Thermotechnical calculation of building envelopes. Thermotechnical calculation of structures: what is it and how is it carried out Thermotechnical calculation of the outer wall of masonry 640

Creating comfortable conditions for living or working is the primary task of construction. A significant part of the territory of our country is located in the northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always relevant. With rising energy tariffs, lower energy consumption for heating comes to the fore.

Climatic characteristics

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

• temperature of the coldest five-day security of 0.92, indicated by Tn;
• average temperature, Thoth is designated;
• duration is indicated by ZOT.

For an example for Murmansk, the values \u200b\u200bhave the following meanings:

• Tn \u003d -30 deg;
• That \u003d -3.4 deg;
• ZOT \u003d 275 days.

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

To perform the thermotechnical calculation of building envelopes, the GSOP value (degree-day of the heating period) is preliminarily calculated:
GSOP \u003d (TV - That) x ZOT.
In our example, GSOP \u003d (20 - (-3.4)) x 275 \u003d 6435.

Main characteristics

For the correct choice of materials for enclosing structures, it is necessary to determine what thermotechnical characteristics they should have. The ability of a substance to conduct heat is characterized by its thermal conductivity, is indicated 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 \u003d d / l.

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

Heat transfer resistance is the main indicator of the outdoor structure. Its value should exceed the normative value. Performing the heat engineering calculation of the building envelope, we must determine the economically sound composition of the walls and roof.

Thermal conductivity

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 the operating conditions “A” or “B”. For our country, the conditions of operation "B" correspond to most regions. Performing the heat engineering calculation of the building envelope, this value should be used. The values \u200b\u200bof thermal conductivity are indicated on the label or in the passport of the material, but if they are not, you can use the reference values \u200b\u200bfrom the Code of Practice. Values \u200b\u200bfor the most popular materials are listed below:

• Masonry from ordinary brick - 0.81 W (m x deg.).
• Masonry made of silicate brick - 0.87 W (m x deg.).
• Gas and foam concrete (with a density of 800) - 0.37 W (m x deg.).
• Coniferous wood - 0.18 W (m x city).
• Extruded polystyrene foam - 0.032 W (m x deg.).
• Mineral wool plates (density 180) - 0.048 W (m x deg.).

Standard value of heat transfer resistance

The calculated value of the heat transfer resistance should not be less than the base value. The base value is determined according to table 3 of SP50.13330.2012 “buildings”. The table defines the coefficients for calculating the basic values \u200b\u200bof resistance to heat transfer of all building envelopes and types of buildings. Continuing the initiated thermal engineering calculation of enclosing structures, an example of calculation can be represented as follows:

• Rsten \u003d 0.00035x6435 + 1.4 \u003d 3.65 (mx deg / W).
• Рпокр \u003d 0.0005х6435 + 2.2 \u003d 5.41 (mx deg / W).
• Rcherd \u003d 0.00045x6435 + 1.9 \u003d 4.79 (mx deg / W).
• Rokna \u003d 0.00005x6435 + 0.3 \u003d x deg / W).

The thermotechnical calculation of the external building envelope is performed for all structures that close the “warm” circuit — ground floor or technical underfloor overlap, external walls (including windows and doors), combined covering or overlapping of an unheated attic. Also, the calculation must be performed for internal structures, if the temperature difference in adjacent rooms is more than 8 degrees.

Thermotechnical calculation of walls

Most walls and ceilings are multilayered and heterogeneous in design. The thermal engineering calculation of building envelopes of a multilayer structure is as follows:
R \u003d 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:

• the outer layer of plaster is 3 cm thick, the thermal conductivity is 0.93 W (m x deg.);
• masonry of full-clay clay brick 64 cm, thermal conductivity 0.81 W (m x deg.);
• the inner layer of plaster is 3 cm thick, the thermal conductivity is 0.93 W (m x deg.).

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

R \u003d 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 0.85 (mx 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 to be insulated. For wall insulation we use a thickness of 150 mm and a thermal conductivity of 0.048 W (m x deg.).

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

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

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

The calculation of ceilings and combined coatings is performed similarly.

Thermotechnical calculation of floors in contact with the soil

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 least the floor design should not allow dew to fall. The calculation of structures in contact with the soil is performed as follows: the floors are divided into strips (zones) of 2 meters wide, starting from the outer border. There are up to three such zones; the remaining area belongs to the fourth zone. If an effective insulation is not provided for in the floor structure, then the heat transfer resistance of the zones is taken as follows:

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

It is easy to see that the farther the floor is from the external wall, the higher its resistance to heat transfer. Therefore, they are often limited to warming the perimeter of the floor. At the same time, the heat transfer resistance of the insulated structure is added to the heat transfer resistance of the zone.
Calculation of resistance to heat transfer to the floor must be included in the general heat engineering calculation of building envelopes. An example of calculating floors on the ground will be considered below. Take a floor area of \u200b\u200b10 x 10, equal to 100 sq. M.

