What is the maximum thermal load. How to correctly calculate the heat loads for heating
The topic of this article is thermal load. We will find out what this parameter is, what it depends on and how it can be calculated. In addition, the article will provide a number of reference values \u200b\u200bof thermal resistance different materials that may be needed for the calculation.
What it is
The term is essentially intuitive. Heat load is the amount of heat energy that is necessary to maintain a comfortable temperature in a building, apartment or separate room.
The maximum hourly heating load is thus the amount of heat that may be required to maintain normalized parameters for an hour under the most unfavorable conditions.
Factors
So, what affects the heat demand of a building?
- Wall material and thickness. It is clear that a wall of 1 brick (25 centimeters) and a wall of aerated concrete under a 15-centimeter foam coat will allow VERY different amounts of thermal energy to pass through.
- Material and structure of the roof. A flat roof made of reinforced concrete slabs and an insulated attic will also differ quite noticeably in terms of heat loss.
- Ventilation is another important factor. Its performance, the presence or absence of a heat recovery system affects how much heat is lost to the exhaust air.
- Glazing area. Significantly more heat is lost through windows and glass facades than through solid walls.
However: triple-glazed windows and glass with energy-saving spraying reduce the difference by several times.
- The level of insolation in your area, the degree of absorption of solar heat by the external coating and the orientation of the planes of the building relative to the cardinal points. Extreme cases are a house that is in the shade of other buildings all day long and a house oriented with a black wall and a black sloping roof with a maximum area to the south.
- temperature delta between indoor and outdoor determines the heat flow through the building envelope at a constant resistance to heat transfer. At +5 and -30 on the street, the house will lose a different amount of heat. It will, of course, reduce the need for thermal energy and lower the temperature inside the building.
- Finally, a project often has to include prospects for further construction. For example, if the current heat load is 15 kilowatts, but in the near future it is planned to attach an insulated veranda to the house, it is logical to purchase it with a margin of thermal power.
Distribution
In the case of water heating, the peak heat output of the heat source must be equal to the sum of the heat output of all heating appliances in the house. Of course, wiring should not become a bottleneck either.
The distribution of heating devices in rooms is determined by several factors:
- The area of the room and the height of its ceiling;
- Location inside the building. Corner and end rooms lose more heat than those located in the middle of the house.
- Distance from heat source. In individual construction, this parameter means the distance from the boiler, in the system central heating apartment building- by the fact that the battery is connected to the supply or return riser and by the floor you live on.
Clarification: in houses with a lower bottling, the risers are connected in pairs. On the supply side, the temperature decreases when you rise from the first floor to the last, on the reverse, respectively, vice versa.
It is also not difficult to guess how the temperatures will be distributed in the case of top bottling.
- Desired room temperature. In addition to filtering heat through external walls, inside the building with an uneven distribution of temperatures, the migration of thermal energy through partitions will also be noticeable.
- For living rooms in the middle of the building - 20 degrees;
- For living rooms in the corner or end of the house - 22 degrees. More heat, among other things, prevents freezing of the walls.
- For the kitchen - 18 degrees. It, as a rule, has a large number of its own heat sources - from a refrigerator to an electric stove.
- For a bathroom and a combined bathroom, the norm is 25C.
In the case of air heating, the heat flow entering a separate room is determined throughput air sleeve. As a rule, the simplest method of adjustment is to manually adjust the positions of adjustable ventilation grilles with temperature control using a thermometer.
Finally, if we are talking about a heating system with distributed heat sources (electric or gas convectors, electric underfloor heating, infrared heaters and air conditioners) required temperature regime simply set on the thermostat. All that is required of you is to provide peak thermal power devices at the peak level of room heat loss.
Calculation methods
Dear reader, do you have a good imagination? Let's imagine a house. Let it be a log house from a 20-centimeter beam with an attic and a wooden floor.
Mentally draw and specify the picture that has arisen in my head: the dimensions of the residential part of the building will be equal to 10 * 10 * 3 meters; in the walls we will cut 8 windows and 2 doors - to the front and inner courtyards. And now let's place our house ... let's say, in the city of Kondopoga in Karelia, where the temperature at the peak of frost can drop to -30 degrees.
Determining the heat load on heating can be done in several ways with varying complexity and reliability of the results. Let's use the three most simple ones.
Method 1
The current SNiP offer us the simplest way to calculate. One kilowatt of thermal power is taken per 10 m2. The resulting value is multiplied by the regional coefficient:
- For southern regions(Black Sea coast, Krasnodar Territory) the result is multiplied by 0.7 - 0.9.
- The moderately cold climate of the Moscow and Leningrad regions will force the use of a coefficient of 1.2-1.3. It seems that our Kondopoga will fall into this climate group.
- Finally, for the Far East of the Far North, the coefficient ranges from 1.5 for Novosibirsk to 2.0 for Oymyakon.
Instructions for calculating using this method are incredibly simple:
- The area of the house is 10*10=100 m2.
- The base value of the heat load is 100/10=10 kW.
- We multiply by the regional coefficient 1.3 and get 13 kilowatts of thermal power needed to maintain comfort in the house.
However: if we use such a simple technique, it is better to make a margin of at least 20% to compensate for errors and extreme cold. Actually, it will be indicative to compare 13 kW with values obtained by other methods.
Method 2
It is clear that with the first method of calculation, the errors will be huge:
- The height of the ceilings in different buildings varies greatly. Taking into account the fact that we have to heat not an area, but a certain volume, and with convection heating, warm air is collected under the ceiling - an important factor.
- Windows and doors let in more heat than walls.
- Finally, it would be a clear mistake to cut a city apartment under the same brush (regardless of its location inside the building) and a private house, which has no space below, above and beyond the walls. warm apartments neighbors, and the street.
Well, let's correct the method.
- For the base value, we take 40 watts per cubic meter of room volume.
- For each door leading to the street, add 200 watts to the base value. 100 per window.
- For corner and end apartments in apartment building we introduce a coefficient of 1.2 - 1.3 depending on the thickness and material of the walls. We also use it for the extreme floors in case the basement and attic are poorly insulated. For a private house, we multiply the value by 1.5.
- Finally, we apply the same regional coefficients as in the previous case.
How is our house in Karelia doing there?
- The volume is 10*10*3=300 m2.
- The base value of thermal power is 300*40=12000 watts.
- Eight windows and two doors. 12000+(8*100)+(2*200)=13200 watts.
- A private house. 13200*1.5=19800. We begin to vaguely suspect that when selecting the power of the boiler according to the first method, we would have to freeze.
- But there is still a regional coefficient! 19800*1.3=25740. In total, we need a 28-kilowatt boiler. Difference with the first value received in a simple way- double.
However: in practice, such power will be required only on a few days of peak frost. It is often a smart decision to limit the power of the main heat source to a lower value and buy a backup heater (for example, an electric boiler or several gas convectors).
Method 3
Do not flatter yourself: the described method is also very imperfect. We very conditionally took into account the thermal resistance of the walls and ceiling; the temperature delta between the internal and external air is also taken into account only in the regional coefficient, that is, very approximately. The price of simplifying calculations is a big error.
Recall that in order to maintain a constant temperature inside the building, we need to provide an amount of thermal energy equal to all losses through the building envelope and ventilation. Alas, here we will have to somewhat simplify our calculations, sacrificing the reliability of the data. Otherwise, the resulting formulas will have to take into account too many factors that are difficult to measure and systematize.
The simplified formula looks like this: Q=DT/R, where Q is the amount of heat lost by 1 m2 of the building envelope; DT is the temperature delta between indoor and outdoor temperatures, and R is the resistance to heat transfer.
Note: we're talking about heat loss through walls, floors, and ceilings. On average, another 40% of heat is lost through ventilation. For the sake of simplifying the calculations, we will calculate the heat loss through the building envelope, and then simply multiply them by 1.4.
Temperature delta is easy to measure, but where do you get data on thermal resistance?
Alas - only from directories. Here is a table for some popular solutions.
- A wall of three bricks (79 centimeters) has a heat transfer resistance of 0.592 m2 * C / W.
- A wall of 2.5 bricks - 0.502.
