Thermal engineering calculation of an external wall 640. How to make a thermal engineering calculation of the external walls of a low-rise building? Thermal calculation of walls

Initial data

Place of construction - Omsk

z ht = 221 days

t ht = -8.4ºС.

t ext = -37ºС.

t int = + 20ºС;

air humidity: = 55%;

Operating conditions of enclosing structures - B. Heat transfer coefficient of the internal surface of the enclosure A i nt = 8.7 W/m 2 °C.

a ext = 23 W/m 2 °C.

The necessary data on the structural layers of the wall for thermal engineering calculations are summarized in the table.

1. Determination of the degree-day of the heating period using formula (2) SP 23-101-2004:

D d = (t int - t ht) z th = (20–(8.4))·221= 6276.40

2. Standardized value of heat transfer resistance of external walls according to formula (1) SP 23-101-2004:

R reg = a · D d + b =0.00035·6276.40+ 1.4 =3.6m 2 ·°С/W.

3. Reduced resistance to heat transfer R 0 r of external brick walls with effective insulation of residential buildings is calculated by the formula

R 0 r = R 0 conditional r,

where R 0 conventional is the heat transfer resistance of brick walls, conventionally determined by formulas (9) and (11) without taking into account heat-conducting inclusions, m 2 °C/W;

R 0 r - reduced heat transfer resistance taking into account the coefficient of thermal uniformity r, which for walls is 0.74.

The calculation is carried out from the condition of equality

hence,

R 0 conventional = 3.6/0.74 = 4.86 m 2 °C / W

R 0 conventional =R si +R k +R se

R k = R reg - (R si + R se) = 3.6- (1/8.7 + 1/23) = 3.45 m 2 °C / W

4. Thermal resistance of outer brick wall a layered structure can be represented as the sum of the thermal resistances of individual layers, i.e.

R k = R 1 + R 2 + R ut + R 4

5. Determine the thermal resistance of the insulation:

R ut = R k + (R 1 + R 2 + R 4) = 3.45– (0.037 + 0.79) = 2.62 m 2 °C/W.

6. Find the thickness of the insulation:

Ri

= · R ut = 0.032 · 2.62 = 0.08 m.

We accept the insulation thickness as 100 mm.

The final wall thickness will be (510+100) = 610 mm.

We check taking into account the accepted thickness of the insulation:

R 0 r = r (R si + R 1 + R 2 + R ut + R 4 + R se) = 0.74 (1/8.7 + 0.037 + 0.79 + 0.10/0.032+ 1/23 ) = 4.1m 2 °C/W.

Condition R 0 r = 4.1> = 3.6m 2 °C/W is satisfied.

Checking compliance with sanitary and hygienic requirements

thermal protection of the building

1. Check if the condition is met :

∆t = (t int – t ext)/ R 0r a int = (20-(37))/4.1 8.7 = 1.60 ºС

According to table. 5SP 23-101-2004 ∆ t n = 4 °С, therefore, the condition ∆ t = 1,60< ∆t n = 4 ºС is satisfied.

2. Check if the condition is met :

] = 20 – =

20 – 1.60 = 18.40ºС

3. According to Appendix SP 23-101–2004 for internal air temperature t int = 20 ºC and relative humidity = 55% dew point temperature t d = 10.7ºС, therefore, the condition τsi = 18.40> t d = performed.

Conclusion. The enclosing structure satisfies regulatory requirements thermal protection of the building.

4.2 Thermal engineering calculation of the attic covering.

Initial data

Determine the thickness of the attic floor insulation, consisting of insulation δ = 200 mm, vapor barrier, prof. sheet

Attic floor:

Combined coverage:

Place of construction - Omsk

Duration of the heating season z ht = 221 days.

Average design temperature of the heating period t ht = -8.4ºС.

Cold five-day temperature t ext = –37ºС.

The calculation was made for a five-story residential building:

indoor air temperature t int = + 20ºС;

air humidity: = 55%;

humidity conditions premises - normal.

Operating conditions of enclosing structures – B.

Heat transfer coefficient of the inner surface of the fence A i nt = 8.7 W/m 2 °C.

Heat transfer coefficient of the outer surface of the fence a ext = 12 W/m 2 °C.

