Design and thermal calculation of a heating system - mandatory stage when arranging home heating. The main task of computing activities is to determine optimal parameters boiler and radiator system.

Agree, at first glance it may seem that carrying out thermotechnical calculation Only an engineer can do it. However, not everything is so complicated. Knowing the algorithm of actions, you will be able to independently perform the necessary calculations.

The article describes in detail the calculation procedure and provides all the necessary formulas. For a better understanding, we have prepared an example of a thermal calculation for a private home.

Classic thermal calculation heating system is a consolidated technical document that includes mandatory step-by-step standard calculation methods.

But before studying these calculations of the main parameters, you need to decide on the concept of the heating system itself.

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The heating system is characterized by forced supply and involuntary removal of heat into the room.

The main tasks of calculating and designing a heating system:

• most reliably determine heat losses;
• determine the amount and conditions of use of the coolant;
• select the elements of generation, movement and heat transfer as accurately as possible.

And here room temperature air in winter period provided by a heating system. Therefore, we are interested in temperature ranges and their deviation tolerances for the winter season.

In the majority regulatory documents The following temperature ranges are specified that allow a person to stay comfortably in the room.

For non-residential premises office type with an area of ​​up to 100 m2:

• 22-24°С — optimal temperature air;
• 1°С— permissible fluctuation.

For office-type premises with an area of ​​more than 100 m2, the temperature is 21-23°C. For non-residential industrial premises, temperature ranges vary greatly depending on the purpose of the room and established standards labor protection.

Each person has their own comfortable room temperature. Some people like it to be very warm in the room, others feel comfortable when the room is cool - it’s all quite individual

As for residential premises: apartments, private houses, estates, etc., there are certain temperature ranges that can be adjusted depending on the wishes of the residents.

And yet, for specific premises of an apartment and house we have:

• 20-22°С- living room, including children's room, tolerance ±2°С -
• 19-21°С— kitchen, toilet, tolerance ±2°С;
• 24-26°С— bathroom, shower, swimming pool, tolerance ±1°С;
• 16-18°С— corridors, hallways, staircases, storage rooms, tolerance +3°C

It is important to note that there are several more basic parameters that affect the temperature in the room and which you need to focus on when calculating the heating system: humidity (40-60%), oxygen concentration and carbon dioxide in the air (250:1), the speed of movement of air masses (0.13-0.25 m/s), etc.

Calculation of heat loss in the house

According to the second law of thermodynamics (school physics), there is no spontaneous transfer of energy from less heated to more heated mini- or macro-objects. A special case of this law is the “striving” to create temperature equilibrium between two thermodynamic systems.

For example, the first system is an environment with a temperature of -20°C, the second system is a building with an internal temperature of +20°C. According to the above law, these two systems will strive to balance through the exchange of energy. This will happen with the help of heat losses from the second system and cooling in the first.

We can definitely say that the ambient temperature depends on the latitude at which it is located. a private house. And the temperature difference affects the amount of heat leakage from the building (+)

Heat loss refers to the involuntary release of heat (energy) from some object (house, apartment). For ordinary apartment this process is not so “noticeable” in comparison with a private house, since the apartment is located inside the building and “adjacent” to other apartments.

In a private house through external walls, floor, roof, windows and doors, heat “leaks” to one degree or another.

Knowing the amount of heat loss for the most unfavorable weather conditions and the characteristics of these conditions, it is possible to calculate the power of the heating system with high accuracy.

So, the volume of heat leakage from the building is calculated using the following formula:

Q=Q floor +Q wall +Q window +Q roof +Q door +…+Q i, Where

Qi— the volume of heat loss from a homogeneous type of building envelope.

Each component of the formula is calculated using the formula:

Q=S*∆T/R, Where

• Q– heat leakage, V;
• S– area of ​​a specific type of structure, sq. m;
• ∆T– difference in ambient and indoor air temperatures, °C;
• R– thermal resistance of a certain type of structure, m 2 *°C/W.

The very value of thermal resistance for real existing materials It is recommended to take it from the auxiliary tables.

Additionally, the thermal resistance can be obtained using the following relationship:

R=d/k, Where

• R– thermal resistance, (m 2 *K)/W;
• k– coefficient of thermal conductivity of the material, W/(m 2 *K);
• d– thickness of this material, m.

In old houses with damp roofing structures, heat leaks occur through the upper part of the building, namely through the roof and attic. Carrying out activities to solve this problem.

