Calculation of heat loss through building envelopes. Calculation of room heat loss

Below is a pretty simple heat loss calculation buildings, which, however, will help to accurately determine the power required to heat your warehouse, shopping center or other similar building. This will make it possible at the design stage to preliminarily estimate the cost of heating equipment and subsequent heating costs, and, if necessary, adjust the project.

Where does the heat go? Heat escapes through walls, floors, roofs and windows. In addition, heat is lost during ventilation of the premises. To calculate heat loss through building envelope, use the formula:

Q - heat loss, W

S – construction area, m2

T - temperature difference between indoor and outdoor air, °C

R is the value of the thermal resistance of the structure, m2 °C/W

The calculation scheme is as follows - we calculate the heat loss of individual elements, summarize and add the heat loss during ventilation. All.

Suppose we want to calculate the heat loss for the object shown in the figure. The height of the building is 5 ... 6 m, width - 20 m, length - 40 m, and thirty windows measuring 1.5 x 1.4 meters. Indoor temperature 20 °C, outside temperature -20 °C.

We consider the area of ​​​​enclosing structures:

floor: 20 m * 40 m = 800 m2

roof: 20.2 m * 40 m = 808 m2

window: 1.5 m * 1.4 m * 30 pcs = 63 m2

walls:(20 m + 40 m + 20 m + 40 m) * 5 m = 600 m2 + 20 m2 (including pitched roof) = 620 m2 - 63 m2 (windows) = 557 m2

Now let's see the thermal resistance of the materials used.

The value of thermal resistance can be taken from the table of thermal resistances or calculated based on the value of the thermal conductivity coefficient using the formula:

R - thermal resistance, (m2 * K) / W

? - coefficient of thermal conductivity of the material, W / (m2 * K)

d – material thickness, m

The value of thermal conductivity coefficients for different materials can be viewed.

floor: concrete screed 10 cm and mineral wool with a density of 150 kg/m3. 10 cm thick.

R (concrete) = 0.1 / 1.75 = 0.057 (m2*K)/W

R (mineral wool) \u003d 0.1 / 0.037 \u003d 2.7 (m2 * K) / W

R (floor) \u003d R (concrete) + R (mineral wool) \u003d 0.057 + 2.7 \u003d 2.76 (m2 * K) / W

roof:

R (roof) = 0.15 / 0.037 = 4.05 (m2*K)/W

window: the value of thermal resistance of windows depends on the type of double-glazed window used
R (windows) \u003d 0.40 (m2 * K) / W for single-chamber glass wool 4–16–4 at? T \u003d 40 ° С

walls: mineral wool panels 15 cm thick
R (walls) = 0.15 / 0.037 = 4.05 (m2*K)/W

Let's calculate the heat loss:

Q (floor) \u003d 800 m2 * 20 ° C / 2.76 (m2 * K) / W \u003d 5797 W \u003d 5.8 kW

Q (roof) \u003d 808 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 7980 W \u003d 8.0 kW

Q (windows) \u003d 63 m2 * 40 ° C / 0.40 (m2 * K) / W \u003d 6300 W \u003d 6.3 kW

Q (walls) \u003d 557 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 5500 W \u003d 5.5 kW

We get that the total heat loss through the building envelope will be:

Q (total) = 5.8 + 8.0 + 6.3 + 5.5 = 25.6 kWh

Now about ventilation losses.

To heat 1 m3 of air from a temperature of -20 °C to +20 °C, 15.5 W will be required.

Q (1 m3 of air) \u003d 1.4 * 1.0 * 40 / 3.6 \u003d 15.5 W, here 1.4 is the air density (kg / m3), 1.0 is the specific heat capacity of air (kJ / ( kg K)), 3.6 is the conversion factor to watts.

It remains to determine the amount of air required. It is believed that with normal breathing, a person needs 7 m3 of air per hour. If you use a building as a warehouse and 40 people work on it, then you need to heat 7 m3 * 40 people = 280 m3 of air per hour, this will require 280 m3 * 15.5 W = 4340 W = 4.3 kW. And if you have a supermarket and on average there are 400 people on the territory, then air heating will require 43 kW.

Final result:

For heating the proposed building, a heating system of the order of 30 kWh is required, and a ventilation system with a capacity of 3000 m3 / h with a heater with a power of 45 kW / h.

Designing a heating system "by eye" with a high probability can lead either to an unjustified overestimation of the cost of its operation, or to underheating of the home.

