Installing a heating system in your home or even an apartment in the city is a very responsible undertaking. Purchasing boiler equipment "by eye," that is, without considering all of the housing’s characteristics, will be wholly unreasonable at the same time. It is quite possible to go in one of two directions: either the boiler’s capacity will be insufficient, causing the machinery to operate "to the fullest," without interruption, but without producing the desired outcome, or, on the other hand, the equipment will be purchased by an extremely costly device, the capabilities of which will go entirely unutilized.

Heating calculation based on room area

That’s not all, though. Acquiring the appropriate heating boiler is not enough; it’s crucial to select and install heat transfer devices such as convectors, radiators, and "warm floors" in the appropriate locations within the rooms. Once more, following your gut feeling or the "sound counsel" of your neighbors is not the best course of action. In other words, you cannot do anything without precise calculations.

Of course, it is ideal for such thermal engineering to employ pertinent specialists, but doing so is frequently very expensive. But isn’t trying to do it yourself really not interesting? This publication will go into great detail about how the area of the premises is used to calculate heating, accounting for a number of significant subtleties. Through an analogy, one can determine the amount of heating in a private home. The calculator included on this page will assist with the necessary computations. Although the method cannot be deemed entirely "sinless," it does enable you to obtain the desired outcome with a level of accuracy that is entirely acceptable.

- The simplest calculation techniques
- Calculations of the necessary thermal power, taking into account the characteristics of the premises
- General principles and calculation formula
- Calculation calculator of the required thermal heating power by rooms
- Assessment of the degree of insulation of the element of the house and the required thickness of thermal insulation
- The general principle of calculation
- Calculator assessment of the need for additional insulation
- Video: an example of calculating a heating system using a special applied program

### The simplest calculation techniques

The heating system has two primary responsibilities that it must fulfill in order to provide cozy living conditions during the winter. These are highly conditional functions, and their separation is highly interdependent.

- The first is to maintain the optimal level of air temperature in the entire volume of the heated room. Of course, in height, the temperature level may change slightly, but this difference should not be significant. The average indicator of +20 ° C is considered quite comfortable conditions – this is precisely the temperature, as a rule, is taken as the initial in heat engineering calculations.

Stated differently, a specific amount of air should be able to be warmed by the heating system.

If we meet all the requirements exactly, GOST 30494-96 specifies the standards for the appropriate microclimate for each individual space in residential buildings. Exposure from the document is shown in the following table:

The purpose of the room | Air temperature, ° C | Relative humidity, % | Air speed, m/s | |||
---|---|---|---|---|---|---|

Optimal | permissible | Optimal | Permissible, max | Optimal, max | Permissible, max | |

For cold season | ||||||

Living room | 20 ÷ 22 | 18 ÷ 24 (20 ÷ 24) | 45 ÷ 30 | 60 | 0.15 | 0.2 |

The same, but for living rooms in regions with minimal temperatures from – 31 ° C and below | 21 ÷ 23 | 20 ÷ 24 (22 ÷ 24) | 45 ÷ 30 | 60 | 0.15 | 0.2 |

Kitchen | 19 ÷ 21 | 18 ÷ 26 | N/n | N/n | 0.15 | 0.2 |

Toilet | 19 ÷ 21 | 18 ÷ 26 | N/n | N/n | 0.15 | 0.2 |

Bathroom, combined bathroom | 24 ÷ 26 | 18 ÷ 26 | N/n | N/n | 0.15 | 0.2 |

Rest and training premises | 20 ÷ 22 | 18 ÷ 24 | 45 ÷ 30 | 60 | 0.15 | 0.2 |

Inter -apartment corridor | 18 ÷ 20 | 16 ÷ 22 | 45 ÷ 30 | 60 | N/n | N/n |

Lobby, staircase | 16 ÷ 18 | 14 ÷ 20 | N/n | N/n | N/n | N/n |

Pantries | 16 ÷ 18 | 12 ÷ 22 | N/n | N/n | N/n | N/n |

For the warm season (the standard only for residential premises. For the rest – not normalized) | ||||||

Living room | 22 ÷ 25 | 20 ÷ 28 | 60 ÷ 30 | 65 | 0.2 | 0.3 |

- Second – compensation of heat losses through the construction of the building.

Heat loss through building structures is the heating system’s most significant "opponent."

Unfortunately, the biggest "rival" to any heating system is heat loss. They can be minimized to some extent, but even the best thermal insulation won’t be able to totally eliminate them. Leaks of thermal energy can occur in any direction; the table below illustrates their general distribution:

The construction element of the building | Approximate value of heat loss |
---|---|

Foundation, floors on soil or over unheated basement (basement) rooms | 5 to 10% |

"Cold bridges" through poorly isolated joints of building structures | 5 to 10% |

Places of input of engineering communications (sewage, water supply, gas pipes, electric cables and t.P.) | up to 5% |

External walls, depending on the degree of insulation | from 20 to 30% |

Low -quality windows and external doors | about 20 ÷ 25%, of which about 10% – through leaky joints between boxes and the wall, and due to ventilation |

Roof | up to 20% |

Ventilation and chimney | up to 25 ÷ 30% |

Naturally, the heating system needs to have a certain amount of heat capacity in order to handle such tasks. This capacity needs to match not only the building’s (or apartment’s) overall needs, but also the proper distribution of the system’s heat throughout the premises based on their area and several other crucial considerations.

The computation and process are typically done "from small to large." To put it simply, each heated room’s required amount of thermal energy is calculated, the results are summarized, and then 10% of the reserve is added (to prevent the equipment from operating at its maximum capacity). The outcome will indicate how much power the heating boiler needs. And the numbers for every room will serve as a basis for figuring out how many radiators are needed.

In an informal setting, the simplest and most widely applied technique is to accept the standard of 100 W of thermal energy per square meter of area:

A 100 W/m2 ratio is the most basic method of counting.

Q is equal to S × 100.

Q is the amount of thermal power required in the space;

S is the room’s area (square meters);

100 is the specific power (W/m2) per unit area.

For instance, the 3.2 x 5.5 m room

S equals 3.2 × 5.5, or 17.6 m^.

Q is equal to 17.6 × 100 = 1760 W − 1.8 kW.

The process is obviously very straightforward but very flawed. Noting right away that it is only conditionally applicable with a standard ceiling height of about 2.7 m (acceptable – in the range from 2.5 to 3.0 m) is important. From this vantage point, the room’s volume rather than its area will yield a more accurate calculation.

Thermal power calculation using the room’s volume

It is evident that the specific power value in this instance is intended for a cubic meter. For a house made of brick or other materials, it is equivalent to 34 W/m³, or 41 W/m³ for a reinforced concrete panel house.

S × h × 41 (or 34) = Q

H, or ceiling height (meters);

The specific power per unit volume (W/m³) is either 41 or 34.

As an illustration, consider the following room in a panel home where the ceiling height is 3.2 meters:

Q is equal to 17.6 × 3.2 × 41 = 2309 W − 2.3 kW.

The outcome is more accurate because it already accounts for the room’s entire linear dimensions as well as, to some extent, the characteristics of the walls.

However, he is still quite distant in reality; many subtleties lie "beyond the brackets." The next section of the publication will explain how to complete the calculations in a way that is more realistic.

Here is some information about bimetallic heating radiators that might be of interest to you.

### Calculations of the necessary thermal power, taking into account the characteristics of the premises

The calculation algorithms discussed above are useful for the initial “estate”, but they still should be relying on them with very great caution. Even a person who does not understand anything in construction heat engineering may probably seem doubtful the indicated average values – they cannot be equal, say, for the Krasnodar Territory and for the Arkhangelsk region. In addition, the room is the room: one is located on the corner of the house, that is, it has two outer walls, and the other on three sides is protected from heat loss by other rooms. In addition, the room can have one or more windows, both small and very overall, sometimes even a panoramic type. And the windows themselves can differ in manufacturing material and other design features. And this is far from a complete list – just such features are visible even by a “naked eye”.

To put it succinctly, there are many subtleties that affect how much heat a given room loses, so it is best to be thorough in your calculations rather than be lazy. I assure you that using the approach suggested in the article won’t be too tough to accomplish.

#### General principles and calculation formula

The same ratio—100 watts per square meter—will serve as the foundation for the computations. However, the formula itself is the only thing that is "surround" by a sizable number of different correction coefficients.

Q is equal to (s × 100) × a, b, c, d, f, g, h, i, j, k, l, and m.

The coefficients are represented by Latin letters that are chosen at random, arranged alphabetically, and have no bearing on any values that are commonly recognized in physics. Each coefficient’s value will be disclosed separately.

- "A" – a coefficient taking into account the number of external walls in a particular room.

Naturally, the area through which thermal losses occur increases with the size of the room’s exterior walls. Furthermore, angles indicate the presence of two or more external walls, which are particularly vulnerable locations for the creation of "cold bridges." The coefficient "A" will change this particular room feature.

The coefficient is assumed to be equal to:

– Interior external walls No.: A = 0.8;

– A = 1.0 for the outer wall one;

– Walls two outside: A = 1.2;

– Three external walls: A = 1.4.

- "B" is a coefficient taking into account the location of the external walls of the room relative to the cardinal points.

The location of the walls in relation to the cardinal points affects the amount of heat loss through them.

Solar energy has an impact on the building’s temperature balance even on the coldest winter days. It is normal for the south-facing side of the house to get some heating from the sun and heat loss through it.

However, the windows and walls that face north "never see." Even though it "grabs" the morning sunlight, the eastern portion of the house does not effectively receive any heating from it.

We present the coefficient "B" based on this:

– The room’s exterior walls face either north or east: B = 1.1;

– The room’s exterior walls face either the west or the south: B = 1.0.

- “C” is a coefficient taking into account the location of the room relative to the winter “wind rose”

Maybe homes that are situated in wind-sheltered areas don’t need to make this modification as much. However, occasionally the dominant winter winds can make "hard adjustments" to the building’s thermal equilibrium. Naturally, compared to the leeward, opposite side, the windward side—that is, the side that the wind has "substituted"—will lose a great deal more than the body.

Predominant winter winds can undergo significant changes.

The so-called "wind rose," a graphic scheme illustrating the predominant wind directions in the winter and summer, is compiled based on the findings of many years of meteorological observations in any region. The hydrometeorological service in your area has this information. Nevertheless, even in the absence of meteorologists, many locals are well aware of the general direction of the winds and which side of the house typically has the deepest snowdrifts.

If more accuracy is desired in the calculations, you can add the correction factor "C" to the formula, making it equal to:

– The house’s windward side: C = 1.2;

– The house’s leeward walls: C = 1.0;

– Wall situated perpendicular to wind direction: C = 1.1.

- "D" is a correction factor taking into account the features of the climatic conditions of the region to build a house

Naturally, the degree of winter temperature will have a significant impact on the quantity of heat loss through all of the building’s structures. It makes sense that the thermometer would "dance" in a particular range throughout the winter, but each region has an average indicator of the lowest temperatures found in the year’s five coldest days (which is typically January). Here is a map of Russia’s territory, for instance, where the colors correspond to approximate values.

Minimum January temperature chart on a map

Although it is generally simple to clarify this value with the local weather service, you could, in theory, concentrate on your own observations.

Thus, we accept the coefficient "D" in the following manner for ours, accounting for the regional climate features:

– D = 1.5 at -35 °C and below;

– D = 1.3 between -30 and -34 °C;

D = 1.2 between – 25 and – 29 degrees Celsius;

Between -20 and -24 degrees Celsius: D = 1.1;

Between -15 and -19 degrees Celsius: D = 1.0;

– D = 0.9 from -10 °C to -14 °C;

Not any colder: D = 0.7 at 10 °C.

- "E" – a coefficient taking into account the degree of insulation of the external walls.

The degree of insulation provided by all building components directly affects the building’s overall thermal loss value. Walls are one of the "leaders" in heat loss. Thus, the quality of their thermal insulation determines the value of the thermal power needed to keep the room at a comfortable temperature.

It matters a lot how well-insulated the exterior walls are.

For our computations, the coefficient’s value can be determined as follows:

– There is no insulation on the exterior walls: E is 1.27.

The average level of insulation, whether it be in the form of two-brick walls or surface thermal insulation from another heater: E is 1.0.

High-quality insulation was installed based on heat engineering calculations: E is equal to 0.85.

Recommendations for assessing the level of insulation in the walls and other building structures are provided below for this publication.

- The coefficient "F" – an amendment to the height of the ceilings

Particularly in private homes, ceilings can vary in height. As a result, this parameter will also change depending on the thermal power needed to heat a specific room in the same area.

Accepting the following values for the correction factor "f" won’t be a major mistake:

– Ceiling height: F = 1.0 up to 2.7 m;

– F = 1.05 for stream height between 2.8 and 3.0 m;

– Ceiling height: F = 1.1 for 3.1 to 3.5 m;

– 3.6 to 4.0 m ceiling heights: F = 1.15;

– Over 4.1 meters in ceiling height: F = 1.2.

- "G " – coefficient taking into account the type of floor or room located under the ceiling.

The floor is one of the main places where heat is lost, as was previously mentioned. Therefore, some modifications to the computation and to this feature of a specific room are required. One can interpret the correction factor "G" as follows:

– a cold floor over an unheated space, such as a basement or soil: G = 1.4;

– an insulated floor above an unheated space or along the ground: G = 1.2;

– The room below is heated: G = 1.0.

- "h " – coefficient taking into account the type of room located on top.

Since heated air always rises, a room with a cold ceiling will inevitably experience higher heat losses, necessitating an increase in the required thermal power. We present the "H" coefficient while accounting for the following aspect of the computed room:

– The top attic is the "cold" attic: h = 1.0;

– H = 0.9 indicates the location of the insulated attic or other insulated room on top.

– The top of any heated room is indicated by h = 0.8.

- "I " – coefficient taking into account the design features of the windows

One of the "main routes" for heat leaks is through windows. Naturally, the quality of the window structure itself plays a major role in this situation. The degree of thermal insulation of older wooden frames, which were once found in every home, is far less than that of contemporary multi-chamber systems with double-glazed windows.

It is evident, even in the absence of words, that these windows’ thermal insulation levels differ greatly.

However, there is some variation amongst PVZH-Okles. A double-glazed window with three chambers, for instance, will have significantly more "warm" air than one with just one chamber.

Thus, a certain coefficient "I" must be introduced while accounting for the kind of windows that are installed in the space:

– Ordinary double-glazed wooden windows: I = 1.27;

– Contemporary window systems featuring a double-glazed, single-chamber window: I = 1.0;

– Contemporary window systems featuring double-glazed windows with two or three chambers and argon filling: I = 0.85.

- "j " – correction factor for the total area of glazing room

It will not be possible to totally prevent heat loss through the windows, no matter how good they are. However, it makes sense that a small window and panoramic glazing that covers nearly the entire wall cannot be compared.

The overall heat loss is more significant the larger the glazing area.

Finding the ratio between the area of each window in the space and the space itself will require some work:

X equals ∑SOK / SP

∑SOK – the room’s total window area;

SP – The room’s area.

The correction factor "j" is calculated based on the value obtained:

– J = 0.8 → x = 0 ÷ 0.1;

– J = 0.9 because x = 0.11 ÷ 0.2;

J = 1.0 → x = 0.21 ÷ 0.3;

J = 1.1 → x = 0.31 ÷ 0.4;

– J = 1.2 → x = 0.41 ÷ 0.5;

- "k " – coefficient giving an amendment for the presence of the front door

An additional "loophole" for the cold is always the door outside or on an unheated balcony.

The thermal balance of the room can be automatically adjusted by the door leading to the street or the open balcony, as each opening results in a significant amount of cold air entering the space. It makes sense to consider its existence as a result; to do this, we will introduce the coefficient "K," which we will accept to be equal to:

Door-free: K = 1.0;

– One door on the balcony or outside: K = 1.3;

– K = 1.7 for two doors leading to the balcony or the street.

- "l " – possible amendments to the connection diagram of heating radiators

Even though this may seem like a minor inconvenience to some, why not consider the intended plan for connecting heating radiators right away? The truth is that when feeding and "return" pipes are inserted differently, their heat transfer and, consequently, their role in preserving a particular temperature balance in the space, alters quite noticeably.

Illustration | Type of the radiator insert | The value of the coefficient "L" |
---|---|---|

Diagonal connection: feed on top, “return” from below | l = 1.0 | |

Connection on one side: feed on top, “return” from below | l = 1.03 | |

Bilateral connection: both feed and “return” from below | l = 1.13 | |

Diagonal connection: feed from below, “return” from above | l = 1.25 | |

Connection on one side: feed from below, “return” from above | l = 1.28 | |

Unilateral connection, and feed and “return” from below | l = 1.28 |

- "m " – correction factor for the features of the installation site of heating radiators

And lastly, the final coefficient, which is also related to the characteristics of radiator connections for heating. It should be obvious that the battery will transfer heat as efficiently as possible if it is installed openly and does not obstruct any airflow from above or from the facade. Nevertheless, it is not always feasible to install radiators in this manner; more frequently, radiators are partially covered by window sills. There are other choices. Additionally, some owners conceal the heating priorities behind fully or partially decorative screens in an attempt to integrate them into the interior ensemble they have created. This has a substantial impact on thermal return.

By adding a unique coefficient "M" to the calculations, it is also possible to account for any "tags" that specify how and where the radiators will be mounted:

Illustration | Features of the installation of radiators | The value of the coefficient "M" |
---|---|---|

The radiator is located on the wall openly or does not overlap on top with the windowsill | m = 0.9 | |

The radiator on top is blocked by a windowsill or shelf | m = 1.0 | |

The radiator on top is blocked by a protruding wall niche | m = 1.07 | |

The radiator is covered with a windowsill (niche) from above, and with a decorative screen from the front part | m = 1.12 | |

The radiator is completely enclosed in a decorative casing | m = 1.2 |

Therefore, clarity is with the calculation formula. They say it’s too hard and laborious, so one of the readers will undoubtedly take his head off right away. On the other hand, if the case is approached methodically, it becomes simplified and lacks complexity altogether.

A competent homeowner will have a thorough graphic plan of their "possessions" that is often oriented toward the card and includes all relevant dimensions. It is simple to understand the region’s climatic characteristics. All that’s left to do is use a tape measure to measure each room and explain a few details. The hosts are the only ones who truly know the features of the house, such as the location of the entry doors, the "neighborhood on vertical" from above and below, and the purported or actual diagram showing how the heating radiators are installed.

It is advised that you create a work table right away and fill it with all the information required for each room. It will be filled in with the computation result. The calculations themselves, however, will facilitate the use of an integrated calculator that has all of the previously mentioned coefficients and ratios "laid down."

Naturally, you cannot include any missing data in the computation; however, in this scenario, the default calculator will compute the outcome by accounting for the least favorable circumstances.

Can be illustrated with an example. We have a house plan, which we took entirely at random.

For instance, a residential building’s entirely arbitrary plan was taken

Area where the lowest temperatures are between -20 and 25 °C. Northeastern winds are predominant during the winter. The home has an insulated attic and is one story. ground level floors with insulation. The radiators that will be positioned beneath the window sills will have the best diagonal connection possible.

We create a table of this kind:

Room, its area, ceiling height. Floor insulation and "neighborhood" up and down | The number of external walls and their main location relative to the cardinal points and the "wind roses". The degree of wall insulation | Quantity, type and size of windows | The presence of entrance doors (per street or on the balcony) | Required thermal capacity (taking into account 10% of the reserve) |
---|---|---|---|---|

Area 78.5 m² | 10.87 kW ≈ 11 kW | |||

1. Hallway. 3.18 m². Ceiling 2.8 m. Dedicated floor on the ground. On top – insulated attic. | One, south, average degree of insulation. The leeward side | No | One | 0.52 kW |

2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. On top – insulated attic | No | No | No | 0.62 kW |

3. The kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well -insulated floor on the ground. Fly – insulated attic | Two. South, west. The average degree of insulation. The leeward side | Two, single -chamber double -glazed window, 1200 × 900 mm | No | 2.22 kW |

4. Children"s room. 18.3 m². Ceiling 2.8 m. Well -insulated floor on the ground. On top – insulated attic | Two, north – West. High degree of insulation. Wind -up | Two, two -chamber double -glazed window, 1400 × 1000 mm | No | 2.6 kW |

5. Sleeping. 13.8 m². Ceiling 2.8 m. Well -insulated floor on the ground. On top – insulated attic | Two, north, east. High degree of insulation. The windward side | One, two -chamber double -glazed window, 1400 × 1000 mm | No | 1.73 kW |

6. Living room. 18.0 m². Ceiling 2.8 m. Well -insulated floor. Above is a dumb attic | Two, east, south. High degree of insulation. In parallel to the direction of the wind | Four, two -chamber double -glazed window, 1500 × 1200 mm | No | 2.59 kW |

7. The bathroom is combined. 4.12 m². Ceiling 2.8 m. Well -insulated floor. Above is a dumb attic. | One, north. High degree of insulation. The windward side | One. Double glazing wooden frame. 400 × 500 mm | No | 0.59 kW |

Total: |

Next, using the calculator for each room that is positioned beneath the calculator (which is already deducting 10% from the reserve). This won’t take long to complete with the suggested application. Subsequently, the obtained values for each room must be smoked; this will determine the required total power of the heating system.

To help determine the appropriate number of heating radiators, the results for each room will, incidentally, be divided into the specific thermal power of one section and round the maximum.

#### Calculation calculator of the required thermal heating power by rooms

Agree that the calculated results may differ significantly from those that would result from the higher ratio of 1 m² mentioned above, particularly if you take the premises into account separately.

By the way, the calculator allows the owners to "play" around a little bit with the initial data to see how it affects the outcome. Maybe this will make it easier to spot "weaknesses" and provide a strange urge to take action to ensure that the house is as insulated as possible. The savings on the heating system will quickly offset the costs of high-quality thermal insulation.

The calculation of the thermal power of the heating system above could raise concerns about the plan’s lack of clarity regarding the requirements for wall insulation. You may agree, but the purpose of this is to make independent calculations simpler while maintaining a fully acceptable degree of error. Starting with the precise "canonical" computation of heat losses will make the algorithm unnecessarily complex and large, making it unmanageable for the majority of visitors.

As a helpful "bonus," however, a straightforward method for evaluating the thermotechnical properties of the walls and other building components will be provided. This will allow any owner to determine how well the walls are performing and how much more thermal insulation they still require.

### Assessment of the degree of insulation of the element of the house and the required thickness of thermal insulation

#### The general principle of calculation

The calculation’s guiding principle states that every residential building’s construction structure needs to have a specific normalized value of heat transfer resistance. Experts created these parameters, which are then reduced in the SNiP tables, individually for every region based on the characteristics of the local climate.

Since the tables are excessively lengthy, we propose using the map scheme below.

A map diagram displaying the normalized values of building structures’ heat transfer resistance

Note that the values of the walls, ceilings (floors or ceilings), and coatings (roofs) are indicated; different shades are used to highlight them.

The walls and other surrounding components of the house typically have multiple layers (this is not a rigid rule; it’s possible to have a single layer, but that would make calculations much simpler). Each layer has unique thermal resistance properties, and when combined, they will provide the final parameter.

The heat transfer resistance value for every single layer is:

RX is equal to hX / λX.

Thickness of layer in meters, or hX

ΛX is the layer material’s thermal conductivity value. It is simple to locate this tabular value for any building, finishing, or insulation material in reference books.

It is therefore simple to compute the total value of the heat transfer resistance and determine the extent to which it deviates from the normalized value when one is aware of the design features of the wall or other fence. The recommended thickness of the insulation will be determined by multiplying the resultant difference by the thermal conductivity of the chosen thermal insulation material, ensuring that the design adheres to the required parameters.

Diagram of a multi-layer enclosing structure simplified

The calculator that is suggested below can be used to calculate a multi-layer structure that includes the main layer (pos. 1), any insulation that may already be present (pos. 2), an internal layer (pos. 3), and external finishes (pos. 4). This calculator item is simply not filled if there are no layers in reality.

Note: Since their thermal resistance has little bearing on overall insulation, the external finishing layers of ventilated structures of the facade or roof (such as siding or roofing material) are not considered.

The calculator’s final paragraph will suggest which kind of insulation to use, and the computations’ output will show the suggested thickness for the thermal insulation layer.

#### Calculator assessment of the need for additional insulation

It will now be easy to determine the required thermal power for heating and to assess the level of insulation in their walls or other building components. You could do something similar like this: introduce all the values that are needed, and at the end, list the material that is being used as insulation, like mineral basalt cotton wool.

- If you get a result striving to zero (less than 10 mm thickness) or even a negative value, then the walls can be considered well -insulated.
- With the recommended thickness of insulation up to 75 ÷ 80 mm, it can be conditionally assumed that the walls have an average degree of insulation.
- In the event that the result is larger, and even worse – “rolls over” for 100 mm – trouble, the level of heat loss is very high, and the heating system will “devour” energy resources for no one’s “heat heating”. And in this case, the main efforts should be concentrated on ensuring reliable thermal insulation.

Of course, if you’d like, you can find more sophisticated, professional-level software online that can calculate the thermotechnical properties of the heating system. A video that demonstrates the steps involved in such a computation is one example. However, we reiterate that the suggested methodology is excellent for carrying out independent computations; the degree of error will be manageable. Click the link to learn more about a long-burning stove.

### Video: an example of calculating a heating system using a special applied program

You might want to know more about what a heating system bypass is.

Afanasyev Evgeny, Chief Editor

The publication’s author on February 11, 2016

Calculating your heating needs according to the size of each room is essential to maintaining a cozy home. You can obtain an accurate assessment of the heating needs and ensure comfort without going over budget by taking into account the size of each room. The square footage of each room is multiplied by a standard heating factor in this simple process. This factor takes the climate, insulation, and other variables that affect heat loss into account. Recall that precise computations result in effective heating, which will save you money and ensure that your house is comfortable all year round.

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