# Calculation of air heating: basic principles + calculation example

Installation of the heating system is not possible without preliminary calculations. The information obtained should be as accurate as possible, therefore, the calculation of air heating is done by experts using specialized programs, taking into account the nuances of the design.

It is possible to calculate the air heating system (hereinafter - NWO) independently, having elementary knowledge in mathematics and physics.

In this article, we will tell you how to calculate the level of heat loss at home and water heat treatment. In order for everything to be as clear as possible, specific examples of calculations will be given.

The content of the article:

## Calculation of heat loss at home

To select the CBO, it is necessary to determine the amount of air for the system, the initial temperature of the air in the duct for optimal heating of the room. To find out this information, you need to calculate the heat loss at home, and start the basic calculations later.

Any building during the cold weather loses thermal energy. Its maximum number leaves the room through the walls, roof, windows, doors and other enclosing elements (hereinafter - OK), facing one side on the street.

To ensure a certain temperature in the house, you need to calculate the thermal power, which is able to compensate for the heat costs and maintain in the house desired temperature.

There is a misconception that heat losses are the same for every home. Some sources claim that 10 kW is enough to heat a small house of any configuration, others are limited to 7-8 kW per sq. meter.

According to the simplified calculation scheme every 10 m^{2} the exploited area in the northern regions and the middle-band areas should be provided with the supply of 1 kW of thermal power. This figure, individual for each building, is multiplied by a factor of 1.15, thereby creating a reserve of thermal power in case of unexpected losses.

However, such estimates are rather rough, in addition, they do not take into account the quality, features of the materials used in the construction of the house, climatic conditions and other factors affecting heat costs.

If the construction of the house used modern construction thermal conductivity materials which are low, then the heat loss of the structure will be less, which means that the heat power will need less.

If you take thermal equipment that generates more power than necessary, then excess heat will appear, which is usually compensated by ventilation. In this case, additional financial expenses appear.

If low-power equipment is selected for the CBO, then a shortage of heat will be felt in the room, since the device will not be able to generate the required amount of energy, which will require the purchase of additional heating units.

Thermal costs of a building depend on:

- the structure of the enclosing elements (walls, ceilings, etc.), their thickness;
- heated surface area;
- orientation relative to cardinal points;
- minimum temperature outside the window in the region or city during 5 winter days;
- the duration of the heating season;
- processes of infiltration, ventilation;
- domestic heat supply;
- heat consumption for domestic needs.

It is impossible to correctly calculate heat loss without taking into account infiltration and ventilation, which significantly affect the quantitative component. Infiltration is a natural process of moving air masses that occurs during the movement of people around the room, opening windows for ventilation and other domestic processes.

Ventilation is a specially installed system through which air is supplied, and air can enter a room with a lower temperature.

Heat enters the room not only through the heating system, but also through heating appliances, incandescent lamps, and people. It is also important to take into account the heat consumption for heating cold items brought from the street, clothes.

Before choosing equipment for water cooling systems, heating system design It is important to calculate the heat loss at home with high accuracy. This can be done using the free program Valtec. In order not to delve into the intricacies of the application, you can use mathematical formulas that give high accuracy of calculations.

To calculate the total heat loss Q of the home, it is necessary to calculate the heat consumption of the building envelope Q_{org.k}, energy consumption for ventilation and infiltration Q_{v}, take into account household expenses Q_{t}. Losses are measured and recorded in watts.

To calculate the total heat consumption Q use the formula:

**Q = Q _{org.k} + Q_{v} - Q_{t}**

Next, we consider the formulas for determining heat costs:

**Q _{org.k }, Q_{v}, Q_{t}**.

### Determination of heat losses of building envelopes

Through the enclosing elements of the house (walls, doors, windows, ceiling and floor), the greatest amount of heat is released. To determine Q_{org.k} it is necessary to separately calculate the heat loss that each structural element bears.

That is Q_{org.k} calculated by the formula:

**Q _{org.k}**

**= Q**

_{pol}+ Q_{st}+ Q_{okn}+ Q_{pt}+ Q_{dv}To determine the Q of each element of the house, it is necessary to find out its structure and coefficient of thermal conductivity or coefficient of thermal resistance, which is indicated in the passport of the material.

Calculation of heat loss occurs for each homogeneous layer of the enclosing element. For example, if a wall consists of two dissimilar layers (insulation and brickwork), then the calculation is made separately for insulation and brickwork.

Calculate the heat consumption of the layer, taking into account the desired temperature in the room by the expression:

**Q _{st} = S × (t_{v} - t_{n}) × B × l / k**

Variables have the following meanings in an expression:

- S - layer area, m
^{2}; - t
_{v} - t
_{n}- the average temperature of the coldest 5 days in the region, ° С; - k is the coefficient of thermal conductivity of the material;
- B is the thickness of each layer of the enclosing element, m;
- l– tabular parameter, takes into account the features of heat consumption for OK located in different parts of the world.

If windows or doors are built into the wall for calculation, then when calculating Q from the total area of OK, it is necessary to subtract the area of the window or door, since their heat consumption will be different.

The coefficient of thermal resistance is calculated by the formula:

**D = B / k**

The heat loss formula for a single layer can be represented as:

**Q _{st} = S × (t_{v} - t_{n}) × D × l**

In practice, to calculate the Q of the floor, walls or ceilings, the D coefficients of each OK layer are separately calculated, summed, and substituted into the general formula, which simplifies the calculation process.

### Accounting for infiltration and ventilation costs

Low temperature air can enter the room from the ventilation system, which significantly affects heat loss. The general formula for this process is as follows:

**Q _{v} = 0.28 × L_{n} × p_{v} × c × (t_{v} - t_{n})**

In an expression, alphabetic characters have the meaning:

- L
_{n}- intake air flow, m^{3}/ h; - p
_{v}- air density in the room at a given temperature, kg / m^{3}; - t
_{v} - t
_{n}- the average temperature of the coldest 5 days in the region, ° С; - c is the heat capacity of air, kJ / (kg * ° C).

Parameter L_{n }taken from the technical characteristics of the ventilation system. In most cases, the supply air has a specific flow rate of 3 m^{3}/ h, based on which L_{n} calculated by the formula:

**L _{n} = 3 × S_{pol}**

In the formula S_{pol} - floor area, m^{2}.

Indoor air density**p _{v}** defined by the expression:

**p _{v} = 353/273 + t_{v}**

Here t_{v} - the set temperature in the house, measured in ° C.

The heat capacity c is a constant physical quantity and is equal to 1.005 kJ / (kg × ° C).

Unorganized ventilation, or infiltration, is determined by the formula:

**Q _{i} = 0.28 × ∑G_{h }×**

**c × (t**

_{v}- t_{n}) × k_{t}In the equation:

- G
_{h}- air flow through each fence is a tabular value, kg / h; - k
_{t}- coefficient of influence of thermal air flow, taken from the table; - t
_{v }, t_{n}- set temperatures indoors and outdoors, ° C.

When the doors are opened, the most significant heat loss occurs, therefore, if the entrance is equipped with air-curtains, they should also be taken into account.

To calculate the heat loss of the doors, the formula is used:

**Q _{ot.d} = Q_{dv} × j × H**

In the expression:

- Q
_{dv}- calculated heat loss of the external doors; - H - building height, m;
- j is a tabular coefficient, depending on the type of doors and their location.

If the house has organized ventilation or infiltration, then the calculations are made according to the first formula.

The surface of the enclosing structural elements may be heterogeneous - there may be gaps or leaks on it, through which air passes. These heat losses are considered negligible, but they can also be determined. This can be done exclusively by software methods, since it is impossible to calculate some functions without using applications.

### Household heat

Through electrical appliances, the human body, lamps, additional heat comes into the room, which is also taken into account when calculating heat losses.

It has been experimentally established that such receipts cannot exceed the mark of 10 W per 1 m^{2}. Therefore, the calculation formula can be of the form:

**Q _{t} = 10 × S_{pol}**

In the expression S_{pol} - floor area, m^{2}.

## The main methodology for calculating NWO

The main principle of operation of any NWO is to transfer thermal energy through the air by cooling the coolant. Its main elements are a heat generator and a heat pipe.

Air is supplied into the room already heated to a temperature t_{r}to maintain the desired temperature t_{v}. Therefore, the amount of accumulated energy should be equal to the total heat loss of the building, that is, Q. There is equality:

**Q = E _{ot} ×**

**c × (t**

_{v}- t_{n})In the formula E - flow rate of heated air kg / s for heating the room. From equality we can express E_{ot}:

**E _{ot }= Q / (c × (t_{v} - t_{n}))**

Recall that the heat capacity of air is c = 1005 J / (kg × K).

The formula determines only the amount of air supplied, used only for heating only in recirculation systems (hereinafter - RSVO).

If CBO is used as ventilation, the amount of air supplied is calculated as follows:

- If the amount of air for heating exceeds the amount of air for ventilation or is equal to it, then the amount of air for heating is taken into account, and the system is selected as direct-flow (hereinafter - PSVO) or with partial recirculation (hereinafter - HRWS).
- If the amount of air for heating is less than the amount of air needed for ventilation, then only the amount of air needed for ventilation is taken into account, the HVAC is introduced (sometimes - HVAC), and the temperature of the supplied air is calculated by the formula: t
_{r}= t_{v}+ Q / c × E_{vent}.

In case of exceeding by t_{r }permissible parameters, the amount of air introduced through ventilation should be increased.

If the room has sources of constant heat, then the temperature of the supplied air is reduced.

For a single room, the indicator t_{r }may be different. Technically, it is possible to realize the idea of supplying different temperatures to individual rooms, but it is much easier to supply air of the same temperature to all rooms.

In this case, the total temperature t_{r} take the one that turned out to be the smallest. Then the amount of air supplied is calculated by the formula defining E_{ot}.

Next, we determine the formula for calculating the volume of incoming air V_{ot} at its heating temperature t_{r}:

**V _{ot} = E_{ot}/ p_{r}**

The answer is written in m^{3}/ h

However, indoor air exchange V_{p} will differ from the value of V_{ot}, since it is necessary to determine it based on the internal temperature t_{v}:

**V _{ot} = E_{ot}/ p_{v}**

In the formula for determining V_{p} and v_{ot} air density indicators p_{r} and p_{v} (kg / m^{3}) are calculated taking into account the temperature of the heated air t_{r} and room temperature t_{v}.

Indicated room temperature t_{r} must be higher than t_{v}. This will reduce the amount of air supplied and will reduce the dimensions of the channels of systems with natural air movement or reduce electricity consumption if mechanical motivation is used to circulate the heated air mass.

Traditionally, the maximum temperature of the air entering the room when it is supplied at a height exceeding the mark of 3.5 m should be 70 ° С. If air is supplied at an altitude of less than 3.5 m, then its temperature is usually equated to 45 ° C.

For residential premises 2.5 m high, the permissible temperature limit is 60 ° C. When the temperature is set higher, the atmosphere loses its properties and is not suitable for inhalation.

If the air-thermal curtains are located at the external gates and openings facing outward, then the temperature of the incoming air is allowed 70 ° C, for curtains located in the outer doors, up to 50 ° C.

The supplied temperature is affected by the air supply methods, the direction of the jet (vertically, along the slope, horizontally, etc.). If people are constantly in the room, then the temperature of the supplied air should be reduced to 25 ° C.

After carrying out preliminary calculations, it is possible to determine the necessary heat consumption for heating the air.

