Thermal load Gcal. Calculation of the heat load for heating

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

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

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

The simplest methods of calculation

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

• The first is maintaining 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 the standards for the necessary microclimate are established - they are defined by GOST 30494-96. An excerpt from this document is in the table below:

Purpose of the room	Air temperature, °С		Relative humidity, %		Air speed, m/s
	optimal	admissible	optimal	admissible, max	optimal, max	admissible, max
For the cold season
Living room	20÷22	18÷24 (20÷24)	45÷30	60	0.15	0.2
Same but for living rooms in regions with minimum temperatures from -31 °C and below	21÷23	20÷24 (22÷24)	45÷30	60	0.15	0.2
Kitchen	19:21	18:26	N/N	N/N	0.15	0.2
Toilet	19:21	18:26	N/N	N/N	0.15	0.2
Bathroom, combined bathroom	24÷26	18:26	N/N	N/N	0.15	0.2
Premises for rest and study	20÷22	18:24	45÷30	60	0.15	0.2
Inter-apartment corridor	18:20	16:22	45÷30	60	N/N	N/N
lobby, stairwell	16÷18	14:20	N/N	N/N	N/N	N/N
Storerooms	16÷18	12÷22	N/N	N/N	N/N	N/N
For the warm season (The standard is only for residential premises. For the rest - it is not standardized)
Living room	22÷25	20÷28	60÷30	65	0.2	0.3

• 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 construction

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:

Building element	Approximate value of heat loss
Foundation, floors on the ground or over unheated basement (basement) premises	from 5 to 10%
"Cold bridges" through poorly insulated joints of building structures	from 5 to 10%
Places of entry of engineering communications (sewerage, water supply, gas pipes, electrical cables, etc.)	up to 5%
External walls, depending on the degree of insulation	from 20 to 30%
Poor quality windows and exterior doors	about 20÷25%, of which about 10% - through non-sealed joints between the boxes and the wall, and due to ventilation
Roof	up to 20%
Ventilation and chimney	up to 25 ÷30%

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 a norm of 100 watts of thermal energy for each square meter area:

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

Q = S× 100

Q- the required thermal power for the room;

S– area of ​​the room (m²);

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

For example, room 3.2 × 5.5 m

S= 3.2 × 5.5 = 17.6 m²

Q= 17.6 × 100 = 1760 W ≈ 1.8 kW

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

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

Q = S × h× 41 (or 34)

h- ceiling height (m);

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

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

Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW

The result is more accurate, since it already takes into account not only all linear dimensions rooms, 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. And the windows themselves may differ in the material of manufacture and other design features. And this is not a complete list - just such features are visible even to the "naked eye".

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

General principles and calculation formula

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

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

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

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

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

The coefficient is taken equal to:

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

- outer wall one: a = 1.0;

- external walls two: a = 1.2;

- external walls three: a = 1.4.

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

You may be interested in information about what are

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

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

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

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

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

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

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

Based on the results of long-term meteorological observations in any region, the so-called "wind rose" is compiled - a graphic diagram showing the prevailing wind directions in winter and summer. 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 the building structures of the building will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer indicators “dance” in a certain range, but for each region there is an average indicator of the most low temperatures, characteristic of the coldest five-day period of the year (usually this is characteristic of January). For example, below is a map-scheme of the territory of Russia, on which approximate values ​​​​are shown in colors.

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

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

— from – 35 °С and below: d=1.5;

— from – 30 °С to – 34 °С: d=1.3;

— from – 25 °С to – 29 °С: d=1.2;

— from – 20 °С to – 24 °С: d=1.1;

— from – 15 °С to – 19 °С: d=1.0;

— from – 10 °С to – 14 °С: d=0.9;

- not colder - 10 ° С: d=0.7.

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

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

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

- external walls are not insulated: e = 1.27;

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

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

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

• coefficient "f" - correction for ceiling height

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

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

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

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

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

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

– ceiling height over 4.1 m: f = 1.2.

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

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

- cold floor on the ground or 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 structure itself. Old wooden frames, which were previously installed everywhere in all houses, are significantly inferior to modern multi-chamber systems with double-glazed windows in terms of their thermal insulation.

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

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

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

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

– modern window systems with single-chamber double-glazed windows: i = 1,0 ;

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

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

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 there is no way to compare a small window with panoramic windows almost the whole wall.

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

x = ∑SOK /SP

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

SP- area of ​​the room.

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

- x \u003d 0 ÷ 0.1 →j = 0,8 ;

- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;

- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;

- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;

- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;

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

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

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

- no door k = 1,0 ;

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

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

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

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

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

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

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

If there are certain “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”:

Illustration	Features of installing radiators	The value of the coefficient "m"
	The radiator is located on the wall openly or is not covered from above by a window sill	m = 0.9
	The radiator is covered from above by a window sill or a shelf	m = 1.0
	The radiator is blocked from above by a protruding wall niche	m = 1.07
	The radiator is covered from above with a window sill (niche), and from the front - with a decorative screen	m = 1.12
	The radiator is completely enclosed in a decorative casing	m = 1.2

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

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

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

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

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

Let's create a table like this:

The room, its area, ceiling height. Floor insulation and "neighborhood" from above and below	The number of external walls and their main location relative to the cardinal points and the "wind rose". Degree of wall insulation	Number, type and size of windows	Existence of entrance doors (to the street or to the balcony)	Required heat output (including 10% reserve)
Area 78.5 m²				10.87 kW ≈ 11 kW
1. Hallway. 3.18 m². Ceiling 2.8 m. Warmed floor on the ground. Above is an insulated attic.	One, South, the average degree of insulation. Leeward side	Not	One	0.52 kW
2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. Above - insulated attic	Not	Not	Not	0.62 kW
3. Kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well insulated floor on the ground. Svehu - insulated attic	Two. South, west. Average degree of insulation. Leeward side	Two, single-chamber double-glazed window, 1200 × 900 mm	Not	2.22 kW
4. Children's room. 18.3 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic	Two, North - West. High degree of insulation. windward	Two, double glazing, 1400 × 1000 mm	Not	2.6 kW
5. Bedroom. 13.8 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic	Two, North, East. High degree of insulation. windward side	One, double-glazed window, 1400 × 1000 mm	Not	1.73 kW
6. Living room. 18.0 m². Ceiling 2.8 m. Well insulated floor. Top - insulated attic	Two, East, South. High degree of insulation. Parallel to wind direction	Four, double glazing, 1500 × 1200 mm	Not	2.59 kW
7. Bathroom combined. 4.12 m². Ceiling 2.8 m. Well insulated floor. Above is an insulated attic.	One, North. High degree of insulation. windward side	One. wooden frame with double glazing. 400 × 500 mm	Not	0.59 kW
TOTAL:

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

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

The calculation of the heat load for heating a house was made according to specific heat loss, the consumer approach to determining the reduced heat transfer coefficients is the main issue that we will consider in this post. Hello dear friends! We will calculate with you the heat load for heating the house (Qо.р) different ways on enlarged meters. So what we know so far: 1. Estimated winter outdoor temperature for heating design tn = -40 °C. 2. Estimated (averaged) air temperature inside the heated house tv = +20 °C. 3. The volume of the house according to the external measurement V = 490.8 m3. 4. Heated area of ​​the house Sot \u003d 151.7 m2 (residential - Szh \u003d 73.5 m2). 5. Degree day of the heating period GSOP = 6739.2 °C * day.

1. Calculation of the heat load for heating the house according to the heated area. Everything is simple here - it is assumed that the heat loss is 1 kW * hour per 10 m2 of the heated area of ​​​​the house, with a ceiling height of up to 2.5 m. For our house, the calculated heat load for heating will be equal to Qо.р = Sot * wud = 151.7 * 0.1 = 15.17 kW. Determining the heat load in this way is not particularly accurate. The question is, where did this ratio come from and how does it correspond to our conditions. Here it is necessary to make a reservation that this ratio is valid for the Moscow region (tn = up to -30 ° C) and the house should be normally insulated. For other regions of Russia, specific heat losses wsp, kW/m2 are given in Table 1.

