Determination of thermal loads for heating. Thermal calculation of the heating system

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 is 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 the excess thermal power will only affect the mode of operation of heating: say, 5 kWh of heat will be given off in more than one hour continuous work with a power of 5 kW, and in 50 minutes 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 is connected with the operation of classic solid fuel boilers and is 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 boiler power or the total power of appliances in a distributed heating system.
• The need for warmth private room.
• The number of sections of the sectional radiator and the size of the register corresponding to a certain value of 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.
• Options circulation pump, which sets in motion the coolant in the circuit with the given parameters.
• The size of the expansion tank that 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:

• Behind base value 60 watts are taken 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 heat source power of 54 kW, the maximum hourly load for heating will be 54/1162.2=0.046 Gcal*h.

Useful: for each region of the country, local authorities standardize heat consumption in gigacalories per square meter 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. At lateral connection for 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. Addiction heat flow from 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 value.

Let's say, when the air temperature in the room is +20C, and the coolant temperature is +55C, the power of an aluminum section of a standard size will be 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 ( horizontal pipe known dimensions) 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 upstream sections warm air, an additional coefficient 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 minimum value internal diameter of the filling pipe or supply pipe 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.

Let's say for a boiler with a power of 20 kW, the minimum inner diameter of the filling at a flow rate of 0.8 m / s will be 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, polypropylene pipe size 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 autonomous system- 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 hydrostatic pressure contour (height top point circuit above the expansion tank).

There is, however, one caveat that greatly simplifies the calculation. If understating the volume of the tank will at best lead to a 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.

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.

Feature structural elements buildings. These are windows, walls, doors, roof and ventilation system.

Housing dimensions. The larger it is, the more powerful the heating system should be. Be sure to take into account the area of ​​window openings, doors, exterior walls and the volume of each interior space.

Availability of rooms special purpose(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, this can be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in 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). Bimetallic radiators with a distance between the axes of 500 mm, on average, 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. The minimum temperature in the cold season is -20 ° 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 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.

When closed loop 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. Lamps daylight and the 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, paying special attention to 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.

First and most milestone in the difficult process of organizing heating of any property (whether Vacation home or an industrial facility) is the competent execution 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.

Performing preliminary calculations is necessary not only in order 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 generators.

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 facility. 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.

Total heat load on modern system heating 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 possible outdoor fences(walls, floors, roofs), sizes of openings (balconies, loggias, doors and windows). 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 you need 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 of maintenance- the presence of hot water supply, such as district heating, ventilation and air conditioning systems;
• The total number of points from which hot water is drawn. It is on this characteristic that special attention should be paid, because what more number points - the greater the heat 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 an 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 even at the design stage of a country cottage or other real estate object - this is due to simplicity and the absence of extra cash costs. At the same time, the requirements of various norms and standards, TCP, SNB and GOST are taken into account.

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 already mentioned earlier, the design parameters of 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 flux from heating devices (all actually located in the building heating batteries), the maximum hourly consumption of heat energy, as well as total costs heat output 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 was built (applied when the design temperature is different from -30C); q0 specific characteristic heating, 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 when selecting equipment), a large number of various thermal loads are taken into account:

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 building's external enclosures, 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 bills, 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. - the calculated temperature of hot and cold water, the density of water, as well as the coefficient in which the values ​​\u200b\u200bare taken into account maximum load hot water supply to the average value established by GOST;

Comprehensive calculation of thermal loads

Except, in fact, theoretical issues 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 during the passage of a certain strictly a certain amount heat 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 building 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

The calculation of thermal loads, as well as, is an important factor, the calculations of which must be made 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.

Before proceeding with the purchase of materials and installation of heat supply systems for a house or apartment, it is necessary to calculate the heating based on the area of ​​\u200b\u200beach room. Basic parameters for heating design and heat load calculation:

• Square;
• Number of window blocks;
• Ceiling height;
• The location of the room;
• Heat loss;
• Heat dissipation of radiators;
• Climatic zone (outside temperature).

The method described below is used to calculate the number of batteries for a room area without additional heating sources (heat-insulated floors, air conditioners, etc.). There are two ways to calculate heating: using a simple and complicated formula.

Before starting the design of heat supply, it is worth deciding which radiators will be installed. The material from which the heating batteries are made:

• Cast iron;
• Steel;
• Aluminum;
• Bimetal.

Aluminum and bimetallic radiators are considered the best option. The highest thermal output of bimetallic devices. Cast iron batteries they heat up for a long time, but after turning off the heating, the temperature in the room lasts for quite a long time.

A simple formula for designing the number of sections in a heating radiator is:

K = Sx(100/R), where:

S is the area of ​​the room;

R - section power.

