Creation comfortable conditions for accommodation or labor activity is the primary task of construction. A significant part of the territory of our country is located in northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always important. With rising energy tariffs, reducing energy consumption for heating comes to the fore.
Climatic characteristics
The choice of wall and roof design depends primarily on the climatic conditions of the construction area. To determine them, you need to refer to SP131.13330.2012 “Building climatology”. The following quantities are used in the calculations:
- the temperature of the coldest five-day period with a probability of 0.92 is designated Tn;
- average temperature, designated Thot;
- duration, denoted by ZOT.
Using the example for Murmansk, the values have the following values:
- Tn=-30 degrees;
- Tot=-3.4 degrees;
- ZOT=275 days.
In addition, it is necessary to set the estimated temperature inside the TV room, it is determined in accordance with GOST 30494-2011. For housing, you can take TV = 20 degrees.
To perform a thermal engineering calculation of enclosing structures, first calculate the GSOP value (degree-day of the heating period):
GSOP = (Tv - Tot) x ZOT.
In our example, GSOP = (20 - (-3.4)) x 275 = 6435.
Key indicators
For the right choice materials of enclosing structures, it is necessary to determine what thermal characteristics they should have. The ability of a substance to conduct heat is characterized by its thermal conductivity, denoted by the Greek letter l (lambda) and measured in W/(m x deg.). The ability of a structure to retain heat is characterized by its resistance to heat transfer R and is equal to the ratio of thickness to thermal conductivity: R = d/l.
If the structure consists of several layers, the resistance is calculated for each layer and then summed up.
Heat transfer resistance is the main indicator of the external structure. Its value must exceed normative meaning. When performing thermal engineering calculations of the building envelope, we must determine the economically justified composition of the walls and roof.
Thermal conductivity values
The quality of thermal insulation is determined primarily by thermal conductivity. Each certified material undergoes laboratory tests, as a result of which this value is determined for operating conditions “A” or “B”. For our country, most regions correspond to operating conditions “B”. When performing thermal engineering calculations of the building envelope, this value should be used. Thermal conductivity values are indicated on the label or in the material passport, but if they are not available, you can use reference values from the Code of Practice. Values for the most popular materials are given below:
- Masonry made of ordinary brick - 0.81 W (m x deg.).
- Sand-lime brickwork - 0.87 W (m x deg.).
- Gas and foam concrete (density 800) - 0.37 W (m x deg.).
- Wood coniferous species- 0.18 W (m x deg.).
- Extruded polystyrene foam - 0.032 W (m x deg.).
- Mineral wool slabs (density 180) - 0.048 W (m x deg.).
Standard value of heat transfer resistance
The calculated value of heat transfer resistance should not be less than the base value. The basic value is determined according to Table 3 SP50.13330.2012 “buildings”. The table defines the coefficients for calculating the basic values of heat transfer resistance of all enclosing structures and types of buildings. Continuing the started thermal engineering calculation of enclosing structures, an example of the calculation can be presented as follows:
- Rsten = 0.00035x6435 + 1.4 = 3.65 (m x deg/W).
- Rpokr = 0.0005x6435 + 2.2 = 5.41 (m x deg/W).
- Rcherd = 0.00045x6435 + 1.9 = 4.79 (m x deg/W).
- Rokna = 0.00005x6435 + 0.3 = x deg/W).
Thermal engineering calculations of the external enclosing structure are performed for all structures that close the “warm” circuit - the floor on the ground or the ceiling of a technical underground, external walls (including windows and doors), a combined covering or the ceiling of an unheated attic. Also, the calculation must be performed for internal structures, if the temperature difference in adjoining rooms is more than 8 degrees.
Thermal calculation of walls
Most walls and ceilings are multi-layered and heterogeneous in their design. Thermal engineering calculation of enclosing structures of a multilayer structure is as follows:
R= d1/l1 +d2/l2 +dn/ln,
where n are the parameters of the nth layer.
If we consider a brick plastered wall, we get the following design:
- outer layer of plaster 3 cm thick, thermal conductivity 0.93 W (m x deg.);
- masonry made of solid clay brick 64 cm, thermal conductivity 0.81 W (m x deg.);
- the inner layer of plaster is 3 cm thick, thermal conductivity 0.93 W (m x deg.).
