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Thermal engineering calculation of an external brick wall. Thermal engineering calculation with an example Thermal engineering calculation of an external wall made of ceramic bricks

If you are planning to build
small brick cottage, then you will certainly have questions: “Which
thickness should the wall be?”, “Do you need insulation?”, “Which side should you put it on?”
insulation? etc. and so on.

In this article we will try in
understand this and answer all your questions.

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

By this calculation we define
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
It will be useful to us later). You decided to cover the outer surface
facing tiles, thickness 1 cm (be sure to find out when purchasing
density), and the inner surface is ordinary plaster, layer thickness 1.5
cm, also do not forget to find out its density. A total of 535mm.

In order for the building not to
collapsed, this is certainly enough, but unfortunately in most cities
Russian winters are cold and therefore such walls will freeze. And so as not
The walls were frozen, we needed another layer of insulation.

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

1. You need to download SNiP on the Internet
II 3-79* —
« Construction heating engineering"and SNiP 23-01-99 - "Building climatology".

2. 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, °C, security
0.98" and lines with your city. For the city of Penza, for example, t n = -32 o C.

3. Estimated indoor air temperature
take

t in = 20 o C.

Heat transfer coefficient for internal wallsa in = 8.7 W/m 2 ˚С

Heat transfer coefficient for external walls in winter conditionsa n = 23W/m2·˚С

Standard temperature difference between internal temperature
air and temperature inner surface enclosing structuresΔ tn = 4 o C.

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

Using formula (1) we calculate:

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

We take the insulation thickness to be 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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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 calculating efficiency, strength, fire resistance, and durability.

Based on thermal engineering standards, climatic characteristics, steam and moisture permeability, materials are selected for the construction of enclosing structures. 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. Wherein relative humidity must stay within comfortable limits, and this 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 differences in atmospheric temperature indicators internal structures 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 in heat flow. This characteristic is called heat resistance.

By performing a thermal calculation, the optimal limits (minimum and maximum) of the dimensions of walls and ceiling thicknesses are determined. 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 warm water is discharged 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, a “reserve” is provided in the project.

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. Numbers 100 or 150 - specific value 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. WITH - specific heat. 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. 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 in climatic region 1 (Russia), subdistrict 1B. All data is taken from table 1 of SNiP 23-01-99. The coldest temperature observed over five days with a probability of 0.92 is tн = -22⁰С.

In accordance with SNiP, the heating period (zop) lasts 148 days. Average temperature during the heating period with average daily temperature indicators air outside 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.

Rotary and plate heat exchangers provide effective ventilation of residential buildings. 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.

Square external wall 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 portion of the heat is lost through them. Multilayer double-glazed windows will reduce losses, heat reflective films, double frames, 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 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:

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 log house it is necessary to insulate 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 equal to 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 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 the thermal engineering calculation is to calculate the thickness of the insulation for a given thickness of the load-bearing part of the outer wall, which 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(pos. 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 in 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 outer surface enclosing structures 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 23-01-99;

GSOP=(22-(-4))*221=5746 °C*day.

Let's determine the reduced heat transfer resistance 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 heat transfer resistance 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 the 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

It is required to determine the thickness of the insulation in a three-layer brick exterior wall in a residential building located in Omsk. Wall construction: inner layer - ordinary brickwork clay brick 250 mm thick and density 1800 kg/m 3, outer layer - brickwork made of facing bricks thickness 120 mm and density 1800 kg/m 3; located between the outer and inner layers effective insulation made of expanded polystyrene with a density of 40 kg/m 3; The outer and inner layers are connected to each other by fiberglass flexible connections with a diameter of 8 mm, located in increments of 0.6 m.

1. Initial data

Purpose of the building – residential building

Construction area - Omsk

Estimated indoor air temperature t int= plus 20 0 C

Estimated outside air temperature t ext= minus 37 0 C

Estimated indoor air humidity – 55%

2. Determination of normalized heat transfer resistance

Determined according to Table 4 depending on the degree-day of the heating period. Degree-days of the heating season, D d , °С×day, determined by formula 1, based on the average outside temperature and the duration of the heating period.

According to SNiP 23-01-99*, we determine that in Omsk the average outdoor air temperature during the heating period is equal to: t ht = -8.4 0 C, duration of the heating season z ht = 221 days. The degree-day value of the heating period is equal to:

D d = (t int - t ht) z ht = (20 + 8.4)×221 = 6276 0 C day.

According to table. 4. standardized heat transfer resistance Rreg external walls for residential buildings corresponding to the value D d = 6276 0 C day equals R reg = a D d + b = 0.00035 × 6276 + 1.4 = 3.60 m 2 0 C/W.

3. Choosing a design solution for the outer wall

Constructive solution the outer wall is proposed in the assignment and is a three-layer fence with an inner layer of brickwork 250 mm thick, an outer layer of brickwork 120 mm thick, with polystyrene foam insulation between the outer and inner layers. The outer and inner layers are connected to each other by flexible fiberglass ties with a diameter of 8 mm, located in increments of 0.6 m.

4. Determining the thickness of insulation

The thickness of the insulation is determined by formula 7:

d ut = (R reg ./r – 1/a int – d kk /l kk – 1/a ext)× l ut

Where Rreg. – standardized heat transfer resistance, m 2 0 C/W; r– coefficient of thermal homogeneity; a int– heat transfer coefficient of the inner surface, W/(m 2 ×°C); a ext– heat transfer coefficient of the outer surface, W/(m 2 ×°C); d kk- thickness of brickwork, m; l kk– calculated thermal conductivity coefficient of brickwork, W/(m×°С); l ut– calculated thermal conductivity coefficient of insulation, W/(m×°С).

The normalized heat transfer resistance is determined: R reg = 3.60 m 2 0 C/W.

The coefficient of thermal uniformity for a three-layer brick wall with fiberglass flexible connections is about r=0.995, and may not be taken into account in the calculations (for information, if steel flexible connections are used, then the coefficient of thermal uniformity can reach 0.6-0.7).

The heat transfer coefficient of the inner surface is determined from the table. 7 a int = 8.7 W/(m 2 ×°C).

The heat transfer coefficient of the outer surface is taken according to Table 8 a e xt = 23 W/(m 2 ×°C).

The total thickness of the brickwork is 370 mm or 0.37 m.

The calculated thermal conductivity coefficients of the materials used are determined depending on the operating conditions (A or B). Operating conditions are determined in the following sequence:

According to the table 1 we determine the humidity regime of the premises: since the calculated temperature of the internal air is +20 0 C, the calculated humidity is 55%, the humidity regime of the premises is normal;

Using Appendix B (map of the Russian Federation), we determine that the city of Omsk is located in a dry zone;

According to the table 2, depending on the humidity zone and the humidity conditions of the premises, we determine that the operating conditions of the enclosing structures are A.

According to adj. D we determine the thermal conductivity coefficients for operating conditions A: for expanded polystyrene GOST 15588-86 with a density of 40 kg/m 3 l ut = 0.041 W/(m×°C); for brickwork made of ordinary clay bricks on cement-sand mortar with a density of 1800 kg/m 3 l kk = 0.7 W/(m×°C).

Let's substitute all the defined values ​​into formula 7 and calculate minimum thickness polystyrene foam insulation:

d ut = (3.60 – 1/8.7 – 0.37/0.7 – 1/23)× 0.041 = 0.1194 m

Round the resulting value to big side with an accuracy of 0.01 m: d ut = 0.12 m. We perform a verification calculation using formula 5:

R 0 = (1/a i + d kk /l kk + d ut /l ut + 1/a e)

R 0 = (1/8.7 + 0.37/0.7 + 0.12/0.041 + 1/23) = 3.61 m 2 0 S/W

5. Limitation of temperature and moisture condensation on the inner surface of the building envelope

Δt o, °C, between the temperature of the internal air and the temperature of the internal surface of the enclosing structure should not exceed the standardized values Δtn, °С, established in table 5, and is defined as follows

Δt o = n(t int – t ext)/(R 0 a int) = 1(20+37)/(3.61 x 8.7) = 1.8 0 C i.e. less than Δt n = 4.0 0 C, determined from table 5.

Conclusion: t The thickness of polystyrene foam insulation in a three-layer brick wall is 120 mm. At the same time, the resistance to heat transfer of the outer wall R 0 = 3.61 m 2 0 C/W, which is greater than the normalized heat transfer resistance Rreg. = 3.60 m 2 0 C/W on 0.01m 2 0 C/W. Estimated temperature difference Δt o, °C, between the internal air temperature and the temperature of the internal surface of the enclosing structure does not exceed the standard value Δtn,.

An example of thermal engineering calculation of translucent enclosing structures

Translucent enclosing structures (windows) are selected according to the following method.

Standardized heat transfer resistance Rreg determined according to Table 4 SNiP 02/23/2003 (column 6) depending on the degree-day of the heating period D d. At the same time, the type of building and D d taken as in the previous example of thermal engineering calculation of light-opaque enclosing structures. In our case D d = 6276 0 C day, then for a residential building window R reg = a D d + b = 0.00005 × 6276 + 0.3 = 0.61 m 2 0 C/W.

The selection of translucent structures is carried out according to the value of the reduced heat transfer resistance R o r obtained as a result of certification tests or according to Appendix L of the Code of Rules. If the reduced heat transfer resistance of the selected translucent structure R o r, more or equal Rreg, then this design satisfies the requirements of the standards.

Conclusion: for a residential building in Omsk we accept windows in PVC frames with double-glazed windows made of hard glass selective coating and filling the interglass space with argon in which R o r = 0.65 m 2 0 C/W more R reg = 0.61 m 2 0 C/W.

LITERATURE

1. SNiP 02/23/2003. Thermal protection buildings.
2. SP 23-101-2004. Design of thermal protection.
3. SNiP 23-01-99*. Construction climatology.
4. SNiP 01/31/2003. Residential multi-apartment buildings.
5. SNiP 2.08.02-89 *. Public buildings and buildings.