The first step in organizing the heating of a private house is the calculation of heat loss. The purpose of this calculation is to find out how much heat escapes outside through walls, floors, roofs and windows (common name - building envelope) during the most severe frosts in a given area. Knowing how to calculate heat loss according to the rules, you can get a fairly accurate result and start selecting a heat source by power.
Basic Formulas
To get a more or less accurate result, it is necessary to perform calculations according to all the rules, a simplified method (100 W of heat per 1 m² of area) will not work here. The total heat loss of a building during the cold season consists of 2 parts:
- heat loss through enclosing structures;
- energy loss for heating ventilation air.
The basic formula for calculating the consumption of thermal energy through external fences is as follows:
Q \u003d 1 / R x (t in - t n) x S x (1+ ∑β). Here:
- Q is the amount of heat lost by a structure of one type, W;
- R is the thermal resistance of the construction material, m²°C / W;
- S is the area of the outer fence, m²;
- t in - internal air temperature, ° С;
- t n - most low temperature environment, °С;
- β - additional heat loss, depending on the orientation of the building.
The thermal resistance of the walls or roof of a building is determined based on the properties of the material from which they are made and the thickness of the structure. For this, the formula R = δ / λ is used, where:
- λ is the reference value of the thermal conductivity of the wall material, W/(m°C);
- δ is the thickness of the layer of this material, m.
If the wall is built from 2 materials (for example, a brick with a mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summed up. The outdoor temperature is selected both according to regulatory documents and according to personal observations, internal - if necessary. Additional heat losses are the coefficients defined by the standards:
- When the wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
- If the structure is facing southeast or west, β = 0.05.
- β = 0 when the outer fence faces south or southwest.
Calculation Order
To take into account all the heat leaving the house, it is necessary to calculate the heat loss of the room, each separately. To do this, measurements are made of all fences adjacent to the environment: walls, windows, roofs, floors and doors.
Important point: measurements should be taken according to outside, capturing the corners of the building, otherwise the calculation of the heat loss of the house will give an underestimated heat consumption.
Windows and doors are measured by the opening they fill.
Based on the measurement results, the area of \u200b\u200beach structure is calculated and substituted into the first formula (S, m²). The value of R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity of the building material. In the case of new metal-plastic windows, the value of R will be prompted by a representative of the installer.
As an example, it is worthwhile to calculate the heat loss through the enclosing walls made of bricks 25 cm thick, with an area of 5 m² at an ambient temperature of -25 ° C. It is assumed that the temperature inside will be +20°C, and the plane of the structure is facing north (β = 0.1). First you need to take from the reference literature the coefficient of thermal conductivity of the brick (λ), it is equal to 0.44 W / (m ° C). Then, according to the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:
R \u003d 0.25 / 0.44 \u003d 0.57 m² ° C / W
To determine the heat loss of a room with this wall, all the initial data must be substituted into the first formula:
Q \u003d 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) \u003d 434 W \u003d 4.3 kW
If the room has a window, then after calculating its area, the heat loss through the translucent opening should be determined in the same way. The same actions are repeated for floors, roofs and front door. At the end, all the results are summarized, after which you can move on to the next room.
Heat metering for air heating
When calculating the heat loss of a building, it is important to take into account the amount of heat energy consumed by the heating system for heating the ventilation air. The share of this energy reaches 30% of the total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss at home through the heat capacity of the air using the popular formula from the physics course:
Q air \u003d cm (t in - t n). In it:
- Q air - heat consumed by the heating system for heating supply air, W;
- t in and t n - the same as in the first formula, ° С;
- m is the mass flow rate of air entering the house from the outside, kg;
- c is the heat capacity of the air mixture, equal to 0.28 W / (kg ° С).
Here, all quantities are known except mass flow air for ventilation. In order not to complicate your task, you should agree with the condition that air environment is updated throughout the house 1 time per hour. Then it is not difficult to calculate the volumetric air flow by adding the volumes of all rooms, and then you need to convert it into mass air through density. Since the density of the air mixture varies with its temperature, you need to take appropriate value from the table:
m = 500 x 1.422 = 711 kg/h
Heating such a mass of air by 45°C will require the following amount of heat:
Q air \u003d 0.28 x 711 x 45 \u003d 8957 W, which is approximately equal to 9 kW.