Table 1

What else should be taken into account when choosing the specific heat loss coefficient? Reputable design organizations require up to 20 additional data from the "Customer" and this is justified, since the correct calculation of heat loss by a house is one of the main factors determining how comfortable it will be in the room. Below are the typical requirements with explanations:
- the severity of the climatic zone - the lower the temperature "overboard", the more you have to heat. For comparison: at -10 degrees - 10 kW, and at -30 degrees - 15 kW;
- the condition of the windows - the more hermetic and the greater the number of glasses, the losses are reduced. For example (at -10 degrees): standard double frame - 10 kW, double glazing- 8 kW, triple glazing - 7 kW;
- the ratio of the areas of windows and the floor - the larger the window, the more losses. At 20% - 9 kW, at 30% - 11 kW, and at 50% - 14 kW;
– wall thickness or thermal insulation directly affect heat loss. So with good thermal insulation and sufficient wall thickness (3 bricks - 800 mm), 10 kW is required, with 150 mm of insulation or a wall thickness of 2 bricks - 12 kW, and with poor insulation or a thickness of 1 brick - 15 kW;
- the number of external walls - is directly related to drafts and the multilateral effects of freezing. If the room has one outer wall, then 9 kW is required, and if - 4, then - 12 kW;
- the height of the ceiling, although not so significant, but still affects the increase in power consumption. At a standard height of 2.5 m, 9.3 kW is required, and at 5 m, 12 kW.
This explanation shows that a rough calculation of the required power of 1 kW of the boiler per 10 m2 of heated area is justified.

2. Calculation of the heat load for heating the house according to aggregated indicators in accordance with § 2.4 of SNiP N-36-73. To determine the heat load for heating in this way, we need to know the living area of ​​​​the house. If it is not known, then it is taken at a rate of 50% of the total area of ​​​​the house. Knowing the estimated outdoor air temperature for heating design, according to Table 2 we determine the aggregated indicator of the maximum hourly heat consumption per 1 m2 of living space.

table 2

For our house, the calculated heat load for heating will be equal to Qо.р = Szh * wsp.zh = 73.5 * 670 = 49245 kJ / h or 49245 / 4.19 = 11752 kcal / h or 11752/860 = 13.67 kW

3. Calculation of the heat load for heating the house according to the specific heating characteristic of the building.Determine heat load on this method we will be according to the specific thermal characteristic (specific heat loss of heat) and the volume of the house according to the formula:

Qo.r \u003d α * qo * V * (tv - tn) * 10-3, kW

Qо.р – estimated heat load on heating, kW;
α is a correction factor that takes into account the climatic conditions of the area and is used in cases where the calculated outdoor temperature tn differs from -30 ° C, is taken according to table 3;
qо – specific heating characteristic buildings, W/m3 * оС;
V is the volume of the heated part of the building according to the external measurement, m3;
tv is the design air temperature inside the heated building, °C;
tn is the calculated outdoor air temperature for heating design, °C.
In this formula, all quantities, except for the specific heating characteristic of the house qo, are known to us. The latter is a thermotechnical assessment of the construction part of the building and shows the heat flow required to increase the temperature of 1 m3 of the building volume by 1 °C. The numerical standard value of this characteristic, for residential building and hotels are shown in Table 4.

Correction factor α

Table 3

tn	-10	-15	-20	-25	-30	-35	-40	-45	-50
α	1,45	1,29	1,17	1,08	1	0,95	0,9	0,85	0,82

Specific heating characteristic of the building, W/m3 * oC

Table 4

So, Qo.r \u003d α * qo * V * (tv - tn) * 10-3 \u003d 0.9 * 0.49 * 490.8 * (20 - (-40)) * 10-3 \u003d 12.99 kW. At the stage of the feasibility study of the construction (project), the specific heating characteristic should be one of the benchmarks. The whole point is that in reference literature, numerical value it is different, because it is given for different time periods, before 1958, after 1958, after 1975, etc. In addition, although not significantly, the climate on our planet has also changed. And we would like to know the value of the specific heating characteristic of the building today. Let's try to define it ourselves.

