IN Lately There are very heated discussions about wall insulation. Some advise insulating, others consider it economically unjustified. It is difficult for an ordinary developer who does not have special knowledge of thermal physics to understand all this. On the one side warm walls are associated with lower heating costs. On the other hand, the “price of the issue” is that warm walls will cost more for the developer.
Why do you need a wall thermal conductivity calculator?
In each individual case, you should consider the required thickness of thermal insulation material for the walls of your house and calculate how much you will save on heating after heating and how long it will take for you to pay for the purchased materials and all the work. We have selected the most convenient and understandable services for calculations required thickness thermal insulation material.
Thermal calculator. Calculation of the dew point in the wall
The online calculator from smartcalc.ru will allow you to calculate optimal thickness insulation for the walls of the house and living quarters. You can calculate the thickness of thermal insulation and calculate the dew point when insulating a house various materials. The smartcalc.ru calculator allows you to clearly see the location of condensation in the wall. This is the most convenient thermal calculator for calculating insulation and dew point.
Insulation thickness calculator for walls, ceilings, floors
Using this calculator, you can calculate the thickness of insulation for walls, roofs, ceilings of a house and other building structures in accordance with the region of your residence, the material and thickness of the walls, as well as other important parameters for thermal insulation. By selecting different thermal insulation materials using a calculator, you can find the optimal insulation thickness for the walls of your home.
KNAUF calculator. Calculation of thermal insulation thickness
This calculator allows you to calculate the thickness of thermal insulation of walls in the main cities of the Russian Federation in various designs on the KNAUF thermal calculator, created by professionals from KNAUF Insulation. All calculations are made according to the requirements of SNiP 02/23/2003 “ Thermal protection buildings." Free online calculator for calculating KNAUF thermal insulation, the service has a convenient and intuitive interface.
Rockwool calculator for calculating the thickness of wall insulation
The calculator was developed by Rockwool specialists to help calculate the required thermal insulation thickness and estimate economic efficiency its installation. Perform thermal engineering calculations, select a suitable brand of thermal insulation and calculate required amount packs of mineral wool are very simple.
How to remove dew point from a wall when insulating
The calculator allows you to determine the type of heat insulating materials for the foundation, calculate the volume necessary materials and get the final cost, including fasteners for the slabs.
Calculator for calculating and choosing insulation for siding.
Using this service, you can determine the types of thermal insulation and waterproofing that are suitable for insulating walls under siding. Moreover, the calculator will allow you to determine the cost and calculate the volume of required materials.
Calculator for calculating thermal insulation for a ventilated facade
In order to choose the right materials for insulating a ventilated facade, select waterproofing and fasteners, use this service. By entering the area of the walls and the thickness of the slabs, you will calculate the required volume of materials and find out their cost.
Online calculator for calculating the cost of a plaster facade.
The service allows you to determine the types of materials, cost and volume. Based on the area of the facade and the thickness of the insulation, you can calculate the approximate cost of a plaster facade.
Calculation of materials for insulating frame walls
If you are faced with a task, isolation frame walls, then this calculator is for you. Knowing the area of the walls and the thickness of the insulation, you can easily calculate the necessary materials.
Calculation of materials for indoor insulation
Online calculation of insulation for floors under screed
For a floor that is planned to be made using cement or any other, special, durable insulating materials are required.
Online calculation of floor insulation by joists
To choose the right insulating materials for the floor, which is laid on wooden joists, use this calculator. He will determine the required density of materials, their quantity and approximate cost.
Calculation of thermal insulation for interior partitions
Choose insulation for interior partitions. You will be able to calculate the quantity and type of insulation, its cost, and also immediately make an application.
Ceiling insulation calculator
Just enter the ceiling area and thermal insulation thickness, get the quantity of materials and their cost.
Determine the cost of materials for insulating interfloor ceilings
To solve such problems, use the online calculation of prices and quantities of required materials.
Calculation of materials for roof insulation
Online attic insulation calculation
To insulate the attic, you should select materials using this service.