• The area of \u200b\u200b1 zone will be 64 square meters.
• The area of \u200b\u200b2 zones will be 32 square meters.
• The area of \u200b\u200b3 zones will be 4 square meters.

The average value of the resistance to heat transfer of the floor through the soil:
Rpola \u003d 100 / (64 / 2.1 + 32 / 4.3 + 4 / 8.6) \u003d 2.6 (mx deg / W).

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

Rpola \u003d 100 / (32 / 2.1 + 32 / (2.1 + 0.05 / 0.032) + 32 / 4.3 + 4 / 8.6) \u003d 4.09 (mx deg / W).

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

Thermotechnical calculation of doors

In a slightly different way, the base value of the heat resistance of the entrance doors is calculated. To calculate it, you will first need to calculate the heat transfer resistance of the wall according to the sanitary-hygienic criterion (dew drop):
Rst \u003d (Tv - Tn) / (Dtn x av).

Here DTn - the temperature difference between the inner surface of the wall and the air temperature in the room, is determined by the Code of Practice and for housing is 4.0.
aB - 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 design of the door, it is required to perform a verification thermal engineering calculation of the enclosing structures. Example of calculating the front door:

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

A door insulated with a 5 cm thick mineral wool slab will correspond to this calculated value. Its heat transfer resistance will be R \u003d 0.05 / 0.048 \u003d 1.04 (mx deg / W), which is more than the calculated one.

Comprehensive requirements

Walls, ceilings, or coatings are calculated to verify item-wise regulatory requirements. The set of rules also established a complete requirement characterizing the quality of insulation of all enclosing structures as a whole. This value is called the "specific heat-shielding characteristic." Not a single thermotechnical calculation of building envelopes can do without its verification. An example of a JV calculation is given below.

Cob \u003d 88.77 / 250 \u003d 0.35, which is less than the normalized value of 0.52. In this case, the area and volume are taken for the house with dimensions of 10 x 10 x 2.5 m. Heat transfer resistance is equal to the basic 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 requirements, to compile the energy passport, they also perform the heat engineering calculation of the building envelope, an example of the design of a passport is given in the appendix to SP50.13330.2012.

Homogeneity coefficient

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

The calculation of this coefficient is quite complicated, therefore, in a simplified form, you can use the approximate values \u200b\u200bfrom the reference literature. For example, for masonry - 0.9, three-layer panels - 0.7.

Effective insulation

Choosing a home insulation system, it is easy to make sure that it is almost impossible to fulfill modern thermal protection requirements without the use of an effective insulation. So, if you use 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 heaters based on polystyrene foam or stone wool allows you to limit yourself to a thickness of 10-20 cm.

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

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

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

The heat engineering 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

Thermotechnical indicators of the used building materials (according to SNiP II-3-79 *)

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

2-brick wall (silicate brick) - 640 mm

3-heater (expanded polystyrene)

4-thin plaster (decorative layer) - 5 mm

When performing the heat engineering calculation, the normal humidity conditions in the premises were adopted — operating conditions (“B”) in accordance with SNiP II-3-79 vol. 1 and adj. 2, i.e. we take the thermal conductivity of the materials used in column “B”.

We calculate the required resistance to heat transfer of the fence, taking into account sanitary and hygienic and comfortable conditions according to the formula:

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

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

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

t n - calculated winter outdoor temperature, ° C, equal to the average temperature of the coldest five-day period, security 0.92 according to SNiP 23-01-99 for the city of Yaroslavl is assumed to be -31 ° C;

n is the coefficient taken according to SNiP II-3-79 * (table 3 *) depending on the position of the outer surface of the building envelope in relation to the outside air and is taken to be n \u003d 1;

Δ t n - the normative and temperature difference between the temperature of the internal air and the temperature of the inner surface of the building envelope - is set according to SNiP II-3-79 * (table 2 *) and is taken equal to Δ t n \u003d 4.0 ° С;

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

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

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

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

t ot.per - average temperature, ° C, of \u200b\u200ba period with an average daily air temperature below or equal to 8 ° C according to SNiP 23-01-99;

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

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

We define the reduced resistance to heat transfer R tr according to the conditions of energy conservation in accordance with the requirements of SNiP II-3-79 * (table 1b *) and sanitary-hygienic and comfortable conditions. Intermediate values \u200b\u200bare determined by interpolation.

table 2

Heat transfer resistance of building envelopes (according to SNiP II-3-79 *)

The heat transfer resistance of the building envelope R (0) is taken as the largest of the values \u200b\u200bcalculated earlier:

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

We write the equation for calculating the actual heat transfer resistance R 0 of the building envelope using the formula in accordance with the specified 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 the estimated 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 is the thickness of the calculated layer of the outer fence in m;

λ x - thermal conductivity coefficient 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 building envelope is taken according to SNiP II-3-79 * (table 4 *) and is taken equal to α in \u003d 8.7 W / m 2 * ° C.

