Example of thermotechnical calculation of an external wall. Thermal calculation of building envelopes How to make a thermal calculation of a wall

A long time ago, buildings and structures were built without thinking about what heat-conducting qualities the enclosing structures have. In other words, the walls were simply made thick. And if you ever happened to be in old merchant houses, then you might have noticed that the outer walls of these houses are made of ceramic bricks, the thickness of which is about 1.5 meters. This thickness brick wall provided and still provides quite a comfortable stay of people in these houses even in the most severe frosts.

At present, everything has changed. And now it is not economically profitable to make the walls so thick. Therefore, materials have been invented that can reduce it. One of them: heaters and gas silicate blocks. Thanks to these materials, for example, the thickness brickwork can be reduced up to 250 mm.

Now walls and ceilings are most often made of 2 or 3 layers, one layer of which is a material with good thermal insulation properties. And in order to determine the optimal thickness of this material, a thermal calculation is carried out and the dew point is determined.

How the calculation is made to determine the dew point, you can find on the next page. Here, the heat engineering calculation will be considered using an example.

Required regulatory documents

For the calculation, you will need two SNiPs, one joint venture, one GOST and one allowance:

• SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings". Updated edition from 2012.
• SNiP 23-01-99* (SP 131.13330.2012). "Construction climatology". Updated edition from 2012.
• SP 23-101-2004. "Design of thermal protection of buildings".
• GOST 30494-96 (replaced by GOST 30494-2011 since 2011). "Residential and public buildings. Indoor microclimate parameters".
• Benefit. E.G. Malyavin "Heat loss of the building. Help Guide" .

Calculated parameters

In the process of performing a heat engineering calculation, the following are determined:

• thermal characteristics building materials enclosing structures;
• reduced heat transfer resistance;
• compliance of this reduced resistance with the standard value.

Example. Thermal engineering calculation of a three-layer wall without an air gap

Initial data

1. The climate of the area and the microclimate of the room

Construction area: Nizhny Novgorod.

Purpose of the building: residential.

Estimated relative humidity of internal air from the condition of no condensation on the internal surfaces of external fences is equal to - 55% (SNiP 23-02-2003 p.4.3. Table 1 for normal humidity conditions).

The optimum air temperature in the living room during the cold season t int = 20°C (GOST 30494-96 table 1).

Estimated outdoor temperature text, determined by the temperature of the coldest five-day period with a security of 0.92 = -31 ° С (SNiP 23-01-99 table 1 column 5);

The duration of the heating period with an average daily outdoor temperature of 8°С is equal to z ht = 215 days (SNiP 23-01-99 table 1 column 11);

The average outdoor temperature during the heating period t ht = -4.1 ° C (SNiP 23-01-99 table. 1 column 12).

2. Wall construction

The wall consists of the following layers:

• Brick decorative (besser) 90 mm thick;
• insulation (mineral wool board), in the figure its thickness is indicated by the sign "X", since it will be found in the calculation process;
• silicate brick 250 mm thick;
• plaster (complex mortar), an additional layer to obtain a more objective picture, since its influence is minimal, but there is.

3. Thermophysical characteristics of materials

The values ​​of the characteristics of the materials are summarized in the table.

Note (*): These characteristics can also be found from manufacturers of thermal insulation materials.

Calculation

4. Determining the thickness of the insulation

To calculate the thickness of the heat-insulating layer, it is necessary to determine the heat transfer resistance of the enclosing structure based on the requirements sanitary norms and energy saving.

4.1. Determination of the norm of thermal protection according to the condition of energy saving

Determination of degree-days of the heating period according to clause 5.3 of SNiP 23-02-2003:

D d = ( t int - tht) z ht = (20 + 4.1)215 = 5182°С×day

Note: also degree-days have the designation - GSOP.

The normative value of the reduced resistance to heat transfer should be taken not less than the normalized values ​​determined by SNIP 23-02-2003 (Table 4) depending on the degree-day of the construction area:

R req \u003d a × D d + b \u003d 0.00035 × 5182 + 1.4 \u003d 3.214m 2 × °С/W,

where: Dd - degree-day of the heating period in Nizhny Novgorod,

a and b - coefficients taken according to table 4 (if SNiP 23-02-2003) or according to table 3 (if SP 50.13330.2012) for the walls of a residential building (column 3).

4.1. Determination of the norm of thermal protection according to the condition of sanitation

In our case, it is considered as an example, since this indicator is calculated for industrial buildings with excess sensible heat of more than 23 W / m 3 and buildings intended for seasonal operation (in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° С and below the given resistance to heat transfer of enclosing structures (with the exception of translucent ones).

Determination of the normative (maximum allowable) resistance to heat transfer according to the condition of sanitation (formula 3 SNiP 23-02-2003):

where: n \u003d 1 - coefficient taken from table 6 for outer wall;

t int = 20°C - value from the initial data;

t ext \u003d -31 ° С - value from the initial data;

Δt n \u003d 4 ° С - normalized temperature difference between the temperature of the indoor air and the temperature of the inner surface of the building envelope, is taken according to table 5 in this case for the outer walls of residential buildings;

α int \u003d 8.7 W / (m 2 × ° С) - heat transfer coefficient of the inner surface of the enclosing structure, taken according to table 7 for external walls.

