Program of heat gains from solar radiation. On-line calculator for calculating the operation of a solar power plant

Room No. 104 (group)

The room has four windows facing NW.

Geographic latitude ts= 56 °N;

window area Fok=10.64 m²;

W/mI, W/mI in the period from 5 to 6 pm according to the table. 2.3 for glazing oriented NW at latitude 56°.

Ac.o.

We accept:

h= 21°; Solar azimuth Ac= 95° according to table. 2.8 for the period 5-6 pm and latitude 56°.

where H is the height of the window; H = 1.4 m; B - width; B = 1.9 m;

L G L G = L IN = 0.13.

TO REGIONAL V= 1 (Fig. 2.6)

TO OBL.G= 1 (Fig. 2.6)

Where TO OTN- relative penetration coefficient solar radiation; for triple glazed windows with light blinds indoors TO OTN = 0.48;

f 2 - coefficient for taking into account window shading;

A coefficient that takes into account the daily variation of outside temperature; (Table 2.9 with e= 0 for the period from 8 to 9 hours);

The amount of heat entering the vertical surface of the SE orientation in the period from 8 to 9 o’clock from direct and diffuse radiation for a latitude of 56°; =460 W/mI; =125 W/mI;

v).

W/(mI.°С)

Room No. 219 (gymnastics and music practice room)

The room has four windows oriented to the southeast.

Geographic latitude ts= 56 °N;

window area Fok=10.08 m²;

1. Maximum amount heat from direct and diffuse solar radiation penetrating through single glazing:

W/mI from 3 to 4 pm for glazing oriented NW at latitude 56°.

Angle between sunbeam and window:

where h is the altitude of the Sun; Ac.o.- solar azimuth of the glazing.

We accept:

h= 37°; Solar azimuth Ac=69° for the period 3-4 pm and latitude 56°. And co =69+45=114.

2. Insolation coefficient of vertical glazing.

where H is the height of the window; H = 1.8 m; B - width; B = 1.4 m;

a = c = 0 - because no external sun visors;

L G- depth of glazing from the outer surface of the facade (accepted 0.13 m, as for brick buildings); L G = L IN = 0.13.

3. COBL irradiation coefficient depends on the angles:

3 0 54"-vertical component TO REGIONAL V= 1 (Fig. 2.6)

5 0 18"-horizontal component TO OBL.G= 1 (Fig. 2.6)

4. Specific heat flux from penetrating solar radiation through the adopted glazing:

Where TO OTN- coefficient of relative penetration of solar radiation; for triple glazed windows with light blinds indoors (Table 2.4) TO OTN = 0.48;

f 2 - coefficient for taking into account window shading; for accepted glazing according to table. 2.5 f 2 = 0.50.

5. External conditional temperature on the window surface:

Where - average temperature hottest month (July); for air-conditioned premises should be taken outside temperature in the warm season according to parameters “B”; t n.sr =28

Average daily amplitude of fluctuations in outside air temperature during the warm period; =10.8°C;

A coefficient that takes into account the daily variation of outside temperature; ;

Reduced radiation absorption coefficient;

The amount of heat entering the vertical surface of the SE orientation in the period from 3 to 4 hours from direct and diffuse radiation for latitude 56° = 0 W/mІ; =86W/mI;

Heat transfer coefficient on the outer surface of the window; for a vertical surface (depending on wind speed v).

In order to choose the right air conditioner, it is necessary to calculate the heat input that it must extinguish. The power of the air conditioner must exceed their maximum value, which is calculated by the formula:

Q = Q1+Q2+Q3+Q4+Q5, where

Q1 – heat gain from solar radiation, and when using electric lighting from artificial light;

Q2 – heat gain from people in the room;

Q3 – heat input from office equipment;

Q4 – heat input from household appliances;

Q5 – heat input from heating.

Heat gain from solar radiation

They primarily depend on the area and location of the windows. In most cases, it is this that accounts for the lion's share of all heat entering the room. Calculation methods are presented in detail in special manuals for SNiP 23-01-99 “Building Climatology” and SNiP II-3-79 “Building Heat Engineering”. Simplified, you can use the following formula for calculation:

Where: S is the area of ​​the room (m2), h is the height of the room (m), q is a coefficient equal to:
- 30 W/m3, if they do not enter the room Sun rays(north side of the building);
-35W/m3 for normal conditions;
- 40 W/m3, if the room has large glazing on the sunny side.
Calculation using this method is applicable for apartments and small offices; in other cases, the errors may be too large.

Heat gain from artificial light can be taken at the rate of 25-30 W per 1 m3.

Heat gain from people in the room

One person, depending on his occupation, identifies:
Rest in a sitting position – 120 W
Easy job in a sitting position – 130 W
Moderately active work in the office – 140 W
Light standing work – 160 W
Light industrial work - 240 W
Slow dancing – 260 W
Medium industrial work – 290 W
Heavy Duty - 440 W

Heat gains from office equipment

Typically they are taken at 30% of power consumption. For example:
Computer – 300-400 W
Laser printer– 400 W
Copier – 500-600 W

Heat gains from household kitchen appliances

Coffee maker with heating surface – 300 W
Coffee machine and electric kettle – 900-1500 W
Electric stove – 900-1500 W per 1 m2 of upper surface
Gas stove – 1800-3000 W 1 m2 of top surface
Deep fryer – 2750-4050 W
Toaster – 1100-1250 W
Waffle iron – 850 W
Grill – 13500 W per 1 m2 of top surface
If there is an exhaust hood, the heat input from the stove is divided by 1.4.

When calculating heat input from household kitchen appliances, it is necessary to take into account that all appliances are never turned on at once. Therefore, the highest combination for a given kitchen is taken. For example, two of the four burners on the stove and an electric kettle.

Heat gains from the heating system

In some cases, in tall buildings with large area glazing and air conditioning may be necessary as early as March, when the heating season is not yet over. In this case, the calculation must take into account the heat excess from the heating system, which can be taken equal to 80-125 W per 1 m2 of area. In this case, it is necessary to take into account not the heat gain from the external walls, but the heat loss, which can be taken equal to 18 W per 1 m2.

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 of all or individual air parameters (temperature, relative humidity, cleanliness, speed of movement) with the aim of ensuring mainly optimal meteorological conditions, the most favorable for the well-being of people, conducting 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 the British number system is used. 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 of the split system 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.

Thousand BTU	7	9	12	14	18	22	24	26	28	30	36	45	54	60	72	90
kW	2,1	2,6	3,5	4,1	5,3	6,4	7,0	7,6	8,2	8,8	10,6	13,2	15,8	17,6	21,1	26,4

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 Sample Microsoft Excel(.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. Temperature difference between inside and outside the room in summer period time is positive, as a result of which, during this period of the year we receive an influx of heat into the room, in winter everything is the other way around - 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 building envelopes, windows, doors), given air has correspondingly different properties in summer and winter period years: 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 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 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.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 air equal to 1 kJ/(kg*⁰С);
tp, ti - estimated air temperatures, °C, respectively, in the room (averagetaking into account the increase for rooms with a height of more than 4 m) and outdoor air in cold period of the year;
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 ⁰С, with a sedentary type of activity, the amount of heat is distributed as 60-65% - obvious and 35-40% latent, with increasing 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 rooms the load from lighting fixtures is significant: trading floors, shops, office space, etc.
It is also necessary to pay attention to the design of the ceilings, for example in ventilated suspended ceilings about 30-40% of total number the heat will be carried away by the exchange 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 , negative meaning not advisable.

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 an office space with an area of ​​24 m2, with a ceiling height of 3.0 m (without suspended ceiling), in which 3 people work at the same time, 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 not big choice in terms of power, these are 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 final decision installation of a particular system can only be accepted on the basis technical and economic justification 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 Novosibirskaverage Maximum temperature air most warm 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 a light gable roof 10%, for a horizontal (flat) roof of a dark color 20%.