondering how to calculate it with a HEAVY structure since I've only found a little about light structures and then you're supposed to calculate average temperatures over periods between 100-300 hours depending on whether you have a cast slab or a crawl space, but I can't find anything on how to calculate it.
Because it's not enough to calculate how much energy is stored in the structure as a garage in leca 10x5x3m with 20cm thick walls and a cast slab of about 10cm will weigh around 23 tons, and the k-value for just the walls will be 1 and 90 sqm, so you lose heat very quickly making it cold (it takes around a few days to lose heat so you're only a few degrees more than the outdoor temperature is, but in reality, it holds heat much more than what one can calculate as stored in the materials, and if it’s insulated externally I can’t find how that should be calculated.
So does anyone have some insights on how to calculate this and how to calculate periods for isoras or the different insulation blocks available?
(hope it wasn't too confusing what I'm asking for)
I assume you are interested in having a structure that retains heat for a very long time and want to quantify it?
Can you try to specify your problem, maybe it will work better?
Heavy structures even out the temperature in short periods, in the longer term it doesn't make much difference.
Can you try to specify your problem, maybe it will work better?
Heavy structures even out the temperature in short periods, in the longer term it doesn't make much difference.
I was trying to figure out how to calculate it since I can't find it in any books I have or anything online, as everything is about static differences and never anything about time perspective except a bit on Boverket where they have different temperatures depending on different frames. But the heaviest thing is having a slab on the ground in a wooden house, and then it was the average temperature over 300 hours you should calculate for areas.
And although Leca is not such good insulation, it's rare for a Leca garage to drop below zero if the roof is insulated and the door is not too bad (if it isn't opened too often). But if you calculate the energy stored in the frame, it should disappear much faster than it does in real life.
A brief example: 1 kg is about 1 KWs/degree, so 3.6 tons is 1 kWh, and if you have 36 tons, then that's 10 kWh. With a K-value of 1 on the walls, that's 100 sqm of wall, and a 10-degree difference means the loss through the wall is such that you'd lose 1 degree in 10 hours. If it's -20 degrees outside and you have 10 degrees inside (after heating), you'd lose about 1 degree every 3 hours, and after 2 days, it would be several degrees below zero there, but cooling doesn't happen so quickly.
If you have high humidity and a lot of wind, the temperature drops faster, and if you have a surface layer and an air gap that stays relatively still, it changes the K-value very little, but still, there is a greater difference as it resists cooling somewhat. The thing is, when you build with stone materials (Leca/lightweight concrete/concrete), you have inertia in the material that I don't think is relative to what you should calculate in the K-value. Even when you add insulation, it takes time for the energy to pass through; it doesn't happen in minutes, it takes time, but the question is how long that time is.
I'm a bit puzzled why I can't find anything about how quickly it cools down, as the slab and ground store large amounts of energy that take energy to maintain a high temperature, but as soon as you stop heating, the materials begin to release energy. But that's a different thing from air and water, where you have movement and high thermal conductivity that can release a large portion of energy quickly (cools fast). However, with foam/mineral wool, I can't find anything about how long it takes for the heat to get through to reach equilibrium, but it's just that you can't find an equilibrium since the outside temperature compared to inside will always differ, and the smaller the temperature difference, the slower the cooling goes.
So, is what I'm asking too confusing since I think someone should know this, and I hope someone here can.
And yes, it's mainly heavy frames and also airtight so there's no unintentional ventilation cooling more than necessary! Also, the ground here is a little bit down at +5.7C, and that means you're heating a construction with the ground, so the lower the temperature goes, the more the ground warms it, but we insulate away much of the contact with the ground, so you have to calculate just the construction itself.
I also hope my dyslexia doesn't make it impossible to understand what I'm asking for.
And although Leca is not such good insulation, it's rare for a Leca garage to drop below zero if the roof is insulated and the door is not too bad (if it isn't opened too often). But if you calculate the energy stored in the frame, it should disappear much faster than it does in real life.
