Calculation of heat loss at home is a necessary step in the design of a heating system. It is carried out according to complex formulas. Incorrectly leads to insufficient heating of the room (if the heat loss indicators are underestimated) or to overpayments for the system and for heating (if the indicators are too high).
Heat supply calculation must be performed at the highest level
Initial data for calculating heat loss at home
To calculate correctly, you need to have a basic dataset. Only with them is it possible to work.
This is the starting data - a mandatory minimum, without which it is impossible to calculate the system. Now proceed to the definition of the characteristics of the future system, as well as your wishes for it.
Defined data for the walls of a residential building
Think about what the future functions of the room are, on the basis of this, draw a conclusion about the desired temperature regime (for example, in storage rooms the temperature may be lower than in those where staff are constantly present, in greenhouses, flower bases have even more specific heating requirements ).
At the next stage, the temperature regime of the room is determined. It is carried out by periodically measuring temperatures. The desired temperatures are determined to be maintained. The heating scheme and the proposed (or desired) installation locations of the risers are selected. The source of heat supply is determined.
When calculating heat loss, the architecture of the building also plays an important role, in particular its shape and geometry. Since 2003, SNiP has taken into account the indicator of the shape of the structure. It is calculated as the ratio of the area of \u200b\u200bthe shell (walls, floor and ceiling) to the volume that it surrounds. Until 2003, the parameter was not taken into account, which led to the fact that the energy was significantly overspending.
Work progress: calculating the percentage of permissible heat loss for a country house from a bar, log, brick, panels
Before proceeding directly to work, the performer conducts some field research at the facility. The premises are examined and measured, the wishes and information from the customer are taken into account. This process involves certain actions:
- Full-scale measurement of premises;
- Their specification according to the customer's data;
- Study of the heating system, if any;
- Ideas for improving or correcting errors in heating (in the existing system);
- Study of the hot water supply system;
- Development of ideas for using it for heating or reducing heat loss (for example, using Valtec equipment (Valtek);
- Calculation of heat loss and others necessary for the development of a heating system plan.
After carrying out these stages, the contractor provides the necessary technical documentation. It includes floor plans, profiles, where each heating device and the general arrangement of the system are displayed, materials for the specifics and type of equipment used.
Calculations: where does the greatest heat loss come from in a frame insulated house and how to reduce them using the device
The most important process in heating design is calculating the future system. Calculation of heat losses through the enclosing structures is carried out, additional losses and heat gains are determined, required amount heating devices of the selected type, etc. The calculation of the coefficient of heat loss at home should be done by an experienced person.
The heat balance equation plays an important role in determining heat losses and developing methods for their compensation. is given below:
V is the volume of the room, calculated taking into account the area of \u200b\u200bthe room and the height of the ceilings. T is the difference between the external and internal temperature of the building. K - coefficient of heat loss.
The heat balance formula does not give the most accurate indicators, therefore it is rarely used.
The main value used in the calculation is the heat load on the heaters. To determine it, the values \u200b\u200bof heat losses and are used. allows you to calculate the amount of heat that the heating system will generate, looks like:
Volume heat loss () is multiplied by 1.2. This is a reserve thermal coefficient - a constant that helps to compensate for some heat losses that are random in nature (prolonged opening of doors or windows, etc.).
It is rather difficult to calculate heat loss. On average, different building envelopes contribute to different energy losses. 10% is lost through the roof, 10% - through the floor, foundation, 40% - walls, 20% - windows and poor insulation, ventilation system, etc. The specific thermal characteristics of different materials are not the same. Therefore, the formula contains coefficients that take into account all the nuances. The table below shows the values \u200b\u200bof the coefficients required to calculate the amount of heat.
The heat loss formula is as follows:
In the formula, the specific heat loss is equal to 100 watts per square meter. m. Pl - the area of \u200b\u200bthe room, also participating in the definition. The formula can now be applied to calculate the amount of heat needed to be released by the boiler.
Count correctly and you will be warm at home
An example of calculating the heat loss coefficient in a private house: the formula for success
The formula for calculating heat for space heating is easily applicable to any building. As an example, consider a hypothetical building with simple glazing, wooden walls and a window-to-floor ratio of 20%. It is located in a temperate climatic zone, where the minimum outside temperature is 25 degrees. It has 4 walls, 3 m high. There is a cold attic above the heated room. The value of the coefficients is found out according to the table K1 - 1.27, K2 - 1.25, K3 - 1, K4 - 1.1, K5 - 1.33, K6 - 1, K7 - 1.05. The area of \u200b\u200bthe premises is 100 sq.m. The formula for the heat balance equation is not complicated and is within the power of every person.
