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

Whether it is an industrial building or a residential building, you need to carry out competent calculations and draw up a diagram of the heating system circuit. At this stage, experts recommend paying special attention to calculating the possible thermal load on the heating circuit, as well as the volume of fuel consumed and heat generated.

Thermal load: what is it?

This term refers to the amount of heat given off. A preliminary calculation of the thermal load will allow you to avoid unnecessary costs for the purchase of heating system components and 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 involved in these calculations. For example, the material from which the building is built, thermal insulation, region, etc. Experts try to take into account as many factors and characteristics as possible to obtain a more accurate result.

Calculation of heat load with errors and inaccuracies leads to inefficient 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 services organizations calculate the cost of services based on data on heat load.

Main Factors

An ideally calculated and designed heating system should maintain the set temperature in the room and compensate for the resulting heat losses. When calculating the heat load on the heating system in a building, you need to take into account:

Purpose of the building: residential or industrial.

Characteristics structural elements buildings. These are windows, walls, doors, roof and ventilation system.

Dimensions of the home. The larger it is, the more powerful the heating system should be. The area must be taken into account window openings, doors, external walls and the volume of each internal room.

Availability of special purpose rooms (bath, sauna, etc.).

Level of equipment technical devices. That is, the availability of hot water supply, ventilation system, air conditioning and type of heating system.

For a separate room. For example, in rooms intended for storage, it is not necessary to maintain a temperature that is comfortable for humans.

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

Area of glazed surfaces. Rooms with French windows lose a significant amount of heat.

Additional terms and conditions. In residential buildings this may be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in a calendar year, shifts, technological chain of the 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 o C outside the window it will require significant expenses.

Features of existing methods

The parameters included in the calculation of the thermal load are found in SNiPs and GOSTs. They also have special heat transfer coefficients. From the passports of the equipment included in the heating system, they take digital specifications, relating to a specific heating radiator, boiler, etc. And also traditionally:

Heat consumption, taken to the maximum per hour of operation of the heating system,

The maximum heat flow emanating from one radiator is

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

The calculations made are compared with the heat transfer area of the entire system. The indicator turns out to be 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. To reduce the error to a minimum, it is necessary to use rather complex calculations. Less accurate schemes are used if the goal is not to optimize the costs of the heating system.

Basic calculation methods

Today, the calculation of the heat load for heating a building can be carried out using one of the following methods.

Three main

For calculations, aggregated indicators are taken.
The indicators of the structural elements of the building are taken as the basis. Here, the calculation of the internal volume of air used for heating will also be important.
All objects included in the heating system are calculated and summed up.

One example

There is also a fourth option. It has a fairly large error, because the indicators taken are very average, or there are not enough of them. This formula is Q from = q 0 * a * V H * (t EN - t NRO), where:

q 0 - 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),
V H is the volume calculated along the external planes.

Example of a simple calculation

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

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

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

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

The heat transfer 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 panel radiator made of steel with a width of 500 mm and a height of 220 mm will amount to 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 external wall.

These calculations show that for every 10 sq. m requires 1,000 W of thermal power. This result is divided by the thermal output of one section. The answer is required amount 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 average calculation is carried out according to the following scheme. If per 1 sq. m requires 100 W of heat flow, then a room of 20 sq. m should receive 2,000 watts. A radiator (popular bimetallic or aluminum) of eight sections produces about Divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.

The exact one looks a little scary. Nothing complicated really. Here's the formula:

Q t = 100 W/m 2 × S(room)m 2 × q 1 × q 2 × q 3 × q 4 × q 5 × q 6 × q 7, Where:

q 1 - type of glazing (regular = 1.27, double = 1.0, triple = 0.85);
q 2 - wall insulation (weak or absent = 1.27, wall laid with 2 bricks = 1.0, modern, high = 0.85);
q 3 - ratio total area window openings to floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
q 4 - street temperature (the minimum value is taken: -35 o C = 1.5, -25 o C = 1.3, -20 o C = 1.1, -15 o C = 0.9, -10 o C = 0.7);
q 5 - number of external walls in the room (all four = 1.4, three = 1.3, corner room= 1.2, one = 1.2);
q 6 - type of calculation room above the calculation room (cold attic = 1.0, warm attic = 0.9, heated residential room = 0.8);
q 7 - 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).

Using any of the described methods, you can calculate the thermal load apartment building.

Approximate calculation

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

Q = 100 W/m 2 × 25 m 2 × 0.85 × 1 × 0.8(12%) × 1.1 × 1.2 × 1 × 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 a room with the specified parameters.

If calculation in gigacalories is required

If there is no thermal energy meter on the open heating circuit calculation of the heat load for heating a building is calculated using the formula Q = V * (T 1 - T 2) / 1000, where:

V - the amount of water consumed by the heating system, calculated in tons or m 3,
T 1 - a number indicating the temperature of hot water, measured in o C and for calculations the temperature corresponding to a certain pressure in the system is taken. This indicator has its own name - enthalpy. If it is not possible to take temperature readings in a practical way, they resort to an averaged reading. It is within 60-65 o C.
T 2 - temperature cold water. It is quite difficult to measure it in the system, so constant indicators have been developed that depend on the temperature 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/hour) is calculated differently:

Q from = α * q o * V * (t in - t n.r.) * (1 + K n.r.) * 0.000001, Where

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

Increasingly, in order to increase the efficiency of the heating system, they are resorting to buildings.

This work is carried out in the dark. For a more accurate result, you need to observe the temperature difference between indoors and outdoors: it should be at least 15 o. Fluorescent and incandescent lamps turn off. It is advisable to remove carpets and furniture as much as possible; they knock down the device, causing some error.

The survey is carried out slowly and data is recorded carefully. The scheme is simple.

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

The second stage is an inspection of the external walls of the building with a thermal imager. The joints are still 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 complete the processing and produce the result.

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

The calculation of the heat load for heating a house is based on specific heat loss, the consumer approach to determining the given heat transfer coefficients - these are the main issues that we will consider in this post. Hello, dear friends! We will calculate with you the heat load for heating the house (Qо.р) different ways according to aggregated measures. So, what we know at the moment: 1. Estimated winter outdoor temperature for heating design tn = -40 oC. 2. Estimated (average) air temperature inside the heated house tв = +20 оС. 3. Volume of the house according to external measurements V = 490.8 m3. 4. Heated area of the house Sfrom = 151.7 m2 (living - Szh = 73.5 m2). 5. Degree day of the heating period GSOP = 6739.2 oC*day.

1. Calculation of the heat load for heating a house based on the heated area. Everything is simple here - it is assumed that heat loss is 1 kW * hour per 10 m2 of heated area of the house, with a ceiling height of up to 2.5 m. For our house, the calculated heat load for heating will be equal to Qo.r = Sot * wud = 151.7 * 0.1 = 15.17 kW. Determining the thermal load using this method is not particularly accurate. The question is, where did this ratio come from and how well does it correspond to our conditions? This is where we need to make a reservation that this ratio is valid for the Moscow region (tn = up to -30 oC) and the house should be properly insulated. For other regions of Russia, specific heat losses wud, kW/m2 are given in Table 1.

Table 1

What else should be taken into account when choosing the specific heat loss coefficient? Reputable design organizations require up to 20 additional data from the “Customer” and this is justified, since the correct calculation of heat loss by a house is one of the main factors determining how comfortable it will be to stay in the room. Below are typical requirements with explanations:
- the severity of the climate zone - the lower the temperature “overboard”, the more you will have to heat it. For comparison: at -10 degrees – 10 kW, and at -30 degrees – 15 kW;
– the condition of the windows – the more airtight and the greater the number of glasses, the lower the losses. For example (at -10 degrees): standard double glazed window - 10 kW, double glazed window - 8 kW, triple glazed window - 7 kW;
– ratio of window and floor areas – than more windows, the more losses. At 20% - 9 kW, at 30% - 11 kW, and at 50% - 14 kW;
– wall thickness or thermal insulation directly affects heat loss. So, with good thermal insulation and sufficient wall thickness (3 bricks - 800 mm), 10 kW is required, with 150 mm of insulation or a wall thickness of 2 bricks - 12 kW, and with poor insulation or a thickness of 1 brick - 15 kW;
– the number of external walls is directly related to drafts and the multilateral effects of freezing. If the room has one external wall, then 9 kW is required, and if it has 4, then 12 kW;
– the ceiling height, although not so significant, still affects the increase in power consumption. At standard height at 2.5 m, 9.3 kW is required, and at 5 m - 12 kW.
This explanation shows that a rough calculation of the required power of 1 kW of a boiler per 10 m2 of heated area is justified.

2. Calculation of the heat load for heating a house using aggregate indicators in accordance with § 2.4 of SNiP N-36-73. To determine thermal load for heating using this method, we need to know the living space of the house. If it is not known, then it is taken as 50% of the total area of the house. Knowing the design temperature of the outside air for heating design, using Table 2 we determine the aggregated indicator of the maximum hourly heat consumption per 1 m2 of living space.

table 2

For our house, the calculated heat load for heating will be equal to Qо.р = Szh * wud.zh = 73.5 * 670 = 49245 kJ/h or 49245/4.19=11752 kcal/h or 11752/860=13.67 kW

3. Calculation of the heat load for heating a house based on the specific heating characteristics of the building.Determine thermal load Using this method, we will use the specific thermal characteristics (specific heat loss) and the volume of the house using the formula:

Qо.р = α * qо * V * (tв – tн) * 10-3, kW

Qо.р – calculated heat load for heating, kW;
α is a correction factor that takes into account the climatic conditions of the area and is used in cases where the estimated outside air temperature tн differs from -30 °C, adopted according to Table 3;
qо – specific heating characteristic buildings, W/m3 * оС;
V – volume of the heated part of the building according to external dimensions, m3;
tв – design air temperature inside the heated building, °C;
tн – design temperature of outside air for heating design, оС.
In this formula, all values, except for the specific heating characteristic of the house qo, are known to us. The latter is a thermal engineering assessment of the construction part of the building and shows the heat flow required to increase the temperature of 1 m3 of building volume by 1 °C. The numerical standard value of this characteristic, for residential buildings and hotels are shown in Table 4.

Correction factor α

Table 3

tн	-10	-15	-20	-25	-30	-35	-40	-45	-50
α	1,45	1,29	1,17	1,08	1	0,95	0,9	0,85	0,82

Specific heating characteristics of the building, W/m3 * оС

Table 4

So, Qо.р = α* qо * V * (tв – tн) * 10-3 = 0.9 * 0.49 * 490.8 * (20 – (-40)) * 10-3 = 12.99 kW. At the stage of feasibility study of construction (project), the specific heating characteristic should be one of the control guidelines. The thing is that in the reference literature its numerical value is different, since it is given for different time periods, before 1958, after 1958, after 1975, etc. In addition, although not significantly, the climate on our planet also changed. And we would like to know the value of the specific heating characteristics of the building today. Let's try to determine it ourselves.

PROCEDURE FOR DETERMINING SPECIFIC HEATING CHARACTERISTICS

1. Prescriptive approach to choosing the heat transfer resistance of external fences. In this case, the consumption of thermal energy is not controlled, and the values of heat transfer resistance individual elements the building must be no less than the standardized values, see Table 5. Here it is appropriate to present Ermolaev’s formula for calculating the specific heating characteristics of the building. This is the formula

qо = [Р/S * ((kс + φ * (kok – kс)) + 1/Н * (kpt + kpl)], W/m3 * оС

φ – glazing coefficient of external walls, take φ = 0.25. This coefficient is taken as 25% of the floor area; P – perimeter of the house, P = 40m; S – area of the house (10 *10), S = 100 m2; H – building height, H = 5m; ks, kok, kpt, kpl – reduced heat transfer coefficients, respectively outer wall, light openings (windows), roof (ceiling), ceiling above the basement (floor). Determination of the given heat transfer coefficients, both with the prescriptive approach and with the consumer approach, see tables 5,6,7,8. Well, we have decided on the building dimensions of the house, but what about the enclosing structures of the house? What materials should the walls, ceiling, floor, windows and doors be made of? Dear friends, you must clearly understand that at this stage we should not be concerned about the choice of material for enclosing structures. The question is, why? Yes, because in the above formula we will put the values of the normalized reduced heat transfer coefficients of enclosing structures. So, regardless of what material these structures will be made of and what their thickness is, the resistance must be certain. (Extract from SNiP II-3-79* Construction heating engineering).

(prescriptive approach)

Table 5

(prescriptive approach)

Table 6

And only now, knowing GSOP = 6739.2 oC*day, using the interpolation method we determine the normalized heat transfer resistance of enclosing structures, see Table 5. The given heat transfer coefficients will be equal, respectively: kpr = 1/ Ro and are given in Table 6. Specific heating characteristics at home qо = = [Р/S * ((kс + φ * (kок – kс)) + 1/Н * (kpt + kpl)] = = 0.37 W/m3 * оС
The calculated heat load for heating with a prescriptive approach will be equal to Qо.р = α* qо * V * (tв – tн) * 10-3 = 0.9 * 0.37 * 490.8 * (20 – (-40)) * 10-3 = 9.81 kW

2. Consumer approach to choosing the heat transfer resistance of external fences. In this case, the heat transfer resistance of external fences can be reduced in comparison with the values indicated in Table 5, until the calculated specific heat energy consumption for heating the house does not exceed the normalized one. The heat transfer resistance of individual fencing elements should not be lower than the minimum values: for the walls of a residential building Rс = 0.63 Ro, for the floor and ceiling Rpl = 0.8 Ro, Rpt = 0.8 Ro, for windows Roк = 0.95 Ro. The calculation results are shown in Table 7. Table 8 shows the given heat transfer coefficients for the consumer approach. As for the specific consumption of thermal energy during the heating period, for our house this value is equal to 120 kJ/m2 * оС * day. And it is determined according to SNiP 02/23/2003. We will determine this value when we calculate the heat load for heating more than in a detailed way– taking into account specific fencing materials and their thermophysical properties (clause 5 of our plan for calculating the heating of a private house).

Standardized heat transfer resistance of enclosing structures
(consumer approach)

Table 7

Determination of reduced heat transfer coefficients of enclosing structures
(consumer approach)

Table 8

Specific heating characteristic of the house qо = = [Р/S * ((kс + φ * (kок – kс)) + 1/Н * (kpt + kpl)] = = 0.447 W/m3 * оС. Estimated heat load for heating at consumer approach will be equal to Qо.р = α * qо * V * (tв – tн) * 10-3 = 0.9 * 0.447 * 490.8 * (20 – (-40)) * 10-3 = 11.85 kW

Main conclusions:
1. Estimated heating load for the heated area of the house, Qо.р = 15.17 kW.
2. Estimated heat load for heating based on aggregated indicators in accordance with § 2.4 of SNiP N-36-73. heated area of the house, Qо.р = 13.67 kW.
3. Estimated heat load for heating the house according to the standard specific heating characteristic of the building, Qо.р = 12.99 kW.
4. Estimated heat load for heating a house using a prescriptive approach to choosing the heat transfer resistance of external fences, Qо.р = 9.81 kW.
5. Estimated heat load for heating a house based on the consumer approach to choosing the heat transfer resistance of external fences, Qo.r = 11.85 kW.
As you can see, dear friends, the calculated heat load for heating a house, with different approaches to its determination, varies quite significantly - from 9.81 kW to 15.17 kW. Which one to choose and not make a mistake? We will try to answer this question in the following posts. Today we completed the 2nd point of our house plan. Who hasn't had time yet, join us!

Best regards, Grigory Volodin

Ask any specialist how to properly organize the heating system in a building. It doesn’t matter whether the property is residential or industrial. And the professional will answer that the main thing is to accurately make calculations and carry out the design correctly. We are talking, in particular, about calculating the heat load for heating. The volume of thermal energy consumption, and therefore fuel, depends on this indicator. That is, economic indicators stand next to technical characteristics.

Performing accurate calculations allows you to get not only full list necessary for carrying out installation work documentation, but also to select the necessary equipment, additional components and materials.

Thermal loads - definition and characteristics

What is usually meant by the term “heating load”? This is the amount of heat that is given off by all heating devices installed in the building. To avoid unnecessary expenses on work, as well as the purchase of unnecessary equipment and materials, a preliminary calculation is necessary. With its help, you can adjust the rules for installing and distributing heat throughout all rooms, and this can be done economically and evenly.

But that's not all. Very often, experts carry out calculations relying on exact indicators. They relate to the size of the house and the nuances of construction, which takes into account the variety of elements of the building and their compliance with the requirements of thermal insulation and other things. It is accurate indicators that make it possible to correctly make calculations and, accordingly, obtain options for distributing thermal energy throughout the premises as close to ideal as possible.

But errors in calculations often occur, which leads to ineffective heating operation as a whole. Sometimes it is necessary to redo not only circuits, but also sections of the system during operation, which leads to additional costs.

What parameters influence the calculation of the thermal load as a whole? Here it is necessary to divide the load into several positions, which include:

System central heating.
Underfloor heating system, if one is installed in the house.
Ventilation system - both forced and natural.
Hot water supply to the building.
Branches for additional household needs. For example, for a sauna or bathhouse, for a swimming pool or shower.

Main characteristics

Professionals do not lose sight of any little detail that may affect the correctness of the calculation. Hence, there is a fairly large list of characteristics of the heating system that should be taken into account. Here are just a few of them:

The purpose of the property or its type. It can be a residential building or an industrial one. Heat energy suppliers have standards that are distributed according to the type of building. They are often the basis for calculations.
Architectural part of the building. This can include enclosing elements (walls, roofing, ceilings, floors), their dimensions, thickness. Be sure to take into account all kinds of openings - balconies, windows, doors, etc. It is very important to take into account the presence of basements and attics.
Temperature conditions for each room separately. This is very important because General requirements to the temperature in the house do not give an accurate picture of heat distribution.
Purpose of premises. This mainly applies to production workshops, in which more strict adherence to temperature conditions is necessary.
Availability of special premises. For example, in private residential buildings these could be baths or saunas.
Degree of technical equipment. The presence of a ventilation and air conditioning system, hot water supply, and the type of heating used are taken into account.
The number of points through which hot water is drawn. And the more such points, the greater the thermal load the heating system is exposed to.
The number of people at the site. Criteria such as indoor humidity and temperature depend on this indicator.
Additional indicators. In residential premises, you can highlight the number of bathrooms, separate rooms, and balconies. IN industrial buildings- the number of shifts of workers, the number of days a year when the workshop itself operates in the technological chain.

What is included in load calculations

Heating scheme

Calculation of thermal loads for heating is carried out at the design stage of the building. But at the same time, the norms and requirements of various standards must be taken into account.

For example, heat loss from the building envelope. Moreover, all rooms separately are taken into account. Next, this is the power that is needed to heat the coolant. Let's add here the amount of thermal energy required for heating supply ventilation. Without this, the calculation will not be very accurate. Let's also add the energy that is spent on heating water for a bathhouse or swimming pool. Experts must take into account the further development of the heating system. Suddenly, in a few years, you will decide to arrange a Turkish hammam in your own private home. Therefore, it is necessary to add a few percent to the loads - usually up to 10%.

Recommendation! It is necessary to calculate thermal loads with a “margin” for country houses. It is the reserve that will allow you to avoid additional financial costs in the future, which are often determined by amounts of several zeros.

Features of calculating thermal load

Air parameters, or more precisely, its temperature, are taken from GOSTs and SNiPs. Heat transfer coefficients are also selected here. By the way, the passport data of all types of equipment (boilers, heating radiators, etc.) must be taken into account.

What is usually included in a traditional heat load calculation?

Firstly, the maximum flow of thermal energy emanating from heating devices (radiators).
Secondly, maximum flow heat per 1 hour of operation of the heating system.
Thirdly, the total heat costs for a certain period of time. Usually the seasonal period is calculated.

If all these calculations are compared and compared with the heat transfer area of the system as a whole, you will get a fairly accurate indicator of the efficiency of heating a house. But small deviations will also have to be taken into account. For example, reducing heat consumption at night. For industrial facilities You will also have to take into account weekends and holidays.

Methods for determining thermal loads

Heated floor design

Currently, experts use three main methods for calculating thermal loads:

Calculation of the main heat losses, where only aggregated indicators are taken into account.
Indicators based on the parameters of enclosing structures are taken into account. Losses for heating the internal air are usually added here.
All systems that are part of the heating networks are calculated. This includes heating and ventilation.

There is another option called enlarged calculation. It is usually used when there are no basic indicators and parameters of the building necessary for standard calculations. That is, the actual characteristics may differ from the design ones.

To do this, experts use a very simple formula:

Q max from.=α x V x q0 x (tв-tн.р.) x 10 -6

α is a correction factor depending on the region of construction (tabular value)
V - volume of the building along external planes
q0 - characteristic of the heating system according to the specific indicator, usually determined by the coldest days of the year

Types of thermal loads

Thermal loads that are used in heating system calculations and equipment selection have several varieties. For example, seasonal loads, which have the following features:

Changes in outdoor temperature throughout the heating season.
Meteorological features of the region where the house was built.
Load surges on the heating system during the day. This indicator usually falls into the “minor load” category, because the enclosing elements prevent high pressure for heating in general.
Everything related to thermal energy associated with a building's ventilation system.
Heat loads that are determined throughout the year. For example, hot water consumption in the summer season is reduced by only 30-40% when compared with winter time of the year.
Dry heat. This feature is inherent specifically in domestic heating systems, where a fairly large number of indicators are taken into account. For example, the number of window and doorways, the number of people living or permanently staying in the house, ventilation, air exchange through all kinds of cracks and gaps. To determine this value, use a dry thermometer.
Hidden thermal energy. There is also a term that is defined by evaporation, condensation, and so on. To determine the indicator, a wet thermometer is used.

Thermal load regulators

Programmable controller, temperature range - 5-50 C

Modern heating units and devices are provided with a set of different regulators, with the help of which you can change the thermal loads, thereby avoiding dips and surges in thermal energy in the system. Practice has shown that with the help of regulators it is possible not only to reduce loads, but also to bring the heating system to rational use fuel. And this is a purely economic side of the issue. This especially applies to industrial facilities, where you have to pay quite large fines for excessive fuel consumption.

If you are not sure of the correctness of your calculations, then use the services of specialists.

Let's look at a couple more formulas that relate to different systems. For example, ventilation and hot water supply systems. Here you will need two formulas:

Qв.=qв.V(tн.-tв.) - this concerns ventilation.
Here:
tn. and tв - air temperature outside and inside
qv. - specific indicator
V - external volume of the building

Qgws.=0.042rv(tg.-tx.)Pgsr - for hot water supply, where

tg.-tx - temperature of hot and cold water
r - water density
c - the ratio of the maximum load to the average, which is determined by GOSTs
P - number of consumers
Gav - average hot water consumption

Complex calculation

In combination with calculation issues, thermotechnical studies must be carried out. For this, various instruments are used that provide accurate indicators for calculations. For example, for this purpose, window and door openings, ceilings, walls, and so on are examined.

It is such an examination that helps to determine the nuances and factors that can have a significant impact on heat loss. For example, thermal imaging diagnostics will accurately show the temperature difference when a certain amount of thermal energy passes through 1 square meter of the building envelope.

So practical measurements are indispensable when carrying out calculations. This is especially true for bottlenecks in the building structure. In this regard, the theory will not be able to show exactly where and what is wrong. And practice will indicate where it is necessary to apply different methods protection against heat loss. And the calculations themselves are becoming more accurate in this regard.

Conclusion on the topic

The calculated heat load is a very important indicator obtained in the process of designing a home heating system. If you approach the matter wisely and carry out everything necessary calculations correctly, we can guarantee that heating system will work great. And at the same time it will be possible to save on overheating and other costs that can simply be avoided.

The topic of this article is thermal load. We will find out what this parameter is, what it depends on and how it can be calculated. In addition, the article will provide a number of reference values for thermal resistance different materials, which may be needed for calculations.

What it is

The term is essentially intuitive. Thermal load means the amount of thermal energy that is necessary to maintain a comfortable temperature in a building, apartment or separate room.

Maximum hourly load for heating, therefore, is the amount of heat that may be required to maintain normal parameters for an hour under the most unfavorable conditions.

Factors

So, what influences a building's heat demand?

Wall material and thickness. It is clear that a wall of 1 brick (25 centimeters) and a wall made of aerated concrete under a 15-centimeter foam coat will transmit VERY different amounts of thermal energy.
Roof material and structure. Flat roof from reinforced concrete slabs and an insulated attic will also differ very noticeably in heat loss.
Ventilation is another important factor. Its performance and the presence or absence of a heat recovery system affect how much heat is lost in the exhaust air.
Glazing area. Significantly more heat is lost through windows and glass facades than through solid walls.

However: triple glazed windows and glass with energy-saving coating reduce the difference several times.

Insolation level in your region, absorption rate solar heat external covering and orientation of the building planes relative to the cardinal directions. Edge Cases- a house located all day long in the shade of other buildings and a house oriented with a black wall and a black sloping roof with a maximum area facing south.

Temperature delta between indoors and outdoors determines the heat flow through the enclosing structures at constant resistance to heat transfer. At +5 and -30 outside, the house will lose different amounts of heat. This will, of course, reduce the need for thermal energy and reduce the temperature inside the building.
Finally, it is often necessary to include in a project prospects for further construction. Let’s say, if the current heat load is 15 kilowatts, but in the near future it is planned to add an insulated veranda to the house, it is logical to purchase one with a reserve of heat power.

Distribution

In the case of water heating, the peak thermal power of the heat source must be equal to the sum of the thermal power of all heating devices in the house. Of course, wiring should not become a bottleneck either.

The distribution of heating devices throughout the premises is determined by several factors:

The area of the room and the height of its ceiling;
Location inside the building. Corner and end rooms lose more heat than those located in the middle of the house.
Remoteness from the heat source. IN individual construction this parameter means the distance from the boiler, in the central heating system of an apartment building - whether the battery is connected to the supply or return riser and what floor you live on.

Clarification: in houses with bottom filling, the risers are connected in pairs. On the supply side, the temperature decreases as you rise from the first floor to the last; on the return side, the opposite is true.

It’s also not difficult to guess how the temperatures will be distributed in the case of top filling.

Desired room temperature. In addition to filtering heat through external walls, inside the building, with uneven temperature distribution, the migration of thermal energy through the partitions will also be noticeable.

For living rooms in the middle of the building - 20 degrees;
For living rooms in the corner or end of the house - 22 degrees. More heat, among other things, prevents walls from freezing.
For the kitchen - 18 degrees. As a rule, it contains a large number of own heat sources - from a refrigerator to an electric stove.
For a bathroom and a combined toilet, the norm is 25C.

When air heating the heat flow entering a separate room is determined by the throughput of the air hose. Usually, simplest method adjustments - manual adjustment of the positions of adjustable ventilation grilles with temperature control using a thermometer.

Finally, if we are talking about a heating system with distributed heat sources (electric or gas convectors, electric heated floors, infrared heaters and air conditioners) necessary temperature regime simply set on the thermostat. All that is required of you is to provide the peak thermal power of the devices at the level of the peak heat loss of the room.

Calculation methods

Dear reader, do you have a good imagination? Let's imagine a house. Let it be a log house made of 20-centimeter timber with an attic and a wooden floor.

Let’s mentally complete and concretize the picture that has arisen in our heads: the dimensions of the residential part of the building will be equal to 10*10*3 meters; We will cut 8 windows and 2 doors in the walls - to the front and inner courtyards. Now let’s place our house... say, in the city of Kondopoga in Karelia, where the temperature at the peak of frost can drop to -30 degrees.

Determining the heat load for heating can be done in several ways with varying complexity and reliability of the results. Let's use the three simplest ones.

Method 1

Current SNiPs offer us the simplest method of calculation. One kilowatt of thermal power is taken per 10 m2. The resulting value is multiplied by the regional coefficient:

For the southern regions (Black Sea coast, Krasnodar region) the result is multiplied by 0.7 - 0.9.
The moderately cold climate of the Moscow and Leningrad regions will force the use of a coefficient of 1.2-1.3. It seems that our Kondopoga will fall into this particular climate group.
Finally, for Far East regions of the Far North, the coefficient ranges from 1.5 for Novosibirsk to 2.0 for Oymyakon.

The instructions for calculating using this method are incredibly simple:

The area of the house is 10*10=100 m2.
The basic value of the thermal load is 100/10=10 kW.
We multiply by the regional coefficient of 1.3 and get 13 kilowatts of thermal power necessary to maintain comfort in the house.

However: if you use such a simple method, it is better to make a reserve of at least 20% to compensate for errors and extreme cold. Actually, it will be indicative to compare 13 kW with values obtained by other methods.

Method 2

It is clear that with the first calculation method the errors will be huge:

Ceiling heights vary greatly between buildings. Taking into account the fact that we have to heat not an area, but a certain volume, and with convection heating warm air going under the ceiling is an important factor.
Windows and doors let in more heat than walls.
Finally, it would be a clear mistake to cut hair with one brush city apartment(and regardless of its location inside the building) and a private house, which has no below, above and behind the walls warm apartments neighbors, and the street.

Well, let's adjust the method.

Let's take 40 watts per cubic meter of room volume as the base value.
For each door leading to the street, add 200 watts to the base value. For each window - 100.
For corner and end apartments in apartment building Let's introduce a coefficient of 1.2 - 1.3 depending on the thickness and material of the walls. We also use it for the outermost floors if the basement and attic are poorly insulated. For a private house, we will multiply the value by 1.5.
Finally, we apply the same regional coefficients as in the previous case.

How is our house in Karelia doing?

The volume is 10*10*3=300 m2.
The basic value of thermal power is 300*40=12000 watts.
Eight windows and two doors. 12000+(8*100)+(2*200)=13200 watts.
A private house. 13200*1.5=19800. We begin to vaguely suspect that when selecting the boiler power using the first method, we would have to freeze.
But there is still a regional coefficient left! 19800*1.3=25740. Total - we need a 28-kilowatt boiler. Difference from the first value obtained in a simple way- double.

However: in practice, such power will be required only on a few days of peak frost. Often, a reasonable solution would be to limit the power of the main heat source to a lower value and buy a backup heater (for example, an electric boiler or several gas convectors).

Method 3

Make no mistake: the described method is also very imperfect. We very roughly took into account the thermal resistance of the walls and ceiling; The temperature delta between internal and external air is also taken into account only in the regional coefficient, that is, very approximately. The price of simplifying calculations is a large error.

Let us remember: to maintain a constant temperature inside the building, we need to provide an amount of thermal energy equal to all losses through the building envelope and ventilation. Alas, here too we will have to somewhat simplify our calculations, sacrificing the reliability of the data. Otherwise, the resulting formulas will have to take into account too many factors that are difficult to measure and systematize.

The simplified formula looks like this: Q=DT/R, where Q is the amount of heat that is lost by 1 m2 of the building envelope; DT is the temperature delta between the internal and external temperatures, and R is the heat transfer resistance.

Please note: we are talking about heat loss through the walls, floor and ceiling. On average, another 40% of heat is lost through ventilation. To simplify the calculations, we will calculate the heat loss through the enclosing structures, and then simply multiply them by 1.4.

Temperature delta is easy to measure, but where do you get thermal resistance data?

Alas, only from reference books. Here is a table for some popular solutions.

A wall of three bricks (79 centimeters) has a heat transfer resistance of 0.592 m2*C/W.
A wall of 2.5 bricks is 0.502.
Wall with two bricks - 0.405.
Brick wall (25 centimeters) - 0.187.
Log house with a log diameter of 25 centimeters - 0.550.
The same, but from logs with a diameter of 20 cm - 0.440.
Log house made of 20 cm timber - 0.806.
Log frame made of timber 10 cm thick - 0.353.
Frame wall 20 centimeters thick with insulation mineral wool — 0,703.
A wall made of foam or aerated concrete with a thickness of 20 centimeters is 0.476.
The same, but with a thickness increased to 30 cm - 0.709.
Plaster 3 centimeters thick - 0.035.
Ceiling or attic floor — 1,43.
Wooden floor - 1.85.
Double door made of wood - 0.21.

Now let's go back to our house. What parameters do we have?

The temperature delta at the peak of frost will be equal to 50 degrees (+20 inside and -30 outside).
Heat loss through a square meter of floor will be 50/1.85 (heat transfer resistance of a wooden floor) = 27.03 watts. Across the entire floor - 27.03*100=2703 watts.
Let's calculate the heat loss through the ceiling: (50/1.43)*100=3497 watts.
The area of the walls is (10*3)*4=120 m2. Since our walls are made of 20-centimeter timber, the R parameter is 0.806. Heat loss through the walls is equal to (50/0.806)*120=7444 watts.
Now let’s add up the resulting values: 2703+3497+7444=13644. This is exactly how much our house will lose through the ceiling, floor and walls.

Note: in order not to calculate fractions square meters, we neglected the difference in thermal conductivity of walls and windows and doors.

Then we add 40% of losses for ventilation. 13644*1.4=19101. According to this calculation, a 20-kilowatt boiler should be enough for us.

Conclusions and problem solving

As you can see, the available methods for calculating the thermal load with your own hands give very significant errors. Fortunately, excess boiler power won't hurt:

Gas boilers operate at reduced power with virtually no drop in efficiency, while condensing boilers even reach the most economical mode at partial load.
The same applies to solar boilers.
Electric heating equipment of any type always has an efficiency of 100 percent (of course, this does not apply to heat pumps). Remember physics: all power not spent on mechanical work (that is, moving mass against the gravity vector) is ultimately spent on heating.

The only type of boilers for which operation at a power less than rated is contraindicated is solid fuel. The power control in them is carried out in a rather primitive way - by limiting the flow of air into the firebox.

What is the result?

If there is a lack of oxygen, the fuel does not burn completely. More ash and soot are produced, which pollute the boiler, chimney and atmosphere.
The consequence of incomplete combustion is a drop in boiler efficiency. It’s logical: after all, fuel often leaves the boiler before it burns.

However, here too there is a simple and elegant way out - including a heat accumulator in the heating circuit. A thermally insulated tank with a capacity of up to 3000 liters is connected between the supply and return pipelines, disconnecting them; in this case, a small contour is formed (between the boiler and the buffer tank) and a large one (between the tank and the heating devices).

How does this scheme work?

After lighting, the boiler operates at rated power. Moreover, due to natural or forced circulation its heat exchanger transfers heat to the buffer tank. After the fuel has burned out, circulation in the small circuit stops.
For the next few hours, the coolant moves along a large circuit. The buffer tank gradually releases the accumulated heat to radiators or water-heated floors.

Conclusion

As always, you will find some additional information on how else the heat load can be calculated in the video at the end of the article. Warm winters!

Hello, dear readers! Today is a short post about calculating the amount of heat for heating using aggregated indicators. In general, the heating load is accepted according to the project, that is, the data calculated by the designer is entered into the heat supply contract.

But often such data is simply not available, especially if the building is small, such as a garage, or some kind of utility room. In this case, the heating load in Gcal/h is calculated using the so-called aggregated indicators. I wrote about this. And this figure is already included in the contract as the calculated heating load. How is this figure calculated? And it is calculated according to the formula:

Qot = α*qо*V*(tв-tн.р)*(1+Kн.р)*0.000001; Where

α is a correction factor that takes into account the climatic conditions of the area; it is applied in cases where the estimated outdoor air temperature differs from -30 °C;

qo is the specific heating characteristic of the building at tn.r = -30 °C, kcal/cub.m*C;

V is the volume of the building according to external measurements, m³;

tв is the design temperature inside the heated building, °C;

tн.р - calculated outside air temperature for heating design, °C;

Kn.r is the infiltration coefficient, which is determined by thermal and wind pressure, that is, the ratio of heat losses by the building with infiltration and heat transfer through external fences at the air temperature outside, which is calculated for heating design.

So, in one formula you can calculate the heat load for heating any building. Of course, this calculation is largely approximate, but it is recommended in the technical literature on heat supply. Heat supply organizations also include this figure for the heating load Qot, in Gcal/h, in heat supply contracts. So the calculation is necessary. This calculation is well presented in the book - V.I. Manyuk, Ya.I. Kaplinsky, E.B. Khizh and others. “Handbook for the setup and operation of water heating networks.” This book is one of my reference books, a very good book.

Also, this calculation of the heat load for heating a building can be done using the “Methodology for determining the amounts of thermal energy and coolant in public water supply systems” of RAO Roskommunenergo of the State Construction Committee of Russia. True, there is an inaccuracy in the calculation in this method (in formula 2 in Appendix No. 1 it is indicated 10 to the minus third power, but it should be 10 to the minus sixth power, this must be taken into account in the calculations), you can read more about this in the comments to this article.

I fully automated this calculation, added reference tables, including a table of climatic parameters for all regions of the former USSR (from SNiP 01/23/99 “Construction Climatology”). You can purchase a calculation in the form of a program for 100 rubles by writing to me by email [email protected].

I will be glad to receive comments on the article.