Formula thermotechnical calculation enclosing structures looks like this:
R=0.03/0.93 + 0.64/0.81 + 0.03/0.93 = 0.85(m x deg/W).
The obtained value is significantly less than the previously determined base value of the heat transfer resistance of the walls of a residential building in Murmansk 3.65 (m x deg/W). The wall doesn't satisfy regulatory requirements and needs insulation. To insulate the wall we use a thickness of 150 mm and a thermal conductivity of 0.048 W (m x deg.).
Having selected an insulation system, it is necessary to perform a verification thermal engineering calculation of the enclosing structures. An example calculation is given below:
R=0.15/0.048 + 0.03/0.93 + 0.64/0.81 + 0.03/0.93 = 3.97(m x deg/W).
The resulting calculated value is greater than the base value - 3.65 (m x deg/W), the insulated wall meets the requirements of the standards.
The calculation of floors and combined coverings is carried out similarly.
Thermal engineering calculation of floors in contact with the ground
Often in private homes or public buildings are carried out on the ground. The heat transfer resistance of such floors is not standardized, but at a minimum the design of the floors should not allow dew to occur. The calculation of structures in contact with the ground is carried out as follows: the floors are divided into strips (zones) 2 meters wide, starting from the outer border. There are up to three such zones; the remaining area belongs to the fourth zone. If the floor design does not provide effective insulation, then the heat transfer resistance of the zones is assumed to be as follows:
- 1 zone - 2.1 (m x deg/W);
- Zone 2 - 4.3 (m x deg/W);
- Zone 3 - 8.6 (m x deg/W);
- Zone 4 - 14.3 (m x deg/W).
It is easy to notice that the further the floor area is from external wall, the higher its resistance to heat transfer. Therefore, they are often limited to insulating the perimeter of the floor. In this case, the heat transfer resistance of the insulated structure is added to the heat transfer resistance of the zone.
The calculation of the heat transfer resistance of the floor must be included in the general thermal engineering calculation of the enclosing structures. We will consider an example of calculating floors on the ground below. Let's take a floor area of 10 x 10 equal to 100 square meters.
- The area of zone 1 will be 64 square meters.
- The area of zone 2 will be 32 square meters.
- The area of zone 3 will be 4 square meters.
Average value of resistance to heat transfer of the floor over the ground:
Rpol = 100 / (64/2.1 + 32/4.3 + 4/8.6) = 2.6 (m x deg/W).
Having insulated the perimeter of the floor with an expanded polystyrene board 5 cm thick, a strip 1 meter wide, we obtain the average value of heat transfer resistance:
Rpol = 100 / (32/2.1 + 32/(2.1+0.05/0.032) + 32/4.3 + 4/8.6) = 4.09 (m x deg/W).
It is important to note that not only floors are calculated in this way, but also wall structures in contact with the ground (walls of a recessed floor, warm basement).
Thermal calculation of doors
The basic value of heat transfer resistance is calculated slightly differently entrance doors. To calculate it, you will first need to calculate the heat transfer resistance of the wall according to the sanitary and hygienic criterion (no dew):
Rst = (Tv - Tn)/(DTn x av).
Here DTn is the temperature difference between the inner surface of the wall and the air temperature in the room, determined according to the Code of Rules and for housing is 4.0.
ab - heat transfer coefficient inner surface walls, according to SP is 8.7.
The basic value of doors is taken equal to 0.6xРst.
For the selected door design, it is necessary to perform a verification thermal engineering calculation of the enclosing structures. An example of calculating an entrance door:
Rdv = 0.6 x (20-(-30))/(4 x 8.7) = 0.86 (m x deg/W).
This calculated value will correspond to an insulated door mineral wool slab 5 cm thick. Its heat transfer resistance will be R=0.05 / 0.048=1.04 (m x deg/W), which is greater than the calculated one.
Comprehensive Requirements
Calculations of walls, floors or coverings are performed to verify the element-by-element requirements of the standards. The set of rules also establishes a comprehensive requirement characterizing the quality of insulation of all enclosing structures as a whole. This value is called “specific thermal protection characteristic”. Not a single thermal engineering calculation of enclosing structures can be done without checking it. An example of calculation for a joint venture is given below.
Kob = 88.77 / 250 = 0.35, which is less than the normalized value of 0.52. In this case, the area and volume are taken for a house with dimensions of 10 x 10 x 2.5 m. Heat transfer resistances are equal to the basic values.