A brief example: 1 kg is about 1 KWs/degree, so 3.6 tons is 1 kWh, and if you have 36 tons, then that's 10 kWh. With a K-value of 1 on the walls, that's 100 sqm of wall, and a 10-degree difference means the loss through the wall is such that you'd lose 1 degree in 10 hours. If it's -20 degrees outside and you have 10 degrees inside (after heating), you'd lose about 1 degree every 3 hours, and after 2 days, it would be several degrees below zero there, but cooling doesn't happen so quickly.
If you have high humidity and a lot of wind, the temperature drops faster, and if you have a surface layer and an air gap that stays relatively still, it changes the K-value very little, but still, there is a greater difference as it resists cooling somewhat. The thing is, when you build with stone materials (Leca/lightweight concrete/concrete), you have inertia in the material that I don't think is relative to what you should calculate in the K-value. Even when you add insulation, it takes time for the energy to pass through; it doesn't happen in minutes, it takes time, but the question is how long that time is.
I'm a bit puzzled why I can't find anything about how quickly it cools down, as the slab and ground store large amounts of energy that take energy to maintain a high temperature, but as soon as you stop heating, the materials begin to release energy. But that's a different thing from air and water, where you have movement and high thermal conductivity that can release a large portion of energy quickly (cools fast). However, with foam/mineral wool, I can't find anything about how long it takes for the heat to get through to reach equilibrium, but it's just that you can't find an equilibrium since the outside temperature compared to inside will always differ, and the smaller the temperature difference, the slower the cooling goes.
So, is what I'm asking too confusing since I think someone should know this, and I hope someone here can.
And yes, it's mainly heavy frames and also airtight so there's no unintentional ventilation cooling more than necessary! Also, the ground here is a little bit down at +5.7C, and that means you're heating a construction with the ground, so the lower the temperature goes, the more the ground warms it, but we insulate away much of the contact with the ground, so you have to calculate just the construction itself.
I also hope my dyslexia doesn't make it impossible to understand what I'm asking for.
What?? Now I'm not following?? To what?? Now it's getting confusing.
Leca itself is a dead material, and dead materials can store heat, but they need to be "charged" first. Are you building a garage and wondering what will be the most energy-efficient? If you want to use the walls to store heat, you should insulate them from the outside and cover them.
Now it's getting confusing for me too.
Don't take it the wrong way.
Leca itself is a dead material, and dead materials can store heat, but they need to be "charged" first. Are you building a garage and wondering what will be the most energy-efficient? If you want to use the walls to store heat, you should insulate them from the outside and cover them.
Now it's getting confusing for me too.
Don't take it the wrong way.
hoped it wouldn't get messy but sometimes it does when I write
the reason I took a garage is to have an example without a heat source where it is charged with the temperature the air has and has had for a while, a bit like how a stone garage never gets warm in the summer as it absorbs the heat from the air into the walls, so my thought was about how much inertia there is and also how to calculate it, with insulation on the outside acting as a barrier
and if you go to southern latitudes, they've built with stone and cement without any insulation at all and those houses are "cool" when it's +35C outside and not so cold when it's cold outside because the season you want to avoid the outdoor climate almost passes before the house reaches a balance compared to outside (and thus avoiding heating much in the winter and no need for cooling in the summer
and it's somewhat the same with an uninsulated basement that it takes energy to heat the ground, walls, and slab, but once you don't need to heat, you still have roughly the same temperature because there is tremendous inertia in the material so it takes quite a long time to heat up but also a long time to cool down, and then I can't reconcile this with the amount of energy stored because it should cool down and heat up heavy houses/buildings quickly
or is it that I calculated wrong on a unit so it should be 1kWh/degree and kg? :blushing:
I am curious about how they manage to factor in passive houses like isorast or whatever they are called as it must be that you have a few degrees warmer indoors in the summer and it is stored in the structure which then cools very slowly, providing good resistance against the cold when winter comes, as I can't reconcile it when calculating an R-value of 0.16 in walls and it should stay warm just by living there because, calculating that there are only two people and with energy-efficient things, I conclude it couldn't handle a temperature difference of more than 10 degrees and if you don't have heat through ventilation (heat pump heat exchanger) which is an auxiliary heat source that shouldn't be OK in my opinion