Since the formula is known, the amount of heat required to heat a room can be calculated as follows:
TP \u003d 100 * 100 * 1.27 * 1.25 * 1 * 1.1 * 1.33 * 1 * 1.05 \u003d 24386.38 W \u003d 24.386 kW
And to calculate the heat energy for heating, the boiler power formula is used as follows:
Mk \u003d 1.2 * 24.386 \u003d 29.2632 kW.
At further stages, the number of required heating elements and the load on each of them, as well as the energy consumption for heating, are determined. Calculation of heat loss at home in our time of savings is very relevant.
Heat loss through enclosing structures
I will give an example of calculation for the outer walls of a two-story house.| 1) We calculate the resistance to heat transfer of the wall, dividing the thickness of the material by its coefficient of thermal conductivity. For example, if a wall is built of warm ceramic 0.5 m thick with a thermal conductivity coefficient of 0.16 W / (m × ° C), then we divide 0.5 by 0.16: 0.5 m / 0.16 W / (m × ° C) \u003d 3.125 m 2 × ° C / W Thermal conductivity coefficients building materials you can take. |
| 2) We calculate the total area of \u200b\u200bthe external walls. Here's a simplified example of a square house: (10 m width × 7 m height × 4 sides) - (16 windows × 2.5 m 2) \u003d 280 m 2 - 40 m 2 \u003d 240 m 2 |
| 3) We divide the unit by the resistance to heat transfer, thereby obtaining heat loss from one square meter walls by one degree difference in temperature. 1 / 3.125 m 2 × ° C / W \u003d 0.32 W / m 2 × ° C |
| 4) We calculate the heat loss of the walls. We multiply the heat loss from one square meter of the wall by the area of \u200b\u200bthe walls and by the difference in temperature inside the house and outside. For example, if the inside is + 25 ° C, and the outside is -15 ° C, then the difference is 40 ° C. 0.32 W / m 2 × ° C × 240 m 2 × 40 ° C \u003d 3072 W This number is the heat loss of the walls. Heat loss is measured in watts, i.e. this is the heat loss power. |
| 5) In kilowatt-hours it is more convenient to understand the meaning of heat loss. In 1 hour, thermal energy goes through our walls at a temperature difference of 40 ° C: 3072 W × 1 h \u003d 3.072 kW × h Energy is consumed in 24 hours: 3072 W × 24 h \u003d 73.728 kW × h |
It is clear that during the heating period the weather is different, i.e. the temperature difference changes all the time. Therefore, in order to calculate the heat loss for the entire heating period, you need to multiply in step 4 by the average temperature difference for all days of the heating period.
For example, for 7 months of the heating period, the average temperature difference in the room and outside was 28 degrees, which means heat loss through the walls during these 7 months in kilowatt-hours:
0.32 W / m 2 × ° C × 240 m 2 × 28 ° C × 7 months × 30 days × 24 h \u003d 10838016 W × h \u003d 10838 kW × h
The number is quite tangible. For example, if the heating was electric, then you can calculate how much money would be spent on heating by multiplying the resulting number by the cost of kWh. You can calculate how much money was spent on gas heating by calculating the cost of kWh of energy from a gas boiler. To do this, you need to know the cost of gas, the heat of combustion of the gas and the efficiency of the boiler.
By the way, in the last calculation, instead of the average temperature difference, the number of months and days (but not hours, we leave the clock), it was possible to use the degree-day of the heating period - GSOP, some information. You can find the already calculated GSOP for different cities of Russia and multiply the heat loss from one square meter by the wall area, by these GSOP and by 24 hours, having received heat loss in kW * h.
Similarly to walls, you need to calculate the values \u200b\u200bof heat loss for windows, entrance doors, roofs, and foundations. Then add everything up and you get the value of heat loss through all the enclosing structures. For windows, by the way, it will not be necessary to find out the thickness and thermal conductivity, usually there is already a ready-made resistance to heat transfer of a glass unit calculated by the manufacturer. For gender (in case slab foundation) the temperature difference will not be too large, the soil under the house is not as cold as the outside air.