- Wall in two bricks - 0.405.
- Brick wall (25 centimeters) - 0.187.
- Log cabin with a log diameter of 25 centimeters - 0.550.
- The same, but from logs with a diameter of 20 cm - 0.440.
- A log house from a 20-centimeter beam - 0.806.
- A log house made of timber 10 cm thick - 0.353.
- Frame wall 20 centimeters thick with mineral wool insulation - 0.703.
- A wall of foam or aerated concrete with a thickness of 20 centimeters - 0.476.
- The same, but with a thickness increased to 30 cm - 0.709.
- Plaster 3 cm thick - 0.035.
- Ceiling or attic floor - 1.43.
- Wooden floor - 1.85.
- Double door made of wood - 0.21.
Now let's get back to our house. What options do we have?
- The temperature delta at the peak of frost will be equal to 50 degrees (+20 inside and -30 outside).
- Heat loss through square meter floors will be 50 / 1.85 (heat transfer resistance of a wooden floor) \u003d 27.03 watts. Through the entire floor - 27.03 * 100 \u003d 2703 watts.
- Let's calculate the heat loss through the ceiling: (50/1.43)*100=3497 watts.
- The area of the walls is (10*3)*4=120 m2. Since our walls are made of a 20 cm beam, the R parameter is 0.806. The heat loss through the walls is (50/0.806)*120=7444 watts.
- Now let's add the obtained values: 2703+3497+7444=13644. This is how much our house will lose through the ceiling, floor and walls.
Note: in order not to calculate the fractions of square meters, we neglected the difference in the thermal conductivity of walls and windows with doors.
- Then add 40% ventilation losses. 13644*1.4=19101. According to this calculation, a 20-kilowatt boiler should be enough for us.
Conclusions and problem solving
As you can see, the available methods for calculating the heat load with your own hands give very significant errors. Fortunately, excess boiler power will not hurt:
- Gas boilers at reduced power operate with virtually no drop in efficiency, and condensing boilers even reach the most economical mode at partial load.
- The same applies to solar boilers.
- Electric heating equipment of any type always has an efficiency of 100 percent (of course, this does not apply to heat pumps). Remember physics: all the power that is not spent on mechanical work (that is, the movement of mass against the gravitational vector) is ultimately spent on heating.
The only type of boilers for which operation at less than nominal power is contraindicated is solid fuel. Power adjustment in them is carried out in a rather primitive way - by limiting the flow of air into the furnace.
What is the result?
- With a lack of oxygen, the fuel does not burn completely. More ash and soot are formed, which pollute the boiler, chimney and atmosphere.
- The consequence of incomplete combustion is a drop in boiler efficiency. It is logical: after all, often the fuel leaves the boiler before it burns out.
However, even here there is a simple and elegant way out - the inclusion of a heat accumulator in the heating circuit. A thermally insulated tank with a capacity of up to 3000 liters is connected between the supply and return pipeline, opening them; in this case, a small circuit is formed (between the boiler and the buffer tank) and a large one (between the tank and heaters).
How does such a scheme work?
- After ignition, the boiler operates at nominal power. At the same time, due to natural or forced circulation, its heat exchanger gives off heat to the buffer tank. After the fuel has burned out, the circulation in the small circuit stops.
- The next few hours, the coolant moves along a large circuit. The buffer tank gradually releases the accumulated heat to radiators or water heated floors.
Conclusion
As usual, you will find some additional information on how the thermal load can be calculated in the video at the end of the article. Warm winters!
Ask any specialist how to properly organize the heating system in the building. It doesn't matter if it's residential or industrial. And the professional will answer that the main thing is to accurately make calculations and correctly carry out the design. We are talking, in particular, about the calculation of the heat load on heating. The volume of consumption of thermal energy, and hence fuel, depends on this indicator. That is, economic indicators are next to the technical characteristics.
Performance accurate calculations allows you to get not only a complete list of the documentation necessary for the installation work, but also to select the necessary equipment, additional components and materials.
Thermal loads - definition and characteristics
What is usually meant by the term "heat load on heating"? This is the amount of heat that all heating devices installed in the building give off. To avoid unnecessary expenses for the production of work, as well as the purchase of unnecessary devices and materials, a preliminary calculation is necessary. With it, you can adjust the rules for installing and distributing heat in all rooms, and this can be done economically and evenly.
But that's not all. Very often, experts carry out calculations, relying on accurate indicators. They relate to the size of the house and the nuances of construction, which takes into account the diversity of the elements of the building and their compliance with the requirements of thermal insulation and other things. It is precisely the exact indicators that make it possible to correctly make calculations and, accordingly, obtain options for the distribution of thermal energy throughout the premises as close to the ideal as possible.
But often there are errors in the calculations, which leads to inefficient operation of the heating as a whole. Sometimes it is necessary to redo during operation not only the circuits, but also sections of the system, which leads to additional costs.
What parameters affect the calculation of the heat load in general? Here it is necessary to divide the load into several positions, which include:
- Central heating system.
- Underfloor heating system, if one is installed in the house.
- Ventilation system - both forced and natural.
- Hot water supply of the building.
- Branches for additional household needs. For example, a sauna or a bath, a pool or a shower.
Main characteristics
Professionals do not lose sight of any trifle that can affect the correctness of the calculation. Hence the rather large list of characteristics of the heating system that should be taken into account. Here are just a few of them:
- The purpose of the property or its type. It can be a residential building or an industrial building. Heat suppliers have standards that are distributed by type of building. They often become fundamental in carrying out calculations.
- The architectural part of the building. This can include enclosing elements (walls, roofs, ceilings, floors), their overall dimensions, thickness. Be sure to take into account all kinds of openings - balconies, windows, doors, etc. It is very important to take into account the presence of basements and attics.
- Temperature regime for each room separately. This is very important because General requirements to the temperature in the house do not give an accurate picture of the distribution of heat.
- Appointment of premises. This mainly applies to production shops, which require stricter compliance with the temperature regime.
- Availability of special premises. For example, in residential private houses it can be baths or saunas.
- Degree of technical equipment. The presence of a ventilation and air conditioning system, hot water supply, and the type of heating used are taken into account.
- Number of points through which sampling is carried out hot water. And the more such points, the greater the heat load the heating system is exposed to.
- The number of people on the site. Criteria such as indoor humidity and temperature depend on this indicator.
- Additional indicators. In residential premises, one can distinguish the number of bathrooms, separate rooms, balconies. In industrial buildings - the number of shifts of workers, the number of days in a year when the workshop itself works in the technological chain.
What is included in the calculation of loads
Heating scheme
The calculation of thermal loads for heating is carried out at the design stage of the building. But at the same time, the norms and requirements of various standards must be taken into account.
For example, the heat loss of the enclosing elements of the building. Moreover, all rooms are taken into account separately. Further, this is the power that is needed to heat the coolant. We add here the amount of thermal energy required to heat the supply ventilation. Without this, the calculation will not be very accurate. We also add the energy that is spent on heating water for a bath or pool. Specialists must take into account the further development of the heating system. Suddenly, in a few years, you will decide to arrange a Turkish hammam in your own private house. Therefore, it is necessary to add a few percent to the loads - usually up to 10%.
Recommendation! It is necessary to calculate thermal loads with a "margin" for country houses. It is the reserve that will allow in the future to avoid additional financial costs, which are often determined by amounts of several zeros.
Features of calculating the heat load
Air parameters, or rather, its temperature, are taken from GOSTs and SNiPs. Here, the heat transfer coefficients are selected. By the way, the passport data of all types of equipment (boilers, heating radiators, etc.) are taken into account without fail.
What is usually included in a traditional heat load calculation?
- Firstly, the maximum flow of thermal energy coming from heating devices (radiators).
- Secondly, the maximum heat consumption for 1 hour of operation of the heating system.
- Thirdly, the total heat costs for a certain period of time. Usually the seasonal period is calculated.
If all these calculations are measured and compared with the heat transfer area of the system as a whole, then a fairly accurate indicator of the efficiency of heating a house will be obtained. But you have to take into account small deviations. For example, reducing heat consumption at night. For industrial facilities, you will also have to take into account weekends and holidays.
