The thickness of the insulation

The thickness of the insulation on your roof is a major factor in keeping your house cool in the summer and warm in the winter. Insulation functions as a barrier, preventing heat from entering in warmer weather and escaping during the colder months. The proper amount of insulation improves comfort and lowers energy costs by increasing the energy efficiency of your house.

The type of roofing material being used, local building codes, and your climate zone all play a role in determining the appropriate thickness of insulation. Greater thermal resistance, also known as an R-value, or the degree to which insulation impedes the flow of heat, is often associated with thicker insulation in colder climates. On the other hand, a marginally lower R-value may be adequate in milder climates to strike a balance between cost-effectiveness and energy efficiency.

It’s crucial to take the kind of insulation material into account as well. Mineral wool, cellulose, fiberglass, and spray foam are examples of common materials. Each has a unique R-value for every inch of thickness. For instance, spray foam can offer greater thermal resistance in less space than fiberglass, which typically has an R-value ranging from 3.5 to 6.7 per inch.

Seeking professional advice when installing or upgrading roof insulation can help you determine the best type and thickness for your particular requirements. To guarantee the insulation works well over time, considerations including the current roof structure, ventilation needs, and moisture control techniques should all be made.

Formula for calculating the thickness of the insulation for walls

What depends on what the thickness depends

Identifying the primary construction technology indicators should be the first step in calculating the thickness of wall insulation. These indicators include things like the building’s design and how well it has held up over time, the interior size of the room, the thickness and material of the current walls, the heat insulator’s composition, the climate in your area, and other, more recent indicators.

Examine the components of the wall thermal insulator calculation in greater detail.

The ZhEK or the management company can provide you with a copy of your housing’s technical passport, which includes information on the wall thickness and the materials used to construct it. These indicators are significant because they show the norms for building and, consequently, heat supply, for every climate zone.

The type of insulation is crucial because it determines how much less heat escapes your apartment as a result. Because every material has a different thermal conductivity coefficient, there will be variations in the minimum amount of insulation that can be used.

Wall construction and wear have an impact on the insulation process as well because, depending on the side (interior or exterior), utilities may need to be consulted during the insulation process. This will allow you to determine the extent of wall damage. In addition to the thicker layer of insulation, puttying joints and cracks and strengthening the ceilings during the installation process will take a significant amount of time if the building has not had cosmetic repairs performed in a long time.

It should be mentioned that exterior wall insulation thickness is not determined with the same meticulousness as interior wall insulation. The inability to forecast the weather is the cause of this neglect. It is impossible to forecast the weather outside if you can only measure the temperature inside the apartment during the winter months for annual indicators during the heating season. Consequently, a thickness greater than the minimum of 1.5 times is used for external insulation. As a result, you won’t have to buy additional materials for wall insulation.

The norms for heat resistance

What indicators are used to form the norms for heat resistance has already been written above. It is important to keep in mind that a material’s thermal conductivity determines how well it conducts heat, and its heat resistance determines how well it delays heat. For this reason, you want to select insulation and wall materials with a high coefficient of heat resistance.

The following formula is used to determine the wall’s coefficient of heat resistance:

The wall’s thermal resistance, or R, is equal to the thickness in meters times the material’s coefficient of thermal conductivity in WT times (m · ºΡ).

Since ready-made tables exist that indicate the heat resistance needed for each region, there’s no need to calculate this coefficient independently. The largest needs are for cities like Tynda, Yakutsk, Anadyr, and Urengoy. The least—for Tuapse and Sochi. The coefficient in Moscow should be 3.0 W/(m · ºΡ), while in the capital of the north it should be 2.9 W/(m · ºΡ).

Heat resistance requirements apply not only to the building’s walls but also to the ceiling and windows. The same formula can be used for calculations, but data can be found online or at a construction company.

We obtain the formula for determining the thickness of the insulation for the interior wall by accounting for all the data. It appears like this:

Rreg is a regional heating indicator (a self-made calculation or ready-made data);

Δ is the heat insulator’s thickness;

K is the WT/m2 · ºΡ coefficient of thermal conductivity.

Let’s now take a closer look at the carrier wall’s coefficients of thermal resistance and the variables that impact thermal insulation.

Thermal coefficients for the carrier wall’s materials:

The walls’ ability to withstand heat is directly influenced by the material used to construct the structure. The external walls of the building comprise the supporting structures situated in between the apartments. Within the apartment, the walls serve as partitions that are not insulated.