α n - heat transfer coefficient (for winter conditions) of the outer surface of the building envelope is taken according to SNiP II-3-79 * (table 6 *) and taken equal to α n \u003d 23 W / m 2 * ° С.

The thermal resistance of the building envelope with successive homogeneous layers should be defined as the sum of the thermal resistances of the individual layers.

For external walls and ceilings, the thickness of the insulating layer of the fence δ x it is calculated from the condition that the value of the actual reduced resistance to heat transfer of the building envelope R 0 must be at least the normalized value of R 0 mp calculated by the formula (2):

R 0 ≥ R 0 tr

Opening the value of R 0, we get:

R 0 \u003d 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 insulating layer

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

δ x \u003d 0.10 m

We take into account the thickness of the insulation (expanded polystyrene) δ x \u003d 0.10 m

Determine the actual resistance to heat transfer calculated enclosing structures R 0, taking into account the adopted thickness of the insulating layer δ x \u003d 0.10 m

R 0 \u003d 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 complied with, R 0 \u003d 3.43 (m 2 * ° C) / W ≥ R 0 mp \u003d 3.41 (m 2 * ° C) / W

During the operation of the building, both overheating and freezing are undesirable. Determine the middle ground will allow thermal engineering calculation, which is no less important than the calculation of profitability, strength, resistance to fire, durability.

Based on heat 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 consider in the article.

Much depends on the thermal characteristics of the capital fencing of the building. This is the humidity of structural elements, and temperature indicators that affect the presence or absence of condensate on the interior partitions and ceilings.

The calculation will show whether stable temperature and humidity characteristics are 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 in the cold period.

You cannot start designing without having all this data. Based on them, choose the thickness of the walls and floors, the sequence of layers.

According to the regulations of GOST 30494-96 temperature values \u200b\u200binside. On average, it is 21⁰. At the same time, relative humidity must remain in a comfortable framework, and this is an average of 37%. The highest velocity of air mass movement - 0.15 m / s

The heat engineering calculation aims to determine:

1. Are the designs identical to the stated requirements 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 rooms. If it is not observed, these surfaces will absorb heat, and inside the temperature will remain very low.

Changes in the heat flux should not significantly affect the internal temperature. This characteristic is called heat resistance.

By performing a thermal calculation, the optimal limits (minimum and maximum) of the dimensions of the walls, floors 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, you need the initial parameters.

They depend on a number of characteristics:

1. Destination of the building and its type.
2. Orientations of vertical building envelopes relative to the orientation to the cardinal points.
3. The geographic parameters of the future home.
4. The volume of the building, its number of floors, area.
5. Types and dimensional data of door, window openings.
6. Type of heating and its technical parameters.
7. The number of permanent residents.
8. Material of vertical and horizontal enclosing structures.
9. Overlapping the top floor.
10. Equipped with hot water.
11. Type of ventilation.

Other structural features of the structure are taken into account in the calculation. The air permeability of building envelopes should not contribute to excessive cooling inside the house and reduce the heat-shielding characteristics of the elements.

Loss of heat causes and waterlogging of the walls, and in addition, this leads to dampness, adversely affecting the durability of the building.

In the calculation process, first of all, the thermal engineering data of building materials are determined, from which the building envelope is made. In addition, the reduced heat transfer resistance and conformity to its normative value is subject to determination.

Formulas for calculating

Leaks of heat lost by the house can be divided into two main parts: losses through building envelopes and losses caused by functioning. In addition, heat is lost when warm water is discharged into the sewer system.

For the materials that make up the enclosing structures, it is necessary to find the value of the thermal conductivity index Kt (W / m x degree). They are in the relevant directories.

Now, knowing the thickness of the layers, according to the formula: R \u003d S / CTcalculate the thermal resistance of each unit. If the design is multilayer, all the obtained values \u200b\u200bare added up.

The dimensions of heat loss are most easily determined by adding the heat flows through the enclosing structures that actually form this building

Guided by this technique, take into account the moment that the materials that make up the structure have a different structure. It is also taken into account that the heat flux passing through them has different specifics.

