The calculator allows you to determine the type of thermal insulation materials for the foundation, calculate the volume of required 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 the task of insulating 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 a floor that 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

Select 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)

Insulating a pitched roof requires, in addition to insulation, a vapor barrier and a wind and moisture barrier 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 a flat roof, we suggest using this calculator. The calculation also includes a waterproofing membrane and telescopic fasteners.

Gutter calculation calculator

The calculator will allow you to make a preliminary calculation of the necessary materials for installing a drainage system. Preliminarily determine the cost/

Even the now popular cottages made of logs or profiled timber must be additionally insulated or built from solid wood 35-40 cm thick that is practically non-existent on the market. 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 in order to improve an already used cottage with facade work. 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. Under a favorable set of circumstances (the presence of a reserve of heat generator performance), the problem can be solved by increasing the power of the heating system, which clearly entails a significant increase in the cost of purchasing 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, building structures and finishing materials are destroyed... What is most unpleasant is the fact that it is impossible to eliminate troubles with little expense. For example, on a facade you will have to dismantle (or “bury”) the finishing layer, then create another barrier from 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 will not cost much, and as the thickness increases, the price will also increase proportionally. Therefore, there is usually no point in creating too much thermal insulation reserve; 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 determine the main reason for 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) is played by various leaks in the building elements - infiltration. Therefore, you should also pay attention to tightness and drafts.

Brick, steel, concrete, glass, wooden beams... - every material used in the construction of buildings, to one degree or another, has the ability to transfer 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 the physical properties of 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 the thermal insulation characteristics of walls, ceilings, windows, doors, roofs...

To calculate 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 the materials and enclosing structures, which, using several tables, will help us calculate the actual thermal resistance of the 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, you don't have to use complicated formulas again. All the necessary information is in SNiP 02/23/2003 “Thermal protection of buildings”. This regulatory document discusses buildings for various purposes, operated in different climatic zones. This is understandable, since the temperature for residential premises and industrial premises does not need to be the same. In addition, individual regions are characterized by their extreme sub-zero temperatures and the duration of the heating season, therefore they distinguish such an average characteristic as degree-days of the heating season.

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 with an average indoor temperature of about 24 degrees) to build a house only from solid brick, then the wall will have to be made with a thickness of more than 2.4 meters (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 locality 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.

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 for the inhabitants is normal, then the current degree-day indicator in this case is 6000. We find the corresponding indicator in the table of thermal resistance standards, 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. They can be obtained with mineral wool or polystyrene foam, which will be installed on the side of the facade in a ventilated curtain structure or with wet 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 to make a fence for a warm glazed balcony from solid sand-lime brick in the same house, 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 ceilings, 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, utilities, street paths and platforms.

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 the relative air humidity in the rooms.

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 of load-bearing walls without thermal insulation, type of cladding...

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 the construction technologies used. Also, do not forget that the thermal insulation of a private house must be dealt with comprehensively; all enclosing structures must be properly insulated.

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, foam plastic, penoplex, mineral wool or ecowool, as well as plaster and other finishing materials are used.

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. Brick and concrete walls require more insulation, while wooden and foam block walls require less. Pay attention to how thick the material chosen for load-bearing structures is, and what its thermal conductivity is. The thinner the supporting structures, the greater the thickness of the insulation should be.

If thick insulation is required, it is better to insulate the house from the outside. This will save 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. Increased air humidity 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

Mineral wool or ecowool is most often chosen as insulation for a frame house.

The required thickness is determined using the same formulas as in traditional construction. Additional layers of a multilayer wall provide 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 are 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, materials with low thermal conductivity are used for roof insulation.

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.

A warm house is the dream of every owner; to achieve this goal, thick walls are built, heating is provided, and high-quality thermal insulation is installed. In order for insulation to be rational, it is necessary to choose the right material and correctly calculate its thickness.

The size of the insulation layer depends on the thermal resistance of the material. This indicator is the reciprocal of thermal conductivity. Each material - wood, metal, brick, foam plastic or mineral wool - has a certain ability to transfer thermal energy. The thermal conductivity coefficient is calculated during laboratory tests, and is indicated on the packaging for consumers.

If the material is purchased without labeling, you can find a summary table of indicators on the Internet.

The thermal resistance of a material ® is a constant value; it is defined as the ratio of the temperature difference at the edges of the insulation to the force of the heat flow passing through the material. Formula for calculating the coefficient: R=d/k, where d is the thickness of the material, k is the thermal conductivity. The higher the value obtained, the more effective the thermal insulation.

Why is it important to correctly calculate insulation indicators?

Thermal insulation is installed to reduce energy loss through the walls, floor and roof of a home. Insufficient insulation thickness will cause the dew point to move inside the building. This means the appearance of condensation, dampness and fungus on the walls of the house. An excess layer of thermal insulation does not significantly change temperature indicators, but requires significant financial costs, and is therefore irrational. This disrupts air circulation and natural ventilation between the rooms of the house and the atmosphere. To save money while ensuring optimal living conditions, an accurate calculation of the thickness of the insulation is required.

Calculation of the thermal insulation layer: formulas and examples

To be able to accurately calculate the amount of insulation, it is necessary to find the heat transfer resistance coefficient of all materials in a wall or other area of ​​the house. It depends on the climatic indicators of the area, therefore it is calculated individually using the formula:

GSOP=(tv-tot)xzot

tв - indoor temperature indicator, usually 18-22ºC;

tot - average temperature value;

zot - duration of the heating season, days.

Values ​​for calculation can be found in SNiP 01/23/99.

When calculating the thermal resistance of a structure, it is necessary to add up the indicators of each layer: R=R1+R2+R3, etc. Based on the average indicators for private and multi-storey buildings, approximate values ​​of the coefficients are determined:

• walls - at least 3.5;
• ceiling - from 6.

The thickness of the insulation depends on the building material and its size; the lower the thermal resistance of the wall or roof, the larger the insulation layer should be.

Example: a wall made of sand-lime brick 0.5 m thick, which is insulated with foam plastic.

Rst.=0.5/0.7=0.71 - thermal resistance of the wall

R- Rst.=3.5-0.71=2.79 - value for foam plastic

For foam plastic, thermal conductivity k=0.038

d=2.79×0.038=0.10 m - foam boards 10 cm thick will be required

Using this algorithm, it is easy to calculate the optimal amount of thermal insulation for all areas of the house except the floor. When making calculations regarding base insulation, you must refer to the soil temperature table in your region of residence. It is from this that the data is taken to calculate the GSOP, and then the resistance of each layer and the required value of insulation are calculated.

Popular ways to insulate a home

Thermal insulation of a building can be done during the construction stage or after its completion. Among the popular methods:

• Monolithic wall of significant thickness (at least 40 cm) made of ceramic brick or wood.
• The construction of enclosing structures by well masonry is the creation of a cavity for insulation between two parts of the wall.
• Installation of external thermal insulation in the form of a multilayer structure made of insulation, lathing, moisture-proof film and decorative finishing.

Using ready-made formulas, you can calculate the optimal thickness of insulation without the help of a specialist. When calculating, the number should be rounded up; a small margin of the thermal insulation layer will be useful for temporary temperature drops below the average.