Heat loss through ventilation
The approximate volume of available air in the house (volume interior walls and do not take into account furniture):10 m х10 m х 7 m \u003d 700 m 3
Air density at a temperature of + 20 ° C 1.2047 kg / m 3. Specific heat capacity of air 1.005 kJ / (kg × ° C). Air mass in the house:
700 m 3 × 1.2047 kg / m 3 \u003d 843.29 kg
Let's say all the air in the house changes 5 times a day (this is an approximate number). With an average difference between the internal and external temperatures of 28 ° C for the entire heating period, heat energy will be spent on average per day to heat the incoming cold air:
5 × 28 ° C × 843.29 kg × 1.005 kJ / (kg × ° C) \u003d 118,650.903 kJ
118,650.903 kJ \u003d 32.96 kWh (1 kWh \u003d 3600 kJ)
Those. during the heating season, with a fivefold air replacement, the house through ventilation will lose on average 32.96 kWh of heat energy per day. For 7 months of the heating period, energy losses will be:
7 x 30 x 32.96 kWh \u003d 6921.6 kWh
Heat loss through the sewer
During the heating season, the water entering the house is rather cold, for example, it has an average temperature of + 7 ° C. Heating of water is required when residents wash their dishes and take baths. Also, the water from the ambient air in the toilet cistern is partially heated. All the heat received by the water is flushed down the drain.Let's say that a family in a house consumes 15 m 3 of water per month. The specific heat capacity of water is 4.183 kJ / (kg × ° C). The density of water is 1000 kg / m 3. Let's say that on average the water entering the house is heated to + 30 ° C, i.e. temperature difference 23 ° C.
Accordingly, per month the heat loss through the sewer will be:
1000 kg / m 3 × 15 m 3 × 23 ° C × 4.183 kJ / (kg × ° C) \u003d 1443135 kJ
1443135 kJ \u003d 400.87 kWh
For 7 months of the heating period, residents pour into the sewer:
7 × 400.87 kWh \u003d 2806.09 kWh
Conclusion
At the end, you need to add the obtained numbers of heat losses through the enclosing structures, ventilation and sewage. This gives the approximate total number of heat losses at home.It must be said that heat losses through ventilation and sewerage are quite stable and difficult to reduce. You will not wash less often in the shower or poorly ventilate the house. Although partly the heat loss through ventilation can be reduced using a recuperator.
If I made a mistake somewhere, write in the comments, but I seem to have double-checked everything several times. It must be said that there are much more complex methods for calculating heat losses, additional coefficients are taken into account there, but their effect is insignificant.
Show all 50 comments.
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Valery (17.06.2015 22:23)
The approach is absolutely correct. This is exactly what is needed for those who are building their own house (cottage, individual housing construction). In this case, the goals can be two: 1 - to determine the structure of the building (for example, the material and thickness of the insulation); 2 - decide on the type and design of the heating system (for example, the power of the boiler, heating appliances …). Taking into account the specific heat loss of 0.32 instead of 0.305, we provide a "safety margin" for the system. In addition, the final result of the calculations is corrected by the available standard-size range of materials and equipment. For instance: 1 - if, as a result of the calculations, the thickness of the insulation plate was obtained, for example, 117mm, then we will put 3 layers of 50mm in the structure (we will not cut the plate in thickness +/- 5% .... Or will we? ...); 2 - if, as a result of the calculations, the boiler power was obtained, for example, 10.3 kW, then we will put 12 in the system, and preferably 14, so it is more reliable. Therefore, insignificant errors in relation to the rigorous method (SP, SNIP, etc.) will not affect the final decision. Thank you Dmitry! P.S. It would be nice to designate materials that are popular in building markets, such as Rockwool, in the thermal conductivity table. (since you saved a lazy developer like me from long wandering around the sites ...) |
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Designing a heating system "by eye" with a high probability can lead either to an unjustified overestimation of the cost of its operation, or to underheating of the home.
So that neither one nor the other happens, it is necessary first of all to correctly calculate the heat loss at home.
And only on the basis of the results obtained, the power of the boiler and radiators is selected. Our conversation will focus on how these calculations are performed and what must be taken into account in this case.
The authors of many articles reduce the calculation of heat loss to one simple action: it is proposed to multiply the area of \u200b\u200bthe heated room by 100 W. The only condition that is put forward relates to the height of the ceiling - it should be 2.5 m (for other values, it is proposed to enter a correction factor).
In fact, such a calculation is so approximate that the figures obtained with its help can be safely equated with "taken from the ceiling." Indeed, a number of factors affect the specific amount of heat loss: the material of the enclosing structures, the outside temperature, the area and type of glazing, the rate of air exchange, etc.
Heat loss at home
Moreover, even for houses with different heated areas, all other things being equal, its value will be different: in a small house - more, in a large - less. This is how the square-cube law manifests itself.