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

Q \u003d (A / R) x dT

• A - area in m².
• R is the design resistance to heat transfer.
• dT is the temperature difference between the outside and the inside. It must 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 parameter dT, taking into account the temperature, not the lowest, but the average.

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

There is a formula for this:

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

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

The disadvantages of calculating the area

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

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

If you nevertheless chose this method for calculation, you need to consider the following nuances:

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

The area formula is:

Q \u003d S x 100 (150) W.

Here Q is a comfortable level of heat in the building, S is the area with heating in m². Numbers 100 or 150 - the specific value of the thermal energy spent for heating 1 m².

Losses through home 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 unity.

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

It provides for a complete update of the air inside the building in one hour. Buildings constructed according to the DIN standard have walls with vapor barrier, so here the air exchange rate is taken to be two.

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

Qw \u003d (V x Qu: 3600) x P x C x dT

Here, the symbols indicate the following:

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

Based on the results of this calculation, you can determine the power of the heat generator of the heating system. If the power value is too high, it can become a way out of the situation. Let's look at a few examples for houses made of different materials.

An example of heat engineering calculation No. 1

We calculate a residential building located in 1 climatic region (Russia), subarea 1B. All data is taken from table 1 of SNiP 23-01-99. The coldest temperature observed for five days with a security of 0.92 - tn \u003d -22⁰С.

In accordance with SNiP, the heating period (zop) lasts 148 days. The average temperature during the heating period with daily average temperature indices of air in the street is 8 t - tot \u003d -2.3⁰. The outside temperature during the heating season is tht \u003d -4.4⁰.

Heat loss at home is 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 strive to ensure that they are as expedient as possible.

The condition is stipulated that the temperature of 22 дома should be provided in the rooms of the house. The house has two floors and walls with a thickness of 0.5 m. Its height is 7 m, the dimensions in the plan are 10 x 10 m. The material of the vertical walling is warm ceramics. For it, 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 CT 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, it is necessary to determine the thermal resistance of both the ceramic wall and the insulation. In the first case, R1 \u003d 0.5: 0.16 \u003d 3.125 sq. mx C / W. In the second - R2 \u003d 0.05: 0.04 \u003d 1.25 sq. mx C / W. In general, for a vertical building envelope: R \u003d R1 + R2 \u003d 3.125 + 1.25 \u003d 4.375 sq. mx C / W.

Since heat loss has a directly proportional relationship with the area of \u200b\u200bthe building envelope, we calculate the wall area:

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 walling are calculated in the same way. As a result, all the results are summarized.

If the basement under the floor of the first floor is heated, the floor can not be insulated. Basement walls are still better sheathed with insulation so that the heat does not go into the ground.

Determination of losses through ventilation

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

V \u003d 10х10х7 \u003d 700 mᶾ.

With a multiplicity of air exchange Kv \u003d 2, the heat loss is:

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

If Kv \u003d 1:

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

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

An example of heat engineering calculation No. 2

It is required to calculate the losses through a 51 cm thick brick wall. 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 external wall of the room is oriented to the south, there are no external doors.

For brick, the thermal conductivity coefficient Kt \u003d 0.58 W / m º C, for mineral wool - 0.04 W / m º C. Thermal resistance:

R1 \u003d 0.51: 0.58 \u003d 0.879 sq. mx C / W. R2 \u003d 0.1: 0.04 \u003d 2.5 sq. mx C / W. In general, for a vertical building envelope: R \u003d R1 + R2 \u003d 0.879 + 2.5 \u003d 3.379 square meters. mx C / W.

External wall area 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 is usually indicated in the passport and it is not necessary to calculate it.

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

If the house’s windows 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 t.

An example of heat engineering calculation No. 3

We will perform a thermal calculation of a wooden log building with a facade erected from pine logs with a thickness of 0.22 m. The coefficient for this material is K \u003d 0.15. In this situation, heat loss will amount to:

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

The lowest temperature of the five-day period is -18⁰, for comfort in the house the temperature is set to 21⁰. The difference is 39⁰. If we proceed from an area of \u200b\u200b120 m², we get the result:

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

For comparison, we determine the loss of a brick house. The coefficient for silicate brick is 0.72.

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

In the same conditions, a wooden house is more economical. Silicate brick for walling is not suitable here at all.

The wooden structure has a high heat capacity. Its enclosing structures keep a comfortable temperature for a long time. Nevertheless, even a log house needs to be insulated and it is better to do this 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 created from foam blocks was taken. How insulation is applied. Finishing construction - plaster on both sides. Its structure is calcareous and sandy.

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;
• polystyrene foam - 0.065 m.

The task is to find the necessary heat transfer resistance and the actual one. The necessary Rtr is determined by substituting the values \u200b\u200bin the expression:

Rtr \u003d a x GSOP + b

where GOSP is the degree-day of the heating season, and 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 \u003d 1.4.

GSOP is calculated by the formula taken from the same SP:

GOSP \u003d (tv - tot) x zot.

In this formula, tv \u003d 20⁰, tf \u003d -2.2⁰, zf - 205 - the heating period in days. Hence:

GSOP \u003d (20 - (-2,2)) x 205 \u003d 4551⁰ С 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 thermal conductivity for each layer of the wall:

• λb1 \u003d 0.81 W / m ⁰С;
• λb2 \u003d 0.26 W / m ⁰С;
• λb3 \u003d 0.041 W / m ⁰С;
• λb4 \u003d 0.81 W / m ⁰С.

The total conditional resistance to heat transfer Ro, equal to the sum of the resistances of all layers. Calculate it by the formula:

Substituting the values \u200b\u200breceive: \u003d 2.54 m2 ° C / W. Rf is determined by multiplying Ro by a coefficient 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 value from the article we recommend.

Conclusions and useful video on the topic

Performing a heat engineering calculation using an online calculator:

The correct heat engineering calculation:

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

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

Please leave comments, ask questions, post a photo 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.

An example of the heat engineering calculation of building envelopes

1. The source data

Technical task. In connection with the unsatisfactory heat and humidity conditions of the building, it is necessary to warm its walls and the attic roof. To this end, perform calculations of thermal resistance, heat resistance, air and vapor permeability of building envelopes with an assessment of the possibility of moisture condensation in the thickness of the enclosures. Set the required thickness of the heat-insulating layer, the need for wind and vapor barrier, the arrangement of layers in the structure. To develop a design solution that meets the requirements of SNiP 23-02-2003 “Thermal protection of buildings” for building envelopes. Calculations are performed in accordance with the set of rules for the design and construction of SP 23-101-2004 "Design of thermal protection of buildings".

General characteristics of the building. A two-story residential building with an attic is located in the village. Shepherd of the Leningrad region. The total area of \u200b\u200bthe external walling - 585.4 m 2; total wall area of \u200b\u200b342.5 m 2; total window area 51.2 m 2; roof area - 386 m 2; basement height - 2.4 m.

The structural scheme of the building includes load-bearing walls, reinforced concrete floors made of multi-hollow panels, 220 mm thick, and a concrete foundation. The external walls are made of masonry and plastered inside and out with a mortar with a layer of about 2 cm.

The building covering has a rafter structure with a steel seam roof, made on the crate with a pitch of 250 mm. The insulation 100 mm thick is made of mineral wool boards laid between the rafters

The building provides stationary electro-heat storage heating. The basement has a technical purpose.

Climatic parameters. According to SNiP 23-02-2003 and GOST 30494-96, the calculated average temperature of the internal air is taken equal

t int \u003d 20 ° C.

According to SNiP 23-01-99 we accept:

1) the estimated outdoor temperature in the cold season for the conditions of pos. Swiritsa of the Leningrad Region

t ext \u003d -29 ° C;

2) the duration of the heating period

z ht \u003d 228 days .;

3) the average outside temperature during the heating period

t ht \u003d -2.9 ° C.

Heat transfer coefficients.The values \u200b\u200bof the heat transfer coefficient of the inner surface of the fences are taken: for walls, floors and smooth ceilings α int \u003d 8.7 W / (m 2 · ºС).

The values \u200b\u200bof the heat transfer coefficient of the outer surface of the fence take: for walls and coatings α ext \u003d 23; floors of attics α ext \u003d 12 W / (m 2 · ºС);

Normalized heat transfer resistance. Degree day of the heating period G d are determined by the formula (1)

G d \u003d 5221 ° C

Because the value G d different from table values, normative value R req determined by the formula (2).

According to SNiP 23-02-2003, for the obtained value of a degree-day, the normalized heat transfer resistance R req , m 2 · ° C / W, is:

For exterior walls 3.23;

Coatings and ceilings over driveways 4.81;

Fencing over unheated underground and basement 4.25;

Windows and balcony doors 0.54.

2. Thermotechnical calculation of external walls

2.1. Heat Resistance of Exterior Walls

Exterior walls made of hollow ceramic bricks and have a thickness of 510 mm. The walls are plastered from the inside with a lime-cement mortar 20 mm thick, and from the outside - with a cement mortar of the same thickness.

