Calculation of the heating load of the building. Calculation of the heat load for heating the building

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, manufacturers will tell you this coefficient. 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 should function in a private house, as a rule, with a natural impulse. Air exchange is created due to the presence of draft in the ventilation ducts and the boiler chimney.

Proposed in normative documentation The method for determining the heat load from ventilation 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 to heat the 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. It is difficult to find out how much it gets inside the house with natural ventilation. 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 volume of air 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 the thermal energy spent on heating water. 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 the cold tap water equal to 10 °C.

The unit of equipment power is always referred to 1 hour, and the resulting 5.2 kW - to the day. But it is impossible to divide this figure by 24, because we want to receive hot water 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 the traditional method by area, although you will 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 task can be solved only by an integrated approach that takes into account all the parameters of the system, the building and the 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 calculation system for 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 the power of heating equipment - boiler, radiators and pipelines. The heat losses of the house are also taken into account.

Ideally heat output 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 losses;
• House dimensions. It is logical to assume that the larger the room, the more intensively the heating system should work. An important factor in this case is not only the total volume of each room, 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 the 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 for heating is necessary to reduce energy consumption and maintain the optimal level of 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 ensure the optimal temperature regime for heating.

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 the technical characteristics of the 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 a house with standard room sizes, ceiling heights and good thermal insulation, a known ratio of room area to required heat output can be applied. 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. His total area be 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 the exact characteristics of the building. General formula to determine 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, but- correction factor, which was mentioned above, Vn- external volume of the building, m³, Tvn And Tnro– temperature values ​​inside the house and outside.

Suppose that it is necessary to calculate the maximum hourly heating load in a house with an external wall volume 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). The 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 heating load according to aggregated indicators makes it possible to determine the optimal power for each radiator in a particular 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 material of manufacture individual elements houses - 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 following characteristics Houses:

• 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, an external insulation was installed - expanded polystyrene with a thickness of 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 a very good indicator of the 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 losses 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 the average temperature during the heating season is -7°C, then 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 using temperature controllers and installed temperature sensors. But at the same time, a two-pipe heating system must be installed in the building.

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

Ask any specialist how to properly organize the heating system in the building. It doesn't matter if it's residential or industrial. And the professional will answer that the main thing is to accurately make calculations and correctly carry out the design. We are talking, in particular, about the calculation of the heat load on heating. The volume of consumption of thermal energy, and hence fuel, depends on this indicator. I.e economic indicators stand next to the technical specifications.

Performing accurate calculations allows you to get not only full list documentation necessary for the installation work, but also to select the necessary equipment, additional components and materials.

Thermal loads - definition and characteristics

What is usually meant by the term "heat load on heating"? This is the amount of heat that all heating devices installed in the building give off. To avoid unnecessary expenses for the production of work, as well as the purchase of unnecessary devices and materials, a preliminary calculation is necessary. With it, you can adjust the rules for installing and distributing heat to all rooms, and this can be done economically and evenly.

But that's not all. Very often, experts carry out calculations, relying on accurate indicators. They relate to the size of the house and the nuances of construction, which takes into account the diversity of building elements and their compliance with the requirements of thermal insulation and other things. It is precisely the exact indicators that make it possible to correctly make calculations and, accordingly, obtain options for the distribution of thermal energy throughout the premises as close to the ideal as possible.

But often there are errors in the calculations, which leads to inefficient operation of the heating as a whole. Sometimes it is necessary to redo during operation not only the circuits, but also sections of the system, which leads to additional costs.

What parameters affect the calculation of the heat load in general? Here it is necessary to divide the load into several positions, which include:

• Central heating system.
• Underfloor heating system, if one is installed in the house.
• Ventilation system - both forced and natural.
• Hot water supply of the building.
• Branches for additional household needs. For example, a sauna or a bath, a pool or a shower.

