Calculation of the heat load for heating a building SNP. About thermal energy in simple language! Example of calculating the thermal load of a building

What is this unit of measurement, how gigacalorie? What does it have to do with traditional kilowatt-hours, in which it is calculated? thermal energy? What information do you need to have in order to correctly calculate Gcal for heating? Finally, what formula should be used during the calculation? This, as well as many other things, will be discussed in today’s article.

What is Gcal?

We should start with a related definition. A calorie refers to the specific amount of energy required to heat one gram of water to one degree Celsius (at atmospheric pressure, of course). And due to the fact that from the point of view of heating costs, say, at home, one calorie is a tiny amount, gigacalories (or Gcal for short), corresponding to one billion calories, are used for calculations in most cases. We've decided on this, let's move on.

The use of this value is regulated by the relevant document of the Ministry of Fuel and Energy, published back in 1995.

Note! On average, the consumption standard in Russia per one square meter equal to 0.0342 Gcal per month. Of course, this figure may change for different regions because everything depends on climatic conditions.

So, what is a gigacalorie if we “transform” it into values ​​that are more familiar to us? See for yourself.

1. One gigacalorie is equal to approximately 1,162.2 kilowatt-hours.

2. One gigacalorie of energy is enough to heat a thousand tons of water to +1°C.

What is all this for?

The problem should be considered from two points of view - from the point of view apartment buildings and private. Let's start with the first ones.

Apartment buildings

There is nothing complicated here: gigacalories are used in thermal calculations. And if you know how much thermal energy remains in the house, then you can present the consumer with a specific bill. Let's give a small comparison: if centralized heating operates in the absence of a meter, then you have to pay according to the area of ​​the heated room. If there is a heat meter, this in itself implies wiring horizontal type(either collector or sequential): two risers are brought into the apartment (for “return” and supply), and the intra-apartment system (more precisely, its configuration) is determined by the residents. This kind of scheme is used in new buildings, thanks to which people regulate the consumption of thermal energy, making a choice between savings and comfort.

Let's find out how this adjustment is carried out.

1. Installation of a general thermostat on the return line. In this case, the flow rate of the working fluid is determined by the temperature inside the apartment: if it decreases, the flow rate will accordingly increase, and if it increases, it will decrease.

2. Throttling of heating radiators. Thanks to the throttle, maneuverability heating device limited, the temperature decreases, which means the consumption of thermal energy is reduced.

Private houses

We continue to talk about calculating Gcal for heating. Owners country houses They are interested, first of all, in the cost of a gigacalorie of thermal energy obtained from one or another type of fuel. The table below may help with this.

Table. Comparison of cost of 1 Gcal (including transport costs)

* - prices are approximate, since tariffs may differ depending on the region, moreover, they are constantly growing.

Heat meters

Now let’s find out what information is needed in order to calculate the heating. It's easy to guess what this information is.

1. Temperature of the working fluid at the outlet/inlet of a specific section of the pipeline.

2. The flow rate of the working fluid that passes through the heating devices.

Consumption is determined through the use of heat metering devices, that is, meters. These can be of two types, let’s get acquainted with them.

Vane meters

Such devices are intended not only for heating systems, but also for hot water supply. Their only difference from those meters that are used for cold water is the material from which the impeller is made - in this case it is more resistant to elevated temperatures.

As for the mechanism of operation, it is almost the same:

• due to the circulation of the working fluid, the impeller begins to rotate;
• the rotation of the impeller is transmitted to the accounting mechanism;
• transmission is carried out without direct interaction, but with the help of a permanent magnet.

Despite the fact that the design of such counters is extremely simple, their response threshold is quite low; moreover, there is also reliable protection from distortion of readings: the slightest attempts to brake the impeller using external magnetic field are prevented thanks to an antimagnetic screen.

Devices with a difference recorder

Such devices operate on the basis of Bernoulli's law, which states that the speed of a gas or liquid flow is inversely proportional to its static movement. But how does this hydrodynamic property apply to calculations of working fluid flow? It’s very simple - you just need to block its path with a retaining washer. In this case, the rate of pressure drop on this washer will be inversely proportional to the speed of the moving flow. And if the pressure is recorded by two sensors at once, then the flow can be easily determined, and in real time.

Note! The design of the meter implies the presence of electronics. The vast majority of these modern models provides not only dry information (temperature of the working fluid, its consumption), but also determines the actual use of thermal energy. The control module here is equipped with a port for connecting to a PC and can be configured manually.

Many readers will probably have a logical question: what to do if we are not talking about a closed heating system, and about the open one, in which selection for hot water supply is possible? How to calculate Gcal for heating in this case? The answer is quite obvious: here pressure sensors (as well as retaining washers) are installed simultaneously on both the supply and the “return”. And the difference in the flow rate of the working fluid will indicate the amount of heated water that was used for domestic needs.

How to calculate the consumed thermal energy?

