Calculation of heat consumption for space heating. Calculation of thermal loads for heating, methodology and calculation formula. Calculation of the specific annual consumption of thermal energy for hot water supply of apartment buildings for the regions

What is it - the specific consumption of thermal energy for heating a building? Is it possible to calculate the hourly heat consumption for heating in a cottage with your own hands? This article will be devoted to terminology and general principles calculation of the need for thermal energy.

The basis of new building projects is energy efficiency.

Terminology

What is specific heat consumption for heating?

We are talking about the amount of heat energy that must be brought inside the building in terms of each square or cubic meter to maintain normalized parameters in it, comfortable for work and living.

Usually, a preliminary calculation of heat losses is carried out according to enlarged meters, that is, based on the average thermal resistance walls, estimated temperature in the building and its total volume.

Factors

What affects the annual heat consumption for heating?

• Duration heating season (). It, in turn, is determined by the dates when the average daily temperature in the street for the last five days falls below (and rises above) 8 degrees Celsius.

Useful: in practice, when planning the start and stop of heating, the weather forecast is taken into account. Long thaws occur in winter, and frosts can strike as early as September.

• Average temperatures of the winter months. Usually when designing heating system the average monthly temperature of the coldest month, January, is taken as a guideline. It is clear that the colder it is outside, the more heat the building loses through the building envelope.

• The degree of thermal insulation of the building greatly affects what will be the rate of thermal power for him. An insulated facade can reduce the need for heat by half relative to a wall made of concrete slabs or brick.
• building glazing factor. Even when using multi-chamber double-glazed windows and energy-saving spraying, noticeably more heat is lost through windows than through walls. The greater part of the facade is glazed, the greater the need for heat.
• The degree of illumination of the building. On a sunny day, a surface oriented perpendicular to the sun's rays is capable of absorbing up to a kilowatt of heat per square meter.

Clarification: in practice, an accurate calculation of the amount of absorbed solar heat will be extremely difficult. Those same glass facades, which lose heat in cloudy weather, will serve as heating in sunny weather. The orientation of the building, the slope of the roof, and even the color of the walls will all affect the ability to absorb solar heat.

Calculations

Theory is theory, but how are the heating costs of a country house calculated in practice? Is it possible to estimate the estimated costs without plunging into the abyss of complex heat engineering formulas?

Consumption of the required amount of thermal energy

The instruction for calculating the approximate amount of heat required is relatively simple. The key phrase is an approximate amount: for the sake of simplifying calculations, we sacrifice accuracy, ignoring a number of factors.

• The base value of the amount of thermal energy is 40 watts per cubic meter of cottage volume.
• To the base value is added 100 watts for each window and 200 watts for each door in the outer walls.

• Further, the obtained value is multiplied by a coefficient, which is determined by the average amount of heat loss through the outer contour of the building. For apartments in the center apartment building a coefficient equal to one is taken: only losses through the facade are noticeable. Three of the four walls of the contour of the apartment border on warm rooms.

For corner and end apartments, a coefficient of 1.2 - 1.3 is taken, depending on the material of the walls. The reasons are obvious: two or even three walls become external.

Finally, in a private house, the street is not only along the perimeter, but also from below and above. In this case, a coefficient of 1.5 is applied.

Please note: for apartments on the extreme floors, if the basement and attic are not insulated, it is also quite logical to use a coefficient of 1.3 in the middle of the house and 1.4 at the end.

• Finally, the received thermal power is multiplied by a regional coefficient: 0.7 for Anapa or Krasnodar, 1.3 for St. Petersburg, 1.5 for Khabarovsk and 2.0 for Yakutia.

In a cold climate zone, there are special requirements for heating.

Let's calculate how much heat is needed for a cottage measuring 10x10x3 meters in the city of Komsomolsk-on-Amur, Khabarovsk Territory.

The volume of the building is 10*10*3=300 m3.

Multiplying the volume by 40 watts/cube will give 300*40=12000 watts.

Six windows and one door is another 6*100+200=800 watts. 1200+800=12800.

Private house. Coefficient 1.5. 12800*1.5=19200.

Khabarovsk region. We multiply the need for heat by another one and a half times: 19200 * 1.5 = 28800. In total - at the peak of frost, we need about a 30-kilowatt boiler.

Calculation of heating costs

The easiest way to calculate the consumption of electricity for heating: when using an electric boiler, it is exactly equal to the cost of thermal power. With continuous consumption of 30 kilowatts per hour, we will spend 30 * 4 rubles (approximate current price of a kilowatt-hour of electricity) = 120 rubles.

Fortunately, the reality is not so nightmarish: as practice shows, the average heat demand is about half the calculated one.

• Firewood - 0.4 kg / kW / h. Thus, the approximate norms for the consumption of firewood for heating in our case will be equal to 30/2 (the rated power, as we remember, can be divided in half) * 0.4 \u003d 6 kilograms per hour.
• Consumption brown coal in terms of kilowatt of heat - 0.2 kg. The consumption rates of coal for heating are calculated in our case as 30/2*0.2=3 kg/h.

Brown coal is a relatively inexpensive heat source.

• For firewood - 3 rubles (the cost of a kilogram) * 720 (hours in a month) * 6 (hourly consumption) \u003d 12960 rubles.
• For coal - 2 rubles * 720 * 3 = 4320 rubles (read others).

Conclusion

You can, as usual, find additional information on cost calculation methods in the video attached to the article. Warm winters!

What is specific heat consumption for heating? In what quantities is the specific consumption of thermal energy for heating a building measured and, most importantly, where do its values ​​\u200b\u200bare taken for calculations? In this article, we will get acquainted with one of the basic concepts of heat engineering, and at the same time study several related concepts. So, let's go.

What it is

Definition

The definition of specific heat consumption is given in SP 23-101-2000. According to the document, this is the name of the amount of heat needed to maintain the normal temperature in the building, referred to a unit of area or volume and to another parameter - degree-days of the heating period.

What is this setting used for? First of all - to assess the energy efficiency of the building (or, what is the same, the quality of its insulation) and planning heat costs.

Actually, SNiP 23-02-2003 explicitly states: the specific (per square or cubic meter) consumption of thermal energy for heating a building should not exceed the given values.
How better thermal insulation, the less energy is required for heating.

Degree day

At least one of the terms used needs clarification. What is a degree day?

This concept directly refers to the amount of heat required to maintain a comfortable climate inside a heated room in winter time. It is calculated by the formula GSOP=Dt*Z, where:

• GSOP is the desired value;
• Dt is the difference between the normalized internal temperature of the building (according to current SNiP it should be from +18 to +22 C) and average temperature the coldest five days of winter.
• Z is the length of the heating season (in days).

As you might guess, the value of the parameter is determined by the climatic zone and for the territory of Russia varies from 2000 (Crimea, Krasnodar region) up to 12000 (Chukotka Autonomous Okrug, Yakutia).

Units

In what quantities is the parameter of interest measured?

• SNiP 23-02-2003 uses kJ / (m2 * C * day) and, in parallel with the first value, kJ / (m3 * C * day).
• Along with the kilojoule, other units of heat can be used - kilocalories (Kcal), gigacalories (Gcal) and kilowatt hours (KWh).

How are they related?

• 1 gigacalorie = 1,000,000 kilocalories.
• 1 gigacalorie = 4184000 kilojoules.
• 1 gigacalorie = 1162.2222 kilowatt-hours.

In the photo - a heat meter. Heat metering devices can use any of the listed units of measurement.

Normalized parameters

For single-family one-story detached houses

For apartment buildings, hostels and hotels

Please note: with an increase in the number of floors, the heat consumption rate decreases.
The reason is simple and obvious: the larger the object, the simpler geometric shape, the greater the ratio of its volume to surface area.
For the same reason, the specific cost of heating a country house decreases with an increase in the heated area.

Computing

It is practically impossible to calculate the exact value of heat loss by an arbitrary building. However, methods of approximate calculations have long been developed, which give fairly accurate average results within the limits of statistics. These calculation schemes are often referred to as aggregated indicator (measurement) calculations.

Along with the thermal power, it often becomes necessary to calculate the daily, hourly, annual consumption of thermal energy or the average power consumption. How to do it? Let's give some examples.

The hourly heat consumption for heating according to enlarged meters is calculated by the formula Qot \u003d q * a * k * (tin-tno) * V, where:

• Qot - the desired value for kilocalories.
• q - specific heating value of the house in kcal / (m3 * C * hour). It is looked up in directories for each type of building.

• a - ventilation correction factor (usually equal to 1.05 - 1.1).
• k is the correction factor for the climatic zone (0.8 - 2.0 for different climatic zones).
• tvn - internal temperature in the room (+18 - +22 C).
• tno - street temperature.
• V is the volume of the building together with the enclosing structures.

To calculate the approximate annual heat consumption for heating in a building with specific consumption at 125 kJ / (m2 * C * day) and an area of ​​​​100 m2, located in a climatic zone with a parameter GSOP = 6000, you just need to multiply 125 by 100 (house area) and 6000 (degree-days of the heating period). 125*100*6000=75000000 kJ or about 18 gigacalories or 20800 kilowatt-hours.

To recalculate the annual consumption into the average heat consumption, it is enough to divide it by the length of the heating season in hours. If it lasts 200 days, the average heating power in the above case will be 20800/200/24=4.33 kW.

Energy carriers

How to calculate energy costs with your own hands, knowing the heat consumption?

enough to know calorific value appropriate fuel.

The easiest way to calculate the electricity consumption for heating a house: it is exactly equal to the amount of heat produced by direct heating.

So, the average in the last case considered by us will be equal to 4.33 kilowatts. If the price of a kilowatt-hour of heat is 3.6 rubles, then we will spend 4.33 * 3.6 = 15.6 rubles per hour, 15 * 6 * 24 = 374 rubles per day, and so on.

It is useful for owners of solid fuel boilers to know that the consumption rates for firewood for heating are about 0.4 kg / kWh. The norms of coal consumption for heating are half as much - 0.2 kg / kWh.

Thus, in order to calculate the average hourly consumption of firewood with your own hands with an average heating power of 4.33 kW, it is enough to multiply 4.33 by 0.4: 4.33 * 0.4 = 1.732 kg. The same instruction is valid for other coolants - you just need to get into the reference books.

Conclusion

We hope that our acquaintance with the new concept, even if somewhat superficial, could satisfy the reader's curiosity. The video attached to this material, as usual, will offer Additional information. Good luck!

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

But that's not all. It is not enough to purchase the necessary heating boiler correctly - it is very important to optimally select and correctly place heat exchange devices in the premises - radiators, convectors or "warm floors". And again, rely only on your intuition or " good advice» neighbors is not the most reasonable option. In a word, certain calculations are indispensable.

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

The simplest methods of calculation

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

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

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

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

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

The main "enemy" of the heating system is heat loss through building structures.

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

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

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

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

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

Q = S× 100

Q- the required thermal power for the room;

S– area of ​​the room (m²);

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

For example, room 3.2 × 5.5 m

S= 3.2 × 5.5 = 17.6 m²

Q= 17.6 × 100 = 1760 W ≈ 1.8 kW

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

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

Q = S × h× 41 (or 34)

h- ceiling height (m);

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

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

Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW

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

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

You may be interested in information about what they are

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

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

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

General principles and calculation formula

The calculations will be based on the same ratio: 100 W per 1 square meter. But that's just the formula itself "overgrown" with a considerable number of various correction factors.

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

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

• "a" - a coefficient that takes into account the number of external walls in a particular room.

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

The coefficient is taken equal to:

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

- outer wall one: a = 1.0;

- external walls two: a = 1.2;

- external walls three: a = 1.4.

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

You may be interested in information about what are

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

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

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

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

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

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

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

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

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

- windward side of the house: c = 1.2;

- leeward walls of the house: c = 1.0;

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

• "d" - a correction factor that takes into account the peculiarities of the climatic conditions of the region where the house was built

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

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

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

— from – 35 °С and below: d=1.5;

— from – 30 °С to – 34 °С: d=1.3;

— from – 25 °С to – 29 °С: d=1.2;

— from – 20 °С to – 24 °С: d=1.1;

— from – 15 °С to – 19 °С: d=1.0;

— from – 10 °С to – 14 °С: d=0.9;

- not colder - 10 ° С: d=0.7.

