Independent calculation of the heat load for heating: hourly and annual indicators.

When equipping a building with a heating system, you need to take into account a lot of points, starting with the quality Supplies and functional equipment and ending with calculations of the required power of the node. So, for example, you will need to calculate the heat load for heating a building, a calculator for which will be very useful. It is carried out according to several methods, which take into account a huge number of nuances. Therefore, we invite you to take a closer look at this issue.

Average values ​​as a basis for calculating heat load

In order to correctly calculate the heating of the room by the volume of the coolant, the following data must be determined:

• the value of the required amount of fuel;
• heating unit performance;
• efficiency of the specified type of fuel resources.

In order to eliminate cumbersome computational formulas, the specialists of housing and communal enterprises have developed a unique methodology and program with which it is possible to calculate the heat load for heating and other data necessary for the design of the heating unit in just a matter of minutes. Moreover, using this technique, it is possible to correctly determine the volume of the coolant for heating a particular room, regardless of the type of fuel resources.

Basics and features of the technique

A technique of this kind, which can be used by using a calculator for calculating heat energy for heating a building, is very often used by employees of cadastral firms to determine the economic and technological efficiency of all kinds of programs aimed at energy saving. In addition, with the help of such computational and computational methods, new functional equipment is introduced into projects and energy-efficient processes are launched.

So, to calculate the heat load on heating a building, experts use the following formula:

• a - coefficient that shows the corrections of the difference in the temperature regime of the outside air when determining the efficiency of the functioning of the heating system;
• t i, t 0 - temperature difference in the room and outside;
• q 0 - specific exponent, which is determined by additional calculations;
• K u.p - coefficient of infiltration, taking into account all kinds of heat loss, from weather conditions to the absence of a heat-insulating layer;
• V is the volume of the structure that needs heating.

How to calculate the volume of a room in cubic meters (m 3)

The formula is very primitive: you just need to multiply the length, width and height of the room. However, this option is suitable only for determining the cubic capacity of a structure that has a square or rectangular shape. In other cases, this value is determined in a slightly different way.

If the room is a room of irregular shape, then the task is somewhat more complicated. In this case, it is necessary to divide the area of ​​the rooms into simple figures and determine the cubic capacity of each of them, having made all measurements in advance. It remains only to add up the resulting figures. Calculations should be carried out in the same units of measurement, for example, in meters.

In the event that the structure for which the aggregated calculation of the building's thermal load is made is equipped with an attic, then the cubic capacity is determined by multiplying the indicator of the horizontal section of the house (we are talking about an indicator that is taken from the level of the floor surface of the first floor) by its full height, taking into account the highest point of the attic insulation layer.

Before calculating the volume of a room, it is necessary to take into account the fact that there are basements or basements. They also need heating and if there are any, then another 40% of the area of ​​these rooms should be added to the cubic capacity of the house.

To determine the infiltration coefficient, K u.p, you can take the following formula as a basis:

where is the root of the total volume of rooms in the structure, and n is the number of rooms in the building.

Possible energy loss

In order for the calculation to be as accurate as possible, absolutely all types of energy losses must be taken into account. So, the main ones include:

• through the attic and the roof, if not properly insulated, the heating unit loses up to 30% of heat energy;
• if there is natural ventilation in the house (chimney, regular ventilation, etc.), up to 25% of heat energy is consumed;
• if the wall ceilings and the floor surface are not insulated, then up to 15% of the energy can be lost through them, the same amount goes through the windows.

The more windows and doorways in the house, the more heat loss. With poor-quality thermal insulation of the house, on average, up to 60% of the heat goes through the floor, ceiling and facade. The largest in terms of heat transfer surface are the window and the facade. First of all, the windows are changed in the house, after which they begin to insulate.

Taking into account possible energy losses, it is necessary either to exclude them by resorting to the help of heat-insulating material, or to add their value while determining the amount of heat for heating the room.

As for the arrangement of stone houses, the construction of which has already been completed, it is necessary to take into account the higher heat losses at the beginning of the heating season. In this case, it is necessary to take into account the completion date of the construction:

• from May to June - 14%;
• September - 25%;
• from October to April - 30%.

Hot water supply

The next step is to calculate the average load of hot water supply during the heating season. For this, the following formula is used:

• a - average daily rate of use hot water(this value is normalized and can be found in the SNiP table, Appendix 3);
• N is the number of residents, employees, students or children (if we are talking about preschool) in the building;
• t_c-value of the water temperature (measured after the fact or taken from the averaged reference data);
• T is the time interval during which hot water is supplied (if we are talking about hourly water supply);
• Q_ (t.n) - coefficient of heat loss in the hot water supply system.

Is it possible to regulate the loads in the heating block?

Just a few decades ago, this was an unrealistic task. Today, almost all modern heating boilers for industrial and domestic use are equipped with thermal load regulators (PTN). Thanks to such devices, the power of the heating units is maintained at a given level, and surges, as well as passes during their operation, are excluded.

Regulators of heat loads allow to reduce the financial costs of paying for the consumption of energy resources for heating the structure.

This is due to the fixed power limit of the equipment, which, regardless of its functioning, does not change. This is especially true for industrial enterprises.

It is not so difficult to make a project on your own and calculate the load of heating units that provide heating, ventilation and air conditioning in a building, the main thing is to have patience and the necessary knowledge base.

VIDEO: Calculation of heating batteries. Rules and errors

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PAYMENT

heat loads and annual quantity

heat and fuel for the boiler room

individual dwelling house

Moscow 2005

OVK Engineering LLC

Moscow 2005

General part and initial data

This calculation has been drawn up to determine the annual consumption of heat and fuel required for a boiler room intended for heating and hot water supply of an individual residential building. The calculation of thermal loads is carried out in accordance with the following regulatory documents:

MDK 4-05.2004 "Methodology for determining the need for fuel, electricity and water in the production and transmission of thermal energy and heat carriers in municipal heat supply systems" (Gosstroy RF 2004); SNiP 23-01-99 "Construction climatology"; SNiP 41-01-2003 "Heating, ventilation and air conditioning"; SNiP 2.04.01-85 * "Internal water supply and sewerage of buildings."

Characteristics of the building:

Construction volume of the building - 1460 m2 Total area - 350.0 m2 Living area - 107.8 m2 Estimated number of residents - 4 people

Climatol Logical data of the construction area:

Place of construction: Russian Federation, Moscow region, Domodedovo

Design temperaturesair:

For designing a heating system: t = -28 ºС For designing a ventilation system: t = -28 ºС In heated rooms: t = +18 C

Correction factor α (at -28 С) - 1.032

Specific heating characteristic of the building - q = 0.57 [Kcal / m · h · C]

Heating period:

Duration: 214 days Average temperature of the heating period: t = -3.1 ºС Average of the coldest month = -10.2 ºС Boiler efficiency - 90%

Initial data for calculating DHW:

Operating mode - 24 hours a day Duration of DHW operation during the heating period - 214 days Duration of DHW operation in the summer period - 136 days Temperature tap water during the heating season - t = +5 C Tap water temperature in summer - t = +15 C Coefficient of change in hot water consumption depending on the season - β = 0.8 Water consumption rate for hot water supply per day - 190 l / person The rate of water consumption for hot water supply per hour is 10.5 l / person. Boiler efficiency - 90% Boiler efficiency - 86%

Humidity zone - "normal"

The maximum hourly loads of consumers are as follows:

For heating - 0.039 Gcal / hour For hot water supply - 0.0025 Gcal / hour For ventilation - no

The total maximum hourly heat consumption, taking into account heat losses in networks and for auxiliary needs - 0.0415 Gcal / hour

For heating a residential building, it is planned to install a boiler room equipped with an Ishma-50 gas boiler (capacity 48 kW). For hot water supply, it is planned to install a storage gas boiler "Ariston SGA 200" 195 l (capacity 10.1 kW)

Heating boiler power - 0.0413 Gcal / hour

Boiler power - 0.0087 Gcal / hour

Fuel - natural gas; the total annual consumption of natural fuel (gas) will amount to 0.0155 million nm³ per year or 0.0177 thousand tons of fuel equivalent. per year of conventional fuel.

The calculation was made by: L.A. Altshuler

SCROLL

Data submitted by regional head offices, enterprises (associations) to the Administration of the Moscow Region, together with a petition to establish the type of fuel for enterprises (associations) and heat-consuming installations.

General issues

Questions	Answers
Ministry (department)	Burlakov V.V.
Enterprise and its location (region, district, settlement, street)	Individual residential building located at: Moscow region, Domodedovo st. Nightingale, 1
The distance of the object to: - a railway station - a gas pipeline - a base of petroleum products - the nearest source of heat supply (CHP, boiler room) with an indication of its capacity, load and accessories
The readiness of the enterprise to use fuel and energy resources (operating, projected, under construction) with an indication of the category	under construction, residential
Documents, approvals (conclusions), date, number, name of the organization: - on the use natural gas, coal; - on the transportation of liquid fuel; - on the construction of an individual or expanded boiler house.	Permission of PO Mosoblgaz No. _______ dated ___________ Permission from the Ministry of Housing and Communal Services, Fuel and Energy of the Moscow Region No. _______ dated ___________
On the basis of which document is the enterprise designed, built, expanded, reconstructed
The type and amount (t.f.) of the currently used fuel and on the basis of which document (date, number, established consumption), for solid fuel, indicate its deposit, and for Donetsk coal - its brand	not used
The type of fuel requested, the total annual consumption (t.f.) and the year of the beginning of consumption	natural gas; 0.0155 thousand tons of fuel equivalent in year; 2005 year
The year the enterprise reached its design capacity, the total annual consumption (thousand tons of fuel equivalent) of fuel this year	2005 year; 0.0177 thousand tce

Boiler plants

a) the need for heat

What needs	Connected maximum heat load (Gcal / hour)		Number of hours of work per year	Annual heat demand (Gcal)		Heat demand coverage (Gcal / year)
	Existing	managed, including			Projected, including	Boiler room	energetic go resources	At the expense of others
Hot water supply
what needs
consumption consumption
natural boiler room
Heat losses

Note: 1. In column 4, indicate in brackets the number of hours of operation per year of technological equipment at maximum loads. 2. In columns 5 and 6, show the supply of heat to third-party consumers.

b) the composition and characteristics of boiler equipment, type and annual

fuel consumption

Boiler type by group			Fuel used			Requested fuel
			Main type leg (reserve	consumption	howling expense	Main type leg (reserve	consumption	howling expense
Operating of them: dismantled
"Ishma-50" "Ariston SGA 200"		0,050						thousand tons of fuel equivalent in year;

Note: 1. Specify the total annual fuel consumption by boiler group. 2. Specify the specific fuel consumption taking into account the own needs of the boiler house. 3. In columns 4 and 7, indicate the method of fuel combustion (layered, chamber, in a fluidized bed).

Heat consumers

	Heat consumers	Maximum heat loads (Gcal / hour)			Technology
		Heating		Hot water supply
	House
	House
	Total forresidential building

Heat demand for production needs

	Heat consumers	Name of production	products	Specific heat consumption per unit products	Annual heat consumption

Technological fuel-consuming installations

a) the capacity of the enterprise for the production of main types of products

Product type	Annual release (indicate unit of measure)		Specific fuel consumption (kg standard fuel / unit of production)
	existing	projected	actual	calculated

b) the composition and characteristics of technological equipment,

type and annual fuel consumption

Technological type technical equipment			Fuel used		Requested fuel
				Annual expense (reporting) thousand tons of fuel equivalent		Annual expense (reporting) from what year thousand tons of fuel equivalent

Note: 1. In addition to the requested fuel, indicate other types of fuel that can be used technological installations.

Use of fuel and thermal secondary resources

Fuel secondary resources				Thermal secondary resources
View, source	thousand tons of fuel equivalent	The amount of fuel used (thous.toe)		View, source	thousand tons of fuel equivalent	The amount of heat used (thousand Gcal / hour)
		Existing				Existence

PAYMENT

hourly and annual consumption of heat and fuel

Maximum hourly heat consumption forheating of consumers is calculated by the formula:

Qfrom. = Vzd. x qfrom. x (Tvn. - Tr.ot.) x α [Kcal / hour]

Where: Vzd. (M³) - the volume of the building; qfrom. (kcal / hour * m³ * ºС) - specific thermal characteristics of the building; α - correction factor for the change in the heating characteristics of buildings at temperatures other than -30 ° C.