Calculation of insulation for a pitched roof (attic)
Insulation pitched roof, requires, in addition to insulation, a vapor barrier and wind and moisture protection membrane; using this online calculator, you can easily determine the materials you need and their estimated cost.
Calculation of insulation for a flat roof
To calculate materials for flat roof, we suggest using this calculator. Also included in the calculation waterproofing membrane and telescopic fasteners.
Gutter calculation calculator
The calculator will allow you to make a preliminary calculation of the necessary materials for installation drainage system. Preliminarily determine the cost/
Correct calculation of thermal insulation will increase the comfort of your home and reduce heating costs. During construction you cannot do without insulation, whose thickness determined by the climatic conditions of the region and the materials used. For insulation, polystyrene foam, penoplex, mineral wool or ecowool, as well as plaster and others are used. Decoration Materials.
To calculate what thickness the insulation should have, you need to know the minimum thermal resistance value. It depends on the climate. When calculating it, the duration of the heating period and the difference between internal and external (average for the same time) temperatures are taken into account. So, for Moscow, the heat transfer resistance for the external walls of a residential building must be no less than 3.28, in Sochi 1.79 is sufficient, and in Yakutsk 5.28 is required.
The thermal resistance of a wall is defined as the sum of the resistance of all layers of the structure, load-bearing and insulating. That's why The thickness of the thermal insulation depends on the material from which the wall is made. For brick and concrete walls More insulation is required, less for wooden and foam blocks. Please note what thickness is selected for load-bearing structures material, and what is its thermal conductivity. The thinner the supporting structures, the greater the thickness of the insulation should be.
If insulation is required large thickness, it is better to insulate the house from the outside. This will provide savings internal space. In addition, external insulation avoids the accumulation of moisture indoors.
Thermal conductivity
The ability of a material to transmit heat is determined by its thermal conductivity. Wood, brick, concrete, foam blocks conduct heat differently. High humidity air increases thermal conductivity. The inverse of thermal conductivity is called thermal resistance. To calculate it, the value of thermal conductivity in a dry state is used, which is indicated in the passport of the material used. You can also find it in tables.
However, it must be taken into account that in corners, joints of load-bearing structures and other special elements of the structure, thermal conductivity is higher than on a flat surface of the walls. “Cold bridges” may arise through which heat will escape from the house. The walls in these places will sweat. To prevent this, the thermal resistance value in such places is increased by about a quarter compared to the minimum allowable.
Example calculation
It is not difficult to calculate the thickness of thermal insulation using a simple calculator. To do this, first calculate the heat transfer resistance for the supporting structure. The thickness of the structure is divided by the thermal conductivity of the material used. For example, foam concrete with a density of 300 has a thermal conductivity coefficient of 0.29. With a block thickness of 0.3 meters, the thermal resistance value is:
The calculated value is subtracted from the minimum allowable value. For Moscow conditions, insulating layers must have a resistance of no less than:
Then, multiplying the thermal conductivity coefficient of the insulation by the required thermal resistance, we obtain the required layer thickness. For example, for mineral wool with a thermal conductivity coefficient of 0.045, the thickness should be no less than:
0.045*2.25=0.1 m
In addition to thermal resistance, the location of the dew point is taken into account. The dew point is the point in the wall where the temperature can drop enough to cause condensation - dew. If this place ends up on the inner surface of the wall, it fogs up and a putrefactive process may begin. The colder it is outside, the closer to the room the dew point moves. The warmer and more humid the room, the higher the dew point temperature.
Insulation thickness in a frame house
As insulation for frame house Most often they choose mineral wool or ecowool.
The required thickness is determined using the same formulas as in traditional construction. Additional layers multilayer wall give approximately 10% of its value. The thickness of the wall of a frame house is less than with traditional technology, and the dew point may be closer to the inner surface. That's why There is no point in saving unnecessarily on the thickness of the insulation.