The walls’ ability to withstand heat is directly influenced by the material used to construct the structure. The external walls of the building comprise the supporting structures situated in between the apartments. The apartment’s walls are partitions; insulation has not been applied to them.

The technologists created a table of heat resistance based on all the data, which included information about the material’s thickness and the thermal conductivity coefficients in both the presence and absence of excess moisture:

When λ is dry, W/(m · o C)

The thermal conductivity calculation coefficients

When it’s humid*

Common clay brick on mortar made of cement and sand

Brick silicate in mortar made of cement and sand

1400 kg/m 3 (gross) of brick ceramic hollow density on a cement-sand mixture

Brick ceramic hollow density on a cement-sand solution of 1000 kg/m 3 (gross)

Spruce and tree pine throughout the fibers

Oak tree spanning the fibers

Oak tree near the fibers

Furthermore, the minimum thickness of insulation for a specific region of the country was determined based on the indicators of the heat transfer standards, as long as the insulation had a minimum thermal conductivity of 0.40 W/(m · ºΡ). You can find these details online or in the records of the building company that built your building.

In general, the coefficients of heat -resistance for load -bearing structures depend on the construction region, each of which has its own requirements for heat -proceedings (which was already discussed). These coefficients are different, but in general, all regions are divided into two large categories – A and B. In addition to the difference in daily temperatures, there is a difference between the formation of the dew point in both groups. Group A include more dry cities with a stable climate, such as Arkhangelsk (3.6), Krasnodar (2.3), Chita (4.1). Group B includes northern cities and cities located in transitional climatic zones – Bryansk (3.0), Kaliningrad (2.7), Khabarovsk (3.6).

The following cities are included in Group B: Samara, Cheboksary, Yaroslavl, Perm, Arkhangelsk, Murmansk, Syktyvkar, Khabarovsk, Blagoveshchensk, Salekhard, Igarka, Moscow, Novgorod, Ryazan, St. Petersburg, Smolensk, Tula, Ivanovo, and Kaluga.

Cities related to group A: Arkhangelsk, Astrakhan, Barnaul, Belgorod, Volgograd, Voronezh, Vladikavkaz, Grozny, Yekaterinburg, Irkutsk, Kemerovo, Krasnodar, Krasnoyarsk, Kurgan, Kyzyl, Lipetsk, Makhachkala, Nalchik, Novosibirsk, Omsk, Orenburg, Penza Rostov-on-Don, Saransk, Saratov, Stavropol, Tambov, Tyumen, Ulyanovsk, Ulan-Ude, Ufa, Chelyabinsk, Chita, Elista, Yakutsk.

Indicators of thermal conductivity for various heaters can be obtained from a building supply store. For instance, the foam indicator is 0, 037 W/m × k, meaning that 160 mm should be the minimum thickness of polystyrene foam used for wall insulation. Additionally, for wall insulation, the thickness of the extruded polystyrene foam, or foam, should be 120 mm because it retains heat in the space more effectively and densely.

Savings parameters for thermal insulation

Other information that influences thermal insulation should be considered in addition to the data mentioned above:

• damage to the supporting structures;
• “Cold bridges” and cracks in the ceilings;
• moisture-, steam-and the eyes of the insulation;
• environmental friendliness and fire safety of materials and other.

Approximate calculation of wall thickness from homogeneous material

If you wish to perform every calculation yourself, then all of these formulas are required. On the other hand, it is simple to locate an online calculator for insulating thickness on the Internet. These calculators provide a more precise result because they take into account information about air cushion and finishing materials in addition to the walls’ and heat insulator’s thermal conductivity.

Let’s say you have a silicate house in Yakutsk that you have chosen to use ordinary polystyrene foam to warm. Drywall is used to complete the walls. You will be given an indication of roughly 150 mm (air gap 20 mm) when you calculate manually. 135 mm is the result of the online calculator’s calculation using all the data.

How and by what formulas to determine the wall insulation thickness? Which criteria need to be considered. current requirements for heat resistance.

How to calculate the thickness of the insulation – methods and methods

Every owner dreams of having a warm house; to this end, thick walls are constructed, heating is installed, and superior thermal insulation is set up. Rational insulation requires the appropriate material to be chosen and its thickness to be calculated accurately.

What data is needed to calculate the thickness of the insulation?

The material’s thermal resistance determines the insulation layer’s thickness. The reverse thermal conductivity value serves as this indicator. Every material, including foam, minvat, brick, metal, and wood, has a specific capacity to transfer thermal energy. Consumers can find the thermal conductivity on the packaging, which is determined through laboratory testing.

Online, a consolidated table of indicators can be found if the material is bought unmarked.

The ratio of the temperature differential at the insulation’s edges to the heat duct’s strength through the material is known as the thermal resistance of material ®, and it is a constant value. The coefficient can be calculated using the following formula: r = d/k, where d is the material’s thickness and k is its thermal conductivity. The effectiveness of the thermal insulation increases with a higher resulting value.

Why is it important to correctly calculate the insulation indicators?

In order to minimize energy losses through the house’s walls, floor, and roof, thermal insulation is installed. The dew point will move inside the building if the insulation is not thick enough. This denotes the growth of fungus, condensation, and moisture on the home’s walls. It is unreasonable to use an excessive amount of thermal insulation because it does not significantly alter temperature indicators and is expensive. Here, there is a disruption in the natural ventilation and air movement between the house’s rooms and the surrounding air. The ideal living conditions can be achieved at the lowest possible cost by precisely calculating the insulation’s thickness.

Calculation of the thermal insulation layer: formulas and examples

Finding each material’s coefficient of heat transfer resistance in a wall or other area of the house is essential to accurately calculating the value of insulation. Since it is dependent on the local climate indicators, it is computed separately using the following formula:

The TV indicates the indoor temperature, which is typically between 18 and 22 degrees Celsius;

The average temperature is represented by tot;

Zot: the number of days during the heating season.

You can find the values for counting in SNiP 23-01-99.

The indicators for each layer must be added in order to determine the structure’s thermal resistance: R = R1+R2+R3 and T. D. The approximate values of the coefficients are calculated using the average indicators for single- and multi-story buildings:

The construction material and size determine the thickness of the insulation; the less heat-resistant the wall or roof, the thicker the isolation layer should be.

An illustration would be a 0.5 m thick silicate brick wall that has polystyrene insulation.

RST is the wall’s thermal resistance, which is equal to 0.5/0.7 = 0.71.

R-RST = 3.5-0.71 = 2.79, which is the foam value.

With all the information, use the formula D = rxk to determine the necessary insulation layer.

Thermal conductivity for foam is k = 0.038.

D = 2.79 × 0.038 = 0.10 m – 10 cm thick foam plate boards will be needed.

With the exception of the floor, this algorithm makes it simple to determine the ideal thermal insulation value for every area of the house. It is essential to consult the soil temperature table in the area of residence when determining the base insulation. It provides the data needed to calculate the GSOP, which in turn provides the resistance values of each layer and the desired level of insulation.

Popular ways to warm the house

Building thermal insulation can be done during the construction phase or after it is finished. Among the widely used techniques are:

• A monolithic wall of a significant thickness (at least 40 cm) from ceramic brick or wood.
• The construction of the enclosing structures by well masonry is the creation of a cavity for insulation between the two parts of the wall.
• Installation of external thermal insulation in the form of a multi -layer structure of insulation, crate, moisture -proof film and decorative decoration.

Without the assistance of an expert, you can determine the ideal insulation thickness using completed formulas. The small supply of the heat insulator layer will be helpful when the temperature falls below the average indicator, so when calculating, the number should be rounded in the larger direction.

Formulas and examples for calculating the thermal insulation layer under different starting conditions. Information required to determine the thermal insulation thickness.

"Knowing how insulation thickness impacts your home’s comfort and energy efficiency is essential to appreciating its significance for your roof. The proper thickness of insulation plays a major role in lowering energy costs and in preserving steady indoor temperatures. An environment that is both comfortable and sustainable throughout the year is produced by having enough insulation, which blocks heat gain in the summer and loss in the winter. A crucial factor in any roofing project is selecting the right insulation thickness based on your climate and type of building. Doing so can save money in the long run and have a positive environmental impact."

Calculation of the thickness of the insulation for walls

When building a house of their own, everyone wants it to be warm. This can be accomplished in a few different ways, such as by adding thick walls, quality insulation, or efficient heating.

