How much heat kW we need to heat our homes – we calculate it ourselves

Efficient heating of our homes is necessary for both economy and comfort. It is essential to know how much heat power, expressed in kilowatts (kW), we require to heat our homes in order to choose the best heating system and guarantee efficient energy use. We’ll simplify the process of figuring out how much heat kW your house needs in this guide, giving you the information you need to make well-informed decisions about insulation and heating.

First of all, it’s critical to understand that the number of heat kW required to warm a house varies depending on a number of variables, such as the size of the property, the degree of insulation, the climate in the area, and the intended indoor temperature. In a colder climate, a larger house with inadequate insulation will need more kW of heating power than a smaller, better-insulated home in a milder climate.

A straightforward formula based on these variables is used to determine the heat kW for your house. Using the heat loss calculation is a popular technique that takes into account heat gain from sunlight, appliances, and occupants as well as heat loss through walls, windows, doors, roof, and floor. You can calculate the total heat kW needed to keep the interior temperature comfortable by evaluating these variables.

It’s critical to evaluate your home’s insulation levels in-depth when calculating heat loss. By decreasing heat loss and the quantity of kW required for heating, proper insulation is like covering your home with a warm blanket. Insulation materials that can lower your home’s heating costs are fiberglass, cellulose, foam board, and spray foam. These materials can greatly increase your home’s energy efficiency.

Moreover, knowing your heating requirements involves more than just calculating kW. It’s also very important to think about what kind of heating system will work best for your house and your tastes. Traditional boilers and furnaces are among the options; contemporary heat pumps and radiant heating systems are also available; each has a different efficiency rating and energy needs.

Area of the house (m²)	Estimated heat loss (kW)
100	10
150	15
200	20
250	25

A bit of general information – what is the required amount of heat?

In short, everything about this is already known; a little systematization is needed.

In order to live comfortably, modern man must create a specific microclimate, of which the room’s air temperature is one of the most crucial elements. We can confidently state that, while "thermal preferences" may differ, most people’s "thermal comfort" zone falls between 18 and 23 degrees Celsius.

However, natural thermodynamic processes tend to bring everything under the "common bar" when the outside temperature drops, which causes heat to start to escape the living area. From a physics perspective, heat losses are a perfectly normal phenomenon. Although it is not possible to totally eliminate such losses, the entire housing insulation system is designed to minimize their occurrence. Thus, the heating system in the house is intended to offset these precise heat losses.

There is nowhere to escape heat losses, but it is crucial to at least make every effort to reduce them as much as possible.

How to quantify them?

The idea that 100 watts of heat are needed for every square meter of floor space serves as the foundation for the most straightforward method of determining the necessary heat output. or – 1 kW for every 10 m^.

However, even if you’re not an expert, you can imagine how this "equalization" is paired with the particulars of individual homes and the rooms within them, with the placement of buildings on the ground, and with the local climate.

Therefore, it is preferable to use an alternative, more "meticulous" calculation method where a variety of factors are considered. The calculator below uses this algorithm.

Crucial: computations are done independently for every heated room in the house or apartment. The total amount of heat energy needed is only computed at the very end. The simplest method would be to create a small table with all the rooms that have the information needed for calculations listed in the rows. Then, computations won’t take long if the owner has a plan of their residential property on hand.

And one last thing. The outcome could appear overly dramatic. However, it’s important to understand that this indicates the total amount of heat needed to offset heat loss in the worst possible circumstances. In other words, to keep the interior temperature at +20 °C at the lowest outside temperatures that are normal for the area in which you live. Put another way, even during the coldest part of winter, the house will be warm.

But these extremely cold temperatures usually only persist for a very short while. In other words, the heating system will typically run at a reduced capacity. And that implies that adding any more reserve is pointless. There will already be a significant operating power reserve.

The calculator is below, and you can find the necessary short instructions on using the program below that.

What a heating unit consists of?

Many of us are used to believing that the heating system consists solely of heat exchangers and a heating boiler that are connected to one another via a pipeline. But the strapping also has additional components:

• pumping unit;
• devices for controlling and monitoring the operation of the installation;
• heat carrier;
• expansion tank (if necessary).

Determining the heating boiler’s performance is a prerequisite to accurately calculating the house’s heating needs. Furthermore, you must figure out in a different room how many heating batteries are used in a private residence.