So that neither one nor the other happens, it is necessary first of all to correctly calculate the heat loss of the house.

And only on the basis of the results obtained, the power of the boiler and radiators is selected. Our conversation will be about how these calculations are made and what should be taken into account.

The authors of many articles reduce the calculation of heat loss to one simple action: it is proposed to multiply the area of ​​\u200b\u200bthe heated room by 100 watts. The only condition that is put forward in this case refers to the height of the ceiling - it should be 2.5 m (for other values, it is proposed to introduce a correction factor).

In fact, such a calculation is so approximate that the figures obtained with its help can be safely equated with "taken from the ceiling." After all, a number of factors influence the specific value of heat loss: the material of the building envelope, the outside temperature, the area and type of glazing, the frequency of air exchange, etc.

Heat loss at home

Moreover, even for houses with different heated areas, other things being equal, its value will be different: in a small house - more, in a large one - less. This is the law of the square-cube.

Therefore, it is extremely important for the owner of the house to master a more accurate method for determining heat loss. Such a skill will allow not only to select heating equipment with optimal power, but also to evaluate, for example, the economic effect of insulation. In particular, it will be possible to understand whether the service life of the heat insulator will exceed its payback period.

The first thing the contractor needs to do is to decompose the total heat loss into three components:

• losses through enclosing structures;
• caused by the operation of the ventilation system;
• associated with the discharge of heated water into the sewer.

Let's consider each of the varieties in detail.

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How to properly insulate the walls of an apartment from the inside without harming the building structure, read.

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Heat loss calculation

Here's how to do the calculations:

Heat loss through building envelopes

For each material that is part of the enclosing structures, in the reference book or the passport provided by the manufacturer, we find the value of the thermal conductivity coefficient Kt (unit - W / m * degree).

For each layer of enclosing structures, we determine the thermal resistance according to the formula: R = S / Kt, where S is the thickness of this layer, m.

For multilayer structures, the resistances of all layers must be added.

We determine the heat loss for each structure according to the formula Q = (A / R)*dT,

• A is the area of ​​the building envelope, sq. m;
• dT - difference between outside and inside temperatures.
• dT should be determined for the coldest five-day period.

Heat loss through ventilation

For this part of the calculation, it is necessary to know the air exchange rate.

In residential buildings built according to domestic standards (the walls are vapor-permeable), it is equal to one, that is, the entire volume of air in the room must be updated in an hour.

In houses built according to European technology (DIN standard), in which the walls are covered with vapor barrier from the inside, the air exchange rate has to be increased to 2. That is, in an hour, the air in the room should be updated twice.

Heat loss through ventilation is determined by the formula:

Qv \u003d (V * Kv / 3600) * p * s * dT,

• V is the volume of the room, cub. m;
• Kv - air exchange rate;
• P - air density, taken equal to 1.2047 kg / cu. m;
• C is the specific heat capacity of air, assumed to be 1005 J/kg*C.

The above calculation allows you to determine the power that the heat generator of the heating system should have. If it turned out to be too high, you can do the following:

• lower the requirements for the level of comfort, that is, set the desired temperature in the coldest period at the minimum mark, say, 18 degrees;
• for a period of severe cold, reduce the air exchange rate: the minimum allowable supply ventilation capacity is 7 cubic meters. m/h for each inhabitant of the house;
• provide for the organization of supply and exhaust ventilation with a heat exchanger.

Note that the heat exchanger is useful not only in winter, but also in summer: in the heat, it allows you to save the cold produced by the air conditioner, although it does not work as efficiently at this time as in frost.

It is most correct when designing a house to perform zoning, that is, assign a different temperature for each room based on the required comfort. For example, in a nursery or a room for an elderly person, a temperature of about 25 degrees should be provided, while 22 will be enough for a living room. On the landing or in a room where residents rarely appear or there are sources of heat release, the design temperature can generally be limited to 18 degrees.

Obviously, the figures obtained in this calculation are relevant only for a very short period - the coldest five-day period. To determine the total amount of energy consumption for the cold season, the parameter dT must be calculated taking into account not the lowest, but the average temperature. Then you need to do the following:

W \u003d ((Q + Qv) * 24 * N) / 1000,

• W is the amount of energy required to replenish heat losses through building envelopes and ventilation, kWh;
• N is the number of days in the heating season.

