Calculation of room heat inflows. Recommendations for the selection of air conditioning systems - Management Company "114 Repair Plant" Calculation of heat inflows from solar radiation online

One of the first stages
designing an air conditioning system is the calculation of heat input into the room. Strictly speaking, this problem comes down to solving differential equations. However, this approach is unacceptable for engineering calculations. In addition, some of the equations may not have an analytical solution and need to be solved
numerical methods, that is, use special software.

Now you can find many programs that calculate heat flow into a room. All of them can be divided into two categories. The first ones are accurate, they solve differential equations, but are not distributed free of charge. You won’t find them on the Internet, and not every company dealing with climate control equipment can afford to purchase such a program.

Other programs are built using the most simplified methods to the detriment of calculation accuracy. These programs are usually free and therefore popular. But the question arises about the scope of their application. For what geographic latitude and longitude are the calculation results valid? Is it really possible to obtain somewhat accurate results for both Murmansk and Krasnodar without specifying the locality?

And there is one more drawback that is common to both categories of programs. They are rarely consistent with domestic ones regulatory documents: building codes and regulations.

On the website of the Moscow representative office of MITSUBISHI ELECTRIC (www.mitsubishi-aircon.ru) in the section “For Specialists / To Help the Designer” a program is provided that implements the methodology set out in Manual 2.91 to SNiP 2.04.05-91 “Calculation of heat gain from solar radiation in premises” . The calculation is based on the following regulatory documentation:

1. SNiP 23-01-99 "Building climatology";
2. SNiP II-3-79 "Construction heating engineering";
3. SNiP 2.04.05-91 (2000) "Heating, ventilation and air conditioning."

Databases containing thermal indicators are connected to the main module of the program building materials and structures, filling of light openings, sun-protection devices, as well as a database on climatic parameters of the warm period of the year in various cities of Russia, Ukraine and the Republic of Belarus. The amount of heat is automatically selected from the tables solar radiation, arriving on vertical and horizontal surfaces, as well as through the glazing of light openings for the corresponding geographical latitude.

The program calculates not only the maximum heat inputs, but also their hourly values. This is especially important when designing multi-zone systems, since it is necessary to know the non-simultaneous heat load in different rooms. Based on the results obtained, we can conclude that it is permissible to use indoor units whose total performance exceeds the capacity of the compressor-condensing unit.

In addition to the total hourly heat inputs, all components are displayed separately: heat inputs from solar radiation absorbed by the room and transferred to the air, heat flows by heat transfer through windows, massive external enclosures (external walls and covering). Thanks to this, it is possible to evaluate the contribution of each component to the total heat input.

The output of intermediate calculation results is also provided, which allows, if necessary, to “manually” check the correctness of the results. When analyzing the calculation results, it should be remembered that the heat release from artificial lighting, technological equipment and materials, the release of heat and moisture by people, as well as the flow of heat with infiltrating air must be calculated independently. As a rule, this does not cause difficulties.

The heat gain from solar radiation calculated by the program heats the air in the room without changing its moisture content (sensible heat). Do not forget that the full cooling capacity of the air conditioner, which is usually indicated in the specifications and catalogs, is spent on reducing the air temperature, as well as on the condensation of excess moisture. Moreover, these costs can be equal for comfort class air conditioners.

On the same information resource you will find another program that determines the release of heat and moisture by people depending on the energy they expended and the temperature in the room. It is easy to carry out such a calculation yourself using tables, but it is more convenient to use a program, since it interpolates intermediate values ​​that are not in the original tables.

The article was prepared by specialists of the company "ARCTIKA"

For each point in Russia, we collected insolation data with an accuracy of 0.1 degrees in latitude and longitude. The data was kindly provided by NASA service where the history of measurements has been conducted since 1984.

To use our calculator, select your location solar power plant by moving the marker on the map or use the search field on the map. Our calculator only works on the territory of Russia.

1. If you know what solar panels you will use, or they are already installed in your solar station - select solar panels of the required power and their quantity.

2. Specify the angle of your roof and installation location. Also, our calculator automatically shows the optimal angle of inclination of the solar panel for the selected location. The angle is shown for winter, the optimal one is average for the whole year, for summer. This is especially important if you are just planning to install a solar station and during its construction you will be able to indicate to the builders the required angle for installing the solar system.

If, for example, you plan to install solar panels on the roof of your house and the installation angle is predetermined by the design, simply indicate it in the arbitrary angle input field.
Our calculator will calculate based on the angle of your roof.

3. It is very important to correctly assess the power of electricity consumers of your solar station when selecting the required quantity solar panels.

In the load calculator for a solar power plant, select the electrical appliances that you will use, specify their number and power in watts, as well as approximately the time of use per day.

For example, for a small house we choose:

• Electric lamps - 3 pieces with a power of 50 W each, work 6 hours a day - a total of 0.9 kW hours / day.
• TV - 1 piece with a power of 150 W, works 4 hours a day - a total of 0.6 kW hours/day.
• Refrigerator - 1 piece with a power of 200 W, works 6 hours a day - a total of 1.2 kW hours/day.
• Computer - 1 piece with a power of 350 W, works 3 hours a day - a total of 1.05 kW hours/day.

The TV is modern with a flat screen, LED consumes from 100 to 200 W, the refrigerator has a compressor and does not work all the time, but only when cold is needed, i.e. The more often you open the refrigerator door, the more electricity it will consume. Usually the refrigerator works 6 hours a day, the rest of the time it rests. For example, you use a computer on average 3 hours a day.

At given conditions consumption, you will receive the necessary power to power your electrical appliances.
For our example, this will be 3.75 kW/hour per day.

Let's select the required number of solar panels for our example, in the St. Petersburg region:

Let's take 250W solar modules and set the optimal tilt angle suggested by the program to be 60 degrees.
By increasing the number of solar panels we will see that when installing 3x solar modules The 250W consumption of our electrical appliances, 3.75 kW per hour per day, begins to overlap on the generation schedule from April to September, which is enough for those people who, for example, stay in the country in the summer.
If you want to operate solar power all year round, then you will need at least 6 solar modules of 250 W each, and preferably 9 pieces. Please also note that in winter from November to mid-January in St. Petersburg there is more likely no sun than there is. And at this time of year you will use a gasoline-diesel generator to recharge the batteries.

Below the output graph there is a summary table with numerical data on the output of the solar power plant in a convenient numerical form.

Fill out the form below, send us your calculation data and receive a commercial offer for your solar power plant.

Calculation of a solar power plant using a calculator is preliminary. Each project is individual; to formulate a final turnkey proposal, taking into account installation and feasibility study, we recommend consulting with our specialists by phone or ordering an engineer to visit you. Based on the results of the conversation, our specialists will prepare and provide a comprehensive proposal for the cost and installation of your solar power plant.

In order for our managers to be able to prepare preliminary calculations for you on the cost of equipment and installation, send us your calculation data. If the information is not enough, our specialist will contact you for clarification.

Power calculation and selection of split systems

ATTENTION!!! All information given below cannot replace an accurate thermal calculation performed by professional specialists and is for advisory purposes only.

Air conditioning- automatic maintenance in indoors all or individual air parameters (temperature, relative humidity, cleanliness, speed of movement) in order to ensure mainly optimal meteorological conditions, the most favorable for the well-being of people, maintaining technological process, ensuring the safety of valuables.
Air conditioning is divided into comfort and technological.
Comfortable hard currency currency designed to create and automatically maintain temperature, relative humidity, cleanliness and air speed that meet optimal sanitary and hygienic requirements.
Technological hard currency designed to provide air parameters that best meet production requirements.
According to standard ASHRAE 55- 56(USA), thermal comfort is defined as “the state of a person being satisfied with the conditions environment, in which he does not know whether he wants to change the environmental conditions, making it warmer or colder."

Marking of split system models

Most often, manufacturers use the cooling capacity of the system not in W, but in BTU (British thermal unit) to label their split systems. BTU - defined as the amount of heat required to increase the temperature of one pound of water by one degree Fahrenheit, for residents of our country this is not the most convenient system of measures. As is known from the history of air conditioning, the era of the birth of climate control technology in the form in which we know it now began in the United States, where it is used British system calculus. 1 BTU/hour = 0.2930710701722 W, respectively 1000 BTU = 293 W = 0.293 kW. Now the numbering of split systems is more clear, because the number split systems corresponds to the number of thousands of BTU/hour, for example split system No. 07 = 7000 BTU/hour; No. 09 = 9000 BTU/hour.
Example: split system number 07, corresponds to 7000 BTU/hour = 7000*0.293 = 2051 W = 2.1 kW; second option: split system number 07, respectively: 7 * 0.293 = 2.1 kW.
Below is a table of the main standard sizes and their corresponding cooling capacity values ​​in kW.

Calculation of cooling capacity of air conditioning system

Unlike the heating system - where, during thermal calculations, it is necessary to determine the amount of heat loss for its subsequent replenishment, in the air conditioning system the task is diametrical - the goal is to determine the amount of heat gain during the warm period of the year.

In addition to the main calculation, there is " Simplified calculation method air conditioning systems based on split systems" - You can download the calculator for selecting split systems in the format Microsoft Excel Template (.xltx)(developed by specialists of UK 114 Repair Plant LLC based on this calculation method - with detailed recommendations) - DOWNLOAD

Heat balance calculation

Thermal loads acting in the room can be divided into two types:

External thermal loads;


Internal thermal loads.

External thermal loads:

heat gain or heat loss through enclosing structures (walls, ceilings, floors, windows, doors) resulting from the temperature difference between inside and outside the room. The temperature difference between inside and outside the room in the summer is positive, as a result of which during this period of the year we receive an influx of heat into the room, in winter the opposite is true - the difference is negative and heat leaves the room;


heat gain from solar radiation (radiation) through glass, this load can manifest itself in the form of perceived heat. Solar radiation always creates a positive load at any time of the year. In summer, this load must be compensated, but in winter it is insignificant and may not be taken into account.


outside air entering the room (due to infiltration - leaks in enclosing structures, windows, doors), this air has correspondingly different properties in the summer and winter seasons: in summer - warm and humid (in some latitudes - dry); in winter - cold and dry (in some latitudes - wet). Accordingly, in summer the amount of heat and moisture brought in by the air must be compensated by the installation; accordingly, in winter the air must be heated and humidified.

External heat loads can be either positive or negative depending on the time of year and time of day.

Internal thermal loads:

the amount of heat generated by people and animals in the room;


heat generated by lamps and lighting fixtures;


heat generated by operating electrical appliances and equipment: stoves, ovens, refrigerators, computers, televisions, printers, etc.

IN production premises additional heat sources can be:

heated production equipment;


hot materials;


products of combustion and chemical reactions.

Internal heat loads are always positive; in summer they must be compensated by the cooling system, and in winter they reduce the load on the heating system.

Calculation of air conditioning systems.

This calculation is performed on the basis and in accordance with the recommendations:
SNiP II - 3- 79 *"Construction heating engineering";
SNiP 23-01-99*(Code of rules - SP 131.13330.2012 - updated version) "Construction climatology";
SNiP 41-01-2003
SNiP II - 33- 75"Heating, ventilation and air conditioning";
SNiP 2.04.05-91*"Heating, ventilation and air conditioning";
Manual 2.91 to SNiP 2.04.05-91"Calculation of heat input from solar radiation into premises";
SNiP 2.11.02-87(Code of rules - SP 109.13330.2012 - updated version) "Refrigerators";
Designer's Handbook Part 3 "Ventilation and Air Conditioning";
SanPiN 2.1.2.2645-10 " Sanitary and epidemiological requirements for living conditions in residential buildings and premises";
Barkalov B.V., Karpis E.E. "Air conditioning in industrial, public and residential buildings";
SNiP 31-01-2003(Code of rules - SP 54.13330.2011 - updated version) "Residential multi-apartment buildings."

The correct calculation of SCR can only be performed by qualified specialists in the field of heating engineering, ventilation and air conditioning.

Calculation of heat loss (heat gain) through enclosing structures.

The amount of heatQ transmitted through enclosing structures with an areaF , having a heat transfer coefficient k ( W/m2*⁰С), is determined by the formula:

Q = F*k* (t out.calc. - t ext.calc. )*Ѱ , Where

t out.calc. - estimated outside air temperature;
t ext.calc. - design temperature of internal air;

Ѱ - a correction factor that takes into account the amount of heat input, the orientation of the fence to the cardinal direction, wind load, number of floors, infiltration, solar radiation absorbed by the fence.

Calculation of heat gain from solar radiation through light openings (windows).

Excess heat from solar radiation is instantly absorbed by the room environment; depending on the glass, up to 90% of solar energy enters the room, the rest is reflected.
Solar radiation consists of two components:

direct radiation;


scattered radiation.

The intensity of solar radiation depends on the latitude of the area and varies depending on the time of day.
Heat input from solar radiation is taken into account for the summer and transition periods, for outdoor temperatures above +10 ⁰С.
The calculation is performed on the basis of Manual 2.91 to SNiP 2.04.05-91 “Calculation of solar radiation heat input into premises.”
To reduce heat gain from solar radiation, it is recommended to use protective anti-insulation devices, curtains, canopies, blinds; as a result of their use, heat gain from solar radiation can be reduced by up to 60%, which will reduce the capacity of the refrigeration unit by 10-15%.
Example of reduction:

For curtains between window sashes - 50%;


For internal curtains on windows - 40%;


When using blinds - 50%.

Calculation of heat gains from infiltration.

Infiltration is the penetration of outside air into a room under the influence of wind and temperature differences through leaks in enclosing structures. It is especially necessary to take this factor into account for windows and doors located on the leeward side.
The mass amount of air infiltrating through cracks and leaks is determined by the formula:

G= ∑(a*m*l), Where

a - coefficient taking into account the nature of the cracks;
m - specific amount of air penetrating through 1 linear meter of length depending on wind speed (kg/g*m.);
l- length of the slit.

Heat consumption Qi, W, for heating the infiltrating air should be determined by the formula:

Qi = 0.28 Σ Gi c(tp - ti)k , Where

Gi - flow rate of infiltrated air, kg/h, through the building envelope;
With - specific heat capacity of air equal to 1 kJ/(kg*⁰С);
tp, ti - calculated air temperature, °C, respectively, indoors (average staking into account the increase for rooms with a height of more than 4 m) and outdoor air during the cold season;
k - factor taking into account the influence of oncoming heat flow in structures, equal to 0.7 for jointswall panels and windows with triple sashes, 0.8 - for windows and balcony doors with separate sashes and 1.0 - for single windows, windows and balcony doors with paired sashes and open openings.

This calculation must be used to account for infiltration in winter time year in air-conditioned (also heated) rooms, at other times of the year it is permissible to use with a sufficient degree of accuracy additional body gains (heat losses) in the amount of 10% to 20%, depending on the nature and orientation of the enclosing structures.
For rooms equipped with SCR, it is recommended that all fences be made with maximum tightness; in these cases, the calculation for infiltration can be neglected.

Calculation of heat gain from people in the room.

Heat gain from people in the room depends on the intensity of the work they perform, as well as the parameters of the ambient air.
The heat generated by a person consists of explicit - transmitted into the air by convection and radiation, and hidden - spent on the evaporation of moisture from the surface of the skin and from the lungs, the ratio between the amount of sensible and latent heat depends on the amount of muscular work performed by a person, as well as on the parameters of the surrounding air.
With increasing work intensity and ambient temperature, the proportion of latent heat increases. At an ambient temperature of 36 ⁰C, all the heat generated by the body is given off through evaporation.
Note:

Regardless of the type of activity, the total amount of heat generated during low temperatures environment is higher than at high;


at low ambient temperatures, the value of sensible heat is higher than that of latent heat, and vice versa;


at air temperatures corresponding to a comfortable 24-26 ⁰C, during a sedentary activity, the amount of heat is distributed as 60-65% - sensible and 35-40% latent; with an increase in physical activity, latent heat begins to predominate;


It is worth remembering that the number of people stated in the calculation will not always correspond to the number of people simultaneously in the room; for this it is necessary to apply the simultaneity coefficient.

Calculation of heat input from lighting fixtures and lamps.

Currently, three types of lighting are most often used: incandescent, fluorescent and less common LED.
Heat gain from lamps is determined by the formula:

Q osv = ղ * N osv, Where

ղ - transition coefficient electrical energy to thermal;
N osv- installed lamp power W/m2
Coefficient value ղ:

for incandescent lamps: 0.92-0.97;


For fluorescent lamps: 0,5- 0,6;


For LED lamps: 0,6- 0,75.

In some premises the load from lighting devices is significant: trading floors, shops, office premises, etc.
It is also necessary to pay attention to the design of the ceilings, for example, in ventilated suspended ceilings, about 30-40% of the total amount of heat will be carried away by exchanged air, the remaining 60%-70% of the heat will enter the room.
For some establishments, lighting occupancy factors may also apply.

Simplified method for calculating split systems - DOWNLOAD

As you can see, calculating hard currency is a rather labor-intensive process that includes many factors that need to be taken into account. In connection with this, a simplified methodology was created for calculating air conditioning systems based on split systems, as well as monoblock air conditioners.
To select an air conditioner based on cooling capacity, it is necessary to calculate heat gains through the enclosing structures from: solar radiation, lighting, people, electrical appliances and office equipment.

The main heat inputs will consist of:
1. heat gain through building envelopes Q 1 , which are calculated by the formula:

Q 1 =V* q beat., Where

V = S*h- volume of the refrigerated space;
S- room area;
h- room height.

q beat- specific heat load, taken in accordance with:
30-35 W/m3 - if there is no sun indoors (northeast, northwest);
35 W/m3 - average value (south, southeast, southwest);
35-40 W/m3 - a large percentage of glazing on the sunny side (east, west).

2. heat gain due to electrical appliances and office equipment located in it Q 2 .
On average, 300 W is accepted for 1 computer, 200 W for 1 TV, or 30% of the power of electrical equipment (stoves, TVs, production equipment, etc.);

3. heat gain from people in the room Q 3 .
Most often when calculating it is accepted:
For apartments and office premises
1 person - 100-120 W
For premises where a person is engaged in physical labor (for example a restaurant):
1 person - 150-300 W.

Total heat input ∑ Qwill be determined by the formula:

∑ Q = Q 1 + Q 2 + Q 3

TO ∑ Q20% is added for unaccounted heat inputs:

∑ Q = (Q 1 +Q 2 + Q 3 )*1.2, W

The power of the selected air conditioner should be in the range from - 5% to +15% of the design power ∑ Q , a negative value is not desirable.

An example of a typical calculation of the cooling capacity of an air conditioner.

Task: Calculate the power of a split system operating on recirculated air for office space, with an area of ​​24 m2, with a ceiling height of 3.0 m (without a suspended ceiling), in which 3 people work simultaneously, there are 3 computers, 1 printer with a power of 570 W, a coffee machine with a power consumption of 800 W, the windows face the sunny side.

Solution:
1. Calculation of heat input through building envelopes:
Q 1 = S * h * q = 24 * 3 * 40 = 2880 W = 2.9 kW;

2. Calculation of heat input from electrical appliances:
3 computers = 300 W *3 = 900 W;
1 printer = 570 W *0.3 = 171 W;
1 coffee machine = 800 W * 0.3 = 240 W.
∑ Q 2 = 900 W + 171 W + 240 W = 1311 W = 1.3 kW;

3. Calculation of heat gains from people:
1 person = 100 W
Q 3 = 120 * 3 = 360 W = 0.36 kW.

∑ Q = Q 1 + Q 2 + Q 3 = 2.9 kW + 1.3 kW + 0.36 kW = 4.56 kW.

reserve for unaccounted heat inputs: 20%
∑ Q = 4.56 * 1.2 = 5.5 kW.

5 % < ∑ Q < + 15%
5 ,5*0,95 < ∑ Q < 5,5 * 1,15
5 ,2 < ∑ Q < 6,3
Now you need to select the split system that is closest in power.
This will be split system No. 18 with a cooling capacity of 5.3 kW.

Taking into account additional parameters when calculating the power of split systems.

A standard calculation in most cases will give fairly accurate results, but it is also worth taking into account those factors that are not taken into account in the standard calculation; it is also advisable to take them into account when calculating the cooling capacity of the system.

Accounting for admixture fresh air in the case of a slightly open window (to organize the flow of fresh air).

The calculation method described above implies that the air conditioner operates with the windows closed (as provided by the manufacturer), and warm air from the street does not enter. Although sometimes this is necessary (especially in offices and apartments where there is no supply ventilation).
Unlike supply ventilation, in order to calculate the amount of heat entering the room through an open window, you can use the formulas for calculating infiltration given above, but this calculation in this situation will be quite complicated (after all, it is impossible to say exactly what the air exchange rate will be, how much the window will be open and etc.).
You can consider the option that the window is constantly slightly open for ventilation + the air conditioner is constantly running.
Don't forget The air conditioner cannot operate with an open window and the efficiency of such operation cannot be 100% guaranteed.
If this option is still necessary, then the following should be considered:

Q 1 should be increased by 20-25% to compensate for the amount of heat received during ventilation with outside air, this number was obtained with outside air parameters (temperature/humidity) 33⁰С / 50%, internal air temperature 22 ⁰С, single air exchange rate. As the air exchange rate increases, the percentage increase in power will increaseQ 1 . For example, with a 2-fold air exchange, it is recommended to increaseQ 1 by 40-45%, with 3-fold air exchange (if you open the window and door - there is a draft)Q 1worth increasing by 65%.


the cost of the split system will increase;


electricity costs will increase by up to 35% (when using a conventional split system) by 10-15% when using an inverter split system;


in some cases, the outside air temperature increases or the air exchange rate increases, the window will have to be closed or closed altogether;


For this mode, it is recommended to use inverter split systems, because in the case of conventional systems, the level of comfort will be reduced, it is possible for people in the room to be blown out (frequent colds), and energy losses will increase.

We, if possible, recommend avoiding the use of this mode of operation of split systems; for this, you can install a split system with a membrane oxygen generator, which can also provide fresh air from the street, an example of such a system could be -Panasonic HI-END SUPER DELUXE with oxygen generator "Panasonic O2air", one of the disadvantages of such a system is that there is not a large choice in terms of power, usually models No. 9 and No. 12 (2.6 kW and 3.5 kW, respectively), or use cassette split systems with the ability to organize the flow of outside air through the indoor unit. But the final decision on the installation of a particular system can be made only on the basis of a feasibility study carried out by qualified specialists.

Guaranteed operating mode of the system to maintain room temperature +20 ⁰С.

The standard SCR calculation is performed to maintain indoor air parameters of 24-26 ⁰С - which are comfortable for most people, but in some cases it is necessary that the system be able to maintain an indoor temperature of +20 ⁰С (for example, for server rooms, or if this value is the temperature comfort for people in the room). The outside air temperature in a typical calculation corresponds toSNiP 23-01-99* (Code of rules - SP 131.13330.2012 - updated version) "Construction climatology"- for Novosibirskthe average maximum air temperature of the warmest month is +25.4⁰С.
Due to the fact that the calculation is made with a small power reserve, in reality the air conditioner will be able to produce parameters of +20 ⁰С, up to an outside air temperature of +30 ⁰С, but when the outside air temperature rises, the system will no longer cope. Therefore, to ensure this mode of operation, it is recommended to increase the powerQ 1 by 25-30%.

Large glazing area.

In a typical calculation, the average value for heat gain from solar radiation is 1 kW per 10 m2 (glazing) or 100 W per 1 m2 (glazing).
The typical calculation takes into account 2.0 m2 of glazing; if the glazing area is larger than the average value, it is necessary to increaseQ 1 Depending on the additional glazing area, for each additional m2 of glazing you need to add:

250-300 W - for strong lighting;


150-200 W - for average value;


100 W - for low light.

In this case, the power of SCR can increase by 10-15%.

Top floor.

If the apartment is located directly under the roof (must be taken into account for cottages and private houses), then additional heat will enter the room through the enclosing structure, namely the roof. In this caseQ 1 it is necessary to increase by 10-20% depending on the angle of the roof and the color of the roof.
For light gable roof 10%, for a horizontal (flat) roof of a dark color 20%.

When choosing any HVAC system equipment, incl. air conditioner, it is very important to correctly calculate the heat flow of the room. After all, not only its microclimate depends on this. Taking into account the intense heat inflows of a room when calculating a heating system, for example, will help save on heating equipment and energy, and underestimating them when calculating a ventilation and, especially, air conditioning system can lead to increased wear and a decrease in the service life of the equipment.

Calculation of heat inflows of a room can be carried out different ways, - there are several methods. Some are more detailed and are used more often when calculating ventilation and air conditioning systems industrial buildings, other very simplified methods for calculating heat inflows are used by managers when selling air conditioners. Such program for approximate calculation and selection of air conditioner, for example, is located.
The calculation of heat inflows given below takes into account all the main heat inflows, the underestimation of which, in our opinion, is undesirable. Respectively, program for calculating heat inflows using this method you can find .

For long-term reliable operation of the air conditioner, it is important that its cooling capacity is slightly greater than the actual heat flow in the room.

First of all, take into account external heat inputs . This is, first of all, solar radiation penetrating through window openings. The amount of thermal energy supplied in this way depends on the location of the window relative to the cardinal directions, its area and the presence / absence of sun protection elements on it:
Q windows = q windows F windows k, Where
q windows- specific thermal power from solar radiation depending on the orientation of the window W/m 2.

F window - area of ​​the glazed part of the window, m2;
k - coefficient taking into account the presence of sun protection elements on the window.

Heat inflows from a heated protective structure:
q ZS - specific thermal power of heat transfer of the protective structure, W/m 2.

F ZS - area of ​​the protective structure, m 2.
For a constantly open external door, the heat input is 300 W.

Second group heat inflows, this heat release from internal sources indoors - from people, lighting, electrical equipment.

Heat emissions from people:
Q l = q l n, Where
n is the number of people in the corresponding state;
q l - heat generation per person, W/person.

Heat emissions from electrical equipment:
Q e = N e m i, Where
m - number of equipment units;
N e - electric power units of equipment, W;
i is the coefficient of conversion of electrical energy into thermal energy.

For a computer, heat dissipation is assumed to be 300 W.
The calculation of the heat inflow of the room can be considered complete.
The total amount of heat inflow in the room will be:
ΣQ = Σ Q windows + ΣQ ZS + ΣQ l + Σ Q e

Then the air conditioner is selected. The cooling capacity of the selected air conditioner should be 10-20% higher than the total amount of heat inflow in the room:
Q cond = (1.1-1.2) ΣQ

Heat release from operating electrically driven equipment due to the transition of mechanical energy into thermal energy is determined from the expression

Q about = 1000 N mouth· n · k isp k V, W, (1)

Where N mouth– the installed power of the electric motor drive per unit of equipment, kW, is determined by the task; k isp– electric motor power utilization factor, it is usually recommended to take 0.8; k V– the coefficient of simultaneity of equipment operation, determined by the task, can be taken equal to 1. Value Q about does not depend on the period of the year.

Heat gains from lighting for the warm and cold periods of the year are calculated

Q oc = 1000 N oc · n k V · a, W, (2)

Where N OS- - power of one lighting installation, kW; n – number of lighting installations; k V– coefficient of simultaneous operation of lighting installations: during the cold period, k can be taken V=1.0, during the warm period k V= 0.5 - 0.6 – as specified; A- coefficient taking into account the type of lighting installation, which is regulated by SNiP and can be determined from the application, table. P-3.

Heat gains from lighting can be calculated in another way

Q oc = F· q oc k V, W, (3)

Where F– floor surface in the room, m2; q OS= 40 W/m2 – illumination standard 1m2 in accordance with SNiP; k V– coefficient of simultaneous operation of lighting installations.

Heat inputs from service personnel for the cold and warm periods of the year are calculated from the expression

where m is the number of employees; Q obviously– sensible heat release from one person, kJ/h; r = 2250 kJ/kg – latent heat of vaporization; W P– moisture release from one person, g/h.

Numerical values Q obviously And W P are determined in accordance with SNiP depending on the air temperature inside the room and the degree of severity of labor and can be determined from the appendix, table. P-4.

Heat gains from solar radiation through light (window) openings are calculated only for the warm period of the year

Q Wed = F ost· q ost · A ost k , W, (5)

Where F ost– total glazing surface, m2; q ost– the density of heat flux transmitted due to solar radiation, depending on the orientation of the light openings to the cardinal points; A ost– empirical coefficient depending on the type of glazing; k is an empirical coefficient depending on the transparency of the glass.

Numerical value of q ost in accordance with SNiP depending on the characteristics of the glazing and geographical location the object can be determined by the application, table. P-5.

Numerical value A ost andk in accordance with SNiP can be determined from the application, respectively, table. P-6 and table. P-7.

Heat gains through external enclosures from the outside due to the higher temperature of the outside air when designing air conditioning systems are calculated for the warm period if the design temperature of the outside air exceeds the design temperature of the indoor air by 5°C or more, i.e. t n T – t V T 5С

Q ogre = F ogre k ogre · (t n T - t V T ) , W, (6)

whereF ogre– surface of the external fence minus the glazing surface, m 2 ;k ogre t n T And t V T- respectively, the calculated temperature of the outside air and indoor air, С.

Not calculated for floors located on the ground or above basements. For a combined roof, heat inputs for the rooms on the upper floor should be calculated separately.

The heat transfer coefficient is calculated taking into account all thermal resistances

, (7)

Where  V And  n- respectively, the coefficient of heat transfer from indoor air to the wall and from the outer surface of the wall to outdoor air, W/(m 2 С);  i– thickness of the individual layers making up the wall, m;  i– thermal conductivity coefficient of the materials from which the wall is made, W/(m С).

Numerical values ​​of heat transfer coefficients can be determined in accordance with SNiP according to the appendix, table. P-8 and P-9. The thermal conductivity coefficients of some materials are given in the appendix, table. P-10.

For rooms on the upper floor in the absence of an attic floor (combined roof), heat gain through the roof is calculated using formulas (6) and (7) separately from the side surfaces of the walls.

The total heat input into the room for the warm period of the year in the general case is

Q T =Q about + Q OS + Q op + Q Wed + Q ogre, W, (8)

for the cold season

Q X =Q about + Q OS + Q op, Tue. (9)

Calculation of heat losses in a room

Heat losses are calculated only for the cold period of the year.

Heat losses through glazed window light openings are determined from the expression

Q ost= F ost· k · (t V X -t n X ) , W, (10)

Where F ost– total glazing surface, m 2 ; k – heat transfer coefficient through window openings, W/(m 2 С); t V X And t n X– respectively, the calculated indoor and outdoor air temperatures for the cold period of the year, С.

The values ​​of the heat transfer coefficient are determined in accordance with SNiP according to the appendix, table. P-11.

Heat losses through external enclosures (side walls, floors, ceilings) are calculated from the expression

Q ogre = F ogre k ogre · (t V X -t n X ) n, W, (11)

Where F ogre– surface of external fences (minus the area of ​​window and door openings), m2; k ogre– heat transfer coefficient through fences, W/(m 2 С); t V X And t n X– respectively, the calculated temperatures of indoor and outdoor air for the cold period, С; n – empirical correction factor, depending on the nature of the fence.

The heat transfer coefficient k is determined by formula (7). Some of the most common fencing designs are shown in Fig. 3.

The value of the empirical coefficient n in formula (11) can be taken in accordance with SNiP according to the appendix, table. P-12.

Rice. 3. The most common fencing designs:

a - side walls; b - roof; c - interfloor ceilings;

For the conditions of the task under consideration, heat losses for the premises of the second floor are calculated only through window openings and side walls. For rooms on the first floor, in addition to the above, you should calculate heat losses through the floor (above the basement), and for rooms on the third floor - through the roof.

The total heat loss by the room for the cold period of the year will be

Q sweat X = Q ost X + Q ogre X, Tue. (12)