so it's both how to factor in such things and how long a building with a heavy structure can stand without heat and how to calculate the cooling then (if you are going to build something, it could be good to have a guideline for keeping water pipes etc. from freezing as one of many things
the reason I took a garage is to have an example without a heat source where it is charged with the temperature the air has and has had for a while, a bit like how a stone garage never gets warm in the summer as it absorbs the heat from the air into the walls, so my thought was about how much inertia there is and also how to calculate it, with insulation on the outside acting as a barrier
and if you go to southern latitudes, they've built with stone and cement without any insulation at all and those houses are "cool" when it's +35C outside and not so cold when it's cold outside because the season you want to avoid the outdoor climate almost passes before the house reaches a balance compared to outside (and thus avoiding heating much in the winter and no need for cooling in the summer
and it's somewhat the same with an uninsulated basement that it takes energy to heat the ground, walls, and slab, but once you don't need to heat, you still have roughly the same temperature because there is tremendous inertia in the material so it takes quite a long time to heat up but also a long time to cool down, and then I can't reconcile this with the amount of energy stored because it should cool down and heat up heavy houses/buildings quickly
or is it that I calculated wrong on a unit so it should be 1kWh/degree and kg? :blushing:
I am curious about how they manage to factor in passive houses like isorast or whatever they are called as it must be that you have a few degrees warmer indoors in the summer and it is stored in the structure which then cools very slowly, providing good resistance against the cold when winter comes, as I can't reconcile it when calculating an R-value of 0.16 in walls and it should stay warm just by living there because, calculating that there are only two people and with energy-efficient things, I conclude it couldn't handle a temperature difference of more than 10 degrees and if you don't have heat through ventilation (heat pump heat exchanger) which is an auxiliary heat source that shouldn't be OK in my opinion
so it's both how to factor in such things and how long a building with a heavy structure can stand without heat and how to calculate the cooling then (if you are going to build something, it could be good to have a guideline for keeping water pipes etc. from freezing as one of many things
I'm starting to understand what you mean. You're right that stone houses, if we can call them that, maintain cooler indoor temperatures in the summer. The stones absorb the heat that tries to enter. However, they lose the heat quite quickly, except for soapstone. I also find it hard to fully believe in passive houses without heating.
I know of a couple of houses outside Umeå that are built using the earth cellar principle, meaning that the house is buried in the ground and only one side of the house is above ground. These houses take advantage of the ground heat, which rarely falls below 5 degrees Celsius, and are very effective at keeping heating costs down.
HOWEVER, they do not manage solely on their own heat but supplement it through the use of ground source heat pumps and stoves. According to my opinion, these houses are as passive as a house can get in our latitudes, considering the number of degrees below zero during the winter months. But the advantage of these is, as I know from the gentleman who lives in one of them, he has what used to be a greenhouse in the visible wall, where he has vegetables year-round and that room in turn provides heat to the house.
With these houses in slightly more southern latitudes, they probably work excellently.
I know of a couple of houses outside Umeå that are built using the earth cellar principle, meaning that the house is buried in the ground and only one side of the house is above ground. These houses take advantage of the ground heat, which rarely falls below 5 degrees Celsius, and are very effective at keeping heating costs down.
HOWEVER, they do not manage solely on their own heat but supplement it through the use of ground source heat pumps and stoves. According to my opinion, these houses are as passive as a house can get in our latitudes, considering the number of degrees below zero during the winter months. But the advantage of these is, as I know from the gentleman who lives in one of them, he has what used to be a greenhouse in the visible wall, where he has vegetables year-round and that room in turn provides heat to the house.
With these houses in slightly more southern latitudes, they probably work excellently.
have they built with inspiration from Earthship?
And that's the thing, if you use soil, stone, concrete, and other heavy materials to store heat and then insulate as well as possible, you achieve a very stable climate. But now my thought is how to calculate it and if you can calculate the different factors like soil, how quickly it transfers heat depending on moisture and thickness, and the same with walls and floors, whether it's dependent on thickness or the heat capacity of the material, and if one should calculate in a similar way as with the U-value of a wall with different layers but with other numbers and units?
