A long time ago, buildings and structures were built without thinking about what heat-conducting qualities enhanced structures. In other words, the walls were made simply thick. And if you ever happened to be in old merchant houses, then you could notice that the outer walls of these houses are made of ceramic bricks, the thickness of which is about 1.5 meters. Such a thickness brick wall Provided and ensures that there are still quite comfortable stay of people in these houses even in the most severe frosts.
Currently, everything has changed. And now it is economically not profitable to make walls with such fat. Therefore, materials that can reduce it were invented. Some of them: insulation and gas-silicate blocks. Thanks to these materials, for example, thickness brick masonry It can be reduced to 250 mm.
Now the walls and floors most often make 2 or 3-layer, one layer of which is material with good thermal insulation properties. And in order to determine the optimal thickness of this material, a heat engineering calculation is carried out and the dew point is determined.
How to calculate the dew point to determine the next page. It will also consider the heat engineering calculation on the example.
For the calculation, two slips will be required, one joint venture, one GOST and one manual:
In the process of performing thermal calculation, they determine:
1. Climate of terrain and microclimate
Construction area: nizhny Novgorod.
Purpose of the building: residential.
The estimated relative humidity of the internal air from the condition is not condensate falling on the inner surfaces of the outer fences is 55% (SNiP 23-02-2003 p.4.3. Table 1 for normal humidity).
The optimal air temperature in the residential room in cold period years T int \u003d 20 ° С (GOST 30494-96 Table 1).
Calculated outdoor air temperature t Extdetermined by temperature of the coldest five-day security of 0.92 \u003d -31 ° C (SNIP 23-01-99 Table 1 column 5);
The duration of the heating period with the average daily temperature of the outer air is 8 ° C. Z HT = 215 days (SNiP 23-01-99 Table. 1 column 11);
The average outdoor temperature for the heating period T HT \u003d -4.1 ° C (SNiP 23-01-99 Table. 1 column 12).
2. Wall design
The wall consists of the following layers:
3. Thermophysical characteristics Materials
The characteristics of the materials are reduced to the table.
Note (*): These characteristics can also be found from manufacturers of thermal insulation materials.
4. Determination of the solar thickness
To calculate the thickness of the heat-insulating layer, it is necessary to determine the heat transfer resistance of the enclosing design on the basis of the requirements of sanitary standards and energy saving.
4.1. Determination of the heat protection rate under the condition of energy saving
Determination of the degree in the heating period according to p.5.3 SNiP 23-02-2003:
D D. = ( t int - t HT) z HT = (20 + 4,1) 215 \u003d 5182 ° C × day
Note: Also, the degree and day have the designation - HSOP.
The regulatory value of the reduced heat transfer resistance should be made of no less normalized values \u200b\u200bdetermined by SNiP 23-02-2003 (Table 4) depending on the degree of the construction region of the construction region:
R REQ \u003d A × D D + B \u003d 0.00035 × 5182 + 1,4 \u003d 3,214m 2 × ° C / W,
where: DD - degree and day of the heating period in Nizhny Novgorod,
a and B are the coefficients accepted according to Table 4 (if SNiP 23-02-2003) or Table 3 (if the SP 50.13330.2012) for the walls of a residential building (column 3).
4.1. Determination of heat protection rate under sanitation
In our case, it is considered as an example, since this indicator is calculated for production buildings with excess of explicit warmth more than 23 W / m 3 and buildings intended for seasonal operation (autumn or spring), as well as buildings with the calculated temperature of internal air 12 ° C And below the resistance of the heat transfer of the enclosing structures (with the exception of translucent).
Determination of the regulatory (maximum permissible) heat transfer resistance under the condition of sanitation (formula 3 SNiP 23-02-2003):
where: n \u003d 1 - the coefficient adopted according to Table 6 for outdoor Wall;
t int \u003d 20 ° C - value from the source data;
t ext \u003d -31 ° C - value from the source data;
Δt n \u003d 4 ° C - the normalized temperature difference between the temperature of the inner air and the temperature of the inner surface of the enclosing structure, is taken according to Table 5 in this case for the exterior walls of residential buildings;
α int \u003d 8.7 W / (m 2 × ° C) - the heat transfer coefficient of the inner surface of the enclosing structure, is accepted according to Table 7 for the outer walls.