4.3. Thermal protection rate

From the above calculations for the required heat transfer resistance, we choose R req from the energy saving condition and denote it now R tr0 \u003d 3.214 m 2 × °С/W .

5. Determining the thickness of the insulation

For each layer of a given wall, it is necessary to calculate the thermal resistance using the formula:

where: δi - layer thickness, mm;

λ i - calculated coefficient of thermal conductivity of the layer material W/(m × °С).

1 layer (decorative brick): R 1 = 0.09 / 0.96 = 0.094 m 2 × °С/W .

3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × °С/W .

4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × °С/W .

Determination of the minimum allowable (required) thermal resistance of a heat-insulating material (formula 5.6 by E.G. Malyavin "Heat loss of a building. Reference manual"):

where: R int = 1/α int = 1/8.7 - resistance to heat transfer on the inner surface;

R ext \u003d 1/α ext \u003d 1/23 - resistance to heat transfer on the outer surface, α ext is taken according to table 14 for external walls;

ΣR i = 0.094 + 0.287 + 0.023 - the sum of thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the coefficients of thermal conductivity of materials taken in column A or B (columns 8 and 9 of Table D1 SP 23-101-2004) in accordance with the humidity conditions of the wall, m 2 ° С /W

The thickness of the insulation is (formula 5.7):

where: λ ut - coefficient of thermal conductivity of the insulation material, W / (m ° C).

Determination of the thermal resistance of the wall from the condition that the total thickness of the insulation will be 250 mm (formula 5.8):

where: ΣR t, i - the sum of thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 ·°С / W.

From the result obtained, it can be concluded that

R 0 \u003d 3.503m 2 × °С/W> R tr0 = 3.214m 2 × °С/W→ therefore, the thickness of the insulation is selected right.

Influence of the air gap

In the case when mineral wool, glass wool or other slab insulation is used as a heater in a three-layer masonry, it is necessary to install an air ventilated layer between outdoor masonry and a heater. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to drain the insulation, which gets wet from condensate.

This air layer is not a closed space, therefore, if it is present in the calculation, it is necessary to take into account the requirements of clause 9.1.2 of SP 23-101-2004, namely:

a) structural layers located between the air gap and the outer surface (in our case, this is a decorative brick (besser)) are not taken into account in the heat engineering calculation;

b) on the surface of the structure facing towards the layer ventilated by the outside air, the heat transfer coefficient α ext = 10.8 W/(m°C) should be taken.

Note: the influence of the air gap is taken into account, for example, in the heat engineering calculation of plastic double-glazed windows.

During the operation of the building, both overheating and freezing are undesirable. To determine the golden mean will allow thermal engineering calculation, which is no less important than the calculation of efficiency, strength, resistance to fire, durability.

Based on thermal engineering standards, climatic characteristics, vapor and moisture permeability, the choice of materials for the construction of enclosing structures is carried out. How to perform this calculation, we will consider in the article.

A lot depends on the thermal engineering features of the capital fences of the building. These are the humidity of structural elements, and temperature indicators that affect the presence or absence of condensate on interior partitions and ceilings.

The calculation will show whether stable temperature and humidity characteristics will be maintained at plus and minus temperatures. The list of these characteristics also includes such an indicator as the amount of heat lost by the building envelope during the cold period.

You can't start designing without all this data. Based on them, choose the thickness of the walls and ceilings, the sequence of layers.

According to the regulation GOST 30494-96 temperature values ​​indoors. On average, it is 21⁰. At the same time, relative humidity must be within comfortable limits, and this is an average of 37%. The highest speed of movement of the mass of air - 0.15 m / s

Thermal engineering calculation aims to determine:

1. Are the designs identical to the stated requests in terms of thermal protection?
2. Is the comfortable microclimate inside the building so fully ensured?
3. Is optimal thermal protection of structures ensured?

The main principle is to maintain a balance of the difference in temperature indicators of the atmosphere of the internal structures of fences and premises. If it is not observed, heat will be absorbed by these surfaces, and inside the temperature will remain very low.

The internal temperature should not be significantly affected by changes in heat flow. This characteristic is called heat resistance.

By performing a thermal calculation, the optimal limits (minimum and maximum) of the dimensions of the walls, ceilings in thickness are determined. This is a guarantee of the operation of the building for a long period, both without extreme freezing of structures and overheating.

Parameters for performing calculations

To perform heat calculation, initial parameters are needed.

They depend on a number of characteristics:

1. Purpose of the building and its type.
2. Orientation of vertical enclosing structures relative to the direction to the cardinal points.
3. Geographic parameters of the future home.
4. The volume of the building, its number of storeys, area.
5. Types and dimensions of door, window openings.
6. Type of heating and its technical parameters.
7. The number of permanent residents.
8. Material of vertical and horizontal protective structures.
9. Top floor ceilings.
10. Hot water facilities.
11. Type of ventilation.

Taken into account in the calculation and other design features buildings. The air permeability of building envelopes should not contribute to excessive cooling inside the house and reduce the heat-shielding characteristics of the elements.

Waterlogging of the walls also causes heat loss, and in addition, this entails dampness, which negatively affects the durability of the building.

In the process of calculation, first of all, the thermal data of the building materials from which the enclosing elements of the structure are made are determined. In addition, the reduced heat transfer resistance and compliance with its standard value are to be determined.