Therefore, it is extremely important for the home owner to master a more accurate method for determining heat loss. Such a skill will allow you not only to choose heating equipment with optimal power, but also to evaluate, for example, the economic effect of insulation. In particular, it will be possible to understand whether the service life of the heat insulator will exceed the payback period.
The first thing that the performer needs to do is to decompose the total heat loss into three components:
- losses through enclosing structures;
- due to the operation of the ventilation system;
- associated with the discharge of heated water into the sewer.
Let's consider each of the varieties in detail.
Basalt insulation is a popular heat insulator, but there are rumors about its harm to human health. and environmental safety.
How to properly insulate the walls of an apartment from the inside without harm to the structure of the building, read.
A cold roof makes it difficult to create a cozy attic. You will learn how to insulate the ceiling under a cold roof and which materials are the most effective.
Calculation of heat loss
Here's how to calculate:
Heat loss through enclosing structures
For each material that is part of the enclosing structures, in the reference book or the passport provided by the manufacturer, we find the value of the thermal conductivity coefficient Kt (unit of measurement - W / m * degree).
For each layer of the enclosing structures, we determine the thermal resistance by the formula: R \u003d S / Kt, where S is the thickness of this layer, m.
For multi-layer structures, the resistances of all layers must be added.
We determine heat loss for each structure using the formula Q \u003d (A / R) * dT,
- A - the area of \u200b\u200bthe enclosing structure, sq. m;
- dT is the difference between outside and inside temperatures.
- dT should be determined for the coldest five days.
Heat loss through ventilation
For this part of the calculation, it is necessary to know the air exchange rate.
In residential buildings erected according to domestic standards (the walls are vapor-permeable), it is equal to one, that is, the entire volume of air in the room must be updated in an hour.
In houses built according to European technology (DIN standard), in which the walls are covered with vapor barrier from the inside, the air exchange rate has to be increased to 2. That is, the air in the room must be renewed twice in an hour.
Heat loss through ventilation is determined by the formula:
Qw \u003d (V * Kw / 3600) * p * s * dT,
- V - room volume, cubic meters m;
- Kv - the frequency of air exchange;
- Р - air density, taken equal to 1.2047 kg / cu. m;
- С - specific heat capacity of air, taken equal to 1005 J / kg * С.
The above calculation allows you to determine the power that the heat generator of the heating system should have. If it turns out to be too high, you can do the following:
- to lower the requirements for the level of comfort, that is, to set the desired temperature in the coldest period at the minimum mark, say, 18 degrees;
- for the period of severe cold weather, reduce the air exchange rate: the minimum permissible supply ventilation capacity is 7 cubic meters. m / h for each inhabitant of the house;
- provide for the organization of supply and exhaust ventilation with a recuperator.
Note that the recuperator is useful not only in winter, but also in summer: in the heat it allows you to save the cold produced by the air conditioner, although it does not work as efficiently at this time as in frost.
It is most correct to perform zoning when designing a house, that is, to assign its own temperature for each room based on the required comfort. For example, in a nursery or an elderly person's room, a temperature of about 25 degrees should be ensured, while 22 degrees will be enough for a living room. On a staircase or in a room where residents rarely appear or there are heat sources, the design temperature can generally be limited to 18 degrees.
Obviously, the figures obtained in this calculation are relevant only for a very short period - the coldest five-day period. To determine the total energy consumption for the cold season, the dT parameter must be calculated taking into account not the lowest, but the average temperature. Then you need to perform the following action:
W \u003d ((Q + Qv) * 24 * N) / 1000,
- W is the amount of energy required to replenish heat loss through enclosing structures and ventilation, kW * h;
- N is the number of days in the heating season.
However, this calculation will be incomplete if heat losses to the sewer system are not taken into account.
To receive hygienic procedures and wash dishes, residents of the house heat the water and the generated heat goes into the sewer pipe.
But in this part of the calculation, one should take into account not only direct heating of water, but also indirect - heat is taken off by water in the cistern and siphon of the toilet, which is also discharged into the sewer.
Based on this, the average temperature of water heating is taken to be only 30 degrees. We calculate heat loss through the sewer using the following formula:
Qk \u003d (Vw * T * p * s * dT) / 3 600 000,
- Vв - monthly volume of water consumption without division into hot and cold, cubic meters. m / month;
- P is the density of water, we take p \u003d 1000 kg / cu. m;
- С - heat capacity of water, we take с \u003d 4183 J / kg * С;
- dT is the temperature difference. Considering that the water at the inlet in winter has a temperature of about +7 degrees, and we agreed to consider the average temperature of heated water equal to 30 degrees, dT \u003d 23 degrees should be taken.