How to calculate the thickness of roof and attic insulation
The formulas for calculating resistance for roofs use the same, but the minimum thermal resistance in this case is slightly higher. Unheated attics covered with bulk insulation. There are no restrictions on thickness here, so it is recommended to increase it by 1.5 times relative to the calculated one. IN attic rooms For roof insulation, materials with low thermal conductivity are used.
How to calculate the thickness of floor insulation
Although the greatest heat loss occurs through the walls and roof, it is equally important to correctly calculate the insulation of the floor. If the base and foundation are not insulated, it is assumed that the temperature in the underground is equal to the outside temperature, and the thickness of the insulation is calculated in the same way as for external walls. If some insulation of the base is done, its resistance is subtracted from the minimum required thermal resistance for the region of construction.
Calculation of foam thickness
The popularity of polystyrene foam is determined by its low cost, low thermal conductivity, light weight and moisture resistance. Polystyrene foam almost does not allow steam to pass through, so it cannot be used for internal insulation. It is located outside or in the middle of the wall.
The thermal conductivity of polystyrene foam, like other materials, depends on density. For example, at a density of 20 kg/m3 the thermal conductivity coefficient is about 0.035. Therefore, a foam thickness of 0.05 m will provide a thermal resistance of 1.5.
Even the now popular cottages made of logs or profiled beams need to be additionally insulated or built from materials that are practically non-existent on the market. solid wood 35-40 cm thick. What can we say about stone buildings (block, brick, monolithic).
What does it mean to “insulate properly”
So, it is impossible to do without thermal insulation layers, the vast majority of homeowners will agree with this. Some of them have to study the issue while building their own nest, others are puzzled by insulation so that facade works improve an already used cottage. In any case, the issue must be approached very scrupulously.
Compliance with insulation technology is one thing, but developers often make mistakes at the stage of purchasing material, in particular, they incorrectly choose the thickness of the insulation layer. If the home turns out to be too cold, then being in it will be, to put it mildly, uncomfortable. If circumstances are favorable (there is a reserve of heat generator performance), the problem can be solved by increasing the power heating system, which clearly entails a significant increase in costs for the purchase of energy resources.
But usually everything ends much worse: with a small thickness of the insulating layer, the enclosing structures freeze. And this causes the dew point to move inside the premises, which causes condensation to form on the internal surfaces of walls and ceilings. Then mold appears and they are destroyed. building construction and finishing materials... What is most unpleasant is the fact that it is impossible to eliminate troubles with little expense. For example, on the facade it will be necessary to dismantle (or “bury”) finishing layer, then create another barrier of insulation, and then finish the walls again. It turns out to be very expensive, it is better to do everything right away as it should be.
Important! Technological modern insulation materials They will not cost much, and as the thickness increases, the price will increase proportionally. Therefore, there is usually no point in creating too large a reserve of thermal insulation; it is a waste of money, especially if only part of the house’s structures are accidentally over-insulated.
Principles for calculating the insulating layer
Thermal conductivity and thermal resistance
First of all, you need to decide main reason cooling the building. In winter, we have a heating system that warms the air, but the generated heat passes through the building envelope and is dissipated into the atmosphere. That is, heat loss occurs - “heat transfer”. It is always there, the only question is whether it is possible to replenish them through heating, so that the house remains at a stable positive temperature, preferably at + 20-22 degrees.
Important! Note that a very important role in the dynamics of the heat balance (in total heat loss) play various leaks in the building elements - infiltration. Therefore, you should also pay attention to tightness and drafts.
Brick, steel, concrete, glass, wooden beam... - every material used in the construction of buildings, to one degree or another, has the ability to transmit thermal energy. And each of them has the opposite ability - to resist heat transfer. Thermal conductivity is a constant value, therefore in the SI system there is an indicator “thermal conductivity coefficient” for each material. These data are important not only for understanding physical properties structures, but also for subsequent calculations.
We present data for some basic materials in the form of a table.
Now about heat transfer resistance. The value of heat transfer resistance is inversely proportional to thermal conductivity. This indicator applies to both enclosing structures and materials as such. It is used to characterize thermal insulation characteristics walls, ceilings, windows, doors, roofing...