Name of material Y 0, kg/m³ δ, m λ, mR, m 2 °C/W

1. Determination of the degree-day of the heating period using formula (2) SP 23-101-2004:

D d = (t int - t ht) z th = (20 –8.4) · 221=6276.4ºСsut

2. Normalization of the heat transfer resistance value of the attic floor according to formula (1) SP 23-101-2004:

R reg = a D d + b, where a and b are selected according to table 4 SP 23-101-2004

R reg = a · D d + b = 0.00045 · 6276.4+ 1.9 = 4.72 m² · ºС / W

3. Thermal engineering calculation is carried out from the condition that the total thermal resistance R 0 is equal to the normalized R reg, i.e.

4. From formula (8) SP 23-100-2004 we determine the thermal resistance of the enclosing structure R k (m² ºС / W)

R k = R reg - (R si + R se)

R reg = 4.72 m² ºС / W

R si = 1 / α int = 1 / 8.7 = 0.115 m² ºС / W

R se = 1 / α ext = 1 / 12 = 0.083 m² ºС / W

R k = 4.72– (0.115 + 0.083) = 4.52 m² ºС / W

5. The thermal resistance of the enclosing structure (attic floor) can be represented as the sum of the thermal resistances of the individual layers:

R c = R reinforced concrete + R pi + R cs + R ut → R ut = R c + (R reinforced concrete + R pi + R cs) = R c - (d/ λ) = 4.52 – 0.29 = 4 .23

6. We use formula (6) SP 23-101-2004 to determine the thickness of the insulating layer:

d ut = R ut λ ut = 4.23 0.032 = 0.14 m

7. We accept the thickness of the insulating layer as 150mm.

8. We calculate the total thermal resistance R 0:

R 0 = 1 / 8.7 + 0.005 / 0.17 + 0.15 / 0.032 + 1 / 12 = 0.115 + 4.69+ 0.083 = 4.89 m² ºС / W

R 0 ≥ R reg 4.89 ≥ 4.72 satisfies the requirement

Checking the fulfillment of conditions

1. Check the fulfillment of the condition ∆t 0 ≤ ∆t n

The value of ∆t 0 is determined by formula (4) SNiP 02/23/2003:

∆t 0 = n ·(t int - t ext) / R 0 · a int where, n is a coefficient that takes into account the dependence of the position of the outer surface to the outside air according to the table. 6

∆t 0 = 1(20+37) / 4.89 8.7 = 1.34ºС

According to table. (5) SP 23-101-2004 ∆t n = 3 ºС, therefore, the condition ∆t 0 ≤ ∆t n is satisfied.

2. Check the fulfillment of condition τ >t d

τ value calculated using formula (25) SP 23-101-2004

t si = t int– [n(t int–t ext)]/(R o a int)

τ = 20- 1(20+26) / 4.89 8.7 = 18.66 ºС

3. According to Appendix R SP 23-01-2004 for internal air temperature t int = +20 ºС and relative humidity φ = 55% dew point temperature t d = 10.7 ºС, therefore, condition τ >t d is fulfilled.

Conclusion: attic floor meets regulatory requirements.

Creation comfortable conditions for accommodation or labor activity is the primary task 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 rising energy tariffs, reducing energy consumption for heating comes to the fore.

Climatic characteristics

The choice of wall and roof design depends primarily on climatic conditions construction area. To determine them, you need to refer to SP131.13330.2012 “Building climatology”. The following values ​​are used in the calculations:

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

In the example for Murmansk, the values ​​are following values:

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

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

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

Basic indicators

For the 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 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 the external structure. Its value must exceed the standard value. When performing thermal engineering calculations 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 operating conditions “B”. When performing thermal engineering calculations of the building envelope, this value should be used. Thermal conductivity values ​​are indicated on the label or in the material passport, but if they are not available, you can use reference values ​​from the Code of Practice. Values ​​for most popular materials are given below:

• Masonry made of ordinary brick - 0.81 W (m x deg.).
• Sand-lime brickwork - 0.87 W (m x deg.).
• Gas and foam concrete (density 800) - 0.37 W (m x deg.).
• Wood coniferous species- 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 heat transfer resistance

The calculated value of heat transfer resistance should not be less than base value. The basic 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 of all enclosing structures and types of buildings. Continuing the started thermal engineering calculation of enclosing structures, an example of the calculation can be presented in the following way:

• Rsten = 0.00035x6435 + 1.4 = 3.65 (m x deg/W).
• Rpokr = 0.0005x6435 + 2.2 = 5.41 (m x deg/W).
• Rcherd = 0.00045x6435 + 1.9 = 4.79 (m x deg/W).
• Rokna = 0.00005x6435 + 0.3 = x deg/W).