If you insulate the attic space and roof, then total losses heat from the house can be significantly reduced

There are several other types of heat loss in a house through cracks in structures, ventilation systems, kitchen hood, opening windows and doors. But it makes no sense to take into account their volume, since they constitute no more than 5% of the total number of main heat leaks.

Determination of boiler power

To support temperature differences between environment and the temperature inside the house requires an autonomous heating system that maintains the desired temperature in each room of a private house.

The basis of the heating system is different: liquid or solid fuel, electric or gas.

A boiler is the central unit of a heating system that generates heat. The main characteristic of a boiler is its power, namely the rate of conversion of the amount of heat per unit of time.

After calculating the heating load, we obtain the required rated power of the boiler.

For an ordinary multi-room apartment, the boiler power is calculated through the area and specific power:

Boiler P = (room S * specific P)/10, Where

• S premises— total area of ​​the heated room;
• R specific— power density regarding climatic conditions.

But this formula does not take into account heat losses, which are sufficient in a private house.

There is another ratio that takes this parameter into account:

Boiler P =(Q losses *S)/100, Where

• Boiler P— boiler power;
• Q losses— heat loss;
• S- heated area.

The design power of the boiler needs to be increased. The reserve is necessary if you plan to use the boiler to heat water for the bathroom and kitchen.

In most heating systems of private houses, it is recommended to use an expansion tank in which the coolant supply will be stored. Every private home needs hot water supply

In order to provide for the boiler power reserve, the safety factor K must be added to the last formula:

Boiler P = (Q losses * S * K)/100, Where

TO- will be equal to 1.25, that is design power the boiler will be increased by 25%.

Thus, the boiler power makes it possible to maintain standard temperature air in the rooms of the building, as well as have an initial and additional volume hot water in the house.

Features of the selection of radiators

Standard components for providing heat in a room are radiators, panels, underfloor heating systems, convectors, etc. The most common parts of a heating system are radiators.

The thermal radiator is a special hollow modular-type structure made of an alloy with high heat dissipation. It is made of steel, aluminum, cast iron, ceramics and other alloys. The principle of operation of a heating radiator is reduced to the radiation of energy from the coolant into the space of the room through the “petals”.

An aluminum and bimetallic heating radiator replaced massive cast iron batteries. Simplicity of production, high heat transfer, successful design and design have made this product a popular and widespread instrument for radiating heat indoors

There are several techniques in the room. The list of methods below is sorted in order of increasing calculation accuracy.

Calculation options:

1. By area. N=(S*100)/C, where N is the number of sections, S is the area of ​​the room (m 2), C is the heat transfer of one section of the radiator (W, taken from the data sheet or certificate for the product), 100 W is the number heat flow which is necessary to heat 1 m2 (empirical value). The question arises: how to take into account the height of the ceiling of the room?
2. By volume. N=(S*H*41)/C, where N, S, C are similar. H is the height of the room, 41 W is the amount of heat flow that is necessary to heat 1 m 3 (empirical value).
3. By odds. N=(100*S*k1*k2*k3*k4*k5*k6*k7)/C, where N, S, C and 100 are the same. k1 - accounting for the number of chambers in a double-glazed window of a room, k2 - thermal insulation of walls, k3 - ratio of window area to room area, k4 - average subzero temperature in the coldest week of winter, k5 - the number of outer walls of the room (which “extend” to the street), k6 - the type of room on top, k7 - the height of the ceiling.

This is the most accurate option for calculating the number of sections. Naturally, fractional calculation results are always rounded to the next integer.

Hydraulic calculation of water supply

Of course, the “picture” of calculating heat for heating cannot be complete without calculating such characteristics as the volume and speed of the coolant. In most cases, the coolant is ordinary water in a liquid or gaseous aggregate state.

It is recommended to calculate the actual volume of coolant by summing all cavities in the heating system. When using a single-circuit boiler, this is best option. When using double-circuit boilers in a heating system, it is necessary to take into account the consumption of hot water for hygienic and other domestic purposes

Calculation of the volume of water heated double-circuit boiler to provide residents hot water and heating of the coolant is carried out by summing the internal volume of the heating circuit and the actual needs of users for heated water.

The volume of hot water in the heating system is calculated by the formula:

W=k*P, Where

• W— volume of coolant;
• P— heating boiler power;
• k- power factor (number of liters per unit of power, equal to 13.5, range - 10-15 liters).