Formulas for the calculation

Loss of heat lost from a home can be divided into two main parts: losses through building envelopes and losses caused by operation. In addition, heat is lost when warm water is discharged into the sewer system.

For the materials from which the enclosing structures are made, it is necessary to find the value of the thermal conductivity index Kt (W / m x degree). They are in the relevant reference books.

Now, knowing the thickness of the layers, according to the formula: R = S/Kt, calculate the thermal resistance of each unit. If the structure is multilayered, all the obtained values ​​​​are added up.

The dimensions of heat losses are easiest to determine by adding heat flows through the building envelope, which actually form this building.

Guided by this technique, it is taken into account that the materials that make up the structure do not have the same structure. It is also taken into account that the heat flux passing through them has different specifics.

For each individual structure, heat loss is determined by the formula:

Q = (A / R) x dT

• A is the area in m².
• R is the resistance of the structure to heat transfer.
• dT is the temperature difference between outside and inside. It must be determined for the coldest 5-day period.

By doing the calculation in this way, you can get the result only for the coldest five-day period. The total heat loss for the entire cold season is determined by taking into account the parameter dT, taking into account the temperature not the lowest, but the average.

The extent to which heat is absorbed, as well as heat transfer, depends on the humidity of the climate in the region. For this reason, moisture maps are used in calculations.

There is a formula for this:

W \u003d ((Q + Qv) x 24 x N) / 1000

In it, N is the duration of the heating period in days.

Disadvantages of calculating by area

The calculation based on the area index is not very accurate. It does not take into account such a parameter as climate, temperature indicators, both minimum and maximum, humidity. Due to ignoring many important points, the calculation has significant errors.

Often trying to block them, the project provides for a "margin".

If, nevertheless, this method is chosen for the calculation, the following nuances must be taken into account:

1. With a vertical fence height of up to three meters and no more than two openings on one surface, it is better to multiply the result by 100 watts.
2. If the project includes a balcony, two windows or a loggia, they multiply by an average of 125 watts.
3. When the premises are industrial or warehouse, a multiplier of 150 watts is used.
4. In the case of radiators located near windows, their design capacity is increased by 25%.

The area formula is:

Q=S x 100 (150) W.

Here Q is the comfortable level of heat in the building, S is the area with heating in m². The numbers 100 or 150 are the specific value of thermal energy consumed to heat 1 m².

Losses through house ventilation

The key parameter in this case is the air exchange rate. Provided that the walls of the house are vapor-permeable, this value is equal to one.

The penetration of cold air into the house is carried out through supply ventilation. Exhaust ventilation contributes to the escape of warm air. Reduces losses through ventilation heat exchanger-recuperator. It does not allow heat to escape along with the outgoing air, and it heats the incoming flows

A complete renewal of the air inside the building is envisaged in one hour. Buildings built according to the DIN standard have walls with vapor barrier, so here the air exchange rate is taken equal to two.

There is a formula by which heat loss through the ventilation system is determined:

Qv \u003d (V x Kv: 3600) x P x C x dT

Here the symbols mean the following:

1. Qw - heat loss.
2. V is the volume of the room in mᶾ.
3. P - air density. its value is taken equal to 1.2047 kg/mᶾ.
4. Kv - the frequency of air exchange.
5. C is the specific heat capacity. It is equal to 1005 J / kg x C.

Based on the results of this calculation, it is possible to determine the power of the heat generator heating system. In the case of a too high power value, the way out of the situation may be. Consider a few examples for houses made of different materials.

Example of thermal engineering calculation No. 1

We calculate a residential building located in the 1st climatic region (Russia), subregion 1B. All data are taken from Table 1 of SNiP 23-01-99. The coldest temperature observed for five days with a security of 0.92 - tn = -22⁰С.

In accordance with SNiP, the heating period (zop) lasts 148 days. The average temperature during the heating period at the average daily air temperature in the street is 8⁰ - tot = -2.3⁰. The temperature outside during the heating season is tht = -4.4⁰.

Heat losses at home - crucial point at the design stage. The choice of building materials and insulation also depends on the results of the calculation. There are no zero losses, but you need to strive to ensure that they are as expedient as possible.

The condition is stipulated that the temperature of 22⁰ must be ensured in the rooms of the house. The house has two floors and walls 0.5 m thick. Its height is 7 m, its dimensions in plan are 10 x 10 m. The material of the vertical enclosing structures is warm ceramics. For her, the thermal conductivity coefficient is 0.16 W / m x C.

Mineral wool was used as an external insulation, 5 cm thick. The value of Kt for her is 0.04 W / m x C. The number of window openings in the house is 15 pcs. 2.5 m² each.

Heat loss through walls

First of all, the thermal resistance must be defined as ceramic wall, and a heater. In the first case, R1 = 0.5: 0.16 = 3.125 sq. m x C/W. In the second - R2 \u003d 0.05: 0.04 \u003d 1.25 square meters. m x C/W. In general, for a vertical building envelope: R = R1 + R2 = 3.125 + 1.25 = 4.375 sq. m x C/W.

Since heat losses are directly proportional to the area of ​​the building envelope, we calculate the area of ​​the walls:

A \u003d 10 x 4 x 7 - 15 x 2.5 \u003d 242.5 m²

Now you can determine the heat loss through the walls:

Qc \u003d (242.5: 4.375) x (22 - (-22)) \u003d 2438.9 W.