Main characteristics

Professionals do not lose sight of any trifle that can affect the correctness of the calculation. Hence the rather large list of characteristics of the heating system that should be taken into account. Here are just a few of them:

1. The purpose of the property or its type. It can be a residential building or an industrial building. Heat suppliers have standards that are distributed by type of building. They often become fundamental in carrying out calculations.
2. The architectural part of the building. This can include enclosing elements (walls, roofs, ceilings, floors), their overall dimensions, thickness. Be sure to take into account all kinds of openings - balconies, windows, doors, etc. It is very important to take into account the presence of basements and attics.
3. Temperature regime for each room separately. This is very important because General requirements to the temperature in the house do not give an accurate picture of the distribution of heat.
4. Appointment of premises. This mainly applies to production shops, which require stricter compliance with the temperature regime.
5. Availability of special premises. For example, in residential private houses it can be baths or saunas.
6. Degree of technical equipment. The presence of a ventilation and air conditioning system, hot water supply, and the type of heating used are taken into account.
7. Number of points through which sampling is carried out hot water. And the more such points, the greater the heat load the heating system is exposed to.
8. The number of people on the site. Criteria such as indoor humidity and temperature depend on this indicator.
9. Additional indicators. In residential premises, one can distinguish the number of bathrooms, separate rooms, balconies. In industrial buildings - the number of shifts of workers, the number of days in a year when the workshop itself works in the technological chain.

What is included in the calculation of loads

Heating scheme

The calculation of thermal loads for heating is carried out at the design stage of the building. But at the same time, the norms and requirements of various standards must be taken into account.

For example, the heat loss of the enclosing elements of the building. Moreover, all rooms are taken into account separately. Further, this is the power that is needed to heat the coolant. We add here the amount of thermal energy required for heating supply ventilation. Without this, the calculation will not be very accurate. We also add the energy that is spent on heating water for a bath or pool. Specialists must take into account the further development of the heating system. Suddenly, in a few years, you will decide to arrange a Turkish hammam in your own private house. Therefore, it is necessary to add a few percent to the loads - usually up to 10%.

Recommendation! It is necessary to calculate thermal loads with a "margin" for country houses. It is the reserve that will allow in the future to avoid additional financial costs, which are often determined by amounts of several zeros.

Features of calculating the heat load

Air parameters, or rather, its temperature, are taken from GOSTs and SNiPs. Here, the heat transfer coefficients are selected. By the way, the passport data of all types of equipment (boilers, heating radiators, etc.) are taken into account without fail.

What is usually included in a traditional heat load calculation?

• Firstly, the maximum flow of thermal energy coming from heating devices (radiators).
• Secondly, the maximum heat consumption for 1 hour of operation of the heating system.
• Thirdly, the total heat costs for a certain period of time. Usually the seasonal period is calculated.

If all these calculations are measured and compared with the heat transfer area of ​​the system as a whole, then a fairly accurate indicator of the efficiency of heating a house will be obtained. But you have to take into account small deviations. For example, reducing heat consumption at night. For industrial facilities Weekends and holidays must also be taken into account.

Methods for determining thermal loads

Underfloor heating design

Currently, experts use three main methods for calculating thermal loads:

1. Calculation of the main heat losses, where only aggregated indicators are taken into account.
2. The indicators based on the parameters of the enclosing structures are taken into account. This is usually added to the losses for heating the internal air.
3. All systems included in heating networks are calculated. This is both heating and ventilation.

There is another option, which is called the enlarged calculation. It is usually used when there are no basic indicators and building parameters required for a standard calculation. That is, the actual characteristics may differ from the design.

To do this, experts use a very simple formula:

Q max from. \u003d α x V x q0 x (tv-tn.r.) x 10 -6

α is a correction factor depending on the region of construction (table value)
V - the volume of the building on the outer planes
q0 - characteristic of the heating system by specific index, usually determined by the coldest days of the year

Types of thermal loads

Thermal loads that are used in the calculations of the heating system and the selection of equipment have several varieties. For example, seasonal loads, for which the following features are inherent:

1. Changes in outdoor temperature throughout the heating season.
2. Meteorological features of the region where the house was built.
3. Jumps in the load on the heating system during the day. This indicator usually falls into the category of "minor loads", because the enclosing elements prevent a lot of pressure on the heating in general.
4. Everything related to the thermal energy associated with the ventilation system of the building.
5. Thermal loads that are determined throughout the year. For example, the consumption of hot water in the summer season is reduced by only 30-40% when compared with winter time of the year.
6. Dry heat. This feature is inherent in domestic heating systems, where a fairly large number of indicators are taken into account. For example, the number of windows and doorways, the number of people living or permanently in the house, ventilation, air exchange through various cracks and gaps. A dry thermometer is used to determine this value.
7. Hidden thermal energy. There is also such a term, which is defined by evaporation, condensation, and so on. A wet bulb thermometer is used to determine the index.

Thermal Load Controllers

Programmable controller, temperature range - 5-50 C

Modern heating units and devices are provided with a set of different regulators, with which you can change the thermal loads, in order to avoid dips and jumps in thermal energy in the system. Practice has shown that with the help of regulators it is possible not only to reduce the load, but also to bring the heating system to rational use fuel. And this is a purely economic side of the issue. This is especially true for industrial facilities, where quite large fines have to be paid for excessive fuel consumption.

If you are not sure about the correctness of your calculations, then use the services of specialists.

Let's look at a couple more formulas that relate to different systems. For example, ventilation and hot water systems. Here you need two formulas:

Qin. \u003d qin.V (tn.-tv.) - this applies to ventilation.
Here:
tn. and tv - air temperature outside and inside
qv. - specific indicator
V - external volume of the building

Qgvs. \u003d 0.042rv (tg.-tx.) Pgav - for hot water supply, where

tg.-tx - temperature of hot and cold water
r - water density
regarding maximum load to the average, which is determined by GOSTs
P - the number of consumers
Gav - average hot water consumption

Complex calculation

In combination with settlement issues, studies of the thermotechnical order are necessarily carried out. For this, various devices are used that give accurate indicators for calculations. For example, for this, window and door openings, ceilings, walls, and so on are examined.

It is this examination that helps to determine the nuances and factors that can have a significant impact on heat loss. For example, thermal imaging diagnostics will accurately show the temperature difference when a certain amount of thermal energy passes through 1 square meter enclosing structure.

So practical measurements are indispensable when making calculations. This is especially true for bottlenecks in the building structure. In this regard, the theory will not be able to show exactly where and what is wrong. And practice will show where to apply different methods protection against heat loss. And the calculations themselves in this regard are becoming more accurate.

Conclusion on the topic

Estimated heat load is a very important indicator obtained in the process of designing a home heating system. If you approach the matter wisely and carry out all the necessary calculations correctly, then you can guarantee that the heating system will work perfectly. And at the same time, it will be possible to save on overheating and other costs that can simply be avoided.

Whether it is an industrial building or a residential building, you need to make competent calculations and draw up a diagram of the heating system circuit. At this stage, experts recommend paying special attention to the calculation of the possible heat load on the heating circuit, as well as the amount of fuel consumed and heat generated.

Thermal load: what is it?

This term refers to the amount of heat given off. The preliminary calculation of the heat load made it possible to avoid unnecessary costs for the purchase of components of the heating system and for their installation. Also, this calculation will help to correctly distribute the amount of heat generated economically and evenly throughout the building.

There are many nuances in these calculations. For example, the material from which the building is built, thermal insulation, region, etc. Experts try to take into account as many factors and characteristics as possible to obtain a more accurate result.

The calculation of the heat load with errors and inaccuracies leads to inefficient operation of the heating system. It even happens that you have to redo sections of an already working structure, which inevitably leads to unplanned expenses. Yes, and housing and communal organizations calculate the cost of services based on data on heat load.

Main Factors

An ideally calculated and designed heating system must maintain the set temperature in the room and compensate for the resulting heat losses. When calculating the indicator of the heat load on the heating system in the building, you need to take into account:

Purpose of the building: residential or industrial.

Characteristics of the structural elements of the structure. These are windows, walls, doors, roof and ventilation system.

Housing dimensions. The larger it is, the more powerful the heating system should be. Area must be taken into account window openings, doors, exterior walls and the volume of each interior space.