For calculation thermal resistance use the following publicly available formula:
The indicator “d” here means the thickness of the layer, and the indicator “k” is the thermal conductivity of the material. It turns out that heat transfer resistance directly depends on the massiveness of materials and enclosing structures, which, when using several tables, will help us calculate the actual thermal resistance existing wall or the correct insulation thickness.
For example: a half-brick (solid) wall has a thickness of 120 mm, that is, the R value will be 0.17 m² K/W (thickness 0.12 meters divided by 0.7 W/(m*K)). Similar masonry in a brick (250 mm) will show 0.36 m²·K/W, and in two bricks (510 mm) - 0.72 m²·K/W.
Let’s say, for mineral wool with a thickness of 50; 100; 150 mm thermal resistance indicators will be as follows: 1.11; 2.22; 3.33 m²·K/W.
Important! Most building envelopes in modern buildings are multilayer. Therefore, in order to calculate, for example, the thermal resistance of such a wall, you need to separately consider all its layers, and then sum up the resulting indicators.
Are there any thermal resistance requirements?
The question arises: what should the heat transfer resistance indicator be for the building envelope so that the rooms are warm and a minimum of energy is consumed during the heating period? Luckily for homeowners, it doesn't have to be reused. complex formulas. All necessary information is in SNiP 02/23/2003 “Thermal protection of buildings”. In this regulatory document buildings are being considered for various purposes, operated in various climatic zones. This is understandable, since the temperature for residential premises and production premises you don't need the same one. Besides, individual regions characterized by their limiting sub-zero temperatures and the duration of the heating season, therefore, such an average characteristic as degree-days of the heating season is distinguished.
Important! Another interesting point is that the main table we are interested in contains standardized indicators for various enclosing structures. In general, this is not surprising, because heat leaves the house unevenly.
Let's try to simplify the table a little regarding the required thermal resistance, here's what we get for residential buildings (m² K/W):
According to this table, it becomes clear that if in Moscow (5800 degree-days at average temperature indoor temperature is about 24 degrees) if you build a house only from solid brick, then the wall will have to be made more than 2.4 meters thick (3.5 X 0.7). Is this technically and financially feasible? Of course it's absurd. That's why you need to use insulating material.
Obviously, for a cottage in Moscow, Krasnodar and Khabarovsk there will be different requirements. All we need is to determine the degree-daily indicators for our settlement and select the appropriate number from the table. Then, using the heat transfer resistance formula, we work with the equation and obtain the optimal thickness of the insulation that needs to be applied.
