How to calculate heating for your home. Calculation of the heating system of a private house: formulas, reference data, examples

What parameters need to be calculated when designing an autonomous heating system? How is the heating system of a private house calculated in each specific case? In the article, we will provide the reader with all the necessary formulas, reference data and accompany the calculations with examples.

We have to find out how difficult it is to calculate the parameters of autonomous heating.

What do we think

What are the steps involved in calculating a heating system for a private house?

The total demand for heat and the corresponding output of the heating boiler.
The need for heat energy in a separate room and, accordingly, the power of the heater in it.

Note: we have to touch upon the methods for determining the heat output for different heating devices.

Expansion tank volume.
Circulation pump parameters.

Thermal power

There are two ways to roughly estimate the need for heat at home:

By area.
By volume.

Area calculation

This technique is extremely simple and is based on SNiP half a century ago: one kilowatt of thermal power is taken per 10 square meters of area. Thus, a house with a total area of 100 m2 can be heated with a 10 kW boiler.

The scheme is good in that it does not require climbing into the jungle and calculating the thermal resistance of the enclosing structures. But, like any simplified calculation scheme, it gives a very approximate result.

Fast, simple and ... imprecise.

There are several reasons:

The boiler heats up the entire volume of air in the room, which depends not only on the area of the house, but also on the height of the ceilings. And this parameter in private housing construction can vary over the widest range.
Windows and doors lose much more heat per unit area than walls. If only because they are much more transparent to infrared radiation.
The climatic zone also greatly affects the heat loss through the building envelope. A doubling of the delta of temperatures between the room and the street will double the cost of heating.

Calculation by volume with regional coefficients

It is for these reasons that it is better to use a slightly more complex, but giving a much more accurate result, the calculation scheme.

The base value is 60 watts of heat per cubic meter of the volume of the heated room.
For each window in the outer wall, 100 watts are added to the calculated thermal power, for each door - 200.
The result is multiplied by the regional coefficient:

Let's take that same 100 square meter house as an example.

However, this time we will stipulate a number of additional conditions:

The height of its ceilings is 3.5 meters.
The house has 10 windows and 2 doors in the outer walls.
It is located in the city of Verkhoyansk (the average January temperature is 45.4 C, the absolute minimum is 67.6 C).

So, let's calculate the heating of a private house for these conditions.

The internal volume of the heated room is 100 * 3.5 = 350 m3.
The base value of the thermal power will be 350 * 60 = 21000 W.
Windows and doors make things worse: 21,000+ (100 * 10) + (200 * 2) = 22,400 watts.
Finally, the refreshing climate of Verkhoyansk will force us to double the already large thermal power of heating: 22400 * 2 = 44800 watts.

As it is easy to see, the difference with the result obtained by the first method is more than fourfold.

Heating devices

The very method of calculating the heat demand for a separate room is completely identical to the one given above.

For example, for a room of 12 m2 with two windows in the house we have described, the calculation will look like this:

The volume of the room is 12 * 3.5 = 42 m3.
The base thermal power will be 42 * 60 = 2520 watts.
Two windows will add another 200 to it.2520 + 200 = 2720.
The regional coefficient will double the heat demand. 2720 * 2 = 5440 watts.

Manufacturers always indicate the heat output for convectors, plate radiators, etc. in the accompanying documentation.

For sectional radiators, the necessary information can usually be found on the websites of dealers and manufacturers. There you can often find a calculator for converting kilowatts in the section.
Finally, if you use sectional radiators of unknown origin, with their standard size of 500 millimeters along the axes of the nipples, you can focus on the following averaged values:

In an autonomous heating system with its moderate and predictable parameters of the coolant, aluminum radiators are most often used. Their reasonable price is very pleasantly combined with a decent appearance and high heat dissipation.

In our case, aluminum sections with a capacity of 200 watts will require 5440/200 = 27 (rounded off).

Placing so many sections in one room is not a trivial task.

As always, there are a couple of subtleties.

With a lateral connection of a multi-section radiator, the temperature of the last sections is much lower than the first; accordingly, the heat flux from the heater decreases. A simple instruction will help to solve the problem: connect the radiators according to the “bottom-down” scheme.
Manufacturers indicate the heat output for a delta of temperatures between the coolant and the room at 70 degrees (for example, 90 / 20C). When it decreases, the heat flux will fall.

A special case

Often, homemade steel registers are used as heating devices in private houses.

Please note: they attract not only by their low cost, but also by their exceptional tensile strength, which is very useful when connecting a house to a heating main.
In an autonomous heating system, their attractiveness is nullified by their unpretentious appearance and low heat transfer per unit volume of the heater.

Let's face it - not the height of aesthetics.

Nevertheless: how to estimate the thermal power of a register of a known size?

For a single horizontal round pipe, it is calculated by the formula of the form Q = Pi * Dн * L * k * Dt, in which:

Q is the heat flow;
Pi - number "pi", taken equal to 3.1415;
Dн - outer diameter of the pipe in meters;
L is its length (also in meters);
k - coefficient of thermal conductivity, which is taken equal to 11.63 W / m2 * C;
Dt is the delta temperature, the difference between the coolant and the air in the room.

In a multisection horizontal register, the heat transfer of all sections, except for the first one, is multiplied by 0.9, since they give off heat to the upward flow of air heated by the first section.

Let's calculate the heat transfer of a four-section register with a section diameter of 159 mm and a length of 2.5 meters at a coolant temperature of 80 C and an air temperature in the room of 18 C.

The heat transfer of the first section is 3.1415 * 0.159 * 2.5 * 11.63 * (80-18) = 900 watts.
The heat transfer of each of the other three sections is 900 * 0.9 = 810 watts.
The total thermal power of the heater is 900+ (810 * 3) = 3330 watts.

Expansion tank

And in this case, there are two calculation methods - simple and accurate.

Simple circuit

A simple calculation is utterly simple: the volume of the expansion tank is taken equal to 1/10 of the volume of the coolant in the circuit.

Where to get the value of the volume of the coolant?

Here are a couple of the simplest solutions:

Fill the circuit with water, bleed the air, and then drain all the water through the vent into any volumetric container.
In addition, the rough volume of a balanced system can be calculated at the rate of 15 liters of coolant per kilowatt of boiler power. So, in the case of a 45 kW boiler, the system will have approximately 45 * 15 = 675 liters of coolant.

Therefore, in this case, a reasonable minimum would be 80 liters (rounded up to the standard value).

Exact layout

More precisely, you can calculate the volume of the expansion tank with your own hands using the formula V = (Vt x E) / D, in which:

V is the desired value in liters.
Vt is the total volume of the coolant.
E is the coefficient of expansion of the coolant.
D is the efficiency factor of the expansion tank.

Obviously, the last two parameters need some comment.

The expansion coefficient of water and poor water-glycol mixtures can be taken from the following table (when heated from an initial temperature of +10 C):

Heating, С	Extension, %
30	0,75
40	1,18
50	1,68
60	2,25
70	2,89
80	3,58
90	4,34
100	5,16

The efficiency factor of the tank can be calculated using the formula D = (Pv - Ps) / (Pv + 1), in which:

Pv is the maximum pressure in the circuit (pressure relief valve).

Hint: usually it is taken equal to 2.5 kgf / cm2.

Ps - static pressure of the circuit (it is also the pressure of the tank charging). It is calculated as 1/10 of the difference in meters between the level of the tank location and the top point of the circuit (an excess pressure of 1 kgf / cm2 raises the water column by 10 meters). A pressure equal to Ps is generated in the tank air chamber before filling the system.

Let's calculate the tank requirements for the following conditions as an example:

The difference in height between the tank and the top point of the contour is 5 meters.
The power of the heating boiler in the house is 36 kW.
The maximum water heating is 80 degrees (from 10 to 90C).

So:

The efficiency factor of the tank will be (2.5-0.5) / (2.5 + 1) = 0.57.

The volume of the coolant at the rate of 15 liters per kilowatt is 15 * 36 = 540 liters.
The expansion coefficient of water when heated to 80 degrees is 3.58%, or 0.0358.
Thus, the minimum tank volume is (540 * 0.0358) / 0.57 = 34 liters.

Circulation pump

How to choose the optimal parameters?

For us, two parameters are important: the head created by the pump and its performance.

In the photo - a pump in the heating circuit.

With pressure, everything is not simple, but very simple: the contour of any length reasonable for a private house will require a pressure of no more than the minimum 2 meters for budget devices.

Reference: a drop of 2 meters makes the heating system of a 40-apartment building circulate.

The simplest way to select the performance is to multiply the volume of the coolant in the system by 3: the circuit must be turned around three times per hour. So, in a system with a volume of 540 liters, a pump with a capacity of 1.5 m3 / h (with rounding) is sufficient.

A more accurate calculation is performed using the formula G = Q / (1.163 * Dt), in which:

G - productivity in cubic meters per hour.
Q is the power of the boiler or the section of the circuit where circulation is to be ensured, in kilowatts.
1.163 is a coefficient tied to the average heat capacity of water.
Dt is the delta temperature between the flow and return of the circuit.

Sometimes the capacity is indicated in liters per minute. It's easy to recount.

Conclusion

We hope that we have provided all the necessary materials at the disposal of the reader. For more information on how to calculate heating in a private house, see the attached video. Good luck!

In the process of building any house, sooner or later the question arises - how to correctly calculate the heating system? This urgent problem will never exhaust its resource, because if you buy a boiler of less power than necessary, you will have to spend a lot of effort to create secondary heating with oil and infrared radiators, heat guns, electric fireplaces.

In addition, the monthly maintenance, due to expensive electricity, will cost you a pretty penny. The same will happen if you buy a boiler of increased power, which will work at half strength, and consume no less fuel.

Our calculator for calculating the heating of a private house will help you avoid the typical mistakes of novice builders. You will receive the value of heat loss and the required heat output of the boiler as close to reality as possible according to the current data from SNiPs and SP (codes of rules).

The main advantage of the calculator on the website is the reliability of the calculated data and the absence of manual calculations, the whole process is automated, the initial parameters are summarized as much as possible, you can easily see their values in the plan of your house or fill out based on your own experience.

Calculation of a boiler for heating a private house

With the help of our calculator for calculating heating for a private house, you can easily find out the required boiler power to heat your cozy "nest".

As you remember, in order to calculate the heat loss indicator, you need to know several values of the main components of the house, which together account for more than 90% of the total losses. For your convenience, we have added only those fields to the calculator that you can fill in without special knowledge:

glazing;
thermal insulation;
the ratio of the area of the windows and the floor;
outside temperature;
the number of walls facing outward;
which room is above the calculated one;
room height;
area of the room.

After you get the value of the heat loss at home, a safety correction factor equal to 1.2 is taken to calculate the required boiler power.

How to use the calculator

Remember that the thicker the glazing and the better the thermal insulation, the less heating power will be required.

To get results, you need to answer yourself the following questions:

Choose one of the offered types of glazing (triple or double glazing, conventional double glazing).
How are your walls insulated? Solid thick insulation from a couple of layers of mineral wool, foam, EPS for the north and Siberia. Maybe you live in Central Russia and one layer of insulation is enough for you. Or are you one of those who are building a house in the southern regions and a double hollow brick will suit him.
What is your window-to-floor ratio, in%. If you do not know this value, then it is calculated very simply: divide the area of the floors by the area of the windows and multiply by 100%.
Specify the minimum winter temperature for a couple of seasons and round up. It is not necessary to use the average temperature in winter, otherwise you risk getting a boiler of lower capacity, and the house will not be heated enough.
Do we count for the whole house or just for one wall?
What is above our premises. If you have a one-story house, choose the type of attic (cold or warm), if the second floor, then a heated room.
The height of the ceilings and the area of the room are necessary to calculate the volume of the apartment, which in turn is the basis for all calculations.

Calculation example:

one-storey house in the Kaliningrad region;
wall lengths 15 and 10 m, insulated with one layer of mineral wool;
ceiling height 3 m;
6 double-glazed windows of 5 m2 each;
the minimum temperature over the past 10 years is 26 degrees;
we count for all 4 walls;
a warm heated attic on top;

The area of our house is 150 m2, and the area of the windows is 30 m2. 30/150 * 100 = 20% ratio between windows and floor.

We know the rest, we select the appropriate fields in the calculator and we get that our house will lose 26.79 kW of heat.

26.79 * 1.2 = 32.15 kW - required heating capacity of the boiler.

DIY heating system

It is impossible to calculate the heating circuit of a private house without assessing the heat loss of the surrounding structures.

In Russia, as a rule, long cold winters, buildings lose heat due to temperature changes inside and outside the premises. The larger the area of the house, enclosing and through structures (roof, windows, doors), the greater the value of heat loss comes out. The material and thickness of the walls, the presence or absence of thermal insulation, have a significant impact.

For example, walls made of wood and aerated concrete have a much lower thermal conductivity than brick. Materials with the highest thermal resistance values are used as insulation (mineral wool, expanded polystyrene).

Before creating a heating system at home, you need to carefully consider all the organizational and technical aspects, so that immediately after the construction of the "box", proceed to the final phase of construction, and not postpone the long-awaited settlement for many months.

Heating in a private house is based on the "three elephants":

heating element (boiler);
pipe system;
radiators.

Which boiler is better for home?

Heating boilers are the main component of the entire system. It is they who will provide the warmth of your home, so you need to be especially careful about their choice. By the type of food they are divided into:

electrical;
solid fuel;
liquid fuel;
gas.

Each of them has a number of significant advantages and disadvantages.

Electric boilersdid not gain much popularity, primarily due to the rather high cost and high cost of maintenance. Electricity tariffs leave much to be desired, there is a possibility of a break in power lines, as a result of which your house may be left without heating.
Solid fuelboilersoften used in remote villages and towns where there is no centralized communication network. They heat the water with wood, briquettes and coal. An important disadvantage is the need for constant monitoring of the fuel, if the fuel burns out and you do not have time to replenish the supplies, the house will no longer be heated. In modern models, this problem is solved, due to the automatic feeder, but the price of such devices is incredibly high.
Oil-fired boilers, in the overwhelming majority of cases, run on diesel fuel. They have excellent performance due to their high fuel efficiency, but the high price of raw materials and the need for diesel tanks limits many buyers.
The most optimal solution for a country house is gas boilers... Due to their small size, low gas prices and high heat dissipation, they have won the trust of the majority of the population.

How to choose pipes for heating?

Heating mains supply all heating devices in the house. Depending on the material of manufacture, they are divided into:

metal;
metal-plastic;
plastic.

Metal pipes the most difficult to install (due to the need to weld seams), are prone to corrosion, are heavy and expensive. The advantages are high strength, resistance to temperature extremes and the ability to withstand high pressures. They are used in apartment buildings; it is impractical to use them in private construction.

Polymer pipes made of metal-plastic and polypropylene are very similar in their parameters. Lightness of material, plasticity, absence of corrosion, suppression of noise and, of course, low price. The only difference between the former is the presence of an aluminum layer between the two layers of plastic, due to which the thermal conductivity index increases. Therefore, metal-plastic pipes are used for heating, and plastic pipes for water supply.

Choosing radiators for your home

The last element of a classic heating system is radiators. They are also divided by material into the following groups:

cast iron;
steel;
aluminum.

Cast iron batteries are familiar to everyone from childhood, because they were installed in almost all apartment buildings. They have high heat capacity (cool down for a long time), are resistant to temperature and pressure drops in the system. The downside is the high price, fragility and complexity of installation.

They were replaced by steel radiators. The wide variety of shapes and sizes, low cost and ease of installation have influenced its ubiquity. However, they also have their drawbacks. Due to the low heat capacity, the batteries cool down quickly, and the thin case does not allow them to be used in high pressure networks.

Recently, heaters from aluminum... Their main advantage is high heat transfer, which allows the room to warm up to an acceptable temperature in 10-15 minutes. However, they are demanding on the coolant, if alkali or acids are contained in large quantities inside the system, then the service life of the radiator is significantly reduced.

Use the proposed tools to calculate the heating of a private house and design a heating system that will heat your home efficiently, reliably and for a long time, even in the most severe winters.

It can be difficult for the owner of a heating network to find an intelligible answer on how to calculate home heating. This happens simultaneously due to the great complexity of the calculation itself, as such, and due to the extreme simplicity of obtaining the desired results, which usually experts do not like to talk about, believing that everything is already clear.

By and large, the calculation process itself should not interest us. It is important for us to somehow get the correct answer to the existing questions about capacities, diameters, quantities ... What equipment to use? There should be no mistake, otherwise there will be a double or triple overpayment. How to correctly calculate the heating system of a private house?

Why is it so difficult

Calculation of the heating system with permissible errors is only possible for a licensed organization. A number of parameters in a domestic environment are simply not definable.

How much energy is lost due to wind blowing? - and when the tree next to it grows up?
How much energy does the sun drive into the windows? - and how much will it be if the windows are not washed for six months?
How much heat goes away with ventilation? - and after the formation of a gap under the door due to the lack of replacement of the seal?
What is the real moisture content of the styrofoam in the attic? - and why is it needed after the mice eat it up….

All questions show the existing dynamics of changes in heat loss over time at any house. Why, then, is precision today? But even at the current moment, it is impossible to calculate exactly the parameters of the heating system based on heat losses in domestic conditions.
The hydraulic calculation is also complicated.

How to determine heat loss

A certain formula is known according to which heat loss directly depends on the heated area. With a ceiling height of up to 2.6 meters in the coldest month in a "normal" house, we lose 1 kW from 10 square meters. The heating power must cover this.

Real heat losses of private houses are more often in the range of 0.5 kW / 10 sq. M. up to 2.0 kW / 10 sq. m. This indicator characterizes the energy-saving qualities of the house in the first place. And it is less dependent on the climate, although its influence remains significant.

What specific heat losses will the house have, kW / 10 sq. M.?

0.5 - energy saving house
0.8 - insulated
1.0 - insulated "more or less"
1.3 - poor thermal insulation
1.5 - without insulation
2.0 - cold thin materials, there are drafts.

The total heat loss for the house can be found out by multiplying the given value by the heated area, m. But all this interests us to determine the power of the heat generator.

Calculation of boiler power

It is unacceptable to take the power of the boiler based on heat losses of more than 100 W / m2. This means heating (littering) nature. A heat-saving house (50 W / m2) is made, as a rule, according to the project, in which the calculation of the heating system is made. For other houses, 1 kW / 10 sq. M. Is accepted, and no more.

If the house does not correspond to the name "insulated", especially for a temperate and cold climate, then it must be brought into such a state, after which heating is already selected according to the same calculation - 100 W per square meter.

The calculation of the boiler power is performed according to the following formula - multiply the heat loss by 1.2,
where 1.2 is the power reserve usually used for heating domestic water.
For a house 100 sq. M. - 12 kW or a little more.

Calculations show that for a non-automated boiler, the reserve may be 2.0, then you need to heat it carefully (without boiling), but you can quickly warm up the house if you have a powerful circulation pump. And if the circuit has a heat accumulator, then 3.0 are permissible realities for heat generation. But won't they turn out to be prohibitive in price? We are no longer talking about the payback of the equipment, only the convenience of use ...

Let's listen to an expert, he will tell you how best to choose a solid fuel boiler for your home, and what power to take ...

When choosing a solid fuel boiler

It is worth considering only solid fuel boilers of classical design, as reliable, simple and cheap and devoid of the drawbacks of barrel-shaped devices called "long burning" ... In an ordinary solid fuel boiler, the upper loading chamber will always give a little smoke into the room. Boilers with a front loading chamber are more preferable, especially if they are installed in a residential building.
Cast iron boilers require protection from cold return flow, they are afraid of a volley injection of cold water, for example, when the electricity is turned on. A high-quality scheme must be foreseen in advance.
Protection against cold return is also desirable for any type of boiler, so that aggressive condensate does not form on the heat exchanger, at its temperature below 60 degrees.
It is advisable to take a solid fuel boiler with increased power, for example, two times the power required. Then it will not be necessary to constantly stand at a low-power boiler and throw up firewood so that it develops the required power. The process with not intensive combustion will be much more comfortable ...
It is advisable to purchase a boiler with a secondary air supply for afterburning CO with low-intensity combustion. We increase the efficiency and comfort of the furnace.

Power distribution around the house

The power generated by the boiler should be evenly distributed throughout the house, not leaving cold zones. Uniform heating of the building will be ensured if the power of the installed radiators in each room compensates for its heat loss.

The total power of all radiators should be slightly higher than that of the boiler. In the future, we will proceed from the following calculations.

Radiators are not installed in the interior rooms, only a warm floor is possible.

The longer the outer walls of the room and the larger the glazing area in them, the more it loses thermal energy. In a room with one window, a correction factor of (approximately) 1.2 is applied to the usual formula for calculating heat loss by area.
With two windows - 1.4, corner with two windows - 1.6, corner with two windows and long outer walls - 1.7, for example.

Calculation of power and selection of parameters of installed radiators

Manufacturers of radiators indicate the rated thermal power of their products. But at the same time, small-unknown ones overestimate the data as they want (the more powerful, the better they will buy), and the large ones indicate values for the coolant temperature of 90 degrees, etc., which are rarely found in a real heating network.

Then the usual 10 section radiator from the store is taken as 1.5 kW. Corner room with two windows, 20 sq. M. must lose 3 kW of energy (2 kW multiplied by a factor of 1.5). Therefore, under each window in a given room, you need to place
at least 10 radiator sections - 1.5 kW each.

For a full-fledged heating system, it is advisable not to take into account the power of the warm floor - the radiators must do it themselves. But more often they reduce the cost of the radiator network by 2 - 4 times, - only for additional. heating and creating thermal curtains.

What is the peculiarity of hydraulic calculation

If the boiler has already been selected based on the area, then why not choose a pump and pipes using a similar method, especially since the step of gradation of their parameters is much greater than the power of the boilers. Rough selection in the store of the nearest larger parameter does not require the most accurate calculations if the network is typical and compact and standardized equipment is used - circulation pumps, radiators and pipes for heating.

So for a house with an area of 100 square meters. to choose a pump 25/40, and pipes 16 mm (inner diameter) for a group of radiators up to 5 pcs. and 12 mm for connection 1 - 2 pcs. radiators. No matter how hard we try to improve our hydraulic calculation, we don't have to choose anything else ...
For a house with an area of 200 sq. - respectively, a 25/60 pump and pipes from the boiler 20 mm (inner diameter) and further along the branches as indicated above….

For completely non-typical long networks (the boiler room is located at a great distance from the house), it is really better to calculate the hydraulic resistance of the pipeline, based on ensuring the delivery of the required amount of heat carrier in terms of power and select a special pump and pipes according to the calculation ...

Selection of pump parameters for home heating

More specifically, about the choice of a pump for a boiler in a house on the basis of thermal hydraulic calculations. For conventional 3-speed circulation pumps, the following standard sizes are selected:

for an area up to 120 sq. - 25-40,
from 120 to 160 - 25-50,
from 160 to 240 - 25-60,
up to 300 - 25-80.

But for pumps under electronic control, Grundfos recommends slightly increasing the size, since these products can rotate too slowly, so they will not be superfluous in small areas. The following pump selection parameters are recommended by the manufacturer for the Grundfos Alpha line.

Calculation of pipe parameters

There are tables for the selection of pipe diameters, depending on the connected heat output. The table shows the amount of heat energy in watts, (below it is the amount of coolant kg / min), provided:
- on the supply +80 degrees, on the return +60 degrees, air +20 degrees.

It is clear that about 4.5 kW will pass through a metal-plastic pipe with a diameter of 12 mm (outer 16 mm) at a recommended speed of 0.5 m / s. Those. we can connect up to 3 radiators with this diameter, in any case, we will make taps for one radiator only with this diameter.

20 mm (25 mm outer) - almost 13 kW - main line from the boiler for a small house - or a floor up to 150 sq.

The next diameter is 26 mm (32 external metal-plastic) - more than 20 kW is already rarely used in the main highways. A smaller diameter is set, since these sections of the pipeline are usually short, the speed can be increased, up to the occurrence of noise in the boiler room, ignoring a slight increase in the total hydraulic resistance of the system, as not significant ...

Selection of polypropylene pipes

Polypropylene pipes for heating are thicker. And the standardization for them is on the outer diameter. The minimum outer diameter is 20 mm. In this case, the internal pipe PN25 (reinforced with fiberglass, for heating, max. +90 degrees) will be approximately 13.2 mm.

Generally, external diameters of 20 and 25 mm are used, which is roughly equated in terms of the transmitted power to metal-plastic 16 and 20 mm (external), respectively.

Polypropylene 32 m and 40 mm are used less often on highways of large houses or in some special projects (gravity heating, for example).

Standard outer diameters of polypropylene pipes PN25 are 20, 25, 32, 40 mm.
Corresponding inner diameters - 13.2, 16.6, 21.2, 26.6 mm

Thus, on the basis of heat engineering and hydraulic calculations, we chose the diameters of the pipelines, in this case from polypropylene. Previously, we calculated the power of the boiler for a particular house, the power of each radiator in each room, and selected the necessary characteristics of a solid fuel boiler pump for this entire economy, i.e. created a complete calculation of the home heating system.

One of the most important issues in creating comfortable living conditions in a house or apartment is a reliable, correctly calculated and installed, well-balanced heating system. That is why the creation of such a system is the most important task when organizing the construction of your own house or when carrying out major repairs in an apartment of a high-rise building.

Despite the modern variety of heating systems of various types, the proven scheme remains the leader in popularity: pipe circuits with a coolant circulating through them, and heat exchange devices - radiators installed in the premises. It would seem that everything is simple, the batteries are under the windows and provide the required heating ... However, you need to know that the heat transfer from the radiators must correspond to the area of the room and a number of other specific criteria. Thermal calculations based on the requirements of SNiP is a rather complicated procedure performed by specialists. Nevertheless, you can do it on your own, of course, with an acceptable simplification. This publication will tell you how to independently calculate the heating batteries for the area of the heated room, taking into account various nuances.

But, for a start, you need to at least briefly familiarize yourself with the existing heating radiators - the results of the calculations will largely depend on their parameters.

Briefly about the existing types of heating radiators

Steel radiators of panel or tubular construction.
Cast iron batteries.
Aluminum radiators of several modifications.
Bimetallic radiators.

Steel radiators

This type of radiator has not gained much popularity, despite the fact that some models are given a very elegant design. The problem is that the disadvantages of such heat exchange devices significantly exceed their advantages - low price, relatively small weight and ease of installation.

The thin steel walls of such radiators do not have enough heat - they quickly heat up, but also cool down just as rapidly. Problems can also arise during water hammer - the welded joints of the sheets sometimes leak. In addition, inexpensive models that do not have a special coating are prone to corrosion, and the service life of such batteries is short - usually manufacturers give them a rather short warranty.

In the overwhelming majority of cases, steel radiators are an integral structure, and it is not possible to vary the heat transfer by changing the number of sections. They have a rated thermal power, which must be immediately selected based on the area and characteristics of the room where they are planned to be installed. The exception is that some tubular radiators have the ability to change the number of sections, but this is usually done on order, during manufacture, and not at home.

Cast iron radiators

Representatives of this type of batteries are probably familiar to everyone from early childhood - these are the kind of accordions that were previously installed literally everywhere.

Perhaps such MC-140-500 batteries did not differ in particular elegance, but they faithfully served more than one generation of residents. Each section of such a radiator provided a heat transfer of 160 watts. The radiator is prefabricated, and the number of sections, in principle, was not limited by anything.

Currently, there are many modern cast iron radiators on sale. They are already distinguished by a more elegant appearance, flat, smooth outer surfaces that make cleaning easier. Exclusive versions are also produced, with an interesting relief pattern of cast iron casting.

With all this, such models fully retain the main advantages of cast iron batteries:

The high heat capacity of cast iron and the massiveness of the batteries contribute to long-term retention and high heat transfer.
Cast iron batteries, with proper assembly and high-quality sealing of joints, are not afraid of water hammer, temperature changes.
Thick cast iron walls are not susceptible to corrosion and abrasion. Almost any heat carrier can be used, so such batteries are equally good for both autonomous and central heating systems.

If you do not take into account the external data of old cast-iron batteries, then among the shortcomings can be noted the fragility of the metal (accentuated blows are unacceptable), the relative complexity of installation, associated more with massiveness. In addition, not all wall partitions will be able to support the weight of such radiators.

Aluminum radiators

Aluminum radiators, having appeared relatively recently, very quickly gained popularity. They are relatively inexpensive, have a modern, rather elegant appearance, and have excellent heat dissipation.

High-quality aluminum batteries can withstand a pressure of 15 or more atmospheres, a high coolant temperature - about 100 degrees. At the same time, the heat output from one section for some models sometimes reaches 200 W. But at the same time, they are small in mass (the weight of the section is usually up to 2 kg) and do not require a large volume of coolant (capacity - no more than 500 ml).

Aluminum radiators are on sale both as stackable batteries, with the ability to change the number of sections, and as solid products designed for a certain power.

Disadvantages of aluminum radiators:

Some types are highly susceptible to oxygen corrosion of aluminum, with a high risk of gassing. This imposes special requirements on the quality of the coolant, therefore, such batteries are usually installed in autonomous heating systems.
Some non-separable aluminum radiators, whose sections are manufactured using extrusion technology, can leak at the joints under certain unfavorable conditions. At the same time, it is simply impossible to carry out repairs, and you will have to change the entire battery as a whole.

Of all the aluminum batteries, the highest quality is made with the use of anodic oxidation of the metal. These products are practically not afraid of oxygen corrosion.

Outwardly, all aluminum radiators are roughly similar, so you need to read the technical documentation very carefully when making a choice.

Bimetallic heating radiators

Such radiators in their reliability compete with cast iron, and in terms of heat output - with aluminum. The reason for this is their special design.

Each of the sections consists of two, upper and lower, steel horizontal collectors (item 1), connected by the same steel vertical channel (item 2). The connection into a single battery is made with high quality threaded couplings (pos. 3). High heat dissipation is ensured by the outer aluminum shell.

Steel inner pipes are made of metal that does not corrode or has a protective polymer coating. Well, the aluminum heat exchanger does not come into contact with the coolant under any circumstances, and corrosion is absolutely not terrible for it.

Thus, a combination of high strength and wear resistance with excellent thermal performance is obtained.

Prices for popular heating radiators

Heating radiators

Such batteries are not afraid of even very large pressure surges, high temperatures. They are, in fact, universal, and are suitable for any heating systems, however, they still show the best performance characteristics in conditions of high pressure of the central system - they are of little use for circuits with natural circulation.

Perhaps their only drawback is the high price compared to any other radiators.

For ease of perception, there is a table showing the comparative characteristics of radiators. Symbols in it:

TS - tubular steel;
Chg - cast iron;
Al - ordinary aluminum;
AA - anodized aluminum;
BM - bimetallic.

	Chg	TS	Al	AA	BM
Maximum pressure (atmospheres)
working	6-9	6-12	10-20	15-40	35
crimping	12-15	9	15-30	25-75	57
destruction	20-25	18-25	30-50	100	75
Limitation on pH (hydrogen index)	6,5-9	6,5-9	7-8	6,5-9	6,5-9
Corrosion susceptibility by:
oxygen	No	Yes	No	No	Yes
stray currents	No	Yes	Yes	No	Yes
electrolytic vapors	No	weak	Yes	No	weak
Section capacity at h = 500 mm; Dt = 70 °, W	160	85	175-200	216,3	up to 200
Warranty, years	10	1	3-10	30	3-10

Video: recommendations for choosing heating radiators

You may be interested in information about what constitutes

How to calculate the required number of heating radiator sections

It is clear that a radiator installed in the room (one or more) must provide heating to a comfortable temperature and compensate for the inevitable heat loss, regardless of the weather outside.

The base value for calculations is always the area or volume of the room. By themselves, professional calculations are very complex, and take into account a very large number of criteria. But for everyday needs, you can use simplified methods.

The easiest ways to calculate

It is generally accepted that 100 W per square meter of floor space is sufficient to create normal conditions in a standard living space. Thus, you just need to calculate the area of the room and multiply it by 100.

Q = S× 100

Q- the required heat transfer from heating radiators.

S- the area of the heated room.

If you plan to install a non-separable radiator, then this value will become a guideline for the selection of the required model. In the case when batteries will be installed that allow a change in the number of sections, one more calculation should be carried out:

N = Q/ Qus

N- the calculated number of sections.

Qus- specific thermal power of one section. This value is necessarily indicated in the technical passport of the product.

As you can see, these calculations are extremely simple, and do not require any special knowledge of mathematics - a tape measure is enough to measure a room and a piece of paper for calculations. In addition, you can use the table below - there are already calculated values for rooms of various sizes and certain capacities of the heating sections.

Section table

However, it must be remembered that these values are for a standard ceiling height (2.7 m) of a high-rise building. If the height of the room is different, then it is better to calculate the number of battery sections based on the volume of the room. For this, an average indicator is applied - 41 V t t thermal power per 1 m³ of volume in a panel house, or 34 W - in a brick one.

Q = S × h× 40 (34)

where h- ceiling height above floor level.

Further calculation is no different from the one presented above.

Detailed calculation taking into account the features premises

Now let's move on to more serious calculations. The simplified calculation technique given above can give the owners of a house or apartment a "surprise". When the installed radiators will not create the required comfortable microclimate in the living quarters. And the reason for this is a whole list of nuances that the considered method simply does not take into account. Meanwhile, such nuances can be very important.

So, the area of the room is again taken as a basis and all the same 100 W per m². But the formula itself already looks somewhat different:

Q = S× 100 × A × B × C ×D× E ×F× G× H× I× J

Letters from A before J the coefficients are conventionally designated, taking into account the features of the room and the installation of radiators in it. Let's consider them in order:

A is the number of external walls in the room.

It is clear that the higher the area of contact between the room and the street, that is, the more external walls in the room, the higher the total heat loss. This dependence is taken into account by the coefficient A:

One outer wall - A = 1.0
Two outer walls - A = 1.2
Three outer walls - A = 1.3
All four walls are external - A = 1.4

B - orientation of the room to the cardinal points.

The maximum heat loss is always in rooms that do not receive direct sunlight. This is, of course, the northern side of the house, and the eastern side can also be attributed here - the rays of the Sun are here only in the morning, when the luminary has not yet “reached full power”.

The southern and western sides of the house are always warmed up by the Sun much more strongly.

Hence, the values of the coefficient V :

The room faces north or east - B = 1.1
South or west rooms - B = 1, that is, it may not be counted.

C is a coefficient that takes into account the degree of wall insulation.

It is clear that the heat loss from the heated room will depend on the quality of the thermal insulation of the outer walls. Coefficient value WITH take equal:

Middle level - the walls are lined with two bricks, or their surface insulation with another material is provided - C = 1.0
External walls are not insulated - C = 1.27
A high level of insulation based on thermal engineering calculations - C = 0.85.

D - features of the climatic conditions of the region.

Naturally, it is impossible to equal all the basic indicators of the required heating power "one size fits all" - they also depend on the level of winter temperatures below zero, typical for a particular area. This takes into account the coefficient D. To select it, the average temperatures of the coldest decade of January are taken - usually this value is easy to check with the local hydrometeorological service.

- 35 ° WITH and below - D = 1.5
- 25 ÷ - 35 ° WITH – D = 1.3
up to - 20 ° WITH – D = 1.1
not lower - 15 ° WITH – D = 0.9
not lower - 10 ° WITH – D = 0.7

E - coefficient of the height of the ceilings of the room.

As already mentioned, 100 W / m² is the average value for a standard ceiling height. If it differs, you should enter a correction factor E:

Up to 2.7 m – E = 1,0
2,8 – 3, 0 m – E = 1,05
3,1 – 3, 5 m – E = 1, 1
3,6 – 4, 0 m – E = 1.15
More than 4.1 m - E = 1.2

F - coefficient taking into account the type of premises located above

Arranging a heating system in rooms with a cold floor is a pointless exercise, and the owners always take action in this matter. But the type of room located above, often does not depend on them in any way. Meanwhile, if the top is a residential or insulated room, then the total demand for thermal energy will significantly decrease:

cold attic or unheated room - F = 1.0
insulated attic (including - and insulated roof) - F = 0.9
heated room - F = 0.8

G - coefficient of account of the type of installed windows.

Different window structures are not equally susceptible to heat loss. This takes into account the coefficient G:

ordinary wooden frames with double glazing - G = 1.27
windows are equipped with a single-chamber double-glazed window (2 glasses) - G = 1.0
single-chamber glass unit with argon filling or double glass unit (3 glasses) - G = 0.85

H - coefficient of the area of the glazing of the room.

The total amount of heat loss also depends on the total area of the windows installed in the room. This value is calculated based on the ratio of the area of the windows to the area of the room. Depending on the result obtained, we find the coefficient N:

Ratio less than 0.1 - H = 0, 8
0.11 ÷ 0.2 - H = 0, 9
0.21 ÷ 0.3 - H = 1, 0
0.31 ÷ 0.4 - H = 1, 1
0.41 ÷ 0.5 - H = 1.2

I - coefficient taking into account the radiator connection diagram.

Their heat transfer depends on how the radiators are connected to the supply and return pipes. This should also be taken into account when planning the installation and determining the required number of sections:

a - diagonal connection, supply from above, return from below - I = 1.0
b - one-way connection, supply from above, return from below - I = 1.03
c - two-way connection, both supply and return from the bottom - I = 1.13
d - diagonal connection, supply from below, return from above - I = 1.25
d - one-way connection, supply from below, return from above - I = 1.28
e - one-sided bottom connection of return and supply - I = 1.28

J - coefficient taking into account the degree of openness of the installed radiators.

Much also depends on how open the installed batteries are for free heat exchange with the room air. Existing or artificially created barriers can significantly reduce the heat transfer of the radiator. This takes into account the coefficient J:

a - the radiator is located openly on the wall or not covered by a window sill - J = 0.9

b - the radiator is covered from above with a window sill or shelf - J = 1.0

c - the radiator is covered from above with a horizontal protrusion of the wall niche - J = 1.07

d - the radiator is covered from above with a window sill, and from the front parties — partswell covered with a decorative cover - J = 1.12

e - the radiator is completely covered with a decorative casing - J = 1.2

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Well, finally, that's all. Now you can substitute the required values and the coefficients corresponding to the conditions into the formula, and the output will be the required thermal power for reliable heating of the room, taking into account all the nuances.

After that, it will remain either to pick up a non-separable radiator with the desired heat output, or to divide the calculated value by the specific thermal power of one section of the battery of the selected model.

Surely, to many, such a calculation will seem excessively cumbersome, in which it is easy to get confused. To facilitate the calculations, we suggest using a special calculator - all the required values are already included in it. The user only needs to enter the requested initial values or select the required items from the lists. The "calculate" button will immediately lead to an accurate result rounded up.

Calculation of heating a private house is one of the important tasks in its construction or major repairs. It is best to do this at the planning stage. A special online calculator can provide some help in calculations. There are many calculators for calculating fuel consumption, furnace power, ventilation system, chimney cross-section, performance of the pumping and mixing unit of the "warm floor" and others. However, it should be borne in mind that they all show only an approximate result, since can only calculate the simplest configurations. In fact, when calculating heating, it is necessary to take into account a lot of additional nuances. This must be done in order to correctly calculate the costs of the entire heating system and in the future not to suffer from cold in the house or vice versa its surplus, and therefore unnecessary fuel costs.

When choosing a boiler for heating a house, one must take into account all the parameters: both heating equipment and a residential building Source baraholka.com.ru

Calculation of heating in a private house - what needs to be calculated

To calculate the heating of a private house, it is necessary to calculate the power of the heating boiler, determine the number and placement of radiators, take into account a number of factors from the weather, to thermal insulation and the material for making pipes and the boiler.

Keep in mind that the comfort of living in the house will depend on this process, since your calculations will directly affect the quality of heating. In addition, these calculations are the basis of the budget for the installation and further operation of the entire heating system. It is at this stage that you will have to decide how much money you will spend in the future on heating your home. When starting the calculations, it is important to remember the climatic conditions in which your region is located and the conditions in which the house will be used.

Video description

In our video, we'll talk about heating in a private country house. Our guest is the author and presenter of the Teplo-Voda channel Vladimir Sukhorukov:

The heating system is not only a stove and batteries. It includes:

Boiler,

Pumping station,

Radiators,

Control devices,

Sometimes an expansion tank is needed.

Something like this looks like a diagram of the heating system of a house. Source lucheeotoplenie.ru

Calculation of the power of heating devices

Before calculating the output of a heating boiler, it is necessary to determine which type of boiler will be used. Heating boilers have different efficiency and this choice will determine not only the level of heat transfer, but also the financial component of subsequent operation when choosing a fuel:

Electric boilers,

Gas boilers,

Solid fuel boilers,

Boilers for liquid fuel,

Combined boiler electricity / solid fuel.

When the choice of the type of boiler has been made, it is necessary to determine its throughput. The functioning of the entire system will depend on this. The calculation of the capacity of a water heating boiler is carried out taking into account the amount of heat energy required per m3. The calculator can help you calculate the volume of heated rooms:

bedroom: 9 m2 3 m = 27 m3,

bedroom: 12 m2 3 m = 36 m3,

bedroom: 15 m2 3 m = 45 m3,

living room: 25 m2 3 m = 75 m3,

corridor: 6 m2 3 m = 18 m3,

kitchen: 12 m2 3 m = 36 m3,

bathroom: 8 m2 3 m = 24 m3.

The calculation takes into account all the premises of the house, even if it is not planned to install radiators in them. Source stroikairemont.com

On our website you can find contacts of construction companies that offer the service of house insulation. You can communicate directly with representatives by visiting the Low-Rise Country exhibition of houses.

Then the results are summed up, and the total volume of the house is obtained - 261 m3. When calculating, rooms and passages must be taken into account, in which it is not planned to install heating devices, for example, a corridor, a pantry, or a hallway. This is done so that the heat from the radiators installed in the house is enough to heat the whole house.

When calculating the heating system, it is imperative to take into account the climatic zone and the outside temperature in winter.

Let's take an arbitrary indicator for the region of 50 W / m3 and the area of the house of 261 m3, which is planned to be heated. Power calculation formula: 50 W 261 m3 = 13050 W. The result is multiplied by a factor of 1.2 and the boiler power is calculated - 15.6 kW. The factor allows you to add 20% of the reserve power to the boiler. It will enable the boiler to work in a saving mode, avoiding special overloads.

Additional temperature sensors will help monitor the process Source qowipa.dopebi.ru.net

The coefficient correction for the climatic conditions of the regions varies from 0.7 in the southern regions of Russia to 2.0 in the northern regions. The coefficient 1.2 is used in the central part of Russia.

Here's another formula used by online calculators:

To get a preliminary result of the required boiler power, the area of the room can be multiplied by the climatic coefficient and the result obtained can be divided by 10.

An example of a formula for calculating the power of a heating boiler for a house with an area of 120 m2 in the northern region of Russia:

Nk = 120 * 2.0 / 10 = 24 kW

Which pipes are best for the heating main

polyethylene,

polypropylene (with or without reinforcement),

steel,

stainless.

You can take different pipes for heating in the house, but it is important to pass the features of the selected type Source ms.decorexpro.com

Each of these types has its own nuances that should be taken into account when developing and calculating the heating of a private house:

Steel pipes are universal in use and can withstand pressures up to 25 atmospheres, but they have a significant drawback - they rust and have a certain service life. In addition, they have difficulties in installation.

Pipes made of polypropylene, composite metal-plastic and XLPE are easy to install and, due to their weight, can be used on thin walls. The advantage of such pipes is that they are not susceptible to rust, decay and do not react to bacteria. An important indicator is that they do not expand from heat and do not deform in the cold. Withstand a constant temperature up to 90 degrees and a short-term rise up to 110 degrees Celsius.

Copper pipes are distinguished by their high price and increased complexity during installation, but in terms of strength they compete with plastic pipes, are not susceptible to rust and are considered the best option. In addition, copper is ductile, conducts heat well and keeps the temperature of the water in the pipes in the range from –200 to 250 degrees Celsius. This ability of copper will protect the system from possible defrosting, which is very important in Siberia and the northern regions.

If the house is located in the north of the country, then copper pipes for the heating system are best suited Source svizzeraenergia.ch

How to calculate the optimal number and volume of heat exchangers

When calculating the number of required radiators, you should take into account what material they are made of. The market now offers three types of metal radiators:

Aluminum,

Bimetallic alloy,

They all have their own characteristics. Cast iron and aluminum have the same heat transfer rate, but at the same time aluminum cools down quickly, and cast iron heats up slowly, but retains heat for a long time. Bimetallic radiators heat up quickly, but cool down much more slowly than aluminum ones.

When calculating the number of radiators, you should also take into account other nuances:

a corner room is cooler than others and requires more radiators,

the use of double-glazed windows on the windows saves 15% of heat energy,

up to 25% of heat energy "leaves" through the roof.

The number of heating radiators and sections in them depends on many factors Source amikta.ru

In accordance with the norms of SNiP, 100 W of heat is required for heating 1 m3. Therefore, 50 m3 will require 5000 watts. If a bimetallic device for 8 sections emits 120 W, then using a simple calculator we count: 5000: 120 = 41.6. After rounding up, we get 42 radiators.

However, in a private house, the temperature is independently regulated. A single battery is believed to generate 150W of heat. We recalculate and get 5000: 150 = 33.3. That is, you need 34 radiators.

You can use the approximate formula for calculating the radiator sections:

The symbol (*) shows that the fractional part is rounded according to general mathematical rules, N is the number of sections, S is the area of the room in m2, and P is the heat transfer of 1 section in W.

Video description

Conclusion

Installation and calculation of the heating system in a private house is the main component of the conditions for a comfortable stay in it. Therefore, the calculation of heating in a private house should be approached with special care, taking into account the many accompanying nuances and factors.

The calculator will help if you need to quickly and averagely compare different construction technologies among themselves. In other cases, it is better to contact a specialist who will competently carry out the calculations, correctly process the results and take into account all errors.

Not a single program can cope with this task, because it contains only general formulas, and calculators for heating a private house and tables offered on the Internet serve only to facilitate calculations and cannot guarantee accuracy. For accurate correct calculations, it is worth entrusting this work to specialists who will be able to take into account all the wishes, capabilities and technical indicators of the selected materials and devices.