Whether it is an industrial building or a residential building, you need to make competent calculations and draw up a contour diagram heating system. At this stage, experts recommend paying special attention to the calculation of the possible heat load on the heating circuit, as well as the amount of fuel consumed and heat generated.
Thermal load: what is it?
This term refers to the amount of heat given off. The preliminary calculation of the heat load made it possible to avoid unnecessary costs for the purchase of components of the heating system and for their installation. Also, this calculation will help to correctly distribute the amount of heat generated economically and evenly throughout the building.
There are many nuances in these calculations. For example, the material from which the building is built, thermal insulation, region, etc. Specialists try to take into account as many factors and characteristics as possible to obtain a more accurate result.
The calculation of the heat load with errors and inaccuracies leads to inefficient operation of the heating system. It even happens that you have to redo sections of an already working structure, which inevitably leads to unplanned expenses. Yes, and housing and communal organizations calculate the cost of services based on data on heat load.
Main Factors
An ideally calculated and designed heating system must maintain the set temperature in the room and compensate for the resulting heat losses. When calculating the indicator of the heat load on the heating system in the building, you need to take into account:
Purpose of the building: residential or industrial.
Feature structural elements buildings. These are windows, walls, doors, roof and ventilation system.
Housing dimensions. The larger it is, the more powerful the heating system should be. Area must be taken into account window openings, doors, exterior walls and the volume of each interior space.
The presence of rooms for special purposes (bath, sauna, etc.).
Degree of equipment technical devices. That is, the presence of hot water supply, ventilation systems, air conditioning and the type of heating system.
For a single room. For example, in rooms intended for storage, it is not necessary to maintain a comfortable temperature for a person.
Number of points with feed hot water. The more of them, the more the system is loaded.
Area of glazed surfaces. Rooms with French windows lose a significant amount of heat.
Additional terms. In residential buildings, this can be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in a calendar year, shifts, the technological chain of the production process, etc.
Climatic conditions of the region. When calculating heat losses, street temperatures are taken into account. If the differences are insignificant, then a small amount of energy will be spent on compensation. While at -40 ° C outside the window it will require significant expenses.
Features of existing methods
The parameters included in the calculation of the heat load are in SNiPs and GOSTs. They also have special heat transfer coefficients. From the passports of the equipment included in the heating system, are taken digital characteristics relating to a specific heating radiator, boiler, etc. And also traditionally:
The heat consumption, taken to the maximum for one hour of operation of the heating system,
The maximum heat flow from one radiator,
Total heat costs in a certain period (most often - a season); if you need an hourly calculation of the load on heating network, then the calculation must be carried out taking into account the temperature difference during the day.
The calculations made are compared with the heat transfer area of the entire system. The index is quite accurate. Some deviations happen. For example, for industrial buildings, it will be necessary to take into account the reduction in heat energy consumption on weekends and holidays, and in residential buildings - at night.
Methods for calculating heating systems have several degrees of accuracy. To reduce the error to a minimum, it is necessary to use rather complex calculations. Less accurate schemes are used if the goal is not to optimize the costs of the heating system.
Basic calculation methods
To date, the calculation of the heat load on the heating of a building can be carried out in one of the following ways.
Three main
- Aggregated indicators are taken for calculation.
- The indicators of the structural elements of the building are taken as the base. Here, the calculation of the internal volume of air going to warm up will also be important.
- All objects included in the heating system are calculated and summarized.
One exemplary
There is also a fourth option. It has a fairly large error, because the indicators are taken very average, or they are not enough. Here is the formula - Q from \u003d q 0 * a * V H * (t EH - t NPO), where:
- q 0 - specific thermal characteristic of the building (most often determined by the coldest period),
- a - correction factor (depends on the region and is taken from ready-made tables),
- V H is the volume calculated from the outer planes.
Example of a simple calculation
For building with standard parameters(ceiling heights, room sizes and good thermal insulation characteristics), a simple ratio of parameters can be applied, adjusted by a coefficient depending on the region.
Suppose that a residential building is located in the Arkhangelsk region, and its area is 170 square meters. m. The heat load will be equal to 17 * 1.6 \u003d 27.2 kW / h.
Such a definition of thermal loads does not take into account many important factors. For example, design features buildings, temperatures, the number of walls, the ratio of the areas of walls and window openings, etc. Therefore, such calculations are not suitable for serious heating system projects.
It depends on the material from which they are made. Most often today, bimetallic, aluminum, steel are used, much less often cast iron radiators. Each of them has its own heat transfer index (thermal power). Bimetallic radiators with a distance between the axes of 500 mm, on average, have 180 - 190 watts. Aluminum radiators have almost the same performance.
The heat transfer of the described radiators is calculated for one section. Steel plate radiators are non-separable. Therefore, their heat transfer is determined based on the size of the entire device. For example, the thermal power of a two-row radiator with a width of 1,100 mm and a height of 200 mm will be 1,010 W, and panel radiator made of steel with a width of 500 mm and a height of 220 mm will be 1,644 watts.
The calculation of the heating radiator by area includes the following basic parameters:
Ceiling height (standard - 2.7 m),
Thermal power (per sq. m - 100 W),
One outer wall.
These calculations show that for every 10 sq. m requires 1,000 W of thermal power. This result is divided by the heat output of one section. The answer is the required number of radiator sections.
For the southern regions of our country, as well as for the northern ones, decreasing and increasing coefficients have been developed.
Average calculation and exact
Given the factors described, the average calculation is carried out according to the following scheme. If for 1 sq. m requires 100 W of heat flow, then a room of 20 square meters. m should receive 2,000 watts. The radiator (popular bimetallic or aluminum) of eight sections allocates about Divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.
The exact one looks a little intimidating. Actually, nothing complicated. Here is the formula:
Q t \u003d 100 W / m 2 × S (rooms) m 2 × q 1 × q 2 × q 3 × q 4 × q 5 × q 6 × q 7, Where:
- q 1 - type of glazing (ordinary = 1.27, double = 1.0, triple = 0.85);
- q 2 - wall insulation (weak or absent = 1.27, 2-brick wall = 1.0, modern, high = 0.85);
- q 3 - the ratio of the total area of window openings to the floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
- q 4 - outdoor temperature (the minimum value is taken: -35 o C = 1.5, -25 o C = 1.3, -20 o C = 1.1, -15 o C = 0.9, -10 o C = 0.7);
- q 5 - the number of external walls in the room (all four = 1.4, three = 1.3, corner room= 1.2, one = 1.2);
- q 6 - type of calculation room above the calculation room (cold attic = 1.0, warm attic = 0.9, residential heated room = 0.8);
- q 7 - ceiling height (4.5 m = 1.2, 4.0 m = 1.15, 3.5 m = 1.1, 3.0 m = 1.05, 2.5 m = 1.3).