If we consider the example with data: room 4 x 5 m, bimetal radiator, power 180 watts. The calculation will look like this:

K = 20*(100/180) = 11.11. So, for a room with an area of ​​20 m 2, a battery with at least 11 sections is required for installation. Or, for example, 2 radiators with 5 and 6 ribs. The formula is used for rooms with a ceiling height of up to 2.5 m in a standard Soviet-built building.

However, such a calculation of the heating system does not take into account the heat loss of the building, the outdoor temperature of the house and the number of window blocks are also not taken into account. Therefore, these coefficients should also be taken into account for the final refinement of the number of ribs.

Calculations for panel radiators

In the case where the installation of a battery with a panel instead of ribs is supposed, the following formula by volume is used:

W \u003d 41xV, where W is the battery power, V is the volume of the room. The number 41 is the norm of the average annual heating capacity of 1 m 2 of a dwelling.

As an example, we can take a room with an area of ​​​​20 m 2 and a height of 2.5 m. The value of the radiator power for a room volume of 50 m 3 will be 2050 W, or 2 kW.

Heat loss calculation

The main heat loss occurs through the walls of the room. To calculate, you need to know the coefficient of thermal conductivity of the external and internal material, from which the house is built, the thickness of the wall of the building, the average outdoor temperature is also important. Basic formula:

Q \u003d S x ΔT / R, where

ΔT is the temperature difference between the outside and the internal optimum value;

S is the area of ​​the walls;

R is the thermal resistance of the walls, which, in turn, is calculated by the formula:

R = B/K, where B is the thickness of the brick, K is the coefficient of thermal conductivity.

Calculation example: the house is built of shell rock, in stone, located in the Samara region. The thermal conductivity of the shell rock is on average 0.5 W / m * K, the wall thickness is 0.4 m. Considering the average range, minimum temperature-30 °C in winter. In the house, according to SNIP, normal temperature is +25 °C, the difference is 55 °C.

If the room is angular, then both of its walls are in direct contact with environment. The area of ​​the outer two walls of the room is 4x5 m and 2.5 m high: 4x2.5 + 5x2.5 = 22.5 m 2.

R = 0.4/0.5 = 0.8

Q \u003d 22.5 * 55 / 0.8 \u003d 1546 W.

In addition, it is necessary to take into account the insulation of the walls of the room. When finishing with foam plastic of the outer area, heat loss is reduced by about 30%. So, the final figure will be about 1000 watts.

Heat Load Calculation (Advanced Formula)

Scheme of heat loss of premises

To calculate the final heat consumption for heating, it is necessary to take into account all the coefficients according to the following formula:

CT \u003d 100xSxK1xK2xK3xK4xK5xK6xK7, where:

S is the area of ​​the room;

K - various coefficients:

K1 - loads for windows (depending on the number of double-glazed windows);

K2 - thermal insulation of the outer walls of the building;

K3 - loads for the ratio of window area to floor area;

K4 – outdoor air temperature regime;

K5 - taking into account the number of external walls of the room;

K6 - loads, based on the upper room above the calculated room;

K7 - taking into account the height of the room.

As an example, we can consider the same room of a building in the Samara region, insulated from the outside with foam plastic, having 1 window with double glazing above which the heated room is located. The heat load formula will look like this:

KT \u003d 100 * 20 * 1.27 * 1 * 0.8 * 1.5 * 1.2 * 0.8 * 1 \u003d 2926 W.

The calculation of heating is focused on this figure.

Heat consumption for heating: formula and adjustments

Based on the above calculations, 2926 watts are needed to heat a room. Considering heat losses, the requirements are: 2926 + 1000 = 3926 W (KT2). The following formula is used to calculate the number of sections:

K = KT2/R, where KT2 is the final value of the heat load, R is the heat transfer (power) of one section. Final figure:

K = 3926/180 = 21.8 (rounded 22)

So, in order to ensure optimal heat consumption for heating, it is necessary to install radiators with a total of 22 sections. It should be borne in mind that the lowest temperature - 30 degrees below zero in time is a maximum of 2-3 weeks, so you can safely reduce the number to 17 sections (- 25%).

If homeowners are not satisfied with such an indicator of the number of radiators, then batteries with a large heat supply capacity should be taken into account initially. Or insulate the walls of the building both inside and outside modern materials. In addition, it is necessary to correctly assess the needs of housing for heat, based on secondary parameters.

There are several other parameters that affect the additional energy wasted, which entails an increase in heat loss:

1. Features of the outer walls. Heating energy should be enough not only for heating the room, but also to compensate for heat losses. The wall in contact with the environment, over time, from changes in the temperature of the outside air, begins to let moisture in. Especially should be well insulated and held high-quality waterproofing for northern directions. It is also recommended to insulate the surface of houses located in humid regions. Tall annual level precipitation will inevitably lead to an increase in heat loss.
2. Place of installation of radiators. If the battery is mounted under a window, then heating energy leaks through its structure. The installation of high-quality blocks will help reduce heat loss. You also need to calculate the power of the device installed in the window sill - it should be higher.
3. Conventional annual heat demand for buildings in different time zones. As a rule, according to SNIPs, the average temperature is calculated (averaged annual rate) for buildings. However, heat demand is significantly lower if, for example, cold weather and low outdoor air values ​​occur for a total of 1 month of the year.

Advice! In order to minimize the need for heat in winter, it is recommended to install additional sources of indoor air heating: air conditioners, mobile heaters, etc.

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

The heat demand of the building

AT different buildings an unequal amount of thermal energy is required to make a person feel comfortable.

Among the factors affecting the need for heat, the following can be distinguished:

Appliance distribution

When it comes to water heating, maximum power source of heat energy should be equal to the sum of the capacities of all heat sources in the building.

The distribution of appliances in the premises of the house depends on the following circumstances:

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

1. Living quarters in the depth of the building - 20 degrees.
2. Living quarters in the corner and end parts of the building - 22 degrees.
3. Kitchen - 18 degrees. AT kitchen area the temperature is higher, since it contains additional heat sources ( electric stove, refrigerator, etc.).
4. Bathroom and toilet - 25 degrees.

If the house is equipped air heating, the amount of heat flow entering the room depends on the capacity of the air sleeve. Flow adjustable by manual adjustment ventilation grilles and controlled by a thermometer.

The house can be heated by distributed sources of thermal energy: electric or gas convectors, electric heated floors, oil batteries, infrared heaters, air conditioners. In this case, the desired temperatures are determined by the thermostat setting. In this case, it is necessary to provide for such equipment power, which would be sufficient for maximum level heat losses.

Calculation methods

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

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

First calculation option

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

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

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

The second calculation option

The method above suffers from significant errors:

1. The height of the ceilings is not taken into account, but you need to heat not square meters, but volume.
2. Lost through window and doorways more heat than through walls.
3. The type of building is not taken into account - this is an apartment building, where there are heated apartments behind the walls, ceiling and floor or this private house where there is only cold air behind the walls.

Correcting the calculation:

1. As a base applicable next indicator- 40 watts per cubic meter.
2. We will provide 200 W for each door, and 100 W for windows.
3. For apartments in the corner and end parts of the house, we use a coefficient of 1.3. Whether it's the highest or lowest floor apartment building, we use a coefficient of 1.3, and for a private building - 1.5.
4. We also apply the climate coefficient again.

Climate coefficient table

We make a calculation:

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

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

The third calculation option

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

Percentage Heat Loss Chart

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

• Q - heat loss per square meter of the building envelope;
• DT - delta between outside and inside temperatures;
• R is the resistance level for heat transfer.

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

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

• a wall of 3 bricks - the resistance level is 0.592 per square meter. m×S/W;
• wall in 2 bricks - 0.406;
• wall in 1 brick - 0.188;
• a log house from a 25-centimeter beam - 0.805;
• log house from a 12-centimeter beam - 0.353;
• frame material with mineral wool insulation - 0.702;
• wood floor - 1.84;
• ceiling or attic - 1.45;
• wooden double door - 0.22.

1. The temperature delta is 50 degrees (20 degrees of heat indoors and 30 degrees of frost outside).
2. Heat loss per square meter of floor: 50 / 1.84 (data for wood floors) = 27.17 W. Losses over the entire floor area: 27.17 × 144 = 3912 W.
3. Heat loss through the ceiling: (50 / 1.45) × 144 = 4965 W.
4. We calculate the area of ​​​​four walls: (12 × 3) × 4 \u003d 144 square meters. m. Since the walls are made of 25-centimeter timber, R is equal to 0.805. Heat loss: (50/0.805)×144=8944 W.
5. Add up the results: 3912+4965+8944=17821. The resulting number is the total heat loss of the house without taking into account the features of losses through windows and doors.
6. Add 40% ventilation losses: 17821×1.4=24.949. Thus, you will need a 25 kW boiler.

findings

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

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

Solid fuel boilers are regulated by a restrictor for the flow of air into the combustion chamber, however, if the oxygen level is insufficient, the fuel does not completely burn out. This leads to the formation a large number ash and reduce efficiency. You can correct the situation with a heat accumulator. The tank with thermal insulation is installed between the supply and return pipes, opening them. Thus, a small circuit (boiler - buffer tank) and a large circuit (tank - heaters) are created.

The scheme functions as follows:

1. After filling the fuel, the equipment operates at rated power. Through natural or forced circulation, heat is transferred to the buffer. After the combustion of the fuel, the circulation in the small circuit stops.
2. During the following hours, the heat carrier circulates along the large circuit. The buffer slowly transfers heat to radiators or underfloor heating.

Increased power will require additional costs. At the same time, the power reserve of the equipment gives an important positive result: the interval between fuel loads is significantly increased.