Heat losses through horizontal enclosing structures are calculated in a similar way. Finally, all results are summed up.

If the basement under the floor of the first floor is heated, the floor may not be insulated. It is still better to sheathe the walls of the basement with insulation so that the heat does not go into the ground.

Determination of losses through ventilation

To simplify the calculation, they do not take into account the thickness of the walls, but simply determine the volume of air inside:

V \u003d 10x10x7 \u003d 700 mᶾ.

With the air exchange rate Kv = 2, the heat loss will be:

Qv \u003d (700 x 2): 3600) x 1.2047 x 1005 x (22 - (-22)) \u003d 20 776 W.

If Kv = 1:

Qv \u003d (700 x 1): 3600) x 1.2047 x 1005 x (22 - (-22)) \u003d 10 358 W.

Efficient ventilation of residential buildings is provided by rotary and plate heat exchangers. The efficiency of the former is higher, it reaches 90%.

Example of thermal engineering calculation No. 2

It is required to calculate the losses through a brick wall 51 cm thick. It is insulated with a 10 cm layer of mineral wool. Outside - 18⁰, inside - 22⁰. Wall dimensions - 2.7 m in height and 4 m in length. The only outer wall of the room is oriented to the south, there are no external doors.

For brick, the coefficient of thermal conductivity is Kt = 0.58 W / mºС, for mineral wool - 0.04 W / mºС. Thermal resistance:

R1 \u003d 0.51: 0.58 \u003d 0.879 sq. m x C/W. R2 \u003d 0.1: 0.04 \u003d 2.5 sq. m x C/W. In general, for a vertical enclosing structure: R = R1 + R2 = 0.879 + 2.5 = 3.379 sq. m x C/W.

Square outer wall A \u003d 2.7 x 4 \u003d 10.8 m²

Heat loss through the wall:

Qc \u003d (10.8: 3.379) x (22 - (-18)) \u003d 127.9 W.

To calculate losses through windows, the same formula is used, but their thermal resistance, as a rule, is indicated in the passport and it is not necessary to calculate it.

In the thermal insulation of a house, windows are the “weakest link”. A lot of heat goes through them. Multilayer double-glazed windows, heat-reflecting films, double frames will reduce losses, but even this will not help to completely avoid heat loss.

If the windows in the house with dimensions of 1.5 x 1.5 m² are energy-saving, oriented to the North, and the thermal resistance is 0.87 m2 ° C / W, then the losses will be:

Qo \u003d (2.25: 0.87) x (22 - (-18)) \u003d 103.4 tons.

Example of thermal engineering calculation No. 3

Let's perform a thermal calculation of a wooden log building with a facade erected from pine logs with a layer of 0.22 m thick. The coefficient for this material is K = 0.15. In this situation, the heat loss will be:

R \u003d 0.22: 0.15 \u003d 1.47 m² x ⁰С / W.

The lowest temperature of the five-day period is -18⁰, for comfort in the house the temperature is set to 21⁰. The difference will be 39⁰. Based on an area of ​​120 m², the result will be:

Qc \u003d 120 x 39: 1.47 \u003d 3184 watts.

For comparison, we define losses brick house. The coefficient for silicate brick is 0.72.

R \u003d 0.22: 0.72 \u003d 0.306 m² x ⁰С / W.
Qc \u003d 120 x 39: 0.306 \u003d 15,294 watts.

under the same conditions wooden house more economical. Silicate brick for building walls is not suitable at all.

The wooden structure has a high heat capacity. Its enclosing structures keep a comfortable temperature for a long time. Yet, even log house you need to insulate and it is better to do it both from the inside and outside

Heat calculation example No. 4

The house will be built in the Moscow region. For the calculation, a wall made of foam blocks was taken. How is the insulation applied? Finishing of the structure - plaster on both sides. Its structure is lime-sand.

Expanded polystyrene has a density of 24 kg/mᶾ.

The relative humidity in the room is 55% at an average temperature of 20⁰. Layer thickness:

• plaster - 0.01 m;
• foam concrete - 0.2 m;
• expanded polystyrene - 0.065 m.

The task is to find the desired resistance to heat transfer and the actual one. The required Rtr is determined by substituting the values ​​into the expression:

Rtr=a x GSOP+b

where GOSP is the degree-day of the heating season, a and b are the coefficients taken from Table No. 3 of the Code of Rules 50.13330.2012. Since the building is residential, a is 0.00035, b = 1.4.

GSOP is calculated according to the formula taken from the same joint venture:

GOSP \u003d (tin - tot) x zot.

In this formula, tv = 20⁰, tot = -2.2⁰, zot - 205 - the heating period in days. Consequently:

GSOP \u003d (20 - (-2.2)) x 205 \u003d 4551⁰ C x day;

Rtr \u003d 0.00035 x 4551 + 1.4 \u003d 2.99 m2 x C / W.

Using table No. 2 SP50.13330.2012, determine the coefficients of thermal conductivity for each layer of the wall:

• λb1 = 0.81 W/m ⁰С;
• λb2 = 0.26 W/m ⁰С;
• λb3 = 0.041 W/m ⁰С;
• λb4 = 0.81 W/m ⁰С.

The total conditional resistance to heat transfer Ro is equal to the sum of the resistances of all layers. It is calculated by the formula:

Substituting the values ​​get: Rо conv. = 2.54 m2°C/W. Rf is determined by multiplying Ro by a factor r equal to 0.9:

Rf \u003d 2.54 x 0.9 \u003d 2.3 m2 x ° C / W.

The result obliges to change the design of the enclosing element, since the actual thermal resistance is less than the calculated one.

There are many computer services that speed up and simplify calculations.

Thermal engineering calculations are directly related to the definition. You will learn what it is and how to find its meaning from the article we recommend.

Conclusions and useful video on the topic

Performing a heat engineering calculation using an online calculator:

Correct thermal calculation:

A competent heat engineering calculation will allow you to evaluate the effectiveness of the insulation of the external elements of the house, determine the power of the necessary heating equipment.

As a result, you can save on the purchase of materials and heating devices. It is better to know in advance whether the equipment will cope with the heating and air conditioning of the building than to buy everything at random.

Please leave comments, ask questions, post photos on the topic of the article in the block below. Tell us about how the heat engineering calculation helped you choose the heating equipment of the required power or the insulation system. It is possible that your information will be useful to site visitors.

If you are going to build
a small brick cottage, then of course you will have questions: “What
should the wall be thick?”, “Do I need insulation?”, “Which side to put
heater? etc. etc.

In this article, we will try to
figure it out and answer all your questions.

Thermal engineering calculation
enclosing structure is needed, first of all, in order to find out which
thickness should be your outer wall.

First, you need to decide how much
floors will be in your building and depending on this, the calculation is made
enclosing structures according to bearing capacity(not in this article).

Based on this calculation, we determine
the number of bricks in your building's masonry.

For example, it turned out 2 clay
bricks without voids, brick length 250 mm,
mortar thickness 10 mm, total 510 mm (brick density 0.67
will be useful to us later). You decide to cover the outer surface
facing tiles, thickness 1 cm (when buying, be sure to find out
density), and the inner surface with ordinary plaster, layer thickness 1.5
cm, also do not forget to find out its density. In total 535mm.

In order for the building to
collapsed, of course, enough, but unfortunately in most cities
Russian winters are cold and therefore such walls will freeze through. And not to
the walls are frozen, need another layer of insulation.

The thickness of the insulation layer is calculated
in the following way:

1. On the Internet you need to download SNiP
II 3-79* —
"Construction heat engineering" and SNiP 23-01-99 - "Construction climatology".

2. We open SNiP construction
climatology and find your city in table 1 *, and look at the value at the intersection
column "Air temperature of the coldest five-day period, ° С, security
0.98" and strings with your city. For the city of Penza, for example, t n \u003d -32 o C.

3. Estimated indoor air temperature
take

t in = 20 o C.

Heat transfer coefficient for interior wallsa c \u003d 8.7 W / m 2 ˚С

Heat transfer coefficient for external walls in winter conditionsa n \u003d 23W / m 2 ˚С

Normative temperature difference between the temperature of the internal
air and the temperature of the inner surface of the enclosing structuresΔ t n \u003d 4 o C.

4. Next
we determine the required resistance to heat transfer according to the formula # G0 (1a) from building heat engineering
GSOP = (t in - t from.per.) z from.per , GSOP=(20+4.5) 207=507.15 (for the city
Penza).

According to formula (1) we calculate:

(where sigma is directly the thickness
material, and lambda density. Itook as a heater
polyurethane foampanels with a density of 0.025)

We take the thickness of the insulation equal to 0.054 m.

Hence the wall thickness will be:

d = d 1 + d 2 + d 3 + d 4 =

0,01+0,51+0,054+0,015=0,589
m.

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The purpose of thermotechnical calculation is to calculate the thickness of the insulation for a given thickness of the bearing part of the outer wall, which meets sanitary and hygienic requirements and energy saving conditions. In other words, we have external walls with a thickness of 640 mm made of silicate bricks and we are going to insulate them with polystyrene foam, but we do not know what thickness it is necessary to choose a heater in order to comply with building codes.

The thermotechnical calculation of the outer wall of the building is carried out in accordance with SNiP II-3-79 "Construction Heat Engineering" and SNiP 23-01-99 "Construction Climatology".

Table 1

Thermal performance of the building materials used (according to SNiP II-3-79*)

No. according to the scheme	Material	Characteristics of the material in the dry state		Design coefficients (subject to operation according to Appendix 2) SNiP II-3-79*
		Density γ 0, kg / m 3	Thermal conductivity coefficient λ, W/m*°C	Thermal conductivity λ, W/m*°С		Heat absorption (with a period of 24 hours) S, m 2 * ° С / W
	Cement-sand mortar (pos. 71)	1800	0.57	0.76	0.93		11.09
	Brickwork from solid silicate brick (GOST 379-79) on a cement-sand mortar (pos. 87)	1800	0.88	0.76	0.87	9.77	10.90
	Expanded polystyrene (GOST 15588-70) (pos. 144)		0.038	0.038	0.041	0.41	0.49
	Cement-sand mortar - thin-layer plaster (pos. 71)	1800	0.57	0.76	0.93		11.09

1-plaster internal ( cement-sand mortar) - 20 mm

2-brick wall (silicate brick) - 640 mm

3-insulation (polystyrene foam)

4-thin-layer plaster (decorative layer) - 5 mm

When performing a heat engineering calculation, a normal humidity regime in the premises was adopted - operating conditions ("B") in accordance with SNiP II-3-79 v.1 and adj. 2, i.e. the thermal conductivity of the materials used is taken according to column "B".

Calculate the required heat transfer resistance of the fence, taking into account sanitary and comfortable conditions using the formula:

R 0 tr \u003d (t in - t n) * n / Δ t n * α in (1)

where t in is the design temperature of the internal air °С, taken in accordance with GOST 12.1.1.005-88 and design standards

relevant buildings and structures, we accept equal to +22 ° С for residential buildings in accordance with Appendix 4 to SNiP 2.08.01-89;

t n is the calculated winter temperature of the outside air, °С, equal to the average temperature of the coldest five-day period, with a security of 0.92 according to SNiP 23-01-99 for the city of Yaroslavl is taken equal to -31°С;

n is the coefficient accepted according to SNiP II-3-79* (Table 3*) depending on the position of the outer surface of the enclosing structure in relation to the outside air and is taken equal to n=1;

Δ t n - normative and temperature difference between the temperature of the internal air and the temperature of the inner surface of the enclosing structure - is set according to SNiP II-3-79 * (table 2 *) and is taken equal to Δ t n = 4.0 ° С;

R 0 tr \u003d (22- (-31)) * 1 / 4.0 * 8.7 \u003d 1.52

We determine the degree-day of the heating period by the formula:

GSOP \u003d (t in - t from.per) * z from.per. (2)

where t in - the same as in the formula (1);

t from.per - average temperature, ° С, of the period with an average daily air temperature below or equal to 8 ° С according to SNiP 23-01-99;

z from.per - duration, days, of the period with an average daily air temperature below or equal to 8 °C according to SNiP 23-01-99;

GSOP \u003d (22-(-4)) * 221 \u003d 5746 ° C * day.

Let us determine the reduced resistance to heat transfer Ro tr according to the conditions of energy saving in accordance with the requirements of SNiP II-3-79 * (Table 1b *) and sanitary and hygienic and comfortable conditions. Intermediate values ​​are determined by interpolation.

table 2

Heat transfer resistance of enclosing structures (according to SNiP II-3-79*)

Buildings and premises	Degree-day of the heating period, ° C * day	Reduced resistance to heat transfer of walls, not less than R 0 tr (m 2 * ° С) / W
Public administrative and domestic, with the exception of premises with a damp or wet regime	5746	3,41

The resistance to heat transfer of enclosing structures R(0) is taken as the largest of the values ​​calculated earlier:

R 0 tr \u003d 1.52< R 0 тр = 3,41, следовательно R 0 тр = 3,41 (м 2 *°С)/Вт = R 0 .

We write an equation for calculating the actual heat transfer resistance R 0 of the enclosing structure using the formula in accordance with the given design scheme and determine the thickness δ x of the design layer of the fence from the condition:

R 0 \u003d 1 / α n + Σδ i / λ i + δ x / λ x + 1 / α in \u003d R 0

where δ i is the thickness of the individual layers of the fence, except for the calculated one, in m;

λ i - thermal conductivity coefficients of individual layers of the fence (except for the calculated layer) in (W / m * ° C) are taken according to SNiP II-3-79 * (Appendix 3 *) - for this calculation table 1;

δ x - thickness of the design layer of the outer fence, m;

λ x - coefficient of thermal conductivity of the calculated layer of the outer fence in (W / m * ° C) are taken according to SNiP II-3-79 * (Appendix 3 *) - for this calculation table 1;

α in - the heat transfer coefficient of the inner surface of the enclosing structures is taken according to SNiP II-3-79 * (table 4 *) and is taken equal to α in \u003d 8.7 W / m 2 * ° С.

α n - heat transfer coefficient (for winter conditions) the outer surface of the enclosing structure is taken according to SNiP II-3-79 * (table 6 *) and is taken equal to α n = 23 W / m 2 * ° С.

The thermal resistance of a building envelope with sequentially located homogeneous layers should be determined as the sum of the thermal resistances of individual layers.

For external walls and ceilings, the thickness of the heat-insulating layer of the fence δ x is calculated from the condition that the value of the actual reduced resistance to heat transfer of the enclosing structure R 0 must not be less than the normalized value R 0 tr calculated by formula (2):

R 0 ≥ R 0 tr

Expanding the value of R 0 , we get:

R0 = 1 / 23 + (0,02/ 0,93 + 0,64/ 0,87 + 0,005/ 0.93) + δx / 0,041 + 1/ 8,7

Based on this, we determine the minimum value of the thickness of the heat-insulating layer

δ x \u003d 0.041 * (3.41 - 0.115 - 0.022 - 0.74 - 0.005 - 0.043)

δx = 0.10 m

We take into account the thickness of the insulation (polystyrene foam) δ x = 0.10 m

Determine the actual resistance to heat transfer calculated enclosing structures R 0, taking into account the accepted thickness of the heat-insulating layer δ x = 0.10 m

R0 = 1 / 23 + (0,02/ 0,93 + 0,64/ 0,87 + 0,005/ 0,93 + 0,1/ 0,041) + 1/ 8,7

R 0 \u003d 3.43 (m 2 * ° C) / W

Condition R 0 ≥ R 0 tr observed, R 0 = 3.43 (m 2 * ° C) / W ≥ R 0 tr \u003d 3.41 (m 2 * ° C) / W

Creation of comfortable conditions for living or labor activity is the primary goal of construction. A significant part of the territory of our country is located in northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always important. With the growth of energy tariffs, the reduction of energy consumption for heating comes to the fore.

