How is the calculation of the heat load for heating. Thermal calculation of the heating system: formulas, reference data and a specific example

Hello dear readers! Today a small post about the calculation of the amount of heat for heating according to aggregated indicators. In general, the heating load is taken according to the project, that is, the data that the designer calculated are entered into the heat supply contract.

But often there is simply no such data, especially if the building is small, such as a garage, or some utility room. In this case, the heating load in Gcal / h is calculated according to the so-called aggregated indicators. I wrote about this. And already this figure is included in the contract as the estimated heating load. How is this number calculated? And it is calculated according to the formula:

Qot \u003d α * qo * V * (tv-tn.r) * (1 + Kn.r) * 0.000001; where

α is a correction factor that takes into account climatic conditions district, it is used in cases where the calculated outdoor air temperature differs from -30 ° C;

qo is the specific heating characteristic of the building at tn.r = -30 °С, kcal/m3*С;

V - the volume of the building according to the external measurement, m³;

tv is the design temperature inside the heated building, °С;

tn.r - design outdoor air temperature for heating design, °C;

Kn.r is the infiltration coefficient, which is due to thermal and wind pressure, that is, the ratio of heat losses from the building with infiltration and heat transfer through external fences at the outdoor air temperature, which is calculated for heating design.

So, in one formula, you can calculate the heat load on the heating of any building. Of course, this calculation is largely approximate, but it is recommended in the technical literature on heat supply. Heat supply organizations also enter this figure of the heating load Qfrom, in Gcal / h, into heat supply contracts. So the calculation is correct. This calculation is well presented in the book - V.I. Manyuk, Ya.I. Kaplinsky, E.B. Khizh and others. This book is one of my desktop books, a very good book.

Also, this calculation of the heat load on the heating of the building can be done according to the "Methodology for determining the amount of thermal energy and coolant in public water supply systems" of RAO Roskommunenergo of the Gosstroy of Russia. True, there is an inaccuracy in the calculation in this method (in formula 2 in Appendix No. 1, 10 to the minus third power is indicated, but it should be 10 to the minus sixth power, this must be taken into account in the calculations), you can read more about this in the comments to this article.

I fully automated this calculation, added reference tables, including a table of climatic parameters for all regions former USSR(from SNiP 23.01.99 "Construction climatology"). You can buy a calculation in the form of a program for 100 rubles by writing to me at e-mail [email protected]

I will be glad to comments on the article.

The thermal calculation of the heating system seems to most to be easy and does not require special attention occupation. Great amount people believe that the same radiators should be chosen based only on the area of \u200b\u200bthe room: 100 W per 1 sq. m. Everything is simple. But this is the biggest misconception. You cannot limit yourself to such a formula. What matters is the thickness of the walls, their height, material and much more. Of course, you need to set aside an hour or two to get the numbers you need, but everyone can do it.

Initial data for designing a heating system

To calculate the heat consumption for heating, you need, firstly, a house project.

The plan of the house allows you to get almost all the initial data that is needed to determine heat loss and load on heating system

Secondly, data on the location of the house in relation to the cardinal points and the construction area will be needed - the climatic conditions in each region are different, and what is suitable for Sochi cannot be applied to Anadyr.

Thirdly, we collect information about the composition and height of the outer walls and the materials from which the floor (from the room to the ground) and the ceiling (from the rooms and outward) are made.

After collecting all the data, you can get to work. Calculation of heat for heating can be performed using formulas in one to two hours. You can, of course, use special program from Valtec.

To calculate the heat loss of heated rooms, the load on the heating system and heat transfer from heating devices, it is enough to enter only the initial data into the program. A huge number of functions make it indispensable assistant both foreman and private developer

It greatly simplifies everything and allows you to get all the data on heat losses and hydraulic calculation of the heating system.

Formulas for calculations and reference data

The calculation of the heat load for heating involves the determination of heat losses (Tp) and boiler power (Mk). The latter is calculated by the formula:

Mk \u003d 1.2 * Tp, where:

Mk - thermal performance of the heating system, kW;
Tp - heat loss at home;
1.2 - safety factor (20%).

