Average gas consumption for heating a house of 200 m2: calculation features, examples and recommendations

When designing a heating system for a home, owners are required to know in advance how much it will cost to heat their living space during operation. Ultimately, it may be that the heating system will simply be unprofitable if the equipment is chosen incorrectly, the design is poorly designed, or the walls are poorly insulated. Therefore, it is important to correctly calculate the gas consumption for heating a house of 200 m2. From the obtained figure you can already start and start or not start designing and purchasing equipment.

Calculation method for natural gas

The approximate gas consumption for heating is calculated based on half the power of the installed boiler. The thing is that when determining the power of a gas boiler, the lowest temperature is set. This is understandable - even when it is very cold outside, the house should be warm.

You can calculate gas consumption for heating yourself

But calculating gas consumption for heating using this maximum figure is completely incorrect - after all, the temperature is generally much higher, which means much less fuel is burned. That’s why it is generally accepted that the average fuel consumption for heating is about 50% of the heat loss or boiler power.

We calculate gas consumption by heat loss

If you don’t have a boiler yet, and you estimate the cost of heating in different ways, you can calculate it from the total heat loss of the building. They are most likely known to you. The technique here is this: they take 50% of the total heat loss, add 10% to provide hot water supply and 10% to remove heat during ventilation. As a result, we obtain the average consumption in kilowatts per hour.

Next, you can find out the fuel consumption per day (multiply by 24 hours), per month (by 30 days), and, if desired, for the entire heating season (multiply by the number of months during which the heating operates). All these figures can be converted into cubic meters (knowing the specific heat of combustion of gas), and then multiply the cubic meters by the price of gas and, thus, find out the heating costs.

Example of heat loss calculation

Let the heat loss of the house be 16 kW/hour. Let's start counting:

• average heat demand per hour - 8 kW/h + 1.6 kW/h + 1.6 kW/h = 11.2 kW/h;
• per day - 11.2 kW * 24 hours = 268.8 kW;
• per month - 268.8 kW * 30 days = 8064 kW.

The actual gas consumption for heating also depends on the type of burner - modulated are the most economical.
Convert to cubic meters. If we use natural gas, we divide the gas consumption for heating per hour: 11.2 kW/h / 9.3 kW = 1.2 m3/h. In calculations, the figure 9.3 kW is the specific heat capacity of natural gas combustion (available in the table).

By the way, you can also calculate the required amount of fuel of any type - you just need to take the heat capacity for the required fuel.

Since the boiler has not 100% efficiency, but 88-92%, you will have to make further adjustments for this - add about 10% of the obtained figure. In total, we get gas consumption for heating per hour - 1.32 cubic meters per hour. Next you can calculate:

• consumption per day: 1.32 m3 * 24 hours = 28.8 m3/day
• monthly demand: 28.8 m3/day * 30 days = 864 m3/month.

The average consumption for the heating season depends on its duration - multiply by the number of months while the heating season lasts.

This calculation is approximate. In some months, gas consumption will be much less, in the coldest month - more, but on average the figure will be about the same.

Boiler power calculation

The calculations will be a little simpler if you have the calculated boiler power - all the necessary reserves (for hot water supply and ventilation) have already been taken into account. Therefore, we simply take 50% of the calculated capacity and then calculate the consumption per day, month, per season.

For example, the design power of the boiler is 24 kW. To calculate gas consumption for heating, we take half: 12 k/W. This will be the average heat demand per hour. To determine fuel consumption per hour, we divide by the calorific value, we get 12 kW/hour / 9.3 k/W = 1.3 m3. Then everything is calculated as in the example above:

• per day: 12 kW/h * 24 hours = 288 kW in terms of the amount of gas - 1.3 m3 * 24 = 31.2 m3
• per month: 288 kW * 30 days = 8640 m3, consumption in cubic meters 31.2 m3 * 30 = 936 m3.

You can calculate gas consumption for heating a house based on the design capacity of the boiler.
Next, add 10% for the imperfection of the boiler, we find that for this case the consumption will be slightly more than 1000 cubic meters per month (1029.3 cubic meters). As you can see, in this case everything is even simpler - fewer numbers, but the principle is the same.

By quadrature

Even more approximate calculations can be obtained based on the square footage of the house. There are two ways:

• You can calculate according to SNiP standards - on average, heating one square meter in Central Russia requires 80 W/m2. This figure can be used if your house is built according to all requirements and has good insulation.
• You can estimate based on average statistical data:
• with good insulation of the house, 2.5-3 cubic meters/m2 is required;

with average insulation, gas consumption is 4-5 cubic meters/m2.

The better the house is insulated, the lower the gas consumption for heating will be.
Each owner can assess the degree of insulation of his house, accordingly, one can estimate what the gas consumption will be in this case. For example, for a house of 100 sq. m. with average insulation, 400-500 cubic meters of gas will be required for heating, for a house of 150 square meters it will take 600-750 cubic meters per month, for heating a house with an area of ​​200 m2 - 800-100 cubic meters of blue fuel. All this is very approximate, but the figures are derived based on many factual data.

Calculation methods with illustrative example

Boiler Zhukovsky AOGV

The amount of gas spent on heating a house depends mainly on the characteristics of the gas boiler and its operating conditions.

Therefore, for calculations you will need to know:

• technical parameters of the boiler;
• its power and efficiency;
• gas consumption specified in the technical passport;
• room area.

Our example will involve a gas boiler AOGV-17.4-3 (JSC Zhukovsky) with a capacity of 17.4 kW with an efficiency of 88%.

Natural gas consumption – 1.87 cubic meters/h, liquefied gas – 1.3 kg/h.

The boiler will heat up to 140 sq. meters of total premises area.

It is necessary to take into account that the values ​​​​indicated in the passport correspond to the continuous operation of the boiler at full power, but in fact the boiler operates 12-14 hours a day, so we will divide the calculated values ​​by two.

Tariffs and fuel prices

Let’s assume that the cost of the natural gas tariff is 3.9 rubles. for 1 cubic meter.

The cost of refilling a standard 50-liter cylinder with liquefied gas is 600 rubles. Such a cylinder is usually filled to 80% (42.5 l), which is about 21 kg of propane-butane mixture.

Accordingly, the price of 1 kilogram of liquefied gas will be equal to 600 / 21 = 28.6 rubles (excluding the cost of transporting the cylinder to the gas station and back).

According to the device datasheet

This is the simplest and most approximate method of calculation.

For natural gas, the consumption is 1.87 cubic meters per hour, hence:

For liquefied gas, the boiler consumption is 1.3 kg/h, hence:

Educational program on gas boilers using liquefied gas.

According to the specific heat of combustion of gas

The specific heat of combustion (calorific value) of gas depends on the type of natural fuel and the quality of the mixture. This value can be found in reference books on heating engineering.

For natural gas, the lowest value of specific heat of combustion is 34.02 MJ/cub.m or 9.45 kW/h of thermal energy. With a device efficiency of 88%, this figure will be adjusted to 9.5*0.88=8.3 kW/h.

How much does a gas boiler consume:

For liquefied gas, the specific heat of combustion is 50.38 MJ/kg or 13.99 kW/h. With an efficiency of 88%, this figure will be adjusted to 13.99*0.88=12.3 kW/h.

LPG boiler consumption:

The gas consumption of a double-circuit boiler is 20-25% higher (add this difference to the final amount).

When comparing the results, it is noticeable that if we calculate by the specific heat of combustion of gas, then the costs and cost of heating are higher. This happens because the minimum value of the specific heat of combustion is taken, which in fact may be higher.

In any case, the cost of heating with natural gas will be approximately 5 times cheaper than heating a room with bottled gas. However, liquefied fuel is still cheaper than autonomous electric heating.

At the same time, one should take into account the possibility of connecting to the gas pipeline and the cost of such a connection, which amounts to a fairly significant amount.

It is also worth noting that such calculations are very rough and approximate, since they do not take into account a number of circumstances that can significantly affect the final amount of expenses. In this case, the amount of calculations can vary significantly up or down

It is best to entrust an accurate calculation, taking into account all the circumstances, to heating engineering specialists.

Calculation of liquefied gas consumption

Many boilers can run on liquefied gas. How profitable is this? What will be the consumption of liquefied gas for heating? All this can also be calculated. The technique is the same: you need to know either the heat loss or the power of the boiler. Next, we convert the required quantity into liters (units of measurement of liquefied gas), and, if desired, we count the number of required cylinders.

Let's look at the calculation using an example. Let the boiler power be 18 kW, respectively, the average heat demand is 9 kW/hour. When burning 1 kg of liquefied gas, we obtain 12.5 kW of heat . This means that to get 9 kW, you will need 0.72 kg (9 kW / 12.5 kW = 0.72 kg).

• per day: 0.72 kg * 24 hours = 17.28 kg;
• per month 17.28 kg * 30 days = 518.4 kg.

Let's add a correction for the boiler efficiency. We need to look at each specific case, but let’s take 90%, that is, add another 10%, it turns out that the monthly consumption will be 570.24 kg.

Liquefied gas is one of the heating options

To calculate the number of cylinders, divide this figure by 21.2 kg (this is the average kg of gas in a 50 liter cylinder).

Mass of liquefied gas in various cylinders

In total, this boiler will require 27 liquefied gas cylinders. Calculate the cost yourself - prices vary in the regions. But don't forget about transportation costs. By the way, they can be reduced by making a gas holder - a sealed container for storing liquefied gas, which can be refilled once a month or less often - depending on the storage volume and demand.

Again, remember that this is just an approximate figure. In cold months, gas consumption for heating will be higher, in warm months it will be much less.

PS If it is more convenient for you to calculate the consumption in liters:

• 1 liter of liquefied gas weighs approximately 0.55 kg and, when burned, produces approximately 6500 kW of heat;
• There are about 42 liters of gas in a 50 liter cylinder.

Costs with an electric boiler

Now let's look at the capabilities of the electric boiler. In this case, you need to understand whether the allocated power limit is enough for you? Secondly, two-tariff electricity metering is already implemented almost everywhere.

Let's look at two options: single-tariff and two-tariff.

Single tariff option

Accordingly, the cost of 1 kW/h of heat from an electric boiler is 4.04 rubles. In this case, savings can be achieved by using automation so as not to overheat the room needlessly.

Two-tariff option

In the two-tariff version, together with an electric boiler, a storage tank of the appropriate volume is installed. In this case, the boiler mainly operates at night, at maximum power, at the night tariff.

It heats the room, and excess heat goes into the storage tank. Then, during the day, the room is heated using the heat accumulated at night at the night tariff.

Sometimes this heat is not enough and the boiler does not heat the house, sometimes this heat is produced in excess, so we take into account that heating occurs only at the night rate. The cost of 1 kW/h of electricity in the Moscow region for 2022 (at night tariff) is 1.26 rubles

The cost of 1 kW/h of electricity in the Moscow region for 2022 (at night tariff) is 1.26 rubles.

The efficiency of an electric boiler is 1. In this case, the cost of 1 kW/h of heat from an electric boiler is 1.26 rubles.

That is, the cost will be a similar figure to the cost of electricity at the night tariff and will be equal to 1.26 rubles.

How to reduce gas consumption

A well-known rule: the better the house is insulated, the less fuel is used to heat the street. Therefore, before starting the installation of the heating system, you should perform high-quality thermal insulation of the house - roof/attic, floors, walls, replacement of windows, airtight sealing loop on the doors.

You can also save fuel due to the heating system itself. By using heated floors instead of radiators, you will get more efficient heating: since heat spreads by convection currents from bottom to top, the lower the heating device is located, the better.

In addition, the standard temperature of floors is 50 degrees, and radiators are on average 90. Obviously, floors are more economical.

Finally, you can save gas by adjusting the heating according to time. There is no point in actively heating a house when it is empty. It is enough to maintain a low positive temperature so that the pipes do not freeze.

Modern boiler automation (types of automation for gas heating boilers) allows remote control: you can give a command to change the mode through a mobile provider before returning home (what are Gsm modules for heating boilers). At night, the comfortable temperature is slightly lower than during the day, etc.

A gas boiler

Having made a preliminary calculation of the fuel consumed, many choose the liquefied type and install a gas boiler in their home. This choice is determined by the following factors:

1. Long service life. Liquefied fuel does not destroy the material from which boilers are made, so they last for quite a long time.
2. Easy installation. If we compare the process of connecting to a gas pipeline and the process of installing individual gas equipment, the second is much easier and faster.
3. Economic benefit. Since heating the same room requires less liquefied gas than mains gas, its use will be more economical (especially in cases with large houses exceeding 200 m²).
4. Properties. Fuel in cylinders has better characteristics. When it burns, no hazardous substances are released. The minimum sulfur content guarantees high combustion efficiency.
5. Safety. Autonomous units are equipped with modern systems that prevent fuel leakage and explosion in the event of a breakdown.

A gas boiler is economical and safe.
With all its advantages, autonomous heating equipment is not without some disadvantages. These include the need to regularly refill cylinders, the influence of atmospheric pressure on the operation of the system, and the inability to operate during a power outage.

The main part of any autonomous gas heating system is a gas tank . This is a container into which liquefied gas is pumped. The amount of thermal energy released by the gas during combustion directly depends on the area for evaporation of the gas tank. This area inside the tank is called a mirror. If the 4.8 ton tank is fully filled, this volume will be enough for a period of about 200 days.

In addition to the size of the mirror, the volume of heat generated is affected by the design of the heating system, the number and area of ​​the radiators included in it, climatic conditions and other factors.

Calculation for liquefied gas

The above formula is also suitable for other types of fuel. Including for liquefied gas in cylinders for a gas boiler. Its calorific value, of course, is different. We accept this figure as 46 MJ per kilogram, i.e. 12.8 kilowatts per kilogram. Let's say the boiler efficiency is 92%. We substitute the numbers into the formula, we get 0.42 kilograms per hour.

Liquefied gas is counted in kilograms, which are then converted to liters. To calculate the gas consumption for heating a 100 m2 house from a gas holder, the figure obtained from the formula is divided by 0.54 (the weight of one liter of gas).

Calculation formula for main fuel

The power of the boiler affects gas consumption in a private home, so the owner must first calculate the fuel consumption, focusing on gas equipment. The area of ​​the cottage is also taken into account. Each room is calculated separately, based on the average annual street temperature and starting from the obtained minimum value.

To calculate gas consumption, the identified number must be divided by 2, since each winter season has different air temperatures. In the northern regions, it can reach serious disadvantages; accordingly, resource consumption will be higher than in the southern regions.

According to all the rules, it is calculated as follows: the ratio is 1 kW per 10 sq. m. area of ​​the house. Divide the resulting number in half, resulting in 50 watts for 60 minutes of heating. A house with an area of ​​100 m² will consume approximately 5 kW. Formula for calculating gas consumption in a private home (A=Q/q*B):

• A - cubic meter of resource consumption in 60 minutes;
• Q is the power of equipment for heating the house, in this case 5 kW;
• q - determines the minimum specific heat, it all depends on the type of fuel, for example, for G20 - 34.02 MJ per 1 cubic meter comes out to almost 10 kW;
• B is the efficiency of the equipment, for example, it works at 90%, which means they take the figure at 0.90.

All that remains is to substitute the required numbers instead of the letters in the formula. In this case, gas consumption per 100 sq. m. will be like this: 0.560 cubic meters per 60 minutes of boiler operation. For heating a room with an area of ​​150 sq. m. it will turn out to be 0.840 cubic meters and 7.5 kW. For suburban real estate, 1,115 cubic meters and approximately 10 kW will be spent on 200 m².

For a room of 250 m² it will be 1,392 cubic meters. A house of 300 m² will use 1,662 cubic meters. resource. The resulting number is multiplied by 24 hours and the consumption per day is determined, and then multiplied by 30 days.

Standard and real consumption

According to the standards, thermal calculations of premises are carried out for extreme conditions - the coldest winter five-day period. The boiler unit must have sufficient power to replenish the heat losses of the house in conditions of prolonged frosts characteristic of a particular climate.

As a result, we have the following situation:

• The boiler power is designed for maximum load;
• boiler equipment is selected with a small power reserve;
• When selecting the power of the unit, the gas consumption for cooking and the use of other equipment operating on this fuel is taken into account.

Accordingly, it makes no sense to calculate gas fuel consumption based on the rated power of the boiler unit. In real conditions, during the heating season (in the main territory of the country it lasts about 7 months), the temperature fluctuates over a significant range.

Attention! According to the above, in order to calculate the average gas consumption for heating, the design, and not the nameplate, power of the boiler should be divided by 2.

Example. for simplified calculations of coolant consumption, the standard value of thermal power is used: 1 kW per 10 m 2 of a private house. This means that the estimated power of the boiler unit for a house of 100 m2 will be 10 kW. Therefore, the thermal power indicator (Q), which we need to calculate the average fuel consumption, is 10/2 = 5 (kW).

Heating with main gas

To calculate gas fuel consumption, use the following formula

:

V = Q / (Hi × ηi)

. Where:

• V (m 3 /hour)
  – the volume of gas that needs to be consumed to obtain a certain amount of thermal energy;
• Q (kW)
  – design thermal power that allows you to maintain a comfortable temperature level in the house;
• Hi (kWh/m3)
  – indicator of the lowest specific heat of combustion of gas, standard tabular value (details below);
• ηi (%)
  – boiler unit efficiency, an indicator of how efficiently a gas boiler uses the generated thermal energy to heat the coolant.

Let's look at the specific heat of combustion of gas. Main networks mainly use G20 gas, but G25 gas can also be used; information can be obtained from your local gas supply organization. G25 gas has a higher nitrogen content, which reduces its energy potential.

Type of natural gas

Hi (lower specific calorific value)

Hs (highest specific heat of combustion)

32.49 MJ/m3 = 9.02 kWh/m3

In addition to the Hi

.
which we need for calculations, the table shows the Hs
- it is used when calculating fuel consumption for condensing boilers. This new generation equipment is more efficient due to the fact that during the steam condensation process, an additional 10% of thermal energy is removed.

Attention! Hi into the formula

in kWh/m3.

Boiler efficiency (ηi)

indicated in the product passport. If the document contains two indicators (for the lower and higher calorific value of gas fuel), use the smaller coefficient for calculations, since it more accurately reflects the real capabilities of the boiler.

Example. Let's calculate the average consumption of main gas G20 for a house with an area of ​​100 m2. We will assume that the house is insulated and the estimated thermal power is 9.6 kW, and the efficiency of the boiler unit is 0.92%.

As we already know, the calculated thermal power should be divided by 2, i.e. Q = 9.6/2 = 4.8 kW.

Thus: V = 4.8 / (9.45 × 0.92) ≈ 0.56 m 3 /hour.

Let's calculate the fuel consumption of G20:

• per day 0.56 × 24 = 13.44 m 3;
• per month (on average) 13.44 × 30.5 = 409.92 m 3;
• during the heating season (7 months) 409.92 × 7 = 2869.44 m 3.

To calculate the annual financial costs of heating, multiply the resulting value by the cost of one m 3 of main gas in your region.

Let's calculate the gas consumption for heating a house of 150 m2

. If the fuel is main gas G25, the efficiency of the boiler is 0.92, and to calculate the design power, a standard indicator of 1 kW per 10 m 2 was used. i.e. Q = 15/5 = 7.5 kW.

V = 7.5 / (8.13 × 0.92) = 1.002 m 3 /hour.

Let's round up to 1 m 3 / hour and calculate the annual consumption: 1 × 24 × 30.5 × 7 = 5124 m 3.

This calculation system helps to obtain an average value - on cold days the intensity of fuel consumption increases, on warm days it decreases relative to the average.

Costs for heating with liquefied gas

The consumption of liquefied gas for heating a house is calculated so that:

• understand the level of financial costs for purchasing fuel;
• determine the optimal size of the gas tank or count the number of gas cylinders, develop a suitable schedule for their delivery.

Calculations are carried out according to the same scheme as in the case of using main gas, but the volume of liquefied gas is measured in liters.

Liquefied hydrocarbon fuel G30, which is mainly used for autonomous gasification systems, is a propane-butane mixture with the following characteristics:

• fuel density 0.524 kg/l;
• specific heat of combustion 45.2 MJ/kg = 23.68 MJ/l = 6.58 kW/l.

Important! To fill the gas tank, fuel with different percentages of propane and butane is used (summer and winter), so carefully consider the choice of liquefied gas and take into account its characteristics when making calculations.

Let us apply the formula V = Q / (Hi × ηi) that is already familiar to us to calculate the volume of liquefied fuel required to heat a house with an area of ​​200 m2

.

We will assume that the calculated consumption corresponds to the standard (1 kW per 10 m2), i.e. Q = 20/2 = 10 kW. Boiler efficiency is 0.92%.

V = 10 / (6.58 × 0.92) = 1.65 l/hour.

Therefore, the approximate annual consumption will be: 1.65 × 24 × 30.5 × 7 = 8454.6 liters.

By adding to the resulting value the fuel consumption for a gas stove, etc. you can determine what size gas tank you need to choose in order to refill it 1-2 times a year.

If gas is supplied in cylinders, we can calculate the amount required for heating. The total volume of the cylinder is 50 liters, but they are not completely filled, so the volume of liquefied fuel is about 42 liters.

8454.6 / 42 = 201.3 cylinders for a house with an area of ​​200 m2 per heating season (7 months).

So, by substituting into the formula the values ​​corresponding to the parameters of your home, the characteristics of the fuel and the boiler unit, you can easily calculate the average gas consumption for heating.

Example for district heating

Note!

The initial data was taken for a gas mixture of grade G 20. It is this that enters houses from a centralized main.

In order to calculate gas consumption for heating, you need to use a fairly simple formula.

V= Q / (Hi x efficiency)

Where:

V – fuel consumption m3/h; Q is the estimated thermal power required to heat the house; Hi is the lowest value of specific heat during fuel combustion. In accordance with the DIN EN 437 standard, for G 20 fuel this value is 34.02 MJ/m3. Efficiency is the efficiency coefficient of the boiler unit, the value of which shows the efficiency of using the thermal energy released during the combustion of the gas mixture to heat the coolant.

Important!

Initial data: house area 100 m2, recommended heat generator power 10 kW. The efficiency of the boiler unit is 95%.

1. The first thing to do is convert joules to watts. To do this you need to know that 1 kW = 3.6 MJ. For G 20 gas, the calorific value will be 34.02/3.6 = 9.45 kW.
2. The value of 10 kW is the amount of heat that will be required to heat a house in the most unfavorable conditions. During the rest of the heating season, significantly less power is required to heat the house. Based on this, for correct calculations it is necessary to use half of the recommended power. In our case, half is 5 kW.

We substitute the obtained data: V = 5 / (9.45 x 0.95) Total: gas consumption for heating a house of 100 m2 is 0.557 m3/h. Based on the data obtained, it is easy to calculate the consumption of main gas per day and for the entire heating season, which in most regions of Russia lasts 7 months.

• In 24 hours 0.557 x 24 = 13.37 m3;
• For 30 days 13.37 x 30 = 401.1 m3;
• For 7 months (heating season) 401.1 x 7 = 2,807.7 m3.

Knowing the tariffs for paying for a cubic meter of “blue fuel”, you can quite accurately calculate the financial costs of heating for the billing period.

Stepping away from the topic, we would like to inform you that we have prepared comparative reviews on gas boilers. You can familiarize yourself with them in the following materials:

• Gas wall-mounted double-circuit boilers - comparison of models from world brands
• Rating of gas wall-mounted single-circuit boilers with a closed combustion chamber
• Russian-made gas floor-standing boilers: ATON, Concord, Lemax
• How to choose a gas floor-standing boiler based on reliability? Rating of proven models

How to save money?

The financial costs of maintaining a comfortable microclimate in the house can be reduced by

:

• additional insulation of all structures, installation of windows with double-glazed windows and door structures without cold bridges;
• installation of high-quality supply and exhaust ventilation (an incorrectly executed system can cause increased heat loss);
• use of alternative energy sources - solar panels, etc.

Separately, it is worth paying attention to the advantages of the collector heating system and automation, thanks to which the optimal temperature level is maintained in each of the rooms. This allows you to reduce the load on the boiler and fuel consumption when the weather warms up outside, and reduce the heating of the coolant that is supplied to radiators or a heated floor system in unused rooms.

If the house has a standard radiator system, a sheet of thin foam heat insulator with an outer foil surface can be glued to the wall behind each heating device. Such a screen effectively reflects heat, preventing it from escaping through the wall to the street.

A set of measures aimed at increasing the thermal efficiency of a home will help minimize energy costs.

How to avoid heat loss

Fuel consumption for heating a house depends on the total area of ​​the heated premises, as well as the heat loss coefficient. Any building loses heat through the roof, walls, window and door openings, and the floor of the lower floor.

Accordingly, the level of heat loss depends on the following factors

:

• climate features;
• wind roses and the location of the house relative to the cardinal directions;
• characteristics of the materials from which building structures and roofing are constructed;
• presence of a basement/ground floor;
• quality of insulation of floors, wall structures, attic floors and roofs;
• quantity and tightness of door and window structures.

Thermal calculation of the house allows you to select boiler equipment with optimal power parameters. In order to determine the heat requirement as accurately as possible, the calculation is performed for each heated room separately. For example, the heat loss coefficient is higher in rooms with two windows, in corner rooms, etc.

Note! The boiler power is selected with some margin relative to the calculated values ​​obtained. The boiler unit wears out faster and fails if it regularly operates at its maximum capacity. At the same time, excessive power reserves result in increased financial costs for the purchase of a boiler and increased fuel consumption.

Finding the amount of energy loss

In order to determine the amount of energy that a house loses, it is necessary to know the climatic characteristics of the area, the thermal conductivity of materials and ventilation standards. And to calculate the required volume of gas, it is enough to know its calorific value. The most important thing in this work is attention to detail.

Heating a building must compensate for heat losses that occur for two main reasons: heat leakage around the perimeter of the house and the influx of cold air through the ventilation system. Both of these processes are described by mathematical formulas, which you can use to carry out your own calculations.

Thermal conductivity and thermal resistance of the material

Any material can conduct heat. The intensity of its transmission is expressed by the thermal conductivity coefficient λ (W / (m × °C)). The lower it is, the better the structure is protected from freezing in winter.

Heating costs depend on the thermal conductivity of the material from which the house will be built. This is especially important for the “cold” regions of the country

However, buildings can be stacked or insulated with material of varying thicknesses. Therefore, in practical calculations, the heat transfer resistance coefficient is used:

R (m2 × °C / W)

It is related to thermal conductivity by the following formula:

R = h/λ,

where h is the thickness of the material (m).

Example. Let's determine the heat transfer resistance coefficient of aerated concrete blocks of grade D700 of different widths at λ = 0.16:

• width 300 mm: R = 0.3 / 0.16 = 1.88;
• width 400 mm: R = 0.4 / 0.16 = 2.50.

For insulating materials and window blocks, both the thermal conductivity coefficient and the heat transfer resistance coefficient can be given.

If the enclosing structure consists of several materials, then when determining the heat transfer resistance coefficient of the entire “pie,” the coefficients of its individual layers are summed up.

Example. The wall is built from aerated concrete blocks (λb = 0.16), 300 mm thick. From the outside it is insulated with extruded polystyrene foam (λp = 0.03) 50 mm thick, and from the inside it is lined with clapboard (λv = 0.18), 20 mm thick.

There are tables for various regions that indicate the minimum values ​​of the total heat transfer coefficient for the perimeter of the house. They are advisory in nature

Now you can calculate the total heat transfer resistance coefficient:

R = 0.3 / 0.16 + 0.05 / 0.03 + 0.02 / 0.18 = 1.88 + 1.66 + 0.11 = 3.65.

The contribution of layers that are insignificant in terms of the “heat saving” parameter can be neglected.

Calculation of heat loss through building envelopes

Heat loss Q (W) through a homogeneous surface can be calculated as follows:

Q = S × dT / R,

Where:

• S – area of ​​the surface under consideration (m2);
• dT – temperature difference between the air inside and outside the room (°C);
• R – coefficient of resistance to heat transfer of the surface (m2 * °C / W).

To determine the total indicator of all heat losses, perform the following steps:

1. select areas that are homogeneous in terms of heat transfer resistance coefficient;
2. calculate their areas;
3. determine thermal resistance indicators;
4. calculate heat loss for each section;
5. summarize the obtained values.

Example. Corner room 3 × 4 meters on the top floor with a cold attic space. The final ceiling height is 2.7 meters. There are 2 windows, measuring 1 × 1.5 m.

Let’s find the heat loss through the perimeter at an air temperature inside “+25 °С”, and outside – “–15 °С”:

1. Let us select areas that are homogeneous in terms of resistance coefficient: ceiling, wall, windows.
2. Ceiling area Sp = 3 × 4 = 12 m2. Window area So = 2 × (1 × 1.5) = 3 m2. Wall area Sc = (3 + 4) × 2.7 – So = 29.4 m2.
3. The coefficient of thermal resistance of the ceiling is composed of the ceiling (board 0.025 m thick), insulation (mineral wool slabs 0.10 m thick) and the wooden floor of the attic (wood and plywood with a total thickness of 0.05 m): Rп = 0.025 / 0.18 + 0.1 / 0.037 + 0.05 / 0.18 = 3.12. For windows, the value is taken from the passport of a double-glazed window: Ro = 0.50. For a wall built as in the previous example: Rс = 3.65.
4. Qp = 12 × 40 / 3.12 = 154 W. Qо = 3 × 40 / 0.50 = 240 W. Qс = 29.4 × 40 / 3.65 = 322 W.
5. The total heat loss of the model room through the enclosing structures is Q = Qп + Qо + Qс = 716 W.

Calculation using the above formulas gives a good approximation, provided that the material meets the declared thermal conductivity qualities and there are no errors that could be made during construction. The problem may also be the aging of materials and the structure of the house as a whole.

Typical wall and roof geometry

When determining heat loss, it is customary to take the linear parameters (length and height) of a structure internal rather than external. That is, when calculating heat transfer through a material, the contact area of ​​warm rather than cold air is taken into account.

When calculating the internal perimeter, it is necessary to take into account the thickness of the interior partitions. The easiest way to do this is using a house plan, which is usually drawn on paper with a scale grid.

Thus, for example, with house dimensions of 8 × 10 meters and a wall thickness of 0.3 meters, the internal perimeter Pin = (9.4 + 7.4) × 2 = 33.6 m, and the external perimeter Pex = (8 + 10) × 2 = 36 m.

The interfloor ceiling usually has a thickness of 0.20 to 0.30 m. Therefore, the height of the two floors from the floor of the first to the ceiling of the second from the outside will be equal to Hexternal = 2.7 + 0.2 + 2.7 = 5.6 m. If you add up only the final height, you will get a smaller value: Hinternal = 2.7 + 2.7 = 5.4 m. The interfloor ceiling, unlike the walls, does not serve as insulation, so for calculations you need to take Hext.

For two-story houses with dimensions of about 200 m2, the difference between the area of ​​the walls inside and outside is from 6 to 9%. Similarly, the internal dimensions take into account the geometric parameters of the roof and ceilings.

Calculating the wall area for cottages with simple geometry is elementary, since the fragments consist of rectangular sections and gables of attic and attic spaces.

The gables of attics and attics in most cases have the shape of a triangle or a vertically symmetrical pentagon. Calculating their area is quite simple

When calculating heat loss through a roof, in most cases it is enough to apply formulas for finding the areas of a triangle, rectangle and trapezoid.

The most popular forms of roofs of private houses. When measuring their parameters, you need to remember that internal dimensions are included in the calculations (without eaves overhangs)

The area of ​​the laid roof cannot be taken into account when determining heat loss, since it also goes to the overhangs, which are not taken into account in the formula. In addition, often the material (for example, roofing felt or profiled galvanized sheet) is placed with a slight overlap.

Sometimes it seems that calculating the roof area is quite difficult. However, inside the house the geometry of the insulated fencing of the upper floor can be much simpler

The rectangular geometry of the windows also does not cause problems in calculations. If the double-glazed windows have a complex shape, then their area can not be calculated, but can be found out from the product passport.

Heat loss through the floor and foundation

Calculation of heat loss into the ground through the floor of the lower floor, as well as through the walls and floor of the basement, is calculated according to the rules prescribed in Appendix “E” of SP 50.13330.2012. The fact is that the speed of heat propagation in the ground is much lower than in the atmosphere, so soils can also be conditionally classified as insulating materials.

But since they tend to freeze, the floor area is divided into 4 zones. The width of the first three is 2 meters, and the fourth includes the remaining part.

The heat loss zones of the floor and basement follow the shape of the foundation perimeter. The main heat loss will go through zone No. 1

For each zone, the heat transfer resistance coefficient added by the soil is determined:

• zone 1: R1 = 2.1;
• zone 2: R2 = 4.3;
• zone 3: R3 = 8.6;
• zone 4: R4 = 14.2.

If the floors are insulated, then to determine the overall coefficient of thermal resistance, the insulation and soil indicators are added.

Example. Let a house with external dimensions of 10 × 8 m and a wall thickness of 0.3 meters have a basement with a depth of 2.7 meters. Its ceiling is located at ground level. It is necessary to calculate heat loss into the ground at an internal air temperature of “+25 °C”, and an external air temperature of “-15 °C”.

Let the walls be made of FBS blocks, 40 cm thick (λf = 1.69). The inside is lined with boards 4 cm thick (λd = 0.18). The basement floor is filled with expanded clay concrete, 12 cm thick (λk = 0.70). Then the thermal resistance coefficient of the plinth walls: Rс = 0.4 / 1.69 + 0.04 / 0.18 = 0.46, and the floor Rп = 0.12 / 0.70 = 0.17.

The internal dimensions of the house will be 9.4 × 7.4 meters.

Scheme of dividing the basement into zones for the task being solved. Calculating areas with such simple geometry comes down to determining the sides of rectangles and multiplying them

Let's calculate the areas and heat transfer resistance coefficients by zone:

• Zone 1 only goes along the wall. It has a perimeter of 33.6 m and a height of 2 m. Therefore, S1 = 33.6 × 2 = 67.2. Rз1 = Rс + R1 = 0.46 + 2.1 = 2.56.
• Zone 2 along the wall. It has a perimeter of 33.6 m and a height of 0.7 m. Therefore, S2c = 33.6 × 0.7 = 23.52. Rз2с = Rс + R2 = 0.46 + 4.3 = 4.76.
• Zone 2 by floor. S2п = 9.4 × 7.4 – 6.8 × 4.8 = 36.92. Rз2п = Rп + R2 = 0.17 + 4.3 = 4.47.
• Zone 3 goes only on the floor. S3 = 6.8 × 4.8 – 2.8 × 0.8 = 30.4. Rз3 = Rп + R3 = 0.17 + 8.6 = 8.77.
• Zone 4 goes only on the floor. S4 = 2.8 × 0.8 = 2.24. Rз4 = Rп + R4 = 0.17 + 14.2 = 14.37.

Heat loss of the basement Q = (S1 / Rz1 + S2c / Rz2c + S2p / Rz2p + S3 / Rz3 + S4 / Rz4) × dT = (26.25 + 4.94 + 8.26 + 3.47 + 0.16) × 40 = 1723 W.

Accounting for unheated premises

Often, when calculating heat loss, a situation arises when the house has an unheated but insulated room. In this case, energy transfer occurs in two stages. Let's consider this situation using the example of an attic.

In an insulated but not heated attic space, during the cold period the temperature is set higher than outside. This occurs due to heat transfer through the interfloor ceiling

The main problem is that the floor area between the attic and the upper floor is different from the roof and gables. In this case, it is necessary to use the heat transfer balance condition Q1 = Q2.

It can also be written in the following way:

K1 × (T1 – T#) = K2 × (T# – T2),

Where:

• K1 = S1 / R1 + … + Sn / Rn for the overlap between the warm part of the house and the cold room;
• K2 = S1 / R1 + … + Sn / Rn for overlap between a cold room and the street.

From the equality of heat transfer, we find the temperature that will be established in a cold room at known values ​​in the house and outside. T# = (K1 × T1 + K2 × T2) / (K1 + K2). After this, we substitute the value into the formula and find the heat loss.

Example. Let the internal size of the house be 8 x 10 meters. Roof angle – 30°. The indoor air temperature is “+25 °C”, and outside – “-15 °C”.

We calculate the thermal resistance coefficient of the ceiling as in the example given in the section for calculating heat loss through building envelopes: Rп = 3.65. The overlap area is 80 m2, so K1 = 80 / 3.65 = 21.92.

Roof area S1 = (10 × / cos(30) = 92.38. We calculate the thermal resistance coefficient, taking into account the thickness of the wood (sheathing and finishing - 50 mm) and mineral wool (10 cm): R1 = 2.98.

Window area for pediment S2 = 1.5. For an ordinary two-chamber double-glazed window, the thermal resistance R2 = 0.4. We calculate the area of ​​the pediment using the formula: S3 = 82 × tg(30) / 4 – S2 = 7.74. The heat transfer resistance coefficient is the same as that of the roof: R3 = 2.98.

Heat loss through windows accounts for a significant portion of all energy losses. Therefore, in regions with cold winters, you should choose “warm” double-glazed windows

Let's calculate the coefficient for the roof (not forgetting that the number of gables is two):

K2 = S1 / R1 + 2 × (S2 / R2 + S3 / R3) = 92.38 / 2.98 + 2 × (1.5 / 0.4 + 7.74 / 2.98) = 43.69.

Let's calculate the air temperature in the attic:

T# = (21.92 × 25 + 43.69 × (–15)) / (21.92 + 43.69) = –1.64 °C.

Let's substitute the obtained value into any of the formulas for calculating heat loss (assuming they are equal in balance) and get the desired result:

Q1 = K1 × (T1 – T#) = 21.92 × (25 – (–1.64)) = 584 W.

Cooling through ventilation

A ventilation system is installed to maintain a normal microclimate in the house. This leads to the flow of cold air into the room, which also must be taken into account when calculating heat loss.

Requirements for the volume of ventilation are specified in several regulatory documents. When designing the intra-house system of a cottage, first of all, you need to take into account the requirements of §7 SNiP 41-01-2003 and §4 SanPiN 2.1.2.2645-10.

Since the generally accepted unit of measurement for heat loss is the watt, the heat capacity of air c (kJ / kg × °C) must be reduced to the dimension “W × h / kg × °C”. For air at sea level, we can take the value c = 0.28 W × h / kg × ° C.

Since the ventilation volume is measured in cubic meters per hour, it is also necessary to know the air density q (kg / m3). At normal atmospheric pressure and average humidity, this value can be taken as q = 1.30 kg/m3.

Household ventilation unit with recuperator. The declared volume that it passes is given with a small error. Therefore, it makes no sense to accurately calculate the density and heat capacity of air in the area down to hundredths.

Energy consumption to compensate for heat loss due to ventilation can be calculated using the following formula:

Q = L × q × c × dT = 0.364 × L × dT,

Where:

• L – air flow (m3/h);
• dT – temperature difference between room and incoming air (°C).

If cold air enters the house directly, then:

dT = T1 – T2,

Where:

• T1 – indoor temperature;
• T2 – outside temperature.

But for large facilities, a recuperator (heat exchanger) is usually integrated into the ventilation system. It allows you to significantly save energy resources, since partial heating of the incoming air occurs due to the temperature of the outlet flow.

The efficiency of such devices is measured in their efficiency k (%). In this case, the previous formula will take the form:

dT = (T1 – T2) × (1 – k / 100).

Calculation of gas consumption for heating a house of 150 m2

When arranging a heating system and choosing an energy source, it is important to find out the future gas consumption for heating a house of 150 m2 or other area. Indeed, in recent years, a clear trend towards an increase in prices for natural gas has been established; the last increase in price by approximately 8.5% occurred recently, on July 1, 2016. This led to a direct increase in heating costs in apartments and cottages with individual heat sources using blue fuel. That is why developers and homeowners who are just choosing a gas boiler should calculate heating costs in advance.

Save money by not heating all rooms equally

Not all rooms and volumes of the house need to be maintained at the same temperature. For example, storerooms, gyms, garages, workshops may have a slightly lower temperature, while children's rooms, showers or bathrooms may have a higher temperature.

To maintain the desired temperature in a particular room, you need to install regulators on each heating radiator. The principle of their operation is simple - they change the working cross-section of the heater pipe and reduce or increase the coolant circulation rate. It is enough to set the required temperature value on the regulator. This measure optimizes the total gas consumption for heating water in the boiler.

Initial data for calculations

To perform a preliminary calculation, you need to find out the following parameters:

• calorific value (calorific value) of natural gas supplied in your area;
• thermal load on the heating system;
• Efficiency of the boiler that is planned to be installed in a house or apartment.

The calorific value of the fuel is taken based on the value of the lower calorific value of the main gas.

Theoretically, when burning 1 m³ of blue fuel, 9.2 kW of thermal energy is released. In practice, this value differs and, as a rule, to a lesser extent. Due to the same rise in price, some unscrupulous suppliers dilute gas with air, which is why its calorific value can decrease to 7.5-8 kW/m³.

To determine gas consumption for heating a house, it is better to find out the caloric value from the management company, and when this fails, use a reserve figure: 8 kW/m³. If they share with you information about the specific heat of combustion and give you a figure expressed in other units, kcal/h, then you can convert it to Watts by multiplying by a factor of 1.163.

Another important indicator that directly affects fuel consumption is the thermal load on the heating system, which consists of heat losses through the building structures and losses due to heating of ventilation air. The best option is to perform or order an accurate calculation of all heat losses, but in the absence of any other option, you can determine the load using enlarged methods:

1. If the ceiling height does not exceed 3 m, then the heat consumption is assumed to be 0.1 kW per 1 m² of heated area of ​​the building. Thus, for a house of 100 m2 you need about 10 kW of heat, 150 m2 - 15 kW and 200 m2 - 20 kW of heat energy.
2. Apply 40-45 W of heat per 1 m³ of heated room volume. The load is determined by multiplying this value by the volume of all heated rooms.

The efficiency of the heat generator, which affects the efficiency of fuel combustion, is indicated in its technical data sheet. If the unit has not yet been purchased, then you can take the efficiency of gas boilers of various types from the list:

• gas convectors - 86%;
• boilers with an open combustion chamber - 88%;
• heat generators with a closed chamber - 92%;
• condensing boilers - 96%.

Example for a cottage of 200 m2

Let's calculate gas consumption for a cottage near Rostov-on-Don. Duration of the heating period: E = 171 × 24 = 4104 hours. Average outdoor temperature T2 = – 0.6 °C. Desired temperature in the house: T1 = 24 °C.

Two-story cottage with an unheated garage. The total area is about 200 m2. The walls are not additionally insulated, which is acceptable for the climate of the Rostov region

Step 1. Let's calculate heat loss through the perimeter without taking into account the garage.

To do this, we select homogeneous areas:

• Window. In total there are 9 windows measuring 1.6 × 1.8 m, one window measuring 1.0 × 1.8 m and 2.5 round windows with an area of ​​0.38 m2 each. Total window area: Swindows = 28.60 m2. According to the product data sheet, Rwindow = 0.55. Then Qwindows = 1279 W.
• Doors. There are 2 insulated doors measuring 0.9 x 2.0 m. Their area: S doors = 3.6 m2. According to the product data sheet, Rdoor = 1.45. Then Qdoors = 61 W.
• Blank wall. Plot “ABVGD”: 36.1 × 4.8 = 173.28 m2. Plot “YES”: 8.7 × 1.5 = 13.05 m2. Plot "DEZH": 18.06 m2. Roof gable area: 8.7 × 5.4 / 2 = 23.49. Total area of ​​the blank wall: Swall = 251.37 – Swindows – Sdoors = 219.17 m2. The walls are made of 40 cm thick aerated concrete and hollow facing bricks. Rwalls = 2.50 + 0.63 = 3.13. Then Qwalls = 1723 W.

Total heat loss through the perimeter:

Qperimeter = Qwindows + Qdoors + Qwalls = 3063 W.

Step 2. Calculate heat loss through the roof.

The insulation is solid lathing (35 mm), mineral wool (10 cm) and lining (15 mm). Rroofs = 2.98. Roof area above the main building: 2 × 10 × 5.55 = 111 m2, and above the boiler room: 2.7 × 4.47 = 12.07 m2. Total Sroofs = 123.07 m2. Then Qroof = 1016 W.

Step 3. Calculate heat loss through the floor.

The zones for the heated room and the garage must be calculated separately. The area can be determined accurately using mathematical formulas, or using vector editors such as Corel Draw

Resistance to heat transfer is provided by rough flooring boards and plywood under the laminate (5 cm in total), as well as basalt insulation (5 cm). Rfloor = 1.72. Then the heat loss through the floor will be equal to:

Qpol = (S1 / (Rpol + 2.1) + S2 / (Rpol + 4.3) + S3 / (Rpol + 2.1)) × dT = 546 W.

Step 4. Let's calculate the heat loss through the cold garage. Its floor is not insulated.

Heat penetrates from a heated house in two ways:

1. Through a load-bearing wall. S1 = 28.71, R1 = 3.13.
2. Through the brick partition with the boiler room. S2 = 11.31, R2 = 0.89.

We get K1 = S1 / R1 + S2 / R2 = 21.88.

Heat escapes from the garage to the outside as follows:

1. Through the window. S1 = 0.38, R1 = 0.55.
2. Through the gate. S2 = 6.25, R2 = 1.05.
3. Through the wall. S3 = 19.68, R3 = 3.13.
4. Through the roof. S4 = 23.89, R4 = 2.98.
5. Through the floor Zone 1. S5 = 17.50, R5 = 2.1.
6. Through the floor Zone 2. S6 = 9.10, R6 = 4.3.

We get K2 = S1 / R1 + … + S6 / R6 = 31.40

Let's calculate the temperature in the garage subject to the balance of heat transfer: T# = 9.2 °C. Then the heat loss will be equal: Qgarage = 324 W.

Step 5. Calculate heat loss due to ventilation.

Let the calculated volume of ventilation for such a cottage with 6 people living in it be equal to 440 m3/hour. The system has a recuperator with an efficiency of 50%. Under these conditions, heat loss: Qvent = 1970 W.

Step. 6. Let's determine the total heat loss by adding up all local values: Q = 6919 W.

Step 7. Let's calculate the volume of gas required to heat a model house in winter with a boiler efficiency of 92%:

• Natural gas. V = 3319 m3.
• Liquefied gas. V = 2450 kg.

After calculations, you can analyze the financial costs of heating and the feasibility of investments aimed at reducing heat loss.

Performing Calculations

A preliminary calculation of gas consumption for heating is made using the formula:

V = Q / (q x efficiency / 100).

• q is the caloric content of fuel, the default is 8 kW/m³;
• V is the required main gas flow rate, m³/h;
• Efficiency is the efficiency of fuel combustion by a heat source, expressed in %;
• Q is the heating load of a private house, kW.

As an example, we offer the calculation of gas consumption in a small cottage with an area of ​​150 m² with a heating load of 15 kW. It is planned that the heating task will be performed by a heating unit with a closed combustion chamber (efficiency 92%). The theoretical fuel consumption per 1 hour in the coldest period will be:

15 / (8 x 92 / 100) = 2.04 m³/h.

During the day, the heat generator will consume 2.04 x 24 = 48.96 m³ (rounded - 49 cubic meters) of natural gas - this is the maximum consumption on the coldest days. But during the heating season, the temperature can fluctuate between 30-40°C (depending on the region of residence), so the average daily gas consumption will be half as much, about 25 cubic meters.

Then, on average, per month a turbocharged boiler uses 25 x 30 = 750 m³ of fuel to heat a house with an area of ​​150 m², located in central Russia. Consumption for cottages of other sizes is calculated in the same way. Based on preliminary calculations, it is possible to carry out measures aimed at reducing consumption even at the construction stage: insulation, selection of more efficient equipment and the use of automatic control devices.

Using a propane-butane mixture

Autonomous heating of private houses with liquefied propane or its mixture with butane has not yet lost its relevance in the Russian Federation, although in recent years it has noticeably increased in price

It is all the more important to calculate the future consumption of this type of fuel for those homeowners who are planning such heating. The same formula is used for the calculation, only instead of the lower calorific value of natural gas, the parameter value for propane is set: 12.5 kW with 1 kg of fuel

The efficiency of heat generators when burning propane remains unchanged.

Below is an example of a calculation for the same building of 150 m², only heated with liquefied fuel. Its consumption will be:

• for 1 hour - 15 / (12.5 x 92 / 100) = 1.3 kg, per day - 31.2 kg;
• on average per day - 31.2 / 2 = 15.6 kg;
• on average per month - 15.6 x 30 = 468 kg.

When calculating the consumption of liquefied gas for heating a house, it is necessary to take into account that fuel is usually sold in volumetric measures: liters and cubic meters, and not by weight. This is how propane is measured when filling cylinders or a gas tank. This means that it is necessary to convert mass into volume, knowing that 1 liter of liquefied gas weighs about 0.53 kg. The result for the example above will be:

468 / 0.53 = 883 liters, or 0.88 m³, propane will have to be burned on average per month for a building with an area of ​​150 m².

Considering that the retail cost of liquefied gas is on average 16 rubles. for 1 liter, heating will cost a considerable amount, about 14 thousand rubles. per month for the same cottage on one and a half hundred square meters. There is reason to think about how best to insulate walls and take other measures aimed at reducing gas consumption.

Many homeowners expect to use fuel not only for heating, but also to provide hot water supply

These are additional costs, they must be calculated, plus it is important to take into account the additional load on the heating equipment

The thermal power required for hot water supply is easy to calculate. You need to determine the required volume of water per day and use the formula:

• c is the heat capacity of water, equal to 4.187 kJ/kg °C;
• t1—initial water temperature, °C;
• t2—final temperature of heated water, °C;
• m is the amount of water consumed, kg.

As a rule, economical heating occurs to a temperature of 55 °C, and this must be substituted into the formula. The initial temperature varies and lies in the range of 4-10 °C. For a day, a family of 4 people requires approximately 80-100 liters for all needs, provided that it is used sparingly. It is not necessary to convert the volume into mass measures, since in the case of water they are almost the same (1 kg = 1 l). It remains to substitute the obtained value of QDHW into the above formula and determine the additional gas consumption for DHW.

The fundamental parameter is the required thermal power of the heating system

So, we found out that when calculating, we start “dancing” from the actually required power of the heating system.

If you ask a search query on the Internet, how much thermal power is needed for heating, then in 90 cases out of 100 you will receive a very simple and easy-to-calculate answer - consider 1 kW of thermal energy for every 10 m² of heated premises.

Let me doubt the validity of this established dogma...

“Wow, how simple,” the reader will be delighted, but, unfortunately, he will be wrong. Why? – judge for yourself!

— For example, will an equal amount of heat be required to heat a 10-meter room in a house, say, built in the Azov region, and in the same one, but in Khanty-Mansiysk?

— Is there a difference in the required amount of heat for, for example, a corner room with a window facing north, and for the same area, but with one external wall, and even facing the sunny side?

— Are the premises in equal conditions, one of which has a cold basement below, and the second has a warm heated room?

- Where is there more heat loss - in a room with one small window or in one with almost panoramic glazing?

If desired, the list of such questions can be continued. But, probably, what has been said in excess is enough to understand the absurdity of such “combing with one brush.”

Moreover, even the connection diagram of heating radiators to the circuit and the features of their location on the wall can quite seriously influence the efficiency of their heat transfer and, accordingly, the required amount of heat for normal heating of the room.

Therefore, the best option seems to be this approach - for each heated room, an individual calculation of the required thermal power is carried out, taking into account all its specific features. And the subsequent summation of all indicators will give the required value of the required total power of the heating system of the house (or apartment).

It’s very difficult, many will probably think, it’s better to calculate approximately...

But why skimp on accuracy? We assure you that with our algorithm and the online calculator created on its basis, any homeowner will be able to calculate the heating system.

This algorithm is described in great detail in a special publication on our portal. There is also a calculation calculator with all the necessary explanations. So there is no point in repeating myself. Follow the link (the article will open in a new browser tab) and carry out the calculation. And then, with a ready-made value of the required total thermal power, come back here - we’ll start directly calculating the predicted gas consumption.

How can you quickly and accurately calculate the actually required thermal power of a heating system?

The calculation may seem cumbersome, but this is only at first glance. If the algorithm is understood by the reader (and the authors tried to “chew” it down to the smallest detail), then using an online calculator the calculations will not require much time and effort. With all the details, see the article on our portal “Calculation of heating by room area” .

By the way, do not rush to throw away the table compiled with power calculations for the premises of the house. It might come in handy. For example, when choosing the optimal models of heating radiators or even electrical heating appliances.

You may be interested in information about what are the best gas boilers for a private home, rating