PROCEDURE FOR DETERMINING SPECIFIC HEATING CHARACTERISTICS

1. A prescriptive approach to the choice of heat transfer resistance of outdoor enclosures. In this case, the consumption of thermal energy is not controlled, and the values ​​of heat transfer resistance individual elements buildings must be at least standardized values, see table 5. Here it is appropriate to give the Ermolaev formula for calculating the specific heating characteristics of the building. Here is the formula

qо = [Р/S * ((kс + φ * (kok – kс)) + 1/Н * (kpt + kpl)], W/m3 * оС

φ is the coefficient of glazing of the outer walls, we take φ = 0.25. This coefficient accepted in the amount of 25% of the floor area; P - the perimeter of the house, P = 40m; S - house area (10 * 10), S = 100 m2; H is the height of the building, H = 5m; ks, kok, kpt, kpl are the reduced heat transfer coefficients, respectively outer wall, light openings (windows), roofing (ceiling), ceilings above the basement (floor). For the determination of the reduced heat transfer coefficients, both for the prescriptive approach and for the consumer approach, see tables 5,6,7,8. Well, we have decided on the building dimensions of the house, but what about the building envelope of the house? What materials should the walls, ceiling, floor, windows and doors be made of? Dear friends, you must clearly understand that at this stage we should not be concerned about the choice of material for enclosing structures. The question is why? Yes, because in the above formula we will put the values ​​​​of the normalized reduced heat transfer coefficients of enclosing structures. So, regardless of what material these structures will be made of and what their thickness is, the resistance must be certain. (Extract from SNiP II-3-79* Building heat engineering).

(prescriptive approach)

Table 5

(prescriptive approach)

Table 6

And only now, knowing GSOP = 6739.2 °C * day, by interpolation we determine the normalized resistance to heat transfer of enclosing structures, see table 5. The given heat transfer coefficients will be equal, respectively: kpr = 1 / Rо and are given in table 6. Specific heating characteristic at home qo \u003d \u003d [P / S * ((kc + φ * (kok - kc)) + 1 / H * (kpt + kpl)] \u003d \u003d 0.37 W / m3 * °C
The calculated heat load on heating with a prescriptive approach will be equal to Qо.р = α* qо * V * (tв - tн) * 10-3 = 0.9 * 0.37 * 490.8 * (20 - (-40)) * 10-3 = 9.81 kW

2. Consumer approach to the choice of resistance to heat transfer of external fences. In this case, the resistance to heat transfer of external fences can be reduced in comparison with the values ​​\u200b\u200bindicated in Table 5, until the calculated specific consumption of thermal energy for heating the house exceeds the normalized one. The heat transfer resistance of individual fencing elements should not be lower than the minimum values: for the walls of a residential building Rc = 0.63Rо, for the floor and ceiling Rpl = 0.8Rо, Rpt = 0.8Rо, for windows Rok = 0.95Rо. The results of the calculation are shown in table 7. Table 8 shows the reduced heat transfer coefficients for the consumer approach. Concerning specific consumption thermal energy for heating season, then for our house this value is 120 kJ / m2 * oC * day. And it is determined according to SNiP 23-02-2003. We will determine this value when we calculate the heat load for heating more than detailed way- taking into account the specific materials of the fences and their thermophysical properties (clause 5 of our plan for calculating the heating of a private house).

Rated resistance to heat transfer of enclosing structures
(consumer approach)

Table 7

Determination of the reduced heat transfer coefficients of enclosing structures
(consumer approach)

Table 8

Specific heating characteristic of the house qo \u003d \u003d [Р / S * ((kс + φ * (kok - kс)) + 1 / N * (kpt + kpl)] \u003d \u003d 0.447 W / m3 * ° C. Estimated heat load for heating at consumer approach will be equal to Qо.р = α * qо * V * (tв - tн) * 10-3 = 0.9 * 0.447 * 490.8 * (20 - (-40)) * 10-3 = 11.85 kW

Main conclusions:
1. Estimated heat load on heating for the heated area of ​​the house, Qo.r = 15.17 kW.
2. Estimated heat load on heating according to aggregated indicators in accordance with § 2.4 of SNiP N-36-73. heated area of ​​the house, Qo.r = 13.67 kW.
3. Estimated heat load for heating the house according to the normative specific heating characteristic of the building, Qo.r = 12.99 kW.
4. Calculated heat load for heating the house according to the prescriptive approach to the choice of heat transfer resistance of external fences, Qo.r = 9.81 kW.
5. Estimated heat load for home heating according to the consumer approach to the choice of heat transfer resistance of external fences, Qo.r = 11.85 kW.
As you can see, dear friends, the calculated heat load for heating a house with a different approach to its definition varies quite significantly - from 9.81 kW to 15.17 kW. What to choose and not to be mistaken? We will try to answer this question in the following posts. Today we have completed the 2nd point of our plan for the house. For those who haven't joined yet!

