Determination of heat consumption for heating a residential building. Heat consumption for heating and ventilation of industrial enterprises. Basic calculation methods

1.1.1. Estimated maximum heat consumption (W) for heating residential, public and administrative buildings is determined by aggregated indicators

= q o ∙ V (t in t n.r.),

=1.07∙0.38∙19008(16-(-25))=239588.2

Where q o - specific heating characteristic of the building at t n.r. \u003d -25С (W / m  С);

  correction factor that takes into account the climatic conditions of the area and is used in cases where the estimated outdoor temperature differs from  25С, V  building volume according to external measurement, m 3; t in calculated air temperature inside the heated building, t n.r.  calculated outdoor air temperature for heating design, С, see Appendix 2.

The calculation was made for the subscriber No. 1 of the school. For all the rest, the calculation was carried out according to the above proposed formula, the results are listed in Table 2.2.

1.1.2.Average heat flow (W) for heating





The calculation was made for the subscriber No. 1 of the school. For all the rest, the calculation was carried out according to the above proposed formula, the results are listed in Table 2.2.

Where t n.r.sr.  estimated average outdoor air temperature for heating design, C (Appendix 2).

1.2. Determination of heat consumption for ventilation.

1.2.1Maximum heat consumption for ventilation, Q in max , W

Q in max = q in  V   (t in  t n.v.)

Q in max =1.07190080.29(16-(-14))

Where q in  the specific characteristic of the building for the design of the ventilation system.

1.2.2. Average heat consumption for ventilation, Q in sr, W

Q in cf = Q in max 

Q in cf = 176945.5 

The calculation was made for the subscriber No. 1 of the school. For all the rest, the calculation was carried out according to the above proposed formula, the results are listed in Table 2.2.

1.3. Determination of heat consumption for hot water supply.

1.3.1 Average heat consumption for hot water supply of industrial buildings, Qavg h.w.s., W

Q hot water cf =

where   consumption rate hot water(l/day) per unit of measurement (SNiP 2.04.01.-85),

m - number of units of measurement;

c - heat capacity of water С = 4187 J/kg  С;

t g, t x - temperature of hot water, respectively, supplied to the hot water supply system and cold water, С;

h - estimated duration of heat supply to hot water supply, C/day, h/day.

1.3.2 Average heat consumption for hot water supply of residential and public buildings, Q DHW, W

The calculation was made for the subscriber No. 1 of the school. For all the rest, the calculation was carried out according to the above proposed formula, the results are listed in Table 2.2.

where m is the number of people,

  rate of water consumption for hot water at a temperature of 55 С per person per day (SNiP 2.04.01-85, appendix 3)

c - the rate of water consumption for hot water supply is taken as 25 l / day per 1 person;

t x - temperature of cold water (tap) during the heating period (in the absence of data, it is assumed to be 5С)

с - heat capacity of water, С = 4.187 kJ/(kgС)

1.3.3. Maximum heat consumption for hot water supply,
,W

134332,9

The calculation was made for the subscriber No. 1 of the school. For all the rest, the calculation was carried out according to the above proposed formula, the results are listed in Table 2.2.

Table 2.1

1.3.4. Annual heat consumption by residential and public buildings

a) heating

;

b) for ventilation

;

c) for hot water supply

where n o, n r - respectively, the duration of the heating period and the duration of the hot water supply system in sec / year, (hour / year).

Usually n r \u003d 30.2 10 5 s-year (8400 h / year);

t r is the hot water temperature.

d) Total annual heat consumption for heating, ventilation and hot water supply

What is specific heat consumption for heating? In what quantities is the specific consumption of thermal energy for heating a building measured and, most importantly, where do its values ​​\u200b\u200bare taken for calculations? In this article, we will get acquainted with one of the basic concepts of heat engineering, and at the same time study several related concepts. So, let's go.

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 explicitly states: specific (per square or cubic meter) the consumption of thermal energy for heating the building should not exceed the given values.
How better thermal insulation, the less energy is required for heating.

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 current SNiP it should be from +18 to +22 C) and average temperature the coldest five days of winter.
• Z - length 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 varies from 2000 (Crimea, Krasnodar region) up to 12000 (Chukotka Autonomous Okrug, Yakutia).

Units

In what quantities is the parameter of interest measured?

• SNiP 23-02-2003 uses kJ / (m2 * C * day) and, in parallel with the first value, kJ / (m3 * C * day).
• 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.

Normalized parameters

For single-family one-story detached houses

For apartment buildings, hostels and hotels

Please note: with an increase in the number of floors, the heat consumption rate decreases.
The reason is simple and obvious: the larger the object, the simpler geometric shape, the greater the ratio of its volume to surface area.
For the same reason, specific heating costs country house decreases with increasing heated area.

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 indicator (measurement) calculations.

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.

• 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 indoors (+18 - +22 C).
• tno - street 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 specific consumption at 125 kJ / (m2 * C * day) and an area of ​​​​100 m2, located in climate zone with GSOP=6000, you just need to multiply 125 by 100 (house area) and by 6000 (heating degree-days). 125*100*6000=75000000 kJ or about 18 gigacalories or 20800 kilowatt-hours.

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 thermal power heating in the above case will be 20800/200/24 ​​= 4.33 kW.

Energy carriers

How to calculate energy costs with your own hands, knowing the heat consumption?

enough to know calorific value appropriate fuel.

The easiest way to calculate the electricity consumption for heating a house: it is exactly equal to the amount of heat produced by direct heating.

So, the average in the last case considered by us will be equal to 4.33 kilowatts. If the price of a kilowatt-hour of heat is 3.6 rubles, then we will spend 4.33 * 3.6 = 15.6 rubles per hour, 15 * 6 * 24 = 374 rubles per day, and so on.

It is useful for owners of solid fuel boilers to know that the consumption rates for firewood for heating are about 0.4 kg / kWh. The norms of coal consumption for heating are half as much - 0.2 kg / kWh.

Thus, in order to calculate the average hourly consumption of firewood with your own hands with an average heating power of 4.33 kW, it is enough to multiply 4.33 by 0.4: 4.33 * 0.4 = 1.732 kg. The same instruction is valid for other coolants - you just need to get into the reference books.

Conclusion

We hope that our acquaintance with the new concept, even if somewhat superficial, could satisfy the reader's curiosity. The video attached to this material, as usual, will offer additional information. Good luck!

A private house can be considered as a thermodynamic system with internal energy and leading heat exchange with the environment. The energy that a house receives or loses during heat exchange is called heat. The source of heat in a private house is a heat generator: boiler, convector, stove, a heating element etc.

The more intense the heat exchange between the house and the environment, the faster the heat of the house “leaves” and the more intensively the source of thermal energy must work to compensate for the losses. It is clear that the intensive operation of the boiler is associated with high fuel consumption, which leads to an increase in heating costs.

But this is not the main thing: the concept of comfort in a home during the cold season is inextricably linked with the heat in the house, which is possible only if there is a balance between the loss of thermal energy and its production.

However, the capabilities of any heat generator are limited by its design features. This means that in order to provide warmth and comfort in the house, a boiler or other source of thermal energy must be selected in accordance with the heat losses of the building, while making some margin (usually 20%) in case of windy weather or severe frosts.

So, we have decided: before choosing a boiler for heating a house, you need to determine it (at home) heat loss.

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 sufficient difficult process requiring specialized knowledge. Less accurate, but still reliable result can be obtained based on online calculator calculation of heat losses.

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

roof surface

floor surface

ventilation system

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 issues general meaning heat loss 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, additional thermal insulation buildings, as well as use the heating of the air entering the room.

Heat loss know what's next?

The next step is to select heating unit(boiler). Its thermal power must exceed the value of thermal losses by at least 20%. If the boiler is also used for hot water supply, a thermal unit with an additional power reserve is selected. To do this, it is necessary to make an additional calculation, taking into account the needs for hot water supply.

Then get picked up heating appliances, the total power of which must correspond to the power of the heating boiler, excluding hot water supply.

Hydraulic calculation of the heating system

Having picked up the equipment, it is necessary to ensure its operation. This requires pipes circulation pump and expansion tank heating.

If the owner of the house decides to select the heating pipes on his own, you can use the reference literature and select the required diameter from the tables. The length of the pipes is calculated by project documentation. To do this, an additional wiring diagram for the heating system is simply laid on the building diagram and the length of the pipeline is calculated.

If for some reason there is no diagram of the house, you will have to draw it yourself, and then, with its help, calculate the length of the pipeline.

Knowing the length of the pipeline, the diameter of the pipes and having the technical data of the heating devices, the internal volume of the heating system is calculated, according to which the expansion tank and the circulation pump are selected.

Correct hydraulic calculation is also necessary so that all the heat generated by the boiler is evenly distributed throughout the house and reaches the consumer in full.

Summing up

The amount of heat needed to heat a house directly depends on its heat loss. It is possible to reduce heat losses with the help of additional thermal insulation, installation quality windows and insulated doors, as well as when using recuperation in the ventilation system.

The amount of heat loss determines the power of the heating boiler. The total power of heating devices must be equal to the power of the boiler. To ensure the high-quality operation of the boiler and radiators, a hydraulic calculation of heating is carried out, during which the diameter of the pipes, their length, and the internal volume of heating are determined. According to these data, a circulation pump and an expansion heating tank are selected.

In case of severe frost the boiler is bought with a power reserve of at least 20%.

Heat loss is due to:

• penetration of cold temperature from the outer walls of the room, through window slots,
• poor sealing of window frames.

When installing heating systems, it is necessary to take into account the regional peculiarity of the temperature outside the window and, based on the parameters obtained, choose one or another type of heating equipment. But even the most efficient heating technique will not give the desired result if you do not get rid of the so-called "heat leakage points". When installing window frames, you should once spend money on high-quality ones that have a high coefficient of heat retention. To effectively carry out insulation work walls, the market for thermal insulation materials presents a large selection.

The heat consumption for heating will decrease several times if the work on sealing the premises is carried out efficiently. Any modern heating equipment can be regulated by controlling the flow of warm air masses into the room. Power heating appliances increases as the cold air supply decreases.

For complete comfort, two conditions must be met:

• ensure the optimum temperature in the room at 20-22 degrees;
• difference between the indoor air temperature and outer wall should be no more than 4 degrees, while the temperature of the wall should be above the dew point temperature.

The dew point is the cooling of the outside air before condensation begins and its vapors turn into dew. This is easy to achieve with a powerful boiler. But it is important to reduce heating costs.

The heat consumption for heating has two options for the consumption rate:

1. First - established norm on the heat transfer resistance of external walls, window frames, etc.
2. The second - the standard of energy consumption for heating the house is determined. The second method allows to reduce the resistance to heat supply of enclosing structures. Thus, one can choose optimal thickness walls of the room.

Professional builders often use the first option. Raising up concrete walls, to them they perform work on additional insulation various thermal insulation materials. This method significantly complicates the process and increases the cost of work.

When building private houses, it is not necessary to insulate the outer walls, it is enough to create a more insulated layer in the attic and underground. You should also give the house a shape that is energy efficient, given the compactness of the structure. For greater insulation, verandas, loggias are attached to the house, window frames make smaller sizes, etc. Thus, the heat consumption for heating is reduced many times over.

Having eliminated all the shortcomings, you can proceed to the choice heating equipment. It is worth paying attention to the parameters heating system to be installed indoors. The temperature in the house also depends on the quality of the materials from which heat carriers, radiators and boilers of heating equipment will be made. Modern systems heating are in reserve big list new technologically equipped devices for saving heat. Automatic controllers to maintain optimal temperature in the room will be the main assistants in terms of heat energy consumption for heating.

When building an energy-saving house or ordering already finished project It is worth considering carefully the issues of building insulation with the involvement of experienced specialists. Job requires integrated approach and only in this case it is possible to build a comfortable, warm and cozy home.

Heating radiators and thermostats

In radiators, the coolant temperature should not exceed 90 degrees. When choosing powerful and resistant radiators, this temperature is quite suitable for cold winters. In order for the atmosphere in the room to be acceptable to everyone, you need to install thermostats. There are two types - mechanical and automatic. Mechanical must be constantly adjusted manually, not missing the moment of changing thermal values. The open position of the regulator provides the maximum mode, the closed position provides the minimum. If the hot water supply is lost, the battery cools down quickly.

The automatic thermostat, in turn, requires less attention. It is enough to fix the required mark on the scale, and the machine itself adjusts the temperature level. The use of a thermostat is possible only if the pipes are parallel, the use of regulators installed one after another blocks the circulation of the coolant in the pipes.

The consumption of thermal energy for heating carries considerable costs if the heating system is installed without taking into account other costs, such as a boiler, kitchen, bathroom.

Find a "leak"

To save more, when summing up the heating system, you need to take into account all the “sick” places of heat leakage. It will not be superfluous to say that the windows must be sealed. The thickness of the walls allows you to keep the heat, warm floors keep the temperature background at a positive level. The consumption of thermal energy for heating the room depends on the height of the ceilings, such as ventilation system, building materials when building a building.

After deducting all the heat losses, you need to seriously approach the choice heating boiler. The main thing here is the budget part of the issue. Depending on the power and versatility, the price of the device also varies. If there is already gas in the house, then there is savings on electricity (the cost of which is considerable), and along with preparing, for example, dinner, the system warms up at the same time.

Another point in preserving heat is the type of heater - convector, radiator, battery, etc. Most suitable solution question - radiator, the number of sections of which is calculated using a simple formula. One section (rib) of the radiator has a power of 150 watts, for a room of 10 meters 1700 watts is enough. By dividing, we get 13 sections necessary for comfortable space heating.

Installing underfloor heating will half solve the problem of energy savings. According to experts, the amount of consumed heat energy is reduced by 2-3 times. The economical consumption of thermal energy for heating is obvious.

When installing the heating system by placing radiators, you can immediately connect the underfloor heating system. Constant circulation of the coolant creates a uniform temperature throughout the room.

The procedure for calculating heating in the housing stock depends on the availability of metering devices and on how the house is equipped with them. There are several options for completing multi-apartment residential buildings with meters, and according to which, heat energy is calculated:

1. the presence of a common house meter, while apartments and non-residential premises are not equipped with metering devices.
2. heating costs are controlled by a common house device, and all or some rooms are equipped with metering devices.
3. there is no general house device for fixing the consumption and consumption of thermal energy.

Before calculating the number of gigacalories spent, it is necessary to find out the presence or absence of controllers in the house and in each individual room, including non-residential ones. Let's consider all three options for calculating thermal energy, for each of which a specific formula has been developed (posted on the website of state authorized bodies).

Option 1

So the house is equipped control device, and some rooms were left without it. Here it is necessary to take into account two positions: the calculation of Gcal for heating an apartment, the cost of thermal energy for general house needs (ODN).

In this case, formula No. 3 is used, which is based on the readings of the general meter, the area of ​​\u200b\u200bthe house and the footage of the apartment.

Calculation example

We will assume that the controller recorded the heating costs of the house at 300 Gcal / month (this information can be obtained from the receipt or by contacting management company). For example, total area house, which consists of the sum of the areas of all premises (residential and non-residential), is 8000 m² (you can also find out this figure from the receipt or from the management company).

Let's take the area of ​​​​an apartment of 70 m² (indicated in the data sheet, rental agreement or registration certificate). The last figure, on which the calculation of payment for consumed heat energy depends, is the tariff established by the authorized bodies of the Russian Federation (indicated on the receipt or found out in the house management company). Today, the heating tariff is 1,400 rubles/gcal.

Substituting the data in formula No. 3, we get the following result: 300 x 70 / 8,000 x 1,400 \u003d 1875 rubles.

Now you can proceed to the second stage of accounting for heating costs spent on the general needs of the house. Two formulas are required here: the search for the volume of services (No. 14) and the payment for the consumption of gigacalories in rubles (No. 10).

In order to correctly determine the volume of heating in this case, it will be necessary to sum up the area of ​​\u200b\u200ball apartments and premises provided for common use(information provided by the management company).

For example, we have a total footage of 7000 m² (including apartments, offices, retail premises.).

Let's start calculating the payment for the consumption of thermal energy according to formula No. 14: 300 x (1 - 7,000 / 8,000) x 70 / 7,000 \u003d 0.375 Gcal.

Using formula No. 10, we get: 0.375 x 1,400 = 525, where:

• 0.375 - volume of service for heat supply;
• 1400 r. – tariff;
• 525 rubles - amount of payment.

We summarize the results (1875 + 525) and find out that the payment for heat consumption will be 2350 rubles.

Option 2

Now we will calculate payments in those conditions when the house is equipped with a common meter for heating, as well as some apartments are equipped with individual meters. As in the previous case, the calculation will be carried out in two positions (thermal energy consumption for housing and ONE).

We will need formulas No. 1 and No. 2 (accrual rules according to the testimony of the controller or taking into account the norms for heat consumption for residential premises in gcal). Calculations will be carried out in relation to the area of ​​​​a residential building and an apartment from the previous version.

• 1.3 gigacalories - readings of an individual counter;
• 1 1820 r. - approved rate.

• 0.025 gcal - standard indicator of heat consumption per 1 m² of area in an apartment;
• 70 m² - area of ​​the apartment;
• 1 400 rubles - tariff for thermal energy.

As it becomes clear, with this option, the payment amount will depend on the availability of a metering device in your apartment.

Formula No. 13: (300 - 12 - 7,000 x 0.025 - 9 - 30) x 75 / 8,000 \u003d 1.425 gcal, where:

• 300 gcal - indications of a common house meter;
• 12 gcal - the amount of thermal energy used for heating non-residential premises;
• 6,000 m² - the sum of the area of ​​​​all residential premises;
• 0.025 - standard (thermal energy consumption for apartments);
• 9 gcal - the sum of indicators from the meters of all apartments that are equipped with metering devices;
• 35 gcal - the amount of heat spent on the supply of hot water in the absence of its centralized supply;
• 70 m² - area of ​​the apartment;
• 8,000 m² - total area (all residential and non-residential premises in the house).

Please note that this option only includes real amounts of energy consumed, and if your house is equipped with a centralized hot water supply, then the amount of heat spent on hot water needs is not taken into account. The same applies to non-residential premises: if they are not in the house, then they will not be included in the calculation.

• 1.425 gcal - the amount of heat (ONE);

1. 1820 + 1995 = 3,815 rubles - with individual counter.
2. 2 450 + 1995 = 4445 rubles. - without individual device.

Option 3

We have left last option, during which we will consider the situation when there is no heat energy meter on the house. The calculation, as in previous cases, will be carried out in two categories (thermal energy consumption for an apartment and ONE).

We will derive the amount for heating using formulas No. 1 and No. 2 (rules on the procedure for calculating thermal energy, taking into account the readings of individual meters or in accordance with the established standards for residential premises in gcal).

Formula No. 1: 1.3 x 1,400 \u003d 1820 rubles, where:

• 1.3 gcal - readings of an individual meter;
• 1 400 rubles - approved rate.

Formula No. 2: 0.025 x 70 x 1,400 = 2,450 rubles, where:

• 1 400 rubles - approved rate.

As in the second option, the payment will depend on whether your housing is equipped with an individual heat meter. Now it is necessary to find out the amount of heat energy that was spent on general house needs, and this must be done according to formula No. 15 (volume of service for one unit) and No. 10 (amount for heating).

Formula No. 15: 0.025 x 150 x 70 / 7000 \u003d 0.0375 gcal, where:

• 0.025 gcal - standard indicator of heat consumption per 1 m² of living space;
• 100 m² - the sum of the area of ​​\u200b\u200bthe premises intended for general house needs;
• 70 m² - the total area of ​​the apartment;
• 7,000 m² - total area (all residential and non-residential premises).

Formula No. 10: 0.0375 x 1,400 = 52.5 rubles, where:

• 0.0375 - volume of heat (ONE);
• 1400 r. - approved rate.

As a result of the calculations, we found out that the full payment for heating will be:

1. 1820 + 52.5 \u003d 1872.5 rubles. - with individual counter.
2. 2450 + 52.5 \u003d 2,502.5 rubles. – without individual counter.

In the above calculations of payments for heating, data on the footage of the apartment, house, as well as on the meter indicators, which may differ significantly from those that you have, were used. All you need to do is plug your values ​​into the formula and make the final calculation.