For RSVO heat costs Q_{1} calculated by the expression:

**Q _{1} = E_{ot} ×**

**(t**

_{r}- t_{v}) × cFor PSVO calculation Q_{2 }produced by the formula:

**Q _{2} = E_{vent} ×**

**(t**

_{r}- t_{v}) × cHeat Consumption Q_{3 }for HRW is found by the equation:

**Q _{3} = [E_{ot} × (t_{r} - t_{v}) + E_{vent} ×**

**(t**

_{r}- t_{v})] × cIn all three expressions:

- E
_{ot }and E_{vent}- air consumption in kg / s for heating (E_{ot}) and ventilation (E_{vent}); - t
_{n}- outdoor temperature in ° C.

The remaining characteristics of the variables are the same.

In CHRSVO the amount of recirculated air is determined by the formula:

**E _{rec} = E_{ot} - E_{vent}**

Variable e_{ot} expresses the amount of mixed air heated to temperature t_{r}.

There is a peculiarity in PSVO with natural motivation - the amount of moving air varies depending on the temperature outside. If the outside temperature drops, the system pressure rises. This leads to an increase in the air entering the house. If the temperature rises, the reverse process occurs.

Also in the air-conditioning system, unlike ventilation systems, air moves with a lower and changing density in comparison with the density of the air surrounding the air ducts.

Because of this phenomenon, the following processes occur:

- Coming from the generator, the air, passing through the air ducts, is noticeably cooled during movement
- During natural movement, the amount of air entering the room changes during the heating season.

The above processes are not taken into account if fans are used in the air conditioning system for air circulation, and it also has a limited length and height.

If the system has many branches, quite long, and the building is large and tall, then it is necessary to reduce the process of cooling the air in the ducts, to reduce the redistribution of air coming under the influence of natural circulation pressure.

To control the process of cooling the air, perform thermal calculation of the ducts. To do this, it is necessary to establish the initial air temperature and specify its flow rate using formulas.

To calculate the heat flux Q_{ohl} through the walls of the duct, the length of which is equal to l, use the formula:

**Q _{ohl} = q_{1} × l**

In the expression, q_{1} denotes the heat flux passing through the walls of the duct 1 m long. The parameter is calculated by the expression:

**q _{1} = k × S_{1} × (t_{sr} - t_{v}) = (t_{sr} - t_{v}) / D_{1}**

In equation D_{1} - heat transfer resistance from heated air with an average temperature t_{sr} across square S_{1} walls of the duct 1 m long indoors at temperature t_{v}.

The heat balance equation looks like this:

**q _{1}l = E_{ot} × c × (t_{nach} - t_{r})**

In the formula:

- E
_{ot}- the amount of air required for heating the room, kg / h; - c is the specific heat of air, kJ / (kg ° C);
- t
_{nac }- air temperature at the beginning of the duct, ° C; - t
_{r}- temperature of air discharged into the room, ° С.

The heat balance equation allows you to set the initial temperature of the air in the duct at a given final temperature and, conversely, find out the final temperature at a given initial temperature, as well as determine the air flow.

Temperature t_{nach} can also be found by the formula:

**t _{nach} = t_{v} + ((Q + (1 - η) × Q_{ohl})) × (t_{r} - t_{v})**

Here η is a part of Q_{ohl}entering the room in the calculations is taken equal to zero. The characteristics of the remaining variables were named above.

The refined hot air flow formula will look like this:

**Eot = (Q + (1 - η) × Q _{ohl}) / (c × (t_{sr} - t_{v}))**

All literal values in the expression are defined above. Let's move on to an example of calculating air heating for a particular house.

## Example of calculating heat loss at home

The considered house is located in the city of Kostroma, where the temperature outside the window on the coldest five-day day reaches -31 degrees, the temperature of the soil - +5 ° С. Desired room temperature - +22 ° С.

We will consider a house with the following dimensions:

- width - 6.78 m;
- length - 8.04 m;
- height - 2.8 m.

Values will be used to calculate the area of the enclosing elements.

The walls of the building consist of:

- aerated concrete with thickness B = 0.21 m, thermal conductivity coefficient k = 2.87;
- polyfoam B = 0.05 m, k = 1.678;
- facing brick B = 0.09 m, k = 2.26.

When determining k, one should use the information from the tables, or better, information from the technical passport, since the composition of materials from different manufacturers may differ, therefore, have different characteristics.

The floor of the house consists of the following layers:

- sand, B = 0.10 m, k = 0.58;
- crushed stone, B = 0.10 m, k = 0.13;
- concrete, B = 0.20 m, k = 1.1;
- ecowool insulation, B = 0.20 m, k = 0.043;
- reinforced screed, B = 0.30 m k = 0.93.

In the above plan of the house, the floor has the same structure throughout the area, there is no basement.

The ceiling consists of:

- mineral wool, B = 0.10 m, k = 0.05;
- drywall, B = 0.025 m, k = 0.21;
- pine shields, B = 0.05 m, k = 0.35.

The ceiling has no access to the attic.

There are only 8 windows in the house, all of them are double-chamber with K-glass, argon, indicator D = 0.6. Six windows are 1.2 × 1.5 m in size, one is 1.2 × 2 m in size, and one is 0.3 × 0.5 m in size. Doors are 1 × 2.2 m in size and the passport D is 0.36.

### Calculation of wall heat loss

We will calculate the heat loss for each wall individually.

First, find the area of the north wall:

S_{sev} = 8.04 × 2.8 = 22.51

There are no doorways and window openings on the wall, so we will use this value S.

Based on the composition of the wall, we find its total heat resistance equal to:

**D _{s.sten} = D_{gb} + D_{pn} + D_{kr}**

To find D, we use the formula:

**D = B / k**

Then, substituting the initial values, we obtain:

D_{s.sten} = 0.21/2.87 + 0.05/1.678 + 0.09/2.26 = 0.14

For calculations we use the formula:

**Q _{st} = S × (t_{v} - t_{n}) × D × l**

Given that the coefficient l for the northern wall is 1.1, we get:

Q_{sev.st} = 22.51 × (22 + 31) × 0.14 × 1.1 = 184

In the south wall there is one window with an area of:

S_{ok3} = 0.5 × 0.3 = 0.15

Therefore, in calculations from the S southern wall, it is necessary to subtract S windows in order to obtain the most accurate results.

S_{yuj.s} = 22.51 – 0.15 = 22.36

The parameter l for the south direction is 1. Then:

Q_{sev.st} = 22.36 × (22 + 31) × 0.14 × 1 = 166

For the eastern and western walls, the refinement coefficient is l = 1.05; therefore, it suffices to calculate the surface area of the OK without taking into account S windows and doors.

S_{ok1} = 1.2 × 1.5 × 6 = 10.8

S_{ok2} = 1.2 × 2 = 2.4

S_{d} = 1 × 2.2 = 2.2

S_{zap + vost} = 2 × 6.78 × 2.8 – 2.2 – 2.4 – 10.8 = 22.56

Then:

Q_{zap + vost} = 22.56 × (22 + 31) × 0.14 × 1.05 = 176

Ultimately, the total Q of the walls is equal to the sum of Q of all the walls, that is:

Q_{sten} = 184 + 166 + 176 = 526

Total, heat leaves through the walls in the amount of 526 watts.

### Heat loss through windows and doors

The plan of the house shows that the doors and 7 windows face east and west, therefore, the parameter l = 1.05. The total area of 7 windows, taking into account the above calculations, is equal to:

S_{okn} = 10.8 + 2.4 = 13.2

For them, Q, taking into account that D = 0.6, will be calculated as follows:

Q_{ok4} = 13.2 × (22 + 31) × 0.6 × 1.05 = 630

We calculate Q of the south window (l = 1).

Q_{ok5} = 0.15 × (22 + 31) × 0.6 × 1 = 5

For doors, D = 0.36, and S = 2.2, l = 1.05, then:

Q_{dv} = 2.2 × (22 + 31) × 0.36 × 1.05 = 43

We summarize the resulting heat loss and get:

Q_{ok + dv} = 630 + 43 + 5 = 678

Next, we define Q for the ceiling and floor.

### Calculation of heat losses of the ceiling and floor

For ceiling and floor l = 1. Calculate their area.

S_{pol} = S_{pot} = 6.78 × 8.04 = 54.51

Given the composition of the floor, we define the total D.

D_{pol} = 0.10/0.58 + 0.10/0.13 + 0.2/1.1 + 0.2/0.043 + 0.3/0.93 =61

Then the heat loss of the floor, taking into account the fact that the temperature of the earth is +5, is equal to:

Q_{pol} = 54.51 × (21 – 5) × 6.1 × 1 = 5320

Calculate the total D ceiling:

D_{pot} = 0.10/0.05 + 0.025/0.21 + 0.05/0.35 = 2.26

Then Q of the ceiling will be equal to:

Q_{pot} = 54.51 × (22 + 31) × 2.26 = 6530

The total heat loss through OK will be equal to:

Q_{ogr.k} = 526 + 678 +6530 + 5320 = 13054

Total, the heat loss of the house will be equal to 13054 W or almost 13 kW.

### Calculation of heat losses of ventilation

The room operates ventilation with a specific air exchange of 3 m^{3}/ h, the entrance is equipped with an air-thermal canopy, so for calculations it is enough to use the formula:

**Q _{v} = 0.28 × L_{n} × p_{v} × c × (t_{v} - t_{n})**

We calculate the density of air in the room at a given temperature of +22 degrees:

p_{v} = 353/(272 + 22) = 1.2

Parameter L_{n} equal to the product of the specific consumption by the floor area, that is:

L_{n} = 3 × 54.51 = 163.53

The heat capacity of air c is 1.005 kJ / (kg × ° C).

Given all the information, we find the ventilation Q:

Q_{v} = 0.28 × 163.53 × 1.2 × 1.005 × (22 + 31) = 3000

Total heat costs for ventilation will be 3000 watts or 3 kW.

### Domestic Heat

Household income is calculated by the formula.

**Q _{t} = 10 × S_{pol} **

That is, substituting the known values, we obtain:

Q_{t }= 54.51 × 10 = 545

Summing up, we can see that the total heat loss Q at home will be equal to:

Q = 13054 + 3000 - 545 = 15509

We take Q = 16000 W or 16 kW as the operating value.

## Examples of calculations for the CBO

Let the temperature of the supplied air (t_{r}) - 55 ° С, the desired room temperature (t_{v}) - 22 ° C, heat loss at home (Q) - 16,000 watts.

### Determining the amount of air for RSVO

To determine the mass of the supplied air at temperature t_{r} the formula is used:

**E _{ot} = Q / (c × (t_{r} - t_{v})) **

Substituting the parameter values in the formula, we obtain:

E_{ot }= 16000/(1.005 × (55 – 22)) = 483

The volumetric amount of air supplied is calculated by the formula:

**V _{ot} = E_{ot }/ p_{r}**

Where:

p_{r} = 353 / (273 + t_{r})

First, we calculate the density p:

p_{r} = 353/(273 + 55) = 1.07

Then:

V_{ot} = 483/1.07 = 451.

The air exchange in the room is determined by the formula:

**Vp =** **E _{ot }/ p_{v}**

Determine the density of air in the room:

p_{v} = 353/(273 + 22) = 1.19

Substituting the values in the formula, we get:

V_{p} = 483/1.19 = 405

Thus, the air exchange in the room is 405 m^{3} per hour, and the volume of air supplied should be equal to 451 m^{3} in an hour.

