Annual heat consumption for heating a country house

Thermal loads of the object

The calculation of thermal loads is carried out in the following sequence.

1. The total volume of buildings according to the external measurement: V=40000 m3.
2. The calculated internal temperature of heated buildings is: tvr = +18 C - for administrative buildings.
3. Estimated heat consumption for heating buildings:

4. Heat consumption for heating at any outdoor temperature is determined by the formula:

where: tvr is the temperature of the internal air, C; tn is the outside air temperature, C; tn0 is the coldest outdoor temperature during the heating period, C.

5. At the outside air temperature tн = 0С, we get:
6. At the outside air temperature tн= tнв = -2С, we get:
7. At the average outdoor air temperature for the heating period (at tn = tnsr.o = +3.2С) we get:
8. At the outside air temperature tн = +8С we get:
9. At the outside air temperature tн = -17С, we get:

10. Estimated heat consumption for ventilation:

where: qv is the specific heat consumption for ventilation, W/(m3 K), we accept qv = 0.21- for administrative buildings.

11. At any outdoor temperature, the heat consumption for ventilation is determined by the formula:

12. At the average outdoor air temperature for the heating period (at tn = tnsr.o = +3.2С) we get:
13. At outdoor air temperature = = 0С, we get:
14. At outdoor air temperature = = + 8C, we get:
15. At outdoor temperature ==-14C, we get:
16. At the outside air temperature tн = -17С, we get:

17. Average hourly heat consumption for hot water supply, kW:

where: m is the number of personnel, people; q - hot water consumption per employee per day, l/day (q = 120 l/day); c is the heat capacity of water, kJ/kg (c = 4.19 kJ/kg); tg is the temperature of hot water supply, C (tg = 60C); ti is the temperature of cold tap water in winter txz and summer tchl periods, С (txz = 5С, tхl = 15С);

- the average hourly heat consumption for hot water supply in winter will be:

— average hourly heat consumption for hot water supply in summer:

18. The results obtained are summarized in Table 2.2.
19. Based on the data obtained, we build the total hourly schedule of heat consumption for heating, ventilation and hot water supply of the facility:

; ; ; ;

20. On the basis of the obtained total hourly schedule of heat consumption, we build an annual schedule for the duration of the heat load.

Table 2.2 Dependence of heat consumption on outdoor temperature

Heat consumption	tnm= -17С	tno \u003d -14С	tnv=-2C	tn= 0С	tav.o \u003d + 3.2С	tnc = +8C
, MW	0,91	0,832	0,52	0,468	0,385	0,26
, MW	0,294	0,269	0,168	0,151	0,124	0,084
, MW	0,21	0,21	0,21	0,21	0,21	0,21
, MW	1,414	1,311	0,898	0,829	0,719	0,554
1,094	1,000	0,625	0,563	0,463	0,313

Annual heat consumption

To determine the heat consumption and its distribution by season (winter, summer), equipment operation modes and repair schedules, it is necessary to know the annual fuel consumption.

1. The annual heat consumption for heating and ventilation is calculated by the formula:

where: - average total heat consumption for heating during the heating period; — average total heat consumption for ventilation during the heating period, MW; - duration of the heating period.

2. Annual heat consumption for hot water supply:

where: - average total heat consumption for hot water supply, W; - the duration of the hot water supply system and the duration of the heating period, h (usually h); - coefficient of reduction of the hourly consumption of hot water for hot water supply in the summer; - respectively, the temperature of hot water and cold tap water in winter and summer, C.

3. Annual heat consumption for heat loads of heating, ventilation, hot water supply and technological load of enterprises according to the formula:

where: - annual heat consumption for heating, MW; — annual heat consumption for ventilation, MW; — annual heat consumption for hot water supply, MW; — annual heat consumption for technological needs, MW.

MWh/year.

What do you need to calculate

The so-called thermal calculation is carried out in several stages:

First you need to determine the heat loss of the building itself. Typically, heat losses are calculated for rooms that have at least one external wall. This indicator will help determine the power of the heating boiler and radiators.
Then the temperature regime is determined. Here it is necessary to take into account the relationship of three positions, or rather, three temperatures - the boiler, radiators and indoor air. The best option in the same sequence is 75C-65C-20C. It is the basis of the European standard EN 442.
Taking into account the heat loss of the room, the power of the heating batteries is determined.
The next step is hydraulic calculation. It is he who will allow you to accurately determine all the metric characteristics of the elements of the heating system - the diameter of pipes, fittings, valves, and so on. Plus, based on the calculation, an expansion tank and a circulation pump will be selected.
The power of the heating boiler is calculated.
And the last stage is the determination of the total volume of the heating system. That is, how much coolant is needed to fill it. By the way, the volume of the expansion tank will also be determined based on this indicator. We add that the volume of heating will help you find out if the volume (number of liters) of the expansion tank that is built into the heating boiler is enough, or you will have to purchase additional capacity.

By the way, about heat losses. There are certain norms that are set by experts as a standard. This indicator, or rather, the ratio, determines the future efficient operation of the entire heating system as a whole. This ratio is - 50/150 W / m². That is, the ratio of the power of the system and the heated area of \u200b\u200bthe room is used here.

Calculation formula

Thermal energy consumption standards

Thermal loads are calculated taking into account the power of the heating unit and the heat losses of the building. Therefore, in order to determine the capacity of the designed boiler, it is necessary to multiply the heat loss of the building by a multiplying factor of 1.2. This is a kind of margin equal to 20%.

Why is this ratio needed? With it, you can:

Predict the drop in gas pressure in the pipeline. After all, in winter there are more consumers, and everyone tries to take more fuel than the rest.
Vary the temperature inside the house.

We add that heat losses cannot be distributed evenly throughout the building structure. The difference in indicators can be quite large. Here are some examples:

Up to 40% of the heat leaves the building through the outer walls.
Through floors - up to 10%.
The same applies to the roof.
Through the ventilation system - up to 20%.
Through doors and windows - 10%.

So, we figured out the design of the building and made one very important conclusion that heat losses that need to be compensated depend on the architecture of the house itself and its location. But much is also determined by the materials of the walls, roof and floor, as well as the presence or absence of thermal insulation.

This is an important factor.

For example, let's determine the coefficients that reduce heat loss, depending on window structures:

Ordinary wooden windows with ordinary glass. To calculate the thermal energy in this case, a coefficient equal to 1.27 is used. That is, through this type of glazing, thermal energy leaks, equal to 27% of the total.
If plastic windows with double-glazed windows are installed, then a coefficient of 1.0 is used.
If plastic windows are installed from a six-chamber profile and with a three-chamber double-glazed window, then a coefficient of 0.85 is taken.

We go further, dealing with the windows. There is a certain relationship between the area of \u200b\u200bthe room and the area of \u200b\u200bwindow glazing. The larger the second position, the higher the heat loss of the building. And here there is a certain ratio:

If the window area in relation to the floor area has only a 10% indicator, then a coefficient of 0.8 is used to calculate the heat output of the heating system.
If the ratio is in the range of 10-19%, then a coefficient of 0.9 is applied.
At 20% - 1.0.
At 30% -2.
At 40% - 1.4.
At 50% - 1.5.

And that's just the windows. And there is also the effect of the materials that were used in the construction of the house on thermal loads.Let's arrange them in a table where wall materials will be located with a decrease in heat losses, which means that their coefficient will also decrease:

Type of building material

As you can see, the difference from the materials used is significant. Therefore, even at the stage of designing a house, it is necessary to determine exactly what material it will be built from. Of course, many developers build a house based on the budget allocated for construction. But with such layouts, it is worth reconsidering it. Experts assure that it is better to invest initially in order to later reap the benefits of savings from the operation of the house. Moreover, the heating system in winter is one of the main items of expenditure.

Room sizes and building heights

Heating system diagram

So, we continue to understand the coefficients that affect the formula for calculating heat. How does room size affect heat loads?

If the ceiling height in your house does not exceed 2.5 meters, then a factor of 1.0 is taken into account in the calculation.
At a height of 3 m, 1.05 is already taken. A slight difference, but it significantly affects heat loss if the total area of \u200b\u200bthe house is large enough.
At 3.5 m - 1.1.
At 4.5 m -2.

But such an indicator as the number of storeys of a building affects the heat loss of a room in different ways. Here it is necessary to take into account not only the number of floors, but also the location of the room, that is, on which floor it is located. For example, if this is a room on the first floor, and the house itself has three or four floors, then a coefficient of 0.82 is used for the calculation.

When moving the room to the upper floors, the rate of heat loss also increases. In addition, you will have to take into account the attic - is it insulated or not.

As you can see, in order to accurately calculate the heat loss of a building, it is necessary to determine various factors. And all of them must be taken into account. By the way, we have not considered all the factors that reduce or increase heat losses. But the calculation formula itself will mainly depend on the area of \u200b\u200bthe heated house and on the indicator, which is called the specific value of heat losses. By the way, in this formula it is standard and equal to 100 W / m². All other components of the formula are coefficients.

Thermal loads of heat supply systems

The concept of heat load defines the amount of heat that is given off by heating devices installed in a residential building or at an object for other purposes. Before installing the equipment, this calculation is performed in order to avoid unnecessary financial costs and other problems that may arise during the operation of the heating system.

Knowing the main operating parameters of the heat supply design, it is possible to organize the efficient functioning of heating devices. The calculation contributes to the implementation of the tasks facing the heating system, and the compliance of its elements with the norms and requirements prescribed in SNiP.

When the heat load for heating is calculated, even the slightest mistake can lead to big problems, because based on the data obtained, the local housing and communal services department approves limits and other consumption parameters that will become the basis for determining the cost of services.

The total amount of heat load on a modern heating system includes several basic parameters:

load on the heat supply structure;
load on the floor heating system, if it is planned to be installed in the house;
load on the natural and/or forced ventilation system;
load on the hot water supply system;
load associated with various technological needs.

Example of a simple calculation

For a building with standard parameters (ceiling heights, room sizes and good thermal insulation characteristics), a simple ratio of parameters can be applied, adjusted for a coefficient depending on the region.

Suppose that a residential building is located in the Arkhangelsk region, and its area is 170 square meters. m.The heat load will be equal to 17 * 1.6 = 27.2 kW / h.

Such a definition of thermal loads does not take into account many important factors. For example, the design features of the structure, temperature, the number of walls, the ratio of the areas of walls and window openings, etc. Therefore, such calculations are not suitable for serious heating system projects.

Other ways to calculate the amount of heat

It is possible to calculate the amount of heat entering the heating system in other ways.

The calculation formula for heating in this case may differ slightly from the above and have two options:

Q = ((V1 * (T1 - T2)) + (V1 - V2) * (T2 - T)) / 1000.
Q = ((V2 * (T1 - T2)) + (V1 - V2) * (T1 - T)) / 1000.

All values of the variables in these formulas are the same as before.

Based on this, it is safe to say that the calculation of kilowatts of heating can be done on your own. However, do not forget about consulting with special organizations responsible for supplying heat to dwellings, since their principles and calculation system can be completely different and consist of a completely different set of measures.

Having decided to design a so-called “warm floor” system in a private house, you need to be prepared for the fact that the procedure for calculating the volume of heat will be much more difficult, since in this case it is necessary to take into account not only the features of the heating circuit, but also provide for the parameters of the electrical network, from which and the floor will be heated. At the same time, the organizations responsible for monitoring such installation work will be completely different.

Many owners often face the problem of converting the required number of kilocalories into kilowatts, which is due to the use of many auxiliary aids of measuring units in the international system called "Ci". Here you need to remember that the coefficient that converts kilocalories to kilowatts will be 850, that is, in simpler terms, 1 kW is 850 kcal. This calculation procedure is much simpler, since it will not be difficult to calculate the required amount of gigacalories - the prefix "giga" means "million", therefore, 1 gigacalorie - 1 million calories.

In order to avoid errors in calculations, it is important to remember that absolutely all modern heat meters have some error, and often within acceptable limits. The calculation of such an error can also be done independently using the following formula: R = (V1 - V2) / (V1 + V2) * 100, where R is the error of the common house heating meter

V1 and V2 are the parameters of water consumption in the system already mentioned above, and 100 is the coefficient responsible for converting the obtained value into a percentage. In accordance with operating standards, the maximum allowable error can be 2%, but usually this figure in modern devices does not exceed 1%.

Computing

It is practically impossible to calculate the exact value of heat loss by an arbitrary building. However, methods of approximate calculations have long been developed, which give fairly accurate average results within the limits of statistics. These calculation schemes are often referred to as aggregated indicators (measurements) calculations.

The building site must be designed in such a way that the energy required for cooling is kept to a minimum. While residential buildings may be excluded from structural cooling energy demand because internal heat loss is minimal, the situation in the non-residential sector is somewhat different. In such buildings, the internal thermal gains that are needed for mechanical cooling are caused by differential masonry to the overall thermal gain. The workplace also needs to provide a hygienic air flow, which is largely enforced and adjustable.

Along with the thermal power, it often becomes necessary to calculate the daily, hourly, annual consumption of thermal energy or the average power consumption. How to do it? Let's give some examples.

The hourly heat consumption for heating according to enlarged meters is calculated by the formula Qot \u003d q * a * k * (tin-tno) * V, where:

Qot - the desired value for kilocalories.
q - specific heating value of the house in kcal / (m3 * C * hour). It is looked up in directories for each type of building.

Such drainage is also needed during the summer period to cool down due to the removal of heat from outside air and the requirement for eventual dehumidification. Shading in the form of overlays or horizontally dwelling elements is the method today, but the effect is limited to the time when the sun is high above the horizon. From this point of view, the most important method is to extinguish outdoor lifts, of course with regard to daylight.

Reducing the internal thermal benefits is somewhat problematic. This will also help reduce the need for artificial lighting. The performance of the personal computer is steadily increasing, but significant progress has been made in this area. The need for cooling is also represented by building structures capable of storing thermal energy. Such structures are especially heavy building structures such as. concrete floor or ceiling, which can also cause internal spur buildup, exterior walls or rooms.

a - ventilation correction factor (usually equal to 1.05 - 1.1).
k is the correction factor for the climatic zone (0.8 - 2.0 for different climatic zones).
tvn - internal temperature in the room (+18 - +22 C).
tno - outdoor temperature.
V is the volume of the building together with the enclosing structures.

To calculate the approximate annual heat consumption for heating in a building with a specific consumption of 125 kJ / (m2 * C * day) and an area of 100 m2, located in a climatic zone with a parameter GSOP = 6000, you just need to multiply 125 by 100 (house area ) and by 6000 (degree-days of the heating period). 125*100*6000=75000000 kJ or about 18 gigacalories or 20800 kilowatt-hours.

Also advantageous is the use of special materials with a phase shift at a suitable temperature. For light residential buildings without cooling, where the storage capacity is minimal, there are problems with maintaining temperature conditions during the summer months.

In terms of air conditioner design, but also the need for cooling energy, it will be necessary to use accurate, affordable calculation methods. In this regard, a particularly clear design of heat sinks can be predicted. As already mentioned, the need for cooling energy will be minimal in zero buildings. Some buildings cannot be cooled without cooling, and providing optimal parameters for the thermal comfort of workers, especially in office buildings, is now the standard.

To recalculate the annual consumption into the average heat consumption, it is enough to divide it by the length of the heating season in hours. If it lasts 200 days, the average heating power in the above case will be 20800/200/24=4.33 kW.

What it is

Definition

The definition of specific heat consumption is given in SP 23-101-2000. According to the document, this is the name of the amount of heat needed to maintain a normal temperature in the building, related to a unit of area or volume and to another parameter - degree-days of the heating period.

What is this setting used for? First of all - to assess the energy efficiency of the building (or, what is the same, the quality of its insulation) and planning heat costs.

Actually, SNiP 23-02-2003 directly states: the specific (per square or cubic meter) consumption of thermal energy for heating a building should not exceed the given values. The better the thermal insulation, the less energy heating requires.

Degree day

At least one of the terms used needs clarification. What is a degree day?

This concept directly refers to the amount of heat required to maintain a comfortable climate inside a heated room in winter. It is calculated by the formula GSOP=Dt*Z, where:

GSOP is the desired value;
Dt is the difference between the normalized internal temperature of the building (according to the current SNiP, it should be from +18 to +22 C) and the average temperature of the coldest five days of winter.
Z is the length of the heating season (in days).

As you might guess, the value of the parameter is determined by the climatic zone and for the territory of Russia it varies from 2000 (Crimea, Krasnodar Territory) to 12000 (Chukotka Autonomous Okrug, Yakutia).

Units

In what quantities is the parameter of interest measured?

In SNiP 23-02-2003, kJ / (m2 * C * day) and, in parallel with the first value, kJ / (m3 * C * day) are used.
Along with the kilojoule, other units of heat can be used - kilocalories (Kcal), gigacalories (Gcal) and kilowatt hours (KWh).

How are they related?

1 gigacalorie = 1,000,000 kilocalories.
1 gigacalorie = 4184000 kilojoules.
1 gigacalorie = 1162.2222 kilowatt-hours.

In the photo - a heat meter. Heat metering devices can use any of the listed units of measurement.

Heat meters

Now let's find out what information is needed in order to calculate the heating. It is easy to guess what this information is.

1. The temperature of the working fluid at the outlet / inlet of a particular section of the line.

2. The flow rate of the working fluid that passes through the heating devices.

The flow rate is determined through the use of thermal metering devices, that is, meters. These can be of two types, let's get acquainted with them.

Vane meters

Such devices are intended not only for heating systems, but also for hot water supply. Their only difference from those meters that are used for cold water is the material from which the impeller is made - in this case it is more resistant to elevated temperatures.

As for the mechanism of work, it is almost the same:

due to the circulation of the working fluid, the impeller begins to rotate;
the rotation of the impeller is transferred to the accounting mechanism;
the transfer is carried out without direct interaction, but with the help of a permanent magnet.

Despite the fact that the design of such counters is extremely simple, their response threshold is quite low, moreover, there is reliable protection against distortion of readings: the slightest attempt to brake the impeller by means of an external magnetic field is stopped thanks to the antimagnetic screen.

Instruments with differential recorder

Such devices operate on the basis of Bernoulli's law, which states that the speed of a gas or liquid flow is inversely proportional to its static movement. But how is this hydrodynamic property applicable to the calculation of the flow rate of the working fluid? Very simple - you just need to block her path with a retaining washer. In this case, the rate of pressure drop on this washer will be inversely proportional to the speed of the moving stream. And if the pressure is recorded by two sensors at once, then you can easily determine the flow rate, and in real time.

Note! The design of the counter implies the presence of electronics. The overwhelming majority of such modern models provide not only dry information (temperature of the working fluid, its consumption), but also determine the actual use of thermal energy.

The control module here is equipped with a port for connecting to a PC and can be configured manually.

Many readers will probably have a logical question: what if we are not talking about a closed heating system, but about an open one, in which selection for hot water supply is possible? How, in this case, to calculate Gcal for heating? The answer is quite obvious: here pressure sensors (as well as retaining washers) are placed simultaneously on both the supply and the “return”. And the difference in the flow rate of the working fluid will indicate the amount of heated water that was used for domestic needs.

Hydraulic calculation

So, we have decided on heat losses, the power of the heating unit has been selected, it remains only to determine the volume of the required coolant, and, accordingly, the dimensions, as well as the materials of the pipes, radiators and valves used.

First of all, we determine the volume of water inside the heating system. This will require three indicators:

The total power of the heating system.
Temperature difference at the outlet and inlet to the heating boiler.
Heat capacity of water. This indicator is standard and equal to 4.19 kJ.

Hydraulic calculation of the heating system

The formula is as follows - the first indicator is divided by the last two. By the way, this type of calculation can be used for any section of the heating system.

Here it is important to break the line into parts so that in each the speed of the coolant is the same. Therefore, experts recommend making a breakdown from one shutoff valve to another, from one heating radiator to another

Now we turn to the calculation of the pressure loss of the coolant, which depends on the friction inside the pipe system. For this, only two quantities are used, which are multiplied together in the formula. These are the length of the main section and specific friction losses.

But the pressure loss in the valves is calculated using a completely different formula. It takes into account indicators such as:

Heat carrier density.
His speed in the system.
The total indicator of all coefficients that are present in this element.

In order for all three indicators, which are derived by formulas, to approach standard values, it is necessary to choose the right pipe diameters. For comparison, we will give an example of several types of pipes, so that it is clear how their diameter affects heat transfer.

Metal-plastic pipe with a diameter of 16 mm. Its thermal power varies in the range of 2.8-4.5 kW. The difference in the indicator depends on the temperature of the coolant. But keep in mind that this is a range where the minimum and maximum values are set.
The same pipe with a diameter of 32 mm. In this case, the power varies between 13-21 kW.
Polypropylene pipe. Diameter 20 mm - power range 4-7 kW.
The same pipe with a diameter of 32 mm - 10-18 kW.

And the last is the definition of a circulation pump. In order for the coolant to be evenly distributed throughout the heating system, it is necessary that its speed be not less than 0.25 m / s and not more than 1.5 m / s. In this case, the pressure should not be higher than 20 MPa. If the coolant velocity is higher than the maximum proposed value, then the pipe system will work with noise. If the speed is lower, then airing of the circuit may occur.

Heating consumption standard per sq m

hot water supply

1
2
3

Multi-apartment residential buildings equipped with centralized heating, cold and hot water supply, sanitation with showers and bathtubs

Length 1650-1700 mm
8,12
2,62

Length 1500-1550 mm
8,01
2,56

Length 1200 mm
7,9
2,51

Multi-apartment residential buildings equipped with centralized heating, cold and hot water supply, sanitation with a shower without baths

7,13
2,13
3. Multi-apartment residential buildings equipped with centralized heating, cold and hot water supply, sanitation without showers and baths
5,34
1,27

Standards for the consumption of utilities in Moscow

No. p / p

Name of company

Tariffs including VAT (rubles/cub.

cold water

drainage

JSC Mosvodokanal

35,40

25,12

Note. Tariffs for cold water and sanitation for the population of the city of Moscow do not include commission fees charged by credit institutions and payment system operators for the services of accepting these payments.

Heating rates per 1 square meter

It should be remembered that it is not necessary to make a calculation for the entire apartment, because each room has its own heating system and requires an individual approach.In this case, the necessary calculations are made using the formula: C * 100 / P \u003d K, where K is the power of one section of your radiator battery, according to its characteristics; C is the area of the room.

How much are the standards for the consumption of utilities in Moscow in 2019

No. 41 “On the transition to a new system of payment for housing and utilities and the procedure for providing citizens with housing subsidies”, the indicator for heat supply is valid:

heat energy consumption for heating an apartment - 0.016 Gcal/sq. m;
water heating - 0.294 Gcal / person.

Residential buildings equipped with sewerage, plumbing, baths with hot central water supply:

water disposal - 11.68 m³ per 1 person per month;
hot water - 4,745.
cold water - 6.935;

Housing equipped with sewerage, plumbing, bathtubs with gas heaters:

water disposal - 9.86;
cold water - 9.86.

Houses with water supply with gas heaters near the baths, sewerage:

9.49 m³ per person per month.
9,49;

Residential buildings of a hotel type, equipped with water supply, hot water supply, gas:

cold water - 4.386;
hot - 2, 924.
water disposal - 7.31;

Utilities Consumption Standards

Payment for electricity, water supply, sewerage and gas is made according to the established norms if an individual metering device is not installed.

From July 1 to December 31, 2015 - 1.2.
From January 1 to June 30, 2019 - 1.4.
From July 1 to December 31, 2019 - 1.5.
Since 2019 - 1.6.
From January 1 to June 30, 2015 - 1.1.

Thus, if you do not have a collective heat meter installed in your house, and you pay, for example, 1 thousand rubles a month for heating, then from January 1, 2015 the amount will increase to 1,100 rubles, and from 2019 - up to 1600 rubles.

Calculation of heating in an apartment building from 01/01/2019

The calculation methods and examples presented below provide an explanation of the calculation of the amount of payment for heating for residential premises (apartments) located in multi-apartment buildings with centralized systems for supplying heat energy.

How Many Gcal Is Needed For Heating 1 Sq M Norm 2019

Be that as it may, heating standards are not observed, therefore consumers have every right to file a corresponding complaint and demand recalculation of tariff plans. The choice of one or another calculation method depends on whether a heat meter is installed in the house and apartment.

In the absence of a common house meter, tariffs are calculated in accordance with the standards, and those, as we have already found out, are determined by local authorities.

This is done through a special decree, which also determines the payment schedule - whether you will pay all year round or only during the heating season.

How is the heating bill calculated in an apartment building

the house-wide heat energy metering unit put into operation failed and was not repaired within 2 months;
the heat meter is stolen or damaged;
the readings of the house appliance are not transferred to the heat supply organization;
the admission of the organization's specialists to the house meter in order to check the technical condition of the equipment is not provided (2 visits or more).

As an example of calculation, let's take our apartment of 36 m² and assume that for a month an individual meter (or a group of individual meters) "twisted" 0.6, a brownie - 130, and a group of devices in all rooms of the building gave a total of 118 Gcal. The remaining indicators remain the same (see previous sections). How much does heating cost in this case:

Determine heat loss

The heat loss of a building can be calculated separately for each room that has an external part in contact with the environment. Then the received data are summarized. For a private house, it is more convenient to determine the heat loss of the entire building as a whole, considering the heat loss separately through the walls, roof, and floor surface.

It should be noted that the calculation of heat losses at home is a rather complicated process that requires special knowledge. A less accurate, but at the same time quite reliable result can be obtained on the basis of an online heat loss calculator.

When choosing an online calculator, it is better to give preference to models that take into account all possible options for heat loss. Here is their list:

outer wall surface

Having decided to use the calculator, you need to know the geometric dimensions of the building, the characteristics of the materials from which the house is made, as well as their thickness. The presence of a heat-insulating layer and its thickness are taken into account separately.

Based on the listed initial data, the online calculator gives the total value of heat losses at home. To determine how accurate the results obtained can be by dividing the result obtained by the total volume of the building and thus obtaining specific heat losses, the value of which should be in the range from 30 to 100 W.

If the numbers obtained using the online calculator go far beyond the specified values, it can be assumed that an error has crept into the calculation. Most often, the cause of errors in calculations is a mismatch in the dimensions of the quantities used in the calculation.

An important fact: the online calculator data is relevant only for houses and buildings with high-quality windows and a well-functioning ventilation system, in which there is no place for drafts and other heat losses.

To reduce heat loss, you can perform additional thermal insulation of the building, as well as use the heating of the air entering the room.