Because I would like to know how these parts should be calculated and how long it takes to release the energy which a few degrees still are.
And then it's the fact that all materials have different inertia and storage capacity, but I didn't think comparing it to a root cellar was right since it cools and warms through the evaporation of water in the floor and walls. However, the fact that groundwater usually has a relatively stable temperature is one thing, but it requires a certain airflow to evaporate the right amount of water to achieve the right temperature there. But if you go by the fact that a bit down the ground has a stable temp, it warms to 5-6C, and then you only need to heat up from that level instead of from -20C as it can get even here, so there's a 25C difference that one needs to heat some areas. But windows and such don't have the same resistance but are better with just them than walls, roofs, and floors.
And do you or anyone else reading this have knowledge of how to calculate these parts?
And that's the thing, if you use soil, stone, concrete, and other heavy materials to store heat and then insulate as well as possible, you achieve a very stable climate. But now my thought is how to calculate it and if you can calculate the different factors like soil, how quickly it transfers heat depending on moisture and thickness, and the same with walls and floors, whether it's dependent on thickness or the heat capacity of the material, and if one should calculate in a similar way as with the U-value of a wall with different layers but with other numbers and units?
Because I would like to know how these parts should be calculated and how long it takes to release the energy which a few degrees still are.
And then it's the fact that all materials have different inertia and storage capacity, but I didn't think comparing it to a root cellar was right since it cools and warms through the evaporation of water in the floor and walls. However, the fact that groundwater usually has a relatively stable temperature is one thing, but it requires a certain airflow to evaporate the right amount of water to achieve the right temperature there. But if you go by the fact that a bit down the ground has a stable temp, it warms to 5-6C, and then you only need to heat up from that level instead of from -20C as it can get even here, so there's a 25C difference that one needs to heat some areas. But windows and such don't have the same resistance but are better with just them than walls, roofs, and floors.
And do you or anyone else reading this have knowledge of how to calculate these parts?
Did I mess it up too much, or is there no one who knows how to calculate with inertia? I can calculate the other parts, but it feels unreasonable with what I come up with when I scribble on an example.
(I've only learned to calculate the k-value in a few different ways, but that's to maintain an even temperature and then you usually calculate based on annual average temperature or average temperature over 4-13 days, or you calculate ventilation, and then it's different methods. But when it comes to calculating how quickly/slowly a house cools down (loses indoor temperature), I can't find anything about it.
Because everything can be calculated, the question is just how, and someone must have knowledge about this, I guess??)
(I've only learned to calculate the k-value in a few different ways, but that's to maintain an even temperature and then you usually calculate based on annual average temperature or average temperature over 4-13 days, or you calculate ventilation, and then it's different methods. But when it comes to calculating how quickly/slowly a house cools down (loses indoor temperature), I can't find anything about it.
Because everything can be calculated, the question is just how, and someone must have knowledge about this, I guess??)
If we take a simple example, a hollow cube of concrete.
Each side, l: 3m
Thickness, d: 0.5m
External area, A = 6 * 3 * 3 = 54 m^2
Internal volume, V = (3 - 2 * 0.5 )^3 = 8 m^3
Density of air, r = 1.293 kg/m^3
Weight of air in the cube, m = 8 * 1.293 = 10.344kg air
Heat capacity of air, c: 1.012 kJ/kg * K
Thermal conductivity of concrete, K = 1.7 W / m * K
Warm side: +20 degrees C
Cold side: -20 degrees C
Temperature difference, dT = 20-(-20) = 40 degrees C
Time, t = 1s
Heat flow Q = K * (A * dT * t) / d
Q = 1.7 * (54 * 40 * 1) / 0.5 = 7344 W
***To lower the temperature of the air by a number of degrees***
Q = c * m * K
K= Q / (c * m)
K = 7344W / (1012 J/kg K * 10.344kg) = 0.7015... ~0.70 degrees lower after 1 second
*** New conditions ***
Warm side: (+20 - 0.7) = +19.3 degrees C
Cold side: -20 degrees C
Temperature difference, dT = +19.3-(-20)=39.3 degrees C
Heat flow Q = K * (A * dT * t) / d
Q = 1.7 * (54 * 39.3 * 1) / 0.5 = 7215.48 W
Temperature decrease:
K = 7215.48W / (1012 J/kg K * 10.344kg) = 0.689547... ~0.69 degrees lower during the 2nd second
*** New conditions ***
Warm side: (+19.3 - 0.69) = 18.61 degrees C
Cold side: -20 degrees C
Temperature difference, dT = +18.61-(-20)=38.61 degrees C
And so on... if you're really, really, really bored you can keep calculating…
But there are other ways to calculate.. just showing the principle as I thought of it…
But it might be completely wrong, I leave no guarantees, it's late.
Can't someone who really knows tell me how it is!?
And there must be a plethora of other parameters to consider if you want to approach reality?
Spontaneously, if I'm to be critical of my own post, I think a 0.7-degree drop is really quick, feels like I thought wrong.. but oh well..
Each side, l: 3m
Thickness, d: 0.5m
External area, A = 6 * 3 * 3 = 54 m^2
Internal volume, V = (3 - 2 * 0.5 )^3 = 8 m^3
Density of air, r = 1.293 kg/m^3
Weight of air in the cube, m = 8 * 1.293 = 10.344kg air
Heat capacity of air, c: 1.012 kJ/kg * K
Thermal conductivity of concrete, K = 1.7 W / m * K
Warm side: +20 degrees C
Cold side: -20 degrees C
Temperature difference, dT = 20-(-20) = 40 degrees C
Time, t = 1s
Heat flow Q = K * (A * dT * t) / d
Q = 1.7 * (54 * 40 * 1) / 0.5 = 7344 W
***To lower the temperature of the air by a number of degrees***
Q = c * m * K
K= Q / (c * m)
K = 7344W / (1012 J/kg K * 10.344kg) = 0.7015... ~0.70 degrees lower after 1 second
*** New conditions ***
Warm side: (+20 - 0.7) = +19.3 degrees C
Cold side: -20 degrees C
Temperature difference, dT = +19.3-(-20)=39.3 degrees C
Heat flow Q = K * (A * dT * t) / d
Q = 1.7 * (54 * 39.3 * 1) / 0.5 = 7215.48 W
Temperature decrease:
K = 7215.48W / (1012 J/kg K * 10.344kg) = 0.689547... ~0.69 degrees lower during the 2nd second
*** New conditions ***
Warm side: (+19.3 - 0.69) = 18.61 degrees C
Cold side: -20 degrees C
Temperature difference, dT = +18.61-(-20)=38.61 degrees C
And so on... if you're really, really, really bored you can keep calculating…
But there are other ways to calculate.. just showing the principle as I thought of it…
But it might be completely wrong, I leave no guarantees, it's late.
Can't someone who really knows tell me how it is!?
And there must be a plethora of other parameters to consider if you want to approach reality?
Spontaneously, if I'm to be critical of my own post, I think a 0.7-degree drop is really quick, feels like I thought wrong.. but oh well..
In your example, it talks about less than a minute before it reaches freezing point in the cube, but now you estimate that all the energy available is in the air. However, if you relate that, you have 10kg of air but the cube is 45 tonnes of concrete, and we have approximately 1kJ/K/kg there, so with a difference of 40 degrees it should emit 1.8MJ or 500kWh. And if you take into account that the temperature drops to 0 degrees during that time, it should therefore lose 250kWh.
But it must also depend on the thickness and not just the K-value, I think.
So something that makes it easier to calculate is that when the air is relatively still, it doesn't cool down as quickly either, instead there will be a cold air fall by the wall, and the rest of the air will be quite warm when the warmest air is in the middle of the room.
But it's a thought I think sounds reasonable
But when I searched around a bit, I found a table with the unit Ws^0.5/m^2 K 10^3
and a size of 1857 for concrete and 37 for lightweight concrete and 0.8 for mineral wool.
Is it the root of the energy stored, or how should that be calculated?