In actual use, all of these techniques are combined, but from an economic perspective, increasing the thickness of the house’s insulation has greater priority.

How to determine how thick the walls and insulation should be in order to make the house warm and sturdy.

We will divide our computation into two primary phases:

1. The resistance to heat transfer of walls, which is necessary for further calculation.
2. Selection of the required thickness of the insulation Depending on the design and material of the walls.

We first recommend watching a brief video in which a specialist explains in detail why it’s necessary to install insulation in the brick house’s exterior walls as well as the kind of insulation that’s used in tandem with it.

Resistance to heat transfer walls

We use the "Thermal protection of buildings" SP 50.13330.2012, which is available for download on our website (link), to determine this parameter.

Several formulas are provided in paragraph 5 under "Thermal Protection of buildings" to assist us. Determine how thick the walls and insulation are. The heat transfer resistance parameter, denoted by the letter R, is used to accomplish this. It depends on the required temperature inside and the local climate of this city or neighborhood.

R tr = a x GSOP + B is the general formula used to calculate it.

Table 3 shows that the walls of residential buildings have coefficients A and B of 0.00035 and 1.4, respectively.

Finding the GSOP’s size is the only thing left to do. It can be understood as a heating period degree. This value needs to be adjusted.

This formula, tIN, indicates the ideal temperature to be indoors. The norms state that it equals 20-22 0 with.

The average outside air temperature and the number of days during the heating season each year are indicated by the values of the parameters tFROM and zFROM. They are identified in "Construction climatology" (SNiP 23-01-99). (Source: ).

The first thing you’ll notice when you look at this SNiP is a sizable table that lists the climate parameters for each city or district.

The column headed "The duration and average air temperature of the period with an average daily air temperature ≤ 8 0 C" will catch our attention.

An illustration of how to compute the parameter r tr

Let’s compute the resistance of the heat transfer of the walls (r tr) for the house built in g. Kazan so that everything makes more sense.

We have two formulas for this:

Do the GSOP calculation first. Find g. Kazan in the right column under SNiP 23-01-99 to accomplish this.

The table indicates that the duration zFROM = 215 Subters/year and the average temperature tFROM = -5.2 0 C.

You now have to determine what indoor air temperature is comfortable for you. As previously mentioned, TIN = 20–22 0 s is the ideal. Depending on your preferred temperature, take that into account when figuring out the GSOP for value tIN. It might be distinct.

Thus, compute tIN = 18 0 C and TIN = 22 0 s for the temperature.

18 0 s-(-5.2 0 s) x 215 days/year = 4988 is the value of GSOP18.

(22 0 s-(-5.2 0 s) x 215 days/year = 5848 is the GSOP22.

Locate the heat resistance now. Table 3 from SP 50.13330.2012 indicates that coefficients A and B for residential building walls are equal to 0,00035 and 1.4, as we already know.

For 18 0 with indoor, R tr (18 0 s) = 0.00035 x 4988 + 1.4 = 3.15 m 2 * 0 s/W.

For 22 0 s, R tr (22 0 c) = 0.00035 x 5848 + 1.4 = 3.45 m 2 * 0 s/W.

To ensure that the house loses as little heat as possible, such resistance should have a wall in addition to insulation.

Thus, we got the first set of data that we needed. Let’s now proceed to the second phase, which involves figuring out how thick the insulation is.

Calculation of the thickness of the insulation

We hope you are interested enough to read our article’s previous section. Let’s now attempt to compute the insulation thickness based on the material and wall thickness.

Every component of the multi-layer wall has a unique thermal resistance, or R. Therefore, our job is to make sure that the thermal resistance of the R TR, which we computed in the previous head, is equal to the total of all the resistances of the materials used in the wall’s design. i.e.:

The relationship between the layer thickness (ΔS) and thermal conductivity (λS) yields the thermal resistance (R) of each individual material.

In order to prevent further formula confusion, let’s look at three examples.

Examples of how to calculate the insulation thickness for aerated concrete and brick walls

Example 1: A wall made of 30 cm-thick aerated concrete blocks, lined with 1000 kg/m 3 ceramic hollow bricks, and insulated with stone wool that has a density of 80–125 kg/m 3. Work on the construction was done in Gazan.

We will require the material’s λS thermal conductivity values in order to determine the insulation thickness. The certificate for the materials should contain these details.