As a result, the final formula looks like this:

W = 13.5*P

Coolant velocity is the final dynamic assessment of the heating system, which characterizes the speed of fluid circulation in the system.

This value helps to evaluate the type and diameter of the pipeline:

V=(0.86*P*μ)/∆T, Where

• P— boiler power;
• μ — boiler efficiency;
• ∆T- temperature difference between supply water and return water.

Using the above methods, it will be possible to obtain real parameters that are the “foundation” of the future heating system.

Example of thermal calculation

As an example of a thermal calculation, we have an ordinary 1-story house with four living rooms, a kitchen, a bathroom, “ winter Garden"and utility rooms.

The foundation is made of a monolithic reinforced concrete slab (20 cm), the external walls are concrete (25 cm) with plaster, the roof is made of wooden beams, roofing - metal tiles and mineral wool (10 cm)

Let us designate the initial parameters of the house necessary for the calculations.

Building dimensions:

• floor height - 3 m;
• small window on the front and rear of the building 1470*1420 mm;
• large facade window 2080*1420 mm;
• entrance doors 2000*900 mm;
• rear doors (exit to the terrace) 2000*1400 (700 + 700) mm.

The total width of the building is 9.5 m2, length 16 m2. Only living rooms (4 units), a bathroom and a kitchen will be heated.

To accurately calculate heat loss on the walls, you need to subtract the area of ​​all windows and doors from the area of ​​the external walls - this is a completely different type of material with its own thermal resistance

We start by calculating the areas of homogeneous materials:

• floor area - 152 m2;
• roof area - 180 m2, taking into account the attic height of 1.3 m and the width of the purlin - 4 m;
• window area - 3*1.47*1.42+2.08*1.42=9.22 m2;
• door area - 2*0.9+2*2*1.4=7.4 m2.

The area of ​​the external walls will be equal to 51*3-9.22-7.4=136.38 m2.

Let's move on to calculating heat loss for each material:

• Q floor =S*∆T*k/d=152*20*0.2/1.7=357.65 W;
• Q roof =180*40*0.1/0.05=14400 W;
• Q window =9.22*40*0.36/0.5=265.54 W;
• Q door =7.4*40*0.15/0.75=59.2 W;

And also Q wall is equivalent to 136.38*40*0.25/0.3=4546. The sum of all heat losses will be 19628.4 W.

As a result, we calculate the boiler power: P boiler =Q losses *S heating_rooms *K/100=19628.4*(10.4+10.4+13.5+27.9+14.1+7.4)*1.25/100=19628.4*83.7*1.25/100=20536.2=21 kW.

We will calculate the number of radiator sections for one of the rooms. For all others, the calculations are similar. For example, a corner room (on the left, lower corner of the diagram) has an area of ​​10.4 m2.

This means N=(100*k1*k2*k3*k4*k5*k6*k7)/C=(100*10.4*1.0*1.0*0.9*1.3*1.2*1.0*1.05)/180=8.5176=9.

This room requires 9 sections of heating radiator with a heat output of 180 W.

Let's move on to calculating the amount of coolant in the system - W=13.5*P=13.5*21=283.5 l. This means that the coolant speed will be: V=(0.86*P*μ)/∆T=(0.86*21000*0.9)/20=812.7 l.

As a result, a complete turnover of the entire volume of coolant in the system will be equivalent to 2.87 times per hour.

A selection of articles on thermal calculation will help you determine the exact parameters of the heating system elements:

Conclusions and useful video on the topic

A simple calculation of a heating system for a private home is presented in the following review:

All the subtleties and generally accepted methods for calculating the heat loss of a building are shown below:

Another option for calculating heat leaks in a typical private house:

This video describes the features of the circulation of energy carriers for heating a home:

Thermal calculation of a heating system is individual in nature and must be performed competently and carefully. The more accurately the calculations are made, the less the owners will have to overpay country house during operation.

Do you have experience performing thermal calculations of a heating system? Or still have questions on the topic? Please share your opinion and leave comments. Block feedback located below.

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

Heat demand of the building

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

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

Device distribution

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

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

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

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

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

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

Calculation methods

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

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

First calculation option

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

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

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

Second calculation option

The method given above suffers from significant errors:

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

We correct the calculation:

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

Climate coefficient table

We make the calculation:

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

So, based on the second calculation, it is clear that the difference with the first calculation method is almost twofold. It should be understood that such power is needed only during the lowest temperatures. In other words, peak power can be provided by additional heating sources, for example, a backup heater.

