How to independently calculate the power of a heating boiler. Calculation of boiler power for heating Calculation of boiler power for home heating
One of the main components of comfortable housing is the presence of a well-thought-out heating system. At the same time, the choice of the type of heating and the required equipment is one of the main questions that must be answered at the stage of designing a house. An objective calculation of the heating boiler power by area will ultimately result in a completely efficient heating system.
We will now tell you about how to carry out this work correctly. At the same time, we will consider the features inherent in different types of heating. After all, they must be taken into account when carrying out calculations and subsequent decision-making on the installation of this or that type of heating.
Basic calculation rules
At the beginning of our story about how to calculate the power of a heating boiler, we will consider the quantities used in the calculations:
- room area (S);
- specific heater power per 10 m² of heated area – (W spec.). This value is determined adjusted for the climatic conditions of a particular region.
This value (W beat) is:
- for the Moscow region - from 1.2 kW to 1.5 kW;
- for the southern regions of the country - from 0.7 kW to 0.9 kW;
- for the northern regions of the country - from 1.5 kW to 2.0 kW.
Power calculation is carried out as follows:
W cat.=(S*Wsp.):10
Advice! For simplicity, you can use a simplified version of this calculation. In it Wsp.=1. Therefore, the heat output of the boiler is determined as 10 kW per 100 m² of heated area. But with such calculations, you must add at least 15% to the resulting value in order to get a more objective figure.
Calculation example
As you can see, the instructions for calculating the heat transfer intensity are simple. But, nevertheless, we will accompany it with a specific example.
The conditions will be as follows. The area of heated premises in the house is 100 m². The specific power for the Moscow region is 1.2 kW. Substituting the available values into the formula, we get the following:
W boiler = (100x1.2)/10 = 12 kilowatts.
Calculation for different types of heating boilers
The degree of efficiency of a heating system depends primarily on the correct choice of its type. And of course, it depends on the accuracy of the calculation of the required performance of the heating boiler. If the calculation of the thermal power of the heating system was not carried out accurately enough, then negative consequences will inevitably arise.
If the boiler heat transfer is less than required, the rooms will be cold in winter. In case of excess productivity, there will be an overconsumption of energy and, accordingly, money spent on heating the building.
To avoid these and other problems, just knowing how to calculate the power of a heating boiler is not enough.
It is also necessary to take into account the features inherent in systems using different types of heaters (you can see photos of each of them below in the text):
- solid fuel;
- electric;
- liquid fuel;
- gas.
The choice of one type or another largely depends on the region of residence and the level of infrastructure development. It is important to have the opportunity to purchase a certain type of fuel. And, of course, its cost.
Solid fuel boilers
Calculation of the power of a solid fuel boiler must be made taking into account the features characterized by the following features of such heaters:
- low popularity;
- relative accessibility;
- the possibility of autonomous operation - it is provided in a number of modern models of these devices;
- efficiency during operation;
- the need for additional space for fuel storage.
Another characteristic feature that should be taken into account when calculating the heating power of a solid fuel boiler is the cyclicity of the resulting temperature. That is, in rooms heated with its help, the daily temperature will fluctuate within 5ºC.
Therefore, such a system is far from the best. And if possible, you should refuse it. But, if this is not possible, there are two ways to smooth out the existing shortcomings:
- Using a thermal balloon, which is needed to regulate the air supply. This will increase the burning time and reduce the number of fireboxes;
- Application of water heat accumulators, having a capacity from 2 to 10 m². They are included in the heating system, allowing you to reduce energy costs and, thereby, save fuel.
All this will reduce the required productivity. Therefore, the effect of these measures must be taken into account when calculating the power of the heating system.
Electric boilers
Characterized by the following features:
- high cost of fuel - electricity;
- possible problems due to network outages;
- environmental friendliness;
- ease of control;
- compactness.
All these parameters should be taken into account when calculating the power of an electric heating boiler. After all, it is not purchased for one year.
Liquid fuel boilers
They have the following characteristic features:
- not environmentally friendly;
- easy to use;
- require additional space for fuel storage;
- have an increased fire hazard;
- They use fuel, the price of which is quite high.
Gas boilers
In most cases, they are the most optimal option for organizing a heating system. have the following characteristic features that must be taken into account when calculating the power of a heating boiler:
- ease of operation;
- do not require space for fuel storage;
- safe to use;
- low cost of fuel;
- efficiency.
Calculation for heating radiators
Let's say you decide to install a heating radiator yourself. But first you need to purchase it. And choose exactly the one that is suitable in terms of power.
- First we determine the volume of the room. To do this, multiply the area of the room by its height. As a result, we get 42m³.
- Next, you should know that heating 1 m³ of room area in central Russia requires spending 41 watts. Therefore, to find out the required radiator performance, we multiply this figure (41 W) by the volume of the room. As a result, we get 1722W.
- Now let's calculate how many sections our radiator should have. It's easy to do. Each element of a bimetallic or aluminum radiator has a heat output of 150 W.
- Therefore, we divide the performance we received (1722W) by 150. We get 11.48. Round up to 11.
- Now you need to add another 15% to the resulting figure. This will help smooth out the increase in required heat transfer during the most severe winters. 15% of 11 is 1.68. Round up to 2.
- As a result, we add 2 more to the existing number (11). We get 13. So, to heat a room with an area of 14 m², we need a radiator with a power of 1722 W, having 13 sections.
Now you know how to calculate the required performance of the boiler, as well as the heating radiator. Use our tips and ensure yourself an efficient and at the same time not wasteful heating system. If you need more detailed information, you can easily find it in the corresponding video on our website.
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 consider 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 the main parameter of gas equipment is calculated, which determines its performance. The thermal power of devices is measured in kilowatts (kW). This value is always indicated in the technical data sheet of 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 calculations are highly complex, as they take into account a huge number of 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 a special program: the number of windows, doors, wall thickness and other information. Based on them, the program produces the result.
- Manual calculations. The most accessible way to find out the optimal heating performance of a heater is to use the elementary ratio of 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.
When calculating the power characteristics of heating devices, developers often take into account the volume of the premises. In the technical documentation of 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 the 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 there is no exactly suitable option on sale, you should take a modification with greater performance. This method of calculation is only suitable for devices with one circuit, without an indirect heating boiler.
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, the required volume of the boiler is determined so that it can satisfy the needs of the household for 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 designs, the formula is used: 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 of 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 are 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 heat output 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. To heat large houses and cottages, more powerful floor heaters are used. 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 you to use the 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.
When choosing a solid fuel boiler, you need to consider power. It determines whether the device can create the required amount of heat for the entire house or not. It is undesirable to choose a boiler that is too powerful, because it will operate in economy mode, and this will affect the reduction in efficiency.
To make the right one, you need to know two indicators:
- The amount of heat required to heat the room and heat the water.
- The real power of the device.
Calculation of power depending on the volume of the room
The calculation formula is:
Q = VxΔTxK/850,
- where Q – The amount of heat, defined in kW/h4;
- V – room volume(unit of measurement cubic m);
- ΔT is the difference between the outside temperature and the indoor temperature;
- TO - correction factor, taking into account heat loss;
- the number 850 is used to convert the product of the above three indicators to kW/hour.
K can have the following meanings:
- 3-4 – for premises that are a simplified wooden structure or a building made of corrugated sheets.
- 2-2,9 – for buildings with little thermal insulation. The design of such houses is simplified, the wall thickness is equal to the length of 1 brick, the windows and roof have a simple structure.
- 1-1,9 – for houses whose design is standard. The brickwork is double, the number of simple windows is small. The roof has a conventional roof.
- 0,6-0,9 – for houses with improved construction, double thermal insulation of brick walls, double glazed windows, thick floor base, roof made of good thermal insulation material.
As an example, let's take a modern house with an area of 200 square meters. m, wall height 3 m and first-class thermal insulation. The house is located in an area where in winter the temperature does not drop below -25 °C. In this case, ΔT = 20 – (-25) = 45 °C. Therefore, to heat a house you need to create Q = 200*3*45*0.9/850 = 28.58 kW/h. The figure should not be rounded, because it is not final and you need to increase it with your own hands by the amount of heat for the hot water supply. If the water is planned to be heated in a different way, then the result obtained is not adjusted, and part of the calculation is completed.
Calculation of heat for hot water supply
- where c is specific heat capacity of water(the indicator is always 4200 J/kg*K);
- m – mass of water in kg;
- Δt temperature difference between heated water from the water supply.
Read also: Cleaning a solid fuel boiler from tar and soot
Example. The average family's need for warm water can reach 150 liters. If the boiler heats the coolant to a temperature of 80 °C, and the water from the pipeline has a temperature of 10 °C, then Δt = 80 – 10 = 70 °C.
Qв = 4200*150*70 = 44,100,000 J or 12.25 kW/h.
- If 150 liters need to be heated at a time, the capacity of the indirect boiler is 150 liters, then 12.25 kW/h is added to 28.58 kW/h. This must be done because if Qzag is less than 40.83, the room will be colder than the calculated 20 °C.
- If the water must be heated in portions, the volume of the indirect boiler is 50 liters, then 12.25 is divided by 3 and added with your own hands to 28.58. Qzag will be equal to 32.67 kW/h. This is the power of the device for the heating system.
Calculation by area
It is more accurate because it takes into account more factors. The calculation is made using the formula:
Q = 0.1*S*k1*k2*k3*k4*k5*k6*k7, Where:
0.1 kW is the heat norm per 1 sq. m;
S – area of the heated house;
k1 demonstrates heat loss caused by window design. Has the meaning:
- 1.27 – if the windows have one glass;
- 1.0 – if there are double-glazed windows;
- 0.85 – if there are windows with triple glass.
k2 demonstrates heat loss caused by window area (Sw). Is the ratio of Sw to the floor area Sf. Its meanings are:
- 0.8 at Sw/Sf = 0.1;
- 0.9 at Sw/Sf = 0.2;
- 1 at Sw/Sf = 0.3;
- 1.1 at Sw/Sf = 0.4;
- 1.2 at Sw/Sf = 0.5.
k3 is coefficient of heat loss through walls. It happens like this:
- 1.27 with very poor thermal insulation;
- 1 in houses with a wall of 2 bricks or insulation, the thickness of which is 15 cm;
- 0.854 with good thermal insulation.
k4 shows heat loss depending on the air temperature outside the house (tz). Has the following meanings:
- 0.7, if tз = -10 °С;
- 0.9 for tз = -15 °С;
- 1.1 for tз = -20 °С;
- 1.3 for tз = -25 °С;
- 1.5 for tз = -30 °С.
Read also: Advantages of the Popov boiler
k5 demonstrates heat loss through external walls. Is like this:
- 1.1 for rooms with one external wall;
- 1.2 for 2 external walls;
- 1.3 for 3 external walls;
- 1.4 for a building with 4 external walls.
K6 shows how much Additional heat required depending on ceiling height (H). Its meanings are:
- 1 for H = 2.5 m;
- 1.05 for H = 3.0 m;
- 1.1 for H = 3.5 m;
- 1.15 for H = 4.0 m;
- 1.2 for H = 4.5 m.
k7 determines heat loss depending on the type of room located above the heated room. It happens like this:
- 0.8 for heated rooms;
- 0.9 for a warm attic;
- 1 for a cold attic.
Example. The conditions of the problem are the same. The windows are triple glazed and make up 30% of the floor area. The number of external walls is 4. There is a cold attic upstairs.
Q = 0.1*200*0.85*1*0.854*1.3*1.4*1.05*1 = 27.74 kW/h. This figure must be increased by adding with your own hands the amount of heat required for hot water supply.
Real power of a long-burning boiler
Many devices are designed for a specific type of fuel. If other types of fuel are burned in them, their efficiency will be lower.
The power calculation will be carried out on the basis of the Viessmann Vitoligno 100-S 60 pyrolysis boiler. Its features are as follows:
- Powered by wood.
- In 1 hour, from 6 to 15 kg of firewood burns in the loading chamber.
- Its rated power is 60 kW.
- The loading chamber volume is 294 liters.
- Efficiency is 87%
Let the owner plan to burn aspen wood in it. 1 kg of such firewood produces 2.82 kW/h. If a boiler burns 15 kg in 1 hour, then it emits 2.82*15*0.87 = 36.801 kW/h of heat (0.87 is efficiency). Such a device is not enough for heating a house with a 150 liter boiler, but it is quite enough for hot water supply with a 50 liter boiler. To get the figure 32.67 kW/h, you need to burn 13.31 kg of aspen firewood in 1 hour (32.67/(2.82*0.87) = 13.31). This is the case if you calculate the heat demand by volume.
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 the 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. Selecting a suitable model “by eye” or based on 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 insufficient calorific value everything is more or less clear. When winter cold sets in, the boiler will begin to operate at full capacity, and it is not a fact that there will be a comfortable microclimate in the rooms. 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 for household needs - an 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 design 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% |
| Areas where utilities pass through building structures (sewage pipes, water supply pipes, gas supply pipes, 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 for 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 need for thermal energy for premises connected to the central system of the 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 involves assessing not the entire house or apartment by total area or volume, but each heated room separately. Agree that rooms of equal area, but differing, say, in the number of external walls, will require different amounts of 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 of the rooms.
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.
The larger the contact area with the 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.
The coefficients 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, one should not discount the possible effects of solar energy. 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, the local residents themselves are well aware of which winds most often bother 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 that the more effective the wall insulation system is, the lower the level of heat loss will be. Ideally, what we should strive for, thermal insulation should generally be complete, carried out on the basis of thermal calculations performed, taking into account the climatic conditions of the region and the design features of the house.
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 - the better the quality of the 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. The glass area is very important. Agree, it’s difficult to compare a small window and a huge panoramic window that almost fills 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 a cold balcony, to 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 regularly used in winter.
You might be interested in information about what it is
* * * * * * *
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.
Autonomous heating for a private home is convenient, affordable and very diverse. Any owner of a private home willingly buys a gas boiler and installs everything necessary so as to no longer depend on the vagaries of the weather or surprises associated with the operation of centralized heating systems.
However, it is important to choose the right equipment. If its power exceeds the actual heat needs of the building, part of the heating costs will simply be thrown away. A device with low performance will not be able to provide the house with enough heat. Therefore, even at the design stage, you need to find an answer to the question: how to calculate the power of a gas boiler?
What quantities are used in the calculations?
The simplest calculation of boiler power by area looks like this: you need to take 1 kW of power for every 10 sq. m. However, it is worth considering that these standards were drawn up during the Soviet Union. They do not take into account modern construction technologies; in addition, they may turn out to be untenable in areas whose climate is noticeably different from the conditions of Moscow and the Moscow region. Such calculations may be suitable for a small building with an insulated attic, low ceilings, excellent thermal insulation, double-glazed windows, etc. Unfortunately, only a few buildings meet these requirements. To make a more detailed calculation of the boiler power, you need to take into account a number of factors, such as:
- climatic conditions in the region;
- dimensions of the living space;
- degree of insulation of the house;
- possible heat loss of the building;
- the amount of heat required to heat water.
In addition, in houses with forced ventilation, the calculation of the heating boiler must take into account the amount of energy required to heat the air. As a rule, you need to use special software for calculations:
When calculating the power of a gas boiler, you should add about 20% more in case of unforeseen situations, such as severe cold or a decrease in gas pressure in the system.
Is it worth buying a boiler that is too powerful?
Modern heating equipment is equipped with automatic systems that allow you to regulate gas flow. This is very convenient because it eliminates unnecessary expenses. It may seem that an accurate calculation of the power of a heating boiler is not so important, because you can simply buy a boiler with high power ratings. But it's not that simple.
Correct selection of heating equipment will extend its service life
Unreasonably exceeding the thermal power of equipment can lead to:
- increasing costs for the acquisition of system elements;
- reducing the efficiency of the boiler;
- failures in the operation of automatic equipment;
- rapid wear of components;
- formation of condensation in the chimney, etc.
Thus, you need to try to “get into” exactly the power that suits your home.
Gas boiler for standard houses
- MK is the design power of the boiler in kW;
- S – total area of the room in sq.m;
- UMC is the specific power of the boiler, which should be for every 10 sq. m.
The last indicator is set depending on the climate zone and is:
- 0.7-0.9 kW for southern regions;
- 1.0-1.2 kW for the middle band;
- 1.2-1.5 kW for regions near Moscow;
- 1.5-2.0 for northern regions.
According to this formula, the estimated boiler power for a house with an area of 200 square meters. m., which is located in the middle zone, will be: 200X1.1/10=22 kW. Please note that this formula shows how to calculate the power of a boiler that is used only for heating a house. If you intend to use a double-circuit system that provides heating of water for domestic needs, the power of the equipment should be increased by another 25%.
How to take ceiling heights into account when calculating?
Since many private houses are being built according to individual projects, the methods for calculating boiler power given above will not be suitable. To make a fairly accurate calculation of a gas heating boiler, you must use the formula: MK = Qt*Kzap, Where:
- MK – design power of the boiler, kW;
- Qt – predicted heat loss of the building, kW;
- Kzap is a safety factor that is 1.15 to 1.2, i.e. 15-20%, by which experts recommend increasing the design power of the boiler.
The main indicator in this formula is the predicted heat loss of the building. To find out their value, you need to use another formula: Qt = V*Pt*k/860, Where:
- V - volume of the room, cubic meters;
- Рt - difference between external and internal temperatures in degrees Celsius;
- k is the dissipation coefficient, which depends on the thermal insulation of the building.
The dissipation coefficient varies depending on the type of building:
- For buildings without thermal insulation, which are simple structures made of wood or corrugated iron, the dissipation coefficient is 3.0-4.0.
- For structures with low thermal insulation, typical for buildings with single brickwork with ordinary windows and a roof, the dissipation coefficient is taken to be 2.0-2.9.
- For houses with an average level of thermal insulation, for example buildings with double brickwork, a standard roof and a small number of windows, take a dispersion coefficient of 1.0-1.9.
- For buildings with increased thermal insulation, well-insulated floors, roofs, walls and double-glazed windows, a dispersion coefficient of 0.6-0.9 is used.
For small buildings with good thermal insulation, the design power of heating equipment can be quite small. It may happen that there is simply no suitable gas boiler on the market with the required characteristics. In this case, you should purchase equipment whose power will be slightly higher than the calculated one. Automatic heating control systems will help smooth out the difference.
Some manufacturers have taken care of the convenience of customers and have placed special services on their Internet resources that make it possible to calculate the required boiler power without any problems. To do this, you need to enter the following data into the calculator program:
- temperature that must be maintained in the room;
- average temperature for the coldest week of the year;
- need for hot water supply;
- presence or absence of forced ventilation;
- number of floors in the house;
- ceiling height;
- information about floors;
- information about the thickness of external walls and the materials from which they are made;
- information about the length of each wall;
- information about the number of windows;
- description of the type of windows: number of cameras, glass thickness, etc.;
- dimensions of each window.
Once all the fields are filled in, you can find out the estimated power of the boiler. Options for detailed calculations of the power of boilers of various types are clearly presented in the table:
This table has already calculated some options, you can use them as the correct ones in advance (click on the picture to enlarge)
Our calculator for quick calculations
To calculate the thermal power of a heating boiler in this calculator, just enter the area of the heated room, select the necessary parameters and click the “Run calculation” button.