| City | Degree-day Dd of the heating period at temperature, + C | |||||
| 24 | 22 | 20 | 18 | 16 | 14 | |
| Abakan | 7300 | 6800 | 6400 | 5900 | 5500 | 5000 |
| Anadyr | 10700 | 10100 | 9500 | 8900 | 8200 | 7600 |
| Arzanas | 6200 | 5800 | 5300 | 4900 | 4500 | 4000 |
| Arkhangelsk | 7200 | 6700 | 6200 | 5700 | 5200 | 4700 |
| Astrakhan | 4200 | 3900 | 3500 | 3200 | 2900 | 2500 |
| Achinsk | 7500 | 7000 | 6500 | 6100 | 5600 | 5100 |
| Belgorod | 4900 | 4600 | 4200 | 3800 | 3400 | 3000 |
| Berezovo (KhMAO) | 9000 | 8500 | 7900 | 7400 | 6900 | 6300 |
| Biysk | 7100 | 6600 | 6200 | 5700 | 5300 | 4800 |
| Birobidzhan | 7500 | 7100 | 6700 | 6200 | 5800 | 5300 |
| Blagoveshchensk | 7500 | 7100 | 6700 | 6200 | 5800 | 5400 |
| Bratsk | 8100 | 7600 | 7100 | 6600 | 6100 | 5600 |
| Bryansk | 5400 | 5000 | 4600 | 4200 | 3800 | 3300 |
| Verkhoyansk | 13400 | 12900 | 12300 | 11700 | 11200 | 10600 |
| Vladivostok | 5500 | 5100 | 4700 | 4300 | 3900 | 3500 |
| Vladikavkaz | 4100 | 3800 | 3400 | 3100 | 2700 | 2400 |
| Vladimir | 5900 | 5400 | 5000 | 4600 | 4200 | 3700 |
| Komsomolsk-on-Amur | 7800 | 7300 | 6900 | 6400 | 6000 | 5500 |
| Kostroma | 6200 | 5800 | 5300 | 4900 | 4400 | 4000 |
| Kotlas | 6900 | 6500 | 6000 | 5500 | 5000 | 4600 |
| Krasnodar | 3300 | 3000 | 2700 | 2400 | 2100 | 1800 |
| Krasnoyarsk | 7300 | 6800 | 6300 | 5900 | 5400 | 4900 |
| Mound | 6800 | 6400 | 6000 | 5600 | 5100 | 4700 |
| Kursk | 5200 | 4800 | 4400 | 4000 | 3600 | 3200 |
| Kyzyl | 8800 | 8300 | 7900 | 7400 | 7000 | 6500 |
| Lipetsk | 5500 | 5100 | 4700 | 4300 | 3900 | 3500 |
| Saint Petersburg | 5700 | 5200 | 4800 | 4400 | 3900 | 3500 |
| Smolensk | 5700 | 5200 | 4800 | 4400 | 4000 | 3500 |
| Magadan | 9000 | 8400 | 7800 | 7200 | 6700 | 6100 |
| Makhachkala | 3200 | 2900 | 2600 | 2300 | 2000 | 1700 |
| Minusinsk | 4700 | 6900 | 6500 | 6000 | 5600 | 5100 |
| Moscow | 5800 | 5400 | 4900 | 4500 | 4100 | 3700 |
| Murmansk | 7500 | 6900 | 6400 | 5800 | 5300 | 4700 |
| Moore | 6000 | 5600 | 5100 | 4700 | 4300 | 3900 |
| Nalchik | 3900 | 3600 | 3300 | 2900 | 2600 | 2300 |
| Nizhny Novgorod | 6000 | 5300 | 5200 | 4800 | 4300 | 3900 |
| Naryan-Mar | 9000 | 8500 | 7900 | 7300 | 6700 | 6100 |
| Velikiy Novgorod | 5800 | 5400 | 4900 | 4500 | 4000 | 3600 |
| Olonets | 6300 | 5900 | 5400 | 4900 | 4500 | 4000 |
| Omsk | 7200 | 6700 | 6300 | 5800 | 5400 | 5000 |
| Eagle | 5500 | 5100 | 4700 | 4200 | 3800 | 3400 |
| Orenburg | 6100 | 5700 | 5300 | 4900 | 4500 | 4100 |
| Novosibirsk | 7500 | 7100 | 6600 | 6100 | 5700 | 5200 |
| Partizansk | 5600 | 5200 | 4900 | 4500 | 4100 | 3700 |
| Penza | 5900 | 5500 | 5100 | 4700 | 4200 | 3800 |
| Permian | 6800 | 6400 | 5900 | 5500 | 5000 | 4600 |
| Petrozavodsk | 6500 | 6000 | 5500 | 5100 | 4600 | 4100 |
| Petropavlovsk-Kamchatsky | 6600 | 6100 | 5600 | 5100 | 4600 | 4000 |
| Pskov | 5400 | 5000 | 4600 | 4200 | 3700 | 3300 |
| Ryazan | 5700 | 5300 | 4900 | 4500 | 4100 | 3600 |
| Samara | 5900 | 5500 | 5100 | 4700 | 4300 | 3900 |
| Saransk | 6000 | 5500 | 5100 | 5700 | 4300 | 3900 |