The presence of rooms for special purposes (bath, sauna, etc.).

Degree of equipment with technical devices. That is, the presence of hot water supply, ventilation systems, air conditioning and the type of heating system.

For a single room. For example, in rooms intended for storage, it is not necessary to maintain a comfortable temperature for a person.

Number of points with hot water supply. The more of them, the more the system is loaded.

Area of ​​glazed surfaces. Rooms with French windows lose a significant amount of heat.

Additional terms. IN residential buildings it can be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in a calendar year, shifts, technological chain production process etc.

Climatic conditions of the region. When calculating heat losses, street temperatures are taken into account. If the differences are insignificant, then a small amount of energy will be spent on compensation. While at -40 ° C outside the window it will require significant expenses.

Features of existing methods

The parameters included in the calculation of the heat load are in SNiPs and GOSTs. They also have special heat transfer coefficients. From the passports of the equipment included in the heating system, digital characteristics are taken regarding a specific heating radiator, boiler, etc. And also traditionally:

The heat consumption, taken to the maximum for one hour of operation of the heating system,

The maximum heat flow from one radiator,

Total heat costs in a certain period (most often - a season); if you need an hourly calculation of the load on heating network, then the calculation must be carried out taking into account the temperature difference during the day.

The calculations made are compared with the heat transfer area of ​​the entire system. The index is quite accurate. Some deviations happen. For example, for industrial buildings, it will be necessary to take into account the reduction in heat energy consumption on weekends and holidays, and in residential buildings - at night.

Methods for calculating heating systems have several degrees of accuracy. To reduce the error to a minimum, it is necessary to use rather complex calculations. Less accurate schemes are used if the goal is not to optimize the costs of the heating system.

Basic calculation methods

To date, the calculation of the heat load on the heating of a building can be carried out in one of the following ways.

Three main

1. Aggregated indicators are taken for calculation.
2. The indicators of the structural elements of the building are taken as the base. Here, the calculation of the internal volume of air going to warm up will also be important.
3. All objects included in the heating system are calculated and summarized.

One exemplary

There is also a fourth option. It has a fairly large error, because the indicators are taken very average, or they are not enough. Here is the formula - Q from \u003d q 0 * a * V H * (t EH - t NPO), where:

• q 0 - specific thermal characteristic of the building (most often determined by the coldest period),
• a - correction factor (depends on the region and is taken from ready-made tables),
• V H is the volume calculated from the outer planes.

Example of a simple calculation

For a building with standard parameters (ceiling heights, room sizes and good thermal insulation characteristics) you can apply a simple ratio of parameters, corrected by a factor depending on the region.

Suppose that a residential building is located in the Arkhangelsk region, and its area is 170 square meters. m. The heat load will be equal to 17 * 1.6 \u003d 27.2 kW / h.

Such a definition of thermal loads does not take into account many important factors. For example, design features buildings, temperatures, the number of walls, the ratio of the areas of walls and window openings, etc. Therefore, such calculations are not suitable for serious heating system projects.

It depends on the material from which they are made. Most often today, bimetallic, aluminum, steel are used, much less often cast iron radiators. Each of them has its own heat transfer index (thermal power). Bimetal radiators with a distance between the axes of 500 mm, on average they have 180 - 190 watts. Aluminum radiators have almost the same performance.

The heat transfer of the described radiators is calculated for one section. Steel plate radiators are non-separable. Therefore, their heat transfer is determined based on the size of the entire device. For example, the thermal power of a two-row radiator 1100 mm wide and 200 mm high will be 1010 W, and a steel panel radiator 500 mm wide and 220 mm high will be 1644 W.

The calculation of the heating radiator by area includes the following basic parameters:

Ceiling height (standard - 2.7 m),

Thermal power (per sq. m - 100 W),

One outer wall.

These calculations show that for every 10 sq. m requires 1,000 W of thermal power. This result is divided by the heat output of one section. The answer is required amount radiator sections.

