Compared to electric heating devices, your own heating system is more profitable in terms of saving money, and maximum convenience when heating rooms.
The efficiency and profitability of the heating system in the house depends on correct calculations and adherence to precise rules and instructions.
Calculating heating based on the area of a house is a labor-intensive and complex process. You shouldn't skimp on materials. Quality equipment and its installation affects financial budget, but then serves the house well and comfortably.
When equipping your home with a heating system, construction works and heating installation must proceed strictly according to the design and taking into account all safety regulations for use.
The following points should be taken into account:
- house building material,
- footage of window openings;
- climatic features of the area where the house is located;
- arrangement of window frames according to the compass;
- What is the structure of the “warm floor” system?
Subject to all the above rules and calculations for heating, some knowledge in the field of engineering is required. But there is also a simplified system - calculating heating by area, which you can do yourself, again, adhering to the rules and complying with all standards.
Choosing a boiler requires an individual approach
If there is gas in the house, then the most the best option- This a gas boiler. In the absence of a centralized gas pipeline, we choose an electric boiler or a heat generator using solid or liquid fuel. Taking into account regional characteristics and access to the supply of materials, a combination boiler can be installed. Combined generator heat will always maintain comfortable temperature, in any emergency and force majeure situations. Here you need to start from a simple type of operation, the heat transfer coefficient.
After determining the type of boiler, it is necessary to calculate the heating according to the area of the room. The formula is simple, but it takes into account the temperature of the cold period, the coefficient of heat loss for large windows and their location, the thickness of the walls and the height of the ceilings.
Each boiler has a certain power. If you make the wrong choice, the room will be either cold or excessively hot. Thus, if the specific power of the boiler is 10 cubic meters. Taking into account the area of the heated room of 100 sq.m., you can choose the most optimal heat generator.
From the formula that engineers use - Wcat = (SxWsp)/10, kW. – it follows that the boiler with a capacity of 10 kW heats a room of 100 sq.m..
Required number of heating radiator sections.
To make it more clear, let’s solve the problem using an example specific numbers. If we assume that room area 14 sq.m. And ceiling height 3 meters, the volume is determined by multiplication.
14 x 3 = 42 cubic meters.
IN middle lane Russia, Ukraine, Belarus thermal power per cubic meter corresponds to 41 W. We determine: 41x 42 = 1722 W. Found out that for a room of 14 sq.m. need a 1700 W radiator. Each individual section (fin) has a power of 150 W. Dividing the results obtained, we obtain the number of sections required for acquisition. Heating calculations by area are not the same everywhere. For premises over 100 sq.m. required installation of a circulation pump, which serves as a “forcer” for the movement of coolant through the pipes. Its installation occurs in the opposite direction from the heating devices to the heat generator. Circulation pump increases the operating life of the heating system, reducing the contact of hot coolants with appliances.
When installing a heating system " warm floor» the heating coefficient of the house increases significantly. Connect the system underfloor heating you can already use existing types of heating. A pipe is taken out from the heating radiators and the floor heating wiring is supplied. This is the most convenient and profitable option, taking into account saving money and time.
Selection gas boiler optimal power is possible only after calculations. In the technical documentation for boiler equipment its thermal power– TMK. This parameter means the power that the boiler is capable of transmitting to external devices (heating, ventilation, domestic hot water preparation), taking into account its efficiency. But this value in no way informs the user what area can be heated using a specific boiler model.
The problem is that any building, even insulated, gives off some of the heat to the outside air through structures such as walls, ceilings, floors, windows and doors. Therefore, without a thermal calculation of the building, it is difficult not to make a mistake in making the right choice boiler
In this article:
What parameters need to be taken into account
Heat loss of a private house
When choosing boiler equipment for heating your home, you must consider:
- climatic conditions of the region (the calculation formula includes the average temperature for the coldest week of the year);
- set air temperature inside heated rooms;
- the need to organize hot water supply;
- heat loss from forced ventilation(if there is one in the house);
- number of storeys of the building;
- ceiling height;
- design and materials of floors;
- the thickness of the external walls and the materials from which they were built;
- geometric dimensions of external walls;
- floor construction (thickness of layers and materials from which they are constructed);
- sizes, number of windows and doors and their type (glass thickness, number of cameras, etc.).
Heat loss at home
The amount of heat loss from a building is greatly influenced by:
- type of attic (insulated, non-insulated);
- the presence or absence of a basement.
To clearly show dependence of house heat loss on materials, used in its construction, we suggest considering a small comparative table.
The table shows that a wooden house loses less heat than a brick house, respectively, and in the first case a boiler will be required less powerful than for a brick house.