But it must also depend on the thickness and not just the K-value, I think.
So something that makes it easier to calculate is that when the air is relatively still, it doesn't cool down as quickly either, instead there will be a cold air fall by the wall, and the rest of the air will be quite warm when the warmest air is in the middle of the room.
But it's a thought I think sounds reasonable
But when I searched around a bit, I found a table with the unit Ws^0.5/m^2 K 10^3
and a size of 1857 for concrete and 37 for lightweight concrete and 0.8 for mineral wool.
Is it the root of the energy stored, or how should that be calculated?
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Yes, if one assumes the entire volume of the cube is +20 degrees C, and the ambient temperature is +20 degrees C. Then the ambient temperature momentarily drops to -20 degrees C.
First, the stored energy in the concrete volume will be extracted into the surroundings. As soon as that happens, the temperature of the entire concrete volume will have decreased slightly, and energy stored in the air inside will be used, though I think this will happen much slower than my previous number-crunching... my number-crunching probably doesn't match at all.
First, the stored energy in the concrete volume will be extracted into the surroundings. As soon as that happens, the temperature of the entire concrete volume will have decreased slightly, and energy stored in the air inside will be used, though I think this will happen much slower than my previous number-crunching... my number-crunching probably doesn't match at all.
The air will move due to convection that occurs when the air closest to the concrete wall cools down.
The K-value, or U-value, is dimension-dependent on the thickness of the wall. Thermal resistance = Share * Thickness * Lambda U(K)-value = 1/Thermal resistance
find a page that provided some answers but then it also states that the calculations needed are too complicated to be done by hand and instead they refer to some software
the page is
http://www.betongvaruindustrin.se/sv/Bygga-med-prefab/?Chapter=148
if more people want to read about it and more opinions or links are welcome but not as necessary anymore :blushing:
the page is
http://www.betongvaruindustrin.se/sv/Bygga-med-prefab/?Chapter=148
if more people want to read about it and more opinions or links are welcome but not as necessary anymore :blushing:
When it comes to solid materials themselves, from a macroscopic perspective, they only have three properties that affect heat transfer: Thermal conductivity, Heat capacity, and density. None of the three parameters are constants, but vary with temperature. This usually makes the calculations a bit more complex. Of course, dimensions are then added, but they are not a material property. Additionally, the material can absorb and release moisture, which takes up or releases heat.
The transport of heat to and from the material can occur through convection, conduction, and thermal radiation. In the case of building materials, convection is likely to dominate. The calculations can become quite complex for laypeople, but the subject is interesting.
According to the prevailing terminology, all building materials are dead. If there is life in the walls, one should call Anticimex or equivalent. Even wood is a dead material the moment the tree is cut down. It is a misconception that the material is alive because it moves as it equilibrates with the surrounding humidity.
The disciplines of heat transfer, fluid mechanics, and mass transport are usually summarized under the concept of "Transport Phenomena." There are several textbooks on the subject. Here is one:
http://www.amazon.co.uk/Transport-P...=sr_1_1?ie=UTF8&s=books&qid=1261978704&sr=1-1
The transport of heat to and from the material can occur through convection, conduction, and thermal radiation. In the case of building materials, convection is likely to dominate. The calculations can become quite complex for laypeople, but the subject is interesting.
According to the prevailing terminology, all building materials are dead. If there is life in the walls, one should call Anticimex or equivalent. Even wood is a dead material the moment the tree is cut down. It is a misconception that the material is alive because it moves as it equilibrates with the surrounding humidity.
The disciplines of heat transfer, fluid mechanics, and mass transport are usually summarized under the concept of "Transport Phenomena." There are several textbooks on the subject. Here is one:
http://www.amazon.co.uk/Transport-P...=sr_1_1?ie=UTF8&s=books&qid=1261978704&sr=1-1
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mattiasp
What I have found is a bit too much to calculate by hand, and an approximation that gives a hint of where you might end up would be good. Could one calculate with mass and take an average temperature of the different masses (now I'm thinking of solid walls where you have whole layers and not framed walls or similar)?