If for one reason or another there is no heat meter, then to calculate thermal energy you must use the following formula:

Vx(T1-T2)/1000=Q

Let's look at what these symbols mean.

1. V denotes the amount of hot water consumed, which can be calculated either in cubic meters or in tons.

2. T1 is temperature indicator the hottest water (traditionally measured in the usual degrees Celsius). In this case, it is preferable to use exactly the temperature that is observed at a certain operating pressure. By the way, the indicator even has a special name - enthalpy. But if the required sensor is missing, then you can take that one as a basis temperature regime, which is extremely close to this enthalpy. In most cases, the average is approximately 60-65 degrees.

3. T2 in the above formula also denotes the temperature, but of cold water. Due to the fact that to penetrate the highway with cold water– the matter is quite difficult; this value is used constants, capable of changing depending on the climatic conditions outside. So, in winter, when the heating season is in full swing, this figure is 5 degrees, and in the summer, when the heating is turned off, 15 degrees.

4. As for 1000, this is the standard coefficient used in the formula in order to obtain the result in gigacalories. It will be more accurate than if you used calories.

5. Finally, Q is the total amount of thermal energy.

As you can see, there is nothing complicated here, so we move on. If the heating circuit is of a closed type (and this is more convenient from an operational point of view), then the calculations must be made slightly differently. The formula that should be used for a building with a closed heating system should look like this:

((V1x(T1-T)-(V2x(T2-T))=Q

Now, accordingly, to the decoding.

1. V1 indicates the flow rate of the working fluid in the supply pipeline (typically, not only water, but also steam can act as a source of thermal energy).

2. V2 is the flow rate of the working fluid in the return pipeline.

3. T is an indicator of the temperature of a cold liquid.

4. T1 – water temperature in the supply pipeline.

5. T2 – temperature indicator that is observed at the outlet.

6. And finally, Q is the same amount of thermal energy.

It is also worth noting that the calculation of Gcal for heating in this case depends on several notations:

• thermal energy that entered the system (measured in calories);
• temperature indicator during the removal of working fluid through the return pipeline.

Other ways to determine the amount of heat

Let us add that there are also other methods by which you can calculate the amount of heat that enters the heating system. In this case, the formula is not only slightly different from those given below, but also has several variations.

((V1x(T1-T2)+(V1- V2)x(T2-T1))/1000=Q

((V2x(T1-T2)+(V1-V2)x(T1-T)/1000=Q

As for the values ​​of the variables, they are the same as in the previous paragraph of this article. Based on all this, we can confidently conclude that it is quite possible to calculate the heat for heating on your own. However, one should not forget about consulting with specialized organizations that are responsible for providing housing with heat, since their methods and principles of calculations may differ, significantly, and the procedure may consist of a different set of measures.

If you intend to equip a “warm floor” system, then prepare for the fact that the calculation process will be more complex, since it takes into account not only the features of the heating circuit, but also the characteristics electrical network, which, in fact, will heat the floor. Moreover, the organizations that install this kind of equipment will also be different.

Note! People often encounter the problem of converting calories into kilowatts, which is explained by the use of a unit of measurement in many specialized manuals, which is called “C” in the international system.

In such cases, it is necessary to remember that the coefficient due to which kilocalories will be converted into kilowatts is equal to 850. In simpler terms, one kilowatt is 850 kilocalories. This option The calculation is simpler than those given above, since the value in gigacalories can be determined in a few seconds, since a Gcal, as noted earlier, is a million calories.

To avoid possible errors, we should not forget that almost all modern heat meters operate with some error, albeit within acceptable limits. This error can also be calculated by hand, for which you need to use the following formula:

(V1- V2)/(V1+ V2)x100=E

Traditionally, now we find out what each of these variable values ​​means.

1. V1 is the flow rate of the working fluid in the supply pipeline.

2. V2 – a similar indicator, but in the return pipeline.

3. 100 is the number by which the value is converted to a percentage.

4. Finally, E is the error of the accounting device.

According to operational requirements and standards, the maximum permissible error should not exceed 2 percent, although in most meters it is somewhere around 1 percent.

As a result, we note that a correctly calculated Gcal for heating can significantly save money spent on heating the room. At first glance, this procedure is quite complicated, but - and you have seen this personally - if you have good instructions, there is nothing difficult about it.

Video - How to calculate heating in a private house

In houses that were commissioned in last years, usually these rules are met, so the calculation heating power equipment is based on standard coefficients. Individual calculations can be carried out at the initiative of the homeowner or the utility structure involved in the supply of heat. This happens when spontaneous replacement of heating radiators, windows and other parameters occurs.

In an apartment served by a utility company, the calculation of the heat load can only be carried out upon transfer of the house in order to track the SNIP parameters in the premises accepted for balance. Otherwise, the apartment owner does this in order to calculate his heat loss in the cold season and eliminate the shortcomings of insulation - use heat-insulating plaster, glue the insulation, install penofol on the ceilings and install metal-plastic windows with a five-chamber profile.

Calculating heat leaks for a utility for the purpose of opening a dispute, as a rule, does not yield results. The reason is that there are heat loss standards. If the house is put into operation, then the requirements are met. At the same time, heating devices comply with the requirements of SNIP. Battery replacement and selection more heat is prohibited, as radiators are installed according to approved building standards.

Private houses are heated autonomous systems, that in this case the load calculation is carried out to comply with SNIP requirements, and heating power adjustments are carried out in conjunction with work to reduce heat loss.

Calculations can be done manually using a simple formula or a calculator on the website. The program helps to calculate required power heating systems and heat leaks characteristic of the winter period. Calculations are carried out for a specific thermal zone.

Basic principles

The methodology includes a number of indicators that together make it possible to assess the level of insulation of a house, compliance with SNIP standards, as well as the power of the heating boiler. How it works:

An individual or average calculation is carried out for the object. The main point of conducting such a survey is that when good insulation and small heat leaks in winter, you can use 3 kW. In a building of the same area, but without insulation, at low winter temperatures the power consumption will be up to 12 kW. Thus, thermal power and the load is assessed not only by area, but also by heat loss.

The main heat losses of a private house:

• windows – 10-55%;
• walls – 20-25%;
• chimney – up to 25%;
• roof and ceiling – up to 30%;
• low floors – 7-10%;
• temperature bridge in the corners – up to 10%

These indicators can vary for the better and for the worse. They are evaluated depending on the types installed windows, thickness of walls and materials, degree of ceiling insulation. For example, in poorly insulated buildings, heat loss through the walls can reach 45% percent; in this case, the expression “we are drowning the street” is applicable to the heating system. Methodology and
The calculator will help you estimate nominal and calculated values.

Specifics of calculations

This technique can also be found under the name “thermal engineering calculation”. The simplified formula is as follows:

Qt = V × ∆T × K / 860, where

V – room volume, m³;

∆T – maximum difference indoors and outdoors, °C;

K – estimated heat loss coefficient;

860 – conversion factor in kW/hour.

The heat loss coefficient K depends on the building structure, thickness and thermal conductivity of the walls. For simplified calculations, you can use the following parameters:

• K = 3.0-4.0 – without thermal insulation (non-insulated frame or metal structure);
• K = 2.0-2.9 – low thermal insulation (masonry in one brick);
• K = 1.0-1.9 – average thermal insulation ( brickwork two bricks);
• K = 0.6-0.9 – good thermal insulation according to the standard.

These coefficients are averaged and do not allow one to estimate heat loss and thermal load per room, so we recommend using an online calculator.

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The heating system in a private home is, most often, a set of autonomous equipment that uses the most appropriate substances for a specific region as energy and coolant. Therefore, for each specific heating scheme, an individual calculation of the heating power of the heating system is required, which takes into account many factors, such as minimum consumption thermal energy for the home, heat consumption for premises - each and every one, helps determine energy consumption per day and over time heating season, etc.

Formulas and coefficients for thermal calculations

The rated thermal power of a heating system for a private facility is determined by the formula (all results are expressed in kW):

• Q = Q 1 x b 1 x b 2 + Q 2 – Q 3 ; Where:
• Q 1 – total heat loss in the building according to calculations, kW;
• b 1 – coefficient of additional thermal energy from radiators in excess of what the calculation showed. The coefficient values ​​are shown in the table below:

The need for thermal calculations for the entire house and individual heated rooms is justified by saving energy resources and the family budget. In what cases are such calculations carried out:

1. To accurately calculate the power of boiler equipment for the most efficient heating of all rooms connected to heating. When purchasing a boiler without preliminary calculations you can install equipment that is completely inappropriate in terms of parameters, which will not cope with its task, and the money will be wasted. The thermal parameters of the entire heating system are determined as a result of the addition of all thermal energy consumption in rooms connected and not connected to the heating boiler, if the pipeline passes through them. A power reserve for heat consumption is also required to reduce wear. heating equipment and minimize the appearance emergency situations under high loads in cold weather;
2. Calculations of the thermal parameters of the heating system are necessary to obtain a technical certificate (TU), without which it will not be possible to approve a project for gasification of a private house, since in 80% of installation cases autonomous heating install a gas boiler and related equipment. For other types of heating units, technical conditions and documentation for connection are not required. For gas equipment it is necessary to know the annual gas consumption, and without appropriate calculations it will not be possible to obtain an exact figure;
3. Get thermal parameters heating system also requires purchasing the right equipment - pipes, radiators, fittings, filters, etc.

Accurate calculations of power and heat consumption for residential premises

The level and quality of insulation depend on the quality of work and architectural features rooms throughout the house. Most of the heat losses (up to 40%) when heating a building occur through the surface of the external walls, through windows and doors (up to 20%), as well as through the roof and floor (up to 10%). The remaining 30% of the heat can escape from the home through vents and ducts.

To obtain updated results, the following reference coefficients are used:

1. Q 1 – used in calculations for rooms with windows. For PVC windows with double-glazed windows Q 1 =1, for windows with single-chamber glazing Q 1 =1.27, for three-chamber windows Q 1 =0.85;
2. Q 2 – used when calculating the insulation coefficient interior walls. For foam concrete Q 2 = 1, for concrete Q 2 – 1.2, for brick Q 2 = 1.5;
3. Q 3 is used when calculating the ratio of floor areas and window openings. For 20% of the wall glazing area, the coefficient Q3 = 1, for 50% of the glazing Q3 is taken as 1.5;
4. The value of the coefficient Q 4 varies depending on the minimum outside temperature for the entire annual heating season. At outside temperature-20 0 C Q 4 = 1, then for every 5 0 C, 0.1 is added or subtracted in one direction or another;
5. Coefficient Q 5 is used in calculations that take into account the total number of walls of the building. With one wall in the calculations Q 5 = 1, with 12 and 3 walls Q 5 = 1.2, for 4 walls Q 5 = 1.33;
6. Q 6 is used if, when calculating heat loss, the functional purpose of the room under the room for which the calculations are being made is taken into account. If there is a residential floor at the top, then the coefficient Q 6 = 0.82, if the attic is heated or insulated, then Q 6 is 0.91, for a cold attic Q 6 = 1;
7. The Q 7 parameter varies depending on the height of the ceilings of the room being examined. If the ceiling height is ≤ 2.5 m, the coefficient Q 7 = 1.0; if the ceiling is higher than 3 m, then Q 7 is taken as 1.05.

After determining all the necessary corrections, the thermal power and heat losses in the heating system are calculated for each individual room using the following formula:

• Q i = q x Si x Q 1 x Q 2 x Q 3 x Q 4 x Q 5 x Q 6 x Q 7, where:
• q =100 W/m²;
• Si is the area of ​​the room being examined.

The parameter results will increase when applying coefficients ≥ 1, and decrease if Q 1- Q 7 ≤1. After calculations specific meaning Calculation results for a specific room can be used to calculate the total thermal power of private autonomous heating using the following formula:

Q = Σ x Qi, (i = 1…N), where: N is the total number of rooms in the building.

Thermal load for heating is the amount of thermal energy required to achieve comfortable temperature in room. There is also the concept of maximum hourly load, which should be understood as greatest number energy that may be needed in certain hours under unfavorable conditions. To understand what conditions can be considered unfavorable, it is necessary to understand the factors on which the heat load depends.

Heat demand of the building

Different buildings will require different amounts of thermal energy to make a person feel comfortable.

Among the factors influencing the need for heat are the following:

Device distribution

If we are talking about water heating, maximum power thermal energy source should be equal to the sum of the powers of all heat sources in the building.

The distribution of devices throughout the premises of the house depends on the following circumstances:

1. Room area, ceiling level.
2. The position of the room in the building. The rooms in the end part in the corners are characterized by increased heat loss.
3. Distance to heat source.
4. Optimal temperature (from the residents' point of view). Room temperature, among other factors, is affected by movement air flow inside the home.

1. Living quarters in the depths of the building - 20 degrees.
2. Living quarters in the corners and end parts of the building - 22 degrees.
3. Kitchen - 18 degrees. IN kitchen area the temperature is higher because there are additional heat sources ( electric stove, refrigerator, etc.).
4. Bathroom and toilet - 25 degrees.

If the house is equipped air heating, the volume of heat flow entering the room depends on the throughput capacity of the air hose. The flow is regulated by manually adjusting the ventilation grilles, and controlled by a thermometer.

The house can be heated by distributed sources of thermal energy: electric or gas convectors, electric heated floors, oil radiators, IR heaters, air conditioners. In this case required temperatures determined by the thermostat setting. In this case, it is necessary to provide such equipment power that would be sufficient at the maximum level of heat loss.

Calculation methods

Calculation of the heat load for heating can be done using the example of a specific room. Let in this case it be a log house made of a 25-centimeter bursa with attic space and wood flooring. Building dimensions: 12×12×3. There are 10 windows and a pair of doors in the walls. The house is located in an area characterized by very low temperatures in winter (up to 30 degrees below zero).

Calculations can be made in three ways, which will be discussed below.

First calculation option

According to existing standards SNiP, 10 square meters require 1 kW of power. This indicator is adjusted taking into account climatic coefficients:

• southern regions - 0.7-0.9;
• central regions - 1.2-1.3;
• Far East and Far North - 1.5-2.0.

First, we determine the area of ​​the house: 12 × 12 = 144 square meters. In this case, the basic heat load indicator is: 144/10 = 14.4 kW. We multiply the result obtained by the climate correction (we will use a coefficient of 1.5): 14.4 × 1.5 = 21.6 kW. So much power is needed to keep the house at a comfortable temperature.

Second calculation option

The method given above suffers from significant errors:

1. The height of the ceilings is not taken into account, but it is not the square meters that need to be heated, but the volume.
2. More heat is lost through window and door openings than through walls.
3. The type of building is not taken into account - is it an apartment building, where behind the walls, ceiling and floor there are heated apartments or is it a private house, where there is only cold air behind the walls.

We correct the calculation:

1. As a base, we use the following indicator - 40 W per cubic meter.
2. For each door we will provide 200 W, and for windows - 100 W.
3. For apartments in the corners and end parts of the house we use a coefficient of 1.3. If we are talking about the highest or lowest floor apartment building, we use a coefficient of 1.3, and for a private building - 1.5.
4. We will also apply the climate factor again.

Climate coefficient table

We make the calculation:

1. We calculate the volume of the room: 12 × 12 × 3 = 432 square meters.
2. The basic power indicator is 432×40=17280 W.
3. The house has a dozen windows and a couple of doors. Thus: 17280+(10×100)+(2×200)=18680W.
4. If we are talking about a private house: 18680 × 1.5 = 28020 W.
5. We take into account the climate coefficient: 28020×1.5=42030 W.

So, based on the second calculation, it is clear that the difference with the first calculation method is almost twofold. At the same time, you need to understand that such power is needed only during the most low temperatures. In other words, peak power can be provided by additional heating sources, for example, a backup heater.

Third calculation option

There is an even more accurate calculation method that takes into account heat loss.

Heat loss percentage diagram

The formula for calculation is: Q=DT/R, ​​where:

• Q - heat loss per square meter of enclosing structure;
• DT - delta between external and internal temperatures;
• R is the level of resistance during heat transfer.

Note! About 40% of the heat goes into the ventilation system.

To simplify the calculations, we will accept the average coefficient (1.4) of heat loss through the enclosing elements. It remains to determine the parameters thermal resistance from reference literature. Below is a table for the most commonly used design solutions:

• wall of 3 bricks - the resistance level is 0.592 per square meter. m×S/W;
• wall of 2 bricks - 0.406;
• wall of 1 brick - 0.188;
• frame made of 25-centimeter timber - 0.805;
• frame made of 12-centimeter timber - 0.353;
• frame material with mineral wool insulation - 0.702;
• wood floor - 1.84;
• ceiling or attic - 1.45;
• wooden double door - 0,22.

1. Temperature delta - 50 degrees (20 degrees Celsius indoors and 30 degrees below zero outside).
2. Heat loss per square meter of floor: 50/1.84 (data for wood floor) = 27.17 W. Losses over the entire floor area: 27.17×144=3912 W.
3. Heat loss through the ceiling: (50/1.45)×144=4965 W.
4. We calculate the area of ​​four walls: (12 × 3) × 4 = 144 square meters. m. Since the walls are made of 25-centimeter timber, R is equal to 0.805. Heat loss: (50/0.805)×144=8944 W.
5. We add up the results: 3912+4965+8944=17821. The resulting number is the total heat loss of the house without taking into account the peculiarities of losses through windows and doors.
6. Add 40% ventilation losses: 17821×1.4=24.949. Thus, you will need a 25 kW boiler.

conclusions

Even the most advanced of the listed methods does not take into account the entire spectrum of heat loss. Therefore, it is recommended to buy a boiler with some power reserve. In this regard, here are a few facts about the efficiency characteristics of different boilers:

1. Gas boiler equipment operate with very stable efficiency, and condensing and solar boilers switch to economical mode at low load.
2. Electric boilers have 100% efficiency.
3. Operation in a mode below the rated power for solid fuel boilers is not allowed.

Solid fuel boilers are regulated by limiting the flow of air into the combustion chamber, however, if the level of oxygen is insufficient, complete combustion of the fuel does not occur. This leads to the formation of a large amount of ash and a decrease in efficiency. The situation can be corrected using a heat accumulator. A tank with thermal insulation is installed between the supply and return pipes, disconnecting them. Thus, a small circuit is created (boiler - buffer tank) and a large circuit (tank - heating devices).

The circuit works as follows:

1. After adding fuel, the equipment operates at rated power. Thanks to natural or forced circulation, heat is transferred to the buffer. After fuel combustion, circulation in the small circuit stops.
2. Over the next few hours, the coolant circulates through a large circuit. The buffer slowly transfers heat to radiators or underfloor heating.

Increased power will require additional costs. At the same time, the equipment power reserve provides an important positive result: The interval between fuel fills increases significantly.

Create a heating system in own home or even in a city apartment - an extremely responsible occupation. It would be completely unreasonable to purchase boiler equipment, as they say, “by eye,” that is, without taking into account all the features of the home. In this case, it is quite possible that you will end up in two extremes: either the boiler power will not be enough - the equipment will work “to the fullest”, without pauses, but still not give the expected result, or, on the contrary, an overly expensive device will be purchased, the capabilities of which will remain completely unchanged. unclaimed.

But that's not all. It is not enough to correctly purchase the necessary heating boiler - it is very important to optimally select and correctly arrange heat exchange devices in the premises - radiators, convectors or “warm floors”. And again, relying only on your intuition or the “good advice” of your neighbors is not the most reasonable option. In a word, it’s impossible to do without certain calculations.

Of course, ideally, such thermal calculations should be carried out by appropriate specialists, but this often costs a lot of money. Isn't it fun to try to do it yourself? This publication will show in detail how heating is calculated based on the area of ​​the room, taking into account many important nuances. By analogy, it will be possible to perform, built into this page, it will help to perform the necessary calculations. The technique cannot be called completely “sinless”, however, it still allows you to obtain results with a completely acceptable degree of accuracy.

The simplest calculation methods

In order for the heating system to create comfortable living conditions during the cold season, it must cope with two main tasks. These functions are closely related to each other, and their division is very conditional.

• The first is maintaining optimal level air temperature in the entire volume of the heated room. Of course, the temperature level may vary somewhat with altitude, but this difference should not be significant. An average of +20 °C is considered quite comfortable conditions - this is the temperature that is usually taken as the initial one in thermal calculations.

In other words, the heating system must be able to warm up a certain volume of air.

If we approach it with complete accuracy, then for separate rooms V residential buildings standards for the required microclimate have been established - they are defined by GOST 30494-96. An excerpt from this document is in the table below:

• The second is compensation of heat losses through building structural elements.

The most important “enemy” of the heating system is heat loss through building structures

Alas, heat loss is the most serious “rival” of any heating system. They can be reduced to a certain minimum, but even with the highest quality thermal insulation it is not yet possible to completely get rid of them. Thermal energy leaks occur in all directions - their approximate distribution is shown in the table:

Naturally, in order to cope with such tasks, the heating system must have a certain thermal power, and this potential must not only correspond common needs buildings (apartments), but also to be correctly distributed among the premises, in accordance with their area and a number of other important factors.

Usually the calculation is carried out in the direction “from small to large”. Simply put, the required amount of thermal energy is calculated for each heated room, the obtained values ​​are summed up, approximately 10% of the reserve is added (so that the equipment does not work at the limit of its capabilities) - and the result will show how much power the heating boiler is needed. And the values ​​​​for each room will become the starting point for the calculation required quantity radiators.

The most simplified and most frequently used method in a non-professional environment is to adopt a norm of 100 W of thermal energy per square meter of area:

The most primitive way of calculating is the ratio of 100 W/m²

Q = S× 100

Q– required heating power for the room;

S– room area (m²);

100 — specific power per unit area (W/m²).

For example, a room 3.2 × 5.5 m

S= 3.2 × 5.5 = 17.6 m²

Q= 17.6 × 100 = 1760 W ≈ 1.8 kW

The method is obviously very simple, but very imperfect. It is worth mentioning right away that it is conditionally applicable only when standard height ceilings - approximately 2.7 m (acceptable - in the range from 2.5 to 3.0 m). From this point of view, the calculation will be more accurate not from the area, but from the volume of the room.

It is clear that in this case the specific power value is calculated per cubic meter. It is taken equal to 41 W/m³ for reinforced concrete panel house, or 34 W/m³ - in brick or made of other materials.

Q = S × h× 41 (or 34)

h– ceiling height (m);

41 or 34 – specific power per unit volume (W/m³).

For example, the same room in panel house, with a ceiling height of 3.2 m:

Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW

The result is more accurate, since it already takes into account not only all the linear dimensions of the room, but even, to a certain extent, the features of the walls.

But still, it is still far from real accuracy - many nuances are “outside the brackets”. How to perform calculations closer to real conditions is in the next section of the publication.

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Carrying out calculations of the required thermal power taking into account the characteristics of the premises

The calculation algorithms discussed above can be useful for an initial “estimate,” but you should still rely on them completely with great caution. Even to a person who does not understand anything about building heating engineering, the indicated average values ​​may certainly seem dubious - they cannot be equal, say, for Krasnodar region and for the Arkhangelsk region. In addition, the room is different: one is located on the corner of the house, that is, it has two external walls ki, and the other is protected from heat loss by other rooms on three sides. In addition, the room may have one or more windows, both small and very large, sometimes even panoramic. And the windows themselves may differ in the material of manufacture and other design features. And this is not a complete list - it’s just that such features are visible even to the naked eye.

In a word, there are quite a lot of nuances that affect the heat loss of each specific room, and it is better not to be lazy, but to carry out a more thorough calculation. Believe me, using the method proposed in the article, this will not be so difficult.

General principles and calculation formula

The calculations will be based on the same ratio: 100 W per 1 square meter. But the formula itself is “overgrown” with a considerable number of various correction factors.

Q = (S × 100) × a × b× c × d × e × f × g × h × i × j × k × l × m

The Latin letters denoting the coefficients are taken completely arbitrarily, in alphabetical order, and are not related to any standard quantities accepted in physics. The meaning of each coefficient will be discussed separately.

• “a” is a coefficient that takes into account the number of external walls in a particular room.

Obviously, the more external walls there are in a room, the larger the area through which heat losses. In addition, the presence of two or more external walls also means corners - extremely vulnerable places from the point of view of the formation of “cold bridges”. Coefficient “a” will correct for this specific feature of the room.

The coefficient is taken equal to:

— external walls No (interior space): a = 0.8;

- external wall one: a = 1.0;

— external walls two: a = 1.2;

— external walls three: a = 1.4.

• “b” is a coefficient that takes into account the location of the external walls of the room relative to the cardinal directions.

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Even on the coldest winter days, solar energy still has an impact on the temperature balance in the building. It is quite natural that the side of the house that faces south receives some heat from the sun's rays, and heat loss through it is lower.

But walls and windows facing north “never see” the Sun. The eastern part of the house, although it “catches” the morning sun’s rays, still does not receive any effective heating from them.

Based on this, we introduce the coefficient “b”:

- the outer walls of the room face North or East: b = 1.1;

- the external walls of the room are oriented towards South or West: b = 1.0.

• “c” is a coefficient that takes into account the location of the room relative to the winter “wind rose”

Perhaps this amendment is not so mandatory for houses located on areas protected from winds. But sometimes the prevailing winter winds can make their own “hard adjustments” to the thermal balance of a building. Naturally, the windward side, that is, “exposed” to the wind, will lose significantly more body compared to the leeward, opposite side.

Based on the results of long-term weather observations in any region, a so-called “wind rose” is compiled - graphic diagram, showing the prevailing wind directions in winter and summer. This information can be obtained from your local weather service. However, many residents themselves, without meteorologists, know very well where the winds predominantly blow in winter, and from which side of the house the deepest snowdrifts usually sweep.

If you want to carry out calculations with higher accuracy, you can include the correction factor “c” in the formula, taking it equal to:

- windward side of the house: c = 1.2;

- leeward walls of the house: c = 1.0;

- walls located parallel to the wind direction: c = 1.1.

• “d” is a correction factor that takes into account the climatic conditions of the region where the house was built

Naturally, the amount of heat loss through all building structures will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer readings “dance” in a certain range, but for each region there is an average indicator of the lowest temperatures characteristic of the coldest five-day period of the year (usually this is typical for January). For example, below is a map diagram of the territory of Russia, on which approximate values ​​are shown in colors.

Usually this value is easy to clarify in the regional weather service, but you can, in principle, rely on your own observations.

So, the coefficient “d”, which takes into account the climate characteristics of the region, for our calculations is taken equal to:

— from – 35 °C and below: d = 1.5;

— from – 30 °С to – 34 °С: d = 1.3;

— from – 25 °С to – 29 °С: d = 1.2;

— from – 20 °С to – 24 °С: d = 1.1;

— from – 15 °С to – 19 °С: d = 1.0;

— from – 10 °С to – 14 °С: d = 0.9;

- no colder - 10 °C: d = 0.7.

• “e” is a coefficient that takes into account the degree of insulation of external walls.

The total value of heat losses of a building is directly related to the degree of insulation of all building structures. One of the “leaders” in heat loss are walls. Therefore, the value of thermal power required to maintain comfortable conditions living indoors depends on the quality of their thermal insulation.

The value of the coefficient for our calculations can be taken as follows:

— external walls do not have insulation: e = 1.27;

- average degree of insulation - walls made of two bricks or their surface thermal insulation is provided with other insulation materials: e = 1.0;

— insulation was carried out with high quality, based on thermal engineering calculations: e = 0.85.

Below in the course of this publication, recommendations will be given on how to determine the degree of insulation of walls and other building structures.

• coefficient "f" - correction for ceiling heights

Ceilings, especially in private homes, may have different heights. Therefore, the thermal power to warm up a particular room of the same area will also differ in this parameter.

It would not be a big mistake to accept the following values ​​for the correction factor “f”:

— ceiling heights up to 2.7 m: f = 1.0;

— flow height from 2.8 to 3.0 m: f = 1.05;

- ceiling heights from 3.1 to 3.5 m: f = 1.1;

— ceiling heights from 3.6 to 4.0 m: f = 1.15;

- ceiling height more than 4.1 m: f = 1.2.

• « g" is a coefficient that takes into account the type of floor or room located under the ceiling.

As shown above, the floor is one of the significant sources of heat loss. This means that it is necessary to make some adjustments to account for this feature of a particular room. The correction factor “g” can be taken equal to:

- cold floor on the ground or above an unheated room (for example, a basement or basement): g= 1,4 ;

- insulated floor on the ground or above an unheated room: g= 1,2 ;

— the heated room is located below: g= 1,0 .

• « h" is a coefficient that takes into account the type of room located above.

The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat loss is inevitable, which will require an increase in the required heating power. Let us introduce the coefficient “h”, which takes into account this feature of the calculated room:

— the “cold” attic is located on top: h = 1,0 ;

— there is an insulated attic or other insulated room on top: h = 0,9 ;

— any heated room is located on top: h = 0,8 .

• « i" - coefficient taking into account the design features of windows

Windows are one of the “main routes” for heat flow. Naturally, much in this matter depends on the quality of the window design. Old wooden frames, which were previously universally installed in all houses, are significantly inferior in terms of their thermal insulation to modern multi-chamber systems with double-glazed windows.

Without words it is clear that the thermal insulation qualities of these windows differ significantly

But there is no complete uniformity between PVH windows. For example, a two-chamber double-glazed window (with three glasses) will be much “warmer” than a single-chamber one.

This means that it is necessary to enter a certain coefficient “i”, taking into account the type of windows installed in the room:

- standard wooden windows with conventional double glazing: i = 1,27 ;

- modern window systems with single-chamber double-glazed windows: i = 1,0 ;

— modern window systems with two-chamber or three-chamber double-glazed windows, including those with argon filling: i = 0,85 .

• « j" - correction factor for the total glazing area of ​​the room

Whatever quality windows No matter how they were, it will still not be possible to completely avoid heat loss through them. But it’s quite clear that you can’t compare a small window with panoramic glazing covering almost the entire wall.

First you need to find the ratio of the areas of all the windows in the room and the room itself:

x = ∑SOK /SP

∑ SOK– total area of ​​windows in the room;

SP– area of ​​the room.

Depending on the obtained value, the correction factor “j” is determined:

— x = 0 ÷ 0.1 →j = 0,8 ;

— x = 0.11 ÷ 0.2 →j = 0,9 ;

— x = 0.21 ÷ 0.3 →j = 1,0 ;

— x = 0.31 ÷ 0.4 →j = 1,1 ;

— x = 0.41 ÷ 0.5 →j = 1,2 ;

• « k" - coefficient that corrects for the presence of an entrance door

A door to the street or to an unheated balcony is always an additional “loophole” for the cold

Door to the street or open balcony is capable of making adjustments to the thermal balance of the room - each opening of it is accompanied by the penetration of a considerable volume of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient “k”, which we take equal to:

- no door: k = 1,0 ;

- one door to the street or to the balcony: k = 1,3 ;

- two doors to the street or balcony: k = 1,7 .

• « l" - possible amendments to the heating radiator connection diagram

Perhaps this may seem like an insignificant detail to some, but still, why not immediately take into account the planned connection diagram for the heating radiators. The fact is that their heat transfer, and therefore their participation in maintaining a certain temperature balance in the room, changes quite noticeably when different types insertion of supply and return pipes.

Illustration	Radiator insert type	The value of the coefficient "l"
	Diagonal connection: supply from above, return from below	l = 1.0
	Connection on one side: supply from above, return from below	l = 1.03
	Two-way connection: both supply and return from below	l = 1.13
	Diagonal connection: supply from below, return from above	l = 1.25
	Connection on one side: supply from below, return from above	l = 1.28
	One-way connection, both supply and return from below	l = 1.28

• « m" - correction factor for the peculiarities of the installation location of heating radiators

And finally, the last coefficient, which is also related to the peculiarities of connecting heating radiators. It is probably clear that if the battery is installed openly and is not blocked by anything from above or from the front, then it will give maximum heat transfer. However, such an installation is not always possible - more often the radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating elements into the created interior ensemble, hide them completely or partially with decorative screens - this also significantly affects the thermal output.

If there are certain “outlines” of how and where radiators will be mounted, this can also be taken into account when making calculations by introducing a special coefficient “m”:

So, the calculation formula is clear. Surely, some of the readers will immediately grab their head - they say, it’s too complicated and cumbersome. However, if you approach the matter systematically and in an orderly manner, then there is no trace of complexity.

Any good homeowner must have a detailed graphic plan of his “possessions” with dimensions indicated, and usually oriented to the cardinal points. Climatic features region is easy to determine. All that remains is to walk through all the rooms with a tape measure and clarify some of the nuances for each room. Features of housing - “vertical proximity” above and below, location entrance doors, the proposed or existing installation scheme for heating radiators - no one except the owners knows better.

It is recommended to immediately create a worksheet where you can enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will be helped by the built-in calculator, which already contains all the coefficients and ratios mentioned above.

If some data could not be obtained, then you can, of course, not take them into account, but in this case the calculator “by default” will calculate the result taking into account the least favorable conditions.

Can be seen with an example. We have a house plan (taken completely arbitrary).

Region with level minimum temperatures within -20 ÷ 25 °C. Predominance of winter winds = northeast. The house is one-story, with an insulated attic. Insulated floors on the ground. The optimal diagonal connection of radiators that will be installed under the window sills has been selected.

Let's create a table something like this:

Then, using the calculator below, we make calculations for each room (already taking into account the 10% reserve). It won't take much time using the recommended app. After this, all that remains is to sum up the obtained values ​​for each room - this will be the required total power of the heating system.

The result for each room, by the way, will help you choose the right number of heating radiators - all that remains is to divide by the specific thermal power of one section and round up.