The normalized value is determined in accordance with the SP depending on the heated volume of the house.
In addition to the complex requirement, to draw up an energy passport, a thermal engineering calculation of the enclosing structures is also performed; an example of how to prepare a passport is given in the appendix to SP50.13330.2012.
Uniformity coefficient
All the above calculations are applicable for homogeneous structures. Which in practice is quite rare. To take into account inhomogeneities that reduce heat transfer resistance, a correction factor for thermal homogeneity - r - is introduced. It takes into account the change in heat transfer resistance introduced by window and doorways, external corners, heterogeneous inclusions (for example, lintels, beams, reinforcing belts), etc.
The calculation of this coefficient is quite complicated, so in a simplified form you can use approximate values from reference literature. For example, for brickwork- 0.9, three-layer panels - 0.7.
Effective insulation
When choosing a home insulation system, it is easy to make sure that modern requirements thermal protection without the use of effective insulation is almost impossible. So, if you use traditional clay bricks, you will need masonry several meters thick, which is not economically feasible. However, low thermal conductivity modern insulation materials based on expanded polystyrene or stone wool allows you to limit yourself to thicknesses of 10-20 cm.
For example, to achieve a basic heat transfer resistance value of 3.65 (m x deg/W), you will need:
- brick wall 3 m thick;
- masonry made of foam concrete blocks 1.4 m;
- mineral wool insulation 0.18 m.
In order to keep your home warm in the most severe frosts, it is necessary to choose the right thermal insulation system - for this, a thermal engineering calculation is performed outer wall.The calculation result shows how effective the real or designed insulation method is.
How to make a thermal engineering calculation of an external wall
First, you should prepare the initial data. The following factors influence the calculated parameter:
- the climatic region in which the house is located;
- purpose of the premises - residential building, industrial building, hospital;
- operating mode of the building – seasonal or year-round;
- presence in the design of door and window openings;
- indoor humidity, difference between indoor and outdoor temperatures;
- number of floors, floor features.
After collecting and recording the initial information, the thermal conductivity coefficients are determined building materials, from which the wall is made. The degree of heat absorption and heat transfer depends on how damp the climate is. In this regard, to calculate the coefficients, humidity maps compiled for Russian Federation. After this, all numerical values necessary for the calculation are entered into the appropriate formulas. 
Thermal engineering calculation of an external wall, example for a foam concrete wall
As an example, the heat-protective properties of a wall made of foam blocks, insulated with expanded polystyrene with a density of 24 kg/m3 and plastered on both sides with lime-sand mortar are calculated. Calculations and selection of tabular data are based on building regulations.
Initial data: construction area - Moscow; relative humidity - 55%, average temperature in the house tв = 20О С. The thickness of each layer is set: δ1, δ4=0.01m (plaster), δ2=0.2m (foam concrete), δ3=0.065m (expanded polystyrene "SP Radoslav" ).
The purpose of the thermal engineering calculation of an external wall is to determine the required (Rtr) and actual (Rph) heat transfer resistance.
Calculation
- According to Table 1 SP 53.13330.2012, under given conditions, the humidity regime is assumed to be normal. The required value of Rtr is found using the formula:
Rtr=a GSOP+b,
where a, b are taken according to table 3 SP 50.13330.2012. For a residential building and an external wall a = 0.00035; b = 1.4.
GSOP – degree-days of the heating period, they are found using formula (5.2) SP 50.13330.2012:
GSOP=(tv-tot)zot,
where tв=20О С; tot – average outside air temperature during the heating period, according to Table 1 SP131.13330.2012 tot = -2.2°C; zfrom = 205 days. (duration of the heating season according to the same table).
Substituting the table values, they find: GSOP = 4551О С*day; Rtr = 2.99 m2*C/W - According to Table 2 SP50.13330.2012 for normal humidity, select the thermal conductivity coefficients of each layer of the “pie”: λB1=0.81 W/(m°C), λB2=0.26 W/(m°C), λB3=0.041 W/(m° C), λB4=0.81 W/(m°C).
Using formula E.6 SP 50.13330.2012, the conditional heat transfer resistance is determined:
R0condition=1/αint+δn/λn+1/αext.
where αext = 23 W/(m2°C) from clause 1 of table 6 SP 50.13330.2012 for external walls.