Climate characteristics

The choice of wall and roof construction depends primarily on the climatic conditions of the construction area. To determine them, it is necessary to refer to SP131.13330.2012 "Construction climatology". The following quantities are used in the calculations:

• the temperature of the coldest five-day period with a security of 0.92 is denoted by Tn;
• average temperature, denoted by Tot;
• duration, denoted ZOT.

On the example for Murmansk, the values ​​have the following values:

• Tn=-30 deg;
• Tot=-3.4 deg;
• ZOT=275 days.

In addition, it is necessary to set the design temperature inside the room Tv, it is determined in accordance with GOST 30494-2011. For housing, you can take Tv \u003d 20 degrees.

To perform a heat engineering calculation of enclosing structures, pre-calculate the value of GSOP (degree-day of the heating period):
GSOP = (Tv - Tot) x ZOT.
In our example, GSOP \u003d (20 - (-3.4)) x 275 \u003d 6435.

Main characteristics

For right choice materials of enclosing structures, it is necessary to determine what thermal characteristics they should have. The ability of a substance to conduct heat is characterized by its thermal conductivity, denoted by the Greek letter l (lambda) and is measured in W / (m x deg.). The ability of a structure to retain heat is characterized by its resistance to heat transfer R and is equal to the ratio of thickness to thermal conductivity: R = d/l.

If the structure consists of several layers, the resistance is calculated for each layer and then summed up.

Heat transfer resistance is the main indicator of outdoor construction. Its value must exceed the standard value. When performing a thermal engineering calculation of the building envelope, we must determine the economically justified composition of the walls and roof.

Thermal conductivity values

The quality of thermal insulation is determined primarily by thermal conductivity. Each certified material undergoes laboratory tests, as a result of which this value is determined for operating conditions "A" or "B". For our country, most regions correspond to the operating conditions "B". When performing a heat engineering calculation of the enclosing structures of a house, this value should be used. The thermal conductivity values ​​are indicated on the label or in the material passport, but if they are not available, you can use the reference values ​​\u200b\u200bfrom the Code of Practice. The values ​​for the most popular materials are given below:

• Ordinary brickwork - 0.81 W (m x deg.).
• Silicate brick masonry - 0.87 W (m x deg.).
• Gas and foam concrete (density 800) - 0.37 W (m x deg.).
• Coniferous wood - 0.18 W (m x deg.).
• Extruded polystyrene foam - 0.032 W (m x deg.).
• Mineral wool slabs (density 180) - 0.048 W (m x deg.).

Standard value of resistance to heat transfer

The calculated value of the heat transfer resistance must not be less than the base value. The base value is determined according to Table 3 SP50.13330.2012 "buildings". The table defines the coefficients for calculating the basic values ​​of heat transfer resistance for all enclosing structures and types of buildings. Continuing the started thermal engineering calculation of enclosing structures, an example of calculation can be presented as follows:

• Рsten \u003d 0.00035x6435 + 1.4 \u003d 3.65 (m x deg / W).
• Рpocr \u003d 0.0005x6435 + 2.2 \u003d 5.41 (m x deg / W).
• Rcherd \u003d 0.00045x6435 + 1.9 \u003d 4.79 (m x deg / W).
• Rockna \u003d 0.00005x6435 + 0.3 \u003d x deg / W).

The thermotechnical calculation of the outer enclosing structure is performed for all structures that close the "warm" contour - the floor on the ground or the floor of the technical underground, the outer walls (including windows and doors), the combined cover or the floor of the unheated attic. Also, the calculation must be performed for internal structures, if the temperature difference in adjacent rooms is more than 8 degrees.

Thermal engineering calculation of walls

Most walls and ceilings are multi-layered and heterogeneous in their design. The thermotechnical calculation of the enclosing structures of a multilayer structure is as follows:
R= d1/l1 +d2/l2 +dn/ln,
where n are the parameters of the nth layer.

If we consider a brick plastered wall, we get the following design:

• outer layer of plaster 3 cm thick, thermal conductivity 0.93 W (m x deg.);
• masonry of solid clay bricks 64 cm, thermal conductivity 0.81 W (m x deg.);
• inner layer of plaster 3 cm thick, thermal conductivity 0.93 W (m x deg.).

The formula for the thermotechnical calculation of enclosing structures is as follows:

R \u003d 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 0.85 (m x deg / W).

The obtained value is significantly less than the previously determined base value of the resistance to heat transfer of the walls of a residential building in Murmansk 3.65 (m x deg/W). The wall does not satisfy regulatory requirements and needs to be warmed up. For wall insulation, we use a thickness of 150 mm and a thermal conductivity of 0.048 W (m x deg.).