Using any of the methods described, it is possible to calculate the heat load of an apartment building.
Approximate calculation
These are the conditions. The minimum temperature in the cold season is -20 ° C. Room 25 sq. m with triple glazing, double-leaf windows, ceiling height of 3.0 m, two-brick walls and an unheated attic. The calculation will be as follows:
Q \u003d 100 W / m 2 × 25 m 2 × 0.85 × 1 × 0.8 (12%) × 1.1 × 1.2 × 1 × 1.05.
The result, 2 356.20, is divided by 150. As a result, it turns out that 16 sections need to be installed in a room with the specified parameters.
If calculation is required in gigacalories
In the absence of a heat energy meter on the open heating circuit calculation of the heat load for heating the building is calculated by the formula Q \u003d V * (T 1 - T 2) / 1000, where:
- V - the amount of water consumed by the heating system, calculated in tons or m 3,
- T 1 - a number showing the temperature of hot water, measured in o C, and for calculations, the temperature corresponding to a certain pressure in the system is taken. This indicator has its own name - enthalpy. If it is not possible to remove temperature indicators in a practical way, they resort to an average indicator. It is in the range of 60-65 o C.
- T 2 - temperature cold water. It is quite difficult to measure it in the system, so constant indicators have been developed that depend on temperature regime on the street. For example, in one of the regions, in the cold season, this indicator is taken equal to 5, in summer - 15.
- 1,000 is the coefficient for obtaining the result immediately in gigacalories.
In the case of a closed circuit, the heat load (gcal/h) is calculated differently:
Q from \u003d α * q o * V * (t in - t n.r.) * (1 + K n.r.) * 0.000001, Where
The calculation of the heat load turns out to be somewhat enlarged, but it is this formula that is given in the technical literature.
Increasingly, in order to increase the efficiency of the heating system, they resort to buildings.
These works are carried out at night. For a more accurate result, you must observe the temperature difference between the room and the street: it must be at least 15 o. Fluorescent and incandescent lamps are switched off. It is advisable to remove carpets and furniture to the maximum, they knock down the device, giving some error.
The survey is carried out slowly, the data are recorded carefully. The scheme is simple.
The first stage of work takes place indoors. The device is moved gradually from doors to windows, giving Special attention corners and other joints.
The second stage - inspection with a thermal imager external walls buildings. The joints are still carefully examined, especially the connection with the roof.
The third stage is data processing. First, the device does this, then the readings are transferred to a computer, where the corresponding programs complete the processing and give the result.
If the survey was conducted by a licensed organization, then it will issue a report with mandatory recommendations based on the results of the work. If the work was carried out personally, then you need to rely on your knowledge and, possibly, the help of the Internet.
Before proceeding with the purchase of materials and the installation of heat supply systems for a house or apartment, it is necessary to calculate the heating based on the area of \u200b\u200beach room. Basic parameters for heating design and heat load calculation:
- Square;
- Number of window blocks;
- Ceiling height;
- The location of the room;
- Heat loss;
- Heat dissipation of radiators;
- Climatic zone (outside temperature).
The method described below is used to calculate the number of batteries for a room area without additional heating sources (heat-insulated floors, air conditioners, etc.). There are two ways to calculate heating: using a simple and complicated formula.
Before starting the design of heat supply, it is worth deciding which radiators will be installed. The material from which the heating batteries are made:
- Cast iron;
- Steel;
- Aluminum;
- Bimetal.
Aluminum and bimetallic radiators are considered the best option. The highest thermal output of bimetallic devices. Cast iron batteries they heat up for a long time, but after turning off the heating, the temperature in the room lasts for quite a long time.
A simple formula for designing the number of sections in a heating radiator is:
K = Sx(100/R), where:
S is the area of the room;
R - section power.
If we consider the example with data: room 4 x 5 m, bimetal radiator, power 180 watts. The calculation will look like this:
K = 20*(100/180) = 11.11. So, for a room with an area of 20 m 2, a battery with at least 11 sections is required for installation. Or, for example, 2 radiators with 5 and 6 ribs. The formula is used for rooms with a ceiling height of up to 2.5 m in a standard Soviet-built building.
However, such a calculation of the heating system does not take into account the heat loss of the building, the outdoor temperature of the house and the number of window blocks are also not taken into account. Therefore, these coefficients should also be taken into account for the final refinement of the number of ribs.
Calculations for panel radiators
In the case where the installation of a battery with a panel instead of ribs is supposed, the following formula by volume is used:
W \u003d 41xV, where W is the battery power, V is the volume of the room. The number 41 is the norm of the average annual heating capacity of 1 m 2 of a dwelling.
As an example, we can take a room with an area of 20 m 2 and a height of 2.5 m. The value of the radiator power for a room volume of 50 m 3 will be 2050 W, or 2 kW.
Heat loss calculation
H2_2The main heat loss occurs through the walls of the room. To calculate, you need to know the coefficient of thermal conductivity of the external and internal material, from which the house is built, the thickness of the wall of the building, the average outdoor temperature is also important. Basic formula:
Q \u003d S x ΔT / R, where
ΔT is the temperature difference between the outside and the internal optimum value;
S is the area of the walls;
R is the thermal resistance of the walls, which, in turn, is calculated by the formula:
R = B/K, where B is the thickness of the brick, K is the coefficient of thermal conductivity.
Calculation example: the house is built of shell rock, in stone, located in the Samara region. The thermal conductivity of the shell rock is on average 0.5 W/m*K, the wall thickness is 0.4 m. Considering the average range, the minimum temperature in winter is -30 °C. In the house, according to SNIP, normal temperature is +25 °C, the difference is 55 °C.
If the room is angular, then both of its walls are in direct contact with environment. The area of the outer two walls of the room is 4x5 m and 2.5 m high: 4x2.5 + 5x2.5 = 22.5 m 2.
R = 0.4/0.5 = 0.8
Q \u003d 22.5 * 55 / 0.8 \u003d 1546 W.
In addition, it is necessary to take into account the insulation of the walls of the room. When finished with foam outdoor area heat losses are reduced by about 30%. So, the final figure will be about 1000 watts.
Heat Load Calculation (Advanced Formula)
Scheme of heat loss of premises
To calculate the final heat consumption for heating, it is necessary to take into account all the coefficients according to the following formula:
CT \u003d 100xSxK1xK2xK3xK4xK5xK6xK7, where:
S is the area of the room;
K - various coefficients:
K1 - loads for windows (depending on the number of double-glazed windows);
K2 - thermal insulation of the outer walls of the building;
K3 - loads for the ratio of window area to floor area;
K4 – outdoor air temperature regime;
K5 - taking into account the number of external walls of the room;
K6 - loads, based on the upper room above the calculated room;
K7 - taking into account the height of the room.
As an example, we can consider the same room of a building in the Samara region, insulated from the outside with foam plastic, having 1 double-glazed window, above which a heated room is located. The heat load formula will look like this:
KT \u003d 100 * 20 * 1.27 * 1 * 0.8 * 1.5 * 1.2 * 0.8 * 1 \u003d 2926 W.
The calculation of heating is focused on this figure.
Heat consumption for heating: formula and adjustments
Based on the above calculations, 2926 watts are needed to heat a room. Considering heat losses, the requirements are: 2926 + 1000 = 3926 W (KT2). The following formula is used to calculate the number of sections:
K = KT2/R, where KT2 is the final value of the heat load, R is the heat transfer (power) of one section. Final figure:
K = 3926/180 = 21.8 (rounded 22)
So, in order to ensure optimal heat consumption for heating, it is necessary to install radiators with a total of 22 sections. It should be borne in mind that the lowest temperature - 30 degrees below zero in time is a maximum of 2-3 weeks, so you can safely reduce the number to 17 sections (- 25%).
If homeowners are not satisfied with such an indicator of the number of radiators, then batteries with a large heat supply capacity should be taken into account initially. Or insulate the walls of the building both inside and outside modern materials. In addition, it is necessary to correctly assess the needs of housing for heat, based on secondary parameters.
There are several other parameters that affect the additional energy wasted, which entails an increase in heat loss:
- Features of the outer walls. Heating energy should be enough not only for heating the room, but also to compensate for heat losses. The wall in contact with the environment, over time, from changes in the temperature of the outside air, begins to let moisture in. Especially it is necessary to insulate well and carry out high-quality waterproofing for the northern directions. It is also recommended to insulate the surface of houses located in humid regions. High annual rainfall will inevitably lead to increased heat losses.
- Place of installation of radiators. If the battery is mounted under a window, then heating energy leaks through its structure. The installation of high-quality blocks will help reduce heat loss. You also need to calculate the power of the device installed in the window sill - it should be higher.
- Conventional annual heat demand for buildings in different time zones. As a rule, according to SNIPs, the average temperature (annual average) for buildings is calculated. However, heat demand is significantly lower if, for example, cold weather and low outdoor air values occur for a total of 1 month of the year.
Advice! In order to minimize the need for heat in winter, it is recommended to install additional sources of indoor air heating: air conditioners, mobile heaters, etc.
Ask any specialist how to properly organize the heating system in the building. It doesn't matter if it's residential or industrial. And the professional will answer that the main thing is to accurately make calculations and correctly carry out the design. We are talking, in particular, about the calculation of the heat load on heating. The volume of consumption of thermal energy, and hence fuel, depends on this indicator. That is, economic indicators are next to the technical characteristics.
Performing accurate calculations allows you to get not only a complete list of the documentation necessary for the installation work, but also to select the necessary equipment, additional components and materials.
Thermal loads - definition and characteristics
What is usually meant by the term "heat load on heating"? This is the amount of heat that all heating devices installed in the building give off. To avoid unnecessary expenses for the production of work, as well as the purchase of unnecessary devices and materials, a preliminary calculation is necessary. With it, you can adjust the rules for installing and distributing heat in all rooms, and this can be done economically and evenly.
But that's not all. Very often, experts carry out calculations, relying on accurate indicators. They relate to the size of the house and the nuances of construction, which takes into account the diversity of the elements of the building and their compliance with the requirements of thermal insulation and other things. It is precisely the exact indicators that make it possible to correctly make calculations and, accordingly, obtain options for the distribution of thermal energy throughout the premises as close to the ideal as possible.
But often there are errors in the calculations, which leads to inefficient operation of the heating as a whole. Sometimes it is necessary to redo during operation not only the circuits, but also sections of the system, which leads to additional costs.
What parameters affect the calculation of the heat load in general? Here it is necessary to divide the load into several positions, which include:
- Central heating system.
- Underfloor heating system, if one is installed in the house.
- Ventilation system - both forced and natural.
- Hot water supply of the building.
- Branches for additional household needs. For example, a sauna or a bath, a pool or a shower.
Main characteristics
Professionals do not lose sight of any trifle that can affect the correctness of the calculation. Hence the rather large list of characteristics of the heating system that should be taken into account. Here are just a few of them:
- The purpose of the property or its type. It can be a residential building or an industrial building. Heat suppliers have standards that are distributed by type of building. They often become fundamental in carrying out calculations.
- The architectural part of the building. This can include enclosing elements (walls, roofs, ceilings, floors), their dimensions, thickness. Be sure to take into account all kinds of openings - balconies, windows, doors, etc. It is very important to take into account the presence of basements and attics.
- Temperature regime for each room separately. This is very important because General requirements to the temperature in the house do not give an accurate picture of the distribution of heat.
- Appointment of premises. This mainly applies to production shops, which require stricter compliance with the temperature regime.
- Availability of special premises. For example, in residential private houses it can be baths or saunas.
- Degree of technical equipment. The presence of a ventilation and air conditioning system, hot water supply, and the type of heating used are taken into account.
- The number of points through which hot water is taken. And the more such points, the greater the heat load the heating system is exposed to.
- The number of people on the site. Criteria such as indoor humidity and temperature depend on this indicator.
- Additional indicators. In residential premises, one can distinguish the number of bathrooms, separate rooms, balconies. IN industrial buildings- the number of working shifts, the number of days in a year when the shop itself works in the technological chain.
What is included in the calculation of loads
Heating scheme
The calculation of thermal loads for heating is carried out at the design stage of the building. But at the same time, the norms and requirements of various standards must be taken into account.
For example, the heat loss of the enclosing elements of the building. Moreover, all rooms are taken into account separately. Further, this is the power that is needed to heat the coolant. We add here the amount of thermal energy required for heating supply ventilation. Without this, the calculation will not be very accurate. We also add the energy that is spent on heating water for a bath or pool. Specialists must take into account the further development of the heating system. Suddenly, in a few years, you will decide to arrange a Turkish hammam in your own private house. Therefore, it is necessary to add a few percent to the loads - usually up to 10%.
Recommendation! It is necessary to calculate thermal loads with a "margin" for country houses. It is the reserve that will allow in the future to avoid additional financial costs, which are often determined by amounts of several zeros.
Features of calculating the heat load
Air parameters, or rather, its temperature, are taken from GOSTs and SNiPs. Here, the heat transfer coefficients are selected. By the way, the passport data of all types of equipment (boilers, heating radiators, etc.) are taken into account without fail.
What is usually included in a traditional heat load calculation?
- Firstly, the maximum flow of thermal energy coming from heating devices (radiators).
- Secondly, maximum flow heat for 1 hour of operation of the heating system.
- Thirdly, the total heat costs for a certain period of time. Usually the seasonal period is calculated.
If all these calculations are measured and compared with the heat transfer area of the system as a whole, then a fairly accurate indicator of the efficiency of heating a house will be obtained. But you have to take into account small deviations. For example, reducing heat consumption at night. For industrial facilities, you will also have to take into account weekends and holidays.
Methods for determining thermal loads
Underfloor heating design
Currently, experts use three main methods for calculating thermal loads:
- Calculation of the main heat losses, where only aggregated indicators are taken into account.
- The indicators based on the parameters of the enclosing structures are taken into account. This is usually added to the losses for heating the internal air.
- All systems included in heating networks are calculated. This is both heating and ventilation.
There is another option, which is called the enlarged calculation. It is usually used when there are no basic indicators and building parameters required for a standard calculation. That is, the actual characteristics may differ from the design.
To do this, experts use a very simple formula:
Q max from. \u003d α x V x q0 x (tv-tn.r.) x 10 -6
α is a correction factor depending on the region of construction (table value)
V - the volume of the building on the outer planes
q0 - characteristic of the heating system by specific index, usually determined by the coldest days of the year
Types of thermal loads
Thermal loads that are used in the calculations of the heating system and the selection of equipment have several varieties. For example, seasonal loads, for which the following features are inherent:
- Changes in outdoor temperature throughout heating season.
- Meteorological features of the region where the house was built.
- Jumps in the load on the heating system during the day. This indicator usually falls into the category of "minor loads" because the protective elements prevent great pressure for heating in general.
- Everything related to the thermal energy associated with the ventilation system of the building.
- Thermal loads that are determined throughout the year. For example, the consumption of hot water in the summer season is reduced by only 30-40% when compared with winter time of the year.
- Dry heat. This feature is inherent in domestic heating systems, where a fairly large number of indicators are taken into account. For example, the number of windows and doorways, the number of people living or permanently in the house, ventilation, air exchange through various cracks and gaps. A dry thermometer is used to determine this value.
- Hidden thermal energy. There is also such a term, which is defined by evaporation, condensation, and so on. A wet bulb thermometer is used to determine the indicator.
Thermal Load Controllers
Programmable controller, temperature range - 5-50 C
Modern heating units and appliances are provided with a set of different regulators, with which you can change the heat loads, in order to avoid dips and jumps in thermal energy in the system. Practice has shown that with the help of regulators it is possible not only to reduce the load, but also to bring the heating system to rational use fuel. And this is a purely economic side of the issue. This applies especially to industrial facilities where you have to pay fairly large fines for excessive fuel consumption.
If you are not sure about the correctness of your calculations, then use the services of specialists.
Let's look at a couple more formulas that relate to different systems. For example, ventilation and hot water systems. Here you need two formulas:
Qin. \u003d qin.V (tn.-tv.) - this applies to ventilation.
Here:
tn. and tv - air temperature outside and inside
qv. - specific indicator
V - external volume of the building
Qgvs. \u003d 0.042rv (tg.-tx.) Pgav - for hot water supply, where
tg.-tx - temperature of hot and cold water
r - water density
c - the ratio of the maximum load to the average, which is determined by GOSTs
P - the number of consumers
Gav - average hot water consumption
Complex calculation
In combination with settlement issues, studies of the thermotechnical order are necessarily carried out. For this, various devices are used that give accurate indicators for calculations. For example, for this, window and door openings, ceilings, walls, and so on are examined.
It is this examination that helps to determine the nuances and factors that can have a significant impact on heat loss. For example, thermal imaging diagnostics will accurately show the temperature difference when a certain amount of thermal energy passes through 1 square meter of the building envelope.
So practical measurements are indispensable when making calculations. This is especially true for bottlenecks in the building structure. In this regard, the theory will not be able to show exactly where and what is wrong. And practice will show where to apply different methods protection against heat loss. And the calculations themselves in this regard are becoming more accurate.
Conclusion on the topic
Estimated heat load is a very important indicator obtained in the process of designing a home heating system. If you approach the matter wisely and spend everything necessary calculations correctly, you can guarantee that the heating system will work perfectly. And at the same time, it will be possible to save on overheating and other costs that can simply be avoided.
Build a heating system own house or even in a city apartment - an extremely responsible occupation. At the same time, it would be completely unreasonable to purchase boiler equipment, as they say, “by eye”, that is, without taking into account all the features of housing. In this, it is quite possible to fall into two extremes: either the power of the boiler will not be enough - the equipment will work “to its fullest”, without pauses, but will not give the expected result, or, conversely, an overly expensive device will be purchased, the capabilities of which will remain completely unclaimed.
But that's not all. It is not enough to purchase the necessary heating boiler correctly - it is very important to optimally select and correctly place heat exchange devices in the premises - radiators, convectors or "warm floors". And again, relying only on your intuition or the "good advice" of your neighbors is not the most reasonable option. In a word, certain calculations are indispensable.
Of course, ideally, such heat engineering calculations should be carried out by appropriate specialists, but this often costs a lot of money. Isn't it interesting to try to do it yourself? This publication will show in detail how heating is calculated by the area of \u200b\u200bthe room, taking into account many important nuances. By analogy, it will be possible to perform, built into this page, will help you perform the necessary calculations. The technique cannot be called completely “sinless”, however, it still allows you to get a result with a completely acceptable degree of accuracy.
The simplest methods of calculation
In order for the heating system to create comfortable living conditions during the cold season, it must cope with two main tasks. These functions are closely related, and their separation is very conditional.
- The first is maintaining an optimal level of air temperature in the entire volume of the heated room. Of course, the temperature level may vary slightly with altitude, but this difference should not be significant. Quite comfortable conditions are considered to be an average of +20 ° C - it is this temperature that, as a rule, is taken as the initial temperature in thermal calculations.
In other words, the heating system must be able to heat a certain volume of air.
If we approach with complete accuracy, then for individual rooms in residential buildings the standards for the required microclimate have been established - they are defined by GOST 30494-96. An excerpt from this document is in the table below:
| Purpose of the premises | Air temperature, °С | Relative humidity, % | Air speed, m/s | |||
|---|---|---|---|---|---|---|
| optimal | admissible | optimal | admissible, max | optimal, max | admissible, max | |
| For the cold season | ||||||
| Living room | 20÷22 | 18÷24 (20÷24) | 45÷30 | 60 | 0.15 | 0.2 |
| Same but for living rooms in regions with minimum temperatures from -31 °C and below | 21÷23 | 20÷24 (22÷24) | 45÷30 | 60 | 0.15 | 0.2 |
| Kitchen | 19:21 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Toilet | 19:21 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Bathroom, combined bathroom | 24÷26 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Premises for rest and study | 20÷22 | 18:24 | 45÷30 | 60 | 0.15 | 0.2 |
| Inter-apartment corridor | 18:20 | 16:22 | 45÷30 | 60 | N/N | N/N |
| lobby, stairwell | 16÷18 | 14:20 | N/N | N/N | N/N | N/N |
| Storerooms | 16÷18 | 12÷22 | N/N | N/N | N/N | N/N |
| For the warm season (The standard is only for residential premises. For the rest - it is not standardized) | ||||||
| Living room | 22÷25 | 20÷28 | 60÷30 | 65 | 0.2 | 0.3 |
- The second is the compensation of heat losses through the structural elements of the building.
The main "enemy" of the heating system is heat loss through building structures.
Alas, heat loss is the most serious "rival" of any heating system. They can be reduced to a certain minimum, but even with the highest quality thermal insulation, it is not yet possible to completely get rid of them. Thermal energy leaks go in all directions - their approximate distribution is shown in the table:
| Building element | Approximate value of heat loss |
|---|---|
| Foundation, floors on the ground or over unheated basement (basement) premises | from 5 to 10% |
| "Cold bridges" through poorly insulated joints building structures | from 5 to 10% |
| Entry places engineering communications(sewerage, plumbing, gas pipes, electrical cables, etc.) | up to 5% |
| External walls, depending on the degree of insulation | from 20 to 30% |
| Poor quality windows and external doors | about 20÷25%, of which about 10% - through non-sealed joints between the boxes and the wall, and due to ventilation |
| Roof | up to 20% |
| Ventilation and chimney | up to 25 ÷30% |
Naturally, in order to cope with such tasks, the heating system must have a certain thermal power, and this potential must not only meet the general needs of the building (apartment), but also be correctly distributed among the premises, in accordance with their area and a number of other important factors.
Usually the calculation is carried out in the direction "from small to large". Simply put, the required amount of thermal energy is calculated for each heated room, the obtained values are summed up, approximately 10% of the reserve is added (so that the equipment does not work at the limit of its capabilities) - and the result will show how much power the heating boiler needs. And the values for each room will be the starting point for the calculation required amount radiators.
The most simplified and most commonly used method in a non-professional environment is to accept the norm of 100 W of thermal energy per square meter of area:
The most primitive way of counting is the ratio of 100 W / m²
Q = S× 100
Q- the required thermal power for the room;
S– area of the room (m²);
100 — specific power per unit area (W/m²).
For example, room 3.2 × 5.5 m
S= 3.2 × 5.5 = 17.6 m²
Q= 17.6 × 100 = 1760 W ≈ 1.8 kW
The method is obviously very simple, but very imperfect. It should be noted right away that it is conditionally applicable only when standard height ceilings - approximately 2.7 m (permissible - in the range from 2.5 to 3.0 m). From this point of view, the calculation will be more accurate not from the area, but from the volume of the room.
It is clear that in this case the value of specific power is calculated per cubic meter. It is taken equal to 41 W / m³ for reinforced concrete panel house, or 34 W / m³ - in brick or made of other materials.
Q = S × h× 41 (or 34)
h- ceiling height (m);
41 or 34 - specific power per unit volume (W / m³).
For example, the same room panel house, with a ceiling height of 3.2 m:
Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW
The result is more accurate, since it already takes into account not only all the linear dimensions of the room, but even, to a certain extent, the features of the walls.
But still, it is still far from real accuracy - many nuances are “outside the brackets”. How to perform calculations closer to real conditions - in the next section of the publication.
You may be interested in information about what they are
Carrying out calculations of the required thermal power, taking into account the characteristics of the premises
The calculation algorithms discussed above are useful for the initial “estimate”, but you should still rely on them completely with very great care. Even to a person who does not understand anything in building heat engineering, the indicated average values \u200b\u200bmay certainly seem doubtful - they cannot be equal, say, for the Krasnodar Territory and for the Arkhangelsk Region. In addition, the room - the room is different: one is located on the corner of the house, that is, it has two external walls, and the other is protected from heat loss by other rooms on three sides. In addition, the room may have one or more windows, both small and very large, sometimes even panoramic. And the windows themselves may differ in the material of manufacture and other design features. And this is not a complete list - just such features are visible even to the "naked eye".
In a word, there are a lot of nuances that affect the heat loss of each particular room, and it is better not to be too lazy, but to carry out a more thorough calculation. Believe me, according to the method proposed in the article, this will not be so difficult to do.
General principles and calculation formula
The calculations will be based on the same ratio: 100 W per 1 square meter. But that's just the formula itself "overgrown" with a considerable number of various correction factors.
Q = (S × 100) × a × b × c × d × e × f × g × h × i × j × k × l × m
The Latin letters denoting the coefficients are taken quite arbitrarily, in alphabetical order, and are not related to any standard quantities accepted in physics. The meaning of each coefficient will be discussed separately.
- "a" - a coefficient that takes into account the number of external walls in a particular room.
Obviously, the more external walls in the room, the larger the area through which heat loss occurs. In addition, the presence of two or more external walls also means corners - extremely vulnerable places in terms of the formation of "cold bridges". The coefficient "a" will correct for this specific feature rooms.
The coefficient is taken equal to:
- external walls No (interior): a = 0.8;
- outer wall one: a = 1.0;
- external walls two: a = 1.2;
- external walls three: a = 1.4.
- "b" - coefficient taking into account the location of the external walls of the room relative to the cardinal points.
You may be interested in information about what are
Even on the coldest winter days solar energy still affects the temperature balance in the building. It is quite natural that the side of the house that faces south receives a certain amount of heat from the sun's rays, and heat loss through it is lower.
But the walls and windows facing north never “see” the Sun. The eastern part of the house, although it "grabs" the morning sun's rays, still does not receive any effective heating from them.
Based on this, we introduce the coefficient "b":
- the outer walls of the room look at North or East: b = 1.1;
- the outer walls of the room are oriented towards South or West: b = 1.0.
- "c" - coefficient taking into account the location of the room relative to the winter "wind rose"
Perhaps this amendment is not so necessary for houses located in areas protected from the winds. But sometimes the prevailing winter winds can make their own “hard adjustments” to the thermal balance of the building. Naturally, the windward side, that is, "substituted" to the wind, will lose much more body, compared to the leeward, opposite.
Based on the results of long-term meteorological observations in any region, the so-called "wind rose" is compiled - a graphic diagram showing the prevailing wind directions in winter and summer. This information can be obtained from the local hydrometeorological service. However, many residents themselves, without meteorologists, know very well where the winds mainly blow from in winter, and from which side of the house the deepest snowdrifts usually sweep.
If there is a desire to carry out calculations with higher accuracy, then the correction factor “c” can also be included in the formula, taking it equal to:
- windward side of the house: c = 1.2;
- leeward walls of the house: c = 1.0;
- wall located parallel to the direction of the wind: c = 1.1.
- "d" - a correction factor that takes into account the peculiarities of the climatic conditions of the region where the house was built
Naturally, the amount of heat loss through all the building structures of the building will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer indicators “dance” in a certain range, but for each region there is an average indicator of the lowest temperatures characteristic of the coldest five-day period of the year (usually this is characteristic of January). For example, below is a map-scheme of the territory of Russia, on which approximate values are shown in colors.
Usually this value is easy to check with the regional meteorological service, but you can, in principle, rely on your own observations.
So, the coefficient "d", taking into account the peculiarities of the climate of the region, for our calculations in we take equal to:
— from – 35 °С and below: d=1.5;
— from – 30 °С to – 34 °С: d=1.3;
— from – 25 °С to – 29 °С: d=1.2;
— from – 20 °С to – 24 °С: d=1.1;
— from – 15 °С to – 19 °С: d=1.0;
— from – 10 °С to – 14 °С: d=0.9;
- not colder - 10 ° С: d=0.7.
- "e" - coefficient taking into account the degree of insulation of external walls.
The total value of the heat loss of the building is directly related to the degree of insulation of all building structures. One of the "leaders" in terms of heat loss are walls. Therefore, the value of thermal power required to maintain comfortable conditions living indoors depends on the quality of their thermal insulation.
The value of the coefficient for our calculations can be taken as follows:
- external walls are not insulated: e = 1.27;
- medium degree of insulation - walls in two bricks or their surface thermal insulation with other heaters is provided: e = 1.0;
– insulation was carried out qualitatively, on the basis of heat engineering calculations: e = 0.85.
Later in the course of this publication, recommendations will be given on how to determine the degree of insulation of walls and other building structures.
- coefficient "f" - correction for ceiling height
Ceilings, especially in private homes, can have different heights. Therefore, the thermal power for heating one or another room of the same area will also differ in this parameter.
It won't be a big mistake to accept the following values correction factor "f":
– ceiling height up to 2.7 m: f = 1.0;
— flow height from 2.8 to 3.0 m: f = 1.05;
– ceiling height from 3.1 to 3.5 m: f = 1.1;
– ceiling height from 3.6 to 4.0 m: f = 1.15;
– ceiling height over 4.1 m: f = 1.2.
- « g "- coefficient taking into account the type of floor or room located under the ceiling.
As shown above, the floor is one of the significant sources of heat loss. So, it is necessary to make some adjustments in the calculation of this feature of a particular room. The correction factor "g" can be taken equal to:
- cold floor on the ground or over an unheated room (for example, basement or basement): g= 1,4 ;
- insulated floor on the ground or over an unheated room: g= 1,2 ;
- a heated room is located below: g= 1,0 .
- « h "- coefficient taking into account the type of room located above.
The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat losses are inevitable, which will require an increase in the required heat output. We introduce the coefficient "h", which takes into account this feature of the calculated room:
- a "cold" attic is located on top: h = 1,0 ;
- an insulated attic or other insulated room is located on top: h = 0,9 ;
- any heated room is located above: h = 0,8 .
- « i "- coefficient taking into account the design features of windows
Windows are one of the "main routes" of heat leaks. Naturally, much in this matter depends on the quality of the window construction. Old wooden frames, which were previously installed everywhere in all houses, are significantly inferior to modern multi-chamber systems with double-glazed windows in terms of their thermal insulation.
Without words, it is clear that the thermal insulation qualities of these windows are significantly different.
But even between PVC-windows there is no complete uniformity. For example, a two-chamber double-glazed window (with three glasses) will be much warmer than a single-chamber one.
This means that it is necessary to enter a certain coefficient "i", taking into account the type of windows installed in the room:
— standard wooden windows with conventional double glazing: i = 1,27 ;
– modern window systems with single pane glass: i = 1,0 ;
– modern window systems with two-chamber or three-chamber double-glazed windows, including those with argon filling: i = 0,85 .
- « j" - correction factor for the total glazing area of the room
Whatever quality windows however they were, it will still not be possible to completely avoid heat loss through them. But it is quite clear that there is no way to compare a small window with panoramic windows almost the whole wall.
First you need to find the ratio of the areas of all the windows in the room and the room itself:
x = ∑SOK /SP
∑ SOK- the total area of windows in the room;
SP- area of the room.
Depending on the value obtained and the correction factor "j" is determined:
- x \u003d 0 ÷ 0.1 →j = 0,8 ;
- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;
- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;
- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;
- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;
- « k" - coefficient that corrects for the presence of an entrance door
The door to the street or to an unheated balcony is always an additional "loophole" for the cold
door to the street or outdoor balcony is able to make its own adjustments to the heat balance of the room - each of its opening is accompanied by the penetration of a considerable amount of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient "k", which we take equal to:
- no door k = 1,0 ;
- one door to the street or balcony: k = 1,3 ;
- two doors to the street or to the balcony: k = 1,7 .
- « l "- possible amendments to the connection diagram of heating radiators
Perhaps this will seem like an insignificant trifle to some, but still - why not immediately take into account the planned scheme for connecting heating radiators. The fact is that their heat transfer, and hence their participation in maintaining a certain temperature balance in the room, changes quite noticeably with different types tie-in supply and return pipes.
| Illustration | Radiator insert type | The value of the coefficient "l" |
|---|---|---|
 | Diagonal connection: supply from above, "return" from below | l = 1.0 |
 | Connection on one side: supply from above, "return" from below | l = 1.03 |
 | Two-way connection: both supply and return from the bottom | l = 1.13 |
 | Diagonal connection: supply from below, "return" from above | l = 1.25 |
 | Connection on one side: supply from below, "return" from above | l = 1.28 |
 | One-way connection, both supply and return from below | l = 1.28 |
- « m "- correction factor for the features of the installation site of heating radiators
And finally, the last coefficient, which is also associated with the features of connecting heating radiators. It is probably clear that if the battery is installed openly, is not obstructed by anything from above and from the front, then it will give maximum heat transfer. However, such an installation is far from always possible - more often, radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating priors into the created interior ensemble, hide them completely or partially. decorative screens- this also significantly affects the heat output.
If there are certain “baskets” on how and where the radiators will be mounted, this can also be taken into account when making calculations by entering a special coefficient “m”:
| Illustration | Features of installing radiators | The value of the coefficient "m" |
|---|---|---|
| The radiator is located on the wall openly or is not covered from above by a window sill | m = 0.9 | |
| The radiator is covered from above by a window sill or a shelf | m = 1.0 | |
| The radiator is blocked from above by a protruding wall niche | m = 1.07 | |
| The radiator is covered from above with a window sill (niche), and from the front - with a decorative screen | m = 1.12 | |
| The radiator is completely enclosed in a decorative casing | m = 1.2 |
So, there is clarity with the calculation formula. Surely, some of the readers will immediately take up their heads - they say, it's too complicated and cumbersome. However, if the matter is approached systematically, in an orderly manner, then there is no difficulty at all.
Any good homeowner must have a detailed graphical plan of their "possessions" with affixed dimensions, and usually oriented to the cardinal points. It is not difficult to specify the climatic features of the region. It remains only to walk through all the rooms with a tape measure, to clarify some of the nuances for each room. Features of housing - "neighborhood vertically" from above and below, location entrance doors, the proposed or already existing scheme for installing heating radiators - no one except the owners knows better.
It is recommended to immediately draw up a worksheet, where you enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will help to carry out the built-in calculator, in which all the coefficients and ratios mentioned above are already “laid”.
If some data could not be obtained, then, of course, they can not be taken into account, but in this case, the “default” calculator will calculate the result, taking into account the least favorable conditions.
It can be seen with an example. We have a house plan (taken completely arbitrary).
Region with level minimum temperatures within -20 ÷ 25 °С. Predominance of winter winds = northeasterly. The house is one-story, with an insulated attic. Insulated floors on the ground. The optimal diagonal connection of radiators, which will be installed under the window sills, has been selected.
Let's create a table like this:
| The room, its area, ceiling height. Floor insulation and "neighborhood" from above and below | The number of external walls and their main location relative to the cardinal points and the "wind rose". Degree of wall insulation | Number, type and size of windows | Existence of entrance doors (to the street or to the balcony) | Required heat output (including 10% reserve) |
|---|---|---|---|---|
| Area 78.5 m² | 10.87 kW ≈ 11 kW | |||
| 1. Hallway. 3.18 m². Ceiling 2.8 m. Warmed floor on the ground. Above is an insulated attic. | One, South, the average degree of insulation. Leeward side | No | One | 0.52 kW |
| 2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. Above - insulated attic | No | No | No | 0.62 kW |
| 3. Kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well insulated floor on the ground. Svehu - insulated attic | Two. South, west. Average degree of insulation. Leeward side | Two, single-chamber double-glazed window, 1200 × 900 mm | No | 2.22 kW |
| 4. Children's room. 18.3 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic | Two, North - West. High degree of insulation. windward | Two, double glazing, 1400 × 1000 mm | No | 2.6 kW |
| 5. Bedroom. 13.8 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic | Two, North, East. High degree of insulation. windward side | One, double-glazed window, 1400 × 1000 mm | No | 1.73 kW |
| 6. Living room. 18.0 m². Ceiling 2.8 m. Well insulated floor. Top - insulated attic | Two, East, South. High degree of insulation. Parallel to wind direction | Four, double glazing, 1500 × 1200 mm | No | 2.59 kW |
| 7. Bathroom combined. 4.12 m². Ceiling 2.8 m. Well insulated floor. Above is an insulated attic. | One, North. High degree of insulation. windward side | One. wooden frame with double glazing. 400 × 500 mm | No | 0.59 kW |
| TOTAL: |
Then, using the calculator below, we make a calculation for each room (already taking into account a 10% reserve). With the recommended app, it won't take long. After that, it remains to sum the obtained values \u200b\u200bfor each room - this will be the required total power of the heating system.
The result for each room, by the way, will help you choose the right number of heating radiators - it remains only to divide by the specific heat output of one section and round up.
Cosiness and comfort of housing do not begin with the choice of furniture, decoration and appearance generally. They start with the heat that heating provides. And just buy an expensive heating boiler for this () and quality radiators not enough - you first need to design a system that will maintain the optimum temperature in the house. But to get good result, you need to understand what and how to do, what are the nuances and how they affect the process. In this article, you will get acquainted with the basic knowledge about this case - what are heating systems, how it is carried out and what factors affect it.
Why is thermal calculation necessary?
Some owners of private houses or those who are just going to build them are interested in whether there is any point in the thermal calculation of the heating system? After all, we are talking about a simple country cottage, and not about apartment building or industrial enterprise. It would seem that it would be enough just to buy a boiler, install radiators and run pipes to them. On the one hand, they are partially right - for private households, the calculation of the heating system is not such a critical issue as for industrial premises or multi-unit residential complexes. On the other hand, there are three reasons why such an event is worth holding. , you can read in our article.
- Thermal calculation greatly simplifies the bureaucratic processes associated with the gasification of a private house.
- Determining the power required for home heating allows you to select a heating boiler with optimal performance. You will not overpay for excessive product features and will not experience inconvenience due to the fact that the boiler is not powerful enough for your home.
- Thermal calculation allows you to more accurately select pipes, valves and other equipment for the heating system of a private house. And in the end, all these rather expensive products will work for as long as is laid down in their design and characteristics.
Initial data for the thermal calculation of the heating system
Before you start calculating and working with data, you need to get them. Here for those owners of country houses who have not previously engaged in project activities, the first problem arises - what characteristics should you pay attention to. For your convenience, they are summarized in a small list below.
- Building area, height to ceilings and internal volume.
- The type of building, the presence of adjacent buildings.
- The materials used in the construction of the building - what and how the floor, walls and roof are made of.
- The number of windows and doors, how they are equipped, how well they are insulated.
- For what purposes will certain parts of the building be used - where the kitchen, bathroom, living room, bedrooms will be located, and where - non-residential and technical premises.
- The duration of the heating season, the average minimum temperature during this period.
- "Wind rose", the presence of other buildings nearby.
- The area where a house has already been built or is just about to be built.
- Preferred room temperature for residents.
- Location of points for connection to water, gas and electricity.
Calculation of the heating system power by housing area
One of the fastest and easiest to understand ways to determine the power of a heating system is to calculate by the area of \u200b\u200bthe room. A similar method is widely used by sellers of heating boilers and radiators. The calculation of the power of the heating system by area occurs in several simple steps.
Step 1. According to the plan or already erected building, the internal area of \u200b\u200bthe building in square meters is determined.
Step 2 The resulting figure is multiplied by 100-150 - that is how many watts of the total power of the heating system are needed for each m 2 of housing.
Step 3 Then the result is multiplied by 1.2 or 1.25 - this is necessary to create a power reserve so that the heating system is able to maintain a comfortable temperature in the house even in the most severe frosts.
Step 4 The final figure is calculated and recorded - the power of the heating system in watts, necessary to heat a particular housing. As an example, to maintain a comfortable temperature in a private house with an area of 120 m 2, approximately 15,000 W will be required.
Advice! In some cases, cottage owners divide the internal area of \u200b\u200bhousing into that part that requires serious heating, and that for which this is unnecessary. Accordingly, different coefficients are used for them - for example, for living rooms it is 100, and for technical rooms - 50-75.
Step 5 According to the already determined calculated data, a specific model of the heating boiler and radiators is selected.
It should be understood that the only advantage similar way thermal calculation heating system is speed and simplicity. However, the method has many disadvantages.
- The lack of consideration of the climate in the area where housing is being built - for Krasnodar, a heating system with a power of 100 W per square meter will be clearly redundant. And for the Far North, it may not be enough.
- The lack of consideration of the height of the premises, the type of walls and floors from which they are built - all these characteristics seriously affect the level of possible heat losses and, consequently, the required power of the heating system for the house.
- The very method of calculating the heating system in terms of power was originally developed for large industrial premises and apartment buildings. Therefore, for a separate cottage it is not correct.
- Lack of accounting for the number of windows and doors facing the street, and yet each of these objects is a kind of "cold bridge".
So does it make sense to apply the calculation of the heating system by area? Yes, but only as a preliminary estimate, allowing you to get at least some idea of the issue. To achieve better and more accurate results, you should turn to more complex techniques.
Imagine the following method for calculating the power of a heating system - it is also quite simple and understandable, but at the same time it has a higher accuracy of the final result. In this case, the basis for the calculations is not the area of \u200b\u200bthe room, but its volume. In addition, the calculation takes into account the number of windows and doors in the building, the average level of frost outside. Let's imagine a small example of the application of this method - there is a house with a total area of 80 m 2, the rooms in which have a height of 3 m. The building is located in the Moscow region. In total there are 6 windows and 2 doors facing the outside. The calculation of the power of the thermal system will look like this. "How to do , you can read in our article".
Step 1. The volume of the building is determined. This can be the sum of each individual room or the total figure. In this case, the volume is calculated as follows - 80 * 3 \u003d 240 m 3.
Step 2 The number of windows and the number of doors facing the street are counted. Let's take the data from the example - 6 and 2, respectively.
Step 3 A coefficient is determined depending on the area in which the house stands and how severe frosts are there.
Table. Values of regional coefficients for calculating the heating power by volume.
Since in the example we are talking about a house built in the Moscow region, the regional coefficient will have a value of 1.2.
Step 4 For detached private cottages, the value of the volume of the building determined in the first operation is multiplied by 60. We make the calculation - 240 * 60 = 14,400.
Step 5 Then the result of the calculation of the previous step is multiplied by the regional coefficient: 14,400 * 1.2 = 17,280.
Step 6 The number of windows in the house is multiplied by 100, the number of doors facing the outside by 200. The results are summed up. The calculations in the example look like this - 6*100 + 2*200 = 1000.
Step 7 The numbers obtained as a result of the fifth and sixth steps are summed up: 17,280 + 1000 = 18,280 W. This is the power of the heating system required to maintain optimum temperature in the building under the conditions specified above.
It should be understood that the calculation of the heating system by volume is also not absolutely accurate - the calculations do not pay attention to the material of the walls and floor of the building and their thermal insulation properties. Also, no correction is made for natural ventilation characteristic of any home.