Sincerely, Grigory Volodin

q - specific heating characteristic of the building, kcal / mh ° С is taken from the reference book, depending on the external volume of the building.

a is a correction factor taking into account the climatic conditions of the region, for Moscow, a = 1.08.

V - the outer volume of the building, m is determined by construction data.

t - the average air temperature inside the room, ° C is taken depending on the type of building.

t - design temperature of outdoor air for heating, °С for Moscow t= -28 °С.

Source: http://vunivere.ru/work8363

Q yh is made up of the thermal loads of devices serviced by water flowing through the site:

(3.1)

For the section of the supply heat pipeline, the thermal load expresses the heat reserve in the flowing hot water, intended for subsequent (on the further path of water) heat transfer to the premises. For the section of the return heat pipeline - the loss of heat by the flowing chilled water during heat transfer to the premises (on the previous water path). The thermal load of the site is designed to determine the flow of water in the site in the process of hydraulic calculation.

Water consumption on the site G uch at the calculated difference in water temperature in the system t g - t x, taking into account additional heat supply to the premises

where Q ych is the thermal load of the section, found by formula (3.1);

β 1 β 2 - correction factors that take into account additional heat supply to the premises;

c - specific mass heat capacity of water, equal to 4.187 kJ / (kg ° C).

To obtain the water flow in the area in kg / h, the heat load in W should be expressed in kJ / h, i.e. multiply by (3600/1000)=3.6.

as a whole is equal to the sum of heat loads of all heating devices (heat losses of premises). According to the total heat demand for heating the building, the water flow in the heating system is determined.

Hydraulic calculation is associated with the thermal calculation of heating appliances and pipes. Multiple repetition of calculations is required to identify the actual flow and temperature of water, the required area of ​​​​devices. When calculating manually, the hydraulic calculation of the system is first performed, taking the average values ​​of the local resistance coefficient (LFR) of the devices, then the thermal calculation of pipes and devices.

If convectors are used in the system, the design of which includes pipes Dy15 and Dy20, then for a more accurate calculation, the length of these pipes is preliminarily determined, and after hydraulic calculation, taking into account the pressure losses in the pipes of the devices, having specified the flow rate and temperature of the water, they make adjustments to the dimensions of the devices.

Source: http://teplodoma.com.ua/1/gidravliheskiy_rashet/str_19.html

In this section, you will be able to get acquainted with the issues related to the calculation of heat losses and heat loads of the building in as much detail as possible.

The construction of heated buildings without heat loss calculation is prohibited!*)

And although most are still building at random, on the advice of a neighbor or godfather. It is right and clear to start at the stage of developing a working draft for construction. How it's done?

The architect (or the developer himself) provides us with a list of "available" or "priority" materials for arranging walls, roofs, bases, which windows, doors are planned.

Already at the design stage of a house or building, as well as for the selection of heating, ventilation, air conditioning systems, it is necessary to know the heat losses of the building.

Calculation of heat loss for ventilation we often use in our practice to calculate the economic feasibility of modernizing and automating the ventilation / air conditioning system, because calculation of heat losses for ventilation gives a clear idea of ​​the benefits and payback period of funds invested in energy-saving measures (automation, use of recuperation, insulation of air ducts, frequency controllers).

Calculation of building heat losses

This is the basis for competent power selection. heating equipment(boiler, boiler) and heating appliances

The main heat losses of a building usually occur in the roof, walls, windows and floors. Enough most of heat leaves the premises through the ventilation system.

Rice. 1 Building heat loss

The main factors affecting heat loss in a building are the temperature difference between indoors and outdoors (the greater the difference, the greater the body loss) and the thermal insulation properties of building envelopes (foundation, walls, ceilings, windows, roofing).

Fig. 2 Thermal imaging survey of building heat losses

Enclosing materials prevent the penetration of heat from the premises to the outside in winter and the penetration of heat into the premises in summer, because the selected materials must have certain thermal insulation properties, which are denoted by a value called - heat transfer resistance.