However, this calculation will be incomplete if heat losses to the sewer system are not taken into account.

To receive hygiene procedures and wash dishes, the residents of the house heat water and the heat produced goes into the sewer pipe.

But in this part of the calculation, one should take into account not only direct water heating, but also indirect - heat is taken off by water in the tank and toilet siphon, which is also discharged into the sewer.

Based on this, the average temperature of water heating is assumed to be only 30 degrees. Heat loss through the sewer is calculated using the following formula:

Qk \u003d (Vv * T * p * s * dT) / 3,600,000,

• Vв - monthly volume of water consumption without division into hot and cold, cubic meters. m/month;
• P is the density of water, we take p \u003d 1000 kg / cu. m;
• C is the heat capacity of water, we take c \u003d 4183 J / kg * C;
• dT - temperature difference. Given that the water at the inlet in winter has a temperature of about +7 degrees, and we agreed to consider the average temperature of the heated water equal to 30 degrees, we should take dT = 23 degrees.
• 3,600,000 - the number of joules (J) in 1 kWh.

An example of calculating the heat loss of a house

Let us calculate the heat loss of a 2-storey house 7 m high, having dimensions in terms of 10x10 m.

The walls are 500 mm thick and built of warm ceramics (Кт = 0.16 W/m*С), outside they are insulated with mineral wool 50 mm thick (Кт = 0.04 W/m*С).

The house has 16 windows with an area of ​​2.5 square meters. m.

The outside temperature in the coldest five-day period is -25 degrees.

The average outdoor temperature during the heating period is (-5) degrees.

Inside the house, it is required to provide a temperature of +23 degrees.

Water consumption - 15 cubic meters. m/month

Duration of the heating period - 6 months.

We determine the heat loss through the building envelope (for example, consider only the walls)

Thermal resistance:

• base material: R1 = 0.5 / 0.16 = 3.125 sq. m*S/W;
• insulation: R2 = 0.05 / 0.04 = 1.25 sq. m*S/W.

The same for the wall as a whole: R = R1 + R2 = 3.125 + 1.25 = 4.375 sq. m*S/W.

We determine the area of ​​\u200b\u200bthe walls: A \u003d 10 x 4 x 7 - 16 x 2.5 \u003d 240 square meters. m.

Heat loss through the walls will be:

Qc \u003d (240 / 4.375) * (23 - (-25)) \u003d 2633 W.

Heat losses through the roof, floor, foundation, windows and front door are calculated in a similar way, after which all the values ​​obtained are summed up. Manufacturers usually indicate the thermal resistance of doors and windows in the product passport.

Please note that when calculating heat loss through the floor and foundation (if there is a basement), the temperature difference dT will be much smaller, since when calculating it, the temperature of the soil, which is much warmer in winter, is not taken into account.

Heat loss through ventilation

We determine the volume of air in the room (to simplify the calculation, the thickness of the walls is not taken into account):

V \u003d 10x10x7 \u003d 700 cu. m.

Taking the air exchange rate Kv = 1, we determine the heat loss:

Qv \u003d (700 * 1 / 3600) * 1.2047 * 1005 * (23 - (-25)) \u003d 11300 W.

Ventilation in the house

Heat loss through the sewer

Taking into account the fact that residents consume 15 cubic meters. m of water per month, and the billing period is 6 months, the heat loss through the sewer will be:

Qk \u003d (15 * 6 * 1000 * 4183 * 23) / 3,600,000 \u003d 2405 kWh

If you do not live in a country house in the winter, in the off-season or in the cold summer, you still need to heat it. in this case is the most appropriate.

You can read about the reasons for the pressure drop in the heating system. Troubleshooting.

Estimation of the total amount of energy costs

To assess the total volume of energy consumption during the heating period, it is necessary to recalculate the heat loss through ventilation and enclosing structures, taking into account the average temperature, that is, dT will not be 48, but only 28 degrees.

Then the average power loss through the walls will be:

Qc \u003d (240 / 4.375) * (23 - (-5)) \u003d 1536 W.

Suppose that an additional 800 W is lost through the roof, floor, windows and doors, then the total average power of heat loss through the building envelope will be Q = 1536 + 800 = 2336 W.

The average power of heat loss through ventilation will be:

Qv \u003d (700 * 1 / 3600) * 1.2047 * 1005 * (23 - (-5)) \u003d 6592 W.

