Calculation of the amount of heat for heating a building formula. Independent calculation of the heating load: hourly and annual indicators

When designing heating systems for all types of buildings, it is necessary to carry out correct calculations and then develop a competent heating circuit diagram. At this stage, special attention should be paid to calculating the heat load for heating. To solve the problem, it is important to use an integrated approach and take into account all the factors affecting the operation of the system.

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Parameter importance

Using the heat load indicator, you can find out the amount of heat energy required to heat a specific room, as well as the building as a whole. The main variable here is the power of all heating equipment that is planned to be used in the system. In addition, it is necessary to take into account the heat loss of the house.

The ideal situation seems to be in which the power of the heating circuit allows not only to eliminate all losses of heat energy from the building, but also to ensure comfortable conditions accommodation. To correctly calculate the specific heat load, it is necessary to take into account all the factors influencing this parameter:

The optimal operating mode of the heating system can only be determined taking these factors into account. The unit of measurement for the indicator can be Gcal/hour or kW/hour.

heating load calculation

Selecting a Method

Before you begin calculating the heating load using aggregate indicators, you need to decide on the recommended temperature conditions for a residential building. To do this, you will have to refer to SanPiN 2.1.2.2645−10. Based on the data specified in this regulatory document, it is necessary to ensure operating modes of the heating system for each room.

The methods used today for calculating the hourly load on the heating system allow obtaining results of varying degrees of accuracy. In some situations, complex calculations may be required to minimize the error.

If, when designing a heating system, optimizing energy costs is not a priority, it is permissible to use less precise methods.

Calculation of thermal load and design of heating systems Audytor OZC + Audytor C.O.

Simple ways

Any method for calculating the thermal load allows you to select optimal parameters heating systems. This indicator also helps determine the need for work to improve the thermal insulation of a building. Today, two fairly simple methods for calculating the heat load are used.

Depending on the area

If all rooms in the building have standard dimensions and have good thermal insulation, you can use the calculation method required power heating equipment depending on the area. In this case, 1 kW of thermal energy should be produced for every 10 m2 of room. Then the result must be multiplied by the correction factor of the climate zone.

This is the simplest method of calculation, but it has one serious drawback - the error is very high. During calculations, only the climate region is taken into account. However, many factors influence the efficiency of a heating system. Therefore, this technique is not recommended in practice.

Aggregated calculations

By applying the methodology for calculating heat using aggregated indicators, the calculation error will be smaller. This method was first often used to determine the heat load in situations where the exact parameters of the structure were unknown. To determine the parameter, the calculation formula is used:

Qot = q0*a*Vn*(tin - tnro),

where q0 is specific thermal performance buildings;

a - correction factor;

Vн - external volume of the building;

tin, tnro - temperature values inside the house and outside.

As an example of calculating heat loads using aggregated indicators, you can calculate the maximum indicator for heating system building on external walls 490 m2. The two-story building with a total area of 170 m2 is located in St. Petersburg.

First you need to install everything using the regulatory document input data required for calculation:

Thermal characteristics of the building are 0.49 W/m³*C.
Clarifying coefficient - 1.
The optimal temperature inside the building is 22 degrees.

Assuming that the minimum temperature in winter will be -15 degrees, we can substitute all known values into the formula - Q = 0.49 * 1 * 490 (22 + 15) = 8.883 kW. Using the simplest method for calculating the basic heat load indicator, the result would be higher - Q = 17*1 = 17 kW/hour. At the same time The enlarged method of calculating the load indicator takes into account significantly more factors:

Optimal temperature parameters in rooms.
The total area of the building.
Outside air temperature.

Also, this technique allows you to calculate with minimal error the power of each radiator installed in a separate room. Its only drawback is the inability to calculate the heat loss of a building.

Calculation of thermal loads, Barnaul

Complex technique

Since even with an integrated calculation the error turns out to be quite high, it is necessary to use a more complex method for determining the load parameter on the heating system. In order for the results to be as accurate as possible, it is necessary to take into account the characteristics of the house. Among them, the most important is the heat transfer resistance ® of the materials used to make each element of the building - the floor, walls, and also the ceiling.

This value is in inverse relationship with thermal conductivity (λ), showing the ability of materials to transfer heat energy. It is quite obvious that the higher the thermal conductivity, the more actively the house will lose heat energy. Since this thickness of materials (d) is not taken into account in thermal conductivity, you must first calculate the heat transfer resistance using a simple formula - R=d/λ.

The method under consideration consists of two stages. First, heat loss through window openings and external walls is calculated, and then through ventilation. As an example, we can take the following structural characteristics:

The area and thickness of the walls is 290 m² and 0.4 m.
The building has windows (double glazing with argon) - 45 m² (R = 0.76 m²*C/W).
The walls are made of solid brick- λ=0.56.
The building was insulated with expanded polystyrene - d = 110 mm, λ = 0.036.

Based on the input data, it is possible to determine the wall transmission resistance indicator - R=0.4/0.56= 0.71 m²*C/W. Then a similar insulation indicator is determined - R=0.11/0.036= 3.05 m²*C/W. These data allow us to determine the following indicator - R total = 0.71 + 3.05 = 3.76 m²*C/W.

The actual heat loss from the walls will be - (1/3.76)*245+(1/0.76)*45= 125.15 W. Temperature parameters remained unchanged compared to the enlarged calculation. The next calculations are carried out in accordance with the formula - 125.15*(22+15)= 4.63 kW/hour.

Calculation of thermal power of heating systems

At the second stage, the heat loss of the ventilation system is calculated. It is known that the volume of the house is 490 m³, and the air density is 1.24 kg/m³. This allows us to find out its mass - 608 kg. During the day, the air in the room is renewed on average 5 times. After this, you can calculate the heat loss of the ventilation system - (490*45*5)/24= 4593 kJ, which corresponds to 1.27 kW/hour. It remains to determine the general heat losses buildings, adding up the available results - 4.63+1.27=5.9 kW/hour.

The topic of this article is determining the thermal load for heating and other parameters that need to be calculated for. The material is aimed primarily at owners of private houses who are far from heating engineering and who need the simplest possible formulas and algorithms.

So, let's go.

Our task is to learn how to calculate the basic heating parameters.

Redundancy and accurate calculation

It is worth mentioning from the very beginning one subtlety of the calculations: it is almost impossible to calculate absolutely accurate values of heat loss through the floor, ceiling and walls, which have to be compensated by the heating system. We can only talk about one degree or another of the reliability of the estimates.

The reason is that heat loss is influenced by too many factors:

Thermal resistance of main walls and all layers of finishing materials.
The presence or absence of cold bridges.
Wind rose and the location of the house on the terrain.
The operation of ventilation (which, in turn, again depends on the strength and direction of the wind).
The degree of insolation of windows and walls.

There are also good news. Almost all modern heating boilers and distributed heating systems (warm floors, electric and gas convectors, etc.) are equipped with thermostats that dose heat consumption depending on the room temperature.

WITH practical side this means that excess thermal power will only affect the heating operating mode: say, 5 kWh of heat will be released not in one hour of continuous operation with a power of 5 kW, but in 50 minutes of operation with a power of 6 kW. The boiler or other heating device will spend the next 10 minutes in standby mode without consuming electricity or energy.

Therefore: in the case of calculating the thermal load, our task is to determine its minimum acceptable value.

The only exception to the general rule is associated with the operation of classic solid fuel boilers and is due to the fact that a decrease in their thermal power is associated with a serious drop in efficiency due to incomplete combustion of the fuel. The problem is solved by installing a heat accumulator in the circuit and throttling the heating devices with thermal heads.

After lighting, the boiler operates at full power and with maximum efficiency until the coal or wood is completely burned out; then the heat accumulated by the heat accumulator is dosed and used to maintain the optimal temperature in the room.

Most of the other parameters that need to be calculated also allow for some redundancy. However, more on this in the relevant sections of the article.

List of parameters

So, what do we actually have to count?

The total heat load for heating the house. It corresponds to the minimum required boiler power or the total power of devices in a distributed heating system.
The need for heat in a separate room.
The number of sections of a sectional radiator and the size of the register corresponding to a certain value of thermal power.

Please note: for finished heating devices (convectors, plate radiators, etc.), manufacturers usually indicate the total thermal power in the accompanying documentation.

The diameter of the pipeline capable of providing the required heat flow in the case of water heating.
Parameters of the circulation pump driving the coolant in a circuit with specified parameters.
The size of the expansion tank that compensates for the thermal expansion of the coolant.

Let's move on to the formulas.

One of the main factors influencing its value is the degree of insulation of the house. SNiP 02/23/2003, which regulates the thermal protection of buildings, normalizes this factor, deriving recommended values for the thermal resistance of building envelopes for each region of the country.

We will present two ways to perform calculations: for buildings that comply with SNiP 23-02-2003, and for houses with non-standardized thermal resistance.

Normalized thermal resistance

Instructions for calculating thermal power in this case look like this:

The base value is 60 watts per 1 m3 of the total (including walls) volume of the house.
For each window, an additional 100 watts of heat is added to this value.. For each door leading to the street - 200 watts.

To compensate for increasing losses in cold regions, an additional coefficient is used.

As an example, let's perform a calculation for a house measuring 12*12*6 meters with twelve windows and two doors to the street, located in Sevastopol (the average January temperature is +3C).

The heated volume is 12*12*6=864 cubic meters.
The basic thermal power is 864*60=51840 watts.
Windows and doors will increase it slightly: 51840+(12*100)+(2*200)=53440.
The exceptionally mild climate due to the proximity of the sea will force us to use a regional coefficient of 0.7. 53440*0.7=37408 W. It is this value that you can focus on.

Non-standardized thermal resistance

What to do if the quality of home insulation is noticeably better or worse than recommended? In this case, to estimate the heat load, you can use a formula of the form Q=V*Dt*K/860.

In it:

Q is the cherished thermal power in kilowatts.
V is the heated volume in cubic meters.
Dt is the temperature difference between the street and the house. Usually the delta between the recommended SNiP value for interior spaces(+18 - +22C) and the average minimum street temperature in the coldest month over the past few years.

Let’s clarify: counting on the absolute minimum is, in principle, more correct; however, this will mean excess costs for the boiler and heating appliances, the full power of which will only be required once every few years. The price of a slight underestimation of the calculated parameters is a slight drop in the temperature in the room during the peak of cold weather, which is easy to compensate by turning on additional heaters.

K is the insulation coefficient, which can be taken from the table below. Intermediate coefficient values are derived by approximation.

Let's repeat the calculations for our house in Sevastopol, specifying that its walls are 40 cm thick masonry made of shell rock (porous sedimentary rock) without external finishing, and the glazing is made of single-chamber double-glazed windows.

Let us take the insulation coefficient equal to 1.2.
We calculated the volume of the house earlier; it is equal to 864 m3.
We will take the internal temperature to be equal to the recommended SNiP for regions with a lower peak temperature above -31C - +18 degrees. The world-famous Internet encyclopedia will kindly provide information about the average minimum: it is equal to -0.4C.
The calculation will thus be Q = 864 * (18 - -0.4) * 1.2 / 860 = 22.2 kW.

As is easy to see, the calculation gave a result that differed from that obtained by the first algorithm by one and a half times. The reason is primarily that the average minimum we used is noticeably different from the absolute minimum (about -25C). An increase in the temperature delta by one and a half times will increase the estimated heat demand of the building by exactly the same amount.

Gigacalories

When calculating the amount of thermal energy received by a building or room, along with kilowatt-hours, another value is used - gigacalorie. It corresponds to the amount of heat required to heat 1000 tons of water by 1 degree at a pressure of 1 atmosphere.

How to convert kilowatts of thermal power into gigacalories of consumed heat? It's simple: one gigacalorie is equal to 1162.2 kWh. Thus, with a peak power of the heat source of 54 kW, the maximum hourly heating load will be 54/1162.2 = 0.046 Gcal*hour.

Useful: for each region of the country, local authorities standardize heat consumption in gigacalories per square meter of area for a month. The average value for the Russian Federation is 0.0342 Gcal/m2 per month.

Room

How to calculate the heat requirement for a separate room? The same calculation schemes are used here as for the house as a whole, with a single amendment. If a room is adjacent to a heated room without its own heating devices, it is included in the calculation.

So, if a room measuring 4*5*3 meters is adjacent to a corridor measuring 1.2*4*3 meters, the thermal power of the heating device is calculated for a volume of 4*5*3+1.2*4*3=60+14, 4=74.4 m3.

Heating devices

Sectional radiators

In general, information about the heat flow per section can always be found on the manufacturer’s website.

If it is unknown, you can rely on the following approximate values:

Cast iron section - 160 W.
Bimetallic section - 180 W.
Aluminum section - 200 W.

As always, there are a number of subtleties. When connecting a radiator with 10 or more sections to the side, the temperature spread between the sections closest to the supply and the end sections will be quite significant.

However: the effect will be nullified if the eyeliners are connected diagonally or from bottom to bottom.

In addition, usually manufacturers of heating devices indicate power for a very specific temperature delta between the radiator and the air, equal to 70 degrees. The dependence of the heat flow on Dt is linear: if the battery is 35 degrees hotter than the air, the thermal power of the battery will be exactly half the declared one.

Let's say, at an air temperature in the room of +20C and a coolant temperature of +55C, the power of a standard size aluminum section will be equal to 200/(70/35)=100 watts. In order to provide a power of 2 kW, you will need 2000/100 = 20 sections.

Registers

Homemade registers stand apart from the list of heating devices.

The photo shows a heating register.

Manufacturers, for obvious reasons, cannot indicate their thermal power; however, it is not difficult to calculate it yourself.

For the first register section ( horizontal pipe known sizes) power is equal to the product of its outer diameter and length in meters, the temperature delta between the coolant and air in degrees and a constant coefficient of 36.5356.
For subsequent sections located in the rising flow of warm air, an additional coefficient of 0.9 is used.

Let's look at another example - let's calculate the heat flow value for a four-row register with a section diameter of 159 mm, a length of 4 meters and a temperature of 60 degrees in a room with an internal temperature of +20C.

The temperature delta in our case is 60-20=40C.
Convert the pipe diameter to meters. 159 mm = 0.159 m.
We calculate the thermal power of the first section. Q = 0.159*4*40*36.5356 = 929.46 watts.
For each subsequent section, the power will be equal to 929.46*0.9=836.5 W.
The total power will be 929.46 + (836.5*3) = 3500 (rounded) watts.

Pipe diameter

How to determine the minimum value of the internal diameter of a filling pipe or supply pipe to a heating device? Let’s not get into the weeds and use a table containing ready-made results for a difference between supply and return of 20 degrees. This value is typical for autonomous systems.

The maximum coolant flow rate should not exceed 1.5 m/s to avoid noise; More often they focus on a speed of 1 m/s.

Inner diameter, mm	Thermal power of the circuit, W at flow speed, m/s
Inner diameter, mm	0,6	0,8	1
8	2450	3270	4090
10	3830	5110	6390
12	5520	7360	9200
15	8620	11500	14370
20	15330	20440	25550
25	23950	31935	39920
32	39240	52320	65400
40	61315	81750	102190
50	95800	127735	168670

Let's say, for a 20 kW boiler, the minimum internal filling diameter at a flow speed of 0.8 m/s will be 20 mm.

Please note: the internal diameter is close to the nominal bore. Plastic and metal-plastic pipes are usually marked with an outer diameter, which is 6-10 mm larger than the inner one. Thus, a polypropylene pipe measuring 26 mm has an internal diameter of 20 mm.

Circulation pump

Two parameters of the pump are important to us: its pressure and performance. In a private house, with any reasonable length of the circuit, the minimum pressure for the cheapest pumps of 2 meters (0.2 kgf/cm2) is quite sufficient: it is this value of the difference that ensures the circulation of the heating system apartment buildings.

The required performance is calculated using the formula G=Q/(1.163*Dt).

In it:

G - productivity (m3/hour).
Q is the power of the circuit in which the pump is installed (kW).
Dt is the temperature difference between direct and return pipelines in degrees (in an autonomous system the typical value is Dt=20С).

For a circuit with a thermal load of 20 kilowatts, with a standard temperature delta, the calculated productivity will be 20/(1.163*20)=0.86 m3/hour.

Expansion tank

One of the parameters that needs to be calculated for autonomous system— volume of the expansion tank.

An accurate calculation is based on a fairly long series of parameters:

Temperature and type of coolant. The expansion coefficient depends not only on the degree of heating of the batteries, but also on what they are filled with: water-glycol mixtures expand more strongly.
Maximum operating pressure in the system.
The charging pressure of the tank, which in turn depends on the hydrostatic pressure of the circuit (the height of the top point of the circuit above the expansion tank).

There is, however, one nuance that allows you to greatly simplify the calculation. If underestimating the volume of the tank will, at best, lead to constant operation safety valve, and at worst - to the destruction of the circuit, then its excess volume will not harm anything.

That is why a tank with a displacement equal to 1/10 of the total amount of coolant in the system is usually taken.

Hint: to find out the volume of the circuit, just fill it with water and pour it into a measuring cup.

Conclusion

We hope that the given calculation schemes will simplify the reader’s life and save him from many problems. As usual, the video attached to the article will offer additional information.

Before you begin purchasing materials and installing heat supply systems for a house or apartment, it is necessary to carry out heating calculations based on the area of each room. Basic parameters for heating design and heat load calculation:

Square;
Number of window blocks;
Ceiling height;
Room location;
Heat loss;
Heat transfer from radiators;
Climate zone (outside air temperature).

The methodology described below is used to calculate the number of batteries for a room area without additional heating sources (warm floors, air conditioners, etc.). Heating can be calculated in two ways: using a simple and complicated formula.

Before starting heat supply design, it is worth deciding which radiators will be installed. Material from which heating batteries are made:

Cast iron;
Steel;
Aluminum;
Bimetal.

Aluminum and bimetallic radiators are considered the best option. The highest thermal output is for bimetallic devices. Cast iron radiators take a long time to heat up, but after turning off the heating, the temperature in the room remains for quite a long time.

A simple formula for designing the number of sections in a heating radiator:

K = Sх(100/R), where:

S – area of the room;

R – section power.

If we look at the example with data: room 4 x 5 m, bimetallic radiator, power 180 W. The calculation will look like this:

K = 20*(100/180) = 11.11. So, for a room with an area of 20 m2, a battery with at least 11 sections is required for installation. Or, for example, 2 radiators with 5 and 6 fins. The formula is used for rooms with a ceiling height of up to 2.5 m in a standard Soviet-built building.

However, such a calculation of the heating system does not take into account the heat loss of the building, nor does it take into account the temperature of the outside air of the house and the number of window units. Therefore, these coefficients should also be taken into account to finalize the number of edges.

Calculations for panel radiators

In the case where it is intended to install a battery with a panel instead of ribs, the following volume formula is used:

W = 41xV, where W is the battery power, V is the volume of the room. Number 41 is the norm for the average annual heating power of 1 m2 of living space.

As an example, we can take a room with an area of 20 m2 and a height of 2.5 m. The radiator power value for a room volume of 50 m3 will be equal to 2050 W, or 2 kW.

Heat loss calculation

H2_2

The main heat losses occur through the walls of the room. To calculate, you need to know the thermal conductivity coefficient of the external and internal material the material from which the house is built, the thickness of the building wall, and the average outside temperature is also important. Basic formula:

Q = S x ΔT /R, where

ΔT – difference between the temperature outside and the internal optimal value;

S – wall area;

R is the thermal resistance of the walls, which, in turn, is calculated by the formula:

R = B/K, where B is the brick thickness, K is the thermal conductivity coefficient.

Calculation example: a house built of shell rock, stone, located in Samara region. The thermal conductivity of shell rock is on average 0.5 W/m*K, the wall thickness is 0.4 m. Considering the average range, the minimum temperature in winter is -30 °C. In the house, according to SNIP, the normal temperature is +25 °C, the difference is 55 °C.

If the room is corner, then both of its walls are in direct contact with the environment. The area of the outer two walls of the room is 4x5 m and 2.5 m high: 4x2.5 + 5x2.5 = 22.5 m2.

R = 0.4/0.5 = 0.8

Q = 22.5*55/0.8 = 1546 W.

In addition, it is necessary to take into account the insulation of the walls of the room. When finishing the outer area with foam plastic, heat loss is reduced by approximately 30%. So the final figure will be around 1000 watts.

Calculation of thermal load (complicated formula)

Scheme of heat loss of premises

To calculate the final heat consumption for heating, it is necessary to take into account all the coefficients using the following formula:

CT = 100xSxK1xK2xK3xK4xK5xK6xK7, where:

S – room area;

K – various coefficients:

K1 – loads for windows (depending on the number of double-glazed windows);

K2 – thermal insulation of the external walls of the building;

K3 – loads for the ratio of window area to floor area;

K4 – temperature regime of outside air;

K5 – taking into account the number of external walls of the room;

K6 – loads based on the upper room above the room being calculated;

K7 – taking into account the height of the room.

As an example, we can consider the same room of a building in the Samara region, insulated from the outside with polystyrene foam, having 1 double-glazed window, above which there is a heated room. The heat load formula will look like this:

KT = 100*20*1.27*1*0.8*1.5*1.2*0.8*1= 2926 W.

Heating calculations are focused specifically on this figure.

Heat consumption for heating: formula and adjustments

Based on the above calculations, 2926 W are needed to heat the room. Taking into account heat losses, the requirements are: 2926 + 1000 = 3926 W (KT2). To calculate the number of sections, use the following formula:

K = KT2/R, where KT2 is the final value of the thermal load, R is the heat transfer (power) of one section. Final figure:

K = 3926/180 = 21.8 (rounded to 22)

So, in order to ensure optimal heat consumption for heating, it is necessary to install radiators with a total of 22 sections. It should be taken into account that the lowest temperature - 30 degrees below zero - lasts a maximum of 2-3 weeks, so you can safely reduce the number to 17 sections (-25%).

If homeowners are not satisfied with this indicator of the number of radiators, then they should initially take into account batteries that have a large heating power. Or insulate the walls of the building both inside and outside with modern materials. In addition, it is necessary to correctly assess the heating needs of housing based on secondary parameters.

There are several other parameters that influence additional wasted energy consumption, which entails an increase in heat loss:

Features of external walls. The heating energy should be enough not only to heat the room, but also to compensate for heat loss. Over time, a wall in contact with the environment begins to let moisture in due to changes in outside air temperature. It is especially necessary to insulate well and carry out high-quality waterproofing for northern directions. It is also recommended to insulate the surface of houses located in humid regions. High annual precipitation will inevitably lead to increased heat loss.
Radiator installation location. If the battery is mounted under a window, then heating energy leaks through its structure. Installing high-quality blocks will help reduce heat loss. You also need to calculate the power of the device installed in the window sill - it should be higher.
Conventional annual heat demand for buildings in different time zones. As a rule, according to SNIPs, the average temperature (average annual indicator) for buildings is calculated. However, heat requirements are significantly lower if, for example, cold weather and low outdoor air conditions occur for a total of 1 month per year.

Advice! To minimize the need for heat in winter, it is recommended to install additional sources of indoor air heating: air conditioners, mobile heaters, etc.

1. Heating

1.1. The calculated hourly heating load should be taken based on standard or individual building designs.

If the design value of the outside air temperature for heating design adopted in the project differs from the current standard value for a specific area, it is necessary to recalculate the design hourly heat load of the heated building given in the project using the formula:

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

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

tj - design air temperature in a heated building, °C; accepted in accordance with Table 1;

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

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

Table 1. Design air temperature in heated buildings

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

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

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

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

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

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

Table 2. Correction factor  for residential buildings

Table 3. Specific heating characteristics of residential buildings

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

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

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

Name of buildings	Building volume V, m3	Specific thermal characteristics
for heating qo, kcal/m3 h °С	for ventilation qv, kcal/m3 h °С

Administrative buildings, offices


	more than 15000


	more than 10000
Cinemas

	more than 10000




	more than 30,000
Stores

	more than 10000
Kindergartens and nurseries

Schools and higher education institutions

	more than 10000
Hospitals


	more than 15000


	more than 10000
Laundries

	more than 10000
Catering establishments, canteens, factory kitchens

	more than 10000
Laboratories

	more than 10000
Fire stations

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

If the building has an attic floor, the value V, m3, is determined as the product of the horizontal sectional area of the building at the level of its first floor (above the ground floor) by the free height of the building - from the level of the finished floor of the first floor to the upper plane of the heat-insulating layer of the attic floor, with roofs combined with attic floors - up to the middle level of the roof top. Architectural details and niches in the walls of a building protruding beyond the wall surfaces, as well as unheated loggias, are not taken into account when determining the estimated hourly heating load.

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

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

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

L - free height of the building, m;

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

It is not necessary to introduce a so-called correction for the effect of wind into the calculation of the estimated hourly heat load for heating a building, because this quantity is already taken into account in formula (3.3).

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

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

In May-June - by 12%;

In July-August - by 20%;

In September - by 25%;

During the heating season - by 30%.

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

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

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

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

Q = k F t, (3.5)

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

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

t is the temperature pressure of the heating device, °C, defined as the difference between the average temperature of the convective-radiative heating device and the air temperature in the heated building.

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

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

1.6. In the absence of design data and determination of the estimated hourly heat load for heating industrial, public, agricultural and other non-standard buildings (garages, underground heated passages, swimming pools, shops, kiosks, pharmacies, etc.) according to aggregate indicators, the values of this load should be clarified by the heat exchange surface area of installed heating devices of heating systems in accordance with the methodology given in. The initial information for calculations is identified by a representative of the heat supply organization in the presence of a representative of the subscriber with the drawing up of a corresponding act.

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

, (3.6)

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

n - number of technological operations.

In turn,

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

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

Qpol + Qprop - thermal energy consumption for heating irrigation water and steaming the soil, Gcal/h;

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

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

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

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

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

tj and to are the technological temperature in the room and the calculated outdoor air for the design of the corresponding agricultural facility, °C.

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

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

where Finv is the inventory area of the greenhouse, m2;

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

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

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

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

, (3.10)

where Fcreep is the useful area of the greenhouse, m2;

n - duration of watering, hours.

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

2. Supply ventilation

2.1. If there is a standard or individual building design and compliance installed equipment supply ventilation system for the project, the calculated hourly thermal load of ventilation can be taken into account according to the project, taking into account the difference in the values of the calculated outside air temperature for the design of ventilation adopted in the project, and the current standard value for the area where the building in question is located.

Recalculation is carried out using a formula similar to formula (3.1):

, (3.1a)

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

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

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

2.2. In the absence of projects or the installed equipment does not comply with the project, the calculated hourly heat load of the supply ventilation must be determined based on the characteristics of the equipment actually installed, in accordance with the general formula describing the heat transfer of heating units:

Q = Lc (2 + 1) 10-6, (3.12)

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

 - density of heated air, kg/m3;

c is the heat capacity of the heated air, kcal/kg;

2 and 1 - calculated values of air temperature at the inlet and outlet of the heating unit, °C.

The method for determining the estimated hourly heat load of supply heating units is set out in.

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

Qv = Vqv (tj - tv) 10-6, (3.2a)

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

3. Hot water supply

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

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

N - the number of units of measurement referred to a day, - the number of residents studying in educational institutions etc.;

tc - temperature tap water during the heating season, °C; in the absence of reliable information, tc = 5 °C is accepted;

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

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

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

, (3.13a)

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

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

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

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

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

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

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

tн and tк - temperature of hot water at the beginning and end of the design section of the pipeline, °C;

tamb - ambient temperature, °C; take into account the type of pipeline laying:

In furrows, vertical channels, communication shafts of sanitary cabins tamb = 23 °C;

In bathrooms tamb = 25 °C;

In kitchens and toilets tamb = 21 °C;

On staircases tamb = 16 °C;

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

In tunnels tamb = 40 °C;

In unheated basements tamb = 5 °C;

In attics tam = -9 °C (at the average outside air temperature of the coldest month of the heating period tn = -11 ... -20 °C);

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

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

Place and method of laying	Pipeline heat losses, kcal/hm, with nominal diameter, mm


The main supply riser in the drain or communication shaft, insulated
Riser without heated towel rails, insulated, in a sanitary cabin shaft, furrow or communication shaft
Same with heated towel rails
Uninsulated riser in a plumbing shaft, furrow or communication shaft or openly in a bathroom, kitchen
Distribution insulated pipelines (supply):
in the basement, on staircase
in a cold attic
in a warm attic
Insulated circulation pipelines:
in the basement
in a warm attic
in a cold attic
Uninsulated circulation pipelines:
in apartments
on the staircase
Circulation risers in the drain of a plumbing cabin or bathroom:
isolated
non-insulated

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

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

Temperature difference, °C

Pipeline heat losses, kcal/h m, with nominal diameter, mm

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

3.4. In the absence of the initial information necessary to calculate heat losses by hot water supply pipelines, heat losses, Gcal/h, can be determined using a special coefficient Kt.p, taking into account the heat losses of these pipelines, according to the expression:

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

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

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

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

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

studfiles.net

How to calculate the heat load for heating a building

In houses that have been put into operation in recent years, these rules are usually met, so the calculation of the heating power of the equipment is based on standard coefficients. Individual calculations can be carried out at the initiative of the homeowner or the utility structure involved in the supply of heat. This happens when heating radiators, windows and other parameters are spontaneously replaced.

Read also: How to calculate the power of a heating boiler based on the area of the house

Calculation of heating standards in an apartment

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

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

Methodology for calculating heating standards in a private house

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

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

Basic principles

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

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

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

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

The main heat losses of a private house:

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

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

Specifics of calculations

This technique can also be found under the name “thermal engineering calculation”. The simplified formula is as follows:

Qt = V × ∆T × K / 860, where

V – room volume, m³;

∆T – maximum difference indoors and outdoors, °C;

K – estimated heat loss coefficient;

860 – conversion factor in kW/hour.

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

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

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

gidpopechi.ru

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

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

This term refers to the amount of heat given off by heating devices. 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 of the structural elements of the building. 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. It is necessary to take into account the area 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.

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

Number of hot water supply points. 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 – number of working days in a calendar year, shifts, technological chain production process etc.

Climatic conditions of the region. When calculating heat loss, street temperatures are taken into account. If the differences are insignificant, then a small amount of energy will be spent on compensation. While at -40°C outside the window 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, digital characteristics relating to a specific heating radiator, boiler, etc. are taken. 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 an hourly calculation of the load on the heating network is required, then the calculation must be carried out taking into account the temperature difference during the day.

The calculations 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 it will also be important to calculate the heat loss used to warm up the internal volume of air.
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 Qot = q0 * a * VH * (tEN – tHRO), where:

q0 – specific thermal characteristic of the building (most often determined by the coldest period),
a – correction factor (depends on the region and is taken from ready-made tables),
VH – volume calculated from 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 of the structure, temperature, number of walls, ratio of wall areas to window openings, etc. Therefore, such calculations are not suitable for serious heating system projects.

Calculation of heating radiator by area

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 a steel panel radiator with a width of 500 mm and a height of 220 mm will be 1,644 W.

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

Ceiling height (standard – 2.7 m),

Thermal power (per sq. m – 100 W),

One 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 quantity 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 150 W. 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:

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

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

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

Approximate calculation

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

Q = 100 W/m2 × 25 m2 × 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

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

V – the amount of water consumed by the heating system, calculated in tons or m3,
T1 is a number indicating the temperature of hot water, measured in °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 60-65oC.
T2 – cold water temperature. 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:

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

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

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

Thermal imaging inspection

Increasingly, in order to increase the efficiency of the heating system, they are resorting to thermal imaging inspections of the structure.

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 15o. 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 special attention to 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.

highlogistic.ru

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

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

Thermal loads

Carrying out preliminary calculations is necessary not only in order 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 of “thermal load for heating” 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 works.

Calculating the heat load on heating will help organize the uninterrupted and efficient operation of the heating system of 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.

A set of instruments for performing calculations

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.

The total thermal load on a modern 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.

Calculation and components of thermal systems at home

Main 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 when absolutely everything is taken into account, even the smallest details 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;
The design and features of external fencing, including the type of materials, thickness, presence of insulating layers;

Physical indicators of room cooling - data for calculating the heat load

The nature of the purpose of the premises. As a rule, it is inherent in industrial buildings, where it is necessary to create some specific thermal conditions and modes;
Availability and parameters of special premises. The presence of the same baths, swimming pools and other similar structures;
Degree of maintenance – availability of hot water supply, such as central heating, ventilation and air conditioning systems;
The total number of points from which hot water is drawn. It is this characteristic that you should pay special attention to, because what larger number points - the greater the heat load on the entire heating system as a whole;
The 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;

Equipment that may affect thermal loads

Other data. For an 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 calculation of the heating load itself is done with your own hands at the design stage country cottage or another piece of real estate - this is due to the simplicity and lack 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 must be determined during the calculation of thermal power:

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

Gcal/hour – unit of measurement of thermal loads of objects

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;

Heat loss in a standard residential building

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 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, the selection of heat transfer coefficients is made from these same sources (the passport data of the 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 actually located in the building heating batteries), maximum hourly heat energy consumption, as well as total heat power consumption for a certain period, for example, a heating season.

Distribution of heat flows from various types of heaters

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 justification for the use of effective 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.

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

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 costs, which are determined by the meteorological characteristics of the region where the object for which the heat loads are calculated is located;

Thermal load regulator for boiler equipment

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 varies 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.

Heat loss of a country house

In any room, humidity is influenced by:

People and their number who are simultaneously in the room;
Technological and other equipment;
Air 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 industrial and household heating boilers and other boiler equipment, they are supplied with special heat load regulators. 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.

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

Advice. Loads on heating, ventilation and air conditioning systems are an important consideration 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:

Qв.=qв.V(tн.-tв.), where

Measuring heat loss in a practical way

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:

Qgvs.=0.042rv(tg.-tx.)Pgsr, 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.

Device for calculations and energy audits

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, as well as hydraulic calculation of the heating system - important factor, calculations of which must be carried out before organizing the 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.

Page 2

Heating boilers

One of the main components of comfortable housing is the presence of a well-thought-out heating system. At the same time, the choice of the type of heating and the required equipment is one of the main questions that must be answered at the stage of designing a house. An objective calculation of the heating boiler power by area will ultimately result in a completely efficient heating system.

We will now tell you about how to carry out this work correctly. At the same time, we will consider the features inherent in different types of heating. After all, they must be taken into account when carrying out calculations and subsequent decision-making on the installation of one or another type of heating.

Basic calculation rules

room area (S);
specific heater power per 10 m² of heated area – (W spec.). This value is determined adjusted for the climatic conditions of a particular region.

This value (W beat) is:

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

Let's do the calculations

Power calculation is carried out as follows:

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

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

Calculation example

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

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

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

Calculation for different types of heating boilers

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

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

Home heating system

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

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

solid fuel;
electric;
liquid fuel;
gas.

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

Solid fuel boilers

Power calculation solid fuel boiler must be carried out taking into account the features characterized by the following features of such heaters:

low popularity;
relative accessibility;
opportunity battery life- it is provided in a number of modern models of these devices;
efficiency during operation;
the need for additional space for fuel storage.

Solid fuel heater

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

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

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

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

Electric boilers

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

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

Electric boiler

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

Liquid fuel boilers

They have the following characteristic features:

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

Oil heater

Gas boilers

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

ease of operation;
do not require space for fuel storage;
safe to use;
low cost of fuel;
efficiency.

Gas boiler

Calculation for heating radiators

Let's say you decide to install a heating radiator yourself. But first you need to purchase it. And choose exactly the one that is suitable in terms of power.

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

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

Page 3

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

When we make a decision to build our own private house, we are primarily guided largely by emotional criteria - we want to have our own separate housing, independent of city utilities, much larger in size and made according to our own ideas. But somewhere in my soul, of course, there is an understanding that I will have to do a lot of counting. The calculations relate not so much to the financial component of all work, but to the technical one. One of the most important types of calculations will be the calculation of the mandatory heating system, without which there is no escape.

First, of course, you need to take on the calculations - a calculator, a piece of paper and a pen will be the first tools

To begin with, decide what is called, in principle, how to heat your home. After all, you have several options for providing heat at your disposal:

Autonomous heating electrical appliances. Perhaps such devices are good, and even popular, as auxiliary means of heating, but they cannot in any way be considered as the main ones.

Electric heating floors. But this heating method can also be used as the main one for a separate living room. But there is no question of providing all the rooms in the house with such floors.

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

Centralized water heating. Having “torn” yourself from the high-rise building, you can nevertheless bring its “spirit” into your home by connecting to centralized system heating. Is it worth it!? Is it worth it to rush “out of the frying pan and into the fire” again? This should not be done, even if such a possibility exists.

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

You can’t do without a detailed house plan with a diagram of the placement of equipment and wiring of all communications

After resolving the issue in principle

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

Installation of reliable window systems that will not simply “release” all your heating successes onto the street;

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

Fireplace installation. IN lately This auxiliary heating method is increasingly being used. It may not replace general heating, but it is such an excellent support for it that in any case it helps to significantly reduce heating costs.

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

Heating boiler as the main element of the entire system;
A circulation pump that provides coolant flow in the system;
Pipelines are like a kind of “blood vessels” of the entire system;
Heating batteries are those devices that have been known to everyone for a long time and which are the final elements of the system and are responsible in our eyes for the quality of its operation;
Devices for monitoring the state of the system. An accurate calculation of the volume of a heating system is unthinkable without the presence of such devices that provide information about the real temperature in the system and the volume of coolant passing through;
Locking and adjustment devices. Without these devices, the work will be incomplete; they will allow you to regulate the operation of the system and configure it according to the readings of control devices;
Various fitting systems. These systems could well be classified as pipelines, but their influence on the successful operation of the entire system is so great that fittings and connectors are separated into a separate group of elements for the design and calculation of heating systems. Some experts call electronics the science of contacts. You can, without fear of making a big mistake, call the heating system - in many ways, the science of the quality of connections that the elements of this group provide.

The heart of the entire water heating system is the heating boiler. Modern boilers– entire systems to provide the entire system with hot coolant

Useful advice! When we talk about the heating system, the word “coolant” often appears in conversation. With some degree of approximation, we can consider ordinary “water” to be the medium that is intended to move through the pipes and radiators of the heating system. But there are some nuances that are associated with the method of supplying water to the system. There are two ways - internal and external. External - from an external cold water supply. In this situation, indeed, the coolant will be plain water, with all its shortcomings. Firstly, in general availability, and secondly, cleanliness. We strongly recommend that when choosing this method of introducing water from the heating system, install a filter at the inlet, otherwise you cannot avoid heavy pollution systems for only one season of operation. If you choose a completely autonomous filling of water into the heating system, then do not forget to “flavor” it with all kinds of additives against hardening and corrosion. It is water with such additives that is called a coolant.

Types of heating boilers

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

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

Electric - and where would we be without electricity now? But you need to understand that the costs of this type of energy in your home when using electric heating boilers will be so large that solving the question “how to calculate the heating system” in your home will lose any meaning - everything will go into electrical wires;

Liquid fuel. Such boilers using gasoline or diesel fuel suggest themselves, but due to their environmental friendliness, they are very disliked by many, and rightly so;

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

Pay special attention to the quality of all materials used, there is no time to save money here; the quality of each component of the system, including pipes, must be ideal

Boiler calculation

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

W = S * Wud / 10,

S – total area of the heated room in square meters;
Wud – specific boiler power per 10 sq.m. premises.

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

For Middle zone it ranges from 1.2 to 1.5 kW;
for areas of Pskov level and above - from 1.5 to 2.0 kW;
for Volgograd and below - from 0.7 - 0.9 kW.

But, our climate of the 21st century has become so unpredictable that, by and large, the only criterion when choosing a boiler is your familiarity with the experience of other heating systems. Perhaps, understanding this unpredictability, for simplicity, it has long been customary in this formula to always take the specific power as one. Although do not forget about the recommended values.

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

Useful advice! This desire - to get acquainted with existing, already operating, autonomous heating systems will be very important. If you decide to set up such a system at home, and even with your own hands, then be sure to get acquainted with the heating methods used by your neighbors. Getting a “heating system calculation calculator” first-hand will be very important. You will kill two birds with one stone - you will acquire a good adviser, and maybe in the future a good neighbor, and even a friend, and you will avoid mistakes that your neighbor may have made at one time.

Circulation pump

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

Active heating systems, in which a circulation pump is used to move the coolant, have become much more widespread. It is often customary to install such pumps on the line from radiators to the boiler, when the water temperature has already dropped and cannot negatively affect the operation of the pump.

There are certain requirements for pumps:

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

Piping and radiators

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

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

steel;
copper;
polymer.

Steel is the patriarch of heating systems, used from time immemorial. Now steel pipes They are gradually disappearing from the scene; they are inconvenient to use, and, in addition, require welding and are susceptible to corrosion.

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

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

Heating of the house is largely ensured by the precise selection of the piping system and the laying of the pipes

Radiator calculations

Thermal engineering calculation of a heating system necessarily includes the calculation of such an irreplaceable network element as a radiator.

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

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

K = S / (W / 100),

S is the area of the heated room in square meters (we heat, of course, not the area, but the volume, but the standard room height is assumed to be 2.7 m);
W – heat transfer of one section in Watts, characteristics of the radiator;
K – number of sections in the radiator.

Providing heat in the house is a solution to a whole range of problems, often unrelated to each other, but serving the same purpose. One of these autonomous tasks could be installing a fireplace.

In addition to calculations, radiators also require compliance with certain requirements during installation:

installation must be carried out strictly under the windows, in the center, long and generally accepted rule, but some manage to break it (this installation prevents the movement of cold air from the window);
The “fins” of the radiator need to be aligned vertically - but this is a requirement that no one particularly claims to violate, it is obvious;
Another thing is not obvious - if there are several radiators in the room, they should be located on the same level;
it is necessary to ensure at least 5-centimeter gaps from above to the window sill and from below to the floor from the radiator; ease of maintenance plays an important role here.

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

Calculation of water in the system

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

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

The ideal, of course, would be to hide all communications behind plasterboard wall, but this is not always possible to do, and it raises questions from the point of view of the convenience of future system maintenance

Useful advice! Calculate accurately required volume water in the system is often not immediately possible with mathematical precision. Therefore, they act a little differently. First, fill the system, presumably to 90% of the volume, and check its performance. They bleed off as they work excess air and continue filling. Hence the need arises for an additional coolant reservoir in the system. As the system operates, there is a natural loss of coolant as a result of evaporation and convection processes, so calculating the replenishment of the heating system consists of tracking the loss of water from the additional reservoir.

Of course, we turn to specialists

Many renovation work You can, of course, do the housework yourself. But creating a heating system requires too much knowledge and skills. Therefore, even after studying all the photos and video materials on our website, even after familiarizing yourself with such essential attributes of each element of the system as “instructions,” we still recommend that you contact professionals for installation of the heating system.

The pinnacle of the entire heating system is the creation of warm heated floors. But the feasibility of installing such floors must be carefully calculated

The cost of mistakes when installing an autonomous heating system is very high. You shouldn't take risks in this situation. The only thing that remains for you is smart maintenance of the entire system and calling specialists to service it.

Page 4

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

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

Calculation stages

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

Pay attention! Next you will need to check the data. Divide the resulting numbers by the square footage of the room. This way you will get the specific heat loss (W/m²). As a rule, this is 50/150 W/m². If the data received is very different from the data indicated, it means you made a mistake. Therefore, the price of assembling the heating system will be too high.

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

Heating scheme.

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

Thermal calculation

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

What you need to get started

House project.

First of all, you will need a construction project. It must indicate the external and internal dimensions of each room, as well as windows and external doorways.
Next, find out information about the location of the building in relation to the cardinal directions, as well as the climatic conditions in your area.
Gather information about the height and composition of external walls.
You will also need to know the parameters of the floor materials (from indoors to ground), as well as the ceiling (from indoors to outdoors).

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

Required formula

Heat loss of the building.

Calculation of thermal loads on the system should determine heat loss and boiler power. In the latter case, the formula for calculating heating is as follows:

Mk = 1.2 ∙ Tp, where:

Mk – heat generator power, in kW;
Тп – heat loss of the building;
1.2 is a margin of 20%.

Pay attention! This safety factor takes into account the possibility of a pressure drop in the gas pipeline system in winter, in addition to unexpected heat losses. For example, as the photo shows, due to broken window, poor thermal insulation of doors, severe frosts. This reserve allows you to widely regulate the temperature regime.

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

external walls lose about 40% of the total figure;
20% escapes through windows;
floors contribute approximately 10%;
10% evaporates through the roof;
20% escapes through ventilation and doors.

Material coefficients

Thermal conductivity coefficients of some materials.

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

As for windows, their heat loss coefficients are equal:

traditional glazing – 1.27;
double-glazed windows – 1;
three-chamber analogues - 0.85.

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

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

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

At subzero temperatures, heat leakage also increases.

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

When you calculate thermal energy, keep in mind that its losses also depend on how many external walls there are in the building:

one external wall – 1%;
2 walls – 1.2;
3 external walls – 1.22;
4 walls – 1.33.

The greater the number of floors, the more complex the calculations.

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

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

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

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

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

Tp = UDtp ∙ Pl ∙ K1 ∙ K2 ∙ K3 ∙ K4 ∙ K5 ∙ K6 ∙ K7. In this case, UDtp is 100 W/m².

Calculation example

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

Windows with double glazing, i.e. K1 is 1.
External walls- foam concrete, the coefficient is the same. 3 of them are external, in other words K5 is 1.22.
The square footage of the windows is equal to 23% of that of the floor - K3 is 1.1.
Outside temperature is -15°, K4 is 0.9.
The attic of the building is not insulated, in other words K6 will be 1.
The ceiling height is three meters, i.e. K7 is 1.05.
The area of the premises is 135 m².

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

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

Mk = 1.2 ∙ 17.1206 = 20.54472 kW.

Hydraulic calculation for a heating system

An example of a hydraulic calculation diagram.

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

High quality circulation pump.

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

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

We find out the coolant flow for the heating system using the formula:

M = Q/Cp ∙ DPt

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

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

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

DPtr = R ∙ L,

R means specific friction loss in the pipe (Pa/m);
L is the length of the section (m);

After this, calculate the pressure loss on the resistances (valves, fittings), the formula is:

Dms = Σξ ∙ V²/2 ∙ P

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

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

If the speed is higher than the specified value, noise will appear in the system. A minimum speed value of 0.25 m/s is recommended by snip No. 2.04.05-91 so that the pipes do not become airborne.

Pipes from different materials, have different properties.

To comply with all the stated conditions, you need to choose the correct pipe diameter. You can do this using the table below, which shows the total power of the batteries.

At the end of the article you can watch a training video on its topic.

Page 5

For installation, heating design standards must be observed

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

How to heat your home

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

Calculation of the number of radiator sections and installation location

Sections can be added and removed by hand

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

Number of sections=Sroom area*100/Ppower of one section

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

Panel radiators

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

Ppanel radiator power = V volume of the heated room * 41 required number of W per 1 cubic meter.

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

The radiator should be hung under the window

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

Selecting a boiler for the project

Forging gas boiler Bosch Gaz 3000W

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

Gas generator boiler

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

Recommendations. There are other types of boilers, but more briefly about them now. So, if you have chosen an oil heater, you can give preference to a unit with a multi-stage burner, thereby increasing the efficiency of the entire system.

Electrode boiler "Galan"

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

Fireplace for heating a country house

The thermal calculation method is the determination of the surface area of each individual heating device that emits heat into the room. Calculation of thermal energy for heating in in this case takes into account the maximum level of coolant temperature, which is intended for those heating elements for which the thermotechnical calculation of the heating system is carried out. That is, if the coolant is water, then its average temperature in the heating system is taken. In this case, the coolant consumption is taken into account. Likewise, if the coolant is steam, then the calculation of heat for heating uses the value highest temperature steam at a certain pressure level in the heating device.

Calculation method

To calculate heat energy for heating, it is necessary to take the heat demand indicators of a separate room. In this case, the heat transfer of the heat pipe, which is located in this room, should be subtracted from the data.

The surface area that gives off heat will depend on several factors - first of all, on the type of device used, on the principle of connecting it to the pipes and on how exactly it is located in the room. It should be noted that all these parameters also affect the heat flux density emanating from the device.

Calculation of heating devices of the heating system - the heat transfer of the heating device Q can be determined using the following formula:

Q pr = q pr* A p .

However, it can be used only if the indicator of the surface density of the thermal device q pr (W/m 2) is known.

From here you can calculate the calculated area A r. It is important to understand that the calculated area of any heating device does not depend on the type of coolant.

A p = Q np /q np ,

in which Q np is the level of heat transfer of the device required for a certain room.

Thermal calculation of heating takes into account that to determine the heat transfer of the device for a specific room, the formula is used:

Q pp = Q p - µ tr *Q tr

in this case, the indicator Q p is the heat demand of the room, Q tr is the total heat transfer of all elements of the heating system located in the room. Calculation of the heat load for heating implies that this includes not only the radiator, but also the pipes that are connected to it, and the transit heat pipeline (if any). In this formula, µtr is the correction factor, which provides for partial heat transfer of the system, designed to maintain constant temperature indoors. In this case, the size of the correction may vary depending on how exactly the pipes of the heating system were laid in the room. In particular, with the open method – 0.9; in the groove of the wall - 0.5; embedded in concrete wall – 1,8.

Calculation of the required heating power, that is, the total heat transfer (Q tr - W) of all elements of the heating system is determined using the following formula:

Q tr = µk tr *µ*d n *l*(t g - t c)

In it, k tr is an indicator of the heat transfer coefficient of a certain section of pipeline located indoors, d n - O.D. pipes, l – length of the segment. Indicators tg and tv show the temperature of the coolant and air in the room.

Formula Q tr = q in *l in + q g *l g used to determine the level of heat transfer of the heat pipe present in the room. To determine indicators, you should refer to special reference literature. In it you can find a definition of the thermal power of a heating system - a definition of heat transfer vertically (q in) and horizontally (q g) of a heat pipe laid in the room. The data found shows the heat transfer of 1 m of pipe.

Before calculating Gcal for heating, for many years, calculations made using the formula A p = Q np /q np and measurements of heat-transferring surfaces of the heating system were carried out using a conventional unit - equivalent square meters. At the same time, the ECM was conditional equal to the surface heating device with a heat output of 435 kcal/h (506 W). The calculation of Gcal for heating assumes that the temperature difference between the coolant and air (t g - t in) in the room was 64.5 ° C, and the relative water flow in the system was equal to Grel = l.0.

Calculation of thermal loads for heating implies that smooth-tube and panel heating devices, which had greater heat transfer than standard radiators from the times of the USSR, had an ecm area that differed significantly from their physical area. Accordingly, the ecm area of less efficient heating devices was significantly lower than their physical area.

However, such dual measurement of the area of heating devices was simplified in 1984, and the ECM was abolished. Thus, from that moment on, the area of the heating device was measured only in m 2.

After the area of the heating device required for the room has been calculated and the thermal power of the heating system has been calculated, you can begin to select the required radiator from the catalog of heating elements.

It turns out that most often the area of the purchased element is slightly larger than that which was obtained by calculation. This is quite easy to explain - after all, such a correction is taken into account in advance by introducing a multiplying factor µ 1 into the formulas.

Today, sectional radiators are very common. Their length directly depends on the number of sections used. In order to calculate the amount of heat for heating - that is, to calculate the optimal number of sections for a certain room, the formula is used:

N = (A p /a 1)(µ 4 / µ 3)

In it, a 1 is the area of one section of the radiator selected for installation indoors. Measured in m2. µ 4 – correction factor that is applied to the method of installing the heating radiator. µ 3 – correction factor, which indicates the actual number of sections in the radiator (µ 3 - 1.0, provided that A p = 2.0 m 2). For standard radiators of the M-140 type, this parameter is determined by the formula:

µ 3 =0.97+0.06/A p

During thermal tests, standard radiators are used, consisting on average of 7-8 sections. That is, the calculation of heat consumption for heating determined by us - that is, the heat transfer coefficient - is real only for radiators of this particular size.

It should be noted that when using radiators with fewer sections, there is a slight increase in the level of heat transfer.

This is due to the fact that in the outer sections the heat flow is somewhat more active. In addition, the open ends of the radiator contribute to greater heat transfer into the room air. If the number of sections is greater, a weakening of the current is observed in the outer sections. Accordingly, to achieve the required level of heat transfer, the most rational option is to slightly increase the length of the radiator by adding sections, which will not affect the power of the heating system.

For those radiators whose area of one section is 0.25 m 2, there is a formula for determining the coefficient µ 3:

µ 3 = 0.92 + 0.16 /A p

But it should be borne in mind that it is extremely rare when using this formula that an integer number of sections is obtained. Most often, the required quantity turns out to be fractional. Calculation of heating devices of the heating system suggests that in order to obtain a more accurate result, a slight (no more than 5%) reduction in the A r coefficient is permissible. This action leads to limiting the level of temperature deviation in the room. When the heat for heating the room is calculated, after receiving the result, a radiator is installed with the number of sections as close as possible to the obtained value.

Calculation of heating power by area assumes that the architecture of the house also imposes certain conditions on the installation of radiators.

In particular, if there is an external niche under the window, then the length of the radiator should be less than the length of the niche - no less than 0.4 m. This condition is valid only when the pipe is connected directly to the radiator. If a duck liner is used, the difference in the length of the niche and the radiator should be at least 0.6 m. In this case, the extra sections should be separated as a separate radiator.

For certain models of radiators, the formula for calculating heat for heating - that is, determining the length - is not applied, since this parameter is pre-determined by the manufacturer. This fully applies to radiators such as RSV or RSG. However, there are often cases when, to increase the area of a heating device of this type, simply parallel installation of two panels next to each other is used.

If a panel radiator is determined to be the only one acceptable for a given room, then to determine the number of radiators required, use:

N = A p / a 1 .

In this case, the radiator area is a known parameter. If two parallel radiator blocks are installed, the A p indicator is increased, determining the reduced heat transfer coefficient.

In the case of using convectors with a casing, the calculation of heating power takes into account that their length is also determined exclusively by the existing model range. In particular, floor convector“Rhythm” is presented in two models with casing lengths of 1 m and 1.5 m. Wall-mounted convectors may also differ slightly from each other.

In the case of using a convector without a casing, there is a formula that helps determine the number of elements of the device, after which you can calculate the power of the heating system:

N = A p / (n*a 1)

Here n is the number of rows and tiers of elements, which make up the area of the convector. In this case, a 1 is the area of one pipe or element. In this case, when determining the estimated area of the convector, it is necessary to take into account not only the number of its elements, but also the method of their connection.

If a smooth-tube device is used in a heating system, the duration of its heating pipe is calculated as follows:

l = А р *µ 4 / (n*a 1)

µ 4 is the correction factor that is introduced if there is a decorative pipe cover; n – number of rows or tiers of heating pipes; and 1 is a parameter characterizing the area of one meter of horizontal pipe with a predetermined diameter.

To obtain a more accurate (and not fractional) number, a slight (no more than 0.1 m2 or 5%) reduction in indicator A is allowed.

Example No. 1

It is necessary to determine the correct number of sections for the M140-A radiator, which will be installed in a room located on top floor. In this case, the wall is external, there is no niche under the window sill. And the distance from it to the radiator is only 4 cm. The height of the room is 2.7 m. Q n = 1410 W, and t = 18 ° C. Conditions for connecting the radiator: connection to a single-pipe riser of a flow-regulated type (D y 20, KRT tap with 0.4 m inlet); The heating system is routed from the top, t = 105°C, and the coolant flow through the riser is G st = 300 kg/h. The temperature difference between the coolant in the supply riser and the one in question is 2°C.

We define average radiator temperature:

t av = (105 - 2) - 0.5x1410x1.06x1.02x3.6 / (4.187x300) = 100.8 °C.

Based on the data obtained, we calculate the heat flux density:

t av = 100.8 - 18 = 82.8 °C

It should be noted that there was a slight change in the level of water consumption (360 to 300 kg/h). This parameter has practically no effect on q np.

Q pr =650(82.8/70)1+0.3=809W/m2.

Next, we determine the level of heat transfer horizontally (1g = 0.8 m) and vertically (1v = 2.7 - 0.5 = 2.2 m) located pipes. To do this you should use the formula Q tr =q in xl in + q g xl g.

We get:

Q tr = 93x2.2 + 115x0.8 = 296 W.

We calculate the area of the required radiator using the formula A p = Q np /q np and Q pp = Q p - µ tr xQ tr:

A p = (1410-0.9x296)/809 = 1.41 m 2.

We calculate the required number of sections of the M140-A radiator, taking into account that the area of one section is 0.254 m2:

m 2 (µ4 = 1.05, µ 3 = 0.97 + 0.06 / 1.41 = 1.01, we use the formula µ 3 = 0.97 + 0.06 / A r and determine:

N=(1.41/0.254)x(1.05/1.01)=5.8.
That is, the calculation of heat consumption for heating showed that in order to achieve the most comfortable temperature in the room, a radiator consisting of 6 sections should be installed.

Example No. 2

It is necessary to determine the brand of an open wall convector with a casing KN-20k “Universal-20”, which is installed on a single-pipe riser flow type. There is no tap near the installed device.

Determines the average water temperature in the convector:

tcp = (105 - 2) - 0.5x1410x1.04x1.02x3.6 / (4.187x300) = 100.9 °C.

In the Universal-20 convectors, the heat flux density is 357 W/m2. Available data: µt cp = 100.9-18 = 82.9 ° C, Gnp = 300 kg/h. Using the formula q pr =q nom (µ t av /70) 1+n (G pr /360) p we recalculate the data:

q np = 357(82.9 / 70)1+0.3(300 / 360)0.07 = 439 W/m2.

We determine the level of heat transfer of horizontal (1 g - = 0.8 m) and vertical (l in = 2.7 m) pipes (taking into account D y 20) using the formula Q tr = q in xl in +q g xl g. We obtain:

Q tr = 93x2.7 + 115x0.8 = 343 W.

Using the formula A p = Q np /q np and Q pp = Q p - µ tr xQ tr, we determine the estimated area of the convector:

A p = (1410 - 0.9x343) / 439 = 2.51 m 2.

That is, the “Universal-20” convector, the casing length of which is 0.845 m, was accepted for installation (model KN 230-0.918, the area of which is 2.57 m2).

Example No. 3

For a steam heating system, it is necessary to determine the number and length of cast iron finned pipes, provided that the installation open type and is produced in two tiers. In this case, the excess steam pressure is 0.02 MPa.

Additional characteristics: t on = 104.25 °C, t on = 15 °C, Q p = 6500 W, Q tr = 350 W.

Using the formula µ t n = t us - t v, we determine the temperature difference:

µ t n = 104.25-15 = 89.25 °C.

We determine the heat flux density using the known transmission coefficient of this type of pipe in the case when they are installed in parallel one above the other - k = 5.8 W/(m2-°C). We get:

q np = k np x µ t n = 5.8-89.25 = 518 W/m2.

The formula A p = Q np /q np helps determine the required area of the device:

A p = (6500 - 0.9x350) / 518 = 11.9 m 2.

To determine the number of pipes needed, N = A p / (nxa 1). In this case, you should use the following data: the length of one tube is 1.5 m, the heating surface area is 3 m 2.

We calculate: N= 11.9/(2x3.0) = 2 pcs.

That is, in each tier it is necessary to install two pipes, each 1.5 m long. In this case, we calculate the total area of this heating device: A = 3.0x*2x2 = 12.0 m 2.

This article is also available in the following languages: Thai

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THANK YOU so much for the very useful information in the article. Everything is presented very clearly. It feels like a lot of work has been done to analyze the operation of the eBay store
- rootsshell
  
  Thank you and other regular readers of my blog. Without you, I would not have been motivated enough to dedicate much time to maintaining this site. My brain is structured this way: I like to dig deep, systematize scattered data, try things that no one has done before or looked at from this angle. It’s a pity that our compatriots have no time for shopping on eBay because of the crisis in Russia. They buy from Aliexpress from China, since goods there are much cheaper (often at the expense of quality). But online auctions eBay, Amazon, ETSY will easily give the Chinese a head start in the range of branded items, vintage items, handmade items and various ethnic goods.
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    What is valuable in your articles is your personal attitude and analysis of the topic. Don't give up this blog, I come here often. There should be a lot of us like that. Email me I recently received an email with an offer that they would teach me how to trade on Amazon and eBay. And I remembered your detailed articles about these trades. area I re-read everything again and concluded that the courses are a scam. I haven't bought anything on eBay yet. I am not from Russia, but from Kazakhstan (Almaty). But we also don’t need any extra expenses yet. I wish you good luck and stay safe in Asia.
rootsshell

It’s also nice that eBay’s attempts to Russify the interface for users from Russia and the CIS countries have begun to bear fruit. After all, the overwhelming majority of citizens of the countries of the former USSR do not have strong knowledge of foreign languages. No more than 5% of the population speak English. There are more among young people. Therefore, at least the interface is in Russian - this is a big help for online shopping on this trading platform. eBay did not follow the path of its Chinese counterpart Aliexpress, where a machine (very clumsy and incomprehensible, sometimes causing laughter) translation of product descriptions is performed. I hope that at a more advanced stage of development of artificial intelligence, high-quality machine translation from any language to any in a matter of seconds will become a reality. So far we have this (the profile of one of the sellers on eBay with a Russian interface, but an English description):
https://uploads.disquscdn.com/images/7a52c9a89108b922159a4fad35de0ab0bee0c8804b9731f56d8a1dc659655d60.png

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