Thermal engineering calculations of the external enclosing structure are performed for all structures that close the “warm” circuit - the floor on the ground or the ceiling of a technical underground, external walls (including windows and doors), a combined covering or the ceiling of an unheated attic. The calculation must also be performed for internal structures, if the temperature difference in adjacent rooms is more than 8 degrees.

Thermal calculation of walls

Most walls and ceilings are multi-layered and heterogeneous in their design. Thermal engineering calculation of 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 made of solid clay brick 64 cm, thermal conductivity 0.81 W (m x deg.);
• the inner layer of plaster is 3 cm thick, thermal conductivity 0.93 W (m x deg.).

The formula for thermal engineering calculation of enclosing structures is as follows:

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

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

Having selected an insulation system, it is necessary to perform a verification thermal engineering calculation of the enclosing structures. An example calculation is given below:

R=0.15/0.048 + 0.03/0.93 + 0.64/0.81 + 0.03/0.93 = 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 floors and combined coverings is carried out similarly.

Thermal engineering calculation of floors in contact with the ground

Often in private homes or public buildings are carried out on the ground. The heat transfer resistance of such floors is not standardized, but at a minimum the design of the floors should not allow dew to occur. 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 border. There are up to three such zones; the remaining area belongs to the fourth zone. If the floor design does not provide effective insulation, then the heat transfer resistance of the zones is assumed to be 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);
• Zone 4 - 14.3 (m x deg/W).

It is easy to notice that the further the floor area is from external wall, the higher its resistance to heat transfer. Therefore, they are often limited to insulating 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 heat transfer resistance of the floor must be included in the general thermal engineering calculation of the enclosing structures. We will consider an example of calculating floors on the ground below. Let's take a floor area of ​​10 x 10 equal to 100 square meters.

• The area of ​​zone 1 will be 64 square meters.
• The area of ​​zone 2 will be 32 square meters.
• The area of ​​zone 3 will be 4 square meters.

Average value of resistance to heat transfer of the floor over the ground:
Rpol = 100 / (64/2.1 + 32/4.3 + 4/8.6) = 2.6 (m x deg/W).

Having insulated the perimeter of the floor polystyrene foam board 5 cm thick, 1 meter wide strip, we obtain the average value of heat transfer resistance:

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

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

Thermal calculation of doors

The basic value of heat transfer resistance is calculated slightly 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 (no dew):
Rst = (Tv - Tn)/(DTn x av).

Here DTn 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.
ab is the heat transfer coefficient of the inner surface of the wall, according to SP is 8.7.
The basic value of doors is taken equal to 0.6xРst.

For the selected door design, it is necessary to perform a verification thermal engineering calculation of the enclosing structures. An example of calculating an entrance door:

Rdv = 0.6 x (20-(-30))/(4 x 8.7) = 0.86 (m x deg/W).

This calculated value will correspond to a door insulated with a 5 cm thick mineral wool slab. Its heat transfer resistance will be R=0.05 / 0.048=1.04 (m x deg/W), which is greater than the calculated one.

Comprehensive Requirements

Calculations of walls, floors or coverings are performed to verify the element-by-element requirements of the standards. The set of rules also establishes a comprehensive requirement characterizing the quality of insulation of all enclosing structures as a whole. This value is called “specific thermal protection characteristic”. Not a single thermal engineering calculation of enclosing structures can be done without checking it. An example of calculation for a joint venture is given below.

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

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

In addition to the comprehensive requirement for drawing up energy passport They also carry out thermal engineering calculations of enclosing structures; an example of obtaining a passport is given in the appendix to SP50.13330.2012.

Uniformity coefficient

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

The calculation of this coefficient is quite complicated, so in a simplified form you can use approximate values ​​from 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 can meet modern thermal protection requirements without using effective insulation almost impossible. So, if you use traditional clay bricks, you will need masonry several meters thick, which is not economically feasible. However, low thermal conductivity modern insulation materials based on expanded polystyrene or stone wool allows you to limit yourself to thicknesses of 10-20 cm.

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

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

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

Based on thermal engineering standards, climatic characteristics, steam and moisture permeability, the selection of materials for the construction of enclosing structures is carried out. We will look at how to perform this calculation in the article.

Much depends on the thermal technical features of the building's permanent enclosures. That and humidity structural elements, and temperature indicators that affect the presence or absence of condensation on interior partitions and floors.

The calculation will show whether stable temperature and humidity characteristics will be maintained at positive and sub-zero temperature. 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 having all this data. Based on them, the thickness of the walls and ceilings and the sequence of layers are chosen.

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

Thermal engineering calculation aims to determine:

1. Are the designs identical to the stated requirements in terms of thermal protection?
2. How fully is a comfortable microclimate inside the building ensured?
3. Is optimal thermal protection of structures provided?

The basic principle is maintaining a balance of the difference in temperature indicators of the atmosphere of internal structures of fences and premises. If this is not followed, heat will be absorbed by these surfaces and the temperature inside will remain very low.

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

By doing thermal calculation determine the optimal limits (minimum and maximum) of the dimensions of walls and ceiling thicknesses. This guarantees the operation of the building over a long period, both without extreme freezing of structures or overheating.

Options for performing calculations

To perform heat calculations, you need initial parameters.

They depend on a number of characteristics:

1. Purpose of the building and its type.
2. Orientations of vertical enclosing structures relative to the cardinal directions.
3. Geographical parameters of the future home.
4. The volume of the building, its number of storeys, area.
5. Door types and dimensions, window openings.
6. Type of heating and its technical parameters.
7. Number of permanent residents.
8. Materials for vertical and horizontal fencing structures.
9. Upper floor ceilings.
10. Hot water supply equipment.
11. Type of ventilation.

Others are also taken into account when calculating design features buildings. The air permeability of enclosing structures should not contribute to excessive cooling inside the house and reduce the thermal protection characteristics of the elements.

Heat loss is also caused by waterlogging of the walls, and in addition, this entails dampness, which negatively affects the durability of the building.

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

Formulas for making calculations

Heat loss from a home can be divided into two main parts: losses through the building envelope and losses caused by operation. In addition, heat is lost when discharging warm water into the sewer system.

For the materials from which the enclosing structures are constructed, 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 multilayer, all obtained values ​​are added together.

The easiest way to determine the size of heat losses is by adding up the thermal flows through the enclosing structures that actually form this building

Guided by this methodology, they take into account the fact that the materials that make up the structure have a different structure. It is also taken into account that the heat flow passing through them has different specifics.

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

Q = (A / R) x dT

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

Performing 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 dT parameter, taking into account not the lowest temperature, but the average one.

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, humidity maps are used in calculations.

There is a formula for this:

W = ((Q + Qв) x 24 x N)/1000

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

Disadvantages of area calculation

Calculation based on the area indicator is not very accurate. Here, such parameters as climate, temperature indicators, both minimum and maximum, and humidity are not taken into account. Due to ignoring many important points, the calculation has significant errors.

Often trying to cover them, the project includes a “reserve”.

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

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

The formula for area is:

Q=S x 100 (150) W.

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

House ventilation losses

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 by supply ventilation. Exhaust ventilation promotes care warm air. The recuperator-heat exchanger reduces losses through ventilation. It does not allow heat to escape along with the outgoing air, and it heats the incoming air flows

It is envisaged that the air inside the building will be completely renewed in one hour. Buildings built according to the DIN standard have walls with vapor barriers, so here the air exchange rate is taken to be two.

There is a formula that determines heat loss through the ventilation system:

Qv = (V x Kv: 3600) x P x C x dT

Here the symbols mean the following:

1. Qв - 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 - air exchange rate.
5. C - 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 of the heating system. If the power value is too high, a way out of the situation may be. Let's look at a few examples for houses made of different materials.

Example of thermal engineering calculation No. 1

Let's calculate a residential building located 1 climatic region(Russia), subdistrict 1B. All data is taken from table 1 of SNiP 23-01-99. Most cold temperature, observed over five days with a probability of 0.92 - tн = -22⁰С.

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

Heat loss at home - the most important moment at the design stage. The choice of building materials and insulation 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 was stipulated that the temperature in the rooms of the house should be 22⁰. The house has two floors and walls 0.5 m thick. Its height is 7 m, dimensions in plan are 10 x 10 m. The material of the vertical enclosing structures is warm ceramics. For it, the thermal conductivity coefficient is 0.16 W/m x C.