Having selected the insulation system, it is necessary to perform a verification thermotechnical calculation of the enclosing structures. An example calculation is shown below:

R \u003d 0.15 / 0.048 + 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 3.97 (m x deg / W).

The resulting calculated value is greater than the base value - 3.65 (m x deg / W), the insulated wall meets the requirements of the standards.

The calculation of overlaps and combined coverings is carried out in a similar way.

Thermal engineering calculation of floors in contact with the ground

Often in private houses or public buildings are carried out on the ground. The resistance to heat transfer of such floors is not standardized, but at a minimum the design of the floors must not allow dew to fall out. The calculation of structures in contact with the ground is carried out as follows: the floors are divided into strips (zones) 2 meters wide, starting from the outer boundary. Up to three such zones are allocated, the remaining area belongs to the fourth zone. If the floor structure does not provide for effective insulation, then the heat transfer resistance of the zones is taken as follows:

• 1 zone - 2.1 (m x deg / W);
• zone 2 - 4.3 (m x deg / W);
• zone 3 - 8.6 (m x deg / W);
• 4 zone - 14.3 (m x deg / W).

It is easy to see that the farther the floor area is from the outer wall, the higher its resistance to heat transfer. Therefore, they are often limited to warming the perimeter of the floor. In this case, the heat transfer resistance of the insulated structure is added to the heat transfer resistance of the zone.
The calculation of the resistance to heat transfer of the floor must be included in the overall heat engineering calculation of enclosing structures. An example of the calculation of floors on the ground will be considered below. Let's take the floor area 10 x 10, equal to 100 square meters.

• The area of ​​1 zone will be 64 sq. m.
• The area of ​​zone 2 will be 32 sq. m.
• The area of ​​the 3rd zone will be 4 sq. m.

The average value of the resistance to heat transfer of the floor on the ground:
Rpol \u003d 100 / (64 / 2.1 + 32 / 4.3 + 4 / 8.6) \u003d 2.6 (m x deg / W).

After insulating the perimeter of the floor with a polystyrene foam plate 5 cm thick, with a strip 1 meter wide, we obtain the average value of heat transfer resistance:

Rpol \u003d 100 / (32 / 2.1 + 32 / (2.1 + 0.05 / 0.032) + 32 / 4.3 + 4 / 8.6) \u003d 4.09 (m x deg / W).

It is important to note that not only floors are calculated in this way, but also the structures of walls in contact with the ground (walls of a recessed floor, a warm basement).

Thermotechnical calculation of doors

The basic value of heat transfer resistance is calculated somewhat differently entrance doors. To calculate it, you will first need to calculate the heat transfer resistance of the wall according to the sanitary and hygienic criterion (non-dew):
Rst \u003d (Tv - Tn) / (DTn x av).

Here DТn is the temperature difference between the inner surface of the wall and the air temperature in the room, determined according to the Code of Rules and for housing is 4.0.
av - heat transfer coefficient of the inner surface of the wall, according to the joint venture is 8.7.
The base value of the doors is taken equal to 0.6xRst.

For the selected door design, it is required to perform a verification thermotechnical calculation of enclosing structures. An example of the calculation of the front door:

Рdv \u003d 0.6 x (20-(-30)) / (4 x 8.7) \u003d 0.86 (m x deg / W).

This design value will correspond to a door insulated with a 5 cm thick mineral wool board.

Complex Requirements

Wall, floor or roof calculations are performed to check the element-by-element requirements of the regulations. The set of rules also establishes a complete requirement that characterizes the quality of insulation of all enclosing structures as a whole. This value is called "specific heat-shielding characteristic". Not a single thermotechnical calculation of enclosing structures can do without its verification. An example of a SP calculation is shown below.

Kob = 88.77 / 250 = 0.35, which is less than the normalized value of 0.52. In this case, the area and volume are taken for a house with dimensions of 10 x 10 x 2.5 m. Heat transfer resistances are equal to the base values.

The normalized value is determined in accordance with the joint venture, depending on the heated volume of the house.

In addition to the complex requirement, in order to draw up an energy passport, a thermal engineering calculation of building envelopes is also performed; an example of a passport is given in the appendix to SP50.13330.2012.

Uniformity coefficient

All the above calculations are applicable for homogeneous structures. Which is quite rare in practice. To take into account the inhomogeneities that reduce the resistance to heat transfer, a correction factor for thermal engineering uniformity, r, is introduced. It takes into account the change in heat transfer resistance introduced by window and doorways, external corners, inhomogeneous inclusions (for example, lintels, beams, reinforcing belts), etc.

The calculation of this coefficient is quite complicated, therefore, in a simplified form, you can use approximate values ​​​​from the reference literature. For example, for brickwork - 0.9, three-layer panels - 0.7.

Effective insulation

When choosing a home insulation system, it is easy to make sure that you modern requirements thermal protection without using effective insulation almost impossible. So, if you use a traditional clay brick, you will need masonry several meters thick, which is not economically feasible. At the same time, the low thermal conductivity of modern insulation based on expanded polystyrene or stone wool allows you to limit yourself to thicknesses of 10-20 cm.

For example, to achieve a base heat transfer resistance value of 3.65 (m x deg/W), you would need:

• brick wall 3 m thick;
• masonry from foam concrete blocks 1.4 m;
• mineral wool insulation 0.18 m.