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

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.
Number of points through which selection is carried out hot water. 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’s enclosing elements. Moreover, all rooms separately are taken into account. Next, this is the power that is required 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 for 1 hour of operation 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 quite large fines have to be paid 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, you can guarantee that the heating system will work perfectly. And at the same time it will be possible to save on overheating and other costs that can simply be avoided.

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 measurements. So, what we know at the moment: 1. Calculated 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 based on aggregated 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 specific heating characteristic 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 of the building, 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 qо, 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, Qo.r = 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

In district heating systems (DHS), heat is supplied to various heat consumers through heating networks. Despite the significant diversity of heat load, it can be divided into two groups according to the nature of its occurrence over time: 1) seasonal; 2) year-round.

Changes in seasonal load depend mainly on climatic conditions: outside temperature, wind direction and speed, solar radiation, air humidity, etc. The outside temperature plays a major role. Seasonal load has a relatively constant daily schedule and a variable annual load schedule. Seasonal heat loads include heating, ventilation, and air conditioning. None of these types of load is year-round. Heating and ventilation are winter heat loads. Air conditioning in summer requires artificial refrigeration. If this artificial cold is produced by the absorption or ejection method, then the thermal power plant receives an additional summer heat load, which helps to increase the efficiency of heating.

Year-round loads include process load and hot water supply. The only exceptions are some industries, mainly related to the processing of agricultural raw materials (for example, sugar), the work of which is usually seasonal.

The technological load schedule depends on the profile of production enterprises and their operating mode, and the hot water supply load schedule depends on the improvement of residential and public buildings, the composition of the population and their working hours, as well as on the operating mode of public utilities - baths, laundries. These loads have a variable daily schedule. Annual charts process load and hot water supply load also depend to some extent on the time of year. As a rule, summer loads are lower than winter ones due to the higher temperature of the processed raw materials and tap water, as well as due to lower heat losses of heat pipes and production pipelines.

One of the primary tasks in the design and development of the operating mode of centralized heat supply systems is to determine the values and nature of heat loads.

In the case when, when designing district heating installations, there is no data on the estimated heat consumption based on the designs of heat-consuming installations of subscribers, the calculation of the heat load is carried out on the basis of aggregated indicators. During operation, the values of the calculated heat loads are adjusted according to actual costs. Over time, this makes it possible to establish a proven thermal characteristic for each consumer.

The main task of heating is to maintain the internal temperature of the premises at a given level. To do this, it is necessary to maintain a balance between the building’s heat losses and heat gain. The condition for thermal equilibrium of a building can be expressed as the equality

Where Q– total heat losses buildings; Q T– heat loss by heat transfer through external fences; QH– heat loss by infiltration due to cold air entering the room through leaks in the external enclosures; Q o– heat supply to the building through the heating system; Q TB – internal heat generation.

The heat loss of a building mainly depends on the first term Q r Therefore, for ease of calculation, the heat losses of the building can be represented as follows:

(5)

where μ= Q And /Q T– infiltration coefficient, which is the ratio of heat loss by infiltration to heat loss by heat transfer through external fences.

The source of internal heat generated by Q TV in residential buildings is usually people, cooking appliances (gas, electric and other stoves), lighting. These heat releases are largely random in nature and cannot be controlled in any way over time.

In addition, heat emissions are not distributed evenly throughout the building.

To ensure normal temperature conditions in residential areas in all heated rooms, the hydraulic and temperature conditions of the heating network are usually set according to the most unfavorable conditions, i.e. according to the heating mode of rooms with zero heat release (Q TB = 0).

To prevent a significant increase in internal temperature in rooms where internal heat release is significant, it is necessary to periodically turn off some heating devices or reduce the coolant flow through them.

A high-quality solution to this problem is possible only with individual automation, i.e. when installing autoregulators directly on heating devices and ventilation heaters.

The source of internal heat generation in industrial buildings is thermal and power plants and mechanisms (furnaces, dryers, engines, etc.) of various kinds. Internal heat dissipation industrial enterprises They are quite stable and often represent a significant share of the design heating load, so they should be taken into account when developing a heat supply regime for industrial areas.

Heat loss by heat transfer through external fences, J/s or kcal/h, can be determined by calculation using the formula

(6)

Where F- surface area of individual external fences, m; To- heat transfer coefficient of external fences, W/(m 2 K) or kcal/(m 2 h °C); Δt is the difference in air temperature from the inside and outside of the enclosing structures, °C.

For a building with a volume along the external dimension V, m, perimeter in plan R, m, plan area S, m, and height L, m, equation (6) can easily be reduced to the formula proposed by prof. N.S. Ermolaev.

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 one. 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 to the 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 a central heating system 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 an 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), the required temperature regime is 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 technique, 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!

The first and most important stage in the difficult process of organizing heating of any property (be it Vacation home or industrial facility) is the competent execution of design and calculations. In particular, it is necessary to calculate the thermal load on the heating system, as well as the volume of heat and fuel consumption.

Performance preliminary calculations necessary not only to obtain the entire range of documentation for organizing the heating of a property, but also to understand the volumes of fuel and heat, and the selection of one or another type of heat generator.

Thermal loads of the heating system: characteristics, definitions

The definition should be understood as the amount of heat that is collectively given off by heating devices installed in a house or other 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 organize uninterrupted and effective work heating systems for the property. Thanks to this calculation, you can quickly complete absolutely all heat supply tasks and ensure their compliance with the standards and requirements of SNiP.

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

Total heat load per modern system heating system consists of several main load parameters:

On common system central heating;
Per system underfloor heating(if it is available in the house) – warm floor;
Ventilation system (natural and forced);
Hot water supply system;
For all kinds of technological needs: swimming pools, baths and other similar structures.

Basic characteristics of the object that are important to take into account when calculating the heat load

The most correct and competent calculation of the heat load for heating will be determined only if absolutely everything is taken into account, even the most small parts and parameters.

This list is quite large and can include:

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

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

Architectural part. The dimensions of all kinds of external fences (walls, floors, roofs), and the sizes of openings (balconies, loggias, doors and windows) are taken into account. The number of floors 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 temperature modes for each room of a residential building or area of an administrative building;
Design and features of external fencing, including the type of materials, thickness, presence of insulating layers;

The nature of the purpose of the premises. As a rule, it is inherent in industrial buildings, where it is necessary to create certain thermal conditions and regimes for a workshop or site;
Availability and parameters of special premises. The presence of the same baths, swimming pools and other similar structures;
Degree Maintenance – availability of hot water supply, such as central heating, ventilation and air conditioning systems;
Total number of points, from which hot water is drawn. It is this characteristic that you should pay attention to Special attention, after all, what larger number points - the greater the heat load on the entire heating system as a whole;
Number of people living in the house or on site. The requirements for humidity and temperature depend on this - factors that are included in the formula for calculating the thermal load;

Other data. For industrial facility Such factors include, for example, the number of shifts, the number of workers per 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

The actual calculation of the heating load with your own hands is carried out at the design stage of a country cottage or other real estate property - this is due to the simplicity and absence of extra cash costs. At the same time, the requirements of various norms and standards, TKP, SNB and GOST are taken into account.

The following factors are required to be determined during the calculation of thermal power:

Heat loss from external enclosures. Includes desired temperature conditions in each of the rooms;
Power required to heat water in the room;
The amount of heat required to heat the air ventilation (in the case where forced forced ventilation is required);
Heat needed to heat water in a swimming pool or sauna;

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

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

Features of calculating thermal load

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

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

The above instructions for calculating thermal 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 most correctly develop a rationale for using efficient heating, as well as energy inspection of houses and buildings.

An ideal method of calculation for emergency heating of an industrial facility, when it is assumed that temperatures will decrease during non-working hours (holidays and weekends are also taken into account).

Methods for determining thermal loads

Currently, thermal loads are calculated in several main ways:

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

Enlarged method for calculating heating loads

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

For a larger calculation of the heating heat load, a fairly simple and uncomplicated formula is used:

Qmax from.=α*V*q0*(tв-tн.р.)*10 -6

The following coefficients are used in the formula: α is a correction factor that takes into account climatic conditions in the region where the building is built (applied 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 week”); V – external volume of the building.

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

When performing calculations (as well as when selecting equipment), a large number of different thermal loads are taken into account:

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

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

Changes in the load on the heating system depending on the time of day. Due to the heat resistance of the building’s external enclosures, such values are accepted as insignificant;
Thermal energy consumption of the ventilation system by hour of the day.

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. For example, in summer, thermal energy consumption is reduced by almost 30-35% compared to winter;
Dry heat– convection heat exchange and thermal radiation from other similar devices. Determined by dry bulb temperature.

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

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

In any room, humidity is influenced by:

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

Regulators of thermal loads as a way out of difficult situations

As you can see in many photos and videos of modern and other boiler equipment, special heat load regulators are included with them. Equipment in this category is designed to provide support for a certain level of loads and eliminate all kinds of surges and dips.

It should be noted that RTN allows 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 surges and excesses of thermal loads are recorded, fines and similar sanctions are possible.

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

Ventilation and hot water loads are one of the factors in thermal systems

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

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

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

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

Qgws.=0.042rv(tg.-tx.)Pgav, Where

r, in, tg.,tx. – design temperature of hot and cold water, water density, as well as a coefficient that takes into account the values of the maximum load of hot water supply to the average value established by GOST;

Comprehensive calculation of thermal loads

In addition to the theoretical calculation issues themselves, some practical work. For example, comprehensive thermal inspections include mandatory thermography of all structures - walls, ceilings, doors and windows. It should be noted that such work makes it possible to identify and record factors that have a significant impact on the heat loss of a building.

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

Practical measurements are an indispensable component of various calculation works. Taken together, such processes will help obtain the most reliable data on thermal loads and heat losses that will be observed in a certain structure over a certain period of time. Practical calculation will help to achieve what theory will not show, namely the “bottlenecks” of each structure.

Conclusion

Calculation of thermal loads, likewise, is an important factor, the calculations of which must be carried out before starting to organize a heating system. If all the work is done correctly and you approach the process wisely, you can guarantee trouble-free heating operation, as well as save money on overheating and other unnecessary costs.