Substituting the numbers, we get R0cond=2.54m2°C/W. It is clarified using the coefficient r = 0.9, depending on the homogeneity of the structures, the presence of ribs, reinforcement, and cold bridges:
Rf=2.54 0.9=2.29m2 °C/W.
The obtained result shows that the actual thermal resistance is less than the required one, so the wall design needs to be reconsidered.
Thermal calculation of an external wall, the program simplifies calculations
Simple computer services speed up computational processes and the search for the required coefficients. It is worth familiarizing yourself with the most popular programs. 
- "TeReMok". The initial data is entered: building type (residential), internal temperature 20O, humidity regime - normal, area of residence - Moscow. IN next window the calculated value of the standard heat transfer resistance is revealed - 3.13 m2*оС/W.
Based on the calculated coefficient, a thermal engineering calculation is made of an external wall made of foam blocks (600 kg/m3), insulated with extruded polystyrene foam “Flurmat 200” (25 kg/m3) and plastered with cement-lime mortar. Select from the menu necessary materials, indicating their thickness (foam block - 200 mm, plaster - 20 mm), leaving the cell with the thickness of the insulation unfilled.
By clicking the “Calculation” button, the required thickness of the heat insulation layer is obtained – 63 mm. The convenience of the program does not eliminate its shortcoming: it does not take into account different thermal conductivities masonry material and solution. Thanks to the author you can say at this address http://dmitriy.chiginskiy.ru/teremok/ - The second program is offered by the site http://rascheta.net/. Its difference from the previous service is that all thicknesses are set independently. The coefficient of thermal uniformity r is introduced into the calculation. It is selected from the table: for foam concrete blocks with wire reinforcement in horizontal joints r = 0.9.
After filling in the fields, the program produces a report on what the actual thermal resistance chosen design, does it meet climatic conditions. In addition, a sequence of calculations with formulas, normative sources and intermediate values is provided.
When building a house or carrying out thermal insulation works It is important to assess the effectiveness of insulation of an external wall: a thermal calculation performed independently or with the help of a specialist allows you to do this quickly and accurately.
Initial data
Place of construction - Omsk
z ht = 221 days
t ht = -8.4ºС.
t ext = -37ºС.
t int = + 20ºС;
air humidity: = 55%;
Operating conditions of enclosing structures - B. Heat transfer coefficient of the internal surface of the enclosure A i nt = 8.7 W/m 2 °C.
a ext = 23 W/m 2 °C.
The necessary data on the structural layers of the wall for thermal engineering calculations are summarized in the table.
1. Determination of the degree-day of the heating period using formula (2) SP 23-101-2004:
D d = (t int - t ht) z th = (20–(8.4))·221= 6276.40
2. Standardized value of heat transfer resistance of external walls according to formula (1) SP 23-101-2004:
R reg = a · D d + b =0.00035·6276.40+ 1.4 =3.6m 2 ·°С/W.
3. Reduced resistance to heat transfer R 0 r external brick walls with effective insulation residential buildings is calculated using the formula
R 0 r = R 0 conditional r,
where R 0 conventional is the heat transfer resistance of brick walls, conventionally determined by formulas (9) and (11) without taking into account heat-conducting inclusions, m 2 °C/W;
R 0 r - reduced heat transfer resistance taking into account the coefficient of thermal uniformity r, which for walls is 0.74.
The calculation is carried out from the condition of equality
hence,
R 0 conventional = 3.6/0.74 = 4.86 m 2 °C / W
R 0 conventional =R si +R k +R se
R k = R reg - (R si + R se) = 3.6- (1/8.7 + 1/23) = 3.45 m 2 °C / W
4. Thermal resistance of outer brick wall layered structure can be represented as the sum of thermal resistances individual layers, i.e.
R k = R 1 + R 2 + R ut + R 4
5. Define thermal resistance insulation:
R ut = R k + (R 1 + R 2 + R 4) = 3.45– (0.037 + 0.79) = 2.62 m 2 °C/W.
6. Find the thickness of the insulation:
| Ri |
We accept the insulation thickness as 100 mm.
The final wall thickness will be (510+100) = 610 mm.
We check taking into account the accepted thickness of the insulation:
R 0 r = r (R si + R 1 + R 2 + R ut + R 4 + R se) = 0.74 (1/8.7 + 0.037 + 0.79 + 0.10/0.032+ 1/23 ) = 4.1m 2 °C/W.
Condition R 0 r = 4.1> = 3.6m 2 °C/W is satisfied.
Checking compliance with sanitary and hygienic requirements
thermal protection of the building
1. Check if the condition is met 
:
∆t = (t int – t ext)/ R 0r a int = (20-(37))/4.1 8.7 = 1.60 ºС
According to table. 5SP 23-101-2004 ∆ t n = 4 °С, therefore, the condition ∆ t = 1,60< ∆t n = 4 ºС is satisfied.
2. Check if the condition is met 
:
] = 20 – =
20 – 1.60 = 18.40ºС
3. According to Appendix SP 23-101–2004 for internal air temperature t int = 20 ºC and relative humidity = 55% dew point temperature t d = 10.7ºС, therefore, the condition τsi = 18.40> t d = 
is running.
Conclusion. The building envelope meets the regulatory requirements for thermal protection of the building.
4.2 Thermal engineering calculation of the attic covering.
Initial data
Determine the thickness of the attic floor insulation, consisting of insulation δ = 200 mm, vapor barrier, prof. sheet
Attic floor:
Combined coverage:
Place of construction - Omsk
Duration of the heating season z ht = 221 days.
Average design temperature of the heating period t ht = -8.4ºС.
Cold five-day temperature t ext = –37ºС.
The calculation was made for a five-story residential building:
indoor air temperature t int = + 20ºС;
air humidity: = 55%;
The humidity level of the room is normal.
Operating conditions of enclosing structures – B.
Heat transfer coefficient of the inner surface of the fence A i nt = 8.7 W/m 2 °C.
Heat transfer coefficient of the outer surface of the fence a ext = 12 W/m 2 °C.
Name of material Y 0, kg/m³ δ, m λ, mR, m 2 °C/W
1. Determination of the degree-day of the heating period using formula (2) SP 23-101-2004:
D d = (t int - t ht) z th = (20 –8.4) 221=6276.4ºСsut
2. Normalization of the heat transfer resistance value of the attic floor according to formula (1) SP 23-101-2004:
R reg = a · D d + b, where a and b are selected according to table 4 SP 23-101-2004
R reg = a · D d + b = 0.00045 · 6276.4+ 1.9 = 4.72 m² · ºС / W
3. Thermal engineering calculation is carried out from the condition that the total thermal resistance R 0 is equal to the normalized R reg, i.e.
4. From formula (8) SP 23-100-2004, we determine the thermal resistance of the enclosing structure R k (m² ºС / W)
R k = R reg - (R si + R se)
R reg = 4.72 m² ºС / W
R si = 1 / α int = 1 / 8.7 = 0.115 m² ºС / W
R se = 1 / α ext = 1 / 12 = 0.083 m² ºС / W
R k = 4.72– (0.115 + 0.083) = 4.52 m² ºС / W
5. The thermal resistance of the enclosing structure (attic floor) can be represented as the sum of the thermal resistances of the individual layers:
R c = R reinforced concrete + R pi + R cs + R ut → R ut = R c + (R reinforced concrete + R pi + R cs) = R c - (d/ λ) = 4.52 – 0.29 = 4 .23
6. We use formula (6) SP 23-101-2004 and determine the thickness of the insulating layer:
d ut = R ut λ ut = 4.23 0.032 = 0.14 m
7. We accept the thickness of the insulating layer as 150mm.
8. We calculate the total thermal resistance R 0:
R 0 = 1 / 8.7 + 0.005 / 0.17 + 0.15 / 0.032 + 1 / 12 = 0.115 + 4.69+ 0.083 = 4.89 m² ºС / W
R 0 ≥ R reg 4.89 ≥ 4.72 satisfies the requirement
Checking the fulfillment of conditions
1. Check the fulfillment of the condition ∆t 0 ≤ ∆t n
The value of ∆t 0 is determined by formula (4) SNiP 02/23/2003:
∆t 0 = n ·(t int - t ext) / R 0 · a int where, n is a coefficient taking into account the dependence of the position of the outer surface to the outside air according to the table. 6
∆t 0 = 1(20+37) / 4.89 8.7 = 1.34ºС
According to table. (5) SP 23-101-2004 ∆t n = 3 ºС, therefore, the condition ∆t 0 ≤ ∆t n is satisfied.
2. Check the fulfillment of condition τ >t d
τ value calculated using formula (25) SP 23-101-2004
t si = t int– [n(t int–t ext)]/(R o a int)
τ = 20- 1(20+26) / 4.89 8.7 = 18.66 ºС
3. According to Appendix R SP 23-01-2004 for internal air temperature t int = +20 ºС and relative humidity φ = 55% dew point temperature t d = 10.7 ºС, therefore, condition τ >t d is fulfilled.
Conclusion: attic floor meets regulatory requirements.
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. etc.
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
of enclosing structures according to load-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
as follows:
1. You need to download SNiP on the Internet
II 3-79* —
“Construction Heat Engineering” and SNiP 23-01-99 - “Construction 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 the temperature of the inner surface of the 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. Thermal engineering calculations, which are no less important than calculating efficiency, strength, fire resistance, and durability, will allow you to determine the golden mean.
Based on thermal engineering standards, climatic characteristics, steam and moisture permeability, materials for the construction of enclosing structures are selected. 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 plus and minus temperatures. The list of these characteristics also includes such an indicator as the amount of heat lost by the building envelope during the cold period.
You can't start designing without having all this data. Based on them, the thickness of the walls and ceilings and the sequence of layers are chosen.
According to GOST 30494-96 regulations, temperature values indoors. On average it is 21⁰. At the same time, the relative humidity must remain within a comfortable range, which is an average of 37%. The highest speed of air mass movement is 0.15 m/s
Thermal engineering calculation aims to determine:
- Are the designs identical to the stated requirements in terms of thermal protection?
- How fully is a comfortable microclimate inside the building ensured?
- Is optimal thermal protection of structures provided?
The basic principle is maintaining a balance of the difference in temperature indicators of the atmosphere of internal structures of fences and premises. If this is not followed, heat will be absorbed by these surfaces and the temperature inside will remain very low.
The internal temperature should not be significantly affected by changes 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:
- Purpose of the building and its type.
- Orientations of vertical enclosing structures relative to the cardinal directions.
- Geographical parameters of the future home.
- The volume of the building, its number of storeys, area.
- Types and dimensions of door and window openings.
- Type of heating and its technical parameters.
- Number of permanent residents.
- Materials for vertical and horizontal fencing structures.
- Upper floor ceilings.
- Hot water supply equipment.
- Type of ventilation.
Others are also taken into account when calculating design features buildings. The air permeability of enclosing structures should not contribute to excessive cooling inside the house and reduce the thermal protection characteristics of the elements.
Heat loss is also caused by waterlogging of the walls, and in addition, this entails dampness, which negatively affects the durability of the building.
In the calculation process, first of all, the thermal technical data of the building materials from which the building’s enclosing elements are made are determined. In addition, the reduced heat transfer resistance and compliance with its standard value are subject to determination.
Formulas for making calculations
Heat loss from a home can be divided into two main parts: losses through the building envelope and losses caused by operation. In addition, heat is lost when discharging warm water into the sewer system.
For the materials from which the enclosing structures are constructed, it is necessary to find the value of the thermal conductivity index Kt (W/m x degree). They are in the relevant reference books.
Now, knowing the thickness of the layers, according to the formula: R = S/Kt, calculate the thermal resistance of each unit. If the structure is multilayer, all obtained values are added together.
The easiest way to determine the size of heat losses is by adding up the thermal flows through the enclosing structures that actually form this building
Guided by this methodology, they take into account the fact that the materials that make up the structure have a different structure. It is also taken into account that the heat flow passing through them has different specifics.
For each individual structure, heat loss is determined by the formula:
Q = (A / R) x dT
- A - area in m².
- R - resistance of the structure to heat transfer.
- dT - temperature difference between outside and inside. It needs to be determined for the coldest 5-day period.
By performing the calculation in this way, you can only get the result for the coldest five-day period. The total heat loss for the entire cold season is determined by taking into account the dT parameter, taking into account not the lowest temperature, but the average one.
The extent to which heat is absorbed, as well as heat transfer, depends on the humidity of the climate in the region. For this reason, humidity maps are used in calculations.
There is a formula for this:
W = ((Q + Qв) x 24 x N)/1000
In it, N is the duration of the heating period in days.
Disadvantages of area calculation
Calculation based on the area indicator is not very accurate. Here, such parameters as climate, temperature indicators, both minimum and maximum, and humidity are not taken into account. Due to ignoring many important points, the calculation has significant errors.
Often trying to cover them, the project includes a “reserve”.
If, nevertheless, this method is chosen for calculation, the following nuances must be taken into account:
- 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.
- If the project includes a balcony, two windows or a loggia, multiply by an average of 125 W.
- When the premises are industrial or warehouse, a multiplier of 150 W is used.
- If radiators are located near windows, their design capacity is increased by 25%.
The formula for area is:
Q=S x 100 (150) W.
Here Q is the comfortable heat level in the building, S is the heated area in m². The numbers 100 or 150 are the specific amount of thermal energy consumed to heat 1 m².
House ventilation losses
The key parameter in this case is the air exchange rate. Provided that the walls of the house are vapor-permeable, this value is equal to one.
The penetration of cold air into the house is carried out by supply ventilation. Exhaust ventilation promotes care warm air. The recuperator-heat exchanger reduces losses through ventilation. It does not allow heat to escape along with the outgoing air, and it heats the incoming air flows
It is envisaged that the air inside the building will be completely renewed in one hour. Buildings built according to the DIN standard have walls with vapor barriers, so here the air exchange rate is taken to be two.
There is a formula that determines heat loss through the ventilation system:
Qv = (V x Kv: 3600) x P x C x dT
Here the symbols mean the following:
- Qв - heat loss.
- V is the volume of the room in mᶾ.
- P - air density. its value is taken equal to 1.2047 kg/mᶾ.
- Kv - air exchange rate.
- C - specific heat capacity. It is equal to 1005 J/kg x C.
Based on the results of this calculation, it is possible to determine the power of the heat generator heating system. In case too high value power may be a way out of the situation. Let's look at a few examples for houses made of different materials.
Example of thermal engineering calculation No. 1
Let's calculate a residential building located 1 climatic region(Russia), subdistrict 1B. All data is taken from table 1 of SNiP 23-01-99. Most cold temperature, observed over five days with a probability of 0.92 - tн = -22⁰С.
In accordance with SNiP, the heating period (zop) lasts 148 days. The average temperature during the heating period with the average daily air temperature outside is 8⁰ - tot = -2.3⁰. Outside temperature in heating season- tht = -4.4⁰.
Heat loss at home - the most important moment at the design stage. The choice of building materials and insulation depends on the results of the calculation. There are no zero losses, but you need to strive to ensure that they are as expedient as possible
The condition is stipulated that the temperature in the rooms of the house must be 22⁰. The house has two floors and walls 0.5 m thick. Its height is 7 m, plan dimensions are 10 x 10 m. The material of the vertical enclosing structures is warm ceramics. For it, the thermal conductivity coefficient is 0.16 W/m x C.
Mineral wool was used as external insulation, 5 cm thick. The Kt value for it is 0.04 W/m x C. The number of window openings in the house is 15 pcs. 2.5 m² each.
Heat loss through walls
First of all, you need to define the thermal resistance as ceramic wall, and insulation. In the first case, R1 = 0.5: 0.16 = 3.125 sq. m x C/W. In the second - R2 = 0.05: 0.04 = 1.25 sq. m x C/W. In general, for a vertical building envelope: R = R1 + R2 = 3.125 + 1.25 = 4.375 sq. m x C/W.
Since heat loss is directly proportional to the area of the enclosing structures, we calculate the area of the walls:
A = 10 x 4 x 7 – 15 x 2.5 = 242.5 m²
Now you can determine heat loss through the walls:
Qс = (242.5: 4.375) x (22 – (-22)) = 2438.9 W.
Heat loss through horizontal enclosing structures is calculated in a similar way. In the end, all the results are summed up.
If the basement under the floor of the first floor is heated, the floor does not need to be insulated. It is still better to line the basement walls with insulation so that the heat does not escape into the ground.
Determination of losses through ventilation
To simplify the calculation, they do not take into account the thickness of the walls, but simply determine the volume of air inside:
V = 10x10x7 = 700 mᶾ.
With an air exchange rate of Kv = 2, the heat loss will be:
Qв = (700 x 2) : 3600) x 1.2047 x 1005 x (22 – (-22)) = 20,776 W.
If Kv = 1:
Qв = (700 x 1) : 3600) x 1.2047 x 1005 x (22 – (-22)) = 10,358 W.
Effective ventilation of residential buildings is provided by rotary and plate recuperators. The efficiency of the former is higher, it reaches 90%.
Example of thermal engineering calculation No. 2
It is required to calculate losses through a 51 cm thick brick wall. It is insulated with a 10 cm layer mineral wool. Outside – 18⁰, inside – 22⁰. The dimensions of the wall are 2.7 m in height and 4 m in length. The only outer wall of the room is oriented to the south; there are no external doors.
For brick, the thermal conductivity coefficient Kt = 0.58 W/mºC, for mineral wool - 0.04 W/mºC. Thermal Resistance:
R1 = 0.51: 0.58 = 0.879 sq. m x C/W. R2 = 0.1: 0.04 = 2.5 sq. m x C/W. In general, for a vertical building envelope: R = R1 + R2 = 0.879 + 2.5 = 3.379 sq. m x C/W.
External wall area A = 2.7 x 4 = 10.8 m²
Heat loss through the wall:
Qс = (10.8: 3.379) x (22 – (-18)) = 127.9 W.
To calculate losses through windows, the same formula is used, but their thermal resistance, as a rule, is indicated in the passport and does not need to be calculated.
In the thermal insulation of a house, windows are the “weak link”. A fairly large portion of the heat is lost through them. Multilayer double-glazed windows, heat-reflecting films, double frames will reduce losses, but even this will not help avoid heat loss completely
If the house has energy-saving windows measuring 1.5 x 1.5 m², oriented to the North, and the thermal resistance is 0.87 m2°C/W, then the losses will be:
Qо = (2.25: 0.87) x (22 – (-18)) = 103.4 t.
Example of thermal engineering calculation No. 3
Let's perform a thermal calculation of a wooden log building with a facade built from pine logs with a layer 0.22 m thick. The coefficient for this material is K = 0.15. In this situation, the heat loss will be:
R = 0.22: 0.15 = 1.47 m² x ⁰С/W.
The most low temperature five-day period - -18⁰, for comfort in the house the temperature is set to 21⁰. The difference will be 39⁰. Based on an area of 120 m², the result will be:
Qс = 120 x 39: 1.47 = 3184 W.
For comparison, let’s define the losses brick house. The coefficient for sand-lime brick is 0.72.
R = 0.22: 0.72 = 0.306 m² x ⁰С/W.
Qс = 120 x 39: 0.306 = 15,294 W.
Under the same conditions wooden house more economical. Sand-lime brick It’s not suitable for building walls here at all.
The wooden structure has a high heat capacity. Its enclosing structures maintain a comfortable temperature for a long time. Still, even log house it is necessary to insulate and it is better to do this both inside and outside
Example of heat calculation 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 required resistance heat transfer and actual. The required Rtr is determined by substituting the values in the expression:
Rtr=a x GSOP+b
where GOSP is the degree-day of the heating season, a and b are coefficients taken from table No. 3 of the Code of Rules 50.13330.2012. Since the building is residential, a is 0.00035, b = 1.4.
GSOP is calculated using a formula taken from the same SP:
GOSP = (tv – tot) x zot.
In this formula tв = 20⁰, tоt = -2.2⁰, zоt - 205 is the heating period in days. Hence:
GSOP = (20 – (-2.2)) x 205 = 4551⁰ C x day;
Rtr = 0.00035 x 4551 + 1.4 = 2.99 m2 x C/W.
Using table No. 2 SP50.13330.2012, determine the thermal conductivity coefficients for each layer of the wall:
- λb1 = 0.81 W/m ⁰С;
- λb2 = 0.26 W/m ⁰С;
- λb3 = 0.041 W/m ⁰С;
- λb4 = 0.81 W/m ⁰С.
The total conditional resistance to heat transfer Ro is equal to the sum of the resistances of all layers. It is calculated using the formula:
Substituting the values we get: Rо arb. = 2.54 m2°C/W. Rф is determined by multiplying Ro by a coefficient r equal to 0.9:
Rf = 2.54 x 0.9 = 2.3 m2 x °C/W.
The result requires changing the design of the enclosing element, since the actual thermal resistance is less than the calculated one.
There are many computer services that speed up and simplify calculations.
Thermal calculations are directly related to the definition. You will learn what it is and how to find its meaning from the article we recommend.
Conclusions and useful video on the topic
Performing thermal engineering calculations using an online calculator:
Correct thermal 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.
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