4.3. Norm of thermal protection
From the above calculations for the required heat transfer resistance to chooseR REQ from the power saving condition and indicate it now R T0 \u003d 3,214m 2 × ° C / W .
5. Determination of the thickness of the insulation
For each layer of a given wall, it is necessary to calculate thermal resistance by the formula:
where: ΔI- thickness of the layer, mm;
λ i is the calculated coefficient of thermal conductivity of the material W / (M × ° C).
1 layer ( decorative brick): R 1 \u003d 0.09 / 0.96 \u003d 0.094 m 2 × ° C / W .
3 layer (silicate brick): R 3 \u003d 0.25 / 0.87 \u003d 0.287 m 2 × ° C / W .
4 layer (plaster): R 4 \u003d 0.02 / 0.87 \u003d 0.023 m 2 × ° C / W .
Determining the minimum permissible (required) thermal resistance heat insulating material (Formula 5.6 E.G. Malyavina "Teplopotieri Building. Reference Manual"):
where: r int \u003d 1 / α int \u003d 1 / 8.7 - resistance to heat exchange on the inner surface;
R EXT \u003d 1 / α EXT \u003d 1/23 - heat exchange resistance on the outer surface, α EXT is accepted according to table 14 for the outer walls;
ΣR i \u003d 0.094 + 0.287 + 0.023 - the sum of the thermal resistances of all the layers of the wall without a layer of insulation defined taking into account the coefficients of the thermal conductivity of the materials adopted according to the column A or B (columns 8 and 9 of the table D1 SP 23-101-2004) in accordance with the humidity conditions of the wall, m 2 · ° C / W.
The thickness of the insulation is equal (formula 5.7):
where: λ Ut - the thermal conductivity coefficient of the insulation material, W / (M · ° C).
Determination of the thermal resistance of the wall from the condition that the overall thickness of the insulation will be 250 mm (Formula 5.8):
where: σr t, i is the sum of the thermal resistances of all layers of fencing, including the layer of the insulation, adopted structural thickness, m 2 · ° C / W.
From the result obtained, we can conclude that
R 0 \u003d 3.503m 2 × ° C / W \u003e R TR30 \u003d 3,214m 2 × ° C / W→ Therefore, the thickness of the insulation is selected right.
In the case when mineral wool, glass wool or other slab insulation are used in a three-layer masonry as a heater, a device of an air ventilated interlayer between the outer masonry and insulation is necessary. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to dry up the insulation that winks from condensate.
This air layer is not closed space, so if it has its presence, it is necessary to take into account the requirements of P.9.1.2 SP 23-101-2004, namely:
a) layers of construction located between the air layer and the outer surface (in our case, it is a decorative brick (Bessen)), not taken into account in thermal calculation;
b) on the surface of the structure facing in the direction of the interlayer ventilated by the outer air, the heat transfer coefficient α ext \u003d 10.8 W / (M ° C) should be taken.
Note: The effect of the air layer is taken into account, for example, with the heat engineering calculation of plastic glass windows.
In order for the housing to be warm in the strongest frost, it is necessary to correctly select the insulation system - for this, the heat engineering calculation of the outer wall is performed. The results of the calculation shows how effective the actual or projected method of insulation is performed.
First, initial data should be prepared. The estimated parameter affect the following factors:
After collecting and writing the source information, the coefficients of the thermal conductivity of the building materials are determined from which the wall is made. The degree of massacre of heat and heat transfer depends on how rawd it is climate. In this regard, for calculating the coefficients use moisture maps drawn up for Russian Federation. After that, all numeric values \u200b\u200brequired for the calculation are introduced into the appropriate formulas. 