A 20% safety factor allows you to take into account the possible pressure drop in the gas pipeline during the cold season and unforeseen heat losses (for example, a broken window, poor-quality thermal insulation entrance doors or extreme cold). It allows you to insure against a number of troubles, and also makes it possible to widely regulate the temperature regime.

As can be seen from this formula, the power of the boiler directly depends on the heat loss. They are not evenly distributed throughout the house: the outer walls account for about 40% of the total value, the windows - 20%, the floor gives 10%, the roof 10%. The remaining 20% disappear through the doors, ventilation.

Poorly insulated walls and floors, a cold attic, ordinary glazing on windows - all this leads to large heat losses, and, consequently, to an increase in the load on the heating system. When building a house, it is important to pay attention to all the elements, because even ill-conceived ventilation in the house will release heat into the street.

The materials from which the house is built have the most direct impact on the amount of heat lost. Therefore, when calculating, you need to analyze what the walls, and the floor, and everything else consist of.

In the calculations, to take into account the influence of each of these factors, the appropriate coefficients are used:

K1 - type of windows;
K2 - wall insulation;
K3 - the ratio of floor area and windows;
K4 - minimum temperature on the street;
K5 - the number of external walls of the house;
K6 - number of storeys;
K7 - the height of the room.

For windows, the heat loss coefficient is:

ordinary glazing - 1.27;
double-glazed window - 1;
three-chamber double-glazed window - 0.85.

Naturally, the last option will keep the heat in the house much better than the previous two.

Properly executed wall insulation is the key not only to a long life of the house, but also to a comfortable temperature in the rooms. Depending on the material, the value of the coefficient also changes:

concrete panels, blocks - 1.25-1.5;
logs, timber - 1.25;
brick (1.5 bricks) - 1.5;
brick (2.5 bricks) - 1.1;
foam concrete with increased thermal insulation - 1.

How more area windows relative to the floor, the more heat loses home:

The temperature outside the window also makes its own adjustments. At low rates of heat loss increase:

Up to -10С - 0.7;
-10C - 0.8;
-15C - 0.90;
-20C - 1.00;
-25C - 1.10;
-30C - 1.20;
-35C - 1.30.

Heat loss also depends on how much external walls at home:

four walls - 1.33;%
three walls - 1.22;
two walls - 1.2;
one wall - 1.

It’s good if a garage, a bathhouse or something else is attached to it. But if it is blown from all sides by winds, then you will have to buy a more powerful boiler.

The number of floors or the type of room that is located above the room determine the coefficient K6 as follows: if the house has two or more floors above, then for calculations we take the value 0.82, but if it is an attic, then for warm - 0.91 and 1 for cold .

As for the height of the walls, the values \u200b\u200bwill be as follows:

4.5 m - 1.2;
4.0 m - 1.15;
3.5 m - 1.1;
3.0 m - 1.05;
2.5 m - 1.

In addition to the above coefficients, the area of \u200b\u200bthe room (Pl) and the specific value of heat loss (UDtp) are also taken into account.

The final formula for calculating the heat loss coefficient:

Tp \u003d UDtp * Pl * K1 * K2 * K3 * K4 * K5 * K6 * K7.

The UDtp coefficient is 100 W/m2.

Analysis of calculations on a specific example

The house for which we will determine the load on the heating system has double-glazed windows (K1 \u003d 1), foam concrete walls with increased thermal insulation (K2 \u003d 1), three of which go outside (K5 \u003d 1.22). The area of windows is 23% of the floor area (K3=1.1), on the street about 15C frost (K4=0.9). The attic of the house is cold (K6=1), the height of the premises is 3 meters (K7=1.05). The total area is 135m2.

Fri \u003d 135 * 100 * 1 * 1 * 1.1 * 0.9 * 1.22 * 1 * 1.05 \u003d 17120.565 (Watts) or Fri \u003d 17.1206 kW

Mk \u003d 1.2 * 17.1206 \u003d 20.54472 (kW).

Calculation of load and heat loss can be done independently and quickly enough. You just need to spend a couple of hours putting the source data in order, and then just substitute the values into the formulas. The numbers that you will receive as a result will help you decide on the choice of a boiler and radiators.

To find out how much power the heat-power equipment of a private house should have, it is necessary to determine the total load on the heating system, for which a thermal calculation is performed. In this article, we will not talk about an enlarged method for calculating the area or volume of a building, but we will present a more accurate method used by designers, only in a simplified form for better perception. So, 3 types of loads fall on the heating system of the house:

compensation for the loss of thermal energy leaving through building construction(walls, floors, roofing);
heating the air required for ventilation of the premises;
heating water for DHW needs (when a boiler is involved in this, and not a separate heater).

Determination of heat loss through external fences

First, let's present the formula from SNiP, which calculates the heat energy lost through building structures that separate the interior of the house from the street:

Q \u003d 1 / R x (tv - tn) x S, where:

Q is the consumption of heat leaving through the structure, W;
R - resistance to heat transfer through the material of the fence, m2ºС / W;
S is the area of this structure, m2;
tv - the temperature that should be inside the house, ºС;
tn is the average outdoor temperature for the 5 coldest days, ºС.

For reference. According to the methodology, heat loss calculation is performed separately for each room. In order to simplify the task, it is proposed to take the building as a whole, assuming an acceptable average temperature of 20-21 ºС.

The area for each type of external fencing is calculated separately, for which windows, doors, walls and floors with a roof are measured. This is done because they are made from different materials different thickness. So the calculation will have to be done separately for all types of structures, and then the results will be summed up. You probably know the coldest street temperature in your area of residence from practice. But the parameter R will have to be calculated separately according to the formula:

R = δ / λ, where:

λ is the coefficient of thermal conductivity of the fence material, W/(mºС);
δ is the thickness of the material in meters.

Note. The value of λ is a reference value, it is easy to find it in any reference literature, and for plastic windows this coefficient will be prompted by the manufacturers. Below is a table with the coefficients of thermal conductivity of some building materials, and for calculations it is necessary to take the operational values of λ.

As an example, let's calculate how much heat will be lost by 10 m2 brick wall 250 mm thick (2 bricks) with a temperature difference outside and inside the house of 45 ºС:

R = 0.25 m / 0.44 W / (m ºС) = 0.57 m2 ºС / W.

Q \u003d 1 / 0.57 m2 ºС / W x 45 ºС x 10 m2 \u003d 789 W or 0.79 kW.

If the wall consists of different materials (structural material plus insulation), then they must also be calculated separately according to the above formulas, and the results summarized. Windows and roofing are calculated in the same way, but the situation is different with floors. First of all, you need to draw a building plan and divide it into zones 2 m wide, as is done in the figure:

Now you should calculate the area of \u200b\u200beach zone and alternately substitute it into the main formula. Instead of parameter R, you need to take the standard values for zone I, II, III and IV, indicated in the table below. At the end of the calculations, the results are added up and we get the total heat loss through the floors.

Ventilation air heating consumption

Uninformed people often do not take into account that the supply air in the house also needs to be heated and this thermal load also applies to the heating system. Cold air still enters the house from the outside, whether we like it or not, and it takes energy to heat it. Moreover, a full-fledged supply and exhaust ventilation usually with a natural urge. Air exchange is created due to the presence of draft in the ventilation ducts and the boiler chimney.

The method for determining the heat load from ventilation proposed in the regulatory documentation is rather complicated. Pretty accurate results can be obtained if this load is calculated using the well-known formula through the heat capacity of the substance:

Qvent = cmΔt, here:

Qvent - the amount of heat required for heating supply air, W;
Δt - temperature difference in the street and inside the house, ºС;
m is the mass of the air mixture coming from outside, kg;
c is the heat capacity of air, assumed to be 0.28 W / (kg ºС).

The complexity of calculating this type of heat load lies in the correct determination of the mass of heated air. Find out how much it gets inside the house, when natural ventilation difficult. Therefore, it is worth referring to the standards, because buildings are built according to projects where the required air exchanges are laid down. And the regulations say that in most rooms the air environment should change 1 time per hour. Then we take the volumes of all rooms and add to them the air flow rates for each bathroom - 25 m3 / h and a kitchen gas stove– 100 m3/h.

To calculate the heat load on heating from ventilation, the resulting air volume must be converted into mass, having learned its density at different temperatures from the table:

Let us assume that the total amount of supply air is 350 m3/h, the outside temperature is minus 20 ºС, and the inside temperature is plus 20 ºС. Then its mass will be 350 m3 x 1.394 kg / m3 = 488 kg, and the heat load on the heating system will be Qvent = 0.28 W / (kg ºС) x 488 kg x 40 ºС = 5465.6 W or 5.5 kW.

Heat load from DHW heating

To determine this load, you can use the same simple formula, only now you need to calculate thermal energy used for water heating. Its heat capacity is known and amounts to 4.187 kJ/kg °С or 1.16 W/kg °С. Considering that a family of 4 people needs 100 liters of water for 1 day, heated to 55 ° C, for all needs, we substitute these numbers into the formula and get:

QDHW \u003d 1.16 W / kg ° С x 100 kg x (55 - 10) ° С \u003d 5220 W or 5.2 kW of heat per day.

Note. By default, it is assumed that 1 liter of water is equal to 1 kg, and the temperature of cold tap water is 10 °C.

The unit of equipment power is always referred to 1 hour, and the resulting 5.2 kW - to the day. But you cannot divide this figure by 24, because hot water we want to receive as soon as possible, and for this the boiler must have a power reserve. That is, this load must be added to the rest as is.

Conclusion

This calculation of home heating loads will give much more accurate results than traditional way on the area, although you have to work hard. The final result must be multiplied by the safety factor - 1.2, or even 1.4, and selected according to the calculated value boiler equipment. Another way to enlarge the calculation of thermal loads according to the standards is shown in the video:

How to optimize heating costs? This problem is solved only integrated approach, taking into account all parameters of the system, buildings and climatic features of the region. At the same time, the most important component is the heat load on heating: the calculation of hourly and annual indicators are included in the system for calculating the efficiency of the system.

Why do you need to know this parameter

What is the calculation of the heat load for heating? It determines the optimal amount of thermal energy for each room and building as a whole. Variables are power heating equipment– boiler, radiators and pipelines. The heat losses of the house are also taken into account.

Ideally thermal power The heating system must compensate for all heat losses and at the same time maintain a comfortable temperature level. Therefore, before calculating the annual heating load, you need to determine the main factors affecting it:

Characteristics of the structural elements of the house. External walls, windows, doors, ventilation system affect the level of heat loss;
House dimensions. It is logical to assume that the larger the room, the more intensively the heating system should work. Quite a few an important factor at the same time, not only the total volume of each room is, but also the area of \u200b\u200bthe outer walls and window structures;
climate in the region. With relatively small drops in outdoor temperature, a small amount of energy is needed to compensate for heat losses. Those. the maximum hourly heating load directly depends on the degree of temperature decrease in a certain period of time and average annual value for heating season.

Considering these factors, the optimal thermal mode of operation of the heating system is compiled. Summarizing all of the above, we can say that determining the heat load on heating is necessary to reduce energy consumption and comply with optimal level heating in the premises of the house.

To calculate the optimal heating load according to aggregated indicators, you need to know the exact volume of the building. It is important to remember that this technique was developed for large structures, so the calculation error will be large.

Choice of calculation method

Before calculating the heating load using aggregated indicators or with higher accuracy, it is necessary to find out the recommended temperature conditions for a residential building.

During the calculation of the heating characteristics, one must be guided by the norms of SanPiN 2.1.2.2645-10. Based on the data in the table, in each room of the house it is necessary to provide optimal temperature regime heating work.

The methods by which the calculation of the hourly heating load is carried out can have a different degree of accuracy. In some cases, it is recommended to use fairly complex calculations, as a result of which the error will be minimal. If the optimization of energy costs is not a priority when designing heating, less accurate schemes can be used.

When calculating the hourly heating load, it is necessary to take into account the daily change in street temperature. To improve the accuracy of the calculation, you need to know specifications building.

Easy Ways to Calculate Heat Load

Any calculation of the heat load is needed to optimize the parameters of the heating system or improve the thermal insulation characteristics of the house. After its implementation, certain methods of regulating the heating load of heating are selected. Consider non-labor-intensive methods for calculating this parameter of the heating system.

The dependence of heating power on the area

For home with standard sizes rooms, ceiling heights and good thermal insulation, you can apply the known ratio of the area of \u200b\u200bthe room to the required heat output. In this case, 1 kW of heat will be required per 10 m². To the result obtained, you need to apply a correction factor depending on the climatic zone.

Let's assume that the house is located in the Moscow region. Its total area is 150 m². In this case, the hourly heat load on heating will be equal to:

15*1=15 kWh

The main disadvantage of this method is the large error. The calculation does not take into account changes in weather factors, as well as building features - heat transfer resistance of walls and windows. Therefore, it is not recommended to use it in practice.

Enlarged calculation of the thermal load of the building

The enlarged calculation of the heating load is characterized by more accurate results. Initially, it was used to pre-calculate this parameter when it was impossible to determine exact specifications building. The general formula for determining the heat load on heating is presented below:

Where q°- specific thermal characteristic of the structure. The values must be taken from the corresponding table, a- correction factor, which was mentioned above, Vн- external volume of the building, m³, Tvn and Tnro– temperature values inside the house and outside.

Suppose we need to calculate the maximum hourly load for heating in a house with a volume on the outer walls of 480 m³ (area 160 m², two-storey house). In this case, the thermal characteristic will be equal to 0.49 W / m³ * C. Correction factor a = 1 (for the Moscow region). Optimum temperature inside the dwelling (Tvn) should be + 22 ° С. The outside temperature will be -15°C. We use the formula to calculate the hourly heating load:

Q=0.49*1*480(22+15)= 9.408 kW

Compared to the previous calculation, the resulting value is less. However, it takes into account important factors - the temperature inside the room, on the street, the total volume of the building. Similar calculations can be made for each room. The method of calculating the load on heating according to aggregated indicators makes it possible to determine the optimal power for each radiator in a single room. For a more accurate calculation, you need to know the average temperature values \u200b\u200bfor a particular region.

This calculation method can be used to calculate the hourly heat load for heating. But the results obtained will not give the optimally accurate value of the heat loss of the building.

Accurate heat load calculations

But still, this calculation of the optimal heat load on heating does not give the required calculation accuracy. It does not take into account the most important parameter - the characteristics of the building. The main one is the heat transfer resistance of the material for the manufacture of individual elements of the house - walls, windows, ceiling and floor. They determine the degree of conservation of thermal energy received from the heat carrier of the heating system.

What is heat transfer resistance? R)? This is the reciprocal of the thermal conductivity ( λ ) - the ability of the material structure to transfer thermal energy. Those. the higher the thermal conductivity value, the higher the heat loss. This value cannot be used to calculate the annual heating load, since it does not take into account the thickness of the material ( d). Therefore, experts use the heat transfer resistance parameter, which is calculated by the following formula:

Calculation for walls and windows

There are normalized values of heat transfer resistance of walls, which directly depend on the region where the house is located.

In contrast to the enlarged calculation of the heating load, you first need to calculate the heat transfer resistance for external walls, windows, the floor of the first floor and the attic. Let's take as a basis the following characteristics of the house:

Wall area - 280 m². It includes windows 40 m²;
Wall material - solid brick ( λ=0.56). The thickness of the outer walls 0.36 m. Based on this, we calculate the TV transmission resistance - R=0.36/0.56= 0.64 m²*S/W;
To improve the thermal insulation properties, a outer insulation- expanded polystyrene thickness 100 mm. For him λ=0.036. Respectively R \u003d 0.1 / 0.036 \u003d 2.72 m² * C / W;
General value R for exterior walls 0,64+2,72= 3,36 which is very a good indicator thermal insulation of the house;
Heat transfer resistance of windows - 0.75 m²*S/W(double glazing with argon filling).