| Saratov | 5600 | 5200 | 4800 | 4400 | 4000 | 3600 |
| Sortavala | 6300 | 5800 | 5400 | 4900 | 4400 | 3900 |
| Sochi | 1600 | 1400 | 1250 | 1100 | 900 | 700 |
| Surgut | 8700 | 8200 | 7700 | 7200 | 6700 | 6100 |
| Stavropol | 3900 | 3500 | 3200 | 2900 | 2500 | 2200 |
| Syktyvkar | 7300 | 6800 | 6300 | 5800 | 5300 | 4900 |
| Taishet | 7800 | 7300 | 6800 | 6300 | 5800 | 5400 |
| Tambov | 5600 | 5200 | 4800 | 4400 | 4000 | 3600 |
| Tver | 5900 | 5400 | 5000 | 4600 | 4100 | 3700 |
| Tikhvin | 6100 | 5600 | 2500 | 4700 | 4300 | 3800 |
| Tobolsk | 7500 | 7000 | 6500 | 6100 | 5600 | 5100 |
| Tomsk | 7600 | 7200 | 6700 | 6200 | 5800 | 5300 |
| Totna | 6700 | 6200 | 5800 | 5300 | 4800 | 4300 |
| Tula | 5600 | 5200 | 4800 | 4400 | 3900 | 3500 |
| Tyumen | 7000 | 6600 | 6100 | 5700 | 5200 | 4800 |
| Ulan-Ude | 8200 | 7700 | 7200 | 6700 | 6300 | 5800 |
| Ulyanovsk | 6200 | 5800 | 5400 | 5000 | 4500 | 4100 |
| Urengoy | 10600 | 10000 | 9500 | 8900 | 8300 | 7800 |
| Ufa | 6400 | 5900 | 5500 | 5100 | 4700 | 4200 |
| Ukhta | 7900 | 7400 | 6900 | 6400 | 5800 | 5300 |
| Khabarovsk | 7000 | 6600 | 6200 | 5800 | 5300 | 4900 |
| Khanty-Mansiysk | 8200 | 7700 | 7200 | 6700 | 6200 | 5700 |
| Cheboksary | 6300 | 5800 | 5400 | 5000 | 4500 | 4100 |
| Chelyabinsk | 6600 | 6200 | 5800 | 5300 | 4900 | 4500 |
| Cherkessk | 4000 | 3600 | 3300 | 2900 | 2600 | 2300 |
| Chita | 8600 | 8100 | 7600 | 7100 | 6600 | 6100 |
| Elista | 4400 | 4000 | 3700 | 3300 | 3000 | 2600 |
| Yuzhno-Kurilsk | 5400 | 5000 | 4500 | 4100 | 3600 | 3200 |
| Yuzhno-Sakhalinsk | 6500 | 600 | 5600 | 5100 | 4700 | 4200 |
| Yakutsk | 11400 | 10900 | 10400 | 9900 | 9400 | 8900 |
| Yaroslavl | 6200 | 5700 | 5300 | 4900 | 4400 | 4000 |
Examples of calculating the thickness of insulation
We propose to consider in practice the process of calculating the insulating layer of the wall and ceiling of a residential attic. For example, let’s take a house in Vologda, built from blocks (foam concrete) 200 mm thick.
So, if a temperature of 22 degrees is normal for the inhabitants, then the current in this case the degree-day indicator is 6000. We find the corresponding indicator in the table of standards for thermal resistance, it is 3.5 m² K/W - we will strive for it.
The wall will be multi-layered, so first we will determine how much thermal resistance a bare foam block will provide. If the average thermal conductivity of foam concrete is about 0.4 W/(m*K), then with a 20 mm thickness this outer wall will give a heat transfer resistance of 0.5 m²·K/W (0.2 meters divided by a thermal conductivity coefficient of 0.4).
That is, for high-quality insulation we lack about 3 m²·K/W. You can get them mineral wool or foam plastic, which will be installed on the facade side in a ventilated curtain structure or wet method bonded thermal insulation. We slightly transform the formula for thermal resistance and obtain the required thickness - that is, we multiply the required (missing) heat transfer resistance by thermal conductivity (take it from the table).
In numbers it will look like this: d thickness of basalt mineral wool = 3 X 0.035 = 0.105 meters. It turns out that we can use the material in mats or rolls 10 centimeters thick. Note that when using foam with a density of 25 kg/m3 and higher, the required thickness will be similar.
By the way, we can consider another example. Let's say we want from a full-bodied sand-lime brick in the same house, make a fence for a warm glazed balcony, then the missing thermal resistance will be about 3.35 m² K/W (0.12X0.82). If you plan to use PSB-S-15 foam for insulation, then its thickness should be 0.144 mm - that is, 15 cm.
For the attic, roof and floors, the calculation technique will be approximately the same, only thermal conductivity and heat transfer resistance of load-bearing structures are excluded. And also the resistance requirements increase somewhat - you will no longer need 3.5 m²·K/W, but 4.6. As a result, wool is suitable up to 20 cm thick = 4.6 X 0.04 (thermal insulator for roofing).
Using calculators
Manufacturers of insulating materials decided to simplify the task for ordinary developers. To do this, they developed simple and understandable programs for calculating the thickness of the insulation.
Let's look at some options:
In each of them, you need to fill in the fields in several steps, after which, by clicking on a button, you can instantly get the result.
Here are some features of using the programs:
1. Everywhere you are asked to select a city/district/region of construction from a drop-down list.
2. Everyone except TechnoNIKOL asks to determine the type of object: residential/industrial, or, as on the Penoplex website - city apartment/loggia/low-rise building/outbuilding.
3. Then we indicate which structures we are interested in: walls, floors, attic floors, roof. The Penoplex program also calculates the insulation of the foundation, engineering communications, street paths and playgrounds.
4. Some calculators have a field for indicating the desired temperature inside the room; on the Rockwool website they are also interested in the dimensions of the building and the type of fuel used for heating, and the number of people living. Knauf also takes into account relative humidity indoor air.
5. On penoplex.ru you need to indicate the type and thickness of the walls, as well as the material from which they are made.
6. Most calculators have the ability to specify the characteristics of individual or additional layers of structures, for example, features load-bearing walls without thermal insulation, cladding type...
7. The Penoplex calculator for some structures (for example, for roof insulation using the “between the rafters” method) can calculate not only extruded polystyrene foam, which the company specializes in, but also mineral wool.
As you understand, there is nothing complicated in calculating the optimal thickness of thermal insulation; you just need to approach this issue with the utmost care. The main thing is to clearly determine the missing heat transfer resistance, and then choose the insulation that will be best suited for the specific elements of the building and used construction technologies. Also, do not forget that the thermal insulation of a private house must be dealt with comprehensively; all enclosing structures must be properly insulated.
September 7, 2016Specialization: master of internal and exterior decoration(plaster, putty, tiles, drywall, lining, laminate and so on). In addition, plumbing, heating, electrical, conventional cladding and balcony extensions. That is, renovations in an apartment or house were done on a turnkey basis with all necessary types works
Of course, the calculation of insulation for walls in own home, this is a very serious job, especially if this was not done initially and the house is cold. And here you will have to face a number of questions.
For example, what kind of insulation should be, which one is better and what thickness of material is needed? Let's try to understand these issues, and also watch the video in this article, which clearly demonstrates the topic.
Wall insulation
Inside or outside
If you decide to use a calculator to calculate the thickness of insulation for walls, then you will not receive accurate data. Manually you can get more accurate and reliable information. In addition, the location of the insulation, which can be laid both inside and outside the building, is important, which must be taken into account when making calculations!
Features of internal and external insulation:
- Imagine that you are using a calculator to calculate insulation for walls, but at the same time you are laying insulation indoors, will the calculation results be correct? Notice the diagram above;
- no matter how thick the insulation in the room is, the wall will still remain cold and this will lead to certain consequences;
- that is, it means that the dew point or the area where warm air When it encounters cold, it turns into condensation and is transported closer to the room. And the more powerful internal insulation, the closer this point will be;
- in some cases, this zone extends to the surface of the wall, where moisture promotes the development of fungal mold. But even if it remains inside the wall, then the service life does not increase in any way;
- therefore, the instructions and common sense indicate that internal insulation should be installed only as a last resort or when sound insulation is needed ;
- with external insulation, the dew point will fall on the insulation zone, which means that you can increase the shelf life of your wall and avoid the occurrence of dampness.
Calculation is a serious matter!
| No. | Wall material | Coefficient of thermal conductivity | Required thickness (mm) |
| 1 | Expanded polystyrene PSB-S-25 | 0,042 | 124 |
| 2 | Mineral wool | 0,046 | 124 |
| 3 | Glued laminated timber or solid spruce and pine across the grain | 0,18 | 530 |
| 4 | Laying ceramic blocks with thermal insulation glue | 0,17 | 575* |
| 5 | Laying gas and foam blocks 400kg/m3 | 0,18 | 610* |
| 6 | Laying polystyrene blocks with glue 500kg/m3 | 0,18 | 643* |
| 7 | Laying gas and foam blocks 600kg/m3 | 0,29 | 981* |
| 8 | Adhesive laying of expanded clay concrete 800kg/m3 | 0,31 | 1049* |
| 9 | Ceramic masonry hollow brick at CPR 1000kg/m3 | 0,52 | 1530 |
| 10 | Ordinary brick masonry at CPR | 0,76 | 2243 |
| 11 | Sand-lime brick masonry at the central processing center | 0,87 | 2560 |
| 12 | Concrete products 2500kg/m3 | 2,04 | 6002 |
Thermal calculation of various materials
Note to the table. The presence of the * sign indicates the need to add a coefficient of 1.15 if the building has lintels and monolithic belts from heavy concrete. There is a diagram at the top for clarity - the numbers coincide with the table.
So, calculating the thickness of the insulation is the determination of its thermal resistance, which we denote by the letter R— a constant value that is calculated separately for each region.
Let's take the average figure for clarity R=2.8(m2*K/W). According to State Building Regulations this value is the minimum acceptable for residential and public buildings.
In cases where thermal insulation consists of several layers, for example, masonry, polystyrene foam and eurolining, then the sum of all indicators adds up - R=R1+R2+R3. And the total or individual thickness of the thermal insulation layer is calculated using the formula R=p/k.
Here p will indicate the layer thickness in meters, and the letter k, this is the thermal conductivity coefficient of this material(W/m*k), the value of which you can take from the table thermotechnical calculations which is given above.
In fact, using these same formulas, you can calculate the energy efficiency of insulating window sills or find out the thickness of floor insulation. Use the R value according to your region.
In order not to be unfounded, I will give an example, let’s take a brickwork of two bricks ( regular wall), and as insulation we will use polystyrene foam boards PSB-25 (twenty-fifth polystyrene foam), the price of which is quite reasonable even for budget construction.
So, thermal resistance, which we need to achieve should be 2.8 (m2*L/W). First, we find out the thermal resistance of a given brickwork. The brick is 250 mm from end to end and the mortar between them is 10 mm thick.
Hence, p=0.25*2+0.01=0.51m. The silicate coefficient is 0.7 (W/m*k), then Rbrick=p/k=0.51/0.7=0.73 (m2*K/W)- we got thermal conductivity brick wall, calculating it with your own hands.
Let's go further, now we need to achieve a general indicator for a layered wall of 2.8 (m2 * K / W), that is, R = 2.8 (m2 * K / W and for this we need to find out the required thickness of the foam. So, R foam = Rtotal-Rbrick=2.8-0.73=2.07 (m2*K/W).
In the photo - local protection with foam plastic
Now, to calculate the thickness of polystyrene foam, we take as a basis general formula and here Pfoam=Rfoam*kfoam= 2?07*0?035=0?072m. Of course, we can’t find 2 cm in PSB-25, but if we take into account interior decoration and an air gap between the bricks, then 70 cm will be enough for us, and this is two layers