Upon completion of the calculations, the results of heat losses through the external enclosures are added to the ventilation heat losses, which gives the total heat load to the heating system of the building.
The presented calculation methods can be simplified if the formulas are entered into the Excel program in the form of tables with data, this will significantly speed up the calculation.
So that your house does not turn out to be a bottomless pit for heating costs, we suggest studying the basic directions of thermal engineering research and calculation methodology.
So that your house does not turn out to be a bottomless pit for heating costs, we suggest studying the basic directions of thermal engineering research and calculation methodology.
Without preliminary calculation thermal permeability and moisture accumulation, the whole essence of housing construction is lost.
Physics of thermal processes
Different areas of physics have much in common in describing the phenomena they study. So it is in heat engineering: the principles describing thermodynamic systems clearly echo the foundations of electromagnetism, hydrodynamics and classical mechanics. After all, we are talking about the description of the same world, so it is not surprising that models of physical processes are characterized by some common features in many areas of research.
The essence of thermal phenomena is easy to understand. The temperature of a body or the degree of its heating is nothing but a measure of the intensity of oscillations of the elementary particles of which this body is composed. Obviously, when two particles collide, the one with the higher energy level will transfer energy to the particle with lower energy, but never vice versa.
However, this is not the only way of energy exchange; transfer is also possible through thermal radiation quanta. Wherein basic principle necessarily conserved: a quantum emitted by a less heated atom is not able to transfer energy to a hotter elementary particle. It is simply reflected from it and either disappears without a trace, or transfers its energy to another atom with less energy.
Thermodynamics is good because the processes occurring in it are absolutely visual and can be interpreted under the guise of various models. The main thing is to observe the basic postulates, such as the law of energy transfer and thermodynamic equilibrium. So if your presentation complies with these rules, you will easily understand the method of thermal engineering calculations from and to.
The concept of resistance to heat transfer
The ability of a material to transfer heat is called thermal conductivity. In the general case, it is always higher, the greater the density of the substance and the better its structure is adapted to transmit kinetic vibrations.
The quantity inversely proportional to thermal conductivity is thermal resistance. For each material, this property takes on unique values depending on the structure, shape, and a number of other factors. For example, the efficiency of heat transfer in the thickness of materials and in the zone of their contact with other media may differ, especially if there is at least a minimal layer of matter between the materials in a different state of aggregation. Quantitatively, thermal resistance is expressed as the temperature difference divided by the heat flow rate:
Rt = (T2 - T1) / P
where:
- Rt - thermal resistance of the section, K / W;
- T2 - temperature of the beginning of the section, K;
- T1 - temperature of the end of the section, K;
- P - heat flux, W.
In the context of calculating heat losses, thermal resistance plays a decisive role. Any enclosing structure can be represented as a plane-parallel barrier to the heat flow. Its total thermal resistance is the sum of the resistances of each layer, while all partitions are added to spatial construction, which is actually a building.
Rt = l / (λ S)
where:
- Rt - thermal resistance of the circuit section, K / W;
- l - length of the section of the thermal circuit, m;
- λ - coefficient of thermal conductivity of the material, W/(m K);
- S - cross-sectional area of the site, m2.
Factors affecting heat loss
Thermal processes correlate well with electrical ones: the temperature difference acts as a voltage, the heat flux can be considered as a current strength, but you don’t even need to come up with your own term for resistance. The concept of least resistance, which appears in heat engineering as cold bridges, is also fully true.
If we consider an arbitrary material in a section, it is quite easy to establish the path of the heat flow both at the micro- and at the macro level. Let us take as the first model concrete wall, in which, due to technological necessity, through fastenings are made with steel rods of arbitrary section. Steel conducts heat somewhat better than concrete, so we can distinguish three main heat fluxes:
- through the concrete
- through steel bars
- from steel bars to concrete
The last heat flow model is the most interesting. Since the steel rod warms up faster, the temperature difference between the two materials will be observed closer to the outer part of the wall. Thus, steel not only "pumps" heat outward by itself, it also increases the thermal conductivity of adjacent masses of concrete.