In fact, heat losses through the walls will be:

(1/3.36)*240+(1/0.75)*40= 124 W at 1°C temperature difference

We take the temperature indicators the same as for the enlarged calculation of the heating load + 22 ° С indoors and -15 ° С outdoors. Further calculation must be done according to the following formula:

124*(22+15)= 4.96 kWh

Ventilation calculation

Then you need to calculate the losses through ventilation. The total air volume in the building is 480 m³. At the same time, its density is approximately equal to 1.24 kg / m³. Those. its mass is 595 kg. On average, the air is renewed five times per day (24 hours). In this case, to calculate the maximum hourly load for heating, you need to calculate the heat loss for ventilation:

(480*40*5)/24= 4000 kJ or 1.11 kWh

Summing up all the obtained indicators, you can find the total heat loss of the house:

4.96+1.11=6.07 kWh

In this way, the exact maximum heating load is determined. The resulting value directly depends on the temperature outside. Therefore, to calculate the annual load on the heating system, it is necessary to take into account the change weather conditions. If a average temperature during the heating season is -7°С, the total heating load will be equal to:

(124*(22+7)+((480*(22+7)*5)/24))/3600)*24*150(heating season days)=15843 kW

By changing the temperature values, you can make an accurate calculation of the heat load for any heating system.

To the results obtained, it is necessary to add the value of heat losses through the roof and floor. This can be done with a correction factor of 1.2 - 6.07 * 1.2 \u003d 7.3 kW / h.

The resulting value indicates the actual cost of the energy carrier during the operation of the system. There are several ways to regulate the heating load of heating. The most effective of them is to reduce the temperature in rooms where there is no constant presence of residents. This can be done with thermostats and installed sensors temperature. But at the same time, the building must be installed two-pipe system heating.

To calculate the exact value of heat loss, you can use specialized program Valtec. The video shows an example of working with it.

Thermal load refers to the amount of thermal energy required to maintain a comfortable temperature in a house, apartment or a separate room. The maximum hourly heating load is the amount of heat required to maintain normalized performance for an hour under the most unfavorable conditions.

Factors affecting heat load

Wall material and thickness. For example, a brick wall of 25 centimeters and aerated concrete wall of 15 centimeters can skip different amount heat.
Material and structure of the roof. For example, heat loss flat roof from reinforced concrete slabs significantly different from the heat loss of an insulated attic.
Ventilation. The loss of thermal energy with exhaust air depends on the performance of the ventilation system, the presence or absence of a heat recovery system.
Glazing area. Windows lose more heat energy than solid walls.
The level of insolation in different regions. It is determined by the degree of absorption of solar heat by external coatings and the orientation of the planes of buildings in relation to the cardinal points.
Temperature difference between outdoor and indoor. It is determined by the heat flow through the enclosing structures under the condition of a constant resistance to heat transfer.

Heat load distribution

With water heating, the maximum heat output of the boiler must be equal to the sum of the heat output of all heating devices in the house. For the distribution of heating devices influenced by the following factors:

Living rooms in the middle of the house - 20 degrees;
Corner and end living rooms - 22 degrees. At the same time, due to more high temperature walls do not freeze;
Kitchen - 18 degrees, because it has its own heat sources - gas or electric stoves etc.
Bathroom - 25 degrees.

At air heating the heat flow that enters a separate room depends on the throughput of the air sleeve. Often the easiest way to adjust it is to manually adjust the position of the ventilation grilles with temperature control.

In a heating system where a distributive heat source is used (convectors, underfloor heating, electric heaters, etc.), the required temperature mode is set on the thermostat.

Calculation methods

To determine the heat load, there are several methods that have different complexity of calculation and reliability of the results. The following are three of the most simple techniques heat load calculation.

Method #1

According to the current SNiP, there is a simple method for calculating the heat load. On 10 square meters take 1 kilowatt of thermal power. Then the obtained data is multiplied by the regional coefficient:

The southern regions have a coefficient of 0.7-0.9;
For a moderately cold climate (Moscow and Leningrad regions), the coefficient is 1.2-1.3;
Far East and regions of the Far North: for Novosibirsk from 1.5; for Oymyakon up to 2.0.

Example calculation:

The building area (10*10) is equal to 100 square meters.
The base heat load is 100/10=10 kilowatts.
This value is multiplied by a regional coefficient of 1.3, resulting in 13 kW of thermal power, which is required to maintain a comfortable temperature in the house.

Note! If you use this technique to determine the heat load, you still need to take into account a 20 percent headroom to compensate for errors and extreme cold.

Method #2

The first way to determine the heat load has many errors:

Various buildings have different height ceilings. Given that it is not the area that is heated, but the volume, this parameter is very important.
More heat passes through doors and windows than through walls.
Can't be compared city apartment with a private house, where from below, above and behind the walls there are not apartments, but a street.

Method correction:

The base thermal load is 40 watts per 1 cubic meter room volume.
Each door leading outside adds 200 watts to the base heat load, each window adds 100 watts.
Corner and end apartments apartment building have a coefficient of 1.2-1.3, which is affected by the thickness and material of the walls. A private house has a coefficient of 1.5.
Regional coefficients are equal: for the Central regions and the European part of Russia - 0.1-0.15; for the Northern regions - 0.15-0.2; for the Southern regions - 0.07-0.09 kW / sq.m.

Example calculation:

Method #3

Do not flatter yourself - the second method of calculating the heat load is also very imperfect. It very conditionally takes into account the thermal resistance of the ceiling and walls; temperature difference between outside air and inside air.

It is worth noting that in order to maintain a constant temperature inside the house, such an amount of thermal energy is needed that will be equal to all losses through the ventilation system and enclosing devices. However, in this method, the calculations are simplified, since it is impossible to systematize and measure all the factors.

For heat loss wall material affects– 20-30 percent heat loss. 30-40 percent goes through ventilation, 10-25 percent through the roof, 15-25 percent through windows, 3-6 percent through the floor on the ground.

To simplify heat load calculations, heat losses through the enclosing devices are calculated, and then this value is simply multiplied by 1.4. Temperature delta is easy to measure, but you can only take data on thermal resistance in reference books. Below are some popular thermal resistance values:

The thermal resistance of a three-brick wall is 0.592 m2 * C / W.
A wall of 2.5 bricks is 0.502.
Walls in 2 bricks is equal to 0.405.
Walls in one brick (thickness 25 cm) is equal to 0.187.
Log cabin, where the diameter of the log is 25 cm - 0.550.
Log cabin, where the diameter of the log is 20 centimeters - 0.440.
Log house, where the thickness of the log house is 20 cm - 0.806.
Log house, where the thickness is 10 cm - 0.353.
Frame wall, the thickness of which is 20 cm, insulated mineral wool – 0,703.
Walls made of aerated concrete, the thickness of which is 20 cm - 0.476.
Walls made of aerated concrete, the thickness of which is 30 cm - 0.709.
Plaster, the thickness of which is 3 cm - 0.035.
Ceiling or attic floor – 1,43.
Wooden floor - 1.85.
Double wooden door – 0,21.

Example calculation:

Conclusion

As can be seen from the calculations, the methods for determining the heat load have significant errors. Fortunately, an excessive boiler power indicator will not harm:

The operation of a gas boiler at reduced power is carried out without a drop in efficiency, and work condensation devices at part load is carried out in economy mode.
The same applies to solar boilers.
The efficiency index of electrical heating equipment is 100 percent.

Note! The operation of solid fuel boilers at power less than the nominal power value is contraindicated.

The calculation of the heat load for heating is an important factor, the calculations of which must be performed before starting to create a heating system. In the case of a wise approach to the process and competent performance of all work, trouble-free operation of heating is guaranteed, and money is also significantly saved on extra costs.