### Calculation of the amount of air for HWAC

To calculate the amount of air for HWRS, we take the information obtained from the previous example, as well as t_{r }= 55 ° C, t_{v }= 22 ° C; Q = 16000 watts. The amount of air required for ventilation, E_{vent}= 110 m^{3}/ h Estimated outdoor temperature t_{n}= -31 ° C.

For the calculation of the HFRS we use the formula:

**Q _{3} = [E_{ot} × (t_{r} - t_{v}) + E_{vent} × p_{v} ×**

**(t**

_{r}- t_{v})] × cSubstituting the values, we get:

Q_{3} = [483 × (55 – 22) + 110 × 1.19 × (55 – 31)] × 1.005 = 27000

The volume of recirculated air will be 405-110 = 296 m^{3} including additional heat consumption is equal to 27000-16000 = 11000 watts.

### Determination of initial air temperature

The resistance of the mechanical duct is D = 0.27 and is taken from its technical characteristics. The length of the duct outside the heated room is l = 15 m. It is determined that Q = 16 kW, the temperature of the internal air is 22 degrees, and the required temperature for heating the room is 55 degrees.

Define E_{ot} according to the above formulas. We get:

E_{ot} = 10 × 3.6 × 1000/ (1.005 × (55 – 22)) = 1085

Heat flux q_{1} will be:

q_{1} = (55 – 22)/0.27 = 122

The initial temperature with a deviation of η = 0 will be:

t_{nach} = 22 + (16 × 1000 + 137 × 15) × (55 – 22)/ 1000 × 16 = 60

Specify the average temperature:

t_{sr} = 0.5 × (55 + 60) = 57.5

Then:

Q_{otkl} = ((574 -22)/0.27) × 15 = 1972

Given the information we find:

t_{nach} = 22 + (16 × 1000 + 1972) × (55 – 22)/(1000 × 16) = 59

It follows from this that when air moves, 4 degrees of heat is lost. To reduce heat loss, it is necessary to insulate the pipes. We also recommend that you familiarize yourself with our other article, which describes in detail the arrangement process. air heating systems.

## Conclusions and useful video on the topic

An informative video about the calculations of CB using the Ecxel program:

Trusting the calculations of NWO is necessary for professionals, because only specialists have experience, relevant knowledge, will take into account all the nuances in the calculations.

Have questions, find inaccuracies in the above calculations, or want to supplement the material with valuable information? Please leave your comments in the block below.

Such calculations of heat loss are made without fail at the design stage of houses. I had to explain to customers how in the future they can save their money on house maintenance if the cost-based ratio of the cost of thermal insulation of the walls and the upcoming heating costs are taken into account by thermal calculations. Only using accurate numbers, we can conclude that it is unreasonable to build walls too bulky and expensive, since these investments can exceed the savings in heating a house even for several decades.

And with a ready-made house, these calculations will help increase efficiency? Unfortunately, at the design and construction stage, I thought that “it will do so”.

An air heating system is actually a very good thing, it is inexpensive and quite effective, but few people have the right idea about it. In Europe, this type of heating has been used for a very long time, we are behind the times. And his advantages are very significant: he quickly warms up the room, costs a lot, and, in fact, this may be the only heating in the house.

In the example, a strange figure of the coefficient of thermal conductivity of aerated concrete. She is very overpriced. Even for d600 it is no more than 0.2

Everything was good until the video came to the end ... It has long been proven that the walls just do not need to be heated, but the air needs to be heated. For this reason, in the drop-off radiator, the radiator itself should not be mounted on the wall, but at a distance from the wall of at least 5 cm + the height from the floor to the beginning of the radiator is not higher than 20 cm, and the window sill above the radiator is at least 10 cm.

Yes, and the wall behind the radiator is covered with a foil foam, so that the heat does not go into the wall, but is reflected.

All this is done so that from the bottom of the room cold air is sucked in by a radiator and thereby ensures its circulation and heating.And if you warm the walls, then the room will be cold and this will be a waste of energy for nothing.