Then for the entire period you will have to spend on heating:

W \u003d ((2336 + 6592) * 24 * 183) / 1000 \u003d 39211 kWh.

To this value, you need to add 2405 kWh of losses through the sewer, so that the total amount of energy consumption for the heating period will be 41616 kWh.

If only gas is used as an energy carrier, from the 1st cu. m of which it is possible to obtain 9.45 kWh of heat, then it will need 41616 / 9.45 = 4404 cubic meters. m.

Related video

To determine heat loss, you must have:

Floor plans with all building dimensions;

A copy from the general plan with the designation of the countries of the world and the wind rose;

The purpose of each room;

Geographic location of the building;

Structures of all external fences.

All premises on the plans indicate:

They are numbered from left to right, stairwells are designated by letters or Roman numerals, regardless of the floor and are considered as one room.

Heat loss in rooms through building envelopes, rounded up to 10 W:

Q limit \u003d (F / R o) (t in - t n B) (1 + ∑β) n = kF (t in - t n B) (1 - ∑ β) n,(3.2)

where F, k, R o- estimated area, heat transfer coefficient, heat transfer resistance of the enclosing structure, m 2, W / (m 2 o C), (m 2 o C) / W; t in- estimated room air temperature, o C; t n B- calculated outdoor air temperature (B) or air temperature of a colder room; P- coefficient taking into account the position of the outer surface of the enclosing structures in relation to the outside air (Table 2.4); β - additional heat losses in shares of the main losses.

Heat transfer through fences between adjacent heated rooms is taken into account if the temperature difference in them is more than 3°C.

squares F, m 2, fences (external walls (NS), windows (O), doors (D), lanterns (F), ceiling (Pt), floor (P)) are measured according to the plans and sections of the building (Fig. 3.1).

1. The height of the walls of the first floor: if the floor is on the ground, - between the levels of the floors of the first and second floors ( h1); if the floor is on logs - from the outer level of floor preparation on logs to the floor level of the second floor ( h 1 1); in an unheated basement or underground - from the level of the lower surface of the floor structure of the first floor to the level of the clean floor of the second floor ( h 1 11), and in one-story buildings with an attic floor, the height is measured from the floor to the top of the insulation layer of the floor.

2. The height of the walls of the intermediate floor - between the levels of clean floors of this and the overlying floors ( h2), and the upper floor - from the level of its clean floor to the top of the insulating layer of the attic floor ( h 3) or non-attic cover.

3. The length of the outer walls in the corner rooms - from the edge of the outer corner to the axes of the inner walls ( l 1 and l 2l 3).

4. The length of the inner walls - from the inner surfaces of the outer walls to the axes of the inner walls ( m 1) or between the axes of the internal walls (t).

5. Areas of windows, doors and lanterns - according to the smallest dimensions of building openings in the light ( a and b).

6. Ceiling and floor areas above basements and undergrounds in corner rooms - from the inner surface of the outer walls to the axes of opposite walls ( m 1 and P), and in non-angular ones - between the axes of the internal walls ( t) and from the inner surface of the outer wall to the axis of the opposite wall ( P).

The error of linear dimensions is ±0.1 m, the area is ±0.1 m 2.

Rice. 3.1. Scheme of measurement of heat transfer fences

Fig 3.2. Scheme for determining heat loss through floors and walls buried below ground level

1 - the first zone; 2 - the second zone; 3 - the third zone; 4 - fourth zone (last).

Heat loss through the floors is determined by zones-strips 2 m wide, parallel to the outer walls (Fig. 5.2).

Reduced resistance to heat transfer R n.p., m 2 K / W, zones of uninsulated floors on the ground and walls below ground level, with thermal conductivity λ > 1.2 W / (m o C): for the 1st zone - 2.1; for the 2nd zone - 4.3; for the 3rd zone - 8.6; for the 4th zone (the remaining floor area) - 14.2.

Formula (3.2) when calculating heat losses Q pl, W, through the floor, located on the ground, takes the form:

Q pl \u003d (F 1 / R 1n.p + F 2 / R 2n.p + F 3 / R 3n.p + F 4 / R 4n.p) (t in - t n B) (1 + ∑β) n,(3.3)

where F 1 - F 4- area 1 - 4 zones-bands, m 2; R 1, n.p. - R 4, n.p.- resistance to heat transfer of floor zones, m 2 K / W; n =1.