Third calculation option

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

Heat loss percentage diagram

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

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

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

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

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

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

conclusions

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

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

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

The circuit works as follows:

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

Increased power will require additional costs. At the same time, the equipment’s power reserve gives an important positive result: the interval between fuel loadings increases significantly.

The first and most important stage in the difficult process of organizing heating of any property (be it Vacation home or industrial facility) is the competent execution of design and calculations. In particular, it is necessary to calculate the thermal load on the heating system, as well as the volume of heat and fuel consumption.

Carrying out 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, and 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 facility. It should be noted that before installing all the equipment, this calculation is made to eliminate any troubles, unnecessary financial costs and work.

Calculating the heat load on heating will help organize the uninterrupted and efficient operation of the heating system of the property. Thanks to this calculation, you can quickly complete absolutely all heat supply tasks and ensure their compliance with the standards and requirements of SNiP.

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

Total heat load per modern system heating system consists of several main load parameters:

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

Main characteristics of the object that are important to take into account when calculating the heat load

The most correct and competent calculation of the heat load for heating will be determined only if absolutely everything is taken into account, even the most small parts and parameters.

This list is quite large and can include:

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

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

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

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

• Other data. For an industrial facility, such factors include, for example, the number of shifts, the number of workers per shift, as well as 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

The calculation of the heating load itself is done with your own hands at the design stage country cottage or another piece of real estate - this is due to the simplicity and lack of extra cash costs. This takes into account the requirements various standards and standards, TKP, SNB and GOST.

The following factors are required to be determined during the calculation of thermal power:

• Heat loss from external enclosures. Includes the desired temperature conditions in each room;
• Power required to heat water in the room;
• The amount of heat required to heat the air ventilation (in the case where forced forced ventilation is required);
• Heat needed to heat water in a swimming pool or sauna;

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

Advice. Thermal loads are calculated with a “margin” in order to eliminate the possibility of unnecessary financial costs. This is especially true for a country house, where additional connection of heating elements without preliminary design and preparation will be prohibitively expensive.

Features of calculating thermal load

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

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

The above instructions for calculating thermal loads taking into account the heat exchange surface area 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 the use of effective heating, as well as energy inspection of houses and buildings.

An ideal method of calculation for emergency heating of an industrial facility, when it is assumed that temperatures will decrease 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 loss using aggregated indicators;
2. Defining parameters via various elements enclosing structures, additional losses due to air heating;
3. Calculation of the heat transfer of all heating and ventilation equipment installed in the building.

Enlarged method for calculating heating loads

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

For a larger calculation of the heating heat load, a fairly simple and uncomplicated formula is used:

Qmax from.=α*V*q0*(tв-tн.р.)*10 -6

The formula uses the following coefficients: α is correction factor, taking into account the climatic conditions in the region where the building is 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-day week”); V – external volume of the building.

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

When performing calculations (as well as when selecting equipment), a large number of different thermal loads are taken into account:

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

• Throughout the year, heat loads change depending on the air temperature outside the room;
• Annual heat costs, which are determined by the meteorological characteristics of the region where the object for which the heat loads are calculated is located;

• Changes in 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 consumption ventilation system by the hour of the day.

1. Year-round heat 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 little. For example, in summer, thermal energy consumption is reduced by almost 30-35% compared to winter;
2. Dry heat– convection heat exchange and thermal radiation from other similar devices. Determined by dry bulb temperature.

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

1. Latent heat– evaporation and condensation. Relies on wet bulb temperature. The volume of latent heat of humidity and its sources in the room is determined.

In any room, humidity is influenced 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.

Regulators of thermal loads 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. Equipment in this category is designed to provide support for a certain level of loads and eliminate all kinds of surges and dips.

It should be noted that RTN allows you to significantly save on heating costs, because in many cases (and especially for industrial enterprises) certain limits are set that cannot be exceeded. Otherwise, if surges 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 yourself, then it is best to entrust it to specialists. At the same time, all the formulas are simple and uncomplicated, and therefore it is not so difficult to calculate all the parameters yourself.

Ventilation and hot water loads are one of the factors in thermal systems

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

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

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

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

Qgws.=0.042rv(tg.-tx.)Pgav, Where

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

Comprehensive calculation of thermal loads

In addition to the theoretical calculation issues themselves, some practical work is also carried out. For example, comprehensive thermal inspections include mandatory thermography of all structures - walls, ceilings, doors and windows. It should be noted that such work makes it possible to identify and record factors that have a significant impact on the heat loss of a building.

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

Practical measurements are an indispensable component of various calculation works. Taken together, such processes will help obtain the most reliable data on thermal loads and heat losses that will be observed in a certain structure over a certain period of time. Practical calculation will help to achieve what theory will not show, namely the “bottlenecks” of each structure.

Conclusion

Calculation of thermal loads, likewise, is an important factor, the calculations of which must be carried out before starting to organize a heating system. If all the work is done correctly and you approach the process wisely, you can guarantee trouble-free heating operation, as well as save money on overheating and other unnecessary costs.

The topic of this article is thermal load. We will find out what this parameter is, what it depends on and how it can be calculated. In addition, the article will provide a number of reference values ​​for thermal resistance different materials, which may be needed for calculations.

What it is

The term is essentially intuitive. Thermal load means the amount of thermal energy that is necessary to maintain a building, apartment or separate room comfortable temperature.

The maximum hourly heating load, therefore, is the amount of heat that may be required to maintain normalized parameters for an hour under the most unfavorable conditions.

Factors

So, what influences a building's heat demand?

• Wall material and thickness. It is clear that a wall of 1 brick (25 centimeters) and a wall made of aerated concrete under a 15-centimeter foam coat will transmit VERY different amounts of thermal energy.
• Roof material and structure. Flat roof from reinforced concrete slabs and an insulated attic will also differ very noticeably in heat loss.
• Ventilation is another important factor. Its performance and the presence or absence of a heat recovery system affect how much heat is lost in the exhaust air.
• Glazing area. Through the windows and glass facades noticeably more heat is lost than through solid walls.

However: triple glazed windows and glass with energy-saving coating reduce the difference several times.

• Insolation level in your region, absorption rate solar heat external covering and orientation of the building planes relative to the cardinal directions. Extreme cases are a house that is in the shadow of other buildings all day long and a house oriented with a black wall and a black sloping roof with a maximum area facing south.

• Temperature delta between indoors and outdoors determines the heat flow through the enclosing structures at constant resistance to heat transfer. At +5 and -30 outside, the house will lose different amounts of heat. This will, of course, reduce the need for thermal energy and reduce the temperature inside the building.
• Finally, it is often necessary to include in a project prospects for further construction. Let’s say, if the current heat load is 15 kilowatts, but in the near future it is planned to add an insulated veranda to the house, it is logical to purchase one with a reserve of heat power.

Distribution

In the case of water heating, the peak thermal power of the heat source must be equal to the sum of the thermal power of all heating devices in the house. Of course, wiring should not become a bottleneck either.

The distribution of heating devices throughout the premises is determined by several factors:

1. The area of ​​the room and the height of its ceiling;
2. Location inside the building. Corner and end rooms lose more heat than those located in the middle of the house.
3. Remoteness from the heat source. In individual construction, this parameter means distance from the boiler, in a central heating system apartment building- whether the battery is connected to the supply or return riser and what floor you live on.

Clarification: in houses with bottom filling, the risers are connected in pairs. On the supply side, the temperature decreases as you rise from the first floor to the last; on the return side, the opposite is true.

It’s also not difficult to guess how the temperatures will be distributed in the case of top filling.

1. Desired room temperature. In addition to heat filtration through external walls, inside the building, with uneven temperature distribution, migration of thermal energy through partitions will also be noticeable.

1. For living rooms in the middle of the building - 20 degrees;
2. For living rooms in the corner or end of the house - 22 degrees. More heat, among other things, prevents walls from freezing.
3. For the kitchen - 18 degrees. As a rule, it has a large number of its own heat sources - from a refrigerator to an electric stove.
4. For a bathroom and a combined toilet, the norm is 25C.

When air heating the heat flow entering a separate room is determined throughput air sleeve. Usually, simplest method adjustments - manual adjustment of the positions of adjustable ventilation grilles with temperature control using a thermometer.

Finally, if we are talking about a heating system with distributed heat sources (electric or gas convectors, electric heated floors, infrared heaters and air conditioners) necessary temperature regime simply set on the thermostat. All that is required of you is to provide the peak thermal power of the devices at the level of the peak heat loss of the room.

