Before designing a heating system, installing heating equipment, it is important to choose a gas boiler capable of generating the necessary amount of heat for the room. Therefore, it is important to choose a device of such power so that its performance is as high as possible and the resource is large.
We will talk about how to calculate the power of a gas boiler with high accuracy and taking into account certain parameters. In the article presented by us, all types of heat loss through openings and building structures are described in detail, and formulas for calculating them are given. With the features of the production of calculations introduces a specific example.
The content of the article:
- Common mistakes when choosing a boiler
- What is room heat loss?
- Formulas for calculating heat loss
-
Example of heat loss calculation
- Calculating Wall Heat Loss
- Calculation of heat loss windows
- Determination of door heat loss
- Calculation of thermal floor resistance
- Calculation of heat loss through the ceiling
- Determination of heat loss taking into account infiltration
- Calculation of boiler power
- Conclusions and useful video on the topic
Common mistakes when choosing a boiler
The correct calculation of the power of the gas boiler will not only save on consumables, but also increase the efficiency of the device. Equipment, the heat output of which exceeds the real needs for heat, will work inefficiently when, as an insufficiently powerful device, it cannot properly heat the room.
There is a modern automated equipment that independently regulates the flow of gas, which eliminates inappropriate costs. But if such a boiler performs its work at the limit of its capacity, then its service life is reduced.
As a result, the efficiency of equipment decreases, parts wear out faster, and condensate forms. Therefore, it becomes necessary to calculate the optimal power.
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The main condition for the installation of a gas boiler is the installation of an internal gas network connected to the centralized gas supply, a group of cylinders or a gas-holder.
When choosing a gas boiler it is necessary to take into account the diameter of the pipes of the gas and heating systems. To install a double-circuit boiler, the house must be equipped with running water, the minimum pressure in which also requires consideration before purchasing
To choose a gas boiler correctly, it is necessary to take into account the pressure in the gas supplying line. If connected to a centralized network, it is indicated by the fuel supplier.
The power of gas equipment is directly related to the size of the unit, the type of installation and design
The wall variant is more compact, but it should be noted that in 1 minute the wall boiler heats only 0.57 liters of water by 25º. This is acceptable for a dacha or an apartment; for heating a large building, a more powerful unit is needed.
Floor gas boilers acquire, if the volume circulating in the coolant system is more than 150 liters. Power ranges from 10 to 55 kW or more
Floor-standing gas boilers can be used both as a heating boiler and as a water heater, capable of simultaneously supplying water up to 4 outlets
Outdoor gas equipment for heating systems produced in a wide range of modifications, the volume of which can reach 280 liters
Conditions for the installation of a gas boiler
Supply pipelines to the equipment
Indoor gas pipeline
Dimensions and constructive type
Power Limitations
Floor boiler for a large house
Boiler as a water heater
Volume of gas boilers
It is believed that the power of the boiler depends solely on the surface area of the room, and for any dwellings optimum calculation of 100 W per 1 sq.m. Therefore, to choose the power of the boiler, for example, at home 100 sq. M. m, you will need equipment that produces 100 * 10 = 10000 W or 10 kW.
Such calculations are fundamentally wrong with the advent of new finishing materials, improved insulation, which reduce the need to purchase high-power equipment.
The power of the gas boiler is selected taking into account the individual characteristics of the home. Properly selected equipment will work as efficiently as possible with minimal fuel consumption.
To calculate the power gas boiler heating is possible in two ways - manually or using a special program Valtec, which is designed for professional high-precision calculations.
The required power of the equipment directly depends on the heat loss of the room. Knowing the rate of heat loss, you can calculate the power of a gas boiler or any other heating device.
What is room heat loss?
Any room has a certain heat loss. Heat comes out of the walls, windows, floors, doors, the ceiling, so the task of the gas boiler is to compensate for the amount of heat produced and to ensure a certain temperature in the room. This requires a certain heat output.
It was experimentally established that the greatest amount of heat goes through the walls (up to 70%). Up to 30% of heat energy can be released through the roof and windows, and up to 40% through the ventilation system. The lowest heat loss at the door (up to 6%) and the floor (up to 15%)
The following factors affect the heat loss at home.
- Location of the house. Each city has its own climatic features. In the calculations of heat losses, it is necessary to take into account the critical negative temperature characteristic of the region, and also the average temperature and duration of the heating season (for accurate calculations using programs).
- The location of the walls relative to the cardinal points. It is known that the wind rose is located on the north side, so the heat loss of the wall located in this area will be greatest. In winter, the cold wind blows from the western, northern and eastern sides with great force, therefore the heat losses of these walls will be higher.
- The area of the heated room. The size of the outgoing heat depends on the size of the room, the area of walls, ceilings, windows, doors.
- Heat engineering building structures. Any material has its own thermal resistance coefficient and heat transfer coefficient - the ability to pass through a certain amount of heat. To get to know them, you need to use tabular data, as well as apply certain formulas. Information on the composition of walls, ceilings, floors, their thickness can be found in the technical plan of housing.
- Window and door openings. Size, modification of doors and double glazing. The larger the area of the window and door openings, the higher the heat loss. It is important to take into account the characteristics of installed doors and double-glazed windows in the calculations.
- Accounting ventilation. Ventilation always exists in the house, regardless of the presence of artificial exhaust. Airing of the room takes place through the open windows, air movement is created during the closing and opening entrance doors, the movement of people from room to room, which contributes to the departure of warm air from the room, its circulation.
Knowing the above parameters, you can not only calculate heat loss at home and determine the power of the boiler, but also to identify places in need of additional insulation.
Formulas for calculating heat loss
These formulas can be used to calculate heat loss not only of a private house, but also of an apartment. Before starting the calculations, it is necessary to draw the floor plan, mark the location of the walls relative to sides of the world, mark the windows, doorways, as well as calculate the dimensions of each wall, window and door openings.
To determine the heat loss it is necessary to know the structure of the wall, as well as the thickness of the materials used. The calculations take into account the laying and insulation
When calculating heat losses, two formulas are used: with the first one, the heat resistance value of the enclosing structures is determined, and with the second, the heat loss.
To determine thermal resistance, use the expression:
R = B / K
Here:
- R - the value of thermal resistance of enclosing structures, measured in (m2* K) / W.
- K - coefficient of thermal conductivity of the material from which the enclosing structure is made, is measured in W / (m * K).
- AT - material thickness, recorded in meters.
The coefficient of thermal conductivity K is a tabular parameter, the thickness B is taken from the technical plan of the house.
The coefficient of thermal conductivity is a tabular value, it depends on the density and composition material may differ from the table, so it is important to familiarize yourself with the technical documentation on material (+)
The basic formula for calculating heat losses is also used:
Q = L × S × dT / R
In terms of:
- Q - heat loss, measured in watts.
- S - area of enclosing structures (walls, floors, ceilings).
- dT - the difference between the desired temperature of the interior and the exterior is measured and recorded in C.
- R - value of thermal resistance of the structure, m2• C / W, which is according to the formula above.
- L - coefficient depending on the orientation of the walls relative to the cardinal points.
Having the necessary information at hand, you can manually calculate the heat loss of a building.
Example of heat loss calculation
As an example, we calculate the heat loss of a house that has specified characteristics.
The figure shows the plan of the house, for which we will calculate the heat loss. In drawing up an individual plan it is important to correctly determine the orientation of the walls relative to the cardinal points, calculate the height, width and length of the structure, as well as note the locations of window and door openings, their dimensions (+)
Based on the plan, the width of the structure is 10 m, length - 12 m, ceiling height - 2.7 m, the walls are oriented to the north, south, east and west. Three windows are built in the western wall, two of them have dimensions of 1.5x1.7 m, one - 0.6x0.3 m.
When calculating the roof is taken into account layer of insulation, finishing and roofing material. Paro- and waterproofing films that do not affect the thermal insulation are not taken into account.
The southern wall has built-in doors with dimensions of 1.3 × 2 m, there is also a small window 0.5 × 0.3 m. On the east side there are two windows 2.1 × 1.5 m and one 1.5 × 1.7 m.
The walls consist of three layers:
- wall covering DVP (isoplite) outside and inside - 1.2 cm each, coefficient - 0.05.
- glass wool located between the walls, its thickness is 10 cm and the coefficient is 0.043.
The thermal resistance of each wall is calculated separately, because the location of the structure relative to the cardinal points, the number and area of openings are taken into account. The results of wall calculations are summarized.
The floor is multi-layered; the whole area is made using the same technology and includes:
- the cut tongue is grooved, its thickness is 3.2 cm, the coefficient of thermal conductivity is 0.15.
- layer of dry leveling chipboard thickness of 10 cm and a coefficient of 0.15.
- insulation - mineral wool 5 cm thick, coefficient 0.039.
Let us assume that the floor has no deterioration in the basement and similar openings to heat engineering. Consequently, the calculation is made for the area of all premises by a single formula.
The ceilings are made of:
- wooden shields 4 cm with a coefficient of 0.15.
- mineral wool 15 cm, its coefficient is 0.039.
- Paro-, waterproofing layer.
Suppose that the ceiling does not have an exit to the attic above a residential or utility room.
The house is located in the Bryansk region, in the city of Bryansk, where the critical negative temperature is -26 degrees. It is experimentally established that the temperature of the earth is +8 degrees. Desired room temperature + 22 degrees.
Calculating Wall Heat Loss
To find the total thermal resistance of a wall, it is first necessary to calculate the thermal resistance of each of its layers.
The glass wool layer is 10 cm thick. This value must be converted to meters, that is:
B = 10 × 0.01 = 0.1
Received value In = 0.1. The thermal conductivity of thermal insulation - 0.043. Substituting the data in the formula of thermal resistance and get:
Rglass=0.1/0.043=2.32
For a similar example, we calculate the resistance to heat of the isoplite:
Risopl=0.012/0.05=0.24
The total thermal resistance of the wall will be equal to the sum of the thermal resistance of each layer, given that we have two layers of fiberboard.
R = Rglass+ 2 × Risopl=2.32+2×0.24=2.8
By determining the total thermal resistance of the wall, you can find the heat loss. For each wall, they are calculated separately. Calculate Q for the north wall.
Additional coefficients allow to take into account in the calculations the features of heat loss of walls located in different directions of the world
Based on the plan, the north wall has no window openings, its length is 10 m, height is 2.7 m. Then the area of wall S is calculated by the formula:
SNorth Wall=10×2.7=27
Calculate the dT parameter. It is known that the critical ambient temperature for Bryansk is -26 degrees, and the desired room temperature is +22 degrees. Then
dT = 22 - (- 26) = 48
For the north side, the additional factor L = 1.1 is taken into account.
The table shows the thermal conductivity of some materials that are used in the construction of walls. As you can see, mineral wool lets through a minimum amount of heat, reinforced concrete - the maximum
Having made preliminary calculations, you can use the formula for calculating heat losses:
QNorth Walls= 27 × 48 × 1.1 / 2.8 = 509 (W)
Calculate the heat loss for the western wall. Based on the data, 3 windows are built into it, two of them have dimensions of 1.5x1.7 m and one is 0.6x0.3 m. We calculate the area.
Szap.steny1=12×2.7=32.4.
It is necessary to exclude the area of windows from the total area of the western wall, because their heat loss will be different. To do this, calculate the area.
Sok1=1.5×1.7=2.55
Swindow2=0.6×0.4=0.24
To calculate the heat loss, we will use the wall area without taking into account the window area, that is:
Szap.steny=32.4-2.55×2-0.24=25.6
For the west side, the added factor is 1.05. The obtained data is substituted into the basic formula for calculating heat losses.
Qzap.steny=25.6×1.05×48/2.8=461.
Similar calculations are done for the east side. Here are 3 windows, one has dimensions of 1.5x1.7 m, the other two are 2.1x1.5 m. We calculate their area.
Swindow3=1.5×1.7=2.55
Swindow4=2.1×1.5=3.15
The area of the east wall is equal to:
Seast walls1=12×2.7=32.4
From the total area of the wall we subtract the values of the area of the windows:
Seast walls=32.4-2.55-2×3.15=23.55
The added factor for the east wall is -1.05. Based on the data, we calculate the heat loss of the eastern wall.
Qeast walls=1.05×23.55×48/2.8=424
A door with the parameters 1.3x2 m and a window 0.5x0.3 m is located on the south wall. We calculate their area.
Sok5=0.5×0.3=0.15
Sthe door=1.3×2=2.6
The area of the southern wall will be equal to:
Ssouth walls1=10×2.7=27
Determine the area of the wall without windows and doors.
Ssouth wall=27-2.6-0.15=24.25
Calculate the heat loss of the southern wall, taking into account the coefficient L = 1.
Qsouth wall=1×24.25×48/2.80=416
Determining the heat loss of each of the walls, you can find their total heat loss by the formula:
Qwalls= Qsouth wall+ Qeast walls+ Qzap.steny+ QNorth Walls
Substituting the values, we get:
Qwalls= 509 + 461 + 424 + 416 = 1810 W
As a result, the loss of heat walls amounted to 1810 watts per hour.
Calculation of heat loss windows
There are 7 windows in the house, three of them have dimensions of 1.5 × 1.7 m, two are 2.1 × 1.5 m, one is 0.6 × 0.3 m and another one is 0.5 × 0.3 m.
Windows with dimensions of 1.5 × 1.7 m is a two-chamber PVC profile with I-glass. From the technical documentation you can find out that its R = 0.53. The windows with dimensions of 2.1 × 1.5 m are double-chamber with argon and I-glass, have thermal resistance R = 0.75, windows 0.6x0.3 m and 0.5 × 0.3 - R = 0.53.
The window area was calculated above.
Sok1=1.5×1.7=2.55
Swindow2=0.6×0.4=0.24
Swindow3=2.1×1.5=3.15
Swindow4=0.5×0.3=0.15
It is also important to consider the orientation of the windows relative to the cardinal points.
Usually, it is not necessary to calculate the thermal resistance for windows; this parameter is specified in the technical documentation for the product.
Calculate the heat loss of the western windows, taking into account the coefficient L = 1.05. On the side there are 2 windows with dimensions of 1.5 × 1.7 m and one with 0.6 × 0.3 m.
Qok1=2.55×1.05×48/0.53=243
Qwindow2=0.24×1.05×48/0.53=23
Total total losses of western windows are
Qzap.okon=243×2+23=509
In the south side is a 0.5 × 0.3 window, its R = 0.53. We calculate its heat loss taking into account the coefficient 1.
Qsouth window=0.15*48×1/0.53=14
On the east side there are 2 windows with dimensions of 2.1 × 1.5 and one window of 1.5 × 1.7. Calculate the heat loss taking into account the coefficient L = 1.05.
Qok1=2.55×1.05×48/0.53=243
Qwindow3=3.15×1.05×48/075=212
We summarize the heat loss of the eastern windows.
QEast window=243+212×2=667.
The total heat loss of windows will be equal to:
Qwindows= QEast window+ Qsouth window+ Qzap.okon=667+14+509=1190
Total through the window goes 1190 watts of thermal energy.
Determination of door heat loss
The house has one door, it is built into the southern wall, has dimensions of 1.3 × 2 m. Based on the passport data, thermal conductivity of the door material is 0.14, its thickness is 0.05 m. Thanks to these indicators, it is possible to calculate the thermal door resistance.
Rthe doors=0.05/0.14=0.36
For calculations, you need to calculate its area.
Sthe doors=1.3×2=2.6
After calculating the thermal resistance and the area you can find the heat loss. The door is located on the south side, so we use an additional factor of 1.
Qthe doors=2.6×48×1/0.36=347.
Total, through the door goes 347 watts of heat.
Calculation of thermal floor resistance
According to the technical documentation, the floor is multi-layered, the whole area is the same, it has dimensions of 10x12 m. We calculate its area.
Ssex=10×12=210.
The composition of the floor includes boards, chipboard and insulation.
From the table you can find the thermal conductivity of some materials used for flooring. This parameter can also be specified in the technical documentation of materials and differ from the table
Thermal resistance must be calculated for each layer of the floor separately.
Rboards=0.032/0.15=0.21
Rchipboard=0.01/0.15= 0.07
Rheat insulation=0.05/0.039=1.28
The total thermal resistance of the floor is:
Rsex= Rboards+ Rchipboard+ Rheat insulation=0.21+0.07+1.28=1.56
Given that in winter the temperature of the earth is kept at +8 degrees, the temperature difference will be equal to:
dT = 22-8 = 14
Using preliminary calculations, it is possible to find heat losses at home through the floor.
When calculating the heat loss of the floor, materials affecting thermal insulation (+) are taken into account
When calculating the heat loss of the floor, we take into account the coefficient L = 1.
Qsex=210×14×1/1.56=1885
Total floor heat loss is 1885 watts.
Calculation of heat loss through the ceiling
When calculating the heat loss of the ceiling, the layer of mineral wool and wooden shields are taken into account. Paro-, waterproofing is not involved in the process of insulation, so it does not take into account. For calculations we need to find the thermal resistance of wooden shields and a layer of mineral wool. We use their coefficients of thermal conductivity and thickness.
Rshield=0.04/0.15=0.27
Rmin.vat=0.05/0.039=1.28
The total heat resistance will be equal to the sum of Rshield and Rmin.vat.
Rroofing=0.27+1.28=1.55
The ceiling area is the same as the floor.
S ceiling = 120
Next, the heat loss of the ceiling is calculated, taking into account the coefficient L = 1.
Qceiling=120×1×48/1.55=3717
Total through the ceiling leaves 3717 watts.
The table shows the popular insulation for ceilings and their thermal conductivity coefficients. Polyurethane foam is the most effective insulation, straw has the highest coefficient of heat loss
To determine the total heat loss at home, it is necessary to add up the heat loss of walls, windows, doors, ceilings and floors.
Qgeneral= 1810 + 1190 + 347 + 1885 + 3717 = 8949 W
To heat the house with the specified parameters requires a gas boiler that supports a power of 8949 W or about 10 kW.
Determination of heat loss taking into account infiltration
Infiltration is a natural process of heat exchange between the external environment, which occurs during the movement of people around the house, when the entrance doors and windows open.
To calculate heat loss on ventilation You can use the formula:
Qinf= 0.33 × K × V × dT
In terms of:
- K - the calculated air exchange rate, for living rooms use a coefficient of 0.3, for rooms with heating - 0.8, for the kitchen and bathroom - 1.
- V - the volume of the room is calculated taking into account the height, length and width.
- dT - the temperature difference between the environment and the dwelling.
A similar formula can be used if ventilation is installed in the room.
In the presence of artificial ventilation in the house, it is necessary to use the same formula as for infiltration, just substitute K for the parameters of the exhaust and calculate dT to take into account the temperature of the incoming of air
The height of the room - 2.7 m, width - 10 m, length - 12 m. Knowing this data, you can find its volume.
V = 2.7 × 10 × 12 = 324
The temperature difference will be equal to
dT = 48
As the coefficient K, we take the index 0.3. Then
Qinf=0.33×0.3×324×48=1540
To the total calculated index Q you need to add Qinf. Eventually
Qgeneral=1540+8949=10489.
Total, taking into account the infiltration of heat loss at home will be 10489 W or 10.49 kW.
Calculation of boiler power
When calculating the boiler power it is necessary to use the safety factor 1.2. That is, the power will be equal to:
W = Q × k
Here:
- Q - heat loss of the building.
- k - safety factor.
In our example, we substitute Q = 9237 W and calculate the required power of the boiler.
W = 10489 × 1.2 = 12587 watts.
Taking into account the safety factor, the required boiler power for heating a house is 120 m2 equal to about 13 kW.
Conclusions and useful video on the topic
Video instruction: how to calculate the heat loss at home and the power of the boiler using the Valtec program.
Competent calculation of heat losses and gas boiler power using formulas or software methods allows you to determine the high accuracy of the necessary parameters of the equipment, which makes it possible to eliminate unreasonable costs fuel.
Please write comments in the block below. Tell us how heat loss was calculated before purchasing heating equipment for your own summer cottage or a country house. Ask questions, share information and photos on the topic.