- 3,600,000 - the number of joules (J) in 1 kW * h.
Example of calculating heat loss at home
Let's calculate the heat loss of a 2-storey building with a height of 7 m, measuring 10x10 m in plan.
The walls are 500 mm thick and built of warm ceramics (Kt \u003d 0.16 W / m * C), outside they are insulated with 50 mm thick mineral wool (Kt \u003d 0.04 W / m * C).
The house has 16 windows with an area of \u200b\u200b2.5 sq. m.
Outside temperature in the coldest five-day period is -25 degrees.
The average outside temperature for the heating period is (-5) degrees.
Inside the house, it is required to provide a temperature of +23 degrees.
Water consumption - 15 cubic meters m / month
The duration of the heating period is 6 months.
Determine heat loss through enclosing structures (for example, consider only walls)
Thermal resistance:
- main material: R1 \u003d 0.5 / 0.16 \u003d 3.125 sq. m * C / W;
- insulation: R2 \u003d 0.05 / 0.04 \u003d 1.25 sq. m * C / W.
The same for the wall as a whole: R \u003d R1 + R2 \u003d 3.125 + 1.25 \u003d 4.375 sq. m * C / W.
Determine the area of \u200b\u200bthe walls: A \u003d 10 x 4 x 7 - 16 x 2.5 \u003d 240 sq. m.
Heat loss through the walls will be:
Qc \u003d (240 / 4.375) * (23 - (-25)) \u003d 2633 W.
Heat losses through the roof, floor, foundation, windows and front door are calculated in a similar way, after which all the obtained values \u200b\u200bare summed up. Manufacturers usually indicate the thermal resistance of doors and windows in the product passport.
Please note that when calculating heat loss through the floor and foundation (in the presence of a basement), the temperature difference dT will be much smaller, since when calculating it, the temperature is not taken into account, but the temperature of the soil, which is much warmer in winter.
Heat loss through ventilation
Determine the volume of air in the room (to simplify the calculation, the thickness of the walls is not taken into account):
V \u003d 10x10x7 \u003d 700 cubic meters m.
Taking the air exchange rate Kw \u003d 1, we determine the heat loss:
Qw \u003d (700 * 1/3600) * 1.2047 * 1005 * (23 - (-25)) \u003d 11300 W.
Ventilation in the house
Heat loss through the sewer
Taking into account the fact that tenants consume 15 cubic meters. m of water per month, and the settlement period is 6 months, heat loss through the sewer will be:
Qк \u003d (15 * 6 * 1000 * 4183 * 23) / 3 600 000 \u003d 2405 kW * h
If you do not live in country house in winter, in the off-season or in cold summer, you still need to heat it. in this case it is most expedient.
You can read about the reasons for the pressure drop in the heating system. Troubleshooting.
Estimation of the total volume of energy consumption
To estimate the total volume of energy consumption during the heating period, it is necessary to recalculate the heat loss through ventilation and enclosing structures, taking into account the average temperature, that is, dT will be not 48, but only 28 degrees.
Then the average power losses through the walls will be:
Qc \u003d (240 / 4.375) * (23 - (-5)) \u003d 1536 W.
Suppose that an additional 800 W is lost through the roof, floor, windows and doors, then the total average heat loss through the enclosing structures will be Q \u003d 1536 + 800 \u003d 2336 W.
The average heat loss through ventilation will be:
Qw \u003d (700 * 1/3600) * 1.2047 * 1005 * (23 - (-5)) \u003d 6592 W.
Then, for the entire period, you will have to spend on heating:
W \u003d ((2336 + 6592) * 24 * 183) / 1000 \u003d 39211 kWh.
To this value, 2405 kWh of losses through the sewer must be added, so that the total amount of energy consumption for the heating period will be 41616 kWh.
If only gas is used as an energy carrier, from 1 cubic meter. m of which it is possible to obtain 9.45 kW * h of heat, then it will be needed 41616 / 9.45 \u003d 4404 cubic meters. m.
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Many, building a country house, forget about the approach of winter cold, which is why the calculation of the heat loss of the building is done in a hurry, and as a result, heating does not create a comfortable microclimate in the premises. But making the house warm is not difficult, you just need to take into account a number of nuances.
1 What is the basis for calculating the heat loss of