Maximum hourly flowHeat rate for ventilation is calculated by the formula:

Qvent. = Vн. x qvent. x (Tvn. - Tp.v.) [Kcal / hour]

Where: qvent. (kcal / hour * m³ * ºС) - specific ventilation characteristics of the building;

The average heat consumption for the heating period for the needs of heating and ventilation is calculated by the formula:

for heating:

Qо.p. = Qfrom. x (Tvn. - Tr. from.) / (Tvn. - Tr. from.) [Kcal / hour]

For ventilation:

Qо.p. = Qvent. x (Tvn. - Tr. from.) / (Tvn. - Tr. from.) [Kcal / hour]

The annual heat consumption for the building is determined by the formula:

Qf.year. = 24 x Qav. From. x P [Gcal / year]

For ventilation:

Qf.year. = 16 x Qav. x P [Gcal / year]

Average hourly heat consumption for the heating periodfor hot water supply of residential buildings is determined by the formula:

Q = 1.2 m х a х (55 - Тх.з.) / 24 [Gcal / year]

Where: 1.2 is the coefficient taking into account the heat transfer in the room from the pipeline of hot water supply systems (1 + 0.2); a - the rate of water consumption in liters at a temperature of 55 ° C for residential buildings per person per day, should be taken in accordance with the chapter of SNiP on the design of hot water supply; Th.z. - temperature cold water(water supply) during the heating period, taken equal to 5 ° C.

Average hourly heat consumption for hot water supply in summer is determined by the formula:

Qav.op.g.v. = Q х (55 - Тх.л.) / (55 - Тх.з.) х В [Gcal / year]

Where: B - coefficient taking into account the decrease in the average hourly water consumption for hot water supply of residential and public buildings in the summer in relation to heating, is taken equal to 0.8; Th.l. - temperature of cold water (tap water) in summer, taken equal to 15 ° C.

Average hourly heat consumption for hot water supply is determined by the formula:

Qyear = 24Qo.p.g.w.po + 24Q.p.g.w. * (350 - Po) * B =

24Q average from year to + 24Q average from year to (55 - Тх.л.) / (55 - Тх.з.) х В [Gcal / year]

Total annual heat consumption:

Qyear = Qyear from. + Qyear vent. + Qyear + Qyear VTZ. + Qyear of those. [Gcal / year]

The calculation of the annual fuel consumption is determined by the formula:

Wu.t. = Qyear x 10ˉ 6 /Qr.n. x η

Where: Qr.n. - the net calorific value of the equivalent fuel, equal to 7000 kcal / kg of fuel equivalent; η - boiler efficiency; Qyear is the total annual heat consumption for all types of consumers.

PAYMENT

heat loads and annual amount of fuel

Calculation of the maximum hourly heating loads:

1.1. House: Maximum hourly consumption for heating:

Qmax from. = 0.57 x 1460 x (18 - (-28)) x 1.032 = 0.039 [Gcal / hour]

Total for residential building: Q max. from = 0.039 Gcal / hour Total, taking into account the own needs of the boiler house: Q max. from = 0.040 Gcal / hour

Calculation of average hourly and annual heat consumption for heating:

2.1. House:

Qmax from. = 0.039 Gcal / hour

Qav. From. = 0.039 x (18 - (-3.1)) / (18 - (-28)) = 0.0179 [Gcal / hour]

Qyear from. = 0.0179 x 24 x 214 = 91.93 [Gcal / year]

Taking into account the own needs of the boiler house (2%) Qyear from. = 93.77 [Gcal / year]

Total for residential building:

Average hourly heat consumption for heating Q Wed from. = 0.0179 Gcal / hour

Total annual heat consumption for heating Q year from. = 91.93 Gcal / year

Total annual heat consumption for heating, taking into account the own needs of the boiler house Q year from. = 93.77 Gcal / year

Calculation of maximum hourly loads on DHW:

1.1. House:

Qmax.gvs = 1.2 x 4 x 10.5 x (55 - 5) x 10 ^ (- 6) = 0.0025 [Gcal / hour]

Residential building total: Q max.gws = 0.0025 Gcal / hour

Calculation of hourly average and year new heat consumption for hot water supply:

2.1. House: Average hourly heat consumption for DHW:

Qav.GVSZ. = 1.2 x 4 x 190 x (55 - 5) x 10 ^ (- 6) / 24 = 0.0019 [Gcal / hour]

Qavg.gvs.l. = 0.0019 x 0.8 x (55-15) / (55-5) / 24 = 0.0012 [Gcal / hour]

Godotheat consumption for hot water supply: Qyear from. = 0.0019 x 24 x 214 + 0.0012 x 24 x 136 = 13.67 [Gcal / year] Total for DHW:

Average hourly heat consumption during the heating season Q avg.gvs = 0.0019 Gcal / hour

Average hourly heat consumption in summer Q avg.gvs = 0.0012 Gcal / hour

Total annual heat consumption Q year gws = 13.67 Gcal / year

Calculation of the annual amount of natural gas

and equivalent fuel :

∑ Qyear = ∑Qyear from. +Qyear gws = 107.44 Gcal / year

The annual fuel consumption will be:

Per year = ∑Qyear x 10ˉ 6 /Qr.n. x η

Annual consumption of natural fuel

(natural gas) for the boiler room will be:

Boiler (efficiency = 86%) : Vgod nat. = 93.77 x 10ˉ 6 / 8000 x 0.86 = 0.0136 million nm³ per year Boiler (efficiency = 90%): per year nat. = 13.67 x 10ˉ 6 / 8000 x 0.9 = 0.0019 million nm³ per year Total : 0.0155 million nm  in year

The annual consumption of conventional fuel for the boiler house will be:

Boiler (efficiency = 86%) : Vgod u.t. = 93.77 x 10ˉ 6 / 7000 x 0.86 = 0.0155 million nm³ per yearBulletin

The index of production of electrical equipment, electronic and optical equipment in November 2009. in comparison with the corresponding period of the previous year amounted to 84.6%, in January-November 2009.

Program of the Kurgan region "Regional energy program of the Kurgan region for the period up to 2010" Basis for the development

Program

In accordance with paragraph 8 of Article 5 of the Law of the Kurgan region "On forecasts, concepts, programs of socio-economic development and target programs of the Kurgan region",

Explanatory note Justification of the draft master plan General Director

Explanatory note

Development of urban planning documentation for territorial planning and the Rules for land use and development of the municipal formation urban settlement Nikel, Pechenga district, Murmansk region

The thermal calculation of the heating system seems to most to be an easy and not requiring special attention. A huge number of people believe that the same radiators should be chosen based only on the area of ​​the room: 100 W per 1 sq. It's simple. But this is the biggest delusion. You cannot limit yourself to such a formula. What matters is the thickness of the walls, their height, material and much more. Of course, you need to allocate an hour or two to get the numbers you want, but everyone can do it.

Initial data for the design of the heating system

To calculate the heat consumption for heating, you need, firstly, a house project.

The house plan allows you to get almost all the initial data that are needed to determine heat loss and load on the heating system

Secondly, you will need data on the location of the house in relation to the cardinal points and the construction area - climatic conditions in each region are different, and what is suitable for Sochi cannot be applied to Anadyr.

Thirdly, we collect information about the composition and height of the outer walls and materials from which the floor (from the room to the ground) and the ceiling (from the rooms to the outside) are made.

After collecting all the data, you can start working. The calculation of heat for heating can be done according to the formulas in one to two hours. You can, of course, use a special program from Valtec.

To calculate the heat loss of heated rooms, the load on the heating system and heat transfer from heating devices, it is enough to enter only the initial data into the program. A huge number of functions make it an indispensable assistant for both the foreman and the private developer

It greatly simplifies everything and allows you to get all the data on heat losses and hydraulic calculation of the heating system.

Calculation formulas and reference data

The calculation of the heat load for heating involves the determination of heat losses (Тп) and boiler power (Мк). The latter is calculated by the formula:

Mk = 1.2 * TP, where:

• Mk - thermal performance of the heating system, kW;
• Тп - heat loss at home;
• 1.2 - safety factor (is 20%).

A 20% safety factor allows one to take into account the possible pressure drop in the gas pipeline during the cold season and unforeseen heat losses (for example, a broken window, poor-quality insulation of entrance doors or unprecedented frosts). It allows you to insure yourself against a number of troubles, and also provides the possibility of wide regulation of the temperature regime.

As you can see from this formula, the boiler power directly depends on the heat loss. They are not evenly distributed throughout the house: the outer walls account for about 40% of the total, the windows - 20%, the floor gives 10%, the roof 10%. The remaining 20% ​​escapes through the doors, ventilation.

Poorly insulated walls and floors, cold attic, ordinary glazing on windows - all this leads to large heat losses, and, consequently, to an increase in the load on the heating system. When building a house, it is important to pay attention to all the elements, because even ill-conceived ventilation in the house will release heat into the street

The materials from which the house is built have the most direct impact on the amount of lost heat. Therefore, when calculating, you need to analyze what the walls, the floor, and everything else are made of.

In the calculations, in order to take into account the influence of each of these factors, the corresponding coefficients are used:

• K1 - type of windows;
• K2 - wall insulation;
• K3 - the ratio of the area of ​​the floor and windows;
• K4 is the minimum outside temperature;
• K5 - the number of exterior walls of the house;
• K6 - number of storeys;
• K7 - the height of the room.

For windows, the heat loss coefficient is:

• ordinary glazing - 1.27;
• double-glazed window unit - 1;
• three-chamber double-glazed window - 0.85.

Naturally, the latter option will keep the warmth in the house much better than the two previous ones.

Correctly executed wall insulation is the key to not only a long life at home, but also a comfortable temperature in the rooms. Depending on the material, the value of the coefficient also changes:

• concrete panels, blocks - 1.25-1.5;
• logs, timber - 1.25;
• brick (1.5 bricks) - 1.5;
• brick (2.5 bricks) - 1.1;
• foam concrete with increased thermal insulation - 1.

The larger the area of ​​the windows relative to the floor, the more heat the house loses:

The temperature outside the window also makes its own adjustments. At low rates, heat loss increases:

• Up to -10C - 0.7;
• -10C - 0.8;
• -15C 0.90;
• -20C - 1.00;
• -25C 1.10;
• -30C 1.20;
• -35C - 1.30.

Heat loss also depends on how many external walls the house has:

• four walls - 1.33;%
• three walls - 1.22;
• two walls - 1.2;
• one wall - 1.

It's good if a garage, bathhouse or something else is attached to it. But if it is blown by the winds from all sides, then you will have to buy a more powerful boiler.

The number of floors or the type of room that is located above the room determine the K6 coefficient as follows: if there are two or more floors above the house, then for calculations we take the value 0.82, but if the attic, then 0.91 for warm and 1 for cold ...

As for the height of the walls, the values ​​will be as follows:

• 4.5 m - 1.2;
• 4.0 m - 1.15;
• 3.5 m - 1.1;
• 3.0 m - 1.05;
• 2.5 m - 1.

In addition to the listed factors, the area of ​​the room (Pl) and the specific value of heat loss (UDtp) are also taken into account.

The final formula for calculating the heat loss coefficient:

Tp = UDtp * Pl * K1 * K2 * K3 * K4 * K5 * K6 * K7.

The UDtp coefficient is 100 W / m2.

Analysis of calculations using a specific example

The house for which we will determine the load on the heating system has double glazing (K1 = 1), foam concrete walls with increased thermal insulation (K2 = 1), three of which go outside (K5 = 1.22). The area of ​​the windows is 23% of the floor area (K3 = 1.1), outside there is about 15C frost (K4 = 0.9). The attic of the house is cold (K6 = 1), the height of the premises is 3 meters (K7 = 1.05). The total area is 135m2.

Fri = 135 * 100 * 1 * 1 * 1.1 * 0.9 * 1.22 * 1 * 1.05 = 17120.565 (Watt) or Fri = 17.1206 kW

Mk = 1.2 * 17.1206 = 20.54472 (kW).

The calculation of the load and heat loss can be done independently and quickly enough. You just need to spend a couple of hours tidying up the initial data, and then just substitute the values ​​into the formulas. The numbers that you receive as a result will help you decide on the choice of a boiler and radiators.

1. Heating

1.1. The calculated hourly heat load of heating should be taken according to standard or individual building designs.

In the event of a difference between the calculated outside air temperature for the design of heating adopted in the project from the current standard value for a specific area, it is necessary to recalculate the calculated hourly heat load of the heated building given in the project according to the formula:

where Qo max is the calculated hourly heat load of the building heating, Gcal / h;

Qo max pr - the same, according to a standard or individual project, Gcal / h;

tj - design air temperature in a heated building, ° С; taken in accordance with table 1;

to is the design temperature of the outside air for the design of heating in the area where the building is located, according to SNiP 23-01-99, ° С;

to.pr - the same, according to a standard or individual project, ° С.

Table 1. Estimated air temperature in heated buildings

In areas with an estimated outside air temperature for heating design of -31 ° C and below, the value of the estimated air temperature inside heated residential buildings should be taken in accordance with the SNiP 2.08.01-85 chapter equal to 20 ° C.

1.2. In the absence of design information, the calculated hourly heat load of heating of a separate building can be determined by aggregated indicators:

where  is a correction factor that takes into account the difference in the design temperature of the outside air for the design of heating to from to = -30 ° С, at which the corresponding value of qo is determined; taken according to table 2;

V is the volume of the building by external measurement, m3;

qo is the specific heating characteristic of the building at to = -30 ° С, kcal / m3 h ° С; taken according to tables 3 and 4;

Ki.r - the calculated coefficient of infiltration due to thermal and wind pressure, i.e. the ratio of heat losses of a building with infiltration and heat transfer through external fences at an outside air temperature calculated for heating design.

Table 2. Correction factor  for residential buildings

Table 3. Specific heating characteristics of residential buildings

External construction volume V, m3	Specific heating characteristic qo, kcal / m3 h ° С
built before 1958	built after 1958

Table 3a. Specific heating characteristics of buildings built before 1930

Table 4. Specific thermal characteristics of administrative, medical, cultural and educational buildings, children's institutions

Name of buildings	Building volume V, m3	Specific thermal characteristics
for heating qo, kcal / m3 h ° С	for ventilation qv, kcal / m3 h ° С
Administrative buildings, offices
	more than 15000
	more than 10,000
Cinemas
	more than 10,000
	more than 30,000
The shops
	more than 10,000
Kindergartens and nurseries
Schools and higher education institutions
	more than 10,000
Hospitals
	more than 15000
	more than 10,000
Laundries
	more than 10,000
Enterprises Catering, canteens, kitchen factories
	more than 10,000
Laboratories
	more than 10,000
Firefighters Depot

The V value, m3, should be taken according to the information of the standard or individual building projects or the Bureau of Technical Inventory (BTI).

If the building has an attic floor, the value V, m3, is defined as the product of the horizontal sectional area of ​​the building at the level of its 1st floor (above the basement) by the free height of the building - from the level of the finished floor of the 1st floor to the upper plane of the thermal insulation layer of the attic floor, with roofs, combined with attic ceilings - up to the middle mark of the roof top. Protruding wall surfaces architectural details and niches in the walls of the building, as well as unheated loggias, when determining the calculated hourly heat load of heating, are not taken into account.

If there is a heated basement in the building, 40% of the volume of this basement must be added to the obtained volume of the heated building. The construction volume of the underground part of the building (basement, ground floor) is defined as the product of the horizontal sectional area of ​​the building at the level of its first floor by the height of the basement (ground floor).

The calculated infiltration coefficient Ki.r is determined by the formula:

where g is the acceleration of gravity, m / s2;

L is the free height of the building, m;

w0 is the calculated wind speed for a given area during the heating season, m / s; adopted according to SNiP 23-01-99.

It is not required to introduce into the calculation of the calculated hourly heat load of heating the building the so-called correction for the effect of wind, because this value has already been taken into account in formula (3.3).

In areas where the calculated value of the outside air temperature for heating design is  -40 ° C, for buildings with unheated basements, additional heat losses through the unheated floors of the first floor should be taken into account in the amount of 5%.

For completed buildings, the calculated hourly heating load should be increased for the first heating period for stone buildings built:

In May-June - by 12%;

In July-August - by 20%;

In September - by 25%;

During the heating season - by 30%.

1.3. The specific heating characteristic of a building qo, kcal / m3 h ° С, in the absence of qo values ​​corresponding to its building volume in tables 3 and 4, can be determined by the formula:

where a = 1.6 kcal / m 2.83 h ° C; n = 6 - for construction buildings before 1958;

a = 1.3 kcal / m 2.875 h ° C; n = 8 - for buildings under construction after 1958

1.4. If a part of a residential building is occupied by a public institution (office, store, pharmacy, laundry reception, etc.), the calculated hourly heat load of the heating must be determined according to the project. If the calculated hourly heat load in the project is indicated only for the building as a whole, or is determined by aggregated indicators, the heat load of individual rooms can be determined from the heat exchange surface area of ​​the installed heating devices using the general equation describing their heat transfer:

Q = k F t, (3.5)

where k is the heat transfer coefficient of the heating device, kcal / m3 h ° С;

F is the area of ​​the heat exchange surface of the heating device, m2;

t is the temperature head of the heating device, ° С, defined as the difference between the average temperature of the convective-radiant heating device and the air temperature in the heated building.

The method for determining the calculated hourly heat load of heating on the surface of installed heating devices of heating systems is given in.

1.5. When connecting heated towel rails to the heating system, the calculated hourly heat load of these heating devices can be defined as the heat transfer of uninsulated pipes in a room with a calculated air temperature tj = 25 ° C according to the method described in Art.

1.6. In the absence of design data and determination of the calculated hourly heat load of heating industrial, public, agricultural and other atypical buildings (garages, underground heated passages, swimming pools, shops, kiosks, pharmacies, etc.) according to aggregated indicators, the values ​​of this load should be specified by the area of ​​the heat exchange surface of the installed heating devices of heating systems in accordance with the methodology given in Art. The initial information for calculations is revealed by a representative of the heat supply organization in the presence of a representative of the subscriber with the preparation of a corresponding act.

1.7. Heat consumption for the technological needs of greenhouses and greenhouses, Gcal / h, is determined from the expression:

, (3.6)

where Qcxi is the heat energy consumption for the i-e technological operations, Gcal / h;

n is the number of technological operations.

In turn,

Qcxi = 1.05 (Qtp + Qv) + Qpol + Qprop, (3.7)

where Qtp and Qw are heat losses through the enclosing structures and during air exchange, Gcal / h;

Qpol + Qprop is the consumption of heat energy for heating irrigation water and steaming the soil, Gcal / h;

1.05 is a coefficient that takes into account the consumption of heat energy for heating household premises.

1.7.1. Heat loss through the enclosing structures, Gcal / h, can be determined by the formula:

Qtp = FK (tj - to) 10-6, (3.8)

where F is the surface area of ​​the enclosing structure, m2;

K is the heat transfer coefficient of the enclosing structure, kcal / m2 h ° С; for single glazing, you can take K = 5.5, single-layer film fencing K = 7.0 kcal / m2 h ° С;

tj and to are the process temperature in the room and the calculated outside air for the design of the corresponding agricultural facility, ° С.

1.7.2. Heat losses during air exchange for glass-covered greenhouses, Gcal / h, are determined by the formula:

Qv = 22.8 Finv S (tj - to) 10-6, (3.9)

where Finv - inventory area of ​​the greenhouse, m2;

S is the volume coefficient, which is the ratio of the volume of the greenhouse and its inventory area, m; can be taken in the range from 0.24 to 0.5 for small greenhouses and 3 or more m - for hangars.

Heat losses during air exchange for greenhouses with a film coating, Gcal / h, are determined by the formula:

Qv = 11.4 Finv S (tj - to) 10-6. (3.9a)

1.7.3. Heat consumption for heating irrigation water, Gcal / h, is determined from the expression:

, (3.10)

where Fpolz is the useful area of ​​the greenhouse, m2;

n is the duration of watering, h.

1.7.4. Heat consumption for soil steaming, Gcal / h, is determined from the expression:

2. Supply ventilation

2.1. In the presence of a typical or individual building projects and the compliance of the installed equipment of the supply ventilation system with the project, the calculated hourly heat load of ventilation can be taken according to the project, taking into account the difference in the values ​​of the calculated outdoor air temperature for the design of ventilation adopted in the project and the current standard value for the area where the considered building.

The recalculation is carried out according to a formula similar to formula (3.1):

, (3.1a)

Qv.pr - the same, according to the project, Gcal / h;

tv.pr is the design temperature of the outside air at which the heat load of the supply ventilation in the project is determined, ° С;

tv is the design temperature of the outside air for the design of supply ventilation in the area where the building is located, ° С; adopted according to the instructions of SNiP 23-01-99.

2.2. In the absence of projects or non-compliance of the installed equipment with the design, the calculated hourly heat load of the supply ventilation must be determined according to the characteristics of the equipment installed in reality, in accordance with the general formula describing the heat transfer of heating installations:

Q = Lc (2 + 1) 10-6, (3.12)

where L is the volumetric flow rate of heated air, m3 / h;

 - density of heated air, kg / m3;

c - heat capacity of heated air, kcal / kg;

2 and 1 are the calculated values ​​of the air temperature at the inlet and outlet of the heating installation, ° С.

The method for determining the calculated hourly heat load of the supply air heaters is described in.

It is permissible to determine the calculated hourly heat load of the supply ventilation of public buildings by aggregated indicators according to the formula:

Qv = Vqv (tj - tv) 10-6, (3.2а)

where qv is the specific thermal ventilation characteristic of the building, depending on the purpose and construction volume of the ventilated building, kcal / m3 h ° С; can be taken according to table 4.

3. Hot water supply

3.1. The average hourly heat load of the hot water supply of the consumer of thermal energy Qhm, Gcal / h, during the heating period is determined by the formula:

where a is the rate of water consumption for hot water supply to the subscriber, l / unit. measurements per day; must be approved by the local government; in the absence of approved norms, it is adopted according to the table of Appendix 3 (mandatory) SNiP 2.04.01-85;

N is the number of units of measurement per day; is the number of residents studying in educational institutions, etc .;

tc is the temperature of tap water during the heating period, ° С; in the absence of reliable information, tc = 5 ° C is taken;

T is the duration of the subscriber's hot water supply system functioning per day, h;

Qt.p - heat losses in the local hot water supply system, in the supply and circulation pipelines external network hot water supply, Gcal / h.

3.2. The average hourly heat load of hot water supply in the non-heating period, Gcal, can be determined from the expression:

, (3.13a)

where Qhm is the average hourly heat load of hot water supply during the heating period, Gcal / h;

 - coefficient taking into account the decrease in the average hourly load of hot water supply during the non-heating period compared to the load during the heating period; if the value of  is not approved by the local government,  is taken equal to 0.8 for the housing and communal sector of cities in central Russia, 1.2-1.5 - for resort, southern cities and settlements, for enterprises - 1.0;

ths, th - hot water temperature during non-heating and heating periods, ° С;

tcs, tc - tap water temperature during non-heating and heating periods, ° С; in the absence of reliable information, tcs = 15 ° С, tc = 5 ° С are taken.

3.3. Heat losses by pipelines of the hot water supply system can be determined by the formula:

where Ki is the heat transfer coefficient of a section of an uninsulated pipeline, kcal / m2 h ° С; you can take Ki = 10 kcal / m2 h ° C;

di and li - diameter of the pipeline in the section and its length, m;

tн and tк ​​- hot water temperature at the beginning and end of the calculated section of the pipeline, ° С;

tamb - ambient temperature, ° С; take by the type of pipelines:

In furrows, vertical canals, communication shafts of sanitary cabins tamb = 23 ° С;

Bathrooms tamb = 25 ° С;

In kitchens and toilets tamb = 21 ° С;

On stairwells tamb = 16 ° С;

In the channels of underground laying of the external hot water supply network tamb = tgr;

In tunnels, tamb = 40 ° С;

In unheated basements tamb = 5 ° С;

In attics tamb = -9 ° С (at an average outside air temperature of the coldest month of the heating period tн = -11 ... -20 ° С);

 - coefficient of efficiency of thermal insulation of pipelines; taken for pipelines with a diameter of up to 32 mm  = 0.6; 40-70 mm  = 0.74; 80-200 mm  = 0.81.

Table 5. Specific heat losses of pipelines of hot water supply systems (by place and method of laying)

Place and method of laying	Heat losses of the pipeline, kcal / hm, at nominal diameter, mm
Main supply riser in the shaft or communication shaft, insulated
Stand without heated towel rails, insulated, in a plumbing shaft, furrow or communication shaft
The same with heated towel rails
Non-insulated riser in the plumbing shaft, furrow or communication shaft or openly in the bathroom, kitchen
Distribution insulated pipelines (supply):
in the basement, in the stairwell
in a cold attic
in a warm attic
Insulated circulation pipelines:
in the basement
in a warm attic
in a cold attic
Non-insulated circulation pipelines:
in apartments
on the staircase
Circulating risers in the strut sanitary cabin or bathroom:
isolated
uninsulated

Note. In the numerator - specific heat losses of pipelines of hot water supply systems without direct water withdrawal in heat supply systems, in the denominator - with direct water withdrawal.

Table 6. Specific heat losses of pipelines of hot water supply systems (by temperature difference)

Temperature drop, ° С

Heat losses of the pipeline, kcal / h m, at nominal diameter, mm

Note. When the temperature difference of hot water differs from its given values, the specific heat losses should be determined by interpolation.

3.4. With absence background information required for calculating heat losses by hot water pipelines, heat losses, Gcal / h, can be determined using a special coefficient Kt.p, taking into account the heat losses of these pipelines, by the expression:

Qt.p = Qhm Kt.p. (3.15)

The heat flow for hot water supply, taking into account heat losses, can be determined from the expression:

Qg = Qhm (1 + Kt.p). (3.16)

To determine the values ​​of the coefficient Kt.p, you can use table 7.

Table 7. Coefficient taking into account heat losses by pipelines of hot water supply systems

studfiles.net

How to calculate the heat load for heating a building

In houses that have been commissioned in recent years, these rules are usually fulfilled, therefore the calculation of the heating capacity of the equipment is based on standard coefficients. An individual calculation can be carried out at the initiative of the homeowner or the communal structure dealing with the supply of heat. This happens when spontaneously replacing heating radiators, windows and other parameters.

See also: How to calculate the power of a heating boiler by the area of ​​the house

Calculation of standards for heating in an apartment

In an apartment serviced by utility company, the calculation of the heat load can be carried out only during the transfer of the house in order to track the parameters of the SNIP in the room received on the balance. Otherwise, the owner of the apartment does this in order to calculate his heat loss in the cold season and eliminate the drawbacks of insulation - use heat-insulating plaster, glue insulation, mount penofol on the ceilings and install metal-plastic windows with a five-chamber profile.

Calculating heat leaks for a utility company to open a dispute usually does not work. The reason is that there are heat loss standards. If the house is put into operation, then the requirements are met. At the same time, heating devices comply with the requirements of SNIP. Replacing the batteries and extracting more heat is prohibited as the radiators are installed according to approved building standards.

Methodology for calculating norms for heating in a private house

Private houses are heated by autonomous systems, which at the same time calculates the load is carried out to comply with the requirements of SNIP, and the correction of heating power is carried out in conjunction with work to reduce heat loss.

Calculations can be done manually using a simple formula or calculator on the site. The program helps to calculate the required power of the heating system and heat leakage typical for the winter period. Calculations are carried out for a specific heat zone.

Basic principles

The methodology includes a number of indicators, which together make it possible to assess the level of insulation of the house, compliance with SNIP standards, as well as the power of the heating boiler. How it works:

• depending on the parameters of walls, windows, insulation of the ceiling and foundation, you calculate thermal leaks. For example, your wall consists of a single layer of clinker bricks and a frame with insulation, depending on the thickness of the walls, they have a certain thermal conductivity in aggregate and prevent heat leakage in winter. Your task is for this parameter to be no less than the one recommended in the SNIP. The same is true for foundations, ceilings and windows;
• find out where heat is lost, bring the parameters to standard;
• calculate the boiler power based on the total volume of rooms - for each 1 cubic meter. m of the room takes 41 W of heat (for example, a 10 m² entrance hall with a ceiling height of 2.7 m requires 1107 W of heating, you need two 600 W batteries);
• you can calculate from the opposite, that is, from the number of batteries. Each section of the aluminum battery provides 170 W of heat and heats 2-2.5 m of the room. If your house requires 30 sections of batteries, then the boiler that can heat the room must have a capacity of at least 6 kW.

The worse the house is insulated, the higher the heat consumption from the heating system

An individual or averaged calculation is carried out for the object. The main point of such a survey is that with good insulation and low heat leakage in winter, 3 kW can be used. In a building of the same area, but without insulation, at low winter temperatures power consumption will be up to 12 kW. Thus, the thermal power and load are assessed not only by area, but also by heat loss.

The main heat losses of a private house:

• windows - 10-55%;
• walls - 20-25%;
• chimney - up to 25%;
• roof and ceiling - up to 30%;
• low floors - 7-10%;
• temperature bridge in corners - up to 10%

These indicators can vary for the better and for the worse. They are assessed depending on the types of windows installed, the thickness of the walls and materials, and the degree of ceiling insulation. For example, in poorly insulated buildings, heat loss through the walls can reach 45%, in this case the expression “we heat the street” is applicable to the heating system. Methodology and the calculator will help you estimate the nominal and calculated values.

Settlement specifics

This technique can still be found under the name "heat engineering calculation". The simplified formula looks like this:

Qt = V × ∆T × K / 860, where

V is the volume of the room, m³;

∆T - maximum difference indoors and outdoors, ° С;

K is the estimated coefficient of heat loss;

860 - conversion factor in kWh / hour.

The heat loss coefficient K depends on the building structure, thickness and thermal conductivity of the walls. For simplified calculations, you can use the following parameters:

• K = 3.0-4.0 - without thermal insulation (non-insulated frame or metal structure);
• K = 2.0-2.9 - low thermal insulation (laying in one brick);
• K = 1.0-1.9 - average thermal insulation ( brickwork in two bricks);
• K = 0.6-0.9 - good thermal insulation according to the standard.

These coefficients are averaged and do not allow us to estimate the heat loss and thermal load on the room, so we recommend using the online calculator.

gidpopechi.ru

Calculation of the heat load for heating a building: formula, examples

When designing a heating system, whether it be an industrial structure or a residential building, it is necessary to carry out competent calculations and draw up a diagram of the heating system circuit. Specialists recommend paying special attention at this stage to calculating the possible heat load on the heating circuit, as well as to the amount of fuel consumed and heat generated.

This term is understood as the amount of heat given off by heating devices. The preliminary calculation of the heat load will allow avoiding unnecessary costs for the purchase of components of the heating system and for their installation. Also, this calculation will help to correctly distribute the amount of heat generated economically and evenly throughout the building.

There are many nuances in these calculations. For example, the material from which the building is built, thermal insulation, region, etc. Specialists try to take into account as many factors and characteristics as possible to obtain a more accurate result.

Calculation of the heat load with errors and inaccuracies leads to ineffective operation of the heating system. It even happens that you have to redo sections of an already working structure, which inevitably leads to unplanned expenses. And housing and communal organizations calculate the cost of services based on heat load data.

The main factors

An ideally designed and designed heating system must maintain the desired room temperature and compensate for the resulting heat loss. When calculating the indicator of the heat load on the heating system in the building, you need to take into account:

Purpose of the building: residential or industrial.

Characteristics of structural elements of the structure. These are windows, walls, doors, roof and ventilation system.

The dimensions of the dwelling. The larger it is, the more powerful the heating system should be. It is imperative to take into account the area window openings, doors, outer walls and the volume of each indoor space.

The presence of special rooms (bath, sauna, etc.).

The degree of equipping with technical devices. That is, the availability of hot water supply, ventilation systems, air conditioning and the type of heating system.

Temperature regime for a single room. For example, storage rooms do not need to be kept at a comfortable temperature.

Number of hot water dispensing points. The more there are, the more the system is loaded.

The area of ​​the glazed surfaces. Rooms with French windows lose a significant amount of heat.

Additional terms. In residential buildings, this can be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in a calendar year, shifts, technological chain production process etc.

Climatic conditions of the region. When calculating heat loss, street temperatures are taken into account. If the differences are insignificant, then a small amount of energy will be spent on compensation. While at -40 ° C outside the window will require significant costs.

Features of existing techniques

The parameters included in the calculation of the heat load are in SNiPs and GOSTs. They also have special heat transfer coefficients. From the passports of the equipment included in the heating system are taken digital characteristics concerning a specific heating radiator, boiler, etc. And also traditionally:

Heat consumption, taken at the maximum for one hour of operation of the heating system,

Maximum heat flux from one radiator

Total heat consumption in a certain period (most often - the season); if an hourly calculation of the load on the heating network is required, then the calculation must be carried out taking into account the temperature difference during the day.

The calculations performed are compared with the heat transfer area of ​​the entire system. The indicator is quite accurate. Some deviations do happen. For example, for industrial buildings, it will be necessary to take into account the reduction in thermal energy consumption on weekends and holidays, and in residential premises - at night.

Methods for calculating heating systems have several degrees of accuracy. Quite complex calculations have to be used to keep the error to a minimum. Less accurate schemes are used if the goal is not to optimize the costs of the heating system.

Basic calculation methods

To date, the calculation of the heat load for heating a building can be carried out in one of the following ways.

Three main

• For the calculation, aggregated indicators are taken.
• The indicators of the structural elements of the building are taken as the base. Calculation of heat losses going to warm up the internal volume of air will also be important here.
• All objects included in the heating system are calculated and summed up.

One exemplary

There is also a fourth option. It has a rather large error, because the indicators are taken very averaged, or they are not enough. Here is the formula - Qfrom = q0 * a * VH * (tHE - tHPO), where:

• q0 is the specific thermal characteristic of the building (most often determined by the coldest period),
• a - correction factor (depends on the region and is taken from ready-made tables),
• VH is the volume calculated from the outer planes.

Simple Calculation Example

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 factor depending on the region.

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

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

Calculation of a heating radiator by area

It depends on the material from which they are made. Most often today bimetallic, aluminum, steel are used, much less often cast iron radiators... Each of them has its own heat transfer rate (heat output). Bimetallic radiators with a distance between the axes of 500 mm, on average, have 180 - 190 watts. Aluminum radiators have almost the same performance.

The heat dissipation of the described radiators is calculated per section. Steel plate radiators are non-separable. Therefore, their heat transfer is determined based on the size of the entire device. For example, the thermal power of a double-row radiator with a width of 1,100 mm and a height of 200 mm will be 1,010 W, and a panel radiator made of steel with a width of 500 mm and a height of 220 mm will be 1,644 W.

The calculation of a heating radiator by area includes the following basic parameters:

Ceiling height (standard - 2.7 m),

Thermal power (per sq. M - 100 W),

One outer wall.

These calculations show that for every 10 sq. m requires 1,000 watts of thermal power. This result is divided by the heat output of one section. The answer is the required number of radiator sections.

For the southern regions of our country, as well as for the northern ones, decreasing and increasing coefficients have been developed.

Average calculation and accurate

Taking into account the described factors, the averaged calculation is carried out according to the following scheme. If for 1 sq. m requires 100 W of heat flow, then a room of 20 sq. m should receive 2,000 watts. The radiator (the popular bimetallic or aluminum) of eight sections emits about 150 watts. We divide 2000 by 150, we get 13 sections. But this is a rather large-scale calculation of the heat load.

The exact one looks a little intimidating. Nothing really complicated. Here is the formula:

Qt = 100 W / m2 × S (premises) m2 × q1 × q2 × q3 × q4 × q5 × q6 × q7, where:

• q1 - glazing type (normal = 1.27, double = 1.0, triple = 0.85);
• q2 - wall insulation (weak or absent = 1.27, 2 bricks wall = 1.0, modern, high = 0.85);
• q3 is the ratio of the total area of ​​window openings to the floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
• q4 - outdoor temperature (the minimum value is taken: -35 ° C = 1.5, -25 ° C = 1.3, -20 ° C = 1.1, -15 ° C = 0.9, -10 ° C = 0.7);
• q5 is the number of outer walls in the room (all four = 1.4, three = 1.3, corner room = 1.2, one = 1.2);
• q6 - type of calculation room above the calculation room (cold attic = 1.0, warm attic = 0.9, heated living room = 0.8);
• q7 - ceiling height (4.5 m = 1.2, 4.0 m = 1.15, 3.5 m = 1.1, 3.0 m = 1.05, 2.5 m = 1.3).

Any of the described methods can be used to calculate the heat load of an apartment building.

Approximate calculation

The conditions are as follows. The minimum temperature in the cold season is -20 ° C. Room 25 sq. m with triple glazing, double-glazed windows, ceiling height 3.0 m, walls in two bricks and unheated attic... The calculation will be as follows:

Q = 100 W / m2 x 25 m2 x 0.85 x 1 x 0.8 (12%) x 1.1 x 1.2 x 1 x 1.05.

The result, 2 356.20, is divided by 150. As a result, it turns out that 16 sections need to be installed in the room with the specified parameters.

If you need to calculate in gigacalories

In the absence of a heat energy meter on an open heating circuit, the calculation of the heat load for heating the building is calculated by the formula Q = V * (T1 - T2) / 1000, where:

• V - the amount of water consumed by the heating system, calculated in tons or m3,
• T1 is a number showing the temperature of hot water, measured in ° C and the temperature corresponding to a certain pressure in the system is taken for calculations. This indicator has its own name - enthalpy. If in a practical way it is not possible to remove the temperature indicators, they resort to the average indicator. It is in the range of 60-65 ° C.
• T2 - cold water temperature. It is rather difficult to measure it in the system, therefore constant indicators have been developed that depend on the temperature regime outside. For example, in one of the regions, in the cold season, this indicator is taken equal to 5, in the summer - 15.
• 1,000 is the coefficient for obtaining the result immediately in gigacalories.

In the case of a closed circuit, the heat load (gcal / h) is calculated in a different way:

Qfrom = α * qо * V * (tv - tn.r) * (1 + Kn.r) * 0.000001, where

• α is a coefficient designed to correct climatic conditions. Taken into account if the outside temperature differs from -30 ° C;
• V is the volume of the building according to external measurements;
• qо is the specific heating index of the structure at a given tн.р = -30оС, measured in kcal / m3 * С;
• tv is the calculated internal temperature in the building;
• tн.р - calculated street temperature for drawing up a project of a heating system;
• Kn.r - coefficient of infiltration. It is caused by the ratio of heat losses of the design building with infiltration and heat transfer through external structural elements at street temperature, which is set within the framework of the project being prepared.

The calculation of the heat load turns out to be somewhat enlarged, but it is this formula that is given in the technical literature.

Inspection with a thermal imager

Increasingly, in order to improve the efficiency of the heating system, they resort to thermal imaging examinations of the building.

These works are carried out in the dark. For a more accurate result, you need to observe the temperature difference between the room and the street: it should be at least 15o. The fluorescent lamps and incandescent lamps turn off. It is advisable to remove carpets and furniture to the maximum, they knock down the device, giving some error.

The survey is slow and the data are recorded carefully. The scheme is simple.

The first stage of work takes place indoors. The device is moved gradually from doors to windows, giving Special attention corners and other joints.

The second stage is the examination of the outer walls of the building with a thermal imager. All the same, the joints are carefully examined, especially the connection with the roof.

The third stage is data processing. First, the device does this, then the readings are transferred to the computer, where the corresponding programs finish processing and give the result.

If the survey was carried out by a licensed organization, then, based on the results of the work, it will issue a report with mandatory recommendations. If the work was carried out personally, then you need to rely on your knowledge and, possibly, the help of the Internet.

highlogistic.ru

Calculation of the heat load for heating: how to do it correctly?

The first and most important stage in the difficult process of organizing the heating of any real estate object (be it a country house or an industrial facility) is the competent design and calculation. In particular, it is imperative to calculate the heat loads on the heating system, as well as the volume of heat and fuel consumption.

Thermal loads

Performing preliminary calculations is necessary not only in order to obtain the entire range of documentation for organizing heating of a real estate object, but also to understand the volume of fuel and heat, and to select one or another type of heat generators.

Heat loads of the heating system: characteristics, definitions

The definition of "heat load on heating" should be understood as the amount of heat, which in the aggregate is given off by heating devices installed in a house or at another facility. It should be noted that before installing all the equipment, this calculation is made to eliminate any troubles, unnecessary financial costs and work.

Calculation of heat loads for heating will help to organize the uninterrupted and efficient operation of the heating system of the property. Thanks to this calculation, it is possible to quickly complete absolutely all the tasks of heat supply, to ensure their compliance with the norms and requirements of SNiP.

A set of instruments for performing calculations

The cost of a calculation error can be quite significant. The thing is that, depending on the calculated data, in the city's housing and communal services department, the maximum expenditure parameters will be allocated, limits and other characteristics are set, from which they are based on when calculating the cost of services.

The total heat load on a modern heating system consists of several main load parameters:

• On common system central heating;
• On the underfloor heating system (if available in the house) - underfloor heating;
• Ventilation system (natural and forced);
• Hot water supply system;
• For all kinds of technological needs: swimming pools, saunas and other similar structures.

Calculation and components of thermal systems at home

The main characteristics of the object, important for accounting when calculating the heat load

The most correct and competently calculated heat load for heating will be determined only when absolutely everything, even the smallest details and parameters, are taken into account.

This list is quite long and you can include in it:

• Type and purpose of real estate objects. Residential or non-residential building, apartment or administrative building - all this is very important for obtaining reliable data on thermal calculation.

Also, the load rate depends on the type of building, which is determined by heat supply companies and, accordingly, heating costs;

• The architectural part. The dimensions of all kinds of external fences (walls, floors, roofs), the dimensions of openings (balconies, loggias, doors and windows) are taken into account. The number of storeys of the building, the presence of basements, attics and their features are important;
• Temperature requirements for each room in the building. This parameter should be understood as the temperature regimes for each room of a residential building or zone of an administrative building;
• The design and features of external fences, including the type of materials, thickness, the presence of insulation layers;

Physical indicators of room cooling - data for calculating heat load

• The nature of the purpose of the premises. As a rule, it is inherent in industrial buildings, where for a shop or site it is necessary to create some specific thermal conditions and modes;
• Availability and parameters of special premises. The presence of the same baths, pools and other similar structures;
• Degree Maintenance- availability of hot water supply, such as centralized heating, ventilation and air conditioning systems;
• The total number of points from which hot water is drawn. It is on this characteristic that you should pay special attention, because the greater the number of points, the greater the heat load on the entire heating system as a whole;
• The number of people living in the home or in the facility. The requirements for humidity and temperature depend on this - factors that are included in the formula for calculating the heat load;

Equipment that can affect thermal loads

• Other data. For an industrial facility, such factors include, for example, the number of shifts, the number of workers in one shift, as well as working days per year.

As for a private house, you need to take into account the number of people living, the number of bathrooms, rooms, etc.

Calculation of heat loads: what is included in the process

Directly the calculation of the heating load with your own hands is carried out even at the design stage of a country cottage or other real estate object - this is due to the simplicity and lack of unnecessary cash costs. This takes into account the requirements of various norms and standards, TCH, SNB and GOST.

The following factors are required to be determined in the course of calculating the heat output:

• Heat loss of external fences. Includes the desired temperature conditions in each of the rooms;
• The power required to heat the water in the room;
• The amount of heat required to heat ventilation air (in the case when forced supply ventilation is required);
• The heat needed to heat the water in the pool or bath;

Gcal / hour - a unit for measuring the thermal loads of objects

• Possible developments of the further existence of the heating system. This implies the possibility of outputting heating to the attic, to the basement, as well as all kinds of buildings and extensions;

Heat loss in a standard residential building

Advice. Thermal loads are calculated with a "margin" in order to exclude the possibility of unnecessary financial costs. Especially relevant for country house, where additional connection of heating elements without preliminary study and preparation will be prohibitively expensive.

Features of calculating the heat load

As discussed earlier, the design parameters of indoor air are selected from the relevant literature. At the same time, the heat transfer coefficients are selected from the same sources (the passport data of the heating units are also taken into account).

The traditional calculation of heat loads for heating requires a sequential determination of the maximum heat flux from heating devices (all actually located in the building heating batteries), the maximum hourly heat energy consumption, as well as the total consumption of heat power for a certain period, for example, the heating season.

Distribution of heat fluxes from various types of heaters

The above instructions for calculating heat loads taking into account the heat exchange surface area can be applied to various real estate objects. It should be noted that this method allows you to competently and correctly develop a justification for the use of effective heating, as well as energy inspection of houses and buildings.

An ideal way of calculating for standby heating of an industrial facility, when it is meant to reduce temperatures during non-working hours (holidays and weekends are also taken into account).

Methods for determining thermal loads

Thermal loads are currently calculated in several main ways:

1. Calculation of heat loss by means of aggregated indicators;
2. Determination of parameters through various elements of enclosing structures, additional losses for air heating;
3. Calculation of heat transfer for all heating and ventilation equipment installed in the building.

An enlarged method for calculating heating loads

Another method for calculating the loads on the heating system is the so-called consolidated method. As a rule, a similar scheme is used in the event that there is no information about projects or such data does not correspond to the actual characteristics.

Examples of heat loads for residential apartment buildings and their dependence on the number of people living and area

For an aggregated calculation of the heat load of heating, a rather simple and uncomplicated formula is used:

Qmax from. = Α * V * q0 * (tv-tn.r.) * 10-6

The formula uses the following coefficients: α is a correction factor that takes into account the climatic conditions in the region where the building is built (used in the case when the design temperature is different from -30C); q0 specific heating characteristic, selected depending on the temperature of the coldest week of the year (the so-called "five-day"); V is the outer volume of the building.

Types of heat loads to be taken into account in the calculation

In the course of calculations (as well as in the selection of equipment), a large number of a wide variety of thermal loads are taken into account:

1. Seasonal loads. As a rule, they have the following features:

• Throughout the year, there is a change in thermal loads depending on the air temperature outside the room;
• Annual heat consumption, which is determined by the meteorological characteristics of the region where the object is located, for which heat loads are calculated;

Thermal load regulator for boiler equipment

• Changing the load on the heating system depending on the time of day. Due to the heat resistance of the building's external fencing, such values ​​are taken as insignificant;
• Heat consumption of the ventilation system by hours of the day.

1. Year-round heat loads. It should be noted that for heating and hot water supply systems, most domestic facilities have heat consumption throughout the year, which changes quite little. So, for example, in summer heat energy consumption is reduced by almost 30-35% in comparison with winter;
2. Dry heat - convection heat exchange and heat radiation from other similar devices. Determined by the dry bulb temperature.

This factor depends on the mass of parameters, including all kinds of windows and doors, equipment, ventilation systems and even air exchange through cracks in walls and ceilings. The number of people who can be in the room is also taken into account;

1. Latent heat - evaporation and condensation. Based on wet bulb temperature. The volume of latent heat of humidity and its sources in the room is determined.

Heat loss of a country house

In any room, humidity is influenced by:

• People and their number who are simultaneously in the room;
• Technological and other equipment;
• Air currents that pass through cracks and crevices in building structures.

Thermal load regulators as a way out of difficult situations

As you can see in many photos and videos of modern industrial and domestic heating boilers and other boiler equipment, they come with special heat load regulators. The technique of this category is designed to provide support for a certain level of loads, to exclude all kinds of jumps and failures.

It should be noted that RTNs allow you to significantly save on heating costs, because in many cases (and especially for industrial enterprises) certain limits are set that cannot be exceeded. Otherwise, if jumps and excess of heat loads are recorded, then fines and similar sanctions are possible.

An example of the total heat load for a certain area of ​​the city

Advice. Loads on heating, ventilation and air conditioning systems - important point in designing a house. If it is impossible to carry out the design work on your own, then it is best to entrust it to specialists. At the same time, all formulas are simple and straightforward, and therefore it is not so difficult to calculate all the parameters yourself.

The load on ventilation and hot water supply is one of the factors of thermal systems

Thermal loads for heating, as a rule, are calculated in conjunction with ventilation. This is a seasonal load, it is intended to replace the exhaust air with clean air, as well as to heat it up to the set temperature.

Hourly heat consumption for ventilation systems is calculated according to a certain formula:

Qv. = Qv.V (tn.-tv.), Where

Measuring heat loss in a practical way

In addition to the ventilation itself, the heat loads on the hot water supply system are also calculated. The reasons for such calculations are similar to ventilation, and the formula is somewhat similar:

Qgvs. = 0.042rv (tg.-tx.) Pgav, where

r, b, tg., tx. - the calculated temperature of hot and cold water, the density of water, as well as the coefficient in which the values ​​are taken into account maximum load hot water supply to the average value established by GOST;

Comprehensive calculation of thermal loads

In addition to, in fact, theoretical issues of calculation, some practical work is also carried out. So, for example, complex heat engineering examinations include mandatory thermography of all structures - walls, ceilings, doors and windows. It should be noted that such works make it possible to determine and fix the factors that have a significant impact on the heat loss of the structure.

Device for calculations and energy audit

Thermal imaging diagnostics will show what the real temperature difference will be when a certain strictly defined amount of heat passes through 1m2 of enclosing structures. Also, it will help to find out the heat consumption at a certain temperature difference.

Practical measurements are an indispensable component of various design work. Together, such processes will help to obtain the most reliable data on heat loads and heat losses that will be observed in a certain structure over a certain period of time. A practical calculation will help to achieve what the theory will not show, namely the "bottlenecks" of each structure.

Conclusion

Calculation of thermal loads, as well as hydraulic calculation of the heating system, is an important factor, the calculations of which must be performed before starting the organization of the heating system. If all the work is done correctly and approached the process wisely, you can guarantee the trouble-free operation of heating, as well as save money on overheating and other unnecessary costs.

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Heating boilers

One of the main components of a comfortable home is a well-thought-out heating system. At the same time, the choice of the type of heating and the required equipment is one of the main questions that need to be answered at the design stage of the house. An objective calculation of the heating boiler power by area will ultimately allow you to get a completely effective heating system.

We will now tell you about the correct conduct of this work. In this case, we will consider the features inherent in different types of heating. After all, they must be taken into account when carrying out calculations and the subsequent decision making on the installation of one or another type of heating.

Basic calculation rules

• room area (S);
• specific power of the heater per 10m² of heated area - (W beats). This value is determined with a correction for the climatic conditions of a particular region.

This value (W beats) is:

• for the Moscow region - from 1.2 kW to 1.5 kW;
• for the southern regions of the country - from 0.7 kW to 0.9 kW;
• for the northern regions of the country - from 1.5 kW to 2.0 kW.

Let's do the calculations

The power is calculated as follows:

W cat. = (S * Wud.): 10

Advice! For simplicity, you can use a simplified version of this calculation. In it, Wud. = 1. Therefore, the heat output of the boiler is defined as 10kW per 100m² of heated area. But with such calculations, at least 15% must be added to the obtained value in order to get a more objective figure.

Calculation example

As you can see, the instructions for calculating the heat transfer rate are simple. But, nevertheless, we will accompany it with a specific example.

The conditions will be as follows. The area of ​​the heated premises in the house is 100m². The specific power for the Moscow region is 1.2 kW. Substituting the available values ​​in the formula, we get the following:

Boiler W = (100x1.2) / 10 = 12 kilowatts.

Calculation for different types of heating boilers

The degree of efficiency of a heating system depends primarily on the correct choice of its type. And of course, on the accuracy of the calculation of the required performance of the heating boiler. If the calculation of the thermal power of the heating system was not carried out accurately enough, then negative consequences will inevitably arise.

If the heat output of the boiler is less than the required one, it will be cold in the rooms in winter. In the case of excess performance, there will be an overconsumption of energy and, accordingly, money spent on heating the building.

House heating system

To avoid these and other problems, it is not enough just knowing how to calculate the power of a heating boiler.

It is also necessary to take into account the features inherent in systems using different types heaters (you can see a photo of each of them further in the text):

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

The choice of one type or another largely depends on the region of residence and the level of infrastructure development. It is important to have the ability to purchase a certain type of fuel. And, of course, its cost.

Solid fuel boilers

The calculation of the power of a solid fuel boiler must be made taking into account the features characterized by the following features of such heaters:

• low popularity;
• relative availability;
• opportunity autonomous work- it is provided in a number of modern models these devices;
• efficiency during operation;
• the need for additional space for storing fuel.

Solid fuel heater

Another characteristic feature that should be taken into account when calculating the heating power of a solid fuel boiler is the cyclicality of the temperature obtained. That is, in rooms heated with its help, the daily temperature will fluctuate within 5 ° C.

Therefore, such a system is far from the best. And if possible, you should refuse it. But, if this is not possible, there are two ways of how to smooth out the existing shortcomings:

1. Using a thermal bulb, which is needed to regulate the air supply. This will increase the burning time and reduce the number of furnaces;
2. The use of water heat accumulators with a capacity of 2 to 10m². They are included in the heating system, allowing you to reduce energy costs and, thereby, save fuel.

All this will reduce the required performance of a solid fuel boiler for heating a private house. Therefore, the effect of the application of these measures must be taken into account when calculating the capacity of the heating system.

Electric boilers

Electric boilers for home heating are characterized by the following features:

• high cost of fuel - electricity;
• possible problems due to network outages;
• environmental friendliness;
• ease of management;
• compactness.

Electric boiler

All these parameters should be taken into account when calculating the power of an electric heating boiler. After all, it is not purchased for one year.

Oil-fired boilers

They have the following characteristic features:

• not environmentally friendly;
• easy to use;
• require additional storage space for fuel;
• have an increased fire hazard;
• use fuel, the price of which is quite high.

Liquid fuel heater

Gas boilers

In most cases, they are the most optimal option for organizing a heating system. Household gas heating boilers have the following characteristic features that must be taken into account when calculating the power of a heating boiler:

• ease of use;
• do not require space for storing fuel;
• safe to operate;
• low fuel cost;
• profitability.

Gas boiler

Calculation for heating radiators

Let's say you decide to install a heating radiator with your own hands. But first, you need to purchase it. Moreover, choose exactly the one that is suitable in terms of power.

• First, we determine the volume of the room. To do this, we multiply the area of ​​the room by its height. As a result, we get 42m³.
• Further, you should know that it takes 41 watts to heat 1m³ of premises in central Russia. Therefore, in order to find out the required performance of the radiator, we multiply this figure (41 W) by the volume of the room. As a result, we get 1722W.
• Now let's count how many sections our radiator should have. This is easy to do. Each element is bimetallic or aluminum radiator heat dissipation is 150W.
• Therefore, we divide the received performance (1722W) by 150. We get 11.48. Round up to 11.
• Now you need to add another 15% to the resulting figure. This will help smooth out the increase in the required heat transfer during the most severe winters. 15% of 11 is 1.68. Round up to 2.
• As a result, to the existing figure (11) we add 2. We get 13. So, to heat a room with an area of ​​14m², we need a 1722W radiator with 13 sections.

Now you know how to calculate the required performance of the boiler, as well as the heating radiator. Take advantage of our advice and provide yourself with an efficient and at the same time not wasteful heating system. If you need more detailed information, then you can easily find it in the corresponding video on our website.

Page 3

All this equipment, indeed, requires a very respectful, prudent attitude - mistakes lead not so much to financial losses as to losses of health and attitude to life.

When we decide to build our own private house, we are primarily guided by largely emotional criteria - we want to have our own separate home, independent of city utilities, much larger in size and made according to our own ideas. But somewhere in the soul, of course, there is also an understanding that you will have to count a lot. Calculations relate not so much to the financial component of all work, but to the technical one. One of the most important types of calculations will be the calculation of the compulsory heating system, without which there is no way to go.

First, of course, you need to tackle the calculations - a calculator, a sheet of paper and a pen will be the first tools

First, decide what is called, in principle, about the methods of heating your home. After all, you have several of the following heat supply options at your disposal:

• Autonomous heating electrical appliances. Perhaps such devices are good, and even popular, as auxiliary heating means, but they can in no way be considered as basic.

• Electric underfloor heating. But this heating method may well be used as the main one for a single living room. But there is no question of providing all rooms in the house with such floors.

• Heating fireplaces. A brilliant option, it warms not only the air in the room, but also the soul, creates an unforgettable atmosphere of comfort. But then again, no one sees fireplaces as a means of providing warmth throughout the house - only in the living room, only in the bedroom, and nothing more.

• Centralized hot water heating. Having “torn” yourself away from a high-rise building, you, nevertheless, can bring its “spirit” into your home by connecting to centralized system heating. Is it worth it !? Is it worth it again to rush "out of the fire, but into the fire." This is not worth doing, even if the possibility exists.

• Autonomous water heating. But this method of providing heat is the most effective, which can be called the main one for private houses.

You cannot do without a detailed plan of the house with a layout of equipment and wiring of all communications

After resolving the issue in principle

When the solution to the fundamental question of how to provide heat in the house using an autonomous water system has taken place, you need to move on and understand that it will be incomplete if you do not think about

• Installation of reliable window systems that will not just "let down" all your heating progress onto the street;

• Additional insulation for both external and interior walls at home. The task is very important and requires a separate serious approach, although it is not directly related to the future installation of the heating system itself;

• Installing a fireplace. Recently, this auxiliary heating method has been increasingly used. It may not replace general heating, but it is such an excellent support for it that in any case it helps to significantly reduce heating costs.

The next step is to create a very accurate diagram of your building with the introduction of all the elements of the heating system into it. Calculation and installation of heating systems without such a scheme is impossible. The elements of this circuit will be:

• Heating boiler, as the main element of the entire system;
• A circulation pump that provides a coolant current in the system;
• Pipelines, as a kind of "blood vessels" of the entire system;
• Heating batteries are those devices that have been known to everyone for a long time and which are the terminal elements of the system and are responsible in our eyes for the quality of its operation;
• Control devices for the state of the system. An accurate calculation of the volume of the heating system is unthinkable without the presence of such devices, which provide information about the real temperature in the system and the volume of the heat carrier passing through;
• Locking and adjusting devices. Without these devices, the work will be incomplete, it is they that will allow you to regulate the operation of the system and adjust according to the readings of control devices;
• Various fitting systems. These systems could well be attributed to pipelines, but their influence on the successful operation of the entire system is so great that fittings and connectors are separated into a separate group of elements for the design and calculation of heating systems. Some experts call electronics - the science of contacts. It is possible, without fear of making a particularly bad mistake, to call the heating system - in many respects, the science of the quality of the compounds, which are provided by the elements of this group.

The heart of the entire hot water heating system is the heating boiler. Modern boilers - whole systems for providing the entire system with hot heat carrier

Helpful advice! When it comes to the heating system, this word "coolant" often appears in the conversation. It is possible, with some degree of approximation, to consider ordinary "water" as the environment that is intended for movement through pipes and radiators of the heating system. But there are some nuances that are associated with the way water is supplied to the system. There are two ways - internal and external. External - from an external cold water supply. In this situation, in fact, ordinary water, with all its disadvantages, will be the coolant. Firstly, in general availability, and, secondly, cleanliness. We strongly recommend that when choosing this method of entering water from the heating system, put a filter at the inlet, otherwise you cannot avoid severe contamination of the system in just one season of operation. If you have chosen a completely autonomous pouring water into the heating system, then do not forget to "flavor" it with all kinds of additives against solidification and corrosion. It is water with such additives that is already called a heat carrier.

Types of heating boilers

Among the heating boilers available for your choice, the following are available:

• Solid fuel - they can be very good in remote areas, in the mountains, in the Far North, where there are problems with external communications. But if access to such communications is not difficult, solid fuel boilers are not used, they lose in the convenience of working with them, if you still need to keep one level of heat in the house;

• Electric - and where now without electricity. But it is necessary to understand that the cost of this type of energy in your home when using electric heating boilers will be so large that the solution to the question of "how to calculate the heating system" in your house will lose any sense - everything will go into electrical wires;

• Liquid fuel. Such boilers on gasoline, solar oil, are asking, but they, due to their non-environmental friendliness, are very unloved by many, and rightly so;

• Household gas heating boilers are the most common types of boilers, very easy to operate and do not require a fuel supply. The efficiency of such boilers is the maximum of all available on the market and reaches 95%.

Pay special attention to the quality of all the materials used, there is no time for savings, the quality of each component of the system, including pipes, must be ideal

Boiler calculation

When they talk about calculating an autonomous heating system, they primarily mean the calculation of a heating gas boiler. Any example of calculating a heating system includes the following formula for calculating the boiler power:

W = S * Wsp / 10,

• S is the total area of ​​the heated room in square meters;
• Wud is the specific power of the boiler per 10 sq. M. premises.

The specific power of the boiler is set depending on the climatic conditions of the region of its use:

• for the Middle band, it is from 1.2 to 1.5 kW;
• for areas of the Pskov level and above - from 1.5 to 2.0 kW;
• for Volgograd and below - from 0.7 - 0.9 kW.

But, after all, our climate of the XXI century has become so unpredictable that, according to by and large, the only criterion when choosing a boiler is your acquaintance with the experience of other heating systems. Perhaps, understanding this unpredictability, for simplicity, it has long been accepted in this formula to always take the specific power as a unit. Don't forget about the recommended values, though.

Calculation and design of heating systems, to a large extent - the calculation of all points of joints, the latest connecting systems, of which there are a huge number on the market, will help here

Helpful advice! This is the desire - to get acquainted with the existing, already working, systems autonomous heating will be very important. If you decide to establish such a system at home, and even with your own hands, then be sure to get acquainted with the heating methods used by your neighbors. It will be very important to get the "calculator for calculating the heating system" first-hand. You will kill two birds with one stone - you will get a good advisor, and maybe in the future, a good neighbor, and even a friend, and you will avoid mistakes that your neighbor may have made in due time.

Circulation pump

The method of supplying the coolant to the system - natural or forced - largely depends on the heated area. Natural does not require any additional equipment and involves the movement of the coolant through the system due to the principles of gravity and heat transfer. Such a heating system can also be called passive.

Much more widespread are active heating systems, in which a circulation pump is used to move the coolant. It is more common to install such pumps on the line from radiators to the boiler, when the water temperature has already subsided and will not be able to negatively affect the operation of the pump.

Certain requirements are imposed on pumps:

• they should be quiet, because they work constantly;
• they must consume little, again due to their constant work;
• they must be very reliable, and this is the most important requirement for pumps in a heating system.

Piping and radiators

The most important component of the entire heating system, which any user constantly encounters, is pipes and radiators.

When it comes to pipes, we have three types of pipes:

• steel;
• copper;
• polymer.

Steel - the patriarchs of heating systems, used from time immemorial. Now steel tubes gradually disappear from the scene, they are inconvenient to use, and, moreover, require welding and are subject to corrosion.

Copper pipes are very popular, especially when carried out hidden wiring... Such pipes are extremely resistant to external influences, but, unfortunately, they are very expensive, which is the main brake on their widespread use.

Polymer - as a solution to problems copper pipes... It is polymer pipes that are a hit of use in modern heating systems. High reliability, resistance to external influences, a huge selection of additional auxiliary equipment specifically for use in heating systems with polymer pipes.

Heating the home is largely ensured by accurate piping and piping.

Radiator calculation

The heat engineering calculation of the heating system necessarily includes the calculation of such an indispensable element of the network as a radiator.

The purpose of calculating a radiator is to obtain the number of its sections for heating a room of a given area.

Thus, the formula for calculating the number of sections in a radiator is:

K = S / (W / 100),

• S is the area of ​​the heated room in square meters (we heat, of course, not the area, but the volume, but the standard height of the room is taken as 2.7 m);
• W - heat transfer of one section in watts, radiator characteristic;
• K is the number of sections in the radiator.

Providing heat in the house is a solution to a whole range of tasks, often not related to each other, but serving the same purpose. One of these autonomous tasks can be the installation of a fireplace.

In addition to the calculation, radiators also require compliance with certain requirements during their installation:

• the installation must be carried out strictly under the windows, in the center, an old and generally accepted rule, but some manage to break it (such an installation prevents the movement of cold air from the window);
• "Ribs" of the radiator must be aligned vertically - but this requirement, somehow no one really pretends to violate, it is obvious;
• the other is not obvious - if there are several radiators in the room, they should be located at the same level;
• it is necessary to provide at least 5-centimeter gaps from top to the window sill and from the bottom to the floor from the radiator; ease of maintenance plays an important role here.

Skillful and precise placement of radiators ensures the success of the entire final result - here you cannot do without diagrams and modeling of the location depending on the size of the radiators themselves

Calculation of water in the system

The calculation of the volume of water in the heating system depends on the following factors:

• the volume of the heating boiler - this characteristic is known;
• pump performance - this characteristic is also known, but it should, in any case, provide the recommended speed of movement of the coolant through the system of 1 m / s;
• the volume of the entire pipeline system - this already needs to be calculated after the fact, after the installation of the system;
• total volume of radiators.

Ideal, of course, looks like hiding all communications behind a plasterboard wall, but this is not always possible to do, and it raises questions from the point of view of the convenience of future maintenance of the system

Helpful advice! It is often not possible to accurately calculate the required volume of water in the system with mathematical precision. Therefore, they act a little differently. First, the system is filled, presumably at 90% of its volume, and its performance is checked. Excess air is vented as work progresses and filling is continued. Hence, there is a need for an additional reservoir with a coolant in the system. As the system is operating, a natural loss of the coolant occurs as a result of evaporation and convection processes, therefore, the calculation of the heating system make-up consists in tracking the loss of water from the additional reservoir.

Of course, we turn to specialists

You can, of course, do many home repairs yourself. But creating a heating system requires too much knowledge and skills. Therefore, even having studied all the photos and video materials on our website, even having familiarized yourself with such indispensable attributes of each element of the system as "instructions", we still recommend that you contact the professionals for installing the heating system.

As the top of the entire heating system - the creation of warm heated floors. But the advisability of installing such floors should be very carefully calculated.

The cost of mistakes when installing an autonomous heating system is very high. It is not worth the risk in this situation. The only thing that remains for you is smart maintenance of the entire system and the call of masters to maintain it.

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Competently made calculations of the heating system for any building - a residential building, workshop, office, store, etc., will guarantee its stable, correct, reliable and quiet operation. In addition, you will avoid misunderstandings with housing workers, unnecessary financial costs and energy losses. Heating can be calculated in several stages.

When calculating heating, many factors must be taken into account.

Calculation stages

• First you need to find out the heat loss of the building. This is necessary to determine the power of the boiler, as well as of each of the radiators. Heat loss is calculated for each room with an external wall.

Note! Next, you will need to check the data. Divide the resulting numbers by the square of the room. This gives you specific heat loss (W / m²). As a rule, it is 50/150 W / m². If the received data is very different from the specified, then you made a mistake. Therefore, the cost of assembling the heating system will be too high.

• Next, you need to select the temperature regime. It is advisable to take the following parameters for calculations: 75-65-20 ° (boiler-radiators-room). This temperature regime, when the heat is calculated, complies with the European heating standard EN 442.

Heating circuit.

• Then it is necessary to select the power of the heating batteries, based on the data on heat loss in the rooms.
• After that, a hydraulic calculation is carried out - heating without it will not be effective. It is needed to determine the diameter of the pipes and the technical properties of the circulation pump. If the house is private, then the cross-section of the pipes can be selected according to the table that will be given below.
• Next, you need to decide on a heating boiler (domestic or industrial).
• Then the volume of the heating system is found. You need to know its capacity in order to choose an expansion tank or to make sure that the volume of the water tank already built into the heat generator is enough. Any online calculator will help you get the data you need.

Thermal calculation

To carry out the thermal engineering stage of the heating system design, you will need initial data.

What you need to get started

House project.

1. First of all, you will need a construction project. It must indicate the external and inner dimensions each of the rooms, as well as windows and external doorways.
2. Next, find out the data on the location of the building in relation to the cardinal points, as well as the climatic conditions in your area.
3. Collect information on the height and composition of the outer walls.
4. You will also need to know the parameters of the floor materials (from the room to the ground), as well as the ceiling (from the premises to the street).

After collecting all the data, you can start calculating the heat consumption for heating. As a result of the work, you will collect information on the basis of which you can carry out hydraulic calculations.

The required formula

Heat loss of the building.

The calculation of the thermal loads on the system should determine the heat loss and the boiler output. In the latter case, the heating calculation formula is as follows:

Мк = 1.2 ∙ Тп, where:

• Мк - power of the heat generator, in kW;
• Тп - heat loss of the building;
• 1.2 is a margin of 20%.

Note! This safety factor takes into account the possibility of a pressure drop in the gas pipeline system in winter, in addition, unforeseen heat losses. For example, as the photo shows, due to a broken window, poor insulation of doors, severe frosts. This margin also allows the temperature regime to be widely regulated.

It should be noted that when the amount of heat energy is calculated, its losses throughout the building are not evenly distributed, on average, the figures are as follows:

• external walls lose about 40% of the total;
• 20% leave through the windows;
• floors give about 10%;
• 10% evaporates through the roof;
• 20% leave through ventilation and doors.

Material ratios

Coefficients of thermal conductivity of some materials.

• K1 - type of windows;
• K2 - wall insulation;
• K3 - means the ratio of the area of ​​windows and floors;
• K4 - the minimum temperature outside;
• K5 - the number of external walls of the building;
• K6 - number of storeys of the structure;
• K7 is the height of the room.

As for the windows, the coefficients of their heat loss are equal:

• traditional glazing - 1.27;
• double-glazed windows - 1;
• three-chamber analogs - 0.85.

The larger the volume of the windows relative to the floors, the large quantity the building loses heat.

When calculating the consumption of thermal energy for heating, keep in mind that the wall material has the following coefficient values:

• concrete blocks or panels - 1.25 / 1.5;
• timber or logs - 1.25;
• masonry 1.5 bricks - 1.5;
• masonry 2.5 bricks - 1.1;
• foam concrete blocks - 1.

Thermal leaks also increase at subzero temperatures.

1. Up to -10 ° the coefficient will be 0.7.
2. From -10 ° it will be 0.8.
3. At -15 °, you need to operate with a figure of 0.9.
4. Up to -20 ° - 1.
5. From -25 ° the value of the coefficient will be 1.1.
6. At -30 ° it will be 1.2.
7. Up to -35 ° this value is 1.3.

When you calculate heat energy, keep in mind that its loss is also dependent on how many external walls in the building:

• one outer wall - 1%;
• 2 walls - 1.2;
• 3 outer walls - 1.22;
• 4 walls - 1.33.

The greater the number of floors, the more difficult the calculations are.

The number of floors or the type of room that is located above the living room affects the K6 coefficient. When the house has two floors and more, the calculation of heat energy for heating takes into account the coefficient 0.82. If at the same time the building has a warm attic, the figure changes to 0.91, if this room is not insulated, then to 1.

The height of the walls affects the level of the coefficient as follows:

• 2.5 m - 1;
• 3 m - 1.05;
• 3.5 m - 1.1;
• 4 m - 1.15;
• 4.5 m - 1.2.

Among other things, the method for calculating the need for heat energy for heating takes into account the area of ​​the room - Pk, as well as the specific value of heat losses - UDtp.

The final formula for the required calculation of the heat loss coefficient looks like this:

Тп = УДтп ∙ Pl ∙ К1 ∙ К2 ∙ К3 ∙ К4 ∙ К5 ∙ К6 ∙ К7. At the same time, UDtp is 100 W / m².

Calculation example

The building for which we will find the load on the heating system will have the following parameters.

1. Double glazed windows, i.e. K1 is 1.
2. External walls are made of foam concrete, the coefficient is the same. 3 of them are external, in other words K5 is 1.22.
3. The square of the windows is 23% of that of the floor - K3 is 1.1.
4. Outside, the temperature is -15 °, K4 is 0.9.
5. The attic of the building is not insulated, in other words, K6 will be 1.
6. The ceiling height is three meters, i.e. K7 is 1.05.
7. The area of ​​the premises is 135 m².

Knowing all the numbers, we substitute them in the formula:

Fri = 135 ∙ 100 ∙ 1 ∙ 1 ∙ 1.1 ∙ 0.9 ∙ 1.22 ∙ 1 ∙ 1.05 = 17120.565 W (17.1206 kW).

Mk = 1.2 ∙ 17.1206 = 20.54472 kW.

Hydraulic calculation for the heating system

An example of a hydraulic calculation scheme.

This design phase will help you choose the right length and diameter of pipes, as well as correctly balance the heating system using radiator valves. This calculation will give you the opportunity to select the power of the electric circulation pump.

High quality circulation pump.

Based on the results of hydraulic calculations, you need to find out the following figures:

• M is the amount of water consumption in the system (kg / s);
• DP - loss of pressure;
• DP1, DP2 ... DPn, - lost head, from heat generator to each battery.

We find out the flow rate of the coolant for the heating system by the formula:

M = Q / Cp ∙ DPt

1. Q means the total heating power, it is taken taking into account the heat losses of the house.
2. Cp is the level of specific heat capacity of water. To simplify calculations, it can be taken as 4.19 kJ.
3. DPt is the temperature difference between the inlet and outlet of the boiler.

In the same way, you can calculate the consumption of water (heat carrier) at any section of the pipeline. Select the areas so that the fluid velocity is the same. According to the standard, the division into sections must be carried out before the reduction or tee. Next, add up the power of all batteries to which water is supplied through each interval of pipes. Then plug the value into the above formula. These calculations need to be made for pipes in front of each of the batteries.

• V is the speed of the coolant advance (m / s);
• M - water consumption in the pipe section (kg / s);
• P is its density (1 t / m³);
  • F is the cross-sectional area of ​​the pipes (m²), it is found by the formula: π ∙ r / 2, where the letter r means the inner diameter.

DPptr = R ∙ L,

• R means specific frictional losses in the pipe (Pa / m);
• L is the length of the section (m);

After that, calculate the pressure loss on the resistances (fittings, fittings), the formula for action:

Dms = Σξ ∙ V² / 2 ∙ P

• Σξ denotes the sum of the coefficients of local resistance in a given section;
• V - water speed in the system
• P is the density of the coolant.

Note! In order for the circulation pump to sufficiently provide all the batteries with heat, the pressure loss on the long branches of the system should not be more than 20,000 Pa. The coolant flow rate should be from 0.25 to 1.5 m / s.

If the speed is higher than the specified value, noise will appear in the system. The minimum speed value of 0, .25 m / s is recommended by SNP # 2.04.05-91, so that the pipes are not airborne.

Pipes made of different materials have different properties.

In order to comply with all the sounded conditions, it is necessary to choose the right pipe diameter. You can do this according to the table below, where the total power of the batteries is indicated.

At the end of the article, you can watch an instructional video on her topic.

Page 5

Heating design standards must be observed for installation

Numerous companies, as well as individuals, offer the population the design of heating with its subsequent installation. But in fact, if you are managing a construction site, you definitely need a specialist in the calculation and installation of heating systems and devices? The fact is that the price of such work is quite high, but with some effort, you can completely cope with it yourself.

How to heat your home

It is impossible to consider the installation and design of heating systems of all types in one article - it is better to pay attention to the most popular ones. Therefore, let's dwell on the calculations of water radiator heating and some features of boilers for heating water circuits.

Calculation of the number of radiator sections and installation location

Sections can be added and removed by hand

• Some Internet users have an obsessive desire to find SNiP for heating calculations in the Russian Federation, but such installations simply do not exist. Such rules are possible for a very small region or country, but not for a country with the most diverse climates. The only thing that can be advised for lovers of printed standards is to contact study guide on the design of water heating systems for the universities of Zaitsev and Lubarets.
• The only standard that deserves attention is the amount of thermal energy that should be emitted by a radiator per 1m2 of the room, with an average ceiling height of 270 cm (but not more than 300 cm). The heat transfer power should be 100W, therefore, the formula is suitable for calculations:

Knumber of sections = Sarea of ​​the room * 100 / P capacity of one section

• For example, you can calculate how many sections are needed for a room of 30m2 with a specific power of one section of 180W. In this case, K = S * 100 / P = 30 * 100/180 = 16.66. Let's round this number in big side for the stock and get 17 sections.

Panel radiators

• And what if the design and installation of heating systems is carried out by panel radiators, where it is impossible to add or remove part of the heating device. In this case, it is necessary to select the power of the battery according to the cubic capacity of the heated room. Now we need to apply the formula:

P power of a panel radiator = V volume of the heated room * 41 the required number of watts per 1 cu.

• Let's take a room of the same size with a height of 270 cm and get V = a * b * h = 5 * 6 * 2? 7 = 81m3. Let's substitute the initial data into the formula: P = V * 41 = 81 * 41 = 3.321kW. But such radiators do not exist, which means we will go to the big side and purchase a device with a power reserve of 4kW.

The radiator must be hung under the window

• Whatever metal the radiators are made of, the rules for designing heating systems provide for their location under the window. The battery heats up the air enveloping it, and as it heats up, it becomes lighter and rises. These warm currents create a natural barrier to cold currents from the window panes, thus increasing the efficiency of the appliance.
• Therefore, if you have calculated the number of sections or calculated the required power of the radiator, this does not mean at all that you can limit yourself to one device if there are several windows in the room (for some panel radiators, the instruction mentions this). If the battery consists of sections, then they can be divided, leaving the same amount under each window, and you just need to purchase several pieces of water from panel heaters, but with less power.

Choosing a boiler for a project

Forging gas boiler Bosch Gaz 3000W

• The terms of reference for the design of a heating system also includes the choice of a domestic heating boiler, and if it runs on gas, then, in addition to the difference in design capacity, it may turn out to be convection or condensing. The first system is quite simple - thermal energy in this case arises only from the combustion of gas, but the second is more complex, because water vapor is also involved there, as a result of which fuel consumption is reduced by 25-30%.
• It is also possible to choose an open or closed combustion chamber. In the first situation, you need a chimney and natural ventilation - this is more cheap way... The second case provides for the forced supply of air into the chamber by a fan and the same removal of combustion products through a coaxial chimney.

Gas generator boiler

• If the design and installation of heating provides for a solid fuel boiler for heating a private house, then it is better to give preference to a gas generator device. The fact is that such systems are much more economical than conventional units, because the combustion of fuel in them occurs almost without residue, and even that evaporates in the form of carbon dioxide and soot. When burning wood or coal from the lower chamber, the pyrolysis gas falls into another chamber, where it already burns to the end, which explains the very high efficiency.

Recommendations. There are still other types of boilers, but now more briefly about them. So, if you opted for a liquid fuel heater, then you can give preference to a unit with a multi-stage burner, thereby increasing the efficiency of the entire system.

Electrode boiler "Galan"

If you prefer electric boilers, then instead of a heating element, it is better to purchase an electrode heater (see photo above). This is a relatively new invention, in which the heat carrier itself serves as a conductor of electricity. But nevertheless, it is completely safe and very economical.

Fireplace for heating a country house

How to optimize heating costs? This task is solved only by an integrated approach that takes into account all the parameters of the system, buildings and climatic features of the region. In this case, the most important component is the heat load on heating: the calculation of hourly and annual indicators is included in the system for calculating the efficiency of the system.

Why you need to know this parameter

What is the calculation of the heat load for heating? It determines the optimal amount of heat energy for each room and the building as a whole. Variables are the power of heating equipment - boiler, radiators and pipelines. The heat losses of the house are also taken into account.

Ideally, the heat output of the heating system should compensate for all heat losses while maintaining a comfortable temperature level. Therefore, before calculating the annual heating load, you need to determine the main factors that affect it:

• Characteristics of the structural elements of the house. External walls, windows, doors, ventilation system affect the level of heat losses;
• House dimensions. It is logical to assume that the larger the room, the more intensively the heating system should work. An important factor in this is not only the total volume of each room, but also the area of ​​the outer walls and window structures;
• The climate in the region. With relatively small drops in temperature outside, a small amount of energy is needed to compensate for heat losses. Those. the maximum hourly heating load directly depends on the degree of temperature decrease in a certain period of time and the average annual value for the heating season.

Taking these factors into account, the optimal thermal mode of operation of the heating system is compiled. Summarizing all of the above, we can say that the determination of the heat load on heating is necessary to reduce energy consumption and maintain the optimal level of heating in the premises of the house.

To calculate the optimal heating load based on aggregated indicators, you need to know the exact volume of the building. It is important to remember that this technique was developed for large structures, so the calculation error will be large.

Choice of calculation method

Before calculating the heating load according to enlarged indicators or with a higher accuracy, it is necessary to find out the recommended temperature conditions for a residential building.

When calculating the characteristics of heating, one must be guided by the norms of SanPiN 2.1.2.2645-10. Based on the data in the table, in each room of the house it is necessary to ensure the optimal temperature mode of heating.

The methods by which the calculation of the hourly heating load is carried out can have varying degrees of accuracy. In some cases, it is recommended to use rather complex calculations, as a result of which the error will be minimal. If the optimization of energy costs is not a priority in the design of heating, less accurate schemes can be used.

When calculating the hourly heating load, the daily change in outdoor temperature must be taken into account. To improve the accuracy of the calculation, you need to know the technical characteristics of the building.

Simple Ways to Calculate Heat Load

Any heat load calculation is needed to optimize the parameters of the heating system or improve thermal insulation characteristics at home. After its completion, certain methods of regulating the heat load of the heating are selected. Consider the easy-to-use methods for calculating this parameter of the heating system.

Dependence of heating power on the area

For a house with standard room sizes, ceiling heights and good thermal insulation, a known ratio of room area to required heat output can be applied. In this case, 10 m² will need to generate 1 kW of heat. To the result obtained, you need to apply a correction factor depending on the climatic zone.

Let's assume that the house is located in the Moscow region. Its total area is 150 m². In this case, the hourly heat load for heating will be equal to:

15 * 1 = 15 kW / hour

The main disadvantage of this method is its large error. The calculation does not take into account changes in weather factors, as well as building features - heat transfer resistance of walls, windows. Therefore, it is not recommended to use it in practice.

Aggregated calculation of the thermal load of a building

The enlarged calculation of the heating load is characterized by more accurate results. Initially, it was used to preliminary calculate this parameter when it was impossible to determine the exact characteristics of the building. The general formula for determining the heat load for heating is presented below:

Where q °- specific thermal characteristics of the structure. The values ​​must be taken from the corresponding table, a- the correction factor mentioned above, Vн- the outer volume of the building, m³, TVn and Tnro- temperature values ​​inside the house and outside.

Suppose you want to calculate the maximum hourly heating load in a house with a volume of 480 m³ along the outer walls (area 160 m², two-storey house). In this case, the thermal characteristic will be equal to 0.49 W / m³ * C. Correction factor a = 1 (for the Moscow region). Optimum temperature inside the dwelling (Tvn) should be + 22 ° С. The temperature outside will be -15 ° C. Let's use the formula to calculate the hourly heating load:

Q = 0.49 * 1 * 480 (22 + 15) = 9.408 kW

Compared to the previous calculation, the resulting value is less. However, it takes into account important factors - the temperature inside the room, outside, the total volume of the building. Similar calculations can be done for every room. The method of calculating the heating load according to the enlarged indicators makes it possible to determine the optimal power for each radiator in a separate room. For a more accurate calculation, you need to know the average temperature values ​​for a particular region.

This calculation method can be used to calculate the hourly heat load for heating. However, the results obtained will not give an optimally accurate value of the building's heat loss.

Accurate heat load calculations

But still, this calculation of the optimal heat load for heating does not give the required calculation accuracy. It does not take into account the most important parameter - the characteristics of the building. The main one is the resistance to heat transfer, the material for the manufacture of individual elements of the house - walls, windows, ceiling and floor. It is they who determine the degree of conservation of thermal energy received from the heat carrier of the heating system.

What is heat transfer resistance ( R)? This is the reciprocal of the thermal conductivity ( λ ) - the ability of the material structure to transfer thermal energy. Those. the higher the value of thermal conductivity, the higher the heat loss. To calculate the annual heating load, you cannot use this value, since it does not take into account the thickness of the material ( d). Therefore, experts use the parameter heat transfer resistance, which is calculated using the following formula:

Calculation for walls and windows

There are normalized values ​​of the heat transfer resistance of walls, which directly depend on the region where the house is located.

In contrast to the aggregated heating load calculation, you first need to calculate the heat transfer resistance for the outer walls, windows, ground floor and attic floor. Let's take the following characteristics of the house as a basis:

• Wall area - 280 m²... It includes windows - 40 m²;
• Wall material - solid brick ( λ = 0.56). External wall thickness - 0.36 m... Based on this, we calculate the resistance of the TV transmission - R = 0.36 / 0.56 = 0.64 m2 * С / W;
• To improve the thermal insulation properties, an external insulation was installed - expanded polystyrene with a thickness 100 mm... For him λ = 0.036... Respectively R = 0.1 / 0.036 = 2.72 m2 * C / W;
• Total value R for external walls is 0,64+2,72= 3,36 which is very a good indicator thermal insulation of the house;
• Heat transfer resistance of windows - 0.75 m² * С / W(double glazing with argon filling).

In fact, heat losses through the walls will be:

(1 / 3.36) * 240 + (1 / 0.75) * 40 = 124 W at a temperature difference of 1 ° C

We take the temperature indicators the same as for the aggregated calculation of the heating load + 22 ° С indoors and -15 ° С outdoors. Further calculation must be done according to the following formula:

124 * (22 + 15) = 4.96 kWh

Ventilation calculation

Then it is necessary to calculate the ventilation losses. The total air volume in the building is 480 m³. Moreover, its density is approximately equal to 1.24 kg / m³. Those. its mass is 595 kg. On average, the air is renewed five times per day (24 hours). In this case, to calculate the maximum hourly load for heating, you need to calculate the heat losses for ventilation:

(480 * 40 * 5) / 24 = 4000 kJ or 1.11 kW / hour

Summing up all the indicators obtained, you can find the total heat loss of the house:

4.96 + 1.11 = 6.07 kWh

In this way, the exact maximum heating load is determined. The resulting value directly depends on the temperature outside. Therefore, to calculate the annual load on the heating system, it is necessary to take into account changes in weather conditions. If the average temperature during the heating season is -7 ° C, then the total heating load will be equal to:

(124 * (22 + 7) + ((480 * (22 + 7) * 5) / 24)) / 3600) * 24 * 150 (days of the heating season) = 15843 kW

By changing the temperature values, you can make an accurate calculation of the heat load for any heating system.

To the results obtained, you need to add the value of heat losses through the roof and floor. This can be done with a correction factor of 1.2 - 6.07 * 1.2 = 7.3 kWh.

The resulting value indicates the actual costs of the energy carrier during the operation of the system. There are several ways to regulate the heating load. The most effective of these is to reduce the temperature in rooms where there is no constant presence of residents. This can be done using thermostats and installed temperature sensors. But at the same time, the building must be installed two-pipe system heating.

To calculate exact value heat loss, you can use the specialized Valtec program. The video material shows an example of working with it.