Selection of the heating element

Boilers can be classified into multiple groups based on the fuel type that is used.

• electric;
• liquid-fuel;
• gas;
• solid fuel;
• combined.

The cost and accessibility of fuel sources directly influence the heater selection.

Of all the models put forth, gas-powered devices are the most widely used. This particular fuel is reasonably advantageous and reasonably priced. Additionally, the efficiency of these units is quite high, something that other units with similar functionality cannot claim, and their maintenance does not require specialized knowledge or skills. Gas boilers, however, are only suitable if your home is connected to a centralized gas main line.

Determining the boiler output

Prior to calculating the heating, you must ascertain the heater’s capacity, as the effectiveness of the thermal installation’s operation is based on this indicator. As a result, a heavy-duty unit will use a lot of fuel, whereas a low-power unit won’t be able to heat the room to the desired level. This makes the heating system calculation a crucial and accountable procedure.

Instead of using intricate formulas to determine the boiler’s performance, you can use the table below. It details the heated building’s area as well as the heater’s power, allowing it to maintain a comfortable temperature throughout.

Total square footage of the house requiring heating, in meters

Calculation of the number and volume of heat exchangers

Three different types of metal are used to make modern radiators: bimetallic alloys, cast iron, and aluminum. While the heat transfer rates of the first two options are equal, cast iron radiators that have been heated cool more slowly than aluminum heat exchangers. Bimetallic radiators cool down more slowly than other radiators while producing a lot of heat. As a result, these kinds of heating devices have become more and more popular in recent years.

What the number of radiators depends on

When determining the quantity of heating radiators in a private residence, the following considerations must be made:

• temperature conditions in the corner room are lower than in other rooms, because it has two walls in contact with the street;
• When the ceiling height is more than 3 meters, to calculate the power of the coolant should be taken not the area of the room, but its volume;
• Thermal insulation of wall slabs and floor surface will save up to 35% of heat energy;
• the lower the air temperature outside in the cold season, the more radiators there should be in the building and, accordingly, the lower it is – the less the number of heat exchangers can be placed in the building;
• Modern glazing with metal-plastic windows will reduce heat loss by 15%;
• single-circuit piping is performed by means of radiators, the size of which does not exceed 10 sections;
• When moving the coolant from the top to the bottom of the main line, it is possible to increase its performance by 20%.

Formula and example of calculation

As per SNiP, one square meter of space requires 100 W of heat, correspondingly, to heat a room measuring 20 square meters.2000 W is required by m. All that is required to calculate the heating radiators on the area is a calculator. Thus, a single eight-section bimetallic heat exchanger outputs roughly 120W. We receive: 2000 / 120 = 17 sections on the final account.

A private home’s heating radiator calculation looks a little different. As we are controlling the coolant temperature separately in this instance, it is considered that a single radiator can generate up to 150 watts. Let’s reassess our assignment: 13,3 is 2000 / 150.

After rounding, we obtain 14 sections. We will require so many heat exchangers to complete the heat circuit plumbing in a 20 square meter space. м.

Regarding radiator placement, it is advised to position them directly against various room walls.

To prevent cold air from coming in through the windows, experts advise putting the majority of the batteries beneath the window sill.

Piped heating system

Pipes composed of the following materials are used to install the thermal circuit:

Every one of these choices has benefits and drawbacks of its own. A metal-plastic pipeline is the ideal choice for piping the heating system. It has a relatively low cost and a 45–60 year service life (assuming proper installation).

Installation of heating devices

The installation of this equipment is done in compliance with SNiP regulations. We want to draw attention to the following crucial considerations that need to be made when installing heating equipment:

1. The gap between the bottom of the appliance and the floor surface must be at least 6 cm. This will not only ensure the possibility of cleaning under the equipment, but also prevent the possibility of heat energy penetration into the floor surface.
2. The gap between the top of the heater and the window sill should not be less than 5 cm. Thanks to this, you will be able to dismantle the heat exchanger without disturbing the window sill.
3. When using radiators with fins, it is extremely important to ensure that they are placed in a vertical position only.
4. The central point of the heating device should coincide with the center of the window frame. In this case, the battery will act as a thermal curtain, preventing the penetration of cold air masses through the double-glazed windows into the room.

Installing all radiators at the same level will improve the effectiveness of the strapping.

You can achieve good heating in your house by following the aforementioned advice.

Many considerations must be made when calculating the kilowatts (kW) of heat required to adequately heat your home. These include your home’s dimensions, the state of the insulation, the outside temperature, and even your own comfort preferences. You can use a straightforward formula based on the dimensions of your space and the local average temperature outside to get an estimate. You can potentially save money and energy in the long run by making better decisions about your heating needs by being aware of these simple calculations. To make determining the heating needs for your home easier for you, let’s take it step by step.

Maximum accurate variant of calculation

We can see from the calculations above that none of them are entirely accurate; therefore, further research is required on the remaining coefficients.Even in cases where the rooms are the same, the outcomes are marginally different.

Use this approach if you want your calculations to be as accurate as possible. It considers a wide range of variables that may impact heating efficiency and other pertinent metrics.

The calculation formula generally takes the following form:

*A * B * C * D * E * F * G * S = 100 W/m2,

• where T is the total amount of heat required to heat the room in question;
• S is the area of the room to be heated.

Further research is required on the other coefficients. Factor A, for instance, considers the room’s glazing.

Peculiarities in the room’s windows

• 1.27 for rooms whose windows are glazed with just two panes of glass;
• 1.0 – for rooms with double-glazed windows;
• 0.85 – if the windows are triple-glazed.

The room’s wall insulation peculiarities are taken into consideration by coefficient B.

Unusualities in the insulation of room walls

• if the insulation is low-performance, the coefficient is taken as 1.27;
• in case of good insulation (for example, if the walls are laid in 2 bricks or purposefully insulated with a high-quality heat insulator), a coefficient equal to 1.0 is used;
• at a high level of insulation – 0.85.

The ratio of the room’s floor area to the total area of window openings is shown by coefficient C.

The proportion between the room’s floor area and the total area of the window openings

This is how the dependence appears:

• if the ratio is equal to 50%, the C coefficient is taken as 1.2;
• if the ratio is 40%, a coefficient equal to 1.1 is used;
• if the ratio is equal to 30%, the value of the coefficient is reduced to 1.0;
• in case of even smaller percentages, use coefficients equal to 0.9 (for 20%) and 0.8 (for 10%).

The average temperature during the coldest time of the year is shown by coefficient D.

When using radiators, the room’s distribution of heat

This is how the dependence appears:

• if the temperature is -35 and below, the coefficient is assumed to be 1.5;
• at temperatures up to -25 degrees, a value of 1.3 is used;
• if the temperature does not fall below -20 degrees, the calculation is performed with a coefficient equal to 1.1;
• residents of regions where the temperature does not fall below -15 should use a coefficient of 0.9;
• if the winter temperature does not fall below -10, count with a coefficient of 0.7.

Indicating the quantity of external walls is coefficient E.

The quantity of external walls

Use a coefficient of 1.1 in the case where there is only one external wall. Raise it to 1.2 in the case of two walls, to 1.3 in the case of three walls, and to 1.4 in the case of four external walls.

The characteristics of the room above it are taken into account by the F coefficient. The following is the dependence:

• if there is an unheated attic above, the coefficient is 1.0;
• if the attic is heated – 0.9;
• if the upstairs neighbor is a heated living room, the coefficient can be reduced to 0.8.

Furthermore, the room’s height is taken into account by the formula’s final coefficient, G.

• in rooms with 2.5 m high ceilings are calculated using a coefficient of 1.0;
• If the room has a 3-meter ceiling, increase the coefficient to 1.05;
• at a ceiling height of 3.5 m, calculate with a coefficient of 1.1;
• rooms with a 4-meter ceiling are calculated with a coefficient of 1.15;
• when calculating the number of battery sections for heating a room with a height of 4.5 m increase the coefficient to 1.2.

This calculation allows you to determine the necessary number of heating unit sections with the least amount of error possible, taking into account nearly all existing nuances. To sum up, all you need to do is divide the computed amount by the heat output of one battery section (refer to the passport that is attached), and naturally, round the result to the next highest whole number.

Heating radiator calculation calculator

To make things easier, a dedicated calculator for heating radiator calculations is used to enter all these parameters. All required parameters only need to be specified, and pressing the "Calculate" button will produce the desired outcome right away:

Heating system piping

These kinds of pipelines can be used for installing a home heating system:

• Pipelines made of polyethylene, polypropylene or metal plastic;
• Copper piping;
• Steel pipes.