Heat transfer resistance of insulated floors on the ground and walls below ground level (λ< 1,2 Вт/(м· о С)) R y .p, m 2 o C / W, also determined for zones according to the formula

R c.p. = R n.p. +∑(δ c.s. /λ c.s.),(3.4)

where R n.a.- heat transfer resistance of non-insulated floor zones (Fig. 3.2), m 2 o C / W; fraction sum- the sum of thermal resistances of insulating layers, m 2 o C / W; δ c.s.- thickness of the insulating layer, m.

Heat transfer resistance of floors on joists R l, m 2 o C / W:

R l.p = 1.18 (R n.p + ∑(δ w.s. /λ w.s.)),(3.5)

Insulating layers - an air layer and a wooden floor on the logs.

When calculating heat losses, floor sections in the corners of the outer walls (in the first two-meter zone) are entered into the calculation twice in the direction of the walls.

Heat losses through the underground part of the outer walls and the floors of the heated basement are also calculated in zones 2 m wide, counting them from the ground level (see Fig. 3.2). Then the floors (when counting zones) are considered as a continuation of the underground part of the outer walls. Heat transfer resistance is determined in the same way as for non-insulated or insulated floors.

Additional heat loss through the fences. In (3.2) the term (1+∑β) takes into account additional heat losses as a fraction of the main heat losses:

1. On orientation in relation to the cardinal points. β external vertical and inclined (vertical projection) walls, windows and doors.

Rice. 3.3. Addition to the main heat losses depending on the orientation of the fences in relation to the cardinal points

2. For ventilation of premises with two or more outer walls. In typical projects through walls, doors and windows facing all countries of the world β = 0.08 with one outer wall and 0.13 for corner rooms and in all living quarters.

3. On the calculated outdoor temperature. For unheated ground floor floors above cold building undergrounds in areas with t n B minus 40°C and below - β = 0,05.

4. For heating the rushing cold air. For exterior doors, without air curtains or air curtains, at building height H, m:

- β = 0,2H- for triple doors with two vestibules between them;

- β = 0,27 H - for double doors with a vestibule between them;

- β = 0,34 H - for double doors without vestibule;

- β = 0,22 H - for single doors.

For external non-equipped gates β =3 without tambour and β = 1 - with a vestibule at the gate. For summer and spare exterior doors and gates β = 0.

Heat losses through the enclosing structures of the premises are entered in the form (form) (Table 3.2).

Table 3.2. Form (form) for calculating heat loss

The areas of the walls in the calculation are measured with the area of ​​the windows, so the area of ​​the windows is taken into account twice, therefore, in column 10, the coefficient k windows are taken as the difference between its values ​​for windows and walls.

Calculation of heat losses is carried out for rooms, floors, building.

Each building, regardless of design features, passes thermal energy through the fences. Heat loss to the environment must be restored using the heating system. The sum of heat losses with a normalized margin is the required power of the heat source that heats the house. In order to create comfortable conditions in a dwelling, heat loss is calculated taking into account various factors: building design and layout of premises, orientation to the cardinal points, wind direction and average mildness of the climate during the cold period, physical qualities of building and heat-insulating materials.

Based on the results of the heat engineering calculation, a heating boiler is selected, the number of battery sections is specified, the power and length of the underfloor heating pipes are considered, a heat generator is selected for the room - in general, any unit that compensates for heat loss. By and large, it is necessary to determine heat losses in order to heat the house economically - without an extra supply of power from the heating system. Calculations are performed manually or a suitable computer program is selected into which data are substituted.

How to make a calculation?

First, you should deal with the manual technique - to understand the essence of the process. To find out how much heat a house loses, determine the losses through each building envelope separately, and then add them up. The calculation is carried out in stages.

1. Form a base of initial data for each room, preferably in the form of a table. In the first column, the pre-calculated area of ​​door and window blocks, external walls, ceilings, and floors is recorded. The thickness of the structure is entered in the second column (these are design data or measurement results). In the third - the coefficients of thermal conductivity of the corresponding materials. Table 1 contains the normative values ​​that will be needed in the further calculation:

The higher λ, the more heat escapes through the meter thickness of the given surface.

2. The heat resistance of each layer is determined: R = v/ λ, where v is the thickness of the building or heat-insulating material.

3. Calculate the heat loss of each structural element according to the formula: Q \u003d S * (T in -T n) / R, where:

• T n - outdoor temperature, ° C;
• T in - indoor temperature, ° C;
• S is the area, m2.

Of course, during the heating period, the weather varies (for example, the temperature ranges from 0 to -25°C), and the house is heated to the desired level of comfort (for example, up to +20°C). Then the difference (T in -T n) varies from 25 to 45.

To make a calculation, you need the average temperature difference for the entire heating season. To do this, in SNiP 23-01-99 "Construction climatology and geophysics" (table 1) find the average temperature of the heating period for a particular city. For example, for Moscow this figure is -26°. In this case, the average difference is 46°C. To determine the heat consumption through each structure, the heat losses of all its layers are added. So, for walls, plaster, masonry material, external thermal insulation, and cladding are taken into account.

4. Calculate the total heat loss, defining them as the sum of Q external walls, floors, doors, windows, ceilings.

5. Ventilation. From 10 to 40% of infiltration (ventilation) losses are added to the result of addition. If high-quality double-glazed windows are installed in the house, and ventilation is not abused, the infiltration coefficient can be taken as 0.1. In some sources, it is indicated that the building does not lose heat at all, since leakages are compensated for by solar radiation and domestic heat emissions.

Counting by hand

Initial data. A one-story house with an area of ​​8x10 m, a height of 2.5 m. The walls, 38 cm thick, are made of ceramic bricks, finished with a layer of plaster from the inside (thickness 20 mm). The floor is made of 30 mm edged board, insulated with mineral wool (50 mm), sheathed with chipboard sheets (8 mm). The building has a cellar, where the temperature in winter is 8°C. The ceiling is covered with wooden panels, insulated with mineral wool (thickness 150 mm). The house has 4 windows 1.2x1 m, an entrance oak door 0.9x2x0.05 m.

Task: determine the total heat loss of the house based on the fact that it is located in the Moscow region. The average temperature difference in the heating season is 46°C (as mentioned earlier). The room and basement have a difference in temperature: 20 – 8 = 12°C.

1. Heat loss through external walls.

Total area (excluding windows and doors): S \u003d (8 + 10) * 2 * 2.5 - 4 * 1.2 * 1 - 0.9 * 2 \u003d 83.4 m2.

The heat resistance of the brickwork and the plaster layer is determined:

• R clade. = 0.38/0.52 = 0.73 m2*°C/W.
• R pieces. = 0.02/0.35 = 0.06 m2*°C/W.
• R total = 0.73 + 0.06 = 0.79 m2*°C/W.
• Heat loss through walls: Q st \u003d 83.4 * 46 / 0.79 \u003d 4856.20 W.

2. Heat loss through the floor.

Total area: S = 8*10 = 80 m2.

The heat resistance of a three-layer floor is calculated.

• R boards = 0.03 / 0.14 = 0.21 m2 * ° C / W.
• R chipboard = 0.008/0.15 = 0.05 m2*°C/W.
• R insulation = 0.05/0.041 = 1.22 m2*°C/W.
• R total = 0.03 + 0.05 + 1.22 = 1.3 m2*°C/W.

We substitute the values ​​\u200b\u200bof the quantities into the formula for finding heat losses: Q floor \u003d 80 * 12 / 1.3 \u003d 738.46 W.

3. Heat loss through the ceiling.

The area of ​​the ceiling surface is equal to the area of ​​the floor S = 80 m2.

When determining the thermal resistance of the ceiling, in this case, wooden panels are not taken into account: they are fixed with gaps and are not a barrier to cold. The thermal resistance of the ceiling coincides with the corresponding parameter of the insulation: R pot. = R ins. = 0.15/0.041 = 3.766 m2*°C/W.

The amount of heat loss through the ceiling: Q sweat. \u003d 80 * 46 / 3.66 \u003d 1005.46 W.

4. Heat loss through windows.

Glazing area: S = 4*1.2*1 = 4.8 m2.

For the manufacture of windows, a three-chamber PVC profile was used (occupies 10% of the window area), as well as a two-chamber double-glazed window with a glass thickness of 4 mm and a distance between glasses of 16 mm. Among the technical characteristics, the manufacturer indicated the thermal resistance of a double-glazed window (R st.p. = 0.4 m2 * ° C / W) and a profile (R prof. = 0.6 m2 * ° C / W). Taking into account the dimensional fraction of each structural element, the average heat resistance of the window is determined:

• R ok. \u003d (R st.p. * 90 + R prof. * 10) / 100 \u003d (0.4 * 90 + 0.6 * 10) / 100 \u003d 0.42 m2 * ° C / W.
• Based on the calculated result, the heat losses through the windows are calculated: Q approx. \u003d 4.8 * 46 / 0.42 \u003d 525.71 W.

Door area S = 0.9 * 2 = 1.8 m2. Thermal resistance R dv. \u003d 0.05 / 0.14 \u003d 0.36 m2 * ° C / W, and Q ext. \u003d 1.8 * 46 / 0.36 \u003d 230 W.

The total amount of heat loss at home is: Q = 4856.20 W + 738.46 W + 1005.46 W + 525.71 W + 230 W = 7355.83 W. Taking into account infiltration (10%), the losses increase: 7355.83 * 1.1 = 8091.41 W.

To accurately calculate how much heat a building loses, use an online heat loss calculator. This is a computer program into which not only the data listed above are entered, but also various additional factors that affect the result. The advantage of the calculator is not only the accuracy of calculations, but also an extensive database of reference data.

Of course, the main sources of heat loss in the house are doors and windows, but when viewing the picture through the screen of a thermal imager, it is easy to see that these are not the only sources of leakage. Heat is also lost through an illiterately mounted roof, a cold floor, and not insulated walls. Heat loss at home today is calculated using a special calculator. This allows you to choose the best option for heating and carry out additional work on the insulation of the building. It is interesting that for each type of building (from timber, logs), the level of heat loss will be different. Let's talk about this in more detail.

Fundamentals of heat loss calculation

Control over heat losses is systematically carried out only for rooms heated in accordance with the season. Premises not intended for seasonal living do not fall under the category of buildings amenable to thermal analysis. The heat loss program at home in this case will not be of practical importance.

In order to conduct a complete analysis, calculate thermal insulation materials and select a heating system with optimal power, it is necessary to have knowledge of the real heat loss of a dwelling. Walls, roofs, windows, and floors are not the only sources of energy leakage from a home. Most of the heat leaves the room through improperly installed ventilation systems.

Factors affecting heat loss

The main factors affecting the level of heat loss are:

• A high level of temperature difference between the internal microclimate of the room and the temperature outside.
• The nature of the thermal insulation properties of enclosing structures, which include walls, ceilings, windows, etc.

Heat loss measurement values

Enclosing structures perform a barrier function for heat and do not allow it to freely go outside. This effect is explained by the thermal insulation properties of products. The value used to measure thermal insulation properties is called heat transfer resistance. Such an indicator is responsible for reflecting the temperature difference during the passage of the nth amount of heat through a section of protective structures with an area of ​​​​1 m 2. So, let's figure out how to calculate the heat loss at home.

The main values ​​\u200b\u200bnecessary for calculating the heat loss of a house include:

• q is a value indicating the amount of heat leaving the room to the outside through 1 m 2 of the barrier structure. Measured in W / m 2.
• ∆T is the difference between indoor and outdoor temperatures. It is measured in degrees (o C).
• R is the resistance to heat transfer. Measured in °C/W/m² or °C m²/W.
• S is the area of ​​the building or surface (used as needed).

Formula for calculating heat loss

The heat loss program of the house is calculated using a special formula:

When calculating, remember that for structures consisting of several layers, the resistance of each layer is summed up. So, how to calculate the heat loss of a frame house lined with bricks from the outside? The resistance to heat loss will be equal to the sum of the resistance of brick and wood, taking into account the air gap between the layers.

Important! Please note that the resistance calculation is carried out for the coldest time of the year, when the temperature difference reaches its peak. Reference books and manuals always indicate exactly this reference value, which is used for further calculations.

Features of calculating the heat loss of a wooden house

The calculation of heat loss at home, the features of which must be taken into account when calculating, is carried out in several stages. The process requires special attention and concentration. You can calculate heat loss in a private house according to a simple scheme as follows:

• Defined through the walls.
• Calculate through window structures.
• Through doorways.
• Calculate through overlaps.
• Calculate the heat loss of a wooden house through the flooring.
• Add up the previously obtained values.
• Considering thermal resistance and energy loss through ventilation: 10 to 360%.

For the results of points 1-5, the standard formula for calculating the heat loss of a house (from timber, brick, wood) is used.

Important! Thermal resistance for window structures is taken from SNIP II-3-79.

Building directories often contain information in a simplified form, that is, the results of calculating the heat loss of a house from a bar are given for different types of walls and floors. For example, they calculate the resistance at a temperature difference for atypical rooms: corner and non-corner rooms, one- and multi-storey buildings.

The need to calculate heat loss

The arrangement of a comfortable home requires strict control of the process at each stage of the work. Therefore, the organization of the heating system, which is preceded by the choice of the method of heating the room itself, cannot be overlooked. When working on the construction of a house, a lot of time will have to be devoted not only to project documentation, but also to calculating the heat loss of the house. If in the future you are going to work in the field of design, then engineering skills in calculating heat loss will definitely come in handy for you. So why not practice doing this work by experience and make a detailed calculation of heat loss for your own home.

Important! The choice of method and power of the heating system directly depends on the calculations you have made. If you calculate the heat loss indicator incorrectly, you risk freezing in cold weather or exhausting from heat due to excessive heating of the room. It is necessary not only to choose the right device, but also to determine the number of batteries or radiators that can heat one room.

Estimation of heat loss on a calculation example

If you do not need to study the calculation of heat loss at home in detail, we will focus on the estimated analysis and determination of heat loss. Sometimes errors occur in the calculation process, so it is better to add the minimum value to the estimated power of the heating system. In order to proceed with the calculations, it is necessary to know the resistance index of the walls. It differs depending on the type of material from which the building is made.

Resistance (R) for houses made of ceramic bricks (with a masonry thickness of two bricks - 51 cm) is 0.73 ° C m² / W. The minimum thickness at this value should be 138 cm. When using expanded clay concrete as the base material (with a wall thickness of 30 cm), R is 0.58 ° C m² / W with a minimum thickness of 102 cm. In a wooden house or a building made of timber with a wall thickness of 15 cm and a resistance level of 0.83 °C m²/W, a minimum thickness of 36 cm is required.

Building materials and their resistance to heat transfer

Based on these parameters, you can easily carry out calculations. You can find the resistance values ​​​​in the reference book. In construction, brick, a log house made of timber or logs, foam concrete, wooden floors, ceilings are most often used.

Heat transfer resistance values ​​for:

• brick wall (thickness 2 bricks) - 0.4;
• a log house made of timber (thickness 200 mm) - 0.81;
• log cabin (diameter 200 mm) - 0.45;
• foam concrete (thickness 300 mm) - 0.71;
• wooden floor - 1.86;
• ceiling overlap - 1.44.

Based on the information provided above, we can conclude that for the correct calculation of heat loss, only two quantities are required: the temperature difference indicator and the level of resistance to heat transfer. For example, a house is made of wood (logs) 200 mm thick. Then the resistance is 0.45 ° C m² / W. Knowing these data, you can calculate the percentage of heat loss. For this, a division operation is carried out: 50 / 0.45 \u003d 111.11 W / m².

The calculation of heat loss by area is performed as follows: heat loss is multiplied by 100 (111.11 * 100 \u003d 11111 W). Taking into account the decoding of the value (1 W \u003d 3600), we multiply the resulting number by 3600 J / h: 11111 * 3600 \u003d 39.999 MJ / h. Having carried out such simple mathematical operations, any owner can find out about the heat loss of his house in an hour.

Calculation of room heat loss online

There are many sites on the Internet that offer the service of online calculation of the heat loss of a building in real time. The calculator is a program with a special form to fill out, where you enter your data and after the automatic calculation you will see the result - a figure that will mean the amount of heat output from the dwelling.

A dwelling is a building in which people live during the entire heating season. As a rule, suburban buildings, where the heating system operates periodically and as needed, do not belong to the category of residential buildings. In order to carry out re-equipment and achieve the optimal heat supply mode, it will be necessary to carry out a number of works and, if necessary, increase the power of the heating system. Such re-equipment can be delayed for a long period. In general, the whole process depends on the design features of the house and the indicators of increasing the power of the heating system.

Many have not even heard of the existence of such a thing as “heat loss at home”, and subsequently, having made a structurally correct installation of the heating system, they suffer all their lives from a lack or excess of heat in the house, without even realizing the true reason. That is why it is so important to take into account every detail when designing a home, to personally control and build, in order to ultimately get a high-quality result. In any case, the dwelling, no matter what material it is built from, should be comfortable. And such an indicator as the heat loss of a residential building will help make staying at home even more pleasant.