IN building codes ah the thermal conductivity indicators for all building materials are described.
Something similar is observed in relation to windows..
Only they are not characterized by thermal conductivity, but, on the contrary, by the heat transfer resistance coefficient: the higher the number, the less heat the window will release from the house (this indicator is also called the R-factor).
As you can see, the more chambers in a window design, the higher its resistance to heat loss. Plays an important role gas mixture, which fills the chambers of double-glazed windows.
How to calculate the TMC of a gas boiler
First of all, the thermal calculation of the building itself
The thermal power of a heating boiler can be calculated in two ways:
- full;
- simplified.
The first method involves carrying out calculations taking into account the thermal properties of all building materials involved in the construction of the house and its finishing. From the data shown in the tables above, you can see how important it is to perform a complete calculation.
But this work is not easy, and in the absence of certain experience, it is difficult to cope with it.
This is usually done by designers design organizations. Although, if you really want to, you can arm yourself with SNiPs and try to do everything yourself.
Thermal conductivity coefficient of building materials
Thermal conductivity coefficients of common building materials
To determine the amount of heat loss through the building envelope, it is necessary to calculate the thermal conductivity coefficient of the building materials from which they are composed.
The initial data for the calculation are:
- a(vn)– coefficient that determines the intensity of heat transfer from the air in the room to the ceiling and walls. This is a constant value equal to 8.7.
- a(nr)– another constant coefficient equal to 23. It characterizes the intensity of heat transfer from the walls and ceiling to the outside air.
- TO– thermal conductivity of building materials that make up the ceiling and walls. Data is taken from building codes. For some materials, thermal conductivity is given in the table of building materials (see above).
- D– thickness of layers of building materials.
After collecting all the initial data, you can begin to calculate the heat transfer coefficient using the formula:
Kt = 1/
CT is calculated for the ceiling and walls separately.
The principle of calculating the floor CT is the same, but there are some nuances: the right approach requires dividing the floor area into 4 zones, located from the outer walls to the center. To simplify calculations, heat loss through the floor structure without heating can be taken equal to 10%.
Calculation of heat loss through windows and doors
The initial data for this part of the calculation are:
- Kst– heat transfer coefficient of a double glazing unit or glass (indicated by the manufacturer).
- F st.– area of the glazed surface of the window.
- Kr- heat transfer coefficient window frame(indicated by the manufacturer).
- F r– area of the window frame.
- R– the perimeter of the glazed surface of the window.
The heat transfer coefficient of windows (Ko) is calculated using the formula:
Kst. x F st. + Kr x F p + P/F, where F is the area of the windows.
Using the same formula, the heat transfer coefficient of doors is calculated.
In this case, instead of the values of glass and frames, the values of the materials from which the doors are made are substituted.
To simplify calculations, you can use the following data:
To determine heat loss, the conditional coefficient is multiplied by the total area of the house.
This method gives only an approximate result. It does not take into account the number of windows, the configuration of the house and its location. But for a preliminary assessment of heat loss it is quite suitable.
Simplified method
The power of a heating boiler is defined as the sum of the power required to heat each heated room. That is, the calculations described in the previous sections are carried out for each room separately.
At the same time, designers are required to take into account the number of lamps, people in the room, and even the operation of household appliances.
Fortunately, in most cases you can do without such complex and expensive thermal calculations. Residential buildings are usually built taking into account the climatic conditions of a particular region, so you can select the required TMC value using a simplified scheme.
The basis for this calculation is the assumption that the specific power of the entire house is equal to the sum of the specific power of each room. In this case, when performing calculations, they operate with experimental values of the specific power of the house, depending on the region.
These tables are valid for well-insulated wooden and reinforced concrete houses with a standard ceiling height of 2.7 meters.
Boiler power per 10 kW. m is calculated by the formula:
- W = S x W beats/10, where
- W – boiler design power
- S - sum of premises areas
- Wud – specific power of the house (see table above)
Example
Typical house plan for 300 sq.m (for example)
For example, let’s calculate the power of a gas boiler for a house located in the Moscow region. The total area of the building is 300 sq. m. m.
Let us take the value of specific power (according to the fourth table) equal to 1.5.
- W = 300 x 1.5/10 = 45 kW
For high ceilings
If the ceiling height differs from the standard values, in this case the power of the heating boiler is calculated using the formula:
- Mk = TxKz, Where
- Mk – boiler power
- T – estimated heat loss
- Кз – safety factor
Heat losses T are calculated using the formula:
- T = VхРхКр/860, Where
- V – volume of the room (in cubic meters)
- P – difference between external and internal temperatures
- Kr – dissipation coefficient
For buildings made of brick, Kr is 2 - 2.9, for poorly insulated buildings - 3-4.
And lastly: if you assume that the boiler will provide the house and hot water, increase the design power by 25%.
The power of a gas boiler is an important parameter on which the comfort of living in the rooms heated by it depends. To choose the best option for a house or apartment, you need to take into account its size. The required performance of heating equipment depends on the area of the heated premises and some other, less significant factors.
What affects the calculated power
The boiler must not only replenish all heat losses of a particular building or room, but also have a certain power reserve. Why is it necessary to take a value greater than the calculated one:
- equipment should not operate at its maximum capacity - this leads to premature wear;
- the likelihood of abnormal temperatures must be taken into account;
- for a private house it is useful to take into account the possibility of expanding the area.
Some buyers do not know in what units it is calculated. main parameter gas equipment, which determines its performance. The thermal power of devices is measured in kilowatts (kW). This value is always indicated in technical passport each model.
What affects heat loss
To find out what equipment performance is needed, in addition to area, you need to take into account other factors:
- climate in a particular region;
- volume of residential building/apartment;
- degree of insulation;
- probable heat loss.
When using turbocharged devices, it is also necessary to take into account the amount of energy spent on heating the air.
To determine the performance of the boiler, you must first calculate the heat loss. Thermal engineering calculation is characterized by increased complexity, as it takes into account great amount components:
- materials from which walls, ceilings, roofing, etc. are constructed;
- type of heating system wiring;
- the presence of a “warm floor” system;
- household appliances that produce heat.
Professionals use thermal cameras and then perform calculations using complex formulas. It is clear that the average user does not have to understand the nuances of heating engineering - there are available methods for them that allow them to quickly and accurately calculate the optimal heating performance of the equipment.
What calculation options are there?
To make the right choice of gas equipment, we suggest using three calculation options:
- Accurate thermal technology - not suitable for ordinary consumers, it is complex and requires the use of a thermal imager.
- On an online calculator - to get the result, the user enters the initial data into special program: number of windows, doors, wall thickness and other information. Based on them, the program produces the result.
- Manual calculations. Most affordable way find out the optimal heating performance of the heater - use elementary attitude area and power. The formula used is: 10 m² = 1,000 W. This simple option is correct for buildings characterized by an average degree of thermal insulation and having ceilings with a height of about 2.7 m.
Developers calculating power characteristics heating devices, often take into account the volume of premises. IN technical documentation In imported models, the parameter “heating in m³” is often found.
Calculation of boiler power with one circuit
Having performed the simplest calculation for a single-circuit wall-mounted or floor-standing boiler using the ratio: 10 kW per 100 m², you need to increase calculated value by 15–20%.
Let's give an example of calculations. It is necessary to equip a house with an area of 80 m². To heat it, you will need a device with 9,600 W = 8,000 W + 20%. If it's not on sale for sure suitable option, you should take a modification with greater performance. This method of calculation is only suitable for devices with one circuit, without a boiler indirect heating.
Calculation of boiler power with two circuits
The calculation is based on the following ratio: 10 m² = 1,000 W + 20% (reserve) + 20% (water heating). If the house has an area of 200 m², then the required value will be: 20,000 W + 40% = 28,000 W.
Determining the power of a model with a boiler
First, determine the required volume of the boiler so that it can satisfy the household needs hot water. Water consumption is calculated taking into account the operation of all water intake points:
- bath - 8–9 l/min;
- shower - 9 l/min;
- toilet - 4 l/min;
- washing - 4 l/min.
The technical documentation for the boiler indicates what boiler performance is required to ensure water heating. For a boiler with a capacity of 200 liters of water, a heater with a power of approximately 30 kW is suitable. Then the performance required for heating is calculated. The results obtained are summarized. At the end of the calculations, you need to subtract 20% from the result obtained, since water heating for hot water supply and heating occurs simultaneously.
Calculation of boiler power for typical houses, taking into account the climate zone
For houses built according to standard projects, apply the formula: M = S*UM/10, where
- M/UM - design/specific power, kW;
- S - area, m².
PA depends on the region, kW:
- south - 0.7–0.9;
- middle band - 1.0–1.2;
- Moscow region - 1.2–1.5;
- North - 1.5–2.0.
Let's perform calculations for a house with an area of 300 m² located in the Moscow region: 300 * 1.3/10 = 39 kW. This result is suitable for installing single-circuit models. To calculate the power of a dual-circuit device, you need to increase the final number by 25%.
Is excess power needed?
You should not buy a model with a performance significantly higher than the maximum (including a 15-20% premium). Excess leads to negative consequences:
- High price. The more powerful the model, the more expensive it is. It is irrational to purchase equipment whose capabilities will not be used.
- Increased costs for consumables.
- Low burner efficiency - this will affect gas consumption.
- At minimal loads, the automation often fails.
- If the equipment is not optimal for a specific area, accelerated wear of components and parts occurs.
How to calculate expenses
Knowing the power characteristics of the equipment, you can calculate gas consumption. The calculation takes into account efficiency. Standard versions have an efficiency of 92-93%, condensing type models - 108-109%. With 100% heat transfer, 10 kW of thermal energy is generated after combustion of 1 m³ natural gas. Thus, to create a power of 10 kW with an efficiency of 92%, the fuel consumption will be 1.12 m³, and with an efficiency of 108% - 0.92 m³.
When calculating the volume of fuel consumed, the performance of the devices is taken into account. A 10 kW model burns 1.12 m³ of gas per hour, and a 40 kW model burns 4.48 m³. Manufacturers often indicate the average fuel consumption in technical documentation, but it is still different for each model.
To find out the upcoming heating costs when using energy-dependent versions, it is also necessary to calculate the energy costs.
How to take into account ceiling heights
The above calculation formulas suitable for buildings whose ceiling height does not exceed 3 meters. If the ceilings are higher, you need to use other formulas: M = Q*K, where:
- M - calculated power, kW;
- Q - heat losses, kW;
- K - safety factor.
K = 1.15-2, or 15–20%.
To calculate heat loss, use the formula:
Q = V*P*k/860, where:
- V - volume of premises, m³;
- P is the difference between the temperatures in the house and outside, °C;
- k is the dissipation coefficient, depending on the thermal insulation characteristics of the structure.
The value of the coefficient is determined by the type of structure:
- not having thermal insulation: wooden structures, buildings made of corrugated iron sheets, - 3.0-4.0;
- with low thermal insulation - 2.0-2.9;
- with average thermal insulation - 1-1.9;
- with high thermal insulation - 0.6-0.9.
If the structure is small and has good thermal insulation characteristics, high boiler output is not required. It happens that there is no option on sale with suitable characteristics. Then you need to choose an option with a heating capacity slightly higher than the calculated value. The difference will be smoothed out by automatic control systems.
Online calculator
The most advanced manufacturers have thought about the comfort of consumers by placing online calculators on their websites that make it easy and quick to find out the required performance of gas equipment. For the calculation, enter the following information:
- the temperature that the consumer wants to have in the house;
- average outdoor temperature in the coldest week;
- availability of hot water supply;
- number of storeys;
- ceiling height;
- floor material;
- the thickness of the walls and the materials from which they are built;
- length of walls;
- number of window openings;
- window features - design details;
- window dimensions.
By filling in the fields, you can quickly calculate the calculated value of the heating capacity.
Choose wall-mounted or floor-standing boilers
The choice of type of heater installation depends not only on consumer preferences, but also on the calculated heat output.
Wall-mounted boilers, unlike floor-mounted ones, have a smaller power range. They are compact and can be placed in the kitchen, attic, or basement.
Floor-standing models are more bulky and are usually installed in separate rooms. Wall-mounted versions are available in a power range of 12-36 kW, while the performance of floor-standing models can reach 160 kW.
The functionality of the wall and floor versions is not very different. Modern devices of both types require manual or automatic control.
As a rule, wall-mounted models are purchased for apartments - they are compact and easily fit into the interior of the kitchen. For heating big houses and cottages use more powerful floor heaters. Atmospheric versions are installed in separate, well-ventilated rooms. The requirements for rooms in which turbocharged devices are installed are much lower.
What else influences the choice
In addition to heating performance, you need to consider:
- number of circuits (only heating or heating and DHW is required);
- installation method (wall or floor);
- combustion chamber (open or closed; in the first case, air is taken from the room, in the second - from the street via a coaxial chimney);
- design - for consumers appearance is not the least important. Modern devices are not only functional, efficient, safe, but also beautiful.
The correct choice of thermal performance of a gas boiler will allow the use of equipment with maximum efficiency. An optimally selected model will not only ensure a comfortable temperature in the house, but will also serve with minimal wear and tear on parts and components.
In any heating system that uses a liquid coolant, its “heart” is the boiler. It is here that the energy potential of fuel (solid, gaseous, liquid) or electricity is converted into heat, which is transferred to the coolant, and is already distributed throughout all the heated rooms of the house or apartment. Naturally, the capabilities of any boiler are not unlimited, that is, they are limited by its technical and operational characteristics specified in the product data sheet.
One of key characteristics is the thermal power of the unit. Simply put, it must be able to generate in a unit of time such an amount of heat that would be sufficient to fully heat all the rooms of a house or apartment. Selection suitable model“by eye” or according to some overly generalized concepts can lead to an error in one direction or another. Therefore, in this publication we will try to offer the reader, although not professional, but still with a fairly high degree of accuracy, an algorithm on how to calculate the power of a boiler for heating a house.
A trivial question - why know the required boiler power?
Despite the fact that the question really seems rhetorical, there is still a need to give a couple of explanations. The fact is that some owners of houses or apartments still manage to make mistakes, going to one extreme or another. That is, purchasing equipment either of obviously insufficient thermal performance, in the hope of saving money, or greatly overestimated, so that, in their opinion, they are guaranteed to provide themselves with heat in any situation with a large margin.
Both of these are completely wrong and have a negative impact on both the provision of comfortable living conditions and the durability of the equipment itself.
- Well, with insufficiency calorific value everything is more or less clear. When advancing winter cold the boiler will operate at full capacity, and it is not a fact that there will be a comfortable microclimate in the premises. This means that you will have to “bring up the heat” with the help of electric heating devices, which will entail significant extra costs. And the boiler itself, operating at the limit of its capabilities, is unlikely to last long. In any case, after a year or two, homeowners will clearly realize the need to replace the unit with a more powerful one. One way or another, the cost of an error is quite impressive.
- Well, why not buy a boiler with a large reserve, what can this hinder? Yes, of course, high-quality heating of the premises will be provided. But now let’s list the “cons” of this approach:
Firstly, a higher-power boiler itself can cost significantly more, and it’s difficult to call such a purchase rational.
Secondly, with increasing power, the dimensions and weight of the unit almost always increase. These are unnecessary difficulties during installation, “stolen” space, which is especially important if the boiler is planned to be placed, for example, in the kitchen or in another room in the living area of the house.
Thirdly, you may encounter uneconomical operation of the heating system - part of the expended energy resources will be spent, in fact, in vain.
Fourthly, excess power means regular long shutdowns of the boiler, which, in addition, are accompanied by cooling of the chimney and, accordingly, abundant formation of condensate.
Fifth, if powerful equipment is never properly loaded, it does not benefit it. Such a statement may seem paradoxical, but it is so - wear becomes higher, the duration of trouble-free operation is significantly reduced.
Prices for popular heating boilers
Excess boiler power will be appropriate only if it is planned to connect a water heating system to it for economic needs– indirect heating boiler. Well, or when it is planned to expand the heating system in the future. For example, the owners plan to build a residential extension to the house.
Methods for calculating the required boiler power
In truth, it is always better to trust specialists to carry out thermal engineering calculations - there are too many nuances to take into account. But, it is clear that such services are not provided free of charge, so many owners prefer to take responsibility for choosing the parameters of boiler equipment.
Let's see what methods of calculating thermal power are most often offered on the Internet. But first, let’s clarify the question of what exactly should influence this parameter. This will make it easier to understand the advantages and disadvantages of each of the proposed calculation methods.
What principles are key when carrying out calculations?
So, the heating system faces two main tasks. Let us immediately clarify that there is no clear division between them - on the contrary, there is a very close relationship.
- The first is creating and maintaining a comfortable temperature in the premises. Moreover, this level of heating should extend to the entire volume of the room. Of course, due to physical laws, temperature gradation in height is still inevitable, but it should not affect the feeling of comfort in the room. It turns out that it should be able to warm up a certain volume of air.
The degree of temperature comfort is, of course, a subjective value, that is, different people can evaluate it in their own way. But it is still generally accepted that this indicator is in the range of +20 ÷ 22 °C. Typically, this is the temperature that is used when carrying out thermal calculations.
This is also evidenced by the standards established by the current GOST, SNiP and SanPiN. For example, the table below shows the requirements of GOST 30494-96:
| Room type | Air temperature level, °C | |
|---|---|---|
| optimal | acceptable | |
| Living spaces | 20÷22 | 18÷24 |
| Residential premises for regions with minimum winter temperatures of - 31 °C and below | 21÷23 | 20÷24 |
| Kitchen | 19÷21 | 18÷26 |
| Toilet | 19÷21 | 18÷26 |
| Bathroom, combined toilet | 24÷26 | 18÷26 |
| Office, recreation and study areas | 20÷22 | 18÷24 |
| Corridor | 18÷20 | 16÷22 |
| Lobby, staircase | 16÷18 | 14÷20 |
| Storerooms | 16÷18 | 12÷22 |
| Residential premises (the rest are not standardized) | 22÷25 | 20÷28 |
- The second task is the constant compensation of possible heat losses. Creating an “ideal” house in which there would be no heat leaks is a problem that is practically unsolvable. You can only reduce them to the bare minimum. And almost all elements of a building’s structure become leakage paths to one degree or another.
| Building structural element | Approximate share of total heat losses |
|---|---|
| Foundation, plinth, floors of the first stage (on the ground or above an unheated basement) | from 5 to 10% |
| Joints of building structures | from 5 to 10% |
| Passage sections engineering communications through building structures (sewage pipes, water supply, gas supply, electrical or communication cables, etc.) | up to 5% |
| External walls, depending on the level of thermal insulation | from 20 to 30% |
| Windows and doors to the street | about 20÷25%, of which about half is due to insufficient sealing of boxes, poor fit of frames or canvases |
| Roof | up to 20% |
| Chimney and ventilation | up to 25÷30% |
Why were all these rather lengthy explanations given? But only so that the reader has complete clarity that when making calculations, willy-nilly, it is necessary to take into account both directions. That is, the “geometry” of the heated rooms of the house, and the approximate level of heat loss from them. And the amount of these heat leaks, in turn, depends on a number of factors. This is the difference in temperatures outside and in the house, and the quality of thermal insulation, and the features of the entire house as a whole and the location of each of its rooms, and other evaluation criteria.
You might be interested in information about which ones are suitable
Now, armed with this preliminary knowledge, let's move on to consider various methods calculating the required thermal power.
Calculation of power based on the area of heated premises
It is proposed to proceed from their conditional relationship that for high-quality heating of one square meter of room area it is necessary to consume 100 W of thermal energy. Thus, it will help to calculate which one:
Q=Stotal / 10
Q- the required thermal power of the heating system, expressed in kilowatts.
Stotal- the total area of the heated premises of the house, square meters.
However, reservations are made:
- First, the ceiling height of the room should be on average 2.7 meters, a range from 2.5 to 3 meters is allowed.
- Secondly, you can make an adjustment for the region of residence, that is, take not a rigid standard of 100 W/m², but a “floating” one:
That is, the formula will take a slightly different form:
Q=Stotal ×Qud / 1000
Qud - The value of the specific thermal power per square meter of area taken from the table shown above.
- Third - the calculation is valid for houses or apartments with an average degree of insulation of enclosing structures.
However, despite the mentioned reservations, such a calculation cannot be called accurate. Agree that it is largely based on the “geometry” of the house and its premises. But heat loss is practically not taken into account, except for the rather “blurred” ranges of specific thermal power by region (which also have very vague boundaries), and remarks that the walls should have an average degree of insulation.
But be that as it may, this method is still popular precisely for its simplicity.
It is clear that the operating power reserve of the boiler must be added to the obtained calculated value. You should not overestimate it - experts advise staying in the range from 10 to 20%. This, by the way, applies to all methods for calculating the power of heating equipment, which will be discussed below.
Calculation of the required thermal power by volume of premises
By and large, this method of calculation largely repeats the previous one. True, the initial value here is not the area, but the volume - essentially the same area, but multiplied by the height of the ceilings.
And the norms of specific thermal power adopted here are:
- for brick houses – 34 W/m³;
- for panel houses – 41 W/m³.
Even based on the proposed values (from their wording), it becomes clear that these standards were established for apartment buildings, and are used mainly to calculate the heat energy demand for premises connected to central system department or to an autonomous boiler station.
It is quite obvious that “geometry” is again being put at the forefront. And the entire system for accounting for heat losses comes down to only differences in the thermal conductivity of brick and panel walls.
In a word, this approach to calculating thermal power is also no different in accuracy.
Calculation algorithm taking into account the characteristics of the house and its individual premises
Description of the calculation method
So, the methods proposed above give only a general idea of the required amount of thermal energy for heating a house or apartment. They have a common weak point - almost complete ignorance of possible heat losses, which are recommended to be considered “average”.
But it is quite possible to carry out more accurate calculations. The proposed calculation algorithm will help with this, which is also embodied in the form of an online calculator, which will be offered below. Just before starting the calculations, it makes sense to consider step by step the very principle of their implementation.
First of all, an important note. The proposed methodology does not involve assessing the entire house or apartment according to total area or volume, and each heated room separately. Agree that rooms of equal area, but differing, say, in the number of external walls, will require different quantities heat. It is impossible to put an equal sign between rooms that have a significant difference in the number and area of windows. And there are many such criteria for evaluating each room.
So it would be more correct to calculate the required power for each room separately. Well, then a simple summation of the obtained values will lead us to the desired indicator of the total thermal power for the entire heating system. That is, in fact, for her “heart” - the cauldron.
One more note. The proposed algorithm does not pretend to be “scientific”, that is, it is not directly based on any specific formulas established by SNiP or other governing documents. However, it has been tested by practical application and shows results with a high degree of accuracy. The differences with the results of professionally carried out thermal engineering calculations are minimal and do not in any way affect the correct choice of equipment based on its rated thermal power.
The “architecture” of the calculation is as follows: the basic, already mentioned above value of specific thermal power, equal to 100 W/m², is taken, and then a whole series of correction factors is introduced, to one degree or another reflecting the amount of heat loss in a particular room.
If we express this in a mathematical formula, it will turn out something like this:
Qк= 0.1 × Sc× k1 × k2 × k3 × k4 × k5 × k6 × k7 × k8 × k9 × k10 × k11
Qк- the required thermal power required for full heating of a specific room
0.1 - conversion of 100 W to 0.1 kW, just for the convenience of obtaining the result in kilowatts.
Sk- area of the room.
k1 ÷k11- correction factors to adjust the result taking into account the characteristics of the room.
Presumably, there should be no problems with determining the area of the room. So let’s immediately move on to a detailed consideration of correction factors.
- k1 is a coefficient that takes into account the height of the ceilings in the room.
It is clear that the height of the ceilings directly affects the volume of air that the heating system must warm up. For the calculation, it is proposed to take the following values of the correction factor:
- k2 is a coefficient that takes into account the number of walls of the room in contact with the street.
How larger area contact with external environment, the higher the level of heat loss. Everyone knows that a corner room is always much cooler than one with only one external wall. And some rooms of a house or apartment may even be internal, having no contact with the street.
In your mind, of course, you should take not only the number of external walls, but also their area. But our calculation is still simplified, so we will limit ourselves to only introducing a correction factor.
Odds for various cases are given in the table below:
We do not consider the case when all four walls are external. This is no longer a residential building, but just some kind of barn.
- k3 is a coefficient that takes into account the position of external walls relative to the cardinal points.
Even in winter, the possible impact of energy should not be discounted sun rays. On a clear day, they penetrate through the windows into the rooms, thereby joining the general heat supply. In addition, the walls also receive a charge of solar energy, which leads to a reduction in the total amount of heat loss through them. But all this is true only for those walls that “see” the Sun. There is no such influence on the northern and northeastern sides of the house, for which a certain correction can also be made.
The values of the correction factor for the cardinal directions are in the table below:
- k4 is a coefficient that takes into account the direction of winter winds.
This amendment may not be mandatory, but for houses located in open areas, it makes sense to take it into account.
You might be interested in information about what they are
In almost any area there is a predominance of winter winds - this is also called the “wind rose”. Local meteorologists are required to have such a diagram - it is compiled based on the results of many years of weather observations. Quite often ourselves local residents are well aware of which winds most often disturb them in winter.
And if the wall of the room is located on the windward side, and is not protected by any natural or artificial barriers from the wind, then it will cool down much more. That is, the heat loss of the room increases. This will be less pronounced near a wall located parallel to the direction of the wind, and to a minimum - located on the leeward side.
If you don’t want to “bother” with this factor, or there is no reliable information about the winter wind rose, then you can leave the coefficient equal to one. Or, on the contrary, take it as maximum, just in case, that is, for the most unfavorable conditions.
The values of this correction factor are in the table:
- k5 is a coefficient that takes into account the level of winter temperatures in the region of residence.
If you carry out thermal engineering calculations according to all the rules, then the assessment of heat losses is carried out taking into account the difference in temperatures indoors and outdoors. It is clear that the colder the climatic conditions of the region, the more heat is required to be supplied to the heating system.
Our algorithm will also take this into account to a certain extent, but with an acceptable simplification. Depending on the level of minimum winter temperatures falling on the coldest ten-day period, the correction factor k5 is selected .
It would be appropriate to make one remark here. The calculation will be correct if temperatures that are considered normal for a given region are taken into account. There is no need to remember the abnormal frosts that happened, say, a few years ago (and that’s why, by the way, they were remembered). That is, the lowest but normal temperature for the given area should be selected.
- k6 is a coefficient that takes into account the quality of thermal insulation of walls.
It is quite clear what more efficient system insulation of walls, the lower the level of heat loss will be. Ideally, which should be strived for, thermal insulation in general should be complete, carried out on the basis of completed thermal engineering calculations, taking into account climatic conditions region and house design features.
When calculating the required thermal power of the heating system, the existing thermal insulation of the walls should also be taken into account. The following gradation of correction factors is proposed:
In theory, an insufficient degree of thermal insulation or its complete absence should not be observed in a residential building. Otherwise, the heating system will be very expensive, and even without a guarantee of creating truly comfortable living conditions.