In porous media, thermal processes proceed in a similar way. Almost all Construction Materials consist of a branched web of solid matter, the space between which is filled with air.
Thus, a solid, dense material serves as the main conductor of heat, but due to the complex structure, the path along which heat spreads turns out to be larger than the cross section. Thus, the second factor that determines thermal resistance is the heterogeneity of each layer and the building envelope as a whole.
The third factor affecting the thermal conductivity, we can name the accumulation of moisture in the pores. Water has a thermal resistance 20–25 times lower than that of air, so if it fills the pores, the overall thermal conductivity of the material becomes even higher than if there were no pores at all. When water freezes, the situation becomes even worse: thermal conductivity can increase up to 80 times. The source of moisture is usually room air and precipitation. Accordingly, the three main methods of dealing with such a phenomenon are outdoor waterproofing walls, the use of vapor protection and the calculation of moisture accumulation, which is necessarily carried out in parallel with the forecasting of heat losses.
Differentiated calculation schemes
The simplest way to determine the size of the heat loss of a building is to sum up the values of the heat flow through the structures that form this building. This technique fully takes into account the difference in the structure various materials, as well as the specifics of the heat flow through them and at the junctions of one plane to another. Such a dichotomous approach greatly simplifies the task, because different enclosing structures can differ significantly in the design of thermal protection systems. Accordingly, in a separate study, it is easier to determine the amount of heat loss, because for this various ways calculations:
- For walls, heat leakage is quantitatively equal to total area multiplied by the ratio of temperature difference to thermal resistance. At the same time, the orientation of the walls to the cardinal points is necessarily taken into account to take into account their heating in the daytime, as well as the ventilation building structures.
- For overlaps, the technique is the same, but the presence of attic space and mode of operation. Also, a value 3–5 °С higher is taken as room temperature, the calculated humidity is also increased by 5–10%.
- Heat losses through the floor are calculated zonal, describing the belts along the perimeter of the building. This is due to the fact that the temperature of the soil under the floor is higher near the center of the building compared to the foundation part.
- The heat flux through the glazing is determined by the nameplate data of the windows; the type of adjoining windows to the walls and the depth of the slopes must also be taken into account.
Q = S (∆T / Rt)
where:
- Q- heat loss, W;
- S - wall area, m2;
- ΔT - temperature difference inside and outside the room, ° С;
- Rt - resistance to heat transfer, m2 °C / W.
Calculation example
Before proceeding to the demo, let's answer the last question: how to correctly calculate the integral thermal resistance of complex multilayer structures? This, of course, can be done manually, fortunately, in modern construction not many types are used load-bearing bases and heating systems. However, take into account the presence decorative finishes, interior and facade plaster, as well as the influence of all transients and other factors is quite difficult, it is better to use automated calculations. One of the best online resources for such tasks is smartcalc.ru, which additionally plots the dew point shift depending on climatic conditions.
For example, let's take an arbitrary building, having studied the description of which the reader will be able to judge the set of initial data necessary for the calculation. There is a one-story house of the correct rectangular shape with dimensions of 8.5x10 m and a ceiling height of 3.1 m, located in the Leningrad region.
The house has an uninsulated floor on the ground with boards on logs with air gap, the floor height is 0.15 m higher than the ground planning mark on the site. The wall material is a slag monolith 42 cm thick with internal cement-lime plaster up to 30 mm thick and external slag-cement plaster of the “fur coat” type up to 50 mm thick. The total glazing area is 9.5 m2, used as windows double glazing in a heat-saving profile with an average thermal resistance of 0.32 m2 °C / W.
The cover was made on wooden beams: plastered on shingles from below, filled with blast-furnace slag and covered from above clay screed, above the ceiling - a cold-type attic. The task of calculating heat loss is the formation of a thermal protection system for walls.
Floor
First of all, heat losses through the floor are determined. Since their share in the total heat outflow is the smallest, and also due to a large number variables (density and type of soil, freezing depth, massiveness of the foundation, etc.), the calculation of heat loss is carried out according to a simplified method using the reduced resistance to heat transfer. Along the perimeter of the building, starting from the line of contact with the ground, four zones are described - encircling strips 2 meters wide.
For each of the zones, its own value of the reduced resistance to heat transfer is taken. In our case, there are three zones with an area of 74, 26 and 1 m2. Let it not bother you total amount areas of zones, which more area building on 16 m2, the reason for this is the double recalculation of the intersecting strips of the first zone in the corners, where heat losses are much higher compared to sections along the walls. Using heat transfer resistance values of 2.1, 4.3, and 8.6 m2 °C/W for zones one through three, we determine the heat flow through each zone: 1.23, 0.21, and 0.05 kW, respectively. .
Walls
Using the terrain data, as well as the materials and thickness of the layers that form the walls, you need to fill in the appropriate fields on the smartcalc.ru service mentioned above. According to the calculation results, the heat transfer resistance is equal to 1.13 m2 °C / W, and the heat flux through the wall is 18.48 W per square meter. With a total wall area (excluding glazing) of 105.2 m2, the total heat loss through the walls is 1.95 kWh. In this case, heat loss through the windows will be 1.05 kW.
Covering and roofing
Calculation of heat loss through attic floor can also be done in the online calculator by selecting desired type enclosing structures. As a result, the resistance of the floor to heat transfer is 0.66 m2 °C/W, and the heat loss is 31.6 W s square meter, that is, 2.7 kW from the entire area of \u200b\u200bthe building envelope.
The total total heat loss according to the calculations is 7.2 kWh. Given the rather low quality of the building structures, this figure is obviously much lower than the real one. In fact, such a calculation is idealized, it does not take into account special coefficients, blowing, the convection component of heat transfer, losses through ventilation and entrance doors.
In fact, due to poor-quality installation of windows, lack of protection at the junction of the roof to the Mauerlat and poor waterproofing of the walls from the foundation, real heat losses can be 2 or even 3 times more than the calculated ones. However, even basic thermal engineering studies help determine whether the structures of the house under construction will comply with sanitary standards at least as a first approximation.
Finally, let's give one important recommendation: If you really want to get a complete understanding of the thermal physics of a particular building, you must use an understanding of the principles described in this overview and specialized literature. For example, a reference manual by Elena Malyavina “Heat loss of a building” can be a very good help in this matter, where the specifics of heat engineering processes are explained in great detail, links are given to the necessary regulations, as well as examples of calculations and all the necessary background information.published
If you have any questions on this topic, ask them to specialists and readers of our project.
Every building, regardless design features, misses thermal energy through the fences. Heat loss in environment must be restored with a heating system. The sum of heat losses with a normalized margin is the required power of the heat source that heats the house. To create in dwelling comfortable conditions, calculation of heat losses is carried out taking into account various factors: building arrangements and layout of premises, orientation to the cardinal points, wind direction and average mildness of the climate during the cold period, physical qualities of building and heat-insulating materials.
Based on the results of the heat engineering calculation, a heating boiler is selected, the number of battery sections is specified, the power and length of the underfloor heating pipes are considered, a heat generator is selected for the room - in general, any unit that compensates for heat loss. By and large, it is necessary to determine heat losses in order to heat the house economically - without an extra supply of power from the heating system. Calculations are performed manually or choose an appropriate computer program into which the data is substituted.
How to make a calculation?
First, you should deal with the manual technique - to understand the essence of the process. To find out how much heat a house loses, determine the losses through each building envelope separately, and then add them up. The calculation is carried out in stages.
1. Form a base of initial data for each room, preferably in the form of a table. In the first column, the pre-calculated area of door and window blocks, external walls, ceilings, and floors is recorded. The thickness of the structure is entered in the second column (these are design data or measurement results). In the third - the coefficients of thermal conductivity of the corresponding materials. Table 1 contains standard values, which will be needed in the further calculation:
The higher λ, the more heat escapes through the meter thickness of the given surface.
2. The heat resistance of each layer is determined: R = v/ λ, where v is the thickness of the building or heat-insulating material.
3. Calculate the heat loss of each structural element according to the formula: Q \u003d S * (T in -T n) / R, where:
- T n - outdoor temperature, ° C;
- T in - indoor temperature, ° C;
- S is the area, m2.
Of course, throughout heating period the weather is different (for example, the temperature ranges from 0 to -25°C), and the house is heated to the desired level of comfort (for example, up to +20°C). Then the difference (T in -T n) varies from 25 to 45.
To make a calculation, you need the average temperature difference for the entire heating season. To do this, in SNiP 23-01-99 "Construction climatology and geophysics" (table 1) find the average temperature of the heating period for a particular city. For example, for Moscow this figure is -26°. In this case, the average difference is 46°C. To determine the heat consumption through each structure, the heat losses of all its layers are added. So, for walls, plaster, masonry material, external thermal insulation, and cladding are taken into account.
4. Calculate the total heat loss, defining them as the sum of Q external walls, floors, doors, windows, ceilings.
5. Ventilation. From 10 to 40% of infiltration (ventilation) losses are added to the result of addition. If high-quality double-glazed windows are installed in the house, and ventilation is not abused, the infiltration coefficient can be taken as 0.1. AT separate sources it is indicated that the building does not lose heat at all, since leakages are compensated for by solar radiation and household heat.
Counting by hand
Initial data. Cottage with an area of 8x10 m, a height of 2.5 m. The walls are 38 cm thick and are made of ceramic brick, from the inside finished with a layer of plaster (thickness 20 mm). The floor is made of 30mm edged board, insulated with mineral wool (50 mm), sheathed chipboard sheets(8 mm). The building has a cellar, where the temperature in winter is 8°C. The ceiling is covered with wooden panels, insulated with mineral wool (thickness 150 mm). The house has 4 windows 1.2x1 m, an entrance oak door 0.9x2x0.05 m.
Task: determine the total heat loss of the house based on the fact that it is located in the Moscow region. The average temperature difference in the heating season is 46°C (as mentioned earlier). The room and basement have a difference in temperature: 20 – 8 = 12°C.
1. Heat loss through external walls.
Total area (excluding windows and doors): S \u003d (8 + 10) * 2 * 2.5 - 4 * 1.2 * 1 - 0.9 * 2 \u003d 83.4 m2.
Thermal resistance is determined brickwork and plaster layer:
- R clade. = 0.38/0.52 = 0.73 m2*°C/W.
- R pieces. = 0.02/0.35 = 0.06 m2*°C/W.
- R total = 0.73 + 0.06 = 0.79 m2*°C/W.
- Heat loss through walls: Q st \u003d 83.4 * 46 / 0.79 \u003d 4856.20 W.
2. Heat loss through the floor.
Total area: S = 8*10 = 80 m2.
The heat resistance of a three-layer floor is calculated.
- R boards = 0.03 / 0.14 = 0.21 m2 * ° C / W.
- R chipboard = 0.008/0.15 = 0.05 m2*°C/W.
- R insulation = 0.05/0.041 = 1.22 m2*°C/W.
- R total = 0.03 + 0.05 + 1.22 = 1.3 m2*°C/W.
We substitute the values \u200b\u200bof the quantities into the formula for finding heat losses: Q floor \u003d 80 * 12 / 1.3 \u003d 738.46 W.
3. Heat loss through the ceiling.
Square ceiling surface equal to the floor area S = 80 m2.
When determining the thermal resistance of the ceiling, in this case they do not take into account wooden shields: they are fixed with gaps and are not a barrier to cold. The thermal resistance of the ceiling coincides with the corresponding parameter of the insulation: R pot. = R ins. = 0.15/0.041 = 3.766 m2*°C/W.
The amount of heat loss through the ceiling: Q sweat. \u003d 80 * 46 / 3.66 \u003d 1005.46 W.
4. Heat loss through windows.
Glazing area: S = 4*1.2*1 = 4.8 m2.
For the manufacture of windows used three-chamber PVC profile(occupies 10% of the window area), as well as a double-glazed window with a glass thickness of 4 mm and a distance between glasses of 16 mm. Among specifications the manufacturer indicated the thermal resistance of the double-glazed window (R st.p. = 0.4 m2*°C/W) and the profile (R prof. = 0.6 m2*°C/W). Taking into account the dimensional fraction of each structural element, the average heat resistance of the window is determined:
- R ok. \u003d (R st.p. * 90 + R prof. * 10) / 100 \u003d (0.4 * 90 + 0.6 * 10) / 100 \u003d 0.42 m2 * ° C / W.
- Based on the calculated result, the heat losses through the windows are calculated: Q approx. \u003d 4.8 * 46 / 0.42 \u003d 525.71 W.
Door area S = 0.9 * 2 = 1.8 m2. Thermal resistance R dv. \u003d 0.05 / 0.14 \u003d 0.36 m2 * ° C / W, and Q ext. \u003d 1.8 * 46 / 0.36 \u003d 230 W.
The total amount of heat loss at home is: Q = 4856.20 W + 738.46 W + 1005.46 W + 525.71 W + 230 W = 7355.83 W. Taking into account infiltration (10%), the losses increase: 7355.83 * 1.1 = 8091.41 W.
To accurately calculate how much heat a building loses, use online calculator heat loss. it computer program, in which not only the data listed above are entered, but also various additional factors that affect the result. The advantage of the calculator is not only the accuracy of calculations, but also an extensive database of reference data.
The exact calculation of heat loss at home is a painstaking and slow task. For its production, initial data are required, including the dimensions of all building envelopes (walls, doors, windows, ceilings, floors).
For single-layer and / or multi-layer walls, as well as floors, the heat transfer coefficient is easy to calculate by dividing the thermal conductivity of the material by the thickness of its layer in meters. For a multilayer structure, the overall heat transfer coefficient will be equal to the reciprocal of the sum of the heat resistances of all layers. For windows, you can use the table of thermal characteristics of windows.
Walls and floors lying on the ground are calculated by zones, so in the table it is necessary to create separate lines for each of them and indicate the corresponding heat transfer coefficient. The division into zones and the values of the coefficients are indicated in the rules for measuring premises.
Column 11. Basic heat loss. Here, the main heat losses are automatically calculated based on the data entered in the previous cells of the line. Specifically, Temperature Difference, Area, Heat Transfer Coefficient and Position Coefficient are used. Formula in cell:
Column 12. Orientation addition. In this column, the additive for orientation is automatically calculated. Depending on the contents of the Orientation cell, the appropriate coefficient is inserted. The formula for calculating a cell looks like this:
IF(H9="E",0.1,IF(H9="SE",0.05,IF(H9="S",0,IF(H9="SW",0,IF(H9="W ";0.05;IF(H9="SW";0.1;IF(H9="S";0.1;IF(H9="SW";0.1;0))))))) )
This formula inserts a factor into a cell as follows:
- East - 0.1
- Southeast - 0.05
- South - 0
- Southwest - 0
- West - 0.05
- Northwest - 0.1
- North - 0.1
- Northeast - 0.1
Column 13. Other additive. Here you enter the addition factor when calculating the floor or doors in accordance with the conditions in the table:
Column 14. Heat loss. Here is the final calculation of the heat loss of the fence according to the line. Cell formula:
As calculations progress, cells can be created with formulas for summing heat losses by rooms and deriving the sum of heat losses from all the fences of the house.
There are also heat losses due to air infiltration. They can be neglected, since they are compensated to some extent by household heat emissions and heat gains from solar radiation. For a more complete, exhaustive calculation of heat loss, you can use the methodology described in the reference manual.
As a result, to calculate the power of the heating system, we increase the amount of heat loss of all the fences of the house by 15 - 30%.
Others, more simple ways heat loss calculation:
- quick calculation in the mind approximate method of calculation;
- somewhat more complex calculation using coefficients;
- the most accurate way to calculate heat loss in real time;
Below is a pretty simple heat loss calculation buildings, which, nevertheless, will help to accurately determine the power required for heating your warehouse, shopping center or other similar building. This will make it possible to preliminarily estimate the cost at the design stage. heating equipment and subsequent heating costs, and if necessary, adjust the project.
Where does the heat go? Heat escapes through walls, floors, roofs and windows. In addition, heat is lost during ventilation of the premises. To calculate heat loss through building envelope, use the formula:
Q - heat loss, W
S – construction area, m2
T - temperature difference between indoor and outdoor air, °C
R - value thermal resistance structure, m2 °C/W
The calculation scheme is as follows - we calculate the heat loss individual elements, summarize and add the heat loss during ventilation. All.
Suppose we want to calculate the heat loss for the object shown in the figure. The height of the building is 5 ... 6 m, width - 20 m, length - 40 m, and thirty windows measuring 1.5 x 1.4 meters. Indoor temperature 20 °C, outside temperature -20 °C.
We consider the area of enclosing structures:
floor: 20 m * 40 m = 800 m2
roof: 20.2 m * 40 m = 808 m2
window: 1.5 m * 1.4 m * 30 pcs = 63 m2
walls:(20 m + 40 m + 20 m + 40 m) * 5 m = 600 m2 + 20 m2 (accounting pitched roof) = 620 m2 - 63 m2 (windows) = 557 m2
Now let's see the thermal resistance of the materials used.
The value of thermal resistance can be taken from the table of thermal resistances or calculated based on the value of the thermal conductivity coefficient using the formula:
R - thermal resistance, (m2 * K) / W
? - coefficient of thermal conductivity of the material, W / (m2 * K)
d – material thickness, m
The value of thermal conductivity coefficients for different materials can be seen.
floor: concrete screed 10 cm and mineral wool with a density of 150 kg/m3. 10 cm thick.
R (concrete) = 0.1 / 1.75 = 0.057 (m2*K)/W
R (mineral wool) \u003d 0.1 / 0.037 \u003d 2.7 (m2 * K) / W
R (floor) \u003d R (concrete) + R (mineral wool) \u003d 0.057 + 2.7 \u003d 2.76 (m2 * K) / W
roof:
R (roof) = 0.15 / 0.037 = 4.05 (m2*K)/W
window: the value of thermal resistance of windows depends on the type of double-glazed window used
R (windows) \u003d 0.40 (m2 * K) / W for single-chamber glass wool 4–16–4 at? T \u003d 40 ° С
walls: panels from mineral wool 15 cm thick
R (walls) = 0.15 / 0.037 = 4.05 (m2*K)/W
Let's calculate the heat loss:
Q (floor) \u003d 800 m2 * 20 ° C / 2.76 (m2 * K) / W \u003d 5797 W \u003d 5.8 kW
Q (roof) \u003d 808 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 7980 W \u003d 8.0 kW
Q (windows) \u003d 63 m2 * 40 ° C / 0.40 (m2 * K) / W \u003d 6300 W \u003d 6.3 kW
Q (walls) \u003d 557 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 5500 W \u003d 5.5 kW
We get that the total heat loss through the building envelope will be:
Q (total) = 5.8 + 8.0 + 6.3 + 5.5 = 25.6 kWh
Now about ventilation losses.
To heat 1 m3 of air from a temperature of -20 °C to +20 °C, 15.5 W will be required.
Q (1 m3 of air) \u003d 1.4 * 1.0 * 40 / 3.6 \u003d 15.5 W, here 1.4 is the air density (kg / m3), 1.0 - specific heat air (kJ / (kg K)), 3.6 - conversion factor to watts.
It remains to determine the number required air. It is believed that with normal breathing, a person needs 7 m3 of air per hour. If you use a building as a warehouse and 40 people work on it, then you need to heat 7 m3 * 40 people = 280 m3 of air per hour, this will require 280 m3 * 15.5 W = 4340 W = 4.3 kW. And if you have a supermarket and on average there are 400 people on the territory, then air heating will require 43 kW.
Final result:
For heating the proposed building, a heating system of the order of 30 kWh is required, and a ventilation system with a capacity of 3000 m3 / h with a heater with a power of 45 kW / h.