For example, an outer wall with panels, air gap, foam plastic, leca, cement/plaster.
1/x -> 0.15+(0.035/0.2)+(0.2/0.2)+(1.7/0.05) = 0.15+5.7+1+0.03=6.88 > k=0.145
Dt =27.5C for simplicity in the example, so the temperature difference in each layer should be double what was used for the K-value.
Thus, the temperature drops 0.12C in the cement at the innermost and 4C in the leca, then 22.7 degrees in the foam plastic, then the panel with an air gap where there's a bit of resistance in the transition there so, as I recall, it becomes 0.6 degrees even there in resistance.
If you say the indoor temperature is 22 degrees, then the cement in the wall should be 22C, and the leca on average +20C (so you would calculate the mass there, how much energy is stored, and then you can ignore the storage in the rest of the wall since the foam plastic doesn't store significant energy, or is this line of thinking too simplistic to set it up this way to get a reference value?
Then there's also an incorrect side where you only calculate that ALL energy goes out through walls/floors/ceilings and not anything that's ventilated out, or that windows and doors lower the indoor temperature so that the wall is cooled from the inside when the building is just standing:
But then you ignore the inertia entirely and calculate pure energy instead, as it didn't seem to be so inert to get the temperature to migrate through a material as I thought initially.
And one aspect I've completely ignored is how moisture is stored in the material and how it, in turn, takes energy when it's supposed to evaporate, as we usually assume that the indoor air is dry and that moisture is not a hindrance in such calculations as long as we maintain an equilibrium, which was what I wanted to step away from to understand how others have come up with different calculations (especially all those claiming about passive houses).
Or can anyone help me understand how, in a house with ventilation at 150-200l/s when it's -20C outside and 60% in heat recovery (Dt 40C * 0.2kg * 0.4 = 3.2kW), which only the controlled ventilation takes, then the uncontrolled perhaps 1-2kW, and if you then add the k-value of 0.15 as an average on the whole house of 300sqm (floor excluded since it doesn't have surface contact), (0.15*300*40C =1.8kW + slab about 150*0.15*10=225W about 2-2.1kW).
So if you add it up, it becomes around 6-7kW that the house requires when it's cold for a few days, and I estimate 4 people + everything generating heat around 1-2kW (depending on how you handle lighting, whether you have energy-consuming or conserving items in the home), so it would lead to when it's cold as it has been for over a month now, the stored heat in the house compensates so that the house needs about 5 times as much energy than what's produced in the house and with today's regulations the difference is so big that it's borderline whether you can have such a powerful electric heater in the house if it's 150sqm which would be needed for heating and hot water (speculating a bit when I'm not completely clear-headed at this time).
But I can't figure out how to get a house that has reasonably good ventilation. If it's Completely sealed and you're not allowed to open any doors or have any air vents open or so, then you can get it down a bit, but surely they can't count on turning off ventilation completely so that the heat required to keep the house warm is just what goes out through walls and roof and floor? Because then you have to count on suffocating the people who live there so they won't lower the temperature too much to live there
Furthermore, I don't know what they consider "normal" energy consumption in a house now (household electricity that is), because I think that 10-30kWh/day sounds like a reasonable figure, but if you say 24kWh, it's just 1kW + those who are there and if the house stands empty most of the day or there's only one person living there then it doesn't heat up much when it's cold.
So I wonder how many will regret deciding to build a passive house now that it has been cold for a while, or if more will build passive, and if winters become even colder than the last month, then I think it will not be so demanded, and I haven’t found, but are there any benefits to building a passive house? like cheaper loans from the state or that you'll never have to pay property tax if it comes back or something else rewarding, or is it just a trend existing right now? (I can understand wanting to build as energy-efficiently as possible but we live in a country where it gets cold for one or a few months a year and then you should have heating in the houses, even if you only need to use it 1-4 months a year, it should be there from the start and not that you must do a solution afterwards that’s not as good and requires more energy than a geothermal heat pump or a furnace or a fireplace)
What I have found is a bit too much to calculate by hand, and an approximation that gives a hint of where you might end up would be good. Could one calculate with mass and take an average temperature of the different masses (now I'm thinking of solid walls where you have whole layers and not framed walls or similar)?
For example, an outer wall with panels, air gap, foam plastic, leca, cement/plaster.
1/x -> 0.15+(0.035/0.2)+(0.2/0.2)+(1.7/0.05) = 0.15+5.7+1+0.03=6.88 > k=0.145
Dt =27.5C for simplicity in the example, so the temperature difference in each layer should be double what was used for the K-value.
Thus, the temperature drops 0.12C in the cement at the innermost and 4C in the leca, then 22.7 degrees in the foam plastic, then the panel with an air gap where there's a bit of resistance in the transition there so, as I recall, it becomes 0.6 degrees even there in resistance.
If you say the indoor temperature is 22 degrees, then the cement in the wall should be 22C, and the leca on average +20C (so you would calculate the mass there, how much energy is stored, and then you can ignore the storage in the rest of the wall since the foam plastic doesn't store significant energy, or is this line of thinking too simplistic to set it up this way to get a reference value?
Then there's also an incorrect side where you only calculate that ALL energy goes out through walls/floors/ceilings and not anything that's ventilated out, or that windows and doors lower the indoor temperature so that the wall is cooled from the inside when the building is just standing:
But then you ignore the inertia entirely and calculate pure energy instead, as it didn't seem to be so inert to get the temperature to migrate through a material as I thought initially.
And one aspect I've completely ignored is how moisture is stored in the material and how it, in turn, takes energy when it's supposed to evaporate, as we usually assume that the indoor air is dry and that moisture is not a hindrance in such calculations as long as we maintain an equilibrium, which was what I wanted to step away from to understand how others have come up with different calculations (especially all those claiming about passive houses).
Or can anyone help me understand how, in a house with ventilation at 150-200l/s when it's -20C outside and 60% in heat recovery (Dt 40C * 0.2kg * 0.4 = 3.2kW), which only the controlled ventilation takes, then the uncontrolled perhaps 1-2kW, and if you then add the k-value of 0.15 as an average on the whole house of 300sqm (floor excluded since it doesn't have surface contact), (0.15*300*40C =1.8kW + slab about 150*0.15*10=225W about 2-2.1kW).
So if you add it up, it becomes around 6-7kW that the house requires when it's cold for a few days, and I estimate 4 people + everything generating heat around 1-2kW (depending on how you handle lighting, whether you have energy-consuming or conserving items in the home), so it would lead to when it's cold as it has been for over a month now, the stored heat in the house compensates so that the house needs about 5 times as much energy than what's produced in the house and with today's regulations the difference is so big that it's borderline whether you can have such a powerful electric heater in the house if it's 150sqm which would be needed for heating and hot water (speculating a bit when I'm not completely clear-headed at this time).
But I can't figure out how to get a house that has reasonably good ventilation. If it's Completely sealed and you're not allowed to open any doors or have any air vents open or so, then you can get it down a bit, but surely they can't count on turning off ventilation completely so that the heat required to keep the house warm is just what goes out through walls and roof and floor? Because then you have to count on suffocating the people who live there so they won't lower the temperature too much to live there
Furthermore, I don't know what they consider "normal" energy consumption in a house now (household electricity that is), because I think that 10-30kWh/day sounds like a reasonable figure, but if you say 24kWh, it's just 1kW + those who are there and if the house stands empty most of the day or there's only one person living there then it doesn't heat up much when it's cold.
So I wonder how many will regret deciding to build a passive house now that it has been cold for a while, or if more will build passive, and if winters become even colder than the last month, then I think it will not be so demanded, and I haven’t found, but are there any benefits to building a passive house? like cheaper loans from the state or that you'll never have to pay property tax if it comes back or something else rewarding, or is it just a trend existing right now? (I can understand wanting to build as energy-efficiently as possible but we live in a country where it gets cold for one or a few months a year and then you should have heating in the houses, even if you only need to use it 1-4 months a year, it should be there from the start and not that you must do a solution afterwards that’s not as good and requires more energy than a geothermal heat pump or a furnace or a fireplace)