Calculation methods

Dear reader, do you have a good imagination? Let's imagine a house. Let it be a log house made of 20-centimeter timber with an attic and a wooden floor.

Let’s mentally complete and concretize the picture that has arisen in our heads: the dimensions of the residential part of the building will be equal to 10*10*3 meters; in the walls we will cut 8 windows and 2 doors - to the front and courtyards. Now let’s place our house... say, in the city of Kondopoga in Karelia, where the temperature at the peak of frost can drop to -30 degrees.

Determining the heat load for heating can be done in several ways with varying complexity and reliability of the results. Let's use the three simplest ones.

Method 1

Current SNiPs offer us the simplest method of calculation. One kilowatt of thermal power is taken per 10 m2. The resulting value is multiplied by the regional coefficient:

• For the southern regions (Black Sea coast, Krasnodar region) the result is multiplied by 0.7 - 0.9.
• The moderately cold climate of the Moscow and Leningrad regions will force the use of a coefficient of 1.2-1.3. It seems that our Kondopoga will fall into this particular climate group.
• Finally, for Far East regions of the Far North, the coefficient ranges from 1.5 for Novosibirsk to 2.0 for Oymyakon.

The instructions for calculating using this method are incredibly simple:

1. The area of ​​the house is 10*10=100 m2.
2. The basic value of the thermal load is 100/10=10 kW.
3. We multiply by the regional coefficient of 1.3 and get 13 kilowatts of thermal power necessary to maintain comfort in the house.

However: if you use such a simple technique, it is better to make a reserve of at least 20% to compensate for errors and extreme cold. Actually, it will be indicative to compare 13 kW with values ​​​​obtained by other methods.

Method 2

It is clear that with the first calculation method the errors will be huge:

• Ceiling heights vary greatly between buildings. Taking into account the fact that we have to heat not an area, but a certain volume, and with convection heating warm air going under the ceiling is an important factor.
• Windows and doors let in more heat than walls.
• Finally, it would be a clear mistake to cut hair with one brush city ​​apartment(and regardless of its location inside the building) and a private house, which has no warm apartments neighbors, and the street.

Well, let's adjust the method.

• Let's take 40 watts per cubic meter of room volume as the base value.
• For each door leading to the street, add 200 watts to the base value. For each window - 100.
• For corner and end apartments in apartment building Let's introduce a coefficient of 1.2 - 1.3 depending on the thickness and material of the walls. We also use it for the outermost floors if the basement and attic are poorly insulated. For a private house, we will multiply the value by 1.5.
• Finally, we apply the same regional coefficients as in the previous case.

How is our house in Karelia doing?

1. The volume is 10*10*3=300 m2.
2. The basic value of thermal power is 300*40=12000 watts.
3. Eight windows and two doors. 12000+(8*100)+(2*200)=13200 watts.
4. A private house. 13200*1.5=19800. We begin to vaguely suspect that when selecting the boiler power using the first method, we would have to freeze.
5. But there is still a regional coefficient left! 19800*1.3=25740. Total - we need a 28-kilowatt boiler. Difference from the first value obtained in a simple way- double.

However: in practice, such power will be required only on a few days of peak frost. Often, a reasonable solution would be to limit the power of the main heat source to a lower value and buy a backup heater (for example, an electric boiler or several gas convectors).

Method 3

Make no mistake: the described method is also very imperfect. We very roughly took into account the thermal resistance of the walls and ceiling; The temperature delta between internal and external air is also taken into account only in the regional coefficient, that is, very approximately. The price of simplifying calculations is a large error.

Let us remember: to maintain a constant temperature inside the building, we need to provide an amount of thermal energy equal to all losses through the building envelope and ventilation. Alas, here too we will have to somewhat simplify our calculations, sacrificing the reliability of the data. Otherwise, the resulting formulas will have to take into account too many factors that are difficult to measure and systematize.

The simplified formula looks like this: Q=DT/R, ​​where Q is the amount of heat that is lost by 1 m2 of the building envelope; DT - temperature delta between internal and outside temperatures, and R is the heat transfer resistance.

Please note: we are talking about heat loss through the walls, floor and ceiling. On average, another 40% of heat is lost through ventilation. To simplify the calculations, we will calculate the heat loss through the enclosing structures, and then simply multiply them by 1.4.