The characteristics of these materials — density γ 0, thermal conductivity coefficient in the dry state  0 and vapor permeability coefficient μ — are taken from Table. Clause 9 of the application. Moreover, in the calculations we use the thermal conductivity of materials  W for operating conditions B, (for wet operating conditions), which are obtained by the formula (2.5). We have:

For lime-cement mortar

γ 0 \u003d 1700 kg / m 3,

 W \u003d 0.52 (1 + 0.168 · 4) \u003d 0.87 W / (m · ° С),

μ \u003d 0.098 mg / (m · h · Pa);

For masonry made of hollow ceramic bricks with cement-sand mortar

γ 0 \u003d 1400 kg / m 3,

 W \u003d 0.41 (1 + 0.207 · 2) \u003d 0.58 W / (m · ° С),

μ \u003d 0.16 mg / (m · h · Pa);

For cement mortar

γ 0 \u003d 1800 kg / m 3

 W \u003d 0.58 (1 + 0.151 · 4) \u003d 0.93 W / (m · ° С),

μ \u003d 0.09 mg / (m · h · Pa).

The heat transfer resistance of the wall without insulation is equal to

R o \u003d 1 / 8.7 + 0.02 / 0.87 + 0.51 / 0.58 + 0.02 / 0.93 + 1/23 \u003d 1.08 m 2 · ° C / W.

In the presence of window openings forming the slopes of the wall, the coefficient of thermotechnical uniformity of brick walls with a thickness of 510 mm is taken r = 0,74.

Then the reduced heat transfer resistance of the walls of the building, determined by the formula (2.7), is

R r o \u003d 0.74 · 1.08 \u003d 0.80 m 2 · ° C / W.

The obtained value is much lower than the normative value of heat transfer resistance, therefore, it is necessary to install external thermal insulation and subsequent plastering with protective and decorative compositions of plaster mortar with fiberglass reinforcement.

To allow the insulation to dry out, the plaster layer covering it must be vapor permeable, i.e. porous with low density. We select a porous cement-perlite mortar having the following characteristics:

γ 0 \u003d 400 kg / m 3,

 0 \u003d 0.09 W / (m · ° С),

 W \u003d 0.09 (1 + 0.067 · 10) \u003d 0.15 W / (m · ° С),

 \u003d 0.53 mg / (m · h · Pa).

Total heat transfer resistance of added insulation layers R t and plastering R w must be at least

R t + R W \u003d 3.23 / 0.74-1.08 \u003d 3.28 m 2 ° C / W.

Previously (with subsequent refinement), we take the thickness of the plaster lining 10 mm, then its resistance to heat transfer is equal to

R W \u003d 0.01 / 0.15 \u003d 0.067 m 2 · ° C / W.

When used for thermal insulation of mineral wool boards manufactured by Mineralnaya Vata CJSC, the brand Fasad Batts  0 \u003d 145 kg / m 3,  0 \u003d 0.033,  W \u003d 0.045 W / (m · ° С) the thickness of the thermal insulation layer will be

δ \u003d 0.045 · (3.28-0.067) \u003d 0.145 m.

Rockwool boards are available in thicknesses from 40 to 160 mm in 10 mm increments. We accept a standard thermal insulation thickness of 150 mm. Thus, the laying of plates will be made in one layer.

Verification of compliance with energy conservation requirements.The design diagram of the wall is shown in Fig. 1. The characteristics of the wall layers and the total wall resistance to heat transfer without taking into account the vapor barrier are given in table. 2.1.

Table 2.1

Characterization of wall layers and total wall resistance to heat transfer

The heat transfer resistance of the walls of the building after insulation will be:

R o = 1 / 8.7 + 4.32 + 1/23 \u003d 4.48 m 2 ° C / W.

Given the coefficient of thermal engineering uniformity of the outer walls ( r \u003d 0.74) we get the reduced heat transfer resistance

R o r \u003d 4.48 · 0.74 \u003d 3.32 m 2 · ° C / W.

Value obtained R o r \u003d 3.32 exceeds the normative R req \u003d 3.23, since the actual thickness of the insulation boards is greater than the calculated one. This situation meets the first requirement of SNiP 23-02-2003 to the thermal resistance of the wall - R o ≥ R req .

Verification of compliance withsanitary-hygienic and comfortable conditions in the room.The calculated difference between the temperature of the internal air and the temperature of the inner surface of the wall Δ t 0 is

Δ t 0 =n(t int – t ext)/(R o r ·α int) \u003d 1.0 (20 + 29) / (3.32.8.7) \u003d 1.7 ºС.

According to SNiP 23-02-2003, for the external walls of residential buildings, a temperature difference of not more than 4.0 ºС is permissible. Thus, the second condition (Δ t 0 ≤Δ t n) performed.

P
we verify the third condition ( τ int >t grew), i.e. Is moisture condensation possible on the inner surface of the wall at the calculated outdoor temperature t ext \u003d -29 ° C. Inner surface temperature τ int building envelope (without heat-conducting inclusion) is determined by the formula

τ int = t int –Δ t 0 \u003d 20–1.7 \u003d 18.3 ° C.

The elasticity of water vapor in the room e int is equal to

Thermotechnical calculation allows you to determine the minimum thickness of the enclosing structures so that there are no cases of overheating or freezing during the operation of the building.

The enclosing structural elements of heated public and residential buildings, with the exception of the requirements of stability and strength, durability and fire resistance, efficiency and architectural design, must first of all meet the thermal engineering standards. Enclosing elements are selected depending on the design solution, climatological characteristics of the built-up area, physical properties, humidity and temperature conditions in the building, as well as in accordance with the requirements of heat transfer resistance, air permeation and vapor permeation.

What is the meaning of the calculation?

1. If during the calculation of the cost of the future structure only strength characteristics are taken into account, then, of course, the cost will be less. However, this is a visible savings: subsequently, much more money will be spent on heating the room.
2. Properly selected materials will create an optimal microclimate in the room.
3. When planning a heating system, a heat engineering calculation is also required. In order for the system to be cost-effective and efficient, it is necessary to have an understanding of the real capabilities of the building.

Thermotechnical Requirements

It is important that the outdoor structures comply with the following heat engineering requirements:

• They had sufficient heat-shielding properties. In other words, overheating of premises should not be allowed in the summer, and excessive heat loss in winter.
• The temperature difference of the internal elements of the fencing and the premises should not be higher than the standard value. Otherwise, excessive cooling of the human body by heat radiation to these surfaces and moisture condensation of the internal air flow on the building envelope may occur.
• In the event of a change in the heat flux, temperature fluctuations in the room should be minimal. This property is called heat resistance.
• It is important that the airtightness of the fencing does not cause strong room cooling and does not impair the heat-shielding properties of structures.
• Fencing should have a normal humidity regime. Since waterlogging fences increases heat loss, causes dampness in the room, reduces the durability of structures.

In order for the structures to meet the above requirements, they perform a heat engineering calculation, as well as calculate the heat resistance, vapor permeability, air permeability and moisture transfer according to the requirements of regulatory documents.

Thermotechnical qualities

From the thermal characteristics of the external structural elements of buildings depends:

• Humidity mode of structural elements.
• The temperature of internal structures, which ensures the absence of condensation on them.
• Constant humidity and temperature in the premises, both in the cold and in the warm season.
• The amount of heat lost by the building during the winter period.

So, based on all of the above, the heat engineering design analysis is considered an important stage in the design process of buildings and structures, both civil and industrial. Design begins with the choice of structures - their thickness and sequence of layers.

Tasks of heat engineering calculation

So, the thermal engineering calculation of enclosing structural elements is carried out with the aim of:

1. Conformity of structures to modern requirements for thermal protection of buildings and structures.
2. Providing a comfortable microclimate in the interior.
3. Ensuring optimal thermal protection of fences.

The main parameters for the calculation

To determine the heat consumption for heating, as well as to make the thermotechnical calculation of a building, it is necessary to take into account many parameters that depend on the following characteristics:

• Purpose and type of building.
• The geographical location of the structure.
• Orientation of the walls to the cardinal points.
• Dimensions of structures (volume, area, number of storeys).
• Type and size of windows and doors.
• Characteristics of the heating system.
• The number of people in the building at the same time.
• The material of the walls, floor and floors of the last floor.
• The presence of a hot water system.
• Type of ventilation systems.
• Other structural design features.

Thermotechnical calculation: program

To date, many programs have been developed that allow this calculation. As a rule, the calculation is carried out on the basis of the methodology described in the normative and technical documentation.

These programs allow you to calculate the following:

• Thermal resistance.
• Heat loss through structures (ceiling, floor, door and window openings, as well as walls).
• The amount of heat required to heat the infiltrating air.
• Selection of sectional (bimetallic, cast iron, aluminum) radiators.
• Selection of panel steel radiators.

Thermotechnical Calculation: Example Calculation for Exterior Walls

For the calculation, it is necessary to determine the following main parameters:

• t in \u003d 20 ° C is the temperature of the air flow inside the building, which is taken to calculate the fencing at the minimum values \u200b\u200bof the most optimal temperature of the corresponding building and structure. It is accepted in accordance with GOST 30494-96.

• According to the requirements of GOST 30494-96, the humidity in the room should be 60%, as a result, the normal humidity mode will be ensured in the room.
• In accordance with Appendix B of SNiPa 23-02-2003, the humidity zone is dry, which means that the operating conditions of the fences are A.
• t n \u003d -34 ° C is the temperature of the external air flow in the winter period of time, which is taken according to SNiP based on the most cold five-day period, having a security of 0.92.
• Z from.per \u003d 220 days - this is the duration of the heating period, which is taken according to SNiP, with an average daily ambient temperature of ≤ 8 ° C.
• T from per. \u003d -5.9 ° C is the ambient temperature (average) in the heating period, which is taken according to SNiP, at a daily ambient temperature of ≤ 8 ° C.

Initial data

In this case, the thermotechnical calculation of the wall will be carried out in order to determine the optimal thickness of the panels and thermal insulation material for them. Sandwich panels (TU 5284-001-48263176-2003) will be used as external walls.

Comfortable conditions

Consider how the thermotechnical calculation of the outer wall is performed. First, you need to calculate the required heat transfer resistance, focusing on comfortable and sanitary conditions:

R 0 mp \u003d (n × (t in - t n)): (Δt n × α in), where

n \u003d 1 is a coefficient that depends on the position of the external structural elements with respect to the outside air. It should be taken according to SNiP 23-02-2003 from table 6.

Δt n \u003d 4.5 ° C is the normalized temperature drop of the internal surface of the structure and internal air. It is taken according to SNiP from table 5.

α b \u003d 8.7 W / m 2 ° C is the heat transfer of internal building envelopes. Data is taken from table 5, according to SNiPu.

Substitute the data in the formula and get:

R 0 mp \u003d (1 × (20 - (-34)): (4.5 × 8.7) \u003d 1.379 m 2 ° C / W.

Energy Saving Terms

Performing the thermotechnical calculation of the wall, based on the conditions of energy conservation, it is necessary to calculate the required heat transfer resistance of the structures. It is determined by GSOP (degree-day of the heating period, ° C) according to the following formula:

GSOP \u003d (t in - t from per.) × Z from per. Where

t in is the temperature of the air flow inside the building, ° C.

Z from per. and t from per. is the duration (days) and temperature (° C) of a period having an average daily air temperature of ≤ 8 ° C.

In this way:

GSOP \u003d (20 - (-5.9)) × 220 \u003d 5698.

Based on the conditions of energy conservation, we determine R 0 tr by interpolation according to SNiP from table 4:

R 0 mp \u003d 2.4 + (3.0 - 2.4) × (5698 - 4000)) / (6000 - 4000)) \u003d 2.909 (m 2 ° C / W)

R 0 \u003d 1 / α in + R 1 + 1 / α n, where

d is the thickness of the insulation, m

l \u003d 0.042 W / m ° C is the thermal conductivity of the mineral wool slab.

α n \u003d 23 W / m 2 ° C - this is the heat transfer of the external structural elements, adopted by SNiPu.

R 0 \u003d 1 / 8.7 + d / 0.042 + 1/23 \u003d 0.158 + d / 0.042.

Insulation thickness

The thickness of the heat-insulating material is determined based on the fact that R 0 \u003d R 0 Tr, while R 0 Tr is taken under energy-saving conditions, thus:

2.0909 \u003d 0.158 + d / 0.042, whence d \u003d 0.116 m.

We select the brand of sandwich panels according to the catalog with the optimal thickness of the insulating material: DP 120, while the total thickness of the panel should be 120 mm. Similarly, the thermal engineering calculation of the building as a whole is performed.

The need to perform the calculation

Designed on the basis of the heat engineering calculation, done competently, the enclosing structures can reduce heating costs, the cost of which is regularly increased. In addition, heat conservation is considered an important environmental task, because it is directly related to reducing fuel consumption, which leads to a decrease in the impact of negative factors on the environment.

In addition, it is worth remembering that improperly performed thermal insulation can lead to waterlogging of structures, which as a result will lead to the formation of mold on the surface of the walls. The formation of mold, in turn, will lead to damage to the interior finish (peeling wallpaper and paint, the destruction of the plaster layer). In especially advanced cases, radical intervention may be required.

Very often, construction companies in their activities seek to use modern technologies and materials. Only a specialist can figure out the need to use this or that material, either separately or in combination with others. It is the heat engineering calculation that will help determine the most optimal solutions that will ensure the durability of structural elements and minimal financial costs.