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
- The main methodology for calculating NWO
- Example of calculating heat loss at home
- Examples of calculations for the CBO
- Conclusions and useful video on the topic
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 m2 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 Qorg.k, energy consumption for ventilation and infiltration Qv, take into account household expenses Qt. Losses are measured and recorded in watts.
To calculate the total heat consumption Q use the formula:
Q = Qorg.k + Qv - Qt
Next, we consider the formulas for determining heat costs:
Qorg.k , Qv, Qt.
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 Qorg.k it is necessary to separately calculate the heat loss that each structural element bears.
That is Qorg.k calculated by the formula:
Qorg.k = Qpol + Qst + Qokn + Qpt + Qdv
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:
Qst = S × (tv - tn) × B × l / k
Variables have the following meanings in an expression:
- S - layer area, m2;
- tv - the desired temperature in the house, ° C; for corner rooms, the temperature is taken 2 degrees higher;
- tn - 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:
Qst = S × (tv - tn) × 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:
Qv = 0.28 × Ln × pv × c × (tv - tn)
In an expression, alphabetic characters have the meaning:
- Ln - intake air flow, m3/ h;
- pv - air density in the room at a given temperature, kg / m3;
- tv - temperature in the house, ° С;
- tn - the average temperature of the coldest 5 days in the region, ° С;
- c is the heat capacity of air, kJ / (kg * ° C).
Parameter Ln taken from the technical characteristics of the ventilation system. In most cases, the supply air has a specific flow rate of 3 m3/ h, based on which Ln calculated by the formula:
Ln = 3 × Spol
In the formula Spol - floor area, m2.
Indoor air densitypv defined by the expression:
pv = 353/273 + tv
Here tv - 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:
Qi = 0.28 × ∑Gh × c × (tv - tn) × kt
In the equation:
- Gh - air flow through each fence is a tabular value, kg / h;
- kt - coefficient of influence of thermal air flow, taken from the table;
- tv , tn - 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:
Qot.d = Qdv × j × H
In the expression:
- Qdv - 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.
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 m2. Therefore, the calculation formula can be of the form:
Qt = 10 × Spol
In the expression Spol - floor area, m2.
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 trto maintain the desired temperature tv. Therefore, the amount of accumulated energy should be equal to the total heat loss of the building, that is, Q. There is equality:
Q = Eot × c × (tv - tn)
In the formula E - flow rate of heated air kg / s for heating the room. From equality we can express Eot:
Eot = Q / (c × (tv - tn))
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: tr = tv + Q / c × Event.
In case of exceeding by tr 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 tr 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 tr take the one that turned out to be the smallest. Then the amount of air supplied is calculated by the formula defining Eot.
Next, we determine the formula for calculating the volume of incoming air Vot at its heating temperature tr:
Vot = Eot/ pr
The answer is written in m3/ h
However, indoor air exchange Vp will differ from the value of Vot, since it is necessary to determine it based on the internal temperature tv:
Vot = Eot/ pv
In the formula for determining Vp and vot air density indicators pr and pv (kg / m3) are calculated taking into account the temperature of the heated air tr and room temperature tv.
Indicated room temperature tr must be higher than tv. 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 Q1 calculated by the expression:
Q1 = Eot × (tr - tv) × c
For PSVO calculation Q2 produced by the formula:
Q2 = Event × (tr - tv) × c
Heat Consumption Q3 for HRW is found by the equation:
Q3 = [Eot × (tr - tv) + Event × (tr - tv)] × c
In all three expressions:
- Eot and Event - air consumption in kg / s for heating (Eot) and ventilation (Event);
- tn - 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:
Erec = Eot - Event
Variable eot expresses the amount of mixed air heated to temperature tr.
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 Qohl through the walls of the duct, the length of which is equal to l, use the formula:
Qohl = q1 × l
In the expression, q1 denotes the heat flux passing through the walls of the duct 1 m long. The parameter is calculated by the expression:
q1 = k × S1 × (tsr - tv) = (tsr - tv) / D1
In equation D1 - heat transfer resistance from heated air with an average temperature tsr across square S1 walls of the duct 1 m long indoors at temperature tv.
The heat balance equation looks like this:
q1l = Eot × c × (tnach - tr)
In the formula:
- Eot - the amount of air required for heating the room, kg / h;
- c is the specific heat of air, kJ / (kg ° C);
- tnac - air temperature at the beginning of the duct, ° C;
- tr - 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 tnach can also be found by the formula:
tnach = tv + ((Q + (1 - η) × Qohl)) × (tr - tv)
Here η is a part of Qohlentering 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) × Qohl) / (c × (tsr - tv))
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:
Ssev = 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:
Ds.sten = Dgb + Dpn + Dkr
To find D, we use the formula:
D = B / k
Then, substituting the initial values, we obtain:
Ds.sten = 0.21/2.87 + 0.05/1.678 + 0.09/2.26 = 0.14
For calculations we use the formula:
Qst = S × (tv - tn) × D × l
Given that the coefficient l for the northern wall is 1.1, we get:
Qsev.st = 22.51 × (22 + 31) × 0.14 × 1.1 = 184
In the south wall there is one window with an area of:
Sok3 = 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.
Syuj.s = 22.51 – 0.15 = 22.36
The parameter l for the south direction is 1. Then:
Qsev.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.
Sok1 = 1.2 × 1.5 × 6 = 10.8
Sok2 = 1.2 × 2 = 2.4
Sd = 1 × 2.2 = 2.2
Szap + vost = 2 × 6.78 × 2.8 – 2.2 – 2.4 – 10.8 = 22.56
Qzap + 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:
Qsten = 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:
Sokn = 10.8 + 2.4 = 13.2
For them, Q, taking into account that D = 0.6, will be calculated as follows:
Qok4 = 13.2 × (22 + 31) × 0.6 × 1.05 = 630
We calculate Q of the south window (l = 1).
Qok5 = 0.15 × (22 + 31) × 0.6 × 1 = 5
For doors, D = 0.36, and S = 2.2, l = 1.05, then:
Qdv = 2.2 × (22 + 31) × 0.36 × 1.05 = 43
We summarize the resulting heat loss and get:
Qok + 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.
Spol = Spot = 6.78 × 8.04 = 54.51
Given the composition of the floor, we define the total D.
Dpol = 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:
Qpol = 54.51 × (21 – 5) × 6.1 × 1 = 5320
Calculate the total D ceiling:
Dpot = 0.10/0.05 + 0.025/0.21 + 0.05/0.35 = 2.26
Then Q of the ceiling will be equal to:
Qpot = 54.51 × (22 + 31) × 2.26 = 6530
The total heat loss through OK will be equal to:
Qogr.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 m3/ h, the entrance is equipped with an air-thermal canopy, so for calculations it is enough to use the formula:
Qv = 0.28 × Ln × pv × c × (tv - tn)
We calculate the density of air in the room at a given temperature of +22 degrees:
pv = 353/(272 + 22) = 1.2
Parameter Ln equal to the product of the specific consumption by the floor area, that is:
Ln = 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:
Qv = 0.28 × 163.53 × 1.2 × 1.005 × (22 + 31) = 3000
Total heat costs for ventilation will be 3000 watts or 3 kW.
Household income is calculated by the formula.
Qt = 10 × Spol
That is, substituting the known values, we obtain:
Qt = 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 (tr) - 55 ° С, the desired room temperature (tv) - 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 tr the formula is used:
Eot = Q / (c × (tr - tv))
Substituting the parameter values in the formula, we obtain:
Eot = 16000/(1.005 × (55 – 22)) = 483
The volumetric amount of air supplied is calculated by the formula:
Vot = Eot / pr
pr = 353 / (273 + tr)
First, we calculate the density p:
pr = 353/(273 + 55) = 1.07
Vot = 483/1.07 = 451.
The air exchange in the room is determined by the formula:
Vp = Eot / pv
Determine the density of air in the room:
pv = 353/(273 + 22) = 1.19
Substituting the values in the formula, we get:
Vp = 483/1.19 = 405
Thus, the air exchange in the room is 405 m3 per hour, and the volume of air supplied should be equal to 451 m3 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 tr = 55 ° C, tv = 22 ° C; Q = 16000 watts. The amount of air required for ventilation, Event= 110 m3/ h Estimated outdoor temperature tn= -31 ° C.
For the calculation of the HFRS we use the formula:
Q3 = [Eot × (tr - tv) + Event × pv × (tr - tv)] × c
Substituting the values, we get:
Q3 = [483 × (55 – 22) + 110 × 1.19 × (55 – 31)] × 1.005 = 27000
The volume of recirculated air will be 405-110 = 296 m3 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 Eot according to the above formulas. We get:
Eot = 10 × 3.6 × 1000/ (1.005 × (55 – 22)) = 1085
Heat flux q1 will be:
q1 = (55 – 22)/0.27 = 122
The initial temperature with a deviation of η = 0 will be:
tnach = 22 + (16 × 1000 + 137 × 15) × (55 – 22)/ 1000 × 16 = 60
Specify the average temperature:
tsr = 0.5 × (55 + 60) = 57.5
Qotkl = ((574 -22)/0.27) × 15 = 1972
Given the information we find:
tnach = 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.