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

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

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

- external walls are not insulated: e = 1.27;

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

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

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

• coefficient "f" - correction for ceiling height

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

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

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

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

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

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

– ceiling height over 4.1 m: f = 1.2.

• « g "- coefficient taking into account the type of floor or room located under the ceiling.

As shown above, the floor is one of the significant sources of heat loss. So, it is necessary to make some adjustments in the calculation of this feature of a particular room. The correction factor "g" can be taken equal to:

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

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

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

• « h "- coefficient taking into account the type of room located above.

The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat losses are inevitable, which will require an increase in the required heat output. We introduce the coefficient "h", which takes into account this feature of the calculated room:

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

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

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

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

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

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

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

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

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

– modern window systems with single pane glass: i = 1,0 ;

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

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

No matter how high-quality the windows are, it will still not be possible to completely avoid heat loss through them. But it is quite clear that it is impossible to compare a small window with panoramic glazing almost on the entire wall.

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

x = ∑SOK /SP

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

SP- area of ​​the room.

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

- x \u003d 0 ÷ 0.1 →j = 0,8 ;

- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;

- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;

- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;

- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;

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

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

door to the street or outdoor balcony is able to make its own adjustments to the heat balance of the room - each of its opening is accompanied by the penetration of a considerable amount of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient "k", which we take equal to:

- no door k = 1,0 ;

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

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

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

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

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

• « m "- correction factor for the features of the installation site of heating radiators

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

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

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

Any good homeowner must have a detailed graphical plan of their "possessions" with affixed dimensions, and usually oriented to the cardinal points. Climatic features region is easy to define. It remains only to walk through all the rooms with a tape measure, to clarify some of the nuances for each room. Features of housing - "vertical neighborhood" from above and below, the location of the entrance doors, the proposed or existing scheme for installing heating radiators - no one except the owners knows better.

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

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

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

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

Let's create a table like this:

Then, using the calculator below, we make a calculation for each room (already taking into account a 10% reserve). With the recommended app, it won't take long. After that, it remains to sum the obtained values ​​\u200b\u200bfor each room - this will be the required total power of the heating system.

The result for each room, by the way, will help you choose the right number of heating radiators - it remains only to divide by the specific heat output of one section and round up.

The thermal calculation of the heating system seems to most to be an easy task that does not require special attention. A huge number of people believe that the same radiators should be chosen based only on the area of ​​\u200b\u200bthe room: 100 W per 1 sq. m. Everything is simple. But this is the biggest misconception. You cannot limit yourself to such a formula. What matters is the thickness of the walls, their height, material and much more. Of course, you need to allocate an hour or two to get the right numbers, but everyone can do it.

Initial data for designing a heating system

To calculate the heat consumption for heating, you need, firstly, a house project.

The plan of the house allows you to get almost all the initial data that is needed to determine the heat loss and the load on the heating system

Secondly, data on the location of the house in relation to the cardinal points and the construction area will be needed - the climatic conditions in each region are different, and what is suitable for Sochi cannot be applied to Anadyr.

Thirdly, we collect information about the composition and height of the outer walls and the materials from which the floor (from the room to the ground) and the ceiling (from the rooms and outward) are made.

After collecting all the data, you can get to work. Calculation of heat for heating can be performed using formulas in one to two hours. You can, of course, use a special program from Valtec.

To calculate the heat loss of heated rooms, the load on the heating system and heat transfer from heating appliances it is enough to enter only the initial data into the program. A huge number of functions make it indispensable assistant both foreman and private developer

It greatly simplifies everything and allows you to get all the data on heat losses and hydraulic calculation heating systems.

Formulas for calculations and reference data

The calculation of the heat load for heating involves the determination of heat losses (Tp) and boiler power (Mk). The latter is calculated by the formula:

Mk \u003d 1.2 * Tp, where:

• Mk - thermal performance of the heating system, kW;
• Tp - heat loss at home;
• 1.2 - safety factor (20%).

A 20% safety factor allows you to take into account the possible pressure drop in the gas pipeline during the cold season and unforeseen heat losses (for example, a broken window, poor-quality thermal insulation of entrance doors or unprecedented frosts). It allows you to insure against a number of troubles, and also makes it possible to widely regulate the temperature regime.

As can be seen from this formula, the power of the boiler directly depends on the heat loss. They are not evenly distributed throughout the house: the outer walls account for about 40% of the total value, the windows - 20%, the floor gives 10%, the roof 10%. The remaining 20% ​​disappear through the doors, ventilation.

Poorly insulated walls and floors, a cold attic, ordinary glazing on windows - all this leads to large heat losses, and, consequently, to an increase in the load on the heating system. When building a house, it is important to pay attention to all the elements, because even ill-conceived ventilation in the house will release heat into the street.

The materials from which the house is built have the most direct impact on the amount of heat lost. Therefore, when calculating, you need to analyze what the walls, and the floor, and everything else consist of.

In the calculations, to take into account the influence of each of these factors, the appropriate coefficients are used:

• K1 - type of windows;
• K2 - wall insulation;
• K3 - the ratio of floor area and windows;
• K4 - the minimum temperature in the street;
• K5 - the number of external walls of the house;
• K6 - number of storeys;
• K7 - the height of the room.

For windows, the heat loss coefficient is:

• ordinary glazing - 1.27;
• double-glazed window - 1;
• three-chamber double-glazed window - 0.85.

Naturally, last option keep the heat in the house much better than the previous two.

Properly executed wall insulation is the key not only to a long life of the house, but also comfortable temperature in the rooms. Depending on the material, the value of the coefficient also changes:

• concrete panels, blocks - 1.25-1.5;
• logs, timber - 1.25;
• brick (1.5 bricks) - 1.5;
• brick (2.5 bricks) - 1.1;
• foam concrete with increased thermal insulation - 1.

The larger the window area relative to the floor, the more heat the house loses:

The temperature outside the window also makes its own adjustments. At low rates of heat loss increase:

• Up to -10С - 0.7;
• -10C - 0.8;
• -15C - 0.90;
• -20C - 1.00;
• -25C - 1.10;
• -30C - 1.20;
• -35C - 1.30.

Heat loss also depends on how many external walls the house has:

• four walls - 1.33;%
• three walls - 1.22;
• two walls - 1.2;
• one wall - 1.

It’s good if a garage, a bathhouse or something else is attached to it. But if it is blown from all sides by winds, then you will have to buy a more powerful boiler.

The number of floors or the type of room that is located above the room determine the K6 coefficient as follows: if the house has two or more floors above, then for calculations we take the value 0.82, but if it is an attic, then for warm - 0.91 and 1 for cold .

As for the height of the walls, the values ​​\u200b\u200bwill be as follows:

• 4.5 m - 1.2;
• 4.0 m - 1.15;
• 3.5 m - 1.1;
• 3.0 m - 1.05;
• 2.5 m - 1.

In addition to the above coefficients, the area of ​​\u200b\u200bthe room (Pl) and the specific value of heat loss (UDtp) are also taken into account.

The final formula for calculating the heat loss coefficient:

Tp \u003d UDtp * Pl * K1 * K2 * K3 * K4 * K5 * K6 * K7.

The UDtp coefficient is 100 W/m2.

Analysis of calculations on a specific example

The house for which we will determine the load on the heating system has double-glazed windows (K1 \u003d 1), foam concrete walls with increased thermal insulation (K2 \u003d 1), three of which go outside (K5 \u003d 1.22). The area of ​​windows is 23% of the floor area (K3=1.1), it is about 15C outside (K4=0.9). The attic of the house is cold (K6=1), the height of the premises is 3 meters (K7=1.05). total area is 135m2.

Fri \u003d 135 * 100 * 1 * 1 * 1.1 * 0.9 * 1.22 * 1 * 1.05 \u003d 17120.565 (Watts) or Fri \u003d 17.1206 kW

Mk \u003d 1.2 * 17.1206 \u003d 20.54472 (kW).

Calculation of load and heat loss can be done independently and quickly enough. You just need to spend a couple of hours putting the source data in order, and then just substitute the values ​​​​into the formulas. The numbers that you will receive as a result will help you decide on the choice of boiler and radiators.