The resulting value will show what the real temperature difference will be when a certain amount of heat passes through 1m² of a particular building envelope, as well as how much heat will leave after 1m² at a certain temperature difference.

#image.jpgHow heat loss is calculated

When calculating the heat losses of a building, we will be mainly interested in all external enclosing structures and the location of internal partitions.

To calculate heat losses along the roof, it is also necessary to take into account the shape of the roof and the presence of an air gap. There are also some nuances in the thermal calculation of the floor of the room.

To get the most exact value The thermal losses of the building must take into account absolutely all enclosing surfaces (foundation, ceilings, walls, roof), their constituent materials and the thickness of each layer, as well as the position of the building relative to the cardinal points and climatic conditions in the region.

To order the calculation of heat losses you need fill out our questionnaire and we will send our commercial offer to the specified postal address as soon as possible (no more than 2 working days).

Scope of work on the calculation of thermal loads of the building

The main composition of the documentation for the calculation of the thermal load of the building:

• building heat loss calculation
• calculation of heat losses for ventilation and infiltration
• permits
• summary table of thermal loads

The cost of calculating the thermal loads of the building

The cost of services for calculating the thermal loads of a building does not have a single price, the price for the calculation depends on many factors:

• heated area;
• availability of project documentation;
• architectural complexity of the object;
• composition of enclosing structures;
• the number of heat consumers;
• the diversity of the purpose of the premises, etc.

It is not difficult to find out the exact cost and order a service for calculating the thermal load of a building, for this you just need to send us a floor plan of the building by e-mail (form), fill out a short questionnaire and after 1 working day you will receive a mailbox our business proposal.

#image.jpgExamples of the cost of calculating thermal loads

Thermal calculations for a private house

Documentation set:

- calculation of heat losses (room by room, floor by floor, infiltration, total)

- calculation of heat load for hot water heating (DHW)

- calculation for heating air from the street for ventilation

A package of thermal documents will cost in this case - 1600 UAH

For such calculations bonus You get:

Recommendations for insulation and elimination of cold bridges

Power selection of the main equipment

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The sports complex is a detached 4-storey building of a typical construction, with a total area of ​​2100 sq.m. with a large gym, heated supply and exhaust ventilation system, radiator heating, a full set of documentation — 4200.00 UAH

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Shop - a premise built into a residential building on the 1st floor, with a total area of ​​240 sq.m. of which 65 sq.m. warehouses, without basement, radiator heating, heated supply and exhaust ventilation with recovery 2600.00 UAH

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Terms of performance of work on the calculation of thermal loads

The term for performing work on the calculation of the thermal loads of the building mainly depends on the following components:

• total heated area of ​​premises or building
• architectural complexity of the object
• complexity or multi-layered enclosing structures
• number of heat consumers: heating, ventilation, hot water, other
• multifunctionality of premises (warehouse, offices, trading floor, residential, etc.)
• organization of a heat energy commercial metering unit
• completeness of the availability of documentation (project of heating, ventilation, executive schemes for heating, ventilation, etc.)
• diversity of use of building envelope materials in construction
• complexity of the ventilation system (recuperation, automatic control system, zone temperature control)

In most cases, for a building with a total area of ​​​​not more than 2000 sq.m. The term for calculating the thermal loads of a building is 5 to 21 business days depending on the above characteristics of the building, provided documentation and engineering systems.

Coordination of calculation of heat loads in heat networks

After completing all the work on the calculation of heat loads and collecting all the necessary documents, we approach the final, but difficult issue of coordinating the calculation of heat loads in urban heating networks. This process is a “classic” example of communication with the state structure, notable for a lot of interesting innovations, clarifications, views, interests of the subscriber (client) or a representative of the contracting organization (which has undertaken to coordinate the calculation of heat loads in heating networks) with representatives of urban heating networks. In general, the process is often difficult, but surmountable.

The list of documents to be submitted for approval looks something like this:

• Application (written directly in thermal networks);
• Calculation of thermal loads (in full);
• License, list of licensed works and services of the contractor performing the calculations;
• Registration certificate for the building or premises;
• The right establishing the documentation for the ownership of the object, etc.