For the southern regions of our country, as well as for the northern ones, decreasing and increasing coefficients have been developed.

Average calculation and exact

Given the factors described, the average calculation is carried out according to the following scheme. If for 1 sq. m requires 100 W of heat flow, then a room of 20 square meters. m should receive 2,000 watts. The radiator (popular bimetallic or aluminum) of eight sections allocates about Divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.

The exact one looks a little intimidating. Actually, nothing complicated. Here is the formula:

Q t \u003d 100 W / m 2 × S (rooms) m 2 × q 1 × q 2 × q 3 × q 4 × q 5 × q 6 × q 7, where:

• q 1 - type of glazing (ordinary = 1.27, double = 1.0, triple = 0.85);
• q 2 - wall insulation (weak or absent = 1.27, 2-brick wall = 1.0, modern, high = 0.85);
• q 3 - the ratio of the total area of ​​window openings to the floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
• q 4 - outdoor temperature (the minimum value is taken: -35 o C = 1.5, -25 o C = 1.3, -20 o C = 1.1, -15 o C = 0.9, -10 o C = 0.7);
• q 5 - the number of external walls in the room (all four = 1.4, three = 1.3, corner room = 1.2, one = 1.2);
• q 6 - type of calculation room above the calculation room (cold attic = 1.0, warm attic = 0.9, residential heated room = 0.8);
• q 7 - ceiling height (4.5 m = 1.2, 4.0 m = 1.15, 3.5 m = 1.1, 3.0 m = 1.05, 2.5 m = 1.3).

Using any of the methods described, it is possible to calculate the heat load of an apartment building.

Approximate calculation

These are the conditions. Minimum temperature in the cold season - -20 o C. Room 25 sq. m with triple glazing, double-leaf windows, ceiling height of 3.0 m, two-brick walls and an unheated attic. The calculation will be as follows:

Q \u003d 100 W / m 2 × 25 m 2 × 0.85 × 1 × 0.8 (12%) × 1.1 × 1.2 × 1 × 1.05.

The result, 2 356.20, is divided by 150. As a result, it turns out that 16 sections need to be installed in a room with the specified parameters.

If calculation is required in gigacalories

In the absence of a heat energy meter on an open heating circuit, the calculation of the heat load for heating the building is calculated by the formula Q \u003d V * (T 1 - T 2) / 1000, where:

• V - the amount of water consumed by the heating system, calculated in tons or m 3,
• T 1 - a number showing the temperature of hot water, measured in o C, and for calculations, the temperature corresponding to a certain pressure in the system is taken. This indicator has its own name - enthalpy. If it is not possible to remove temperature indicators in a practical way, they resort to an average indicator. It is in the range of 60-65 o C.
• T 2 - temperature of cold water. It is quite difficult to measure it in the system, so constant indicators have been developed that depend on the temperature regime on the street. For example, in one of the regions, in the cold season, this indicator is taken equal to 5, in summer - 15.
• 1,000 is the coefficient for obtaining the result immediately in gigacalories.

In the case of a closed circuit, the heat load (gcal/h) is calculated differently:

Q from \u003d α * q o * V * (t in - t n.r.) * (1 + K n.r.) * 0.000001, where

The calculation of the heat load turns out to be somewhat enlarged, but it is this formula that is given in the technical literature.

Increasingly, in order to increase the efficiency of the heating system, they resort to buildings.

These works are carried out at night. For a more accurate result, you must observe the temperature difference between the room and the street: it must be at least 15 o. Fluorescent and incandescent lamps are switched off. It is advisable to remove carpets and furniture to the maximum, they knock down the device, giving some error.

The survey is carried out slowly, the data are recorded carefully. The scheme is simple.

The first stage of work takes place indoors. The device is moved gradually from doors to windows, giving Special attention corners and other joints.

The second stage - inspection with a thermal imager external walls buildings. The joints are still carefully examined, especially the connection with the roof.

The third stage is data processing. First, the device does this, then the readings are transferred to a computer, where the corresponding programs complete the processing and give the result.

If the survey was conducted by a licensed organization, then it will issue a report with mandatory recommendations based on the results of the work. If the work was carried out personally, then you need to rely on your knowledge and, possibly, the help of the Internet.

The topic of this article is to determine the heat load for heating and other parameters that need to be calculated for. The material is aimed primarily at owners of private houses, far from heat engineering and in need of the simplest formulas and algorithms.

So, let's go.

Our task is to learn how to calculate the main parameters of heating.

Redundancy and Accurate Calculation

It is worth specifying one subtlety of calculations from the very beginning: absolutely exact values heat loss through the floor, ceiling and walls, which the heating system has to compensate for, is almost impossible to calculate. It is possible to speak only about this or that degree of reliability of estimates.

The reason is that too many factors affect heat loss:

• Thermal resistance of main walls and all layers of finishing materials.
• The presence or absence of cold bridges.
• The wind rose and the location of the house on the terrain.
• The work of ventilation (which, in turn, again depends on the strength and direction of the wind).
• The degree of insolation of windows and walls.

There are also good news. Almost all modern heating boilers and distributed heating systems (warm floors, electric and gas convectors, etc.) are equipped with thermostats that measure the heat consumption depending on the temperature in the room.

From a practical point of view, this means that excess thermal power will only affect the heating operation mode: say, 5 kWh of heat will be given off not in one hour of continuous operation with a power of 5 kW, but in 50 minutes of operation with a power of 6 kW. The next 10 minutes the boiler or other heating device will spend in standby mode, without consuming electricity or energy carrier.

Therefore: in the case of calculating the thermal load, our task is to determine its minimum allowable value.

The only exception to general rule associated with the operation of classic solid fuel boilers and due to the fact that a decrease in their thermal power is associated with a serious drop in efficiency due to incomplete combustion of the fuel. The problem is solved by installing a heat accumulator in the circuit and throttling heating devices with thermal heads.

The boiler, after kindling, operates at full power and with maximum efficiency until the coal or firewood is completely burned out; then the heat accumulated by the heat accumulator is dosed out to maintain optimum temperature in room.

Most of the other parameters that need to be calculated also allow some redundancy. However, more about this in the relevant sections of the article.

Parameter List

So, what do we actually have to consider?

• The total heat load for home heating. It corresponds to the minimum required power boiler or the total power of devices in a distributed heating system.
• The need for heat in a separate room.
• The number of sections of the sectional radiator and the size of the register corresponding to certain value thermal power.

Please note: for finished heating devices (convectors, plate radiators, etc.), manufacturers usually indicate the total heat output in the accompanying documentation.

• The diameter of the pipeline capable of providing the necessary heat flow in the case of water heating.
• Parameters circulation pump, which sets in motion the coolant in the circuit with the given parameters.
• Size expansion tank, which compensates for the thermal expansion of the coolant.

Let's move on to formulas.

One of the main factors affecting its value is the degree of insulation of the house. SNiP 23-02-2003, which regulates the thermal protection of buildings, normalizes this factor, deriving the recommended values ​​​​of thermal resistance of enclosing structures for each region of the country.

We will give two ways to perform calculations: for buildings that comply with SNiP 23-02-2003, and for houses with non-standardized thermal resistance.

Normalized thermal resistance

The instruction for calculating the thermal power in this case looks like this:

• The base value is 60 watts per 1 m3 of the total (including walls) volume of the house.
• For each of the windows, an additional 100 watts of heat is added to this value.. For each door leading to the street - 200 watts.

• An additional coefficient is used to compensate for losses that increase in cold regions.

Let's, as an example, perform a calculation for a house measuring 12 * 12 * 6 meters with twelve windows and two doors to the street, located in Sevastopol (the average temperature in January is + 3C).

1. The heated volume is 12*12*6=864 cubic meters.
2. The basic thermal power is 864*60=51840 watts.
3. Windows and doors will slightly increase it: 51840+(12*100)+(2*200)=53440.
4. The exceptionally mild climate due to the proximity of the sea will force us to use a regional factor of 0.7. 53440 * 0.7 = 37408 W. It is on this value that you can focus.

Unrated thermal resistance

What to do if the quality of home insulation is noticeably better or worse than recommended? In this case, to estimate the heat load, you can use a formula like Q=V*Dt*K/860.

In it:

• Q is the cherished thermal power in kilowatts.
• V - heated volume in cubic meters.
• Dt is the temperature difference between the street and the house. Usually, a delta is taken between the value recommended by SNiP for interior spaces(+18 - +22С) and the average minimum of outdoor temperature in the coldest month over the past few years.

Let us clarify: it is in principle more correct to count on an absolute minimum; however, this will mean excessive costs for the boiler and heating appliances, the full capacity of which will be required only once every few years. The price of a slight underestimation of the calculated parameters is a slight drop in the temperature in the room at the peak of cold weather, which is easy to compensate by turning on additional heaters.

• K is the insulation coefficient, which can be taken from the table below. Intermediate coefficient values ​​are derived by approximation.

Let's repeat the calculations for our house in Sevastopol, specifying that its walls are 40 cm thick masonry of shell rock (porous sedimentary rock) without exterior finish, and the glazing is made of single-chamber double-glazed windows.

1. We take the coefficient of insulation equal to 1.2.
2. We calculated the volume of the house earlier; it is equal to 864 m3.
3. We will take the internal temperature equal to the recommended SNiP for regions with a lower peak temperature above -31C - +18 degrees. Information about the average minimum will kindly be prompted by the world-famous Internet encyclopedia: it is equal to -0.4C.
4. The calculation, therefore, will look like Q \u003d 864 * (18 - -0.4) * 1.2 / 860 \u003d 22.2 kW.

As you can easily see, the calculation gave a result that differs from that obtained by the first algorithm by one and a half times. The reason, first of all, is that the average minimum used by us differs markedly from the absolute minimum (about -25C). An increase in the temperature delta by one and a half times will increase the estimated heat demand of the building by exactly the same number of times.

gigacalories

In calculating the amount of thermal energy received by a building or room, along with kilowatt-hours, another value is used - gigacalorie. It corresponds to the amount of heat required to heat 1000 tons of water by 1 degree at a pressure of 1 atmosphere.

How to convert kilowatts of thermal power into gigacalories of heat consumed? It's simple: one gigacalorie is equal to 1162.2 kWh. Thus, with a peak power of a heat source of 54 kW, the maximum hourly heating load will be 54/1162.2=0.046 Gcal*h.

Useful: for each region of the country, local authorities normalize heat consumption in gigacalories per square meter of area during the month. The average value for the Russian Federation is 0.0342 Gcal/m2 per month.

Room

How to calculate the heat demand for a separate room? The same calculation schemes are used here as for the house as a whole, with a single amendment. If a heated room without its own heating devices adjoins the room, it is included in the calculation.

So, if a corridor measuring 1.2 * 4 * 3 meters adjoins a room measuring 4 * 5 * 3 meters, the heat output of the heater is calculated for a volume of 4 * 5 * 3 + 1.2 * 4 * 3 \u003d 60 + 14, 4=74.4 m3.

Heating appliances

Sectional radiators

In the general case, information on the heat flux per section can always be found on the manufacturer's website.

If it is unknown, you can focus on the following approximate values:

• Cast iron section - 160 watts.
• Bimetal section - 180 W.
• Aluminum section - 200W.

As always, there are a number of subtleties. With a side connection of a radiator with 10 or more sections, the temperature spread between the closest to the inlet and end sections will be very significant.

However: the effect will be nullified if the eyeliners are connected diagonally or from the bottom down.

In addition, usually manufacturers of heating devices indicate the power for a very specific temperature delta between the radiator and the air, equal to 70 degrees. The dependence of the heat flux on Dt is linear: if the battery is 35 degrees hotter than the air, the thermal power of the battery will be exactly half the declared one.

Let's say, at an air temperature in the room equal to + 20C, and a coolant temperature of + 55C, the power of the aluminum section standard size will be equal to 200/(70/35)=100 watts. In order to provide a power of 2 kW, you need 2000/100=20 sections.

Registers

Self-made registers stand apart in the list of heating devices.

In the photo - the heating register.

Manufacturers, for obvious reasons, cannot specify their heat output; however, it is easy to calculate it yourself.

• For the first section of the register (a horizontal pipe of known dimensions), the power is equal to the product of its outer diameter and length in meters, the temperature delta between the coolant and air in degrees, and a constant coefficient of 36.5356.
• For subsequent sections located in the upward flow of warm air, an additional factor of 0.9 is used.

Let's take another example - calculate the value of the heat flux for a four-row register with a section diameter of 159 mm, a length of 4 meters and a temperature of 60 degrees in a room with an internal temperature of + 20C.

1. The temperature delta in our case is 60-20=40C.
2. Convert pipe diameter to meters. 159 mm = 0.159 m.
3. We calculate the thermal power of the first section. Q \u003d 0.159 * 4 * 40 * 36.5356 \u003d 929.46 watts.
4. For each subsequent section, the power will be equal to 929.46 * 0.9 = 836.5 watts.
5. The total power will be 929.46 + (836.5 * 3) \u003d 3500 (rounded) watts.

Pipeline diameter

How to determine the minimum value of the inner diameter of the filling pipe or supply pipe to the heater? Let's not get into the jungle and use a table containing ready-made results for the difference between supply and return of 20 degrees. This value is typical for autonomous systems.

The maximum flow rate of the coolant should not exceed 1.5 m/s to avoid noise; more often they are guided by a speed of 1 m / s.

Inner diameter, mm	Thermal power of the circuit, W at flow rate, m/s
	0,6	0,8	1
8	2450	3270	4090
10	3830	5110	6390
12	5520	7360	9200
15	8620	11500	14370
20	15330	20440	25550
25	23950	31935	39920
32	39240	52320	65400
40	61315	81750	102190
50	95800	127735	168670

Say, for a 20 kW boiler, the minimum inner diameter filling at a flow rate of 0.8 m / s will be equal to 20 mm.

Please note: the inner diameter is close to DN (nominal diameter). Plastic and metal-plastic pipes are usually marked with an outer diameter that is 6-10 mm larger than the inner one. So, a polypropylene pipe with a size of 26 mm has an inner diameter of 20 mm.

Circulation pump

Two parameters of the pump are important to us: its pressure and performance. In a private house, for any reasonable length of the circuit, the minimum pressure of 2 meters (0.2 kgf / cm2) for the cheapest pumps is quite sufficient: it is this value of the differential that circulates the heating system of apartment buildings.

The required performance is calculated by the formula G=Q/(1.163*Dt).

In it:

• G - productivity (m3 / h).
• Q is the power of the circuit in which the pump is installed (KW).
• Dt is the temperature difference between the direct and return pipelines in degrees (in an autonomous system, Dt = 20С is typical).

For a circuit with a thermal load of 20 kilowatts, at a standard temperature delta, the calculated capacity will be 20 / (1.163 * 20) \u003d 0.86 m3 / h.

Expansion tank

One of the parameters that needs to be calculated for an autonomous system is the volume of the expansion tank.

The exact calculation is based on a rather long series of parameters:

• Temperature and type of coolant. The expansion coefficient depends not only on the degree of heating of the batteries, but also on what they are filled with: water-glycol mixtures expand more.
• The maximum working pressure in the system.
• Tank charging pressure, which in turn depends on the hydrostatic pressure of the circuit (the height of the top point of the circuit above the expansion tank).

There is, however, one caveat that greatly simplifies the calculation. If understating the volume of the tank will lead to best case to permanent operation safety valve, and at worst - to the destruction of the circuit, then its excess volume will not hurt anything.

That is why a tank with a displacement equal to 1/10 of the total amount of coolant in the system is usually taken.

Hint: to find out the volume of the contour, it is enough to fill it with water and pour it into a measuring dish.

Conclusion

We hope that the above calculation schemes will simplify the life of the reader and save him from many problems. As usual, the video attached to the article will offer additional information to his attention.