Polyethylene pipes	Polypropylene pipes
Copper pipes	Steel pipes

Each of these pipe systems has benefits as well as drawbacks. Polymer pipes are resistant to corrosion and require less installation effort. Copper pipes can tolerate higher pressures and temperatures better than other materials. One drawback that sets steel pipes apart is the requirement to perform certain welding procedures. This is one of the many details that the program used to determine how much heat a private home needs to account for.

Calculating the capacity of the heating boiler

Prior to calculating the heating system, it is important to ascertain the boiler’s output requirement as precisely as possible, as this indicator will determine how well the system performs in terms of maintaining the proper temperature in each room of the house.

Less than necessary power will result in an uncomfortably cold house, and excessive boiler output will cause unwarranted fuel consumption that will incur extra expenses.

In order to ascertain the ideal power for the heating boiler, the boiler should be aware of:

• the total area of the premises that are supposed to be heated, denoted by the letter S;
• specific boiler output for every 10 cubic meters of space, denoted W yd.

Additionally, this figure needs to be modified to account for the local climate in the building’s location.

It should be noted that in actuality, the ranges established for particular climatic zones are typically used to determine the value of a given power. Therefore, the specific power for southern regions should be between 0.7 and 0.9 kW. Between 1.2 and 1.5 kW for the middle zone and between 1.5 and 2.0 kW for the north

Accordingly, the specific power for the Middle Belt should be between 1.2 and 1.5 kW, the North should be between 1.5 and 2.0 kW, and the South should be between 0.7 and 0.9 kW.

You can take the specific power equal to one to make calculations simpler. As a result, we obtain a rule for determining the necessary boiler output, which states that 10 kW are needed for every 100 square meters of heated space in the heated room.

The kind of water heating system that is selected for a residential building is primarily determined by the heating calculation.

The size of the house should serve as a guide when making a selection.

If the heating system is installed in a meter and the area is larger than 100 square feet, the only way to guarantee forced circulation of the heat carrier—such as water—is to install a circulation pump.

Installation of a pump is not necessary for smaller houses because natural circulation heating systems can be used in these types of homes.

How to calculate heating without large errors

Seldom do homeowners who have chosen to install an autonomous heating system give up on the possibility of allowing the coolant—typically water, but less frequently antifreeze—to circulate naturally. It is advisable to convert all calculations into Watts because the installation of a boiler and pump implies continuous electricity consumption in the future. Nonetheless, the system’s heat capacity is typically expressed in J/(kg).°Ρ), and the heat generated by radiators is expressed in calories. How can I integrate all of these measurement units? Everything is easy to understand.

To begin with, one calorie is equal to the amount of heat used to heat one gram of water by 1 degree centigrade. If we refer to the heat capacity, 1 calorie is equal to approximately 4.2 J, to be more precise, 4.1868 J. Accordingly, for one liter of water, due to the fact that it weighs 1 kilogram, this value will correspond to 4.2 kJ. In this case, 1 calorie equals 0.001163 watts . hour, which means that 1 kCal will be 1.163 watts . hour. That"s really all it takes to find the relationship between the radiated heat and the wattage of the power consumer.

Now, so that there are no other options but how to correctly calculate heating, let"s turn to the facts. For heating of 1 square meter of room it is necessary to spend 90-125 W (as a rule, this is the power of one section of radiator), depending on climatic features of the area. According to SNiP the power of each section of the radiator must correspond to 100 kW. And this is provided that the height of the ceiling does not exceed three meters, otherwise the power input will increase. Also, the power will have to be increased or decreased by about 15 degrees for every 10 degrees of deviation in a greater or lesser direction from the average 70 degrees of the heater temperature.

Also, for example, the system will be 10% less efficient if the water inflow to the radiators is through the lower openings and the outflow through the upper openings. Based on all of the above, it is not difficult to derive a formula for calculating the heat loss of the heating circuit, which, in fact, and serve for the effective heating of the room, as it occurs within its limits. Let"s take on the task of determining the amount of heat input for a boiler. There are always two pipes to the heat generator, the supply pipe, i.e. the one through which the hot water runs to the radiators, and the return pipe, in which the already cooled water flows back to the boiler.

Suppose the supply requires a temperature of 75 degrees, and on the return, due to heat loss, will be 50 ° C, what is in this case the power of the boiler, the water flow rate of which is 16 liters per minute? We already know that to heat a liter of water at 1 degree it is necessary to spend 1.163 watts per hour. During this time through the boiler will pass 16 . 60 = 960 liters. Consequently, taking into account the temperature difference T = t1 – t2 = 75 – 50 = 25 ° C, we get the boiler output 1.163 . 25 . 960 = 27912 watts . hour or 27.912 kW.

There is another way to calculate the heating system, based on the specific power required to heat 10 square meters, depending on the characteristics of the region. By definition, in the Northern regions the specific boiler power Wud should be 1.2-1.5 kW per 10 m2, in the Central regions this value is equal to 1.2-1.5 kW per the same area, and in the Southern regions – 0.7-0.9 kW. As a rule, calculations are made for the above-mentioned 10 squares with an average ceiling height of 2.7 meters, the boiler output is determined by the formula Wkot = S .Wud / 10, where S is the area of the room. For typical houses the data can be taken from the table.

House area, m2	Boiler output, kW
60 – 200	up to 25
200 – 300	25 – 35
300 – 600	35 – 60
600 – 1200	60 – 100

Which pipes are best suited for heating mains

Very little is needed to determine the boiler’s power; instead, accurate pipe selection and radiator heating calculations for private homes are needed.

Currently, the market offers a variety of pipes for heating systems made of materials like:

• Polypropylene (without reinforcement, with reinforcement).
• Polyethylene.
• Steel.
• Copper.
• Stainless steel

Yes, you can use different pipes for heating throughout the house, but you must consider the characteristics of the chosen species.

When calculating heating and creating a heating plan for a private home, each of them has unique characteristics that need to be considered:

1. Steel pipes are universal in use and can withstand pressure up to 25 atm, but they have one big, significant disadvantage – they are subject to corrosion and have a limited service life. In addition, there are often difficulties in their installation.
2. Polypropylene pipes, as well as elements of cross-linked polyethylene and composite metal plastic are not difficult to install, and due to their low weight they can be installed even on thin walls. The advantage of such pipes is that they are not subject to rot, rust and do not react to bacteria. Another indicator – they do not begin to expand from heat and do not deteriorate even from frost. These pipes are capable of withstanding a constant temperature of 90 degrees and short-term temperature increases up to 110 degrees.
3. Pipes made of copper are characterized by increased requirements for installation and high price, but in durability they will compete with pipes made of plastic, since they are not subject to corrosion and are considered the best option. In addition, copper is a plastic metal, which conducts heat well and can maintain the temperature of water in pipes from -200 to +250 degrees. This ability of the material helps to protect the system from possible freezing, and this is very important in Siberia and northern regions.

Performance reserve depending on the type of boiler

Depending on the house’s insulation and the temperature during the coldest decade, we always advise having a 15–25% capacity reserve for a standard single-circuit boiler, regardless of the fuel type used. But occasionally, a marginally bigger reserve is needed:

• 20-30% reserve, if the boiler is a two-circuit boiler . Most models work on the principle of priority DHW, this means that at the moment of activation of the hot water consumption point the boiler does not heat the heating circuit, to work on two circuits requires a higher capacity;
• 20-25% reserve if the house is organized or planned supply and exhaust ventilation without heat recovery.

Another common scheme is the indirect boiler connection (particularly when solid fuel boilers are involved). In this instance, the excess capacity might be greater than 40–50% (the calculation depends on the circumstances). It should be noted that in all of the scenarios where reserve is used, it is not "idle"; rather, it is put to use when heating hot water, offsetting increased heat losses, or heating buffer tanks.

An indirect storage boiler, located on the boiler’s right side, is the tall, white tank that continuously maintains a significant amount of hot water.

Calculation of the operating parameters of the heating system

After deciding on a boiler type, the heating system calculation for a private residence can be started. It is vital to compute the necessary boiler capacity and other crucial factors before setting up a heating system. Because it is done using a fairly straightforward formula, calculating the heating for a private home won’t be difficult for anyone, even if they live far from the heat supply problems. To calculate the temperature of the room, just multiply its area by the unit’s power and divide the result by ten.

With the information provided about the room sizes, this formula can be used to determine the necessary boiler output.

Important: All rooms with at least one external wall in contact with the outside environment must be included when calculating the total area of rooms, in addition to those where radiators will be installed.

That is, you must add the areas of the rooms with exterior walls and a small power reserve to the result in order to calculate the heating system. A correction for climate peculiarities is the second parameter required for computations. It is determined by taking into account the region and the climate zone that the heated house is situated in. The climate power factor will therefore be between 1.3 and 1.6 kW for the central regions with relatively mild winters, even less for the southern regions (between 0.8 and 0.95 kW), and between 1.6 and 2.2 kW for the northern regions.

You can do the calculation if you know the area of every room with exterior walls and the coefficient of climatic power. Assume for the moment that our home has 100 square meters of total floor space and is situated in a region with a moderate climate:

= 13 kW × 100 × 1.3 / 10

Thus, a boiler with a 15–16 kW capacity is required. A small power reserve is set aside in case the house’s area grows as a result of additions or during particularly "severe" winters.

You can always choose a boiler by getting in touch with the managers of the company "Teplodar" if you have any doubts about the accuracy of the calculations. Just indicate the room’s dimensions, fuel type, and any extra features, and the expert will choose the options best suited to your needs. Another way to narrow the options is by price.

How to calculate heating in the house: formulas and tips

Our article’s focus is on calculating a private home’s heating system. It makes sense that during the dead of winter, frosts would prefer not to deal with the absence of heat; it is also unpleasant to waste money by purchasing more equipment than you actually need. Now let’s begin the calculation.

Our objective is to supply heat to the house. Ideally at the lowest possible cost.

Radiators of what material is better to choose?

The cost of setting up the home’s heating circuit as well as the structural features of the heating system are dependent on the materials used in the production of the heating battery.

1. The most affordable option is batteries made of steel. They are cheap, but they have a small capacity, so they do not cope well with the heating of spacious rooms.
2. Cast iron radiators are durable and reliable in operation. Besides, they serve as interior decorations due to their aesthetic appearance. Cast iron radiators are an excellent choice if your house has brick walls. But the walls of a wooden or cinder-block structure may not be able to cope with the load: such radiators are very heavy.
3. You can also find aluminum and bimetallic radiators on sale. Aluminum radiators are not the best option in apartment buildings, as they are prone to premature wear and tear due to the poor quality of the coolant in the system. But in a country house such radiators will serve for a long time. The main thing is to use only clean water.
4. When buying a radiator should pay attention to anodized models, which have increased protection against corrosion, such radiators are more expensive, but have a longer service life. Service life can reach 30 years, which means that you do not have to spend on new batteries and repair work in the near future.

A large selection of radiators in various models will not only enable you to select a heating appliance that best complements the interior design of the room, but also one with the ideal number of sections.

Which heating system of a private house is better and why

The structural components of an autonomous heating system for a private home consist of a boiler, radiators, and a closed circular pipeline that transports coolant (apart from air). The following types of heating are differentiated based on the type of coolant:

Heat carrier	Advantages	Disadvantages
1. Water (water or antifreeze is used)	Economy, availability of heat carrier, its cheapness and safety of the system.	Premises are heated for quite a long time. In winter, it is impossible to allow neither planned nor emergency shutdown of the system with water, because at minus temperatures will burst the pipes.
2. Steam	Low inertia (rooms are heated immediately after switching on), energy efficiency.	Noisiness, difficulties in regulating the temperature in the room, the need to cover pipes and radiators, high requirements to the quality of pipes and radiators.
3. Air	High efficiency, no costs for pipes and radiators, low inertia. It is an ideal option for summer houses.	Air drying, there are difficulties with air supply (warm air rises up, but the temperature below remains cold).

Fuel types differ in boilers. One could debate for a very long time about the best type of heating for a private residence, weighing the pros and cons of each option. We suggest looking at a comparative table to better summarize the data and present the information.

Heat carrier	Advantages	Disadvantages
1. Gas	Comfortable operation (fully automatic system), a wide choice of boilers (single and double-circuit, wall and floor, convection and condensing), low operating costs, high efficiency, durability.	Limited availability (not everywhere has gas supply), complexity of system installation, need for design and paperwork, high level of danger (leakage cannot be excluded), maintenance costs.
2. Electric	Availability of heat source, low cost of equipment and installation, no chimney and environmental friendliness, economical, comfort in operation, safety, high efficiency.	There is always a possibility of power outages (it is desirable to have an alternative source of heating), it is necessary to comply with the requirements for the power grid, the cost of electricity in some regions of Russia is quite high.
3. Solid fuel	Low cost of energy carrier, a wide choice of fuels (coal, wood, pellets, briquettes), availability of fuel in any region of Russia.	Necessity to load fuel manually, low efficiency, costs for cleaning and maintenance of the boiler and chimney, there must be a room for fuel storage.
4. Liquid-fuel	Low cost of fuel, can operate on diesel, fuel oil, waste oil, autonomy of the system, good efficiency.	Needs a separate boiler room with a fuel storage tank, combustion products may enter the room (depends on the boiler and the project), needs regular maintenance and cleaning.
5. Combined	Universality. Cost-effectiveness and the ability to use the most profitable and practical energy carrier, quick payback. It is possible to choose a single-circuit or double-circuit boiler, to connect a boiler or a floor heating system.	Bulky boiler, a technically complex unit with a lot of additional equipment. High cost of the system and installation.

Calculation of the number of heater sections

If you do not determine the ideal number of radiator sections, the heating system will not operate effectively. An inaccurate calculation will result in the boiler operating at maximum capacity, causing the rooms to heat unevenly or, conversely, "idling" and wasting fuel.

A few homeowners think that having more batteries is preferable. This increases the coolant’s path, which causes it to progressively cool down. As a result, there is a chance that the system’s final rooms won’t receive any heat. This issue can be partially resolved by forcing the coolant to circulate. However, don’t ignore the boiler’s capacity as it might not be able to "pull" the system.

You will require the following values in order to determine the number of sections:

• the area of the heated room (plus the adjacent room without radiators);
• capacity of one radiator (indicated in the technical data);

Consider that, for the average strip of Russia, 100 W of power will be needed per square meter of living area (per SNiPa requirements).

The area of the room is multiplied by 100, and the product is divided by the radiator’s power specifications that need to be installed.

An example for a 25 square meter room with a 120 watt radiator would be (20×100)/185=10,8=11.

This is the simplest formula; alternative values are used when a room’s height is not standard or when its configuration is complex.

In the event that the radiator’s capacity is unknown, how can one accurately compute the amount of heating in a private residence? The 200 W average static power is used by default. For some types of radiators, average values can be obtained. This value is – 185 W for bimetallic and – 190 W for aluminum. The value is substantially less for cast-iron radiators—120 W.

A good tip is to adjust the number of radiator sections by leaving a small amount as a buffer to account for unanticipated heat loss.

The result obtained can be safely multiplied by the coefficient 1.2 if the calculation is done for corner rooms.

expansion tank

Additionally, there are two calculation methods available in this instance: accurate and simple.

simple scheme

A straightforward computation is nearly impossible: the expansion tank’s volume is assumed to be one-tenth that of the coolant in the circuit.

Where can I obtain the heat carrier’s volume value?

A few easy fixes are as follows:

1. Fill the circuit with water, bleed off the air, and then drain all the water through the vent into any measuring cup.
2. In addition, roughly the volume of a balanced system can be calculated from the calculation of 15 liters of coolant per kilowatt of boiler output. So, in the case of a 45 KW boiler in the system will be approximately 45*15=675 liters of coolant.

Thus, in this instance, an 80-liter expansion tank for a heating system will be a reasonable minimum (rounded up to the nearest standard value).

Standard expansion tank volumes.

Exact scheme

With greater accuracy, you can use your hands to calculate the expansion tank’s volume using the formula V = (Vt x E)/D, which is:

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

Comments are obviously needed for the final two parameters.

When heated from an initial temperature of +10 C, the expansion coefficient of water and poor water-glycol mixes can be obtained from the following table:

And these are the coefficients for high-glycol heat transfer fluids.

The following formula can be used to determine a cistern’s efficiency factor: D = (Pv – Ps) / (Pv + 1)

• Pv – maximum pressure in the circuit (pressure at which the safety valve operates).

As a hint, the standard measurement is 2.5 kgf/cm2.

• Ps- static pressure of the circuit (the same as tank charging pressure). It is calculated as 1/10th of the difference in meters between the level of the tank and the top point of the circuit (overpressure of 1 kgf/cm2 raises the water column by 10 meters). Pressure equal to Ps is created in the air chamber of the tank before filling the system.

For illustration, let us determine the cistern’s requirements under the subsequent scenarios:

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

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

The coefficient can be obtained directly from the table, saving computation time.

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

Calculator for calculating the required heat output for space heating

Explanation of calculations

Take the data and enter it sequentially into the calculator’s fields.

• The first thing to determine the climatic features – indicating the approximate minimum temperature inherent in the region of residence in the coldest decade of winter. Naturally, we are talking about the normal temperature for the region, and not about any "records" in one direction or another.

It is evident, by the way, that this field will remain the same when the calculations are made for every room in the house. There is room for variation in the other fields.

• Next comes a group of two fields, in which the area of the room (exactly) and the height of the ceilings (selection from a list) are indicated.

• The next group of data takes into account the peculiarities of the location of the room:

– The number of external walls, or those that face the street (choose from a list with values ranging from 0 to 3).

The outer wall’s position with respect to the light-facing side. The sun’s rays periodically charge certain walls with thermal energy. However, the north wall, for instance, receives no sunlight at all.

It can also be considered if there is a stable wind rose (or other predominant winter wind direction) in the vicinity of the house. State whether the external wall is parallel to the direction of the wind, leeward, or windward. In the event that no such data is available, we leave it at default and the software calculates as if there are the worst possible circumstances.

– The degree of wall insulation is also indicated. chosen from a trio of alternatives. More specifically, of the two, because it is completely absurd to turn on the heating in a home with walls that are typically not insulated.

– What is "vertically" adjacent to the room, that is, what is above and below the room, is indicated by two similar fields. This will assist in calculating the quantity of heat lost through ceilings and floors.

• The next group concerns the windows in the room. Their number, size and type are important here, including the features of double-glazed windows. Based on the combination of these data, the program will work out a correction factor to the result of calculations.
• Finally, the amount of heat loss is seriously influenced by the presence of doors in the room that open to the street, to the balcony, to a cold entranceway, etc.п. If the doors are regularly used during the day, then any opening of them is accompanied by an influx of cold air. Clearly, this requires reimbursement in the form of additional heat output.

Now that all the data has been entered, "press the button." As a result, the user will receive the desired heat output value for a specific room right away.

As previously stated, the result for the entire house (apartment), expressed in kilowatts, will be obtained by adding up all of the values.

By the way, choose a heating boiler based on this value, taking it as the minimum. And when it comes time to figure out the actual financial costs of running the heating system, this total value will be required.

We suggest that you study more in-depth information about choosing a heating boiler for a private residence as well as information about the most cost-effective fuel to use to heat the home.

Additionally, the information for every room is very helpful in selecting the right electric heater model or in determining the placement and design of the heating radiators.

It is essential to know how much heat your home requires to stay warm in order to maintain a cozy and economical living environment. You can learn more about the specific heating requirements of your house and make well-informed choices regarding insulation and heating systems by calculating this requirement yourself.

First and foremost, it’s critical to evaluate the elements affecting the heat loss in your house. The amount of heat your home needs to stay warm depends on a number of factors, including its size, insulation levels, local climate, and even the number of occupants.

After you have a firm grasp on these variables, you can start the computation. To calculate the overall heat loss, you usually need to estimate the heat loss for each room or space in your house and then add up all of the values. Energy efficiency experts can offer formulas or online calculators as helpful tools to streamline this process.

It is important to take into account both passive and active heating methods when determining how much heat your home needs. Enhancing insulation and caulking drafts are two examples of passive strategies that can dramatically lower heat loss and, in turn, the energy required to heat your house. Reducing energy use also requires taking proactive steps, like choosing a suitable heating system and adjusting its temperature to save energy.

It’s critical to periodically reevaluate your home’s heating needs, particularly when making big improvements like upgrading your heating system or adding insulation. You can make sure that your house stays cozy, economical to heat, and energy-efficient for many years to come by being watchful and proactive.

Video on the topic

What and how to heat your home properly. Mark Solonin"s technical school

Permitted input power of electricity kW to the house. How to find out?

How much does it cost to heat a house cheaply??/ CALCULATING THE COST OF ALL TYPES OF HEATING