Mineral wool was used as external insulation, 5 cm thick. The Kt value for it 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, you need to define the thermal resistance as ceramic wall, and insulation. In the first case, R1 = 0.5: 0.16 = 3.125 sq. m x C/W. In the second - R2 = 0.05: 0.04 = 1.25 sq. 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 loss is directly proportional to the area of ​​the enclosing structures, we calculate the area of ​​the walls:

A = 10 x 4 x 7 – 15 x 2.5 = 242.5 m²

Now you can determine heat loss through the walls:

Qс = (242.5: 4.375) x (22 – (-22)) = 2438.9 W.

Heat loss through horizontal enclosing structures is calculated in a similar way. In the end, all the results are summed up.

If the basement under the floor of the first floor is heated, the floor does not need to be insulated. It is still better to line the basement walls with insulation so that the heat does not escape 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 = 10x10x7 = 700 mᶾ.

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

Qв = (700 x 2) : 3600) x 1.2047 x 1005 x (22 – (-22)) = 20,776 W.

If Kv = 1:

Qв = (700 x 1) : 3600) x 1.2047 x 1005 x (22 – (-22)) = 10,358 W.

Effective ventilation residential buildings provide rotary and plate recuperators. The efficiency of the former is higher, it reaches 90%.

Example of thermal engineering calculation No. 2

It is required to calculate losses through a 51 cm thick brick wall. It is insulated with a 10 cm layer mineral wool. Outside – 18⁰, inside – 22⁰. The dimensions of the wall are 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 thermal conductivity coefficient Kt = 0.58 W/mºC, for mineral wool - 0.04 W/mºC. Thermal Resistance:

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

External wall area A = 2.7 x 4 = 10.8 m²

Heat loss through the wall:

Qс = (10.8: 3.379) x (22 – (-18)) = 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 does not need to be calculated.

In the thermal insulation of a house, windows are the “weak link”. A fairly large proportion of heat is lost through them. Multilayer double-glazed windows, heat-reflecting films, double frames will reduce losses, but even this will not help avoid heat loss completely

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

Qо = (2.25: 0.87) x (22 – (-18)) = 103.4 t.

Example of thermal engineering calculation No. 3

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

R = 0.22: 0.15 = 1.47 m² x ⁰С/W.

The most low temperature five-day period - -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:

Qс = 120 x 39: 1.47 = 3184 W.

For comparison, let’s define the losses brick house. The coefficient for sand-lime brick is 0.72.

R = 0.22: 0.72 = 0.306 m² x ⁰С/W.
Qс = 120 x 39: 0.306 = 15,294 W.

Under the same conditions wooden house more economical. Sand-lime brick is not suitable for building walls here at all.

The wooden structure has a high heat capacity. Its enclosing structures are stored for a long time comfortable temperature. Still, even a log house needs to be insulated and it is better to do this both 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 the insulation is applied. The finishing of the structure is plaster on both sides. Its structure is limestone-sand.

Expanded polystyrene has a density of 24 kg/mᶾ.

Relative air 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 required heat transfer resistance and the actual one. The required Rtr is determined by substituting the values ​​in the expression:

Rtr=a x GSOP+b

where GOSP is the degree-day of the heating season, a and b are 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 using a formula taken from the same SP:

GOSP = (tv – tot) x zot.

In this formula tв = 20⁰, tоt = -2.2⁰, zоt - 205 is the heating period in days. Hence:

GSOP = (20 – (-2.2)) x 205 = 4551⁰ C x day;

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

Using table No. 2 SP50.13330.2012, determine the thermal conductivity coefficients 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 using the formula:

Substituting the values ​​we get: Rо arb. = 2.54 m2°C/W. Rф is determined by multiplying Ro by a coefficient r equal to 0.9:

Rf = 2.54 x 0.9 = 2.3 m2 x °C/W.

The result requires changing 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 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 thermal engineering calculations using an online calculator:

Correct thermal engineering calculation:

A competent thermotechnical calculation will allow you to evaluate the effectiveness of insulating the external elements of the house and determine the power of the necessary heating equipment.

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

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

The purpose of thermal engineering calculation is to calculate the thickness of the insulation at given thickness load-bearing part of the outer wall that meets sanitary and hygienic requirements and energy saving conditions. In other words, we have external walls 640 mm thick made of sand-lime brick and we are going to insulate them with polystyrene foam, but we don’t know what thickness of insulation we need to choose in order to comply with building standards.

Thermal calculation outer wall 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 indicators of the building materials used (according to SNiP II-3-79*)

Scheme no.	Material	Characteristics of the material in a 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*°С	Thermal conductivity λ, W/m*°С		Heat absorption (with a period of 24 hours) S, m 2 *°C/W
	Cement-sand mortar (item 71)	1800	0.57	0.76	0.93		11.09
	Brickwork made of solid silicate brick (GOST 379-79) on cement-sand mortar (item 87)	1800	0.88	0.76	0.87	9.77	10.90
	Expanded polystyrene (GOST 15588-70) (item 144)		0.038	0.038	0.041	0.41	0.49
	Cement-sand mortar - thin-layer plaster (item 71)	1800	0.57	0.76	0.93		11.09

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

2-brick wall ( sand-lime brick) - 640 mm

3-insulation (expanded polystyrene)

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

When performing thermal engineering calculations, the normal humidity regime in the premises was adopted - operating conditions (“B”) in accordance with SNiP II-3-79 t.1 and adj. 2, i.e. We take the thermal conductivity of the materials used according to column “B”.

Let's calculate the required heat transfer resistance of the fence, taking into account sanitary, hygienic and comfortable conditions using the formula:

R 0 tr = (t in – t n) * n / Δ t n *α in (1)

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

corresponding buildings and structures, we take equal to +22 °C for residential buildings in accordance with Appendix 4 to SNiP 2.08.01-89;

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

n – 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 - standard and temperature difference between the temperature of the internal air and the temperature of the internal surface of the enclosing structure - is established according to SNiP II-3-79* (Table 2*) and is taken equal to Δ t n = 4.0 °C;

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

Let us determine the degree-day of the heating period using the formula:

GSOP= (t in – t from.trans.)*z from.trans. (2)

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

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

GSOP=(22-(-4))*221=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, 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-days of the heating period, ° C*days	Reduced resistance to heat transfer of walls, not less than R 0 tr (m 2 *°C)/W
Public administrative and domestic, with the exception of rooms with damp or wet conditions	5746	3,41

We take the heat transfer resistance of enclosing structures R(0) as the greatest of the values ​​​​calculated earlier:

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

Let us 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 enclosure from the condition:

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

where δ i is the thickness of individual layers of the fence other than the calculated one in m;

λ i – thermal conductivity coefficients of individual fencing layers (except for the design 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 in m;

λ x – thermal conductivity coefficient of the design 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 internal surface of enclosing structures is taken according to SNiP II-3-79* (Table 4*) and is taken equal to α in = 8.7 W/m 2 *°C.

α n - heat transfer coefficient (for winter conditions) of 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 *°C.

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

For external walls and ceilings, the thickness of the thermal insulation 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 be no less than the standardized 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) + δ x / 0,041 + 1/ 8,7

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

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

δ x = 0.10 m

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

Determine the actual heat transfer resistance calculated enclosing structures R 0, taking into account the accepted thickness of the thermal insulation 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 = 3.43 (m 2 *°C)/W

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

In the climatic conditions of northern latitudes, a correctly made thermal calculation of a building is extremely important for builders and architects. The obtained indicators will give for design necessary information, including information about the materials used for construction, additional insulation, floors and even finishing.

In general, heat calculation affects several procedures:

• taking into account by designers when planning the layout of rooms, load-bearing walls and fencing;
• creation of a heating system and ventilation structure project;
• selection of building materials;
• analysis of the operating conditions of the building.

All this is connected by single values ​​​​obtained as a result of settlement operations. In this article we will tell you how to make a thermal calculation of the outer wall of a building, and also give examples of using this technology.

Objectives of the procedure

A number of goals are relevant only for residential buildings or, conversely, industrial premises, but most of the problems being solved are suitable for all buildings:

• Maintaining comfortable climatic conditions inside the rooms. The term “comfort” includes both the heating system and the natural conditions for heating the surface of the walls, roof, and the use of all heat sources. The same concept includes the air conditioning system. Without proper ventilation, especially in production, the premises will be unsuitable for work.
• Saving electricity and other heating resources. The following meanings apply here:
  • specific heat capacity of the materials and cladding used;
  • climate outside the building;
  • heating power.

It is extremely uneconomical to carry out heating system, which simply will not be used to the proper extent, but will be difficult to install and expensive to maintain. The same rule can be applied to expensive building materials.

Thermal engineering calculation - what is it?

Heat calculation allows you to set the optimal (two boundaries - minimum and maximum) thickness of the enclosing walls and load-bearing structures, which will ensure long-term operation without freezing and overheating of ceilings and partitions. In other words, this procedure allows you to calculate the real or expected, if it is carried out at the design stage, the thermal load of the building, which will be considered the norm.

The analysis is based on the following data:

• room design - presence of partitions, heat-reflecting elements, ceiling height, etc.;
• features of the climate regime in a given area - maximum and minimum temperature limits, the difference and rapidity of temperature changes;
• the location of the building in the cardinal directions, that is, taking into account the absorption of solar heat, at what time of day there is maximum susceptibility of heat from the sun;
• mechanical influences And physical properties construction site;
• indicators of air humidity, the presence or absence of protection of walls from moisture penetration, the presence of sealants, including sealing impregnations;
• operation of natural or artificial ventilation, presence of " greenhouse effect", vapor permeability and much more.

At the same time, the assessment of these indicators must comply with a number of standards - the level of resistance to heat transfer, air permeability, etc. Let's consider them in more detail.

Requirements for thermal engineering calculations of the premises and related documentation

State inspection bodies that govern the organization and regulation of construction, as well as checking the implementation of safety regulations, have drawn up SNiP No. 23-02-2003, which sets out in detail the standards for carrying out measures for the thermal protection of buildings.

The document proposes engineering solutions that will ensure the most economical consumption of heat energy, which is spent on heating premises (residential or industrial, municipal) during the heating period. These recommendations and requirements have been developed taking into account ventilation, air conversion, and the location of heat entry points.

SNiP is a bill at the federal level. Regional documentation is presented in the form of TSN - territorial construction standards.

Not all buildings are within the jurisdiction of these codes. In particular, those buildings that are heated irregularly or are constructed without heating are not checked according to these requirements. Heat calculations are mandatory for the following buildings:

• residential - private and apartment buildings;
• public, municipal - offices, schools, hospitals, kindergartens, etc.;
• industrial – factories, concerns, elevators;
• agricultural - any heated buildings for agricultural purposes;
• warehouses – barns, warehouses.

The text of the document specifies standards for all those components that are included in thermal analysis.

Design requirements:

• Thermal insulation. This is not only the preservation of heat in the cold season and the prevention of hypothermia and freezing, but also protection from overheating in the summer. Isolation, therefore, must be two-way - preventing influences from the outside and the release of energy from within.
• The permissible value of the temperature difference between the atmosphere inside the building and the thermal regime of the interior of the enclosing structures. This will lead to the accumulation of condensation on the walls, as well as negative influence on the health of people in the premises.
• Thermal stability, that is, temperature stability, preventing sudden changes in the heated air.
• Breathability. Balance is important here. On the one hand, the building cannot be allowed to cool down due to active heat transfer; on the other hand, it is important to prevent the occurrence of the “greenhouse effect”. It happens when synthetic, “non-breathable” insulation is used.
• No dampness. High humidity– this is not only the reason for the appearance of mold, but also an indicator due to which serious losses of heat energy occur.

How to do thermal engineering calculations of the walls of a house - basic parameters

Before proceeding with direct heat calculations, you need to collect detailed information about the construction. The report will include answers to the following points:

• The purpose of the building is residential, industrial or public premises, specific purpose.
• The geographic latitude of the area where the facility is or will be located.
• Climatic features of the area.
• The direction of the walls is to the cardinal points.
• Dimensions entrance structures And window frames- their height, width, permeability, type of windows - wooden, plastic, etc.
• Power of heating equipment, layout of pipes, batteries.
• Average number of residents or visitors, workers, if this industrial premises, which are inside the walls at the same time.
• Building materials from which floors, ceilings and any other elements are made.
• Presence or absence of supply hot water, the type of system that is responsible for this.
• Features of ventilation, both natural (windows) and artificial - ventilation shafts, air conditioning.
• The configuration of the entire building - the number of floors, the total and individual area of ​​​​the premises, the location of the rooms.

Once this data has been collected, the engineer can begin calculations.

We offer you three methods that are most often used by specialists. You can also use combined method, when the facts are taken from all three possibilities.

Options for thermal calculation of enclosing structures

Here are three indicators that will be taken as the main ones:

• building area from the inside;
• volume outside;
• specialized thermal conductivity coefficients of materials.

Heat calculation by area of ​​premises

Not the most economical, but the most frequent, especially in Russia, method. It involves primitive calculations based on the area indicator. This does not take into account climate, band, minimum and maximum temperature values, humidity, etc.

Also, the main sources of heat loss are not taken into account, such as:

• Ventilation system – 30-40%.
• Roof slopes – 10-25%.
• Windows and doors – 15-25%.
• Walls – 20-30%.
• Floor on the ground – 5-10%.

These inaccuracies are due to failure to take into account the majority important elements lead to the fact that the heat calculation itself may have a strong error in both directions. Usually, engineers leave a “reserve”, so they have to install heating equipment that is not fully used or threatens severe overheating. There are often cases when heating and air conditioning systems are installed at the same time, because they cannot correctly calculate heat loss and heat gain.

“Bigger” indicators are used. Disadvantages of this approach:

• expensive heating equipment and materials;
• uncomfortable indoor microclimate;
• additional installation of automated control for temperature conditions;
• possible freezing of walls in winter.

Q=S*100 W (150 W)

• Q is the amount of heat required for a comfortable climate in the entire building;
• W S – heated area of ​​the room, m.

The value of 100-150 Watts is a specific indicator of the amount of thermal energy required to heat 1 m2.

If you choose this method, then listen to the following tips:

• If the height of the walls (to the ceiling) is no more than three meters, and the number of windows and doors per surface is 1 or 2, then multiply the result by 100 W. Usually everything residential buildings, both private and multi-family, use this value.
• If the design contains two window openings or a balcony, loggia, then the indicator increases to 120-130 W.
• For industrial and warehouse premises, a coefficient of 150 W is more often taken.
• When choosing heating devices(radiators), if they are located near a window, it is worth increasing their designed power by 20-30%.

Thermal calculation of enclosing structures according to the volume of the building

Typically this method is used for those buildings where high ceilings– more than 3 meters. That is, industrial facilities. The disadvantage of this method is that air conversion is not taken into account, that is, the fact that it is always warmer at the top than at the bottom.

Q=V*41 W (34 W)

• V – external volume of the building in cubic meters;
• 41 W is the specific amount of heat required to heat one cubic meter of a building. If construction is carried out using modern building materials, then the figure is 34 W.

• Glass in windows:
  • double package – 1;
  • binding – 1.25.
• Insulation materials:
  • new modern developments – 0,85;
  • standard brickwork in two layers – 1;
  • small wall thickness – 1.30.
• Air temperature in winter:
  • -10 – 0,7;
  • -15 – 0,9;
  • -20 – 1,1;
  • -25 – 1,3.
• Percentage of windows compared to total surface area:
  • 10% – 0,8;
  • 20% – 0,9;
  • 30% – 1;
  • 40% – 1,1;
  • 50% – 1,2.

All these errors can and should be taken into account, however, they are rarely used in real construction.

An example of thermal engineering calculation of the external building envelope by analyzing the insulation used

If you are building a residential building or cottage yourself, we strongly recommend that you think through everything down to the smallest detail in order to ultimately save money, create an optimal climate inside, and ensure long-term operation of the facility.

To do this, you need to solve two problems:

• make the correct heat calculation;
• install a heating system.

Example data:

• corner living room;
• one window – 8.12 sq. m;
• region – Moscow region;
• wall thickness – 200 mm;
• area according to external parameters – 3000*3000.

It is necessary to find out how much power is needed to heat 1 square meter of space. The result will be Qsp = 70 W. If the insulation (wall thickness) is smaller, the values ​​will also be lower. Let's compare:

• 100 mm – Qsp = 103 W.
• 150 mm – Qsp = 81 W.

This indicator will be taken into account when installing heating.

Software for heating system design

By using computer programs from the ZVSOFT company you can calculate all the materials spent on heating, as well as make a detailed floor plan of communications with radiators displayed, specific heat capacity, energy consumption, nodes.

The company offers basic CAD for design work of any complexity - . In it you can not only design a heating system, but also create detailed diagram for the construction of the entire house. This can be realized thanks to the large functionality, number of tools, as well as work in two- and three-dimensional space.

You can install an add-on to the base software. This program is designed to design all engineering systems, including for heating. WITH using easy line tracing and plan layering functions, you can design several communications on one drawing - water supply, electricity, etc.

Before building a house, do a thermal engineering calculation. This will help you not make a mistake with the choice of equipment and the purchase of building materials and insulation.