Temperature delta is easy to measure, but where do you get thermal resistance data?

Alas, only from reference books. Here is a table for some popular solutions.

• A wall of three bricks (79 centimeters) has a heat transfer resistance of 0.592 m2*C/W.
• A wall of 2.5 bricks is 0.502.
• Wall with two bricks - 0.405.
• Brick wall (25 centimeters) - 0.187.
• A log house with a log diameter of 25 centimeters is 0.550.
• The same, but from logs with a diameter of 20 cm - 0.440.
• Log house made of 20 cm timber - 0.806.
• Log frame made of timber 10 cm thick - 0.353.
• Frame wall 20 centimeters thick with insulation mineral wool — 0,703.
• A wall made of foam or aerated concrete with a thickness of 20 centimeters is 0.476.
• The same, but with a thickness increased to 30 cm - 0.709.
• Plaster 3 centimeters thick - 0.035.
• Ceiling or attic floor - 1.43.
• Wooden floor - 1.85.
• Double door made of wood - 0.21.

Now let's go back to our house. What parameters do we have?

• The temperature delta at the peak of frost will be equal to 50 degrees (+20 inside and -30 outside).
• Heat loss through a square meter of floor will be 50/1.85 (heat transfer resistance of a wooden floor) = 27.03 watts. Across the entire floor - 27.03*100=2703 watts.
• Let's calculate the heat loss through the ceiling: (50/1.43)*100=3497 watts.
• The area of ​​the walls is (10*3)*4=120 m2. Since our walls are made of 20-centimeter timber, the R parameter is 0.806. Heat loss through the walls is equal to (50/0.806)*120=7444 watts.
• Now let’s add up the resulting values: 2703+3497+7444=13644. This is exactly how much our house will lose through the ceiling, floor and walls.

Please note: in order not to calculate fractions of square meters, we neglected the difference in thermal conductivity of walls and windows with doors.

• Then we add 40% of losses for ventilation. 13644*1.4=19101. According to this calculation, a 20-kilowatt boiler should be enough for us.

Conclusions and problem solving

As you can see, the available methods for calculating the thermal load with your own hands give very significant errors. Fortunately, excess boiler power won't hurt:

• Gas boilers operate at reduced power with virtually no drop in efficiency, while condensing boilers even reach the most economical mode at partial load.
• The same applies to solar boilers.
• Electric heating equipment of any type always has an efficiency of 100 percent (of course, this does not apply to heat pumps). Remember physics: all power not spent on mechanical work (that is, moving mass against the gravity vector) is ultimately spent on heating.

The only type of boilers for which operation at a power less than rated is contraindicated is solid fuel. The power control in them is carried out in a rather primitive way - by limiting the flow of air into the firebox.

What is the result?

1. If there is a lack of oxygen, the fuel does not burn completely. More ash and soot are produced, which pollute the boiler, chimney and atmosphere.
2. The consequence of incomplete combustion is a drop in boiler efficiency. It’s logical: after all, fuel often leaves the boiler before it burns.

However, here too there is a simple and elegant way out - including a heat accumulator in the heating circuit. A thermally insulated tank with a capacity of up to 3000 liters is connected between the supply and return pipelines, disconnecting them; in this case a small circuit is formed (between the boiler and buffer capacity) and large (between the tank and heating devices).

How does this scheme work?

• After lighting, the boiler operates at rated power. At the same time, due to natural or forced circulation, its heat exchanger transfers heat to the buffer tank. After the fuel has burned out, circulation in the small circuit stops.
• For the next few hours, the coolant moves along a large circuit. The buffer tank gradually releases the accumulated heat to radiators or water-heated floors.

Conclusion

As always, you will find some additional information on how else the heat load can be calculated in the video at the end of the article. Warm winters!

In houses that have been put into operation in recent years, these rules are usually met, so the calculation of the heating power of the equipment is based on standard coefficients. Individual calculations can be carried out at the initiative of the homeowner or the utility structure involved in the supply of heat. This happens when heating radiators, windows and other parameters are spontaneously replaced.

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

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

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

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

Basic principles

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

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

The main heat losses of a private house:

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

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

Specifics of calculations

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

Qt = V × ∆T × K / 860, where

V – room volume, m³;

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

K – estimated heat loss coefficient;

860 – conversion factor in kW/hour.

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

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

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

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