Calculation of thermal loads for heating, methodology and calculation formula. Calculation of the heat load on the heating of a building: formula, examples
Whether it is an industrial building or a residential building, you need to make competent calculations and draw up a contour diagram heating system. At this stage, experts recommend paying special attention to the calculation of the possible heat load on the heating circuit, as well as the amount of fuel consumed and heat generated.
Thermal load: what is it?
This term refers to the amount of heat given off. The preliminary calculation of the heat load made it possible to avoid unnecessary costs for the purchase of components of the heating system and for their installation. Also, this calculation will help to correctly distribute the amount of heat generated economically and evenly throughout the building.
There are many nuances in these calculations. For example, the material from which the building is built, thermal insulation, region, etc. Specialists try to take into account as many factors and characteristics as possible to obtain a more accurate result.
The calculation of the 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. Yes, and housing and communal organizations calculate the cost of services based on data on heat load.
Main Factors
An ideally calculated and designed heating system must maintain the set temperature in the room and compensate for the resulting heat losses. When calculating the indicator of the heat load on the heating system in the building, you need to take into account:
Purpose of the building: residential or industrial.
Characteristics of the structural elements of the structure. These are windows, walls, doors, roof and ventilation system.
Housing dimensions. The larger it is, the more powerful the heating system should be. Be sure to take into account the area of window openings, doors, exterior walls and the volume of each interior space.
The presence of rooms for special purposes (bath, sauna, etc.).
Degree of equipment with technical devices. That is, the presence of hot water supply, ventilation systems, air conditioning and the type of heating system.
For a single room. For example, in rooms intended for storage, it is not necessary to maintain a comfortable temperature for a person.
Number of points with hot water supply. The more of them, the more the system is loaded.
Area of glazed surfaces. Rooms with French windows lose a significant amount of heat.
Additional terms. In residential buildings, this can be the number of rooms, balconies and loggias and bathrooms. In industrial - the number of working days in a calendar year, shifts, the technological chain of the production process, etc.
Climatic conditions of the region. When calculating heat losses, 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 heat load are in SNiPs and GOSTs. They also have special heat transfer coefficients. From the passports of the equipment included in the heating system, digital characteristics are taken regarding a specific heating radiator, boiler, etc. And also traditionally:
The heat consumption, taken to the maximum for one hour of operation of the heating system,
The maximum heat flow from one radiator,
Total heat costs in a certain period (most often - a season); if you need an hourly calculation of the load on heating network, then the calculation must be carried out taking into account the temperature difference during the day.
The calculations made are compared with the heat transfer area of the entire system. The index is quite accurate. Some deviations happen. For example, for industrial buildings, it will be necessary to take into account the reduction in heat energy consumption on weekends and holidays, and in residential buildings - 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
To date, the calculation of the heat load on the heating of a building can be carried out in one of the following ways.
Three main
- Aggregated indicators are taken for calculation.
- The indicators of the structural elements of the building are taken as the base. Here, the calculation of the internal volume of air going to warm up will also be important.
- All objects included in the heating system are calculated and summarized.
One exemplary
There is also a fourth option. It has a fairly large error, because the indicators are taken very average, or they are not enough. Here is the formula - Q from \u003d q 0 * a * V H * (t EH - t NPO), where:
- q 0 - specific thermal characteristic of the building (most often determined by the coldest period),
- a- correction factor(depends on the region and is taken from ready-made tables),
- V H is the volume calculated from the outer planes.
Example of a simple calculation
For building with standard parameters(ceiling heights, room sizes and good thermal insulation characteristics) you can apply a simple ratio of parameters, corrected by a factor depending on the region.
Suppose 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 \u003d 27.2 kW / h.
Such a definition of thermal loads does not take into account many important factors. For example, design features structures, temperatures, number of walls, ratio of wall areas and window openings etc. Therefore, such calculations are not suitable for serious projects of the heating system.
It depends on the material from which they are made. Most often today, bimetallic, aluminum, steel are used, much less often cast iron radiators. Each of them has its own heat transfer index (thermal power). Bimetal radiators with a distance between the axes of 500 mm, on average they have 180 - 190 watts. Aluminum radiators have almost the same performance.
The heat transfer of the described radiators is calculated for one section. Steel plate radiators are non-separable. Therefore, their heat transfer is determined based on the size of the entire device. For example, thermal power a two-row radiator 1,100 mm wide and 200 mm high will be 1,010 W, and a steel panel radiator 500 mm wide and 220 mm high will be 1,644 W.
The calculation of the heating radiator by area includes the following basic parameters:
Ceiling height (standard - 2.7 m),
Thermal power (per sq. m - 100 W),
One outer wall.
These calculations show that for every 10 sq. m requires 1,000 W of thermal power. This result is divided by the heat output of one section. The answer is the required number of radiator sections.
For the southern regions of our country, as well as for the northern ones, decreasing and increasing coefficients have been developed.
Average calculation and exact
Given the factors described, the average calculation is carried out according to the following scheme. If for 1 sq. m requires 100 W of heat flow, then a room of 20 square meters. m should receive 2,000 watts. The radiator (popular bimetallic or aluminum) of eight sections allocates about Divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.
The exact one looks a little intimidating. Actually, nothing complicated. Here is the formula:
Q t \u003d 100 W / m 2 × S (rooms) m 2 × q 1 × q 2 × q 3 × q 4 × q 5 × q 6 × q 7, Where:
- q 1 - type of glazing (ordinary = 1.27, double = 1.0, triple = 0.85);
- q 2 - wall insulation (weak or absent = 1.27, 2-brick wall = 1.0, modern, high = 0.85);
- q 3 - the ratio of the total area of window openings to the floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
- q 4 - outdoor temperature (the minimum value is taken: -35 o C = 1.5, -25 o C = 1.3, -20 o C = 1.1, -15 o C = 0.9, -10 o C = 0.7);
- q 5 - the number of external walls in the room (all four = 1.4, three = 1.3, corner room= 1.2, one = 1.2);
- q 6 - type of calculation room above the calculation room (cold attic = 1.0, warm attic = 0.9, residential heated room = 0.8);
- q 7 - ceiling height (4.5 m = 1.2, 4.0 m = 1.15, 3.5 m = 1.1, 3.0 m = 1.05, 2.5 m = 1.3).
Using any of the methods described, it is possible to calculate the heat load of an apartment building.
Approximate calculation
These are the conditions. The minimum temperature in the cold season is -20 ° C. Room 25 sq. m with triple glazing, double-leaf windows, ceiling height of 3.0 m, two-brick walls and an unheated attic. The calculation will be as follows:
Q \u003d 100 W / m 2 × 25 m 2 × 0.85 × 1 × 0.8 (12%) × 1.1 × 1.2 × 1 × 1.05.
The result, 2 356.20, is divided by 150. As a result, it turns out that 16 sections need to be installed in a room with the specified parameters.
If calculation is required in gigacalories
In the absence of a heat energy meter on an open heating circuit, the calculation of the heat load for heating the building is calculated by the formula Q \u003d V * (T 1 - T 2) / 1000, where:
- V - the amount of water consumed by the heating system, calculated in tons or m 3,
- T 1 - a number showing the temperature of hot water, measured in o C, and for calculations, the temperature corresponding to a certain pressure in the system is taken. This indicator has its own name - enthalpy. If it is not possible to remove temperature indicators in a practical way, they resort to an average indicator. It is in the range of 60-65 o C.
- T 2 - temperature cold water. It is quite difficult to measure it in the system, so constant indicators have been developed that depend on the temperature regime on the street. For example, in one of the regions, in the cold season, this indicator is taken equal to 5, in summer - 15.
- 1,000 is the coefficient for obtaining the result immediately in gigacalories.
In the case of a closed circuit thermal load(gcal/hour) is calculated differently:
Q from \u003d α * q o * V * (t in - t n.r.) * (1 + K n.r.) * 0.000001, Where
The calculation of the heat load turns out to be somewhat enlarged, but it is this formula that is given in the technical literature.
Increasingly, in order to increase the efficiency of the heating system, they resort to buildings.
These works are carried out in dark time days. For a more accurate result, you must observe the temperature difference between the room and the street: it must be at least 15 o. Fluorescent and incandescent lamps are switched off. It is advisable to remove carpets and furniture to the maximum, they knock down the device, giving some error.
The survey is carried out slowly, the data are recorded carefully. The scheme is simple.
The first stage of work takes place indoors. The device is moved gradually from doors to windows, giving Special attention corners and other joints.
The second stage is the examination of the 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 a computer, where the corresponding programs complete the processing and give the result.
If the survey was conducted by a licensed organization, then it will issue a report with mandatory recommendations based on the results of the work. If the work was carried out personally, then you need to rely on your knowledge and, possibly, the help of the Internet.
The thermal calculation method is the determination of the surface area of each individual heater which releases heat into the room. Calculation of thermal energy for heating in this case takes into account the maximum temperature level of the coolant, which is intended for those heating elements for which the heat engineering 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 flow rate of the coolant is taken into account. Similarly, if the heat carrier is steam, then the calculation of heat for heating uses the value highest temperature steam at a certain pressure level in the heater.
Method of calculation
To calculate the 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 pipes and on how exactly it is located in the room. It should be noted that all these parameters also affect the density of the heat flux coming from the device.
Calculation of heaters of the heating system - the heat output of the heater Q can be determined by the following formula:
Q pr \u003d q pr * A p.
However, it can only be used if the surface density index is known thermal device q pr (W / m 2).
From here it is also possible to calculate the estimated area A p. It is important to understand that the calculated area of any heating device does not depend on the type of coolant.
A p \u003d Q np / q np,
in which Q np is the level of heat transfer of the device required for a certain room.
The thermal calculation of heating takes into account that the formula is used to determine the heat transfer of the device for a certain room:
Q pp = Q p - µ tr *Q tr
while 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. The 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 pipe (if any). In this formula, µ tr is the correction factor, which provides for the partial heat transfer of the system, designed to maintain a constant temperature in the room. In this case, the size of the amendment may vary depending on how exactly the pipes of the heating system were laid in the room. In particular, at open method– 0.9; in the furrow of the wall - 0.5; embedded in a concrete wall - 1.8.
The 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 the pipeline located in the room, d n - outside diameter pipes, l is the length of the segment. Indicators t g and t in show the temperature of the coolant and air in the room.
Formula Q tr \u003d 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 the indicators, refer to the special reference literature. In it you can find the definition of the thermal power of the heating system - the definition of heat transfer vertically (q in) and horizontally (q g) of a heat pipeline laid in the room. The data found show the heat transfer of 1m of pipe.
Before calculating Gcal for heating, for many years, calculations made using the formula A p = Q np / q np and measurements of the heat-releasing surfaces of the heating system were carried out using a conventional unit - equivalent square meters. At the same time, ekm was conditionally equal to the surface of the heating device with a heat transfer of 435 kcal/h (506 W). The calculation of Gcal for heating assumes that in this case 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 G rel = l.0.
The calculation of heat loads for heating implies that smooth-tube and panel heaters, which had a greater heat transfer than the reference radiators of the times of the USSR, had an ekm area that differed significantly from their physical area indicator. Accordingly, the area of less efficient heaters was significantly lower than their physical area.
However, such a dual measurement of the area of \u200b\u200bheating devices in 1984 was simplified, and the ekm was canceled. Thus, from that moment on, the area of the heater was measured only in m 2 .
After the area of the heater required for the room and the calculation of the heat output of the heating system are calculated, you can proceed to the selection of the necessary radiator according to the catalog of heating elements.
It turns out that most often the area of the purchased element is somewhat 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.
Very common today sectional radiators. Their length directly depends on the number of sections used. In order to calculate the amount of heat for heating - that is, calculate optimal amount sections for a specific room, the formula is used:
N = (Ap /a 1)(µ 4 / µ 3)
In it, a 1 is the area of \u200b\u200bone section of the radiator selected for installation in the room. Measured in m 2. µ 4 is the 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 \u003d 2.0 m 2). For standard M-140 type radiators, this parameter is determined by the formula:
µ 3 \u003d 0.97 + 0.06 / A p
During thermal tests, standard radiators are used, consisting of an 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 a smaller number of sections, a slight increase in the level of heat transfer is observed.
This is due to the fact that in the extreme sections the heat flow is somewhat more active. In addition, the open ends of the radiator contribute to greater heat transfer to the room air. If the number of sections is greater, there is a weakening of the current in the extreme sections. Accordingly, to achieve the required level of heat transfer, the most rational is a slight increase in the length of the radiator by adding sections, which will not affect the power of the heating system.
For those radiators, the area of one section of which is 0.25 m 2, there is a formula for determining the coefficient µ 3:
µ 3 \u003d 0.92 + 0.16 / A p
But it should be borne in mind that it is extremely rare when using this formula, an integer number of sections is obtained. Most often, the desired amount is fractional. The calculation of the heating devices of the heating system assumes that in order to obtain a more accurate result, a slight (no more than 5%) decrease in the A p coefficient is acceptable. This action leads to limiting the level of deviation of the temperature indicator in the room. When the calculation of heat for space heating is made, after receiving the result, a radiator is installed with the number of sections as close as possible to the obtained value.
The calculation of heating power by area assumes that certain conditions the architecture of the house also imposes on the installation of radiators.
In particular, if there is an external niche under the window, then the length of the radiator must be less than the length of the niche - not less than 0.4 m. This condition is valid only with a direct pipe connection to the radiator. If a duckbill connection is used, the difference between 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 individual models of radiators, the formula for calculating heat for heating - that is, determining the length - does not apply, since this parameter is predetermined by the manufacturer. This fully applies to radiators such as RSV or RSG. However, it is not uncommon for cases when, to increase the area of \u200b\u200ba heating device of this type, simply parallel installation of two panels side by side is used.
If a panel radiator is defined as the only one allowed for a given room, then the following is used to determine the number of radiators required:
N \u003d Ap / a 1.
In this case, the radiator area is a known parameter. If two parallel blocks of radiators 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 the heating output takes into account that their length is also determined exclusively by the existing model range. In particular, the floor convector "Rhythm" is presented in two models with a casing length of 1 m and 1.5 m. Wall convectors may also slightly differ from each other.
In the case of using a convector without a casing, there is a formula that helps to determine the number of elements of the device, after which it is possible to calculate the power of the heating system:
N \u003d A p / (n * a 1)
Here n is the number of rows and tiers of elements that make up the area of the convector. In this case, a 1 is the area of one pipe or element. At the same time, when determining the calculated 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 the heating system, the duration of its heating pipe is calculated as follows:
l \u003d A p * µ 4 / (n * a 1)
µ 4 is the correction factor that is introduced in the presence of a decorative pipe cover; n is the number of rows or tiers of heating pipes; and 1 is a parameter characterizing the area of one meter horizontal pipe with a predetermined diameter.
To obtain a more accurate (rather than a fractional number), a slight (no more than 0.1 m 2 or 5%) decrease in A is allowed.
Example #1
It is necessary to determine the correct number of sections for the M140-A radiator, which will be installed in the room located on the top floor. At the same time, the wall is external, there is no niche under the windowsill. And the distance from it to the radiator is only 4 cm. The height of the room is 2.7 m. Q n \u003d 1410 W, and t in \u003d 18 ° С. Radiator connection conditions: connection to a single-pipe riser of a flow-controlled type (D y 20, KRT tap with 0.4 m inlet); the wiring of the heating system is upper, t g \u003d 105 ° C, and the coolant flow through the riser is G st \u003d 300 kg / h. The difference between the temperature of the coolant of the supply riser and the one under consideration is 2 ° C.
We define average temperature in the radiator:
t cf \u003d (105 - 2) - 0.5x1410x1.06x1.02x3.6 / (4.187x300) \u003d 100.8 ° С.
Based on the obtained data, we calculate the heat flux density:
t cf \u003d 100.8 - 18 \u003d 82.8 ° С
At the same time, 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 \u003d 650 (82.8 / 70) 1 + 0.3 \u003d 809 W / m2.
Next, we determine the level of heat transfer horizontally (1r \u003d 0.8 m) and vertically (1v \u003d 2.7 - 0.5 \u003d 2.2 m) located pipes. To do this, use the formula Q tr \u003d q in xl in + q g xl g.
We get:
Q tr \u003d 93x2.2 + 115x0.8 \u003d 296 watts.
We calculate the area of the required radiator according to the formula A p \u003d Q np / q np and Q pp \u003d Q p - µ tr xQ tr:
And p \u003d (1410-0.9x296) / 809 \u003d 1.41m 2.
We calculate the required number of sections of the M140-A radiator, given that the area of one section is 0.254 m 2:
m 2 (µ4 = 1.05, µ 3 \u003d 0.97 + 0.06 / 1.41 \u003d 1.01, we use the formula µ 3 \u003d 0.97 + 0.06 / A p and determine:
N \u003d (1.41 / 0.254) x (1.05 / 1.01) \u003d 5.8.
That is, the calculation of heat consumption for heating showed that in order to achieve the maximum comfortable temperature a radiator consisting of 6 sections should be installed.
Example #2
It is necessary to determine the brand of an open wall-mounted convector with a casing KN-20k "Universal-20", which is installed on a single-pipe riser flow type. There is no crane near the installed device.
Determines the average water temperature in the convector:
tcp \u003d (105 - 2) - 0.5x1410x1.04x1.02x3.6 / (4.187x300) \u003d 100.9 ° C.
In "Universal-20" convectors, the heat flux density is 357 W/m 2 . Available data: µt cp =100.9-18=82.9°С, Gnp=300kg/h. According to the formula q pr \u003d q nom (µ t cf / 70) 1 + n (G pr / 360) p recalculate the data:
q np \u003d 357 (82.9 / 70) 1 + 0.3 (300 / 360) 0.07 \u003d 439 W / m 2.
We determine the level of heat transfer of horizontal (1 g - \u003d 0.8 m) and vertical (l in \u003d 2.7 m) pipes (taking into account D y 20) using the formula Q tr \u003d q in xl in + q g xl g. We get:
Q tr \u003d 93x2.7 + 115x0.8 \u003d 343 watts.
Using the formula A p \u003d Q np / q np and Q pp \u003d Q p - µ tr xQ tr, we determine the estimated area of the convector:
And p \u003d (1410 - 0.9x343) / 439 \u003d 2.51 m 2.
That is, the convector "Universal-20" was accepted for installation, the length of the casing of which is 0.845 m (model KN 230-0.918, the area of \u200b\u200bwhich is 2.57 m 2).
Example #3
For a steam heating system, it is necessary to determine the number and length of cast iron finned tubes, 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 nac \u003d 104.25 ° С, t v \u003d 15 ° С, Q p \u003d 6500 W, Q tr \u003d 350 W.
Using the formula µ t n \u003d t us - t in, we determine the temperature difference:
µ t n \u003d 104.25-15 \u003d 89.25 ° С.
We determine the heat flux density using the known transfer coefficient of this type of pipes in the case when they are installed in parallel one above the other - k = 5.8 W / (m2 - ° C). We get:
q np \u003d k np x µ t n \u003d 5.8-89.25 \u003d 518 W / m 2.
The formula A p \u003d Q np / q np helps to determine the required area of \u200b\u200bthe device:
A p \u003d (6500 - 0.9x350) / 518 \u003d 11.9 m 2.
To determine the amount necessary pipes, N = A p / (nхa 1). In this case, you should use the following data: the length of one tube is 1.5 m, the area of \u200b\u200bthe heating surface is 3 m 2.
We calculate: N \u003d 11.9 / (2x3.0) \u003d 2 pcs.
That is, in each tier it is necessary to install two pipes 1.5 m long each. In this case, we calculate the total area of \u200b\u200bthis heater: A \u003d 3.0x * 2x2 \u003d 12.0 m 2.
To find out how much power the heat-power equipment of a private house should have, it is necessary to determine the total load on the heating system, for which a thermal calculation is performed. In this article, we will not talk about an enlarged method for calculating the area or volume of a building, but we will present more exact way, used by designers, only in a simplified form for better understanding. So, 3 types of loads fall on the heating system of the house:
- compensation for the loss of thermal energy leaving through building construction(walls, floors, roofing);
- heating the air required for ventilation of the premises;
- heating water for DHW needs (when a boiler is involved in this, and not a separate heater).
Determination of heat loss through external fences
To begin with, we present a formula from SNiP, according to which the calculation of the thermal energy lost through building structures separating inner space houses from the street:
Q \u003d 1 / R x (tv - tn) x S, where:
- Q is the consumption of heat leaving through the structure, W;
- R - resistance to heat transfer through the material of the fence, m2ºС / W;
- S is the area of this structure, m2;
- tv - the temperature that should be inside the house, ºС;
- tn is the average outdoor temperature for the 5 coldest days, ºС.
For reference. According to the methodology, heat loss calculation is performed separately for each room. In order to simplify the task, it is proposed to take the building as a whole, assuming an acceptable average temperature of 20-21 ºС.
The area for each type of external fencing is calculated separately, for which windows, doors, walls and floors with a roof are measured. This is done because they are made from different materials different thickness. So the calculation will have to be done separately for all types of structures, and then the results will be summed up. You probably know the coldest street temperature in your area of residence from practice. But the parameter R will have to be calculated separately according to the formula:
R = δ / λ, where:
- λ is the coefficient of thermal conductivity of the fence material, W/(mºС);
- δ is the thickness of the material in meters.
Note. The value of λ is a reference value, it is not difficult to find it in any reference literature, and for plastic windows this coefficient will be prompted by the manufacturers. Below is a table with the coefficients of thermal conductivity of some building materials, and for calculations it is necessary to take the operational values of λ.
As an example, let's calculate how much heat will be lost by 10 m2 brick wall 250 mm thick (2 bricks) with a temperature difference outside and inside the house of 45 ºС:
R = 0.25 m / 0.44 W / (m ºС) = 0.57 m2 ºС / W.
Q \u003d 1 / 0.57 m2 ºС / W x 45 ºС x 10 m2 \u003d 789 W or 0.79 kW.
If the wall consists of different materials ( structural material plus insulation), then they must also be calculated separately according to the above formulas, and the results summarized. Windows and roofing are calculated in the same way, but the situation is different with floors. First of all, you need to draw a building plan and divide it into zones 2 m wide, as is done in the figure:
Now you should calculate the area of \u200b\u200beach zone and alternately substitute it into the main formula. Instead of parameter R, you need to take the standard values for zone I, II, III and IV, indicated in the table below. At the end of the calculations, the results are added up and we get the total heat loss through the floors.
Ventilation air heating consumption
Uninformed people often do not take into account that the supply air in the house also needs to be heated, and this heat load also falls on the heating system. Cold air still enters the house from the outside, whether we like it or not, and it takes energy to heat it. Moreover, a full-fledged supply and exhaust ventilation usually with a natural urge. Air exchange is created due to the presence of traction in ventilation ducts and boiler chimney.
Proposed in normative documentation The method for determining the heat load from ventilation is rather complicated. Pretty accurate results can be obtained if this load is calculated using the well-known formula through the heat capacity of the substance:
Qvent = cmΔt, here:
- Qvent - the amount of heat required for heating supply air, W;
- Δt - temperature difference in the street and inside the house, ºС;
- m is the mass of the air mixture coming from outside, kg;
- c is the heat capacity of air, assumed to be 0.28 W / (kg ºС).
The complexity of calculating this type of heat load lies in the correct determination of the mass of heated air. Find out how much it gets inside the house, when natural ventilation difficult. Therefore, it is worth referring to the standards, because buildings are built according to projects where the required air exchanges are laid down. And the regulations say that in most rooms the air environment should change 1 time per hour. Then we take the volumes of all rooms and add to them the air flow rates for each bathroom - 25 m3 / h and a kitchen gas stove– 100 m3/h.
To calculate the heat load on heating from ventilation, the resulting volume of air must be converted into mass, knowing its density at different temperatures from the table:
Let us assume that the total amount of supply air is 350 m3/h, the outside temperature is minus 20 ºС, and the inside temperature is plus 20 ºС. Then its mass will be 350 m3 x 1.394 kg / m3 = 488 kg, and the heat load on the heating system will be Qvent = 0.28 W / (kg ºС) x 488 kg x 40 ºС = 5465.6 W or 5.5 kW.
Heat load from DHW heating
To determine this load, you can use the same simple formula, only now you need to calculate thermal energy used for water heating. Its heat capacity is known and amounts to 4.187 kJ/kg °С or 1.16 W/kg °С. Considering that a family of 4 people needs 100 liters of water for 1 day, heated to 55 ° C, for all needs, we substitute these numbers into the formula and get:
QDHW \u003d 1.16 W / kg ° С x 100 kg x (55 - 10) ° С \u003d 5220 W or 5.2 kW of heat per day.
Note. By default, it is assumed that 1 liter of water is equal to 1 kg, and the temperature of the cold tap water equal to 10 °C.
The unit of equipment power is always referred to 1 hour, and the resulting 5.2 kW - to the day. But you cannot divide this figure by 24, because hot water we want to receive as soon as possible, and for this the boiler must have a power reserve. That is, this load must be added to the rest as is.
Conclusion
This calculation of home heating loads will give much more accurate results than traditional way on the area, although you have to work hard. The end result must be multiplied by the safety factor - 1.2, or even 1.4, and according to the calculated value, select boiler equipment. Another way to enlarge the calculation of thermal loads according to the standards is shown in the video:
When designing heating systems for all types of buildings, you need to make the right calculations, and then develop a competent heating circuit diagram. At this stage, special attention should be paid to the calculation of the heat load on heating. To solve the problem, it is important to use A complex approach and take into account all the factors affecting the operation of the system.
- The thermal characteristic of the building is 0.49 W / m³ * C.
- Refinement coefficient - 1.
- The optimum temperature indicator inside the building is 22 degrees.
- Optimum temperature parameters in the premises.
- The total area of the building.
- Air temperature outside.
- Wall area and thickness - 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.
Show all
Parameter Importance
Using the heat load indicator, you can find out the amount of heat energy needed to heat a particular room, as well as the building as a whole. The main variable here is the power of everything heating equipment, which is planned to be used in the system. In addition, it is required to take into account the heat loss of the house.
The ideal situation seems to be in which the capacity of the heating circuit allows not only to eliminate all losses of heat energy from the building, but also to ensure comfortable conditions residence. To correctly calculate the specific heat load, it is required to take into account all the factors influencing this parameter:
The optimal mode of operation of the heating system can only be compiled taking into account these factors. The unit of measurement of the indicator can be Gcal / hour or kW / hour.
heating load calculation
Choice of method
Before starting the calculation of the heating load according to aggregated indicators, it is necessary to determine 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 the operating modes of the heating system for each room.
The methods used today for calculating the hourly load on the heating system make it possible to obtain results of varying degrees of accuracy. In some situations, complex calculations are required to minimize the error.
If, when designing a heating system, optimizing energy costs is not a priority, less accurate methods can be used.
Heat load calculation and heating system design Audytor OZC + Audytor C.O.
Simple Ways
Any method of calculating the heat load allows you to choose the optimal parameters of the heating system. Also, this indicator helps to determine the need for work to improve the thermal insulation of the building. Today, two fairly simple methods for calculating the heat load are used.
Depending on area
If all rooms in the building have standard sizes and have good thermal insulation, you can use the method of calculating the required power of heating equipment, depending on the area. In this case, 1 kW of thermal energy should be produced for every 10 m 2 of the room. Then the result obtained must be multiplied by the correction factor for the climatic zone.
This is the simplest calculation method, but it has one serious drawback - the error is very high. During the calculations, only the climatic region is taken into account. However, many factors affect the efficiency of the heating system. Thus, it is not recommended to use this technique in practice.
Upscale Computing
Applying the methodology for calculating heat according to aggregated indicators, the calculation error will be smaller. This method was first often used to determine the heat load in a situation where the exact parameters of the structure were unknown. To determine the parameter, the calculation formula is used:
Qot \u003d q0 * a * Vn * (tvn - tnro),
where q0 is the specific thermal characteristic of the structure;
a - correction factor;
Vн - external volume of the building;
tvn, tnro - temperature values inside the house and outside.
As an example of calculating thermal loads using aggregated indicators, you can calculate the maximum indicator for the heating system of a building along the outer walls of 490 m 2. Two storey building with with total area 170 m 2 located in St. Petersburg.
First you need to use normative document install all input data required for the calculation:
Assuming that minimum temperature V winter period will be -15 degrees, all known values \u200b\u200bcan be substituted into the formula - Q \u003d 0.49 * 1 * 490 (22 + 15) \u003d 8.883 kW. Using the most a simple technique calculation of the base indicator of heat load, the result would be higher - Q = 17 * 1 = 17 kW / h. Wherein the enlarged method for calculating the load indicator takes into account much more factors:
Also, this technique allows, with a minimum error, to calculate the power of each radiator installed in a single room. Its only drawback is the inability to calculate the heat loss of the building.
Calculation of thermal loads, Barnaul
Complex technique
Since even with an enlarged calculation, the error turns out to be quite high, it is necessary to use more complicated method 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 these, the most important is the heat transfer resistance ® of the materials used to make each element of the building - the floor, the walls, and the ceiling.
This value is inversely related to thermal conductivity (λ), which shows 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, it is first necessary to calculate the heat transfer resistance using a simple formula - R \u003d d / λ.
The proposed method consists of two stages. First, heat losses are calculated for window openings and external walls, and then for ventilation. As an example, we can take the following characteristics of the structure:
Based on the input data, it is possible to determine the TV transmission resistance index of the walls - R \u003d 0.4 / 0.56 \u003d 0.71 m² * C / W. Then a similar indicator of insulation is determined - R \u003d 0.11 / 0.036 \u003d 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 of the walls will be - (1 / 3.76) * 245 + (1 / 0.76) * 45 = 125.15 W. The temperature parameters remained unchanged in comparison with the integrated calculation. The next calculations are carried out in accordance with the formula - 125.15 * (22 + 15) \u003d 4.63 kW / h.
Calculation of the thermal power of heating systems
At the second stage, heat losses are calculated ventilation system. It is known that the volume of the house is 490 m³, and the air density is 1.24 kg/m³. This allows you to find out its mass - 608 kg. During the day, the air in the room is updated on average 5 times. After that, you can calculate the heat loss of the ventilation system - (490 * 45 * 5) / 24 = 4593 kJ, which corresponds to 1.27 kW / h. It remains to determine the general heat loss buildings, adding up the available results, - 4.63 + 1.27 = 5.9 kW / h.
Start of the preparation of the heating project, both residential country houses, and industrial complexes, follows from the heat engineering calculation. A heat gun is assumed as a heat source.
What is a thermal calculation?
The calculation of heat losses is a fundamental document designed to solve such a problem as the organization of heat supply to a structure. It determines the daily and annual heat consumption, the minimum requirement of a residential or industrial facility for thermal energy and heat losses for each room.
When solving such a problem as a heat engineering calculation, one should take into account a set of object characteristics:
- Object type ( a private house, one-story or multi-story building, administrative, industrial or warehouse).
- The number of people living in the building or working in one shift, the number of hot water supply points.
- Architectural part (dimensions of the roof, walls, floors, dimensions of door and window openings).
- Special data, e.g. number of working days per year (for productions), duration heating season(for objects of any type).
- Temperature conditions in each of the premises of the facility (they are determined by CHiP 2.04.05-91).
- Functional purpose (storage production, residential, administrative or household).
- Roof structures, external walls, floors (type of insulation layers and materials used, thickness of floors).
Why do you need a thermal calculation?
- To determine the power of the boiler.
Suppose you have decided to supply Vacation home or enterprise system autonomous heating. To determine the choice of equipment, first of all, you will need to calculate the power of the heating installation, which will be needed for the uninterrupted operation of hot water supply, air conditioning, ventilation systems, as well as efficient heating building. The power of the autonomous heating system is determined as total amount heat costs for heating all rooms, as well as heat costs for other technological needs. The heating system must have a certain power reserve so that operation at peak loads does not shorten its service life. - To carry out approval for the gasification of the facility and obtain technical specifications.
It is necessary to obtain a permit for the gasification of an object if natural gas is used as a fuel for the boiler. To obtain TS, you will need to provide values annual expense fuel ( natural gas), as well as the total power of heat sources (Gcal/h). These indicators are determined as a result of thermal calculation. Coordination of the project for the implementation of gasification of the facility is a more expensive and time-consuming method of organizing autonomous heating, in relation to the installation of heating systems operating on waste oils, the installation of which does not require approvals and permits. - To select the right equipment.
Thermal calculation data are the determining factor when choosing devices for heating objects. Many parameters should be taken into account - orientation to the cardinal points, dimensions of door and window openings, dimensions of rooms and their location in the building.
How is the thermal calculation
You can use simplified formula to determine the minimum allowable power of thermal systems:
Q t (kW / h) \u003d V * ΔT * K / 860, where
Q t is the heat load on a certain room;
K is the heat loss coefficient of the building;
V - the volume (in m 3) of the heated room (the width of the room for the length and height);
ΔT is the difference (marked C) between required temperature air inside and temperature outside.
Such an indicator as the heat loss coefficient (K) depends on the insulation and type of construction of the room. You can use simplified values calculated for objects of different types:
- K = from 0.6 to 0.9 (increased degree of thermal insulation). Few double glazed windows, double insulated brick walls, high quality material roof, solid floor base;
- K \u003d from 1 to 1.9 (medium thermal insulation). Double brickwork, a roof with a conventional roof, a small number of windows;
- K = 2 to 2.9 (low thermal insulation). The construction of the structure is simplified, single brickwork.
- K = 3 - 4 (lack of thermal insulation). A structure made of metal or corrugated sheet or a simplified wooden structure.
When determining the difference between the required temperature inside the heated volume and the outside temperature (ΔT), you should proceed from the degree of comfort that you want to receive from the thermal installation, as well as from the climatic features of the region in which the object is located. The values defined by CHiP 2.04.05-91 are accepted as default parameters:
- +18 – public buildings and production shops;
- +12 - high-rise storage complexes, warehouses;
- + 5 - garages, as well as warehouses without constant maintenance.
| City | City | Estimated outdoor temperature, °C | |
| Dnepropetrovsk | - 25 | Kaunas | - 22 |
| Ekaterinburg | - 35 | Lviv | - 19 |
| Zaporozhye | - 22 | Moscow | - 28 |
| Kaliningrad | - 18 | Minsk | - 25 |
| Krasnodar | - 19 | Novorossiysk | - 13 |
| Kazan | - 32 | Nizhny Novgorod | - 30 |
| Kyiv | - 22 | Odessa | - 18 |
| Rostov | - 22 | Saint Petersburg | - 26 |
| Samara | - 30 | Sevastopol | - 11 |
| Kharkiv | - 23 | Yalta | - 6 |
Calculation according to a simplified formula does not allow taking into account differences in the heat losses of a building depending on the type of enclosing structures, insulation and placement of premises. So, for example, rooms with big windows, high ceilings and corner rooms. At the same time, rooms that do not have external fences are distinguished by minimal heat losses. It is advisable to use the following formula when calculating such a parameter as the minimum thermal power:
Qt (kW / h) \u003d (100 W / m 2 * S (m 2) * K1 * K2 * K3 * K4 * K5 * K6 * K7) / 1000, where
S - room area, m 2;
W / m 2 - specific value of heat loss (65-80 watt / m 2). This indicator includes heat leakage through ventilation, absorption by walls, windows and other types of leakage;
K1 - coefficient of heat leakage through windows:
- in the presence of triple glazing K1 = 0.85;
- if the double-glazed window is double, then K1 = 1.0;
- with standard glazing K1 = 1.27;
K2 - coefficient of heat loss of walls:
- high thermal insulation (K2 = 0.854);
- insulation with a thickness of 150 mm or walls in two bricks (K2 = 1.0);
- low thermal insulation (K2=1.27);
K3 - an indicator that determines the ratio of areas (S) of windows and floor:
- 50% short circuit=1.2;
- 40% SC=1.1;
- 30% short circuit=1.0;
- 20% short circuit=0.9;
- 10% short circuit=0.8;
K4 - outdoor temperature coefficient:
- -35°C K4=1.5;
- -25°C K4=1.3;
- -20°C K4=1.1;
- -15°C K4=0.9;
- -10°C K4=0.7;
K5 - the number of outward-facing walls:
- four walls K5=1.4;
- three walls K5=1.3;
- two walls K5=1.2;
- one wall K5=1.1;
K6 - type of thermal insulation of the room, which is located above the heated one:
- heated K6-0.8;
- warm attic K6=0.9;
- unheated attic K6=1.0;
K7 - ceiling height:
- 4.5 meters K7=1.2;
- 4.0 meters K7=1.15;
- 3.5 meters K7=1.1;
- 3.0 meters K7=1.05;
- 2.5 meters K7=1.0.
Let us give as an example the calculation of the minimum power of an autonomous heating installation (according to two formulas) for a separate service station service room (ceiling height 4 m, area 250 m 2, volume 1000 m3, large windows with ordinary glazing, no thermal insulation of the ceiling and walls, simplified design ).
Simplified calculation:
Q t (kW / h) \u003d V * ΔT * K / 860 \u003d 1000 * 30 * 4 / 860 \u003d 139.53 kW, where
V is the volume of air in the heated room (250 * 4), m 3;
ΔT is the difference between the air temperature outside the room and the required air temperature inside the room (30°C);
K - heat loss coefficient of the building (for buildings without thermal insulation K = 4.0);
860 - conversion to kWh.
More accurate calculation:
Q t (kW / h) \u003d (100 W / m 2 * S (m 2) * K1 * K2 * K3 * K4 * K5 * K6 * K7) / 1000 \u003d 100 * 250 * 1.27 * 1.27 * 1.1*1.5*1.4*1*1.15/1000=107.12 kWh, where
S - area of the room for which the calculation is performed (250 m 2);
K1 is the parameter of heat leakage through windows (standard glazing, K1 index is 1.27);
K2 - the value of heat leakage through the walls (poor thermal insulation, the K2 indicator corresponds to 1.27);
K3 - the parameter of the ratio of the dimensions of the windows to the floor area (40%, the K3 indicator is 1.1);
K4 - outside temperature value (-35 °C, K4 index corresponds to 1.5);
K5 - the number of walls that go outside (in this case, four K5 is 1.4);
K6 - an indicator that determines the type of room located directly above the heated one (attic without insulation K6 \u003d 1.0);
K7 - an indicator that determines the height of the ceilings (4.0 m, the K7 parameter corresponds to 1.15).
As can be seen from the calculation, the second formula is preferable for calculating the power of heating installations, since it takes into account much large quantity parameters (especially if it is necessary to determine the parameters of low-power equipment designed for use in small spaces). To the result obtained, it is necessary to add a small margin of power to increase the service life of thermal equipment.
By performing simple calculations, you can determine without the help of specialists required power autonomous heating system for equipping residential or industrial facilities.
You can buy a heat gun and other heaters on the company's website or by visiting our retail store.
