Thermal engineering calculation of a building: specifics and formulas for performing calculations + practical examples

Thermal engineering calculation online (calculator overview)

Thermal engineering calculation can be done on the Internet online. Let's take a quick look at how to work with it.

Going to the website of the online calculator, the first step is to select the standards for which the calculation will be made. I choose the 2012 rulebook as it is a newer document.

Next, you need to specify the region in which the object will be built. If your city is not available, choose the nearest big city. After that, we indicate the type of buildings and premises.Most likely you will calculate a residential building, but you can choose public, administrative, industrial and others. And the last thing you need to choose is the type of enclosing structure (walls, ceilings, coatings).

We leave the calculated average temperature, relative humidity and thermal uniformity coefficient the same if you do not know how to change them.

In the calculation options, set all two checkboxes except the first one.

In the table, we indicate the wall cake starting from the outside - we select the material and its thickness. On this, in fact, the whole calculation is completed. Below the table is the result of the calculation. If any of the conditions is not met, we change the thickness of the material or the material itself until the data complies with regulatory documents.

If you want to see the calculation algorithm, then click on the "Report" button at the bottom of the site page.

5.1 The general sequence of performing thermal calculation

1. AT
   in accordance with paragraph 4 of this manual
   determine the type of building and conditions, according to
   which should be counted R_abouttr.

2. Define
   R_abouttr:

• on
  formula (5), if the building is calculated
  for sanitary and hygienic and comfortable
  conditions;

• on
  formula (5a) and table. 2 if the calculation should
  be conducted on the basis of energy saving conditions.

1. Compose
   total resistance equation
   enclosing structure with one
   unknown by formula (4) and equate
   his R_abouttr.

2. Calculate
   unknown thickness of the insulation layer
   and determine the overall thickness of the structure.
   In doing so, it is necessary to take into account typical
   outer wall thicknesses:

• thickness
  brick walls should be a multiple
  brick size (380, 510, 640, 770 mm);

• thickness
  exterior wall panels is accepted
  250, 300 or 350 mm;

• thickness
  sandwich panels are accepted
  equal to 50, 80 or 100 mm.

Factors affecting TN

Thermal insulation - internal or external - significantly reduces heat loss

Heat loss is influenced by many factors:

• Foundation - the insulated version retains heat in the house, the non-insulated one allows up to 20%.
• Wall - porous concrete or wood concrete has a much lower throughput than a brick wall. Red clay brick retains heat better than silicate brick. The thickness of the partition is also important: a brick wall 65 cm thick and foam concrete 25 cm thick have the same level of heat loss.
• Warming - thermal insulation significantly changes the picture. External insulation with polyurethane foam - a sheet 25 mm thick - is equal in efficiency to the second brick wall 65 cm thick. Cork inside - a sheet 70 mm - replaces 25 cm of foam concrete. It is not in vain that experts say that effective heating begins with proper insulation.
• Roof - pitched construction and insulated attic reduce losses. A flat roof made of reinforced concrete slabs transmits up to 15% of heat.
• Glazing area - the thermal conductivity of glass is very high. No matter how tight the frames are, heat escapes through the glass. The more windows and the larger their area, the higher the thermal load on the building.
• Ventilation - the level of heat loss depends on the performance of the device and the frequency of use. The recovery system allows you to somewhat reduce losses.
• The difference between the temperature outside and inside the house - the larger it is, the higher the load.
• The distribution of heat within the building - affects the performance for each room. The rooms inside the building cool down less: in calculations, the comfortable temperature here is considered to be +20 C.The end rooms cool down faster - the normal temperature here will be +22 C. In the kitchen, it is enough to heat the air up to +18 C, since there are many other heat sources here: stove, oven, refrigerator.

Influence of the air gap

In the case when mineral wool, glass wool or other slab insulation is used as a heater in a three-layer masonry, it is necessary to install an air ventilated layer between the outer masonry and the insulation. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to drain the insulation, which gets wet from condensate.