You may be interested in information about the heating system
If the reader wants to independently assess the level of thermal insulation of his home, he can use the information and calculator that are posted in the last section of this publication.
- k7 andk8 – coefficients taking into account heat loss through the floor and ceiling.
The next two coefficients are similar - their introduction into the calculation takes into account the approximate level of heat loss through the floors and ceilings of the premises. There is no need to describe in detail here - both the possible options and the corresponding values of these coefficients are shown in the tables:
For starters, the k7 coefficient, which adjusts the result depending on the characteristics of gender:
Now - the coefficient k8, which corrects for the proximity from above:
- k9 is a coefficient that takes into account the quality of windows in the room.
Here, too, everything is simple - than better quality windows, the less heat loss through them. Old wooden frames, as a rule, do not have good thermal insulation characteristics. This situation is better with modern window systems equipped with double-glazed windows. But they can also have a certain gradation - according to the number of cameras in a double-glazed window and according to other design features.
For our simplified calculation, we can apply the following values of the k9 coefficient:
- k10 is a coefficient that corrects for the glazing area of the room.
The quality of windows does not yet fully reveal all the volumes of possible heat loss through them. Very great importance has a glass area. Agree, it’s difficult to compare a small window and a huge one panoramic window almost the entire wall.
To make adjustments for this parameter, you first need to calculate the so-called glazing coefficient of the room. This is not difficult - you simply find the ratio of the glazing area to the total area of the room.
kw =sw/S
kw- room glazing coefficient;
sw- total area of glazed surfaces, m²;
S- room area, m².
Anyone can measure and sum up the area of windows. And then it’s easy to find the required glazing coefficient by simple division. And it, in turn, makes it possible to go into the table and determine the value of the correction factor k10 :
| Glazing coefficient value kw | k10 coefficient value |
|---|---|
| - up to 0.1 | 0.8 |
| - from 0.11 to 0.2 | 0.9 |
| - from 0.21 to 0.3 | 1.0 |
| - from 0.31 to 0.4 | 1.1 |
| - from 0.41 to 0.5 | 1.2 |
| - over 0.51 | 1.3 |
- k11 is a coefficient that takes into account the presence of doors to the street.
The last of the considered coefficients. The room may have a door leading directly to the street, to cold balcony, in an unheated corridor or entrance, etc. Not only is the door itself often a very serious “cold bridge” - when it is opened regularly, a fair amount of cold air will penetrate into the room each time. Therefore, an allowance should be made for this factor: such heat losses, of course, require additional compensation.
The values of the coefficient k11 are given in the table:
This coefficient should be taken into account if the doors are winter time use regularly.
You might be interested in information about what it is
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So, all correction factors have been considered. As you can see, there is nothing super complicated here, and you can safely move on to the calculations.
One more tip before starting the calculations. Everything will be much simpler if you first draw up a table, in the first column of which you sequentially indicate all the sealed rooms of the house or apartment. Next, place the data required for calculations in columns. For example, in the second column - the area of the room, in the third - the height of the ceilings, in the fourth - the orientation to the cardinal points - and so on. It’s not difficult to create such a sign if you have a plan of your residential property in front of you. It is clear that the calculated values of the required thermal power for each room will be entered in the last column.
The table can be compiled in an office application, or even simply drawn on a piece of paper. And do not rush to part with it after carrying out the calculations - the obtained thermal power indicators will still be useful, for example, when purchasing heating radiators or electric heating devices used as a backup heat source.
To make the task of carrying out such calculations extremely simple for the reader, a special online calculator is located below. With it, with the initial data pre-collected in a table, the calculation will take literally a matter of minutes.
Calculator for calculating the required heating power for the premises of a house or apartment.
Before designing a heating system or installing heating equipment, it is important to select a gas boiler capable of generating required amount heat for the room. Therefore, it is important to choose a device of such power that its performance is as high as possible and its resource is long.
We will tell you how to calculate the power of a gas boiler with high accuracy and taking into account certain parameters. The article we presented describes in detail all types of heat loss through openings and building construction, formulas for their calculation are given. Introduces the features of calculations specific example.
Correct calculation of the power of a gas boiler will not only save on consumables, but will also increase the efficiency of the device. Equipment whose heat output exceeds the actual heat requirements will work ineffectively when, as an insufficiently powerful device, it cannot heat the room properly.
There is modern automated equipment that independently regulates the gas supply, which eliminates unnecessary costs. But if such a boiler performs its work to the limit of its capabilities, then its service life is reduced.
As a result, the efficiency of the equipment decreases, parts wear out faster, and condensation forms. Therefore, there is a need to calculate optimal power.
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