As an example, the thermal protection properties of the wall laid out of foam blocks, insulated with polystyrene foam with a density of 24 kg / m3 and plastered with a lime-sandy solution on both sides. Calculations and selection of tabular data are based on construction Rules.
Initial data: construction area - Moscow; Relative humidity - 55%, average temperature In the house TB \u003d 20O C. The thickness of each layer is set: Δ1, δ4 \u003d 0.01m (plaster), Δ2 \u003d 0.2M (foam concrete), Δ3 \u003d 0.065m (expanded polystyrene "JV Radoslav").
The purpose of the heat engineering calculation of the outer wall is to determine the necessary (RT) and the actual (RF) heat transfer resistance.
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The result obtained shows that the actual heat resistance is less than the required, so it is necessary to revise the wall structure.
Uncomplicated computer services accelerate computing processes and search for the desired coefficients. It is worth familiar with the most popular programs. 
When building a house or conducting heat-insulating work, an assessment of the effectiveness of the insulation of the outer wall is important: the heat engineering calculation, made independently or with the help of a specialist make it possible to make it quickly and accurately.
When operating the building is undesirable both overheating and freezing. Determine golden middle It will allow a heat engineering calculation, which is no less important than calculating the cost-effectiveness, strength, fastness resistance, durability.
Based on the heat engineering norms, climatic characteristics, paro and moisture permeability, the choice of materials for the construction of enclosing structures is carried out. How to fulfill this calculation, consider in the article.
Much depends on the heat engineering features of capital fences. This is humidity constructive elements, and temperature indicators that affect the presence or absence of condensate on interior partitions And overlaps.
The calculation will show whether stable temperature and humidity characteristics will be maintained at the positive and minus temperature. The list of these characteristics includes such an indicator as the amount of heat leaving enclosing structures in the cold period.
It is impossible to start the design without having all this data. Relying on them, choose the thickness of the walls and overlaps, the sequence of layers.
According to the regulation GOST 30494-96 temperature values \u200b\u200bindoors. On average, it is equal to 21⁰. At the same time, the relative humidity is obliged to stay in a comfortable framework, and this is an average of 37%. The highest speed of air mass movement - 0.15 m / s
The heat engineering sets goals to determine:
The basic principle is the observance of the balance of the temperature indicators of the atmosphere of the internal structures of fences and premises. If it is not observed, heat will absorb these surfaces, and inside the temperature will remain very low.
On the inner temperature should not significantly influence the changes heat flux. This characteristic is called heat resistance.
By performing thermal calculation, the optimal limits (minimum and maximum) dimensions of walls, overlaps in thickness determine. This is a guarantee of the operation of the building throughout the long period of both extreme designs of structures and overheating.
To perform the heat meters, the source parameters are needed.
They depend on a number of characteristics:
Taken into account when calculating others constructive features Buildings. The air permeability of the enclosing structures should not contribute to excessive cooling inside the house and reduce the heat shield characteristics of the elements.
The loss of heat causes the overalling of the walls, and besides, it entails dampness, negatively affecting the durability of the building.
In the calculation process, first of all, the heat engineering data of the building materials, of which the enclosure elements of the structure are manufactured. In addition, the determination is subject to the resistance of heat transfer and the suggestion of its regulatory value.
The heat leakage lost by the house can be divided into two main parts: loss through fencing structures and losses caused by functioning. In addition, heat is lost when resetting warm water into the sewer system.
For materials from which enveloped structures are arranged, it is necessary to find the value of the thermal conductivity of the CT (W / m x degree). They are in the relevant reference books.
Now, knowing the thickness of the layers, by the formula: R \u003d S / CT, calculated the thermal resistance of each unit. If the design is multilayer, all the values \u200b\u200bare folded.
The size of heat loss is the easiest way to determine by adding thermal currents through the enclosing structures that this building actually form
Guided by such a technique, to take into account the time that the materials that make up the structures have an unequal structure. It is also taken into account that the heat flow passing through them has different specifics.
For each individual heat loss design, they are determined by the formula:
Q \u003d (A / R) x DT
Performing the calculation in this way, you can get the result only for the coldest five-day period. General heat loss for the entire cold season is determined by accounting the DT parameter, taking into account the temperature is not the lowest, but the average.
To what extent is the heat, as well as heat transfer depends on the humidity of the climate in the region. For this reason, when calculations use moisture maps
For this there is a formula:
W \u003d ((q + qv) x 24 x n) / 1000
In it n - the duration of the heating period in days.
The calculation based on the area is not highly accurate. This parameter is not taken into account such as climate, temperature indicators both minimal and maximum, humidity. Due to ignoring many important points, the calculation has significant errors.
Often trying to overlap them, the project includes "stock".
If it is still chosen to calculate this method, you need to consider the following nuances:
The area of \u200b\u200bthe area has the form:
Q \u003d S x 100 (150) W.
Here Q is a comfortable level of heat in the building, S is an area with heating in m². Numbers 100 or 150 - specific thermal energy consumed for heating 1 m².
The key parameter in this case is the multiple of air exchange. 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 at the supply ventilation. Exhaust ventilation Promotes the care of warm air. Reduces losses through ventilation heat exchanger ventilation. He does not allow heat to go along with the outgoing air, and the incoming streams it heats
It is envisaged to fully update the air inside the building in one hour. Buildings built according to DIN standard have wines with vapor barrier, so the multiplicity of the air exchange is taken equal to two.
There is a formula for which heat loss through the ventilation system is determined:
QB \u003d (V x sq: 3600) x r x with x dt
Here, the characters indicate the following:
Following this calculation, it is possible to determine the power of the heat generator heating system. In the case of a too high power value, the output from the situation may become. Consider several examples for houses from different materials.
Calculate the residential building located in the 1st climatic area (Russia), Subaryon 1B. All data are taken from Table 1 SNiP 23-01-99. The coldest temperature, which is observed for five days by security 0.92 - TN \u003d -22⁰.
In accordance with SNIP, the heating period (ZOP) lasts 148 days. The averaged temperature over the heating period under the average daily temperature indicators of the air on the street 8⁰ - Tot \u003d -2,3⁰. Temperature outside to the heating season - THT \u003d -4,4⁰.
Heat loss houses - the most important moment At the stage of its design. The choice of building materials, and insulation depends on the calculation results. Zero losses does not happen, but it is necessary to strive to ensure that they are most appropriate
The condition is agreed that the houses should be provided in the houses of the house 22⁰. The house has two floors and walls with a thickness of 0.5 m. The height of it is 7 m, the dimensions in terms of 10 x 10 m. The material of vertical enclosing structures is warm ceramics. For it, the coefficient of thermal conductivity - 0.16 W / m x S.
As an outdoor insulation, a thickness of 5 cm, mineral wool used. The value of the CT for it is 0.04 W / m X S. The number of window openings in the house - 15 pcs. 2.5 m² each.
First of all, you need to determine the thermal resistance as ceramic Walland insulation. In the first case, R1 \u003d 0.5: 0.16 \u003d 3,125 square meters. M X C / W. In the second - R2 \u003d 0.05: 0.04 \u003d 1.25 square meters. M X C / W. In general, for the vertical enclosing structure: R \u003d R1 + R2 \u003d 3.125 + 1.25 \u003d 4.375 square meters. M X C / W.
Since heat loss has a direct proportional relationship with an area of \u200b\u200benclosing structures, we calculate the walls of the walls:
A \u003d 10 x 4 x 7 - 15 x 2,5 \u003d 242.5 m²
Now you can determine the heat loss through the walls:
QC \u003d (242.5: 4.375) x (22 - (-22)) \u003d 2438.9 W.
Heat loss through horizontal enclosing structures are calculated similarly. 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 inspired. The walls of the basement are still better to entee insulation so that the heat does not go into the ground.
To simplify the calculation, do not take into account the thickness of the walls, and simply determine the volume of the air inside:
V \u003d 10x10x7 \u003d 700 Mᶾ.
With the multiplicity of air exchange KV \u003d 2, the weight loss will be:
QB \u003d (700 x 2): 3600) x 1,2047 x 1005 x (22 - (-22)) \u003d 20,776 W.
If kv \u003d 1:
QB \u003d (700 x 1): 3600) x 1,2047 x 1005 x (22 - (-22)) \u003d 10 358 watts.
Effective ventilation residential houses Provide rotary and lamellar recuperators. The efficiency of the first above, it reaches 90%.
It is required to calculate losses through a brick wall 51 cm thick. It is insulated with a 10-centimeter layer mineral Wat. Outside - 18⁰, inside - 22⁰. Wall dimensions - 2.7 m in height and 4 m in length. The only outer wall of the room is oriented to the south, there is no external doors.
For bricks, the thermal conductivity coefficient of CT \u003d 0.58 W / m ºС, for mineral wool - 0.04 W / m ºС. Thermal resistance:
R1 \u003d 0.51: 0.58 \u003d 0.879 square meters. M X C / W. R2 \u003d 0.1: 0.04 \u003d 2.5 kV. M X C / W. In general, for the vertical enclosing structure: R \u003d R1 + R2 \u003d 0.879 + 2.5 \u003d 3.379 square meters. M X C / W.
Area external Wall A \u003d 2.7 x 4 \u003d 10.8 m²
Heat loss through the wall:
QC \u003d (10.8: 3.379) x (22 - (-18)) \u003d 127.9 W.
To calculate losses through the windows, the same formula is used, but the thermal resistance of them is usually indicated in the passport and it is not necessary to count it.
In the thermal insulation of the house windows - "weak link". Through them, there is a rather large share of heat. Reduce the loss of multi-layered double-glazed windows, heat-pressing films, double frames, but even it will not help avoid heat loss completely
If the house has 1.5 x 1.5 m² windows, oriented to the north, and the thermal resistance is 0.87 m2 ° C / W, then the losses will be:
Qo \u003d (2.25: 0.87) x (22 - (-18)) \u003d 103.4 tons.
We will perform a thermal calculation of a wooden log building with a facade, erected from pine logs with a layer with a thickness of 0.22 m. The coefficient for this material is K \u003d 0.15. In this situation, the heat loss will be:
R \u003d 0.22: 0.15 \u003d 1.47 m² x ⁰С / W.
SAMI low temperature Five days - -18⁰, for comfort in the house, a temperature is set 21⁰. The difference will be 39⁰. If you proceed from the area of \u200b\u200b120 m², the result will result:
QC \u003d 120 x 39: 1.47 \u003d 3184 W.
For comparison, we define the loss brick house. The coefficient for silicate brick is 0.72.
R \u003d 0.22: 0.72 \u003d 0.306 m² x ⁰С / W.
QC \u003d 120 x 39: 0.306 \u003d 15 294 W.
In the same conditions wooden house More economical. Silicate brick for the construction of walls is not suitable at all.
The wooden structure has a high heat capacity. Its enclosing structures have long stored a comfortable temperature. Yet even log house need to warm and better do it from the inside and outside
The house will be built in the Moscow region. To calculate the wall, created from foam blocks. As the insulation is applied. Finishing design - plaster on both sides. The structure of it is lime-sand.
Polystyrene foam has a density of 24 kg / m.
The relative indicators of air humidity in the room - 55% at averaged temperature 20⁰. Layer thickness:
The task is to find the desired resistance to heat transfer and the actual one. The required RT is determined by substituting the values \u200b\u200bin the expression:
RT \u003d A x HSOP + B
where the m is the degree and day of the heating season, a and b - the coefficients taken from the table number 3 of the rules of the rules of 50.13330.2012. Since the residential building, A is 0.00035, B \u003d 1.4.
HSOP is calculated by the formula taken from the same joint venture:
GR \u003d (TB - TOT) x Zot.
In this formula TB \u003d 20⁰, TOT \u003d -2.2⁰, ZOT - 205 - the heating period in days. Hence:
HSOP \u003d (20 - (-2,2)) x 205 \u003d 4551⁰ with x day;
RT \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 coefficients for each layer of the wall:
Complete conditional resistance to the heat transfer RO, equal to the sum of the resistance of all layers. Calculate it by the formula:
Substituting the values \u200b\u200bget: RO SIL. \u003d 2.54 m2 ° C / W. RF is determined by multiplying RO to the R coefficient, 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 accelerate and simplify calculations.
Heat engineering calculations are directly related to the definition. What it is and how to find her meaning will recognize from the articles recommended by us.
Performing heat engineering with an online calculator:
Proper heat engineering:
Competent heat engineering will allow to evaluate the effectiveness of the insulation of the external elements of the house, determine the capacity of the necessary heating equipment.
As a result, you can save when buying materials and heating devices. It is better to know in advance, whether the technique with heating and air conditioning is cope than buying everything at random.
Please leave comments, ask questions, post photos on the topic of the article below the block. Tell us about how heat engineering helped you choose the heating equipment of the desired power or insulation system. It is possible that your information is useful to site visitors.
Creating comfortable conditions for residence or labor activity is the primary task of construction. A significant part of our country is located in northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always relevant. With an increase in energy tariffs, the reduction in energy consumption for heating is on the fore.
The choice of wall structures and roof depends primarily from the climatic conditions of the construction area. To determine them, you need to contact the SP131.13330.2012 "Construction climatology". The following values \u200b\u200bare used in the calculations:
On the example for Murmansk, the following values \u200b\u200bare:
In addition, it is necessary to set the estimated temperature inside the room of the TV, it is determined in accordance with GOST 30494-2011. For housing you can take TV \u003d 20 degrees.
To perform the heat engineering calculation of the enclosing structures, pre-calculate the value of the HSOP (degree and day of the heating period):
HSOP \u003d (TV - one) x Zot.
On our example HSOP \u003d (20 - (-3.4)) x 275 \u003d 6435.
For right choice Materials of enclosing structures need to determine what heat-technical characteristics they must have. The ability of the substance is carried out heat characterized by its thermal conductivity, is indicated by the Greek letter L (lambda) and is measured in W / (m x deg.). The ability of the structure to retain heat is characterized by its heat transfer resistance R and equals the ratio of thickness to thermal conductivity: R \u003d D / L.
In case the design consists of several layers, the resistance is calculated for each layer and then summed.
The heat transfer resistance is the main indicator of the outer construction. Its value must exceed regulatory value. Performing the heat engineering calculation of the enclosing structures of the building, we must define the economically justified composition of the walls and roofs.
The quality of thermal insulation is determined primarily with thermal conductivity. Each certified material undergoes laboratory studies, as a result of which this value is determined for the operating conditions "A" or "B". For our country, most regions correspond to the conditions of operation "b". By performing the heat engineering calculation of the enclosing house designs, this value should be used. The thermal conductivity values \u200b\u200bindicate the label or in the material passport, but if they are not, you can use the reference values \u200b\u200bfrom the rules. Values \u200b\u200bfor the most popular materials are shown below:
The estimated heat transfer resistance value should not be less than the base value. The basic value is determined by table 3 SP50.13330.2012 "buildings". The table defines coefficients for calculating the basic values \u200b\u200bof the heat transfer resistance of all enclosing structures and types of buildings. Continuing the started heat engineering calculation of the enclosing structures, an example of calculation can be represented as follows:
The heat engineering calculation of the outer enclosing structure is performed for all designs, closing the "warm" circuit - the floor of the soil or the overlap of techpool, the outer walls (including windows and the doors), combined coating or overlapping a unwanted attic. Also, the calculation must be carried out for internal structures, if the temperature difference in adjacent rooms is more than 8 degrees.
Most walls and overlaps in their design are multi-layered and inhomogeneous. The heat engineering calculation of the enclosing structures of the multilayer structure is as follows:
R \u003d d1 / l1 + d2 / l2 + dn / ln,
where n is the parameters of the N-th layer.
If we consider a brick plastered wall, then we get the following design:
The formula of the heat engineering calculation of the enclosing structures is as follows:
R \u003d 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 0.85 (m x / h).
The value obtained is significantly less than a certain basic value of the heat transfer resistance of the walls of a residential building in Murmansk 3.65 (m x / ha / W). The wall does not satisfy regulatory requirements And needs insulation. For the insulation of the wall, use a thickness of 150 mm and a thermal conductivity of 0.048 W (m x x hail.).
Featuring the insulation system, it is necessary to perform the test heat engineering calculation of the enclosing structures. An example of calculation is shown below:
R \u003d 0.15 / 0.048 + 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 3.97 (m x / w).
The resulting calculated value is larger than the base - 3.65 (m x / h), the insulated wall satisfies the requirements of the norms.
Calculation of overlaps and combined coatings is performed similarly.
Often in private homes or public buildings are performed on the ground. The heat transfer resistance of such floors is not normalized, but at least the design of the floors should not allow dew drops. The calculation of the structures in contact with the soil is performed as follows: the floors are divided into stripes (zones) of 2 meters wide, starting from the outer border. Such zones stand out to three, the remaining area belongs to the fourth zone. If the floor design does not provide an effective insulation, the resistance of heat transfer zones is taken as follows:
It is easy to notice that the further the floor section is located on the outer wall, the higher its heat transfer resistance. Therefore, it is often limited to the insulation of the perimeter of the floor. At the same time, the heat transfer resistance of the warmed design 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 overall heat engineering calculation of the enclosing structures. An example of calculating floors on the ground Consider below. We will take a floor area of \u200b\u200b10 x 10, equal to 100 sq. M.
The average temperature of the heat transfer resistance of the floor on the soil:
RPOL \u003d 100 / (64 / 2.1 + 32 / 4.3 + 4 / 8.6) \u003d 2.6 (m x / h).
After carrying out the insulation of the perimeter of the floor by polystyrene foam slab with a thickness of 5 cm, a strip of 1 meter width, we obtain the average heat transfer resistance value:
RPOL \u003d 100 / (32 / 2,1 + 32 / (2.1 + 0.05 / 0.032) + 32 / 4,3 + 4 / 8.6) \u003d 4.09 (m x / h).
It is important to note that not only the floors, but also the designs of the walls in contact with the soil (walls of a ruble floor, warm basement) are calculated in a similar way.
A somewhat different is calculated by the basic value of heat transfer resistance entrance doors. To calculate it, you will need to first calculate the heat transfer resistance of the walls on a sanitary and hygienic criterion (dew misses):
PCT \u003d (TV - TN) / (DNH X AV).
Here dton is the temperature difference between the inner surface of the wall and the air temperature in the room, is determined in terms of the rules and for housing is 4.0.
aB - the heat transfer coefficient of the inner surface of the wall, according to the joint venture is 8.7.
The basic value of the doors is taken equal to 0.6HRS.
For the selected design, the door is required to perform the test heat engineering calculation of the enclosing structures. An example of calculating the input door:
RDV \u003d 0.6 x (20 - (- 30)) / (4 x 8.7) \u003d 0.86 (m x / h).
This calculated value will correspond to the door insulated with a mineral wool plate with a thickness of 5 cm. Its heat transfer resistance will be R \u003d 0.05 / 0.048 \u003d 1.04 (m x / h), which is more calculated.
Calculations of walls, overlap or coating are performed to verify the element requirements of standards. The set of rules also has a complete requirement that characterizes the quality of the insulation of all enclosing structures in general. This value is called "Specific heat protection characteristic". Without its verification, no heat engineering calculation of the enclosing structures is required. An example of calculating the joint venture is shown 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 home with dimensions of 10 x 10 x 2.5 m. Heat transfer resistance is equal to basic quantities.
The normalized value is determined in accordance with the joint venture, depending on the heated volume of the house.
In addition to the comprehensive requirement, the heat engineering calculation of the enclosing structures is also performed for the preparation of the energy passport, the example of the passport is given in the Appendix to the SP50.13330.2012.
All the above calculations are applicable for homogeneous structures. What is in practice is quite rare. To take into account heterogeneities that reduce heat transfer resistance, introduced correction factor Heat engineering homogeneity - r. It takes into account the change in heat transfer resistance made by window and doorways, external corners, inhomogeneous inclusions (for example, jumpers, beams, reinforcing belts), etc.
The calculation of this coefficient is quite complicated, so in a simplified form you can use exemplary values \u200b\u200bfrom reference literature. For example, for brickwork - 0.9, three-layer panels - 0.7.
Choosing a home insulation system, it is easy to make sure that the current requirements of thermal protection without using an effective insulation is almost impossible. So, if you use traditional clay brickIt will take masonry a few meters thick, which is economically inexpedient. However, the low thermal conductivity of modern insulation based on polystyrene foam or stone wati Allows you to limit the thicknesses of 10-20 cm.
For example, to achieve the basic value of the heat transfer resistance of 3.65 (m x / h / W), it will be necessary:
If you are going to build
A small brick cottage, then you will certainly have questions: "What
Thicks should be a wall? "," Does the insulation need? "," From which side to put
insulation? " etc. etc.
In this article we will try in
It is sorting out and answer all your questions.
Heat engineering
Fencing design is needed, first of all, in order to find out what
Thicks should be your outer wall.
First, you need to decide how much
Floors will be in your building and, depending on this, calculates
Fencing structures on the bearing capacity (not in this article).
On this calculation we define
The number of bricks in the masonry of your building.
For example, it turned out 2 clay
bricks without voids, brick length 250 mm,
the thickness of the solution is 10 mm, the total is 510 mm (brick density 0.67
In the future, we will be useful). Outdoor surface you decided to cover
facing tiles, thickness 1 cm (when buying it is necessary to know it
Density), and the inner surface of ordinary plaster, thickness of layer 1.5
See also Do not forget to learn its density. In the amount of 535mm.
In order for the building is not
collapsed this is of course enough, but unfortunately in most cities
Winter is cold and therefore such walls will freeze. And not
The walls were frozen, needed a layer of insulation.
The thickness of the insulation layer is calculated
in the following way:
1. On the Internet you need to download SNiP
II 3-79 * -
"Construction heat engineering" and SNiP 23-01-99 - "Construction climatology".
2. Open SNIP building
Climatology and find your city in Table 1 *, and we look at the intersection
The column "The air temperature is the coldest five days, ° C, secure
0.98 "and rows with your city. For the city of Penza, for example t n \u003d -32 about S.
3. Calculated internal air temperature
Take
t B \u003d 20 o C.
The heat transfer coefficient for inner wallsa. B \u003d 8.7W / m 2 · ˚С
Heat transfer coefficient for exterior walls in winter conditionsa. H \u003d 23W / m 2 · ˚С
Regulatory temperature difference between the temperature of the internal
air and temperature of the inner surface of the enclosing structuresΔt N \u003d 4 O S.
4. Next
Determine the required heat transfer resistance by formula # G0 (1a) from construction heat engineering
HSOP \u003d (T B - T OT.PER.) Z OTP , HSOP \u003d (20 + 4.5) · 207 \u003d 507,15 (for the city
Penza).
By formula (1), we calculate:
(where sigma is directly thick
Material, and lambda density. Itook as insulation
Polyurethanistanovapanel with density 0.025)
We accept the thickness of the insulation of 0.054 m.
From here the thickness of the wall will be:
d. = d. 1 + d. 2 + d. 3 + d. 4 =
0,01+0,51+0,054+0,015=0,589
m.
The repair season came up. Head broke: how to make good repair For less money. There is no thoughts about the loan. Support only on available ...
Instead of postponing the main repair from year to year, you can get ready for him to survive it in moderation ...
For the beginning you need to remove everything that remains from the old company that worked there. We break the artificial partition. After that, I'm moving away everything ...