This air layer is not a closed space, therefore, if it is present in the calculation, it is necessary to take into account the requirements of clause 9.1.2 of SP 23-101-2004, namely:

a) structural layers located between the air gap and the outer surface (in our case, this is a decorative brick (besser)) are not taken into account in the heat engineering calculation;

b) on the surface of the structure facing towards the layer ventilated by the outside air, the heat transfer coefficient αext = 10.8 W/(m°C) should be taken.

Parameters for performing calculations

To perform heat calculation, initial parameters are needed.

They depend on a number of characteristics:

1. Purpose of the building and its type.
2. Orientation of vertical enclosing structures relative to the direction to the cardinal points.
3. Geographic parameters of the future home.
4. The volume of the building, its number of storeys, area.
5. Types and dimensional data of door and window openings.
6. Type of heating and its technical parameters.
7. The number of permanent residents.
8. Material of vertical and horizontal protective structures.
9. Top floor ceilings.
10. Hot water facilities.
11. Type of ventilation.

Other design features of the structure are also 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.

Waterlogging of the walls also causes heat loss, and in addition, this entails dampness, which negatively affects the durability of the building.

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

Thermal load concepts

Calculation of heat loss is carried out separately for each room, depending on the area or volume

Space heating is compensation for heat loss. Through the walls, foundation, windows and doors, heat is gradually removed to the outside. The lower the outside temperature, the faster the heat transfer to the outside. To maintain a comfortable temperature inside the building, heaters are installed. Their performance must be high enough to cover the heat loss.

The heat load is defined as the sum of heat losses of the building, equal to the required heating power. Having calculated how much and how the house loses heat, they will find out the power of the heating system. The total value is not enough. A room with 1 window loses less heat than a room with 2 windows and a balcony, so the indicator is calculated for each room separately.

When calculating, be sure to take into account the height of the ceiling. If it does not exceed 3 m, the calculation is performed by the size of the area. If the height is from 3 to 4 m, the flow rate is calculated by volume.

Typical wall designs

We will analyze options from various materials and various variations of the “pie”, but for starters, it is worth mentioning the most expensive and extremely rare option today - a solid brick wall. For Tyumen, the wall thickness should be 770 mm or three bricks.

bar

In contrast, a fairly popular option is a 200 mm beam. From the diagram and from the table below, it becomes obvious that one beam for a residential building is not enough. The question remains, is it enough to insulate the outer walls with one sheet of mineral wool 50 mm thick?

Material name	Width, m	λ1, W/(m × °С)	R1, m2×°С/W
Softwood lining	0,01	0,15	0,01 / 0,15 = 0,066
Air	0,02	—	—
Ecover Standard 50	0,05	0,04	0,05 / 0,04 = 1,25
Pine beam	0,2	0,15	0,2 / 0,15 = 1,333

Substituting into the previous formulas, we obtain the required thickness of the insulation δ_ut = 0.08 m = 80 mm.

It follows that insulation in one layer of 50 mm mineral wool is not enough, it is necessary to insulate in two layers with an overlap.

For lovers of chopped, cylindered, glued and other types of wooden houses. You can take into account any thickness of wooden walls available to you and make sure that without external insulation during cold periods you will either freeze at equal costs of thermal energy, or spend more on heating. Unfortunately, miracles do not happen.

It is also worth noting the imperfection of the joints between the logs, which inevitably leads to heat loss. In the picture of the thermal imager, the corner of the house was taken from the inside.

Expanded clay block

The next option has also gained popularity recently, a 400 mm expanded clay block with a brick lining. Find out how thick the insulation is needed in this option.

Material name	Width, m	λ1, W/(m × °С)	R1, m2×°С/W
Brick	0,12	0,87	0,12 / 0,87 = 0,138
Air	0,02	—	—
Ecover Standard 50	0,05	0,04	0,05 / 0,04 = 1,25
Expanded clay block	0,4	0,45	0,4 / 0,45 = 0,889

Substituting into the previous formulas, we obtain the required thickness of the insulation δ_ut = 0.094 m = 94 mm.

For masonry made of expanded clay block with brick facing, mineral insulation 100 mm thick is required.

gas block

Gas block 400 mm with insulation and plastering using the "wet facade" technology. The size of the external plaster is not included in the calculation due to the extreme smallness of the layer. Also, due to the correct geometry of the blocks, we will reduce the layer of internal plaster to 1 cm.

Material name	Width, m	λ1, W/(m × °С)	R1, m2×°С/W
Ecover Standard 50	0,05	0,04	0,05 / 0,04 = 1,25
Porevit BP-400 (D500)	0,4	0,12	0,4 / 0,12 = 3,3
Plaster	0,01	0,87	0,01 / 0,87 = 0,012

Substituting into the previous formulas, we obtain the required thickness of the insulation δ_ut = 0.003 m = 3 mm.

Here the conclusion suggests itself: the Porevit block with a thickness of 400 mm does not require insulation from the outside, external and internal plastering or finishing with facade panels is sufficient.

Determining the thickness of the wall insulation

Determination of the thickness of the building envelope. Initial data:

1. Construction area - Sredny
2. Purpose of the building - Residential.
3. Construction type - three-layer.
4. Standard room humidity - 60%.
5. The temperature of the internal air is 18°C.

layer number	Layer name	thickness
1	Plaster	0,02
2	Masonry (cauldron)	X
3	Insulation (polystyrene)	0,03
4	Plaster	0,02

2 Calculation procedure.

I carry out the calculation in accordance with SNiP II-3-79 * “Design standards. Construction heat engineering”

A) I determine the required thermal resistance R_{o(tr) according to the formula:}

R_{o(tr)=n(tv-tn)/(Δtn*αv) , where n is the coefficient that is chosen taking into account the location of the outer surface of the enclosing structure in relation to the outside air.}

n=1

tn is the calculated winter t of outside air, taken in accordance with paragraph 2.3 of SNiPa “Construction heating engineering”.

I accept conditionally 4

I determine that tн for a given condition is taken as the calculated temperature of the coldest first day: tн=t_{x(3) ; tx(1)=-20°C; tx(5)=-15°С.}

t_{x(3)=(tx(1) + tx(5))/2=(-20+(-15))/2=-18°C; tn=-18°С.}

Δtn is the standard difference between tin air and tin the surface of the enclosing structure, Δtn=6°C according to the table. 2

αv - heat transfer coefficient of the inner surface of the fence structure

αv=8.7 W/m2°C (according to Table 4)

R_{o(tr)=n(tv-tn)/(Δtn*αv)=1*(18-(-18)/(6*8.7)=0.689(m2°C/W)}

B) Determine R_about=1/αv+R₁+R₂+R₃+1/αn , where αn is the heat transfer factor, for winter conditions of the outer enclosing surface. αн=23 W/m2°С according to the table. 6#layer

	Material name	item number	ρ, kg/m3	σ, m	λ	S
1	Lime-sand mortar	73	1600	0,02	0,7	8,69
2	Kotelets	98	1600	0,39	1,16	12,77
3	Styrofoam	144	40	X	0,06	0,86
4	Complex mortar	72	1700	0,02	0,70	8,95

To fill in the table, I determine the operating conditions of the enclosing structure, depending on the zones of humidity and the wet regime in the premises.

1 The humidity regime of the premises is normal according to the table. one

2 Humidity zone - dry

I determine the operating conditions → A

R₁₌σ₁/λ₁\u003d 0.02 / 0.7 \u003d 0.0286 (m2 ° C / W)

R₂=σ₂/λ₂=0,39/1,16= 0,3362

R₃=σ₃/λ₃ =X/0.06 (m2°C/W)

R₄=σ₄/λ₄ \u003d 0.02 / 0.7 \u003d 0.0286 (m2 ° C / W)

R_about=1/αv+R₁+R₂+1/αn = 1/8.7+0.0286 + 0.3362+X/0.06 +0.0286+1/23 = 0.518+X/0.06

I accept R_about= R_{o(tr)=0.689m2°C/W}

0.689=0.518+X/0.06

X_tr\u003d (0.689-0.518) * 0.06 \u003d 0.010 (m)

I accept constructively σ_{1(f)=0.050 m}

R_{1(φ)= σ1(f)/ λ1=0.050/0.060=0.833 (m2°C/W)}

3 I determine the inertia of the building envelope (massiveness).

D=R₁*S₁+ R₂*S₂+ R₃*S₃=0,029*8,69+0,3362*12,77+0,833*0,86+0,0286*8,95 = 5,52

Conclusion: the enclosing structure of the wall is made of limestone ρ = 2000kg / m3, 0.390 m thick, insulated with foam plastic 0.050 m thick, which ensures the normal temperature and humidity conditions of the premises and meets the sanitary and hygienic requirements for them.

Losses through house 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 one.

The penetration of cold air into the house is carried out through the supply ventilation. Exhaust ventilation helps warm air escape. Reduces losses through ventilation heat exchanger-recuperator. It does not allow heat to escape along with the outgoing air, and it heats the incoming flows

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

Qv \u003d (V x Kv: 3600) x P x C x dT

Here the symbols mean the following:

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

Based on the results of this calculation, it is possible to determine the power of the heat generator of the heating system. In case of too high power value, a ventilation device with a heat exchanger can become a way out of the situation. Consider a few examples for houses made of different materials.

Regulatory documents required for calculation:

• SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings". Updated edition of 2012.
• SNiP 23-01-99* (SP 131.13330.2012). "Construction climatology". Updated edition of 2012.
• SP 23-101-2004."Design of thermal protection of buildings".
• GOST 30494-2011 Residential and public buildings. Indoor microclimate parameters.

Initial data for calculation:

1. We determine the climatic zone in which we are going to build a house. We open SNiP 23-01-99 *. "Construction climatology", we find table 1. In this table we find our city (or the city located as close as possible to the construction site), for example, for construction in a village located near the city of Murom, we will take indicators of the city of Murom! from column 5 - "Air temperature of the coldest five-day period, with a security of 0.92" - "-30 ° C";
2. We determine the duration of the heating period - open table 1 in SNiP 23-01-99 * and in column 11 (with an average daily outdoor temperature of 8 ° C) the duration is zht = 214 days;
3. We determine the average outdoor temperature for the heating period, for this, from the same table 1 SNIP 23-01-99 *, select the value in column 12 - tht \u003d -4.0 ° С.
4. The optimum indoor temperature is taken according to table 1 in GOST 30494-96 - tint = 20 ° C;

Then, we need to decide on the design of the wall itself. Since earlier houses were built from one material (brick, stone, etc.), the walls were very thick and massive. But, with the development of technology, people have new materials with very good thermal conductivity, which made it possible to significantly reduce the thickness of the walls from the main (bearing material) by adding a heat-insulating layer, thus multilayer walls appeared.

There are at least three main layers in a multilayer wall:

• 1 layer - load-bearing wall - its purpose is to transfer the load from the overlying structures to the foundation;
• 2 layer - thermal insulation - its purpose is to retain heat inside the house as much as possible;
• 3rd layer - decorative and protective - its purpose is to make the facade of the house beautiful and at the same time protect the insulation layer from the effects of the external environment (rain, snow, wind, etc.);

Consider for our example the following wall composition:

• 1st layer - we accept the load-bearing wall of aerated concrete blocks 400 mm thick (we accept constructively - taking into account the fact that floor beams will rest on it);
• 2nd layer - we carry out from a mineral wool plate, we will determine its thickness by thermotechnical calculation!
• 3rd layer - we accept facing silicate brick, layer thickness 120 mm;
• 4th layer - since from the inside our wall will be covered with a layer of cement-sand mortar plaster, we will also include it in the calculation and set its thickness to 20mm;

Calculation of thermal power based on the volume of the room

This method of determining the heat load on heating systems is less universal than the first one, since it is intended for calculating rooms with high ceilings, but it does not take into account that the air under the ceiling is always warmer than in the lower part of the room and, therefore, the amount of heat loss will be vary regionally.

The heat output of the heating system for a building or room with ceilings above the standard is calculated based on the following condition:

Q=V*41W (34W),

where V is the external volume of the room in m?,

And 41 W is the specific amount of heat required to heat one cubic meter of a standard building (in a panel house). If construction is carried out using modern building materials, then the specific heat loss indicator is usually included in the calculations with a value of 34 watts.

When using the first or second method of calculating the heat loss of a building by an enlarged method, you can use correction factors that to some extent reflect the reality and dependence of heat loss by a building depending on various factors.

1. Glazing type:

• triple package 0.85,
• double 1.0,
• double binding 1.27.

1. The presence of windows and entrance doors increases the amount of heat loss at home by 100 and 200 watts, respectively.
2. Thermal insulation characteristics of external walls and their air permeability:

• modern thermal insulation materials 0.85
• standard (two bricks and insulation) 1.0,
• low thermal insulation properties or insignificant wall thickness 1.27-1.35.

1. The percentage of window area to the area of ​​the room: 10% -0.8, 20% -0.9, 30% -1.0, 40% -1.1, 50% -1.2.
2. The calculation for an individual residential building should be made with a correction factor of about 1.5, depending on the type and characteristics of the floor and roof structures used.
3. Estimated outdoor temperature in winter (each region has its own, determined by the standards): -10 degrees 0.7, -15 degrees 0.9, -20 degrees 1.10, -25 degrees 1.30, -35 degrees 1, 5.
4. Heat losses also grow depending on the increase in the number of external walls according to the following relationship: one wall - plus 10% of the heat output.

But, nevertheless, it is possible to determine which method will give an accurate and really true result of the thermal power of heating equipment only after an accurate and complete thermal calculation of the building has been performed.

Types of thermal loads

The calculations take into account the average seasonal temperatures

Thermal loads are of different nature.There is a certain constant level of heat loss associated with the thickness of the wall, the roof structure. There are temporary ones - with a sharp decrease in temperature, with intensive ventilation. The calculation of the entire heat load takes this into account as well.

Seasonal loads

So called heat loss associated with the weather. These include:

• the difference between the temperature of the outdoor air and indoors;
• wind speed and direction;
• the amount of solar radiation - with high insolation of the building and a large number of sunny days, even in winter the house cools less;
• air humidity.

The seasonal load is distinguished by a variable annual schedule and a constant daily schedule. Seasonal heat load is heating, ventilation and air conditioning. The first two species are referred to as winter.

Permanent thermal

Industrial refrigeration equipment generates large amounts of heat

Year-round hot water supply and technological devices are included. The latter is important for industrial enterprises: digesters, industrial refrigerators, steaming chambers emit a huge amount of heat.

In residential buildings, the load on hot water supply becomes comparable to the heating load. This value changes little during the year, but varies greatly depending on the time of day and day of the week. In summer, the consumption of DHW is reduced by 30%, since the temperature of the water in the cold water supply is 12 degrees higher than in winter. During the cold season, hot water consumption increases, especially on weekends.

dry heat

Comfort mode is determined by air temperature and humidity.These parameters are calculated using the concepts of dry and latent heat. Dry is a value measured with a special dry thermometer. It is affected by:

• glazing and doorways;
• sun and heat loads for winter heating;
• partitions between rooms with different temperatures, floors above empty space, ceilings under attics;
• cracks, crevices, gaps in walls and doors;
• air ducts outside heated areas and ventilation;
• equipment;
• people.

Floors on a concrete foundation, underground walls are not taken into account in the calculations.

Latent heat

Humidity in the room raises the temperature inside

This parameter determines the humidity of the air. The source is:

• equipment - heats the air, reduces humidity;
• people are a source of moisture;
• air currents passing through cracks and crevices in the walls.

Room temperature standards

Before carrying out any calculations of system parameters, it is necessary, at a minimum, to know the order of expected results, and also to have standardized characteristics of some tabular values ​​that must be substituted into formulas or be guided by them.

By performing parameter calculations with such constants, one can be confident in the reliability of the desired dynamic or constant parameter of the system.

For premises of various purposes, there are reference standards for the temperature regimes of residential and non-residential premises. These norms are enshrined in the so-called GOSTs.

For a heating system, one of these global parameters is the room temperature, which must be constant regardless of the period of the year and environmental conditions.

According to the regulation of sanitary standards and rules, there are differences in temperature relative to the summer and winter periods of the year. The air conditioning system is responsible for the temperature regime of the room in the summer season, the principle of its calculation is described in detail in this article.

But the room temperature in winter is provided by the heating system. Therefore, we are interested in temperature ranges and their deviation tolerances for the winter season.

Most regulatory documents stipulate the following temperature ranges that allow a person to be comfortable in a room.

For non-residential premises of office type up to 100 m2:

• 22-24°C - optimal air temperature;
• 1°C - allowable fluctuation.

For office-type premises with an area of ​​more than 100 m2, the temperature is 21-23°C. For non-residential premises of an industrial type, the temperature ranges vary greatly depending on the purpose of the premises and the established labor protection standards.

Comfortable room temperature for each person is “own”. Someone likes to be very warm in the room, someone is comfortable when the room is cool - it's all quite individual

As for residential premises: apartments, private houses, estates, etc., there are certain temperature ranges that can be adjusted depending on the wishes of the residents.

And yet, for specific premises of an apartment and a house, we have:

• 20-22°С - residential, including children's, room, tolerance ± 2°С -
• 19-21°C - kitchen, toilet, tolerance ± 2°C;
• 24-26°С - bathroom, shower room, swimming pool, tolerance ±1°С;
• 16-18°С - corridors, hallways, stairwells, storerooms, tolerance +3°С

It is important to note that there are a few more basic parameters that affect the temperature in the room and that you need to focus on when calculating the heating system: humidity (40-60%), the concentration of oxygen and carbon dioxide in the air (250: 1), the speed of movement of air masses (0.13-0.25 m/s), etc.

Calculation of the normalized and specific heat-shielding characteristics of the building

Before proceeding to the calculations, we highlight a few excerpts from the regulatory literature.

Clause 5.1 of SP 50.13330.2012 states that the heat-shielding shell of the building must meet the following requirements:

1. Reduced resistance to heat transfer of individual enclosing
   structures should not be less than the normalized values ​​(element-by-element
   requirements).
2. The specific heat-shielding characteristic of the building should not exceed
   normalized value (complex requirement).
3. The temperature on the internal surfaces of the enclosing structures should
   be not lower than the minimum allowable values ​​(sanitary and hygienic
   requirement).
4. The requirements for the thermal protection of the building will be met while
   fulfillment of conditions 1,2 and 3.

Clause 5.5 of SP 50.13330.2012. The normalized value of the specific heat-shielding characteristic of the building, k(tr ⁄ vol), W ⁄ (m³ × °С), should be taken depending on the heated volume of the building and degree-days of the heating period of the construction area according to Table 7, taking into account
notes.

Table 7. Normalized values ​​of the specific heat-shielding characteristics of the building:

Heated volume buildings, Vot, m³	Values ​​k(tr ⁄ vol), W ⁄ (m² × °C), at GSOP values, °C × day ⁄ year
1000	3000	5000	8000	12000
150	1,206	0,892	0,708	0,541	0,321
300	0,957	0,708	0,562	0,429	0,326
600	0,759	0,562	0,446	0,341	0,259
1200	0,606	0,449	0,356	0,272	0,207
2500	0,486	0,360	0,286	0,218	0,166
6000	0,391	0,289	0,229	0,175	0,133
15 000	0,327	0,242	0,192	0,146	0,111
50 000	0,277	0,205	0,162	0,124	0,094
200 000	0,269	0,182	0,145	0,111	0,084

We launch the "Calculation of the specific heat-shielding characteristics of the building":

As you can see, part of the initial data is saved from the previous calculation.In fact, this calculation is a part of the previous calculation. The data can be changed.

Using the data from the previous calculation, for further work it is necessary:

1. Add a new building element (Add New button).
2. Or select a ready-made element from the directory (button "Select from directory"). Let's choose Construction No. 1 from the previous calculation.
3. Fill in the column "Heated volume of the element, m³" and "Area of ​​the fragment of the enclosing structure, m²".
4. Press the button "Calculation of the specific heat-shielding characteristic".

We get the result: