Determination of the calculated hourly loads of heating, forced ventilation and hot water supply calculated heat loads. Calculation of Gcal for heating Calculation of heating load

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 losses of thermal energy leaving through building structures (walls, floors, roofs);
• 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 not difficult to find it in any reference literature, and for plastic windows this coefficient will be prompted by the manufacturers. Below is a table with the coefficients of thermal conductivity of some building materials, and for calculations it is necessary to take the operational values ​​of λ.

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

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

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

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

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

Ventilation air heating consumption

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

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

Qvent = cmΔt, here:

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

The complexity of calculating this type of heat load lies in the correct determination of the mass of heated air. Find out how much it gets inside the house, when natural ventilation difficult. Therefore, it is worth referring to the standards, because buildings are built according to projects where required air exchanges. And the regulations say that in most rooms air environment should be changed once 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, knowing its density at different temperatures from the table:

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

Heat load from DHW heating

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

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

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

The unit of equipment power is always referred to 1 hour, and the resulting 5.2 kW - to the day. But 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 traditional way on the area, although you have to work hard. Final result it is necessary to multiply by the safety factor - 1.2, or even 1.4 and select 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:

The procedure for calculating heating in the housing stock depends on the availability of metering devices and on how the house is equipped with them. There are several options for completing multi-apartment residential buildings with meters, and according to which, heat energy is calculated:

1. the presence of a common house meter, while apartments and non-residential premises are not equipped with metering devices.
2. heating costs are controlled by a common house device, and all or some rooms are equipped with metering devices.
3. there is no general house device for fixing the consumption and consumption of thermal energy.

Before calculating the number of gigacalories spent, it is necessary to find out the presence or absence of controllers in the house and in each individual room, including non-residential ones. Let's consider all three options for calculating thermal energy, for each of which a specific formula has been developed (posted on the website of state authorized bodies).

Option 1

So, the house is equipped with a control device, and some rooms were left without it. Here it is necessary to take into account two positions: the calculation of Gcal for heating an apartment, the cost of thermal energy for general house needs (ODN).

In this case, formula No. 3 is used, which is based on the readings of the general meter, the area of ​​\u200b\u200bthe house and the footage of the apartment.

Calculation example

We will assume that the controller recorded the heating costs of the house at 300 Gcal / month (this information can be obtained from the receipt or by contacting management company). For example, the total area of ​​the house, which consists of the sum of the areas of all premises (residential and non-residential), is 8000 m² (you can also find this figure from the receipt or from the management company).

Let's take the area of ​​​​an apartment of 70 m² (indicated in the data sheet, rental agreement or registration certificate). last digit, on which the calculation of payment for consumed heat energy depends, this is the tariff established by the authorized bodies of the Russian Federation (indicated in the receipt or found out in the house management company). Today, the heating tariff is 1,400 rubles/gcal.

Substituting the data in formula No. 3, we get the following result: 300 x 70 / 8,000 x 1,400 \u003d 1875 rubles.

Now you can proceed to the second stage of accounting for heating costs spent on the general needs of the house. Two formulas are required here: the search for the volume of services (No. 14) and the payment for the consumption of gigacalories in rubles (No. 10).

In order to correctly determine the volume of heating in this case, it will be necessary to sum up the area of ​​\u200b\u200ball apartments and premises provided for common use(information provided by the management company).

For example, we have a total footage of 7000 m² (including apartments, offices, retail premises.).

Let's start calculating the payment for the consumption of thermal energy according to formula No. 14: 300 x (1 - 7,000 / 8,000) x 70 / 7,000 \u003d 0.375 Gcal.

Using formula No. 10, we get: 0.375 x 1,400 = 525, where:

• 0.375 - volume of service for heat supply;
• 1400 r. – tariff;
• 525 rubles - amount of payment.

We summarize the results (1875 + 525) and find out that the payment for heat consumption will be 2350 rubles.

Option 2

Now we will calculate payments in those conditions when the house is equipped with a common meter for heating, as well as some apartments are equipped with individual meters. As in the previous case, the calculation will be carried out in two positions (thermal energy consumption for housing and ONE).

We will need formulas No. 1 and No. 2 (accrual rules according to the testimony of the controller or taking into account the norms for heat consumption for residential premises in gcal). Calculations will be carried out in relation to the area of ​​​​a residential building and an apartment from the previous version.

• 1.3 gigacalories - readings of an individual counter;
• 1 1820 r. - approved rate.

• 0.025 gcal - standard indicator of heat consumption per 1 m² of area in an apartment;
• 70 m² - area of ​​the apartment;
• 1 400 rubles - tariff for thermal energy.

As it becomes clear, with this option, the payment amount will depend on the availability of a metering device in your apartment.

Formula No. 13: (300 - 12 - 7,000 x 0.025 - 9 - 30) x 75 / 8,000 \u003d 1.425 gcal, where:

• 300 gcal - indications of a common house meter;
• 12 gcal - the amount of thermal energy used for heating non-residential premises;
• 6,000 m² - the sum of the area of ​​​​all residential premises;
• 0.025 - standard (thermal energy consumption for apartments);
• 9 gcal - the sum of indicators from the meters of all apartments that are equipped with metering devices;
• 35 gcal - the amount of heat spent on supply hot water in the absence of its centralized supply;
• 70 m² - area of ​​the apartment;
• 8,000 m² - total area (all residential and non-residential premises in the house).

note that this option includes only real amounts of energy consumed and if your house is equipped with a centralized hot water supply, then the amount of heat spent on hot water needs is not taken into account. The same applies to non-residential premises: if they are not in the house, then they will not be included in the calculation.

• 1.425 gcal - the amount of heat (ONE);

1. 1820 + 1995 = 3,815 rubles - with individual counter.
2. 2 450 + 1995 = 4445 rubles. - without individual device.

Option 3

We have left last option, during which we will consider the situation when there is no heat energy meter on the house. The calculation, as in previous cases, will be carried out in two categories (thermal energy consumption for an apartment and ONE).

We will deduce the amount for heating using formulas No. 1 and No. 2 (rules on the procedure for calculating thermal energy, taking into account the readings of individual metering devices or in accordance with the established standards for residential premises in gcal).

Formula No. 1: 1.3 x 1,400 \u003d 1820 rubles, where:

• 1.3 gcal - readings of an individual meter;
• 1 400 rubles - approved rate.

Formula No. 2: 0.025 x 70 x 1,400 = 2,450 rubles, where:

• 1 400 rubles - approved rate.

As in the second option, the payment will depend on whether your housing is equipped with an individual heat meter. Now it is necessary to find out the amount of heat energy that was spent on general house needs, and this must be done according to formula No. 15 (volume of service for one unit) and No. 10 (amount for heating).

Formula No. 15: 0.025 x 150 x 70 / 7000 \u003d 0.0375 gcal, where:

• 0.025 gcal - standard indicator of heat consumption per 1 m² of living space;
• 100 m² - the sum of the area of ​​\u200b\u200bthe premises intended for general house needs;
• 70 m² - the total area of ​​the apartment;
• 7,000 m² - total area (all residential and non-residential premises).

Formula No. 10: 0.0375 x 1,400 = 52.5 rubles, where:

• 0.0375 - volume of heat (ONE);
• 1400 r. - approved rate.

As a result of the calculations, we found out that the full payment for heating will be:

1. 1820 + 52.5 \u003d 1872.5 rubles. - with individual counter.
2. 2450 + 52.5 \u003d 2,502.5 rubles. – without individual counter.

In the above calculations of payments for heating, data on the footage of the apartment, house, as well as on the meter indicators, which may differ significantly from those that you have, were used. All you need to do is plug your values ​​into the formula and make the final calculation.

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

Why do you need to know this parameter

What is the calculation of the heat load for heating? He defines optimal amount thermal energy for each room and the building as a whole. Variables are power heating equipment– boiler, radiators and pipelines. Also taken into account heat loss Houses.

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 loss;
• House dimensions. It is logical to assume that more 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.

For calculation optimal load for heating 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 you calculate the heating load using aggregated indicators or with higher accuracy, you need 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, the daily shift must be taken into account outdoor temperature. To improve the accuracy of the calculation, you need to know specifications building.

Easy Ways to Calculate Heat Load

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

The dependence of heating power on the area

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

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

15*1=15 kWh

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

Enlarged calculation of the thermal load of the building

The enlarged calculation of the heating load is characterized by more accurate results. It was originally used for preliminary calculation this parameter when it is 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, A- 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). Optimum temperature inside the dwelling (Tvn) should be + 22 ° С. The outside temperature will be -15°C. Let's use the formula to calculate the hourly heating load:

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

Compared to the previous calculation, the resulting value is less. However, it takes into account important factors - the temperature inside the room, on the street, the total volume of the building. Similar calculations can be made for each room. The method of calculating the load on heating according to aggregated indicators makes it possible to determine the optimal power for each radiator in a single room. For more exact calculation you need to know the average temperatures for 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. He doesn't take into account the most important parameter- 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. how more value thermal conductivity - the higher the heat loss. This value cannot be used to calculate the annual heating load, since it does not take into account the thickness of the material ( d). Therefore, experts use the heat transfer resistance parameter, which is calculated by the following formula:

Calculation for walls and windows

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

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

• Wall area - 280 m². It includes windows 40 m²;
• Wall material - solid brick (λ=0.56). The thickness of the outer walls 0.36 m. Based on this, we calculate the TV transmission resistance - R=0.36/0.56= 0.64 m²*S/W;
• To improve the thermal insulation properties, 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 very a good indicator thermal insulation of the house;
• Heat transfer resistance of windows - 0.75 m²*S/W(double glazing with argon filling).

In fact, heat losses through the walls will be:

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

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

124*(22+15)= 4.96 kWh

Ventilation calculation

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

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

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

4.96+1.11=6.07 kWh

In this way, the exact maximum heating load is determined. The resulting value directly depends on the temperature outside. Therefore, to calculate the annual load on the heating system, it is necessary to take into account changes in 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, the building must be installed two-pipe system heating.

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

First and most milestone in the difficult process of organizing the heating of any real estate object (whether it be a country house or an industrial facility) is the competent implementation of design and calculation. In particular, it is necessary to calculate thermal loads on the heating system, as well as the volume of heat and fuel consumption.

Performance preliminary calculation is necessary not only to obtain the entire range of documentation for organizing the heating of a property, but also to understand the volumes of fuel and heat, the selection of one or another type of heat generator.

Thermal loads of the heating system: characteristics, definitions

The definition should be understood as the amount of heat that is collectively given off by heating devices installed in a house or other object. It should be noted that before installing all the equipment, this calculation is made to exclude any troubles, unnecessary financial costs and work.

Calculation of heat loads for heating will help to organize uninterrupted and efficient work real estate heating systems. Thanks to this calculation, you can quickly complete absolutely all the tasks of heat supply, ensure their compliance with the norms and requirements of SNiP.

The cost of an error in the calculation can be quite significant. The thing is that, depending on the calculated data received, the maximum expenditure parameters will be allocated in the housing and communal services department of the city, limits and other characteristics will be set, from which they are repelled when calculating the cost of services.

The total heat load on a modern heating system consists of several main load parameters:

• For a common central heating system;
• per system floor heating(if it is available in the house) - underfloor heating;
• Ventilation system (natural and forced);
• Hot water supply system;
• For all kinds of technological needs: swimming pools, baths and other similar structures.

The main characteristics of the object, important to take into account when calculating the heat load

The most correctly and competently calculated heat load on heating will be determined only when absolutely everything, even the smallest details and parameters, is taken into account.

This list is quite large and can include:

• Type and purpose of real estate objects. A residential or non-residential building, an apartment or an administrative building - all this is very important for obtaining reliable thermal calculation data.

Also, the load rate, which is determined by heat supplier companies and, accordingly, heating costs, depends on the type of building;

• Architectural part. The dimensions of all kinds of external fences (walls, floors, roofs), the dimensions of openings (balconies, loggias, doors and windows) are taken into account. The number of storeys of the building, the presence of basements, attics and their features are important;
• Temperature requirements for each of the premises of the building. This parameter should be understood as temperature regimes for each room of a residential building or zone of an administrative building;
• The design and features of external fences, including the type of materials, thickness, the presence of insulating layers;

• The nature of the premises. As a rule, it is inherent in industrial buildings, where for a workshop or site it is necessary to create some specific thermal conditions and modes;
• Availability and parameters of special premises. The presence of the same baths, pools and other similar structures;
• Degree Maintenance - the presence of hot water supply, such as central heating, ventilation and air conditioning systems;
• The total number of points from which hot water is drawn. It is this characteristic that should be addressed Special attention, because the greater the number of points, the greater will be the thermal load on the entire heating system as a whole;
• The number of people living in the house or located at the facility. The requirements for humidity and temperature depend on this - factors that are included in the formula for calculating the heat load;

• Other data. For industrial facility such factors include, for example, the number of shifts, the number of workers per shift, and working days per year.

As for a private house, you need to take into account the number of people living, the number of bathrooms, rooms, etc.

Calculation of heat loads: what is included in the process

Do-it-yourself calculation of the heating load itself is carried out at the design stage country cottage or another property - this is due to the simplicity and lack of extra cash costs. This takes into account the requirements various norms and standards, TKP, SNB and GOST.

The following factors are mandatory for determination during the calculation of thermal power:

• Heat losses of external protections. Includes the desired temperature conditions in each of the rooms;
• The power required to heat the water in the room;
• The amount of heat required to heat the air ventilation (in the case when forced ventilation is required);
• The heat needed to heat the water in the pool or bath;

• Possible developments of the further existence of the heating system. It implies the possibility of outputting heating to the attic, to the basement, as well as all kinds of buildings and extensions;

Advice. With a "margin", thermal loads are calculated in order to exclude the possibility of unnecessary financial costs. Especially relevant for country house, where additional connection of heating elements without preliminary study and preparation will be prohibitively expensive.

Features of calculating the heat load

As previously stated, design parameters indoor air are selected from the relevant literature. At the same time, heat transfer coefficients are selected from the same sources (passport data of heating units are also taken into account).

The traditional calculation of heat loads for heating requires a consistent determination of the maximum heat flow from heating devices (all heating batteries actually located in the building), the maximum hourly consumption of heat energy, as well as the total cost of heat power for a certain period, for example, the heating season.

The above instructions for calculating thermal loads, taking into account the surface area of ​​​​heat exchange, can be applied to various real estate objects. It should be noted that this method allows you to competently and most correctly develop a justification for using efficient heating as well as energy inspections of houses and buildings.

An ideal calculation method for the standby heating of an industrial facility, when temperatures are expected to drop during non-working hours (holidays and weekends are also taken into account).

Methods for determining thermal loads

Currently, thermal loads are calculated in several main ways:

1. Calculation of heat losses by means of enlarged indicators;
2. Determination of parameters via various elements enclosing structures, additional losses for air heating;
3. Calculation of heat transfer of all heating and ventilation equipment installed in the building.

Enlarged method for calculating heating loads

Another method for calculating the loads on the heating system is the so-called enlarged method. As a rule, such a scheme is used in the case when there is no information about projects or such data does not correspond to the actual characteristics.

For an enlarged calculation of the heat load of heating, a rather simple and uncomplicated formula is used:

Qmax from. \u003d α * V * q0 * (tv-tn.r.) * 10 -6

The following coefficients are used in the formula: α is a correction factor that takes into account the climatic conditions in the region where the building is built (used when the design temperature is different from -30C); q0 specific heating characteristic, selected depending on the temperature of the coldest week of the year (the so-called "five days"); V is the outer volume of the building.

Types of thermal loads to be taken into account in the calculation

In the course of calculations (as well as in the selection of equipment), it is taken into account a large number of a wide variety of thermal loads:

1. seasonal loads. As a rule, they have the following features:

• Throughout the year, there is a change in thermal loads depending on the air temperature outside the premises;
• Annual heat consumption, which is determined by the meteorological features of the region where the facility is located, for which heat loads are calculated;

• Changing the load on the heating system depending on the time of day. Due to the heat resistance of the external fences of the building, such values ​​are accepted as insignificant;
• Thermal energy costs ventilation system by hours of the day.

1. Year-round thermal loads. It should be noted that for heating and hot water supply systems, most domestic facilities have heat consumption throughout the year, which changes quite a bit. So, for example, in summer the cost of thermal energy in comparison with winter is reduced by almost 30-35%;
2. dry heat– convection heat exchange and thermal radiation from other similar devices. Determined by dry bulb temperature.

This factor depends on the mass of parameters, including all kinds of windows and doors, equipment, ventilation systems and even air exchange through cracks in the walls and ceilings. It also takes into account the number of people who can be in the room;

1. Latent heat- Evaporation and condensation. Based on wet bulb temperature. The amount of latent heat of humidity and its sources in the room is determined.

In any room, humidity is affected by:

• People and their number who are simultaneously in the room;
• Technological and other equipment;
• Air flows that pass through cracks and crevices in building structures.

Thermal load regulators as a way out of difficult situations

As you can see in many photos and videos of modern and other boiler equipment, special heat load regulators are included with them. The technique of this category is designed to provide support for a certain level of loads, to exclude all kinds of jumps and dips.

It should be noted that RTN can significantly save on heating costs, because in many cases (and especially for industrial enterprises) certain limits are set that cannot be exceeded. Otherwise, if jumps and excesses of thermal loads are recorded, fines and similar sanctions are possible.

Advice. Loads on heating, ventilation and air conditioning systems - important point in home design. If it is impossible to carry out the design work on your own, then it is best to entrust it to specialists. At the same time, all formulas are simple and uncomplicated, and therefore it is not so difficult to calculate all the parameters by yourself.

Loads on ventilation and hot water supply - one of the factors of thermal systems

Thermal loads for heating, as a rule, are calculated in combination with ventilation. This is a seasonal load, it is designed to replace the exhaust air with clean air, as well as heat it up to the set temperature.

Hourly heat consumption for ventilation systems is calculated according to a certain formula:

Qv.=qv.V(tn.-tv.), Where

In addition to, in fact, ventilation, thermal loads are also calculated on the hot water supply system. The reasons for such calculations are similar to ventilation, and the formula is somewhat similar:

Qgvs.=0.042rv(tg.-tkh.)Pgav, Where

r, in, tg., tx. is the design temperature of the hot and cold water, water density, as well as a coefficient that takes into account the values ​​​​of the maximum load of hot water supply to the average value established by GOST;

Comprehensive calculation of thermal loads

In addition to, in fact, theoretical issues of calculation, some practical work. So, for example, comprehensive thermal surveys include mandatory thermography of all structures - walls, ceilings, doors and windows. It should be noted that such works make it possible to determine and fix the factors that have a significant impact on the heat loss of the building.

Thermal imaging diagnostics will show what the real temperature difference will be when a certain strictly defined amount of heat passes through 1m2 of enclosing structures. Also, it will help to find out the heat consumption at a certain temperature difference.

Practical measurements are an indispensable component of various computational works. In combination, such processes will help to obtain the most reliable data on thermal loads and heat losses that will be observed in a particular structure over a certain period of time. A practical calculation will help to achieve what the theory does not show, namely the "bottlenecks" of each structure.

Conclusion

Calculation of thermal loads, as well as - important factor, which must be calculated before starting the organization of the heating system. If all the work is done correctly and the process is approached wisely, you can guarantee trouble-free operation of heating, as well as save money on overheating and other unnecessary costs.

Build a heating system own house or even in a city apartment - an extremely responsible occupation. At the same time, it would be completely unreasonable to purchase boiler equipment, as they say, “by eye”, that is, without taking into account all the features of housing. In this, it is quite possible to fall into two extremes: either the power of the boiler will not be enough - the equipment will work “to its fullest”, without pauses, but will not give the expected result, or, conversely, an overly expensive device will be purchased, the capabilities of which will remain completely unclaimed.

But that's not all. It is not enough to purchase the necessary heating boiler correctly - it is very important to optimally select and correctly place 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, certain calculations are indispensable.

Of course, ideally, such heat engineering calculations should be carried out by appropriate specialists, but this often costs a lot of money. Isn't it interesting to try to do it yourself? This publication will show in detail how heating is calculated by the area of ​​\u200b\u200bthe room, taking into account many important nuances. By analogy, it will be possible to perform, built into this page, will help you perform the necessary calculations. The technique cannot be called completely “sinless”, however, it still allows you to get a result with a completely acceptable degree of accuracy.

The simplest methods of calculation

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, and their separation is very conditional.

• The first is maintaining an optimal level of air temperature throughout the entire volume of the heated room. Of course, the temperature level may vary slightly with altitude, but this difference should not be significant. Quite comfortable conditions are considered to be an average of +20 ° C - it is this temperature that, as a rule, is taken as the initial temperature in thermal calculations.

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

If we approach with complete accuracy, then for individual rooms in residential buildings the 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 the compensation of heat losses through the structural elements of the building.

The main "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 go 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 to common needs buildings (apartments), but also 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 needs. And the values ​​​​for each room will be the starting point for the calculation required amount radiators.

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

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

Q = S× 100

Q- the required thermal power for the room;

S– area of ​​the room (m²);

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

For example, 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 should be noted right away that it is conditionally applicable only when standard height ceilings - approximately 2.7 m (permissible - 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 value of the specific power is calculated for cubic meter. It is taken equal to 41 W / m³ for a 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 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 - in the next section of the publication.

You may be interested in information about what they are

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

The calculation algorithms discussed above are useful for the initial “estimate”, but you should still rely on them completely with very great care. Even to a person who does not understand anything in building heat engineering, the indicated average values ​​\u200b\u200bmay certainly seem doubtful - they cannot be equal, say, for the Krasnodar Territory and for the Arkhangelsk Region. In addition, the room - the room is different: one is located on the corner of the house, that is, it has two external walls ki, and the other on three sides is protected from heat loss by other rooms. 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 - just such features are visible even to the "naked eye".

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

General principles and calculation formula

The calculations will be based on the same ratio: 100 W per 1 square meter. But that's just the formula itself "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 quite 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" - a coefficient that takes into account the number of external walls in a particular room.

Obviously, the more external walls in the room, the larger the area through which heat loss occurs. In addition, the presence of two or more external walls also means corners - extremely vulnerable places in terms of the formation of "cold bridges". The coefficient "a" will correct for this specific feature of the room.

The coefficient is taken equal to:

- external walls No(indoor): a = 0.8;

- outer wall one: a = 1.0;

- external walls two: a = 1.2;

- external walls three: a = 1.4.

• "b" - coefficient taking into account the location of the external walls of the room relative to the cardinal points.

You may be interested in information about what are

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

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

Based on this, we introduce the coefficient "b":

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

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

• "c" - coefficient taking into account the location of the room relative to the winter "wind rose"

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

Based on the results of long-term meteorological observations in any region, the so-called "wind rose" is compiled - graphic scheme, showing the prevailing wind directions in winter and summer time of the year. This information can be obtained from the local hydrometeorological service. However, many residents themselves, without meteorologists, know very well where the winds mainly blow from in winter, and from which side of the house the deepest snowdrifts usually sweep.

If there is a desire to carry out calculations with higher accuracy, then the correction factor “c” can also be included in the formula, taking it equal to:

- windward side of the house: c = 1.2;

- leeward walls of the house: c = 1.0;

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

• "d" - correction factor that takes into account the features climatic conditions home building region

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

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

So, the coefficient "d", taking into account the peculiarities of the climate of the region, for our calculations in we take equal to:

— from – 35 °С 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;

- not colder - 10 ° С: d=0.7.

• "e" - coefficient taking into account the degree of insulation of external walls.

The total value of the heat loss of the building is directly related to the degree of insulation of all building structures. One of the "leaders" in terms of 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 are not insulated: e = 1.27;

- medium degree of insulation - walls in two bricks or their surface thermal insulation with other heaters is provided: e = 1.0;

– insulation was carried out qualitatively, on the basis of heat engineering calculations: e = 0.85.

Later 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 height

Ceilings, especially in private homes, can have different heights. Therefore, the thermal power for heating one or another room of the same area will also differ in this parameter.

It will not be a big mistake to accept the following values ​​​​of the correction factor "f":

– ceiling height up to 2.7 m: f = 1.0;

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

– ceiling height from 3.1 to 3.5 m: f = 1.1;

– ceiling height from 3.6 to 4.0 m: f = 1.15;

– ceiling height over 4.1 m: f = 1.2.

• « g "- coefficient taking 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. So, it is necessary to make some adjustments in the calculation of this feature of a particular room. The correction factor "g" can be taken equal to:

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

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

- a heated room is located below: g= 1,0 .

• « h "- coefficient taking 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 losses are inevitable, which will require an increase in the required heat output. We introduce the coefficient "h", which also takes into account this feature of the calculated room:

- a "cold" attic is located on top: h = 1,0 ;

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

- any heated room is located above: h = 0,8 .

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

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

Without words, it is clear that the thermal insulation qualities of these windows are significantly different.

But even between PVC-windows there is no complete uniformity. 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 however they were, it will still not be possible to completely avoid heat loss through them. But it is quite clear that it is impossible to compare a small window with panoramic glazing almost on 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- the total area of ​​windows in the room;

SP- area of ​​the room.

Depending on the value obtained and the correction factor "j" is determined:

- x \u003d 0 ÷ 0.1 →j = 0,8 ;

- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;

- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;

- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;

- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;

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

The door to the street or to an unheated balcony is always an additional "loophole" for the cold

The door to the street or to an open balcony is able to make its own adjustments to the heat balance of the room - each of its opening is accompanied by the penetration of a considerable amount 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 balcony: k = 1,3 ;

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

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

Perhaps this will seem like an insignificant trifle to some, but still - why not immediately take into account the planned scheme for connecting heating radiators. The fact is that their heat transfer, and hence their participation in maintaining a certain temperature balance in the room, changes quite noticeably with different types tie-in 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 the bottom	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 features of the installation site of heating radiators

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

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

So, there is clarity with the calculation formula. Surely, some of the readers will immediately take up their heads - they say, it's too complicated and cumbersome. However, if the matter is approached systematically, in an orderly manner, then there is no difficulty at all.

Any good landlord must have a detailed graphic plan of their "possessions" with affixed dimensions, and usually oriented to the cardinal points. It is not difficult to specify the climatic features of the region. It remains only to walk through all the rooms with a tape measure, to clarify some of the nuances for each room. Features of housing - "neighborhood vertically" from above and below, location entrance doors, the proposed or already existing scheme for installing heating radiators - no one except the owners knows better.

It is recommended to immediately draw up a worksheet, where you enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will help to carry out the built-in calculator, in which all the coefficients and ratios mentioned above are already “laid”.

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

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

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

Let's create a table like this:

Then, using the calculator below, we make a calculation for each room (already taking into account a 10% reserve). With the recommended app, it won't take long. After that, it remains to sum the obtained values ​​\u200b\u200bfor 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 - it remains only to divide by specific thermal power one section and round up.