You can view them in Appendix C to SP 50.13330.2012, which we previously used, if they are absent for any reason.

The thermal conductivity of aerated concrete is λSg = 0.14 W/m* 0 C.

The insulation’s thermal conductivity, expressed as λS = 0.045 W/m* 0 C.

Brick has a thermal conductivity of λSC = 0.52 W/m* 0 C.

Next, we determine the value of R for each material, taking into account that the half-brick’sexternalmasonry has δSC = 12 cm and the layer of aerated concrete has ΔSg = 30 cm in thickness.

The thermal resistance of aerated concrete is expressed as RG = δSg/λSg = 0.3/0.14 = 2.14 m 2 * 0 s/W;

The brick’s thermal resistance is expressed as RTO = δSC/λSC = 0.12/0.52 = 0.23 m 2 * 0 s/c.

T.O. If our wall has three layers, then the following equation holds true:

We discovered the value of R TR (22 0 C) for g. Kazan in the previous chapter. It is employed in our computations.

We ascertained the appropriate level of thermal resistance for the insulation. The following formula will be used to determine the insulation’s thickness:

We were informed that there is sufficient insulation, 5 cm thick, for the conditions specified.

Using R TR (18 0 C) = 3.15 m 2 * 0 s/W as our value, we obtain:

As you can see, there has only been a 0.5 centimeter change in the insulation’s thickness.

Example 2: Take into consideration a scenario in which 1800 kg/m 3 of silicate brick is laid in place of aerated concrete blocks. The thickness of the masonry is 38 cm.

We determine the thermal conductivity values based on the table by using an analogy with the earlier calculations:

The thermal conductivity of silicate brick with a density of 1800 kg/m 3 is expressed as λSK1 = 0.87 W/m* 0 C.

The insulation’s thermal conductivity, expressed as λS = 0.045 W/m* 0 C.

Brick with a density of 1000 kg/m 3 has a thermal conductivity of λSK2 = 0.52 W/m* 0 C.

We then determine the values R:

RK1 = δSK1/λSK1 = 0.38/0.87 = 0.44 m 2 * 0 s/W – the brick’s 1800 kg/m 3 thermal resistance;

RK2 = δSK2/λSK2 = 0.12/0.52 = 0.23 m 2 * 0 s/V is the brick’s 1000 kg/m 3 thermal resistance.

We ascertain the insulation’s thermal resistance:

RU is equal to 2.78 m 2 * 0 s/WT – 3.45 – 0.44 – 0.23.

We now compute the insulation’s thickness:

T.e. The insulation is sufficiently thick—12 cm—for these circumstances.

Example 3: To illustrate the significance of insulation, let’s look at a wall made entirely of aerated concrete D600.

Given that λSg = 0.14 W/m* 0 C is the thermal conductivity of aerated concrete blocks, we can compute the required wall thickness of t.To with ease. The wall is one piece.

In order to adhere to all SNiP regulations, we must install a wall that is 0.5 meters thick.

You have two options in this situation: either build the wall to the exact thickness that is required right away, or build it thinner and add insulation on top.

We believe that the first option is more dependable and less expensive because it doesn’t require installing insulation. For already constructed homes, the second choice makes more sense.

These examples all demonstrate how the wall’s thickness affects the wall material’s thickness. By using them as an analogy, you can calculate for any kind of material.

Material Type	Recommended Thickness (inches)
Fiberglass Batt Insulation	12 – 15
Spray Foam Insulation	3 – 5
Cellulose Insulation	12 – 15

Maintaining energy efficiency and comfort in your home depends on selecting the appropriate insulation thickness for your roof. How well your house retains heat in the winter and stays cool in the summer is directly impacted by the thickness of its insulation.

Take into account your local building codes, your budget, and your climate when choosing the thickness of the insulation. Thickner insulation helps keep heat from escaping in colder climates, which lowers heating expenses. On the other hand, sufficient insulation thickness lowers air conditioning costs in warmer climates by keeping your home cooler.

Additionally, by limiting noise transmission and preventing moisture buildup, which can result in mold and mildew, thicker insulation can enhance soundproofing and indoor air quality.

The perfect amount of insulation ultimately depends on your personal requirements and objectives regarding comfort, environmental impact, and energy efficiency. To ensure that the insulation thickness you choose for your roof is in line with your long- and short-term objectives, consult an expert in the field.

Video on the topic

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