Methods for determining thermal loads
Underfloor heating design
Currently, experts use three main methods for calculating thermal loads:
- Calculation of the main heat losses, where only aggregated indicators are taken into account.
- The indicators based on the parameters of the enclosing structures are taken into account. This is usually added to the losses for heating the internal air.
- All systems included in heating networks are calculated. This is both heating and ventilation.
There is another option, which is called the enlarged calculation. It is usually used when there are no basic indicators and building parameters required for a standard calculation. That is, the actual characteristics may differ from the design.
To do this, experts use a very simple formula:
Q max from. \u003d α x V x q0 x (tv-tn.r.) x 10 -6
α is a correction factor depending on the region of construction (table value)
V - the volume of the building on the outer planes
q0 - characteristic of the heating system by specific index, usually determined by the coldest days of the year
Types of thermal loads
Thermal loads that are used in the calculations of the heating system and the selection of equipment have several varieties. For example, seasonal loads, for which the following features are inherent:
- Changes in outdoor temperature throughout the heating season.
- Meteorological features of the region where the house was built.
- Jumps in the load on the heating system during the day. This indicator usually falls into the category of "minor loads", because the enclosing elements prevent a lot of pressure on the heating in general.
- Everything related to the thermal energy associated with the ventilation system of the building.
- Thermal loads that are determined throughout the year. For example, the consumption of hot water in the summer season is reduced by only 30-40% when compared with winter time of the year.
- Dry heat. This feature is inherent in domestic heating systems, where a fairly large number of indicators are taken into account. For example, the number of window and door openings, the number of people living or permanently in the house, ventilation, air exchange through various cracks and gaps. A dry thermometer is used to determine this value.
- Latent thermal energy. There is also such a term, which is defined by evaporation, condensation, and so on. A wet bulb thermometer is used to determine the index.
Thermal Load Controllers
Programmable controller, temperature range - 5-50 C
Modern heating units and appliances are provided with a set of different regulators, with which you can change the heat loads, in order to avoid dips and jumps in thermal energy in the system. Practice has shown that with the help of regulators it is possible not only to reduce the load, but also to bring the heating system to the rational use of fuel. And this is a purely economic side of the issue. This is especially true for industrial facilities, where quite large fines have to be paid for excessive fuel consumption.
If you are not sure about the correctness of your calculations, then use the services of specialists.
Let's look at a couple more formulas that relate to different systems. For example, ventilation and hot water systems. Here you need two formulas:
Qin. \u003d qin.V (tn.-tv.) - this applies to ventilation.
Here:
tn. and tv - air temperature outside and inside
qv. - specific indicator
V - external volume of the building
Qgvs. \u003d 0.042rv (tg.-tx.) Pgav - for hot water supply, where
tg.-tx - temperature of hot and cold water
r - water density
c - the ratio of the maximum load to the average, which is determined by GOSTs
P - the number of consumers
Gav - average hot water consumption
Complex calculation
In combination with settlement issues, studies of the thermotechnical order are necessarily carried out. For this, various devices are used that give accurate indicators for calculations. For example, for this, window and door openings, ceilings, walls, and so on are examined.
It is this examination that helps to determine the nuances and factors that can have a significant impact on heat loss. For example, thermal imaging diagnostics will accurately show the temperature difference when a certain amount of thermal energy passes through 1 square meter of the building envelope.
So practical measurements are indispensable when making calculations. This is especially true for bottlenecks in the building structure. In this regard, the theory will not be able to show exactly where and what is wrong. And practice will indicate where it is necessary to apply different methods of protection against heat loss. And the calculations themselves in this regard are becoming more accurate.
Conclusion on the topic
Estimated heat load is a very important indicator obtained in the process of designing a home heating system. If you approach the matter wisely and carry out all the necessary calculations correctly, then you can guarantee that heating system will work great. And at the same time, it will be possible to save on overheating and other costs that can simply be avoided.
Types of thermal loads
In district heating systems, heat is consumed for heating in building heating systems, supply air heating in ventilation systems for hot water supply in DHW systems, as well as technological processes industrial enterprises.
Heating system - a complex of consumer engineering and technical devices that use thermal energy supplied by an energy supply organization for heating.
Ventilation system - a complex of consumer engineering and technical devices that use thermal energy supplied by the energy supply organization for ventilation.
Hot water supply system (DHW) - a complex of consumer engineering and technical devices that use thermal energy supplied by an energy supply organization for hot water supply.
In heating and ventilation systems, heat is not consumed continuously throughout the year, but only when relatively low temperatures outdoor air in heating season. It is customary to call such consumers of thermal energy seasonal, and their thermal loads -
Thermal energy in hot water supply systems and in technological processes of industrial enterprises, it is consumed continuously throughout the year and little depends on the outside temperature.
Heat loads for hot water supply and process needs are considered year-round thermal loads.
When designing heat supply systems, calculated data on seasonal heat loads should be taken from heating and ventilation projects of buildings. With prospective construction settlement costs heat is recommended to be taken from standard projects with appropriate adjustment for the climatic conditions of the construction area.
In the absence of design data, the heating thermal loads of buildings are determined by one of the following methods:
calculation of heat losses through the elements of enclosing structures and adding losses for heating infiltration air;
calculation of thermal loads according to aggregated indicators;
determining the heat exchange of the heating and ventilation equipment installed in the building.
In the absence of design data, heating thermal loads, as a rule, are determined by aggregated indicators.
The ultimate goal of the calculations is to determine the heat loads (maximum, average for the heating period, etc.) of the heat supply system objects for heating, ventilation and hot water supply, calculation and plotting of heat loads.
^ Seasonal heat loads
The magnitude and nature of the change in seasonal load depend mainly on climatic conditions: outdoor air temperature, wind direction and speed, solar radiation, air humidity, etc. The main influence on the amount of heat load is exerted by outdoor temperature. The seasonal load has a relatively constant daily pattern and a variable annual load pattern.
R- resistance to heat transfer of the fence structure, m 2 o C / W (m 2 0 C h / kcal);
- the estimated temperature of the indoor air, o С;
- estimated outdoor air temperature for heating design or air temperature of a colder room (when calculating heat losses through internal enclosures), o С;
- additional heat losses in shares of the main ones for various types of premises and enclosures; = 0.6 - for the European part north of 52 north latitude; for the central regions of Western (up to 68 N) and Eastern (up to 70 N) Siberia, Khabarovsk and Primorsky Territories, with the exception of coastal regions up to a height of 500 m, as well as for the regions of Central Asia and Transcaucasia;
n- coefficient taken depending on the position of the outer surface of the enclosing structures in relation to the outside air.
Additional heat losses for heating the infiltrating outside air through the enclosing structures of the premises
where G and - the flow rate of infiltrating air through the enclosing structures of the room
| , kg/h, | (2.12) |
where BUT 1 , BUT 2 - areas of external enclosing structures, light openings (windows, balcony doors, lanterns) and other fences, m 2;
^ A 3 - the area of cracks, leaks and openings in the outer enclosing structures, m 2;
p i , p 1 - calculated difference between pressures on the outer and inner surfaces of the enclosing structures, respectively, on the calculated floor and at the floor level of the first floor, Pa;
^R and - resistance to air penetration of openings, m 2 h Pa / kg;
G n - normative air permeability of external enclosing structures, kg / m 2 h.
l- the length of the joints of wall panels, m.
The calculated pressure difference is determined by the formula
| , Pa | (2.13) |
where ^H- the height of the building from the level of the average planning mark of the earth to the top of the eaves, the center of the exhaust holes of the lantern or the mouth of the mine, m;
h i- estimated height from ground level to the top of windows, balcony doors, gates, openings or to the axis of horizontal and mid-vertical joints of wall panels, m;
n, in - specific gravity, respectively, of outdoor air and indoor air, determined by the formula
| kg / (m 2 s 2), | (2.14) |
where i – outside air density, kg/m 3 ;
w- wind speed, m/s;
With ns, With ps - aerodynamic coefficients, respectively, for the windward and leeward surfaces of the building fences;
to 1 - coefficient for taking into account changes in wind speed pressure depending on the height of the building;
p c - conditionally constant air pressure in the building, Pa.
To determine the ratio of heat loss by infiltration to heat loss by heat transfer through external enclosures, an approximate formula can be used
, | (2.15) |
where L- estimated height for the middle floor of the building
, | m, | (2.16) |
where ^H– building height, m;
T in, T n – temperature of internal and external air, K;
To a is the aerodynamic coefficient;
w- estimated wind speed for the cold period for the corresponding area, m/s;
- correction factor;
= 1.2 - for coastal areas of Primorsky Krai up to a height of 500 m above sea level; = 1.0 – for other territories;
b is a constant value ( b= 0.0350.040 - for detached industrial buildings with large light openings; b= 0.0080.010 - for residential and public buildings with double glazing in the continuous development of quarters).
The heat consumption for ventilation of residential buildings, which, as a rule, do not have a special supply system, is small. It usually does not exceed 5-10% of heat consumption for heating and is taken into account by the value of specific heat loss q about.
The heat consumption for ventilation of industrial and municipal enterprises, as well as public and cultural institutions, is a significant share of the total heat consumption of the facility. In industrial enterprises, the heat consumption for ventilation often exceeds the heating consumption.
Approximately the maximum heat flux for the ventilation of public buildings is determined by aggregated indicators: total area F or outside of the building Vн according to formulas (2.17) and (2.18):
, | MW (Gcal/h), | (2.17) |
, | MW (Gcal/h), | (2.18) |
where
- coefficient taking into account the heat flow for the ventilation of public buildings, is taken for buildings before 1985 - 0.4, after 1985 - 0.6;
q c - specific ventilation characteristic kJ / (m 3. h K) (kcal / (m 3. h o C)) ;
- design temperature of the outside air for ventilation, o C.
The average heat flow to ventilation for the average air temperature for heating season
Year-round heat loads
In industry, technological devices often consume heat in large quantities and very varied over time. These are, for example, various drying and evaporating plants, steaming chambers, digesters, galvanizing baths, distillation apparatuses, etc.
Specific norms of technological heat consumption are referred to a unit of production. Therefore, heat consumption for production needs should be determined according to the materials of technological projects or according to departmental design standards.
Improvement and rationalization of the technological process can significantly affect the size and nature of the heat load.
Heat consumption for the purposes of hot water supply during the heating period varies relatively little, but is highly uneven by the hours of the day. In summer, heat consumption in hot water supply systems of residential buildings decreases by 30-35% compared to winter. This is due to the fact that in summer the water temperature in a cold water supply is 10-12 С higher than in winter period. In addition, a significant part of the urban population in the summer, on Saturdays and Sundays, travels to suburban areas, i.e. in those days when the maximum analysis of hot water is observed in the residential sector in winter.
In terms of its significance, in many residential areas of large cities, the load on hot water supply becomes comparable to the heating load. In a number of districts, the annual heat supply for hot water supply reaches 40% of the total heat supply in a residential area.
Average heat flow for hot water supply (DHW) of residential and public buildings
where m- the estimated number of consumers of hot water;
a- the rate of water consumption for hot water at a temperature of 55 C per person per day, living in a building with hot water supply, taken depending on the degree of comfort, l / day;
b- the rate of water consumption for hot water supply in public buildings at a temperature of 55 C, taken in the amount of 25 l / day per 1 person;
The specific heat capacity of water is 4.187 kJ/(kg C) (1 kcal/(kg C));
- temperature of cold (tap) water during the heating period (in the absence of other data, it is assumed to be 5 С);
Maximum heat flow to the hot water supply of residential and public buildings
Average heat flow to hot water supply in the non-heating (summer) period
where t h, t l - respectively, the temperature of cold (tap) water during the heating period (in the absence of data, it is assumed to be 5 С) and the non-heating (summer) period (assumed to be 15 С);
- coefficient taking into account the change in water consumption for hot water supply. It is accepted in the absence of data for the housing and communal sector - 0.8 (for resort and southern cities = 1.5), for enterprises - 1.0.