Usually for term for approval of the calculation of thermal loads accepted - 2 weeks (14 working days) subject to the submission of documentation in full and in the required form.

Services for calculating the thermal loads of the building and related tasks

When concluding or re-executing an agreement on the supply of heat from city heating networks or designing and installing a commercial heat metering unit, heating networks notify the owner of the building (premises) of the need to:

• receive specifications(THAT);
• provide a calculation of the thermal load of the building for approval;
• project for the heating system;
• project for the ventilation system;
• and etc.

We offer our services in carrying out the necessary calculations, designing heating systems, ventilation and subsequent approvals in urban heating networks and other regulatory authorities.

You can order both a separate document, project or calculation, as well as execution of all necessary documents on a turnkey basis from any stage.

Discuss the topic and leave feedback: "CALCULATION OF HEAT LOSSES AND LOADS" on FORUM #image.jpg

We will be glad to continue cooperation with you by offering:

Supply of equipment and materials at wholesale prices

Design work

Assembly / installation / commissioning

Further maintenance and provision of services at reduced prices (for regular customers)

To find out how much power the heat-power equipment of a private house should have, it is necessary to determine the total load on the heating system, for which a thermal calculation is performed. In this article, we will not talk about an enlarged method for calculating the area or volume of a building, but we will present a more accurate method used by designers, only in a simplified form for better perception. So, 3 types of loads fall on the heating system of the house:

• compensation for losses of thermal energy leaving through building structures (walls, floors, roofs);
• heating the air required for ventilation of the premises;
• water heating for DHW needs(when the boiler is involved, and not a separate heater).

Determination of heat loss through external fences

To begin with, we present a formula from SNiP, according to which the calculation of the thermal energy lost through building structures separating inner space houses from the street:

Q \u003d 1 / R x (tv - tn) x S, where:

• Q is the consumption of heat leaving through the structure, W;
• R - resistance to heat transfer through the material of the fence, m2ºС / W;
• S is the area of ​​this structure, m2;
• tv - the temperature that should be inside the house, ºС;
• tn is the average outdoor temperature for the 5 coldest days, ºС.

For reference. According to the methodology, heat loss calculation is performed separately for each room. In order to simplify the task, it is proposed to take the building as a whole, assuming an acceptable average temperature of 20-21 ºС.

The area for each type of external fencing is calculated separately, for which windows, doors, walls and floors with a roof are measured. This is done because they are made from different materials different thickness. So the calculation will have to be done separately for all types of structures, and then the results will be summed up. You probably know the coldest street temperature in your area of ​​​​residence from practice. But the parameter R will have to be calculated separately according to the formula:

R = δ / λ, where:

• λ is the coefficient of thermal conductivity of the fence material, W/(mºС);
• δ is the thickness of the material in meters.

Note. The value of λ is a reference value, it is not difficult to find it in any reference literature, and for plastic windows this coefficient will be prompted by the manufacturers. Below is a table with the coefficients of thermal conductivity of some building materials, and for calculations it is necessary to take the operational values ​​of λ.

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

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

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

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

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

Ventilation air heating consumption

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

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

Qvent = cmΔt, here:

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

The complexity of calculating this type of heat load lies in the correct determination of the mass of heated air. It is difficult to find out how much it gets inside the house with natural ventilation. Therefore, it is worth referring to the standards, because buildings are built according to projects where the required air exchanges are laid down. And the regulations say that in most rooms air environment should be changed once per hour. Then we take the volumes of all rooms and add to them the air flow rates for each bathroom - 25 m3 / h and a kitchen gas stove– 100 m3/h.

To calculate the heat load on heating from ventilation, the resulting volume of air must be converted into mass, knowing its density at different temperatures from the table:

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

Heat load from DHW heating

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

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

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

The unit of equipment power is always referred to 1 hour, and the resulting 5.2 kW - to the day. But you cannot divide this figure by 24, because hot water we want to receive as soon as possible, and for this the boiler must have a power reserve. That is, this load must be added to the rest as is.

Conclusion

This calculation of home heating loads will give much more accurate results than traditional way on the area, although you have to work hard. The end result must be multiplied by the safety factor - 1.2, or even 1.4, and according to the calculated value, select boiler equipment. Another way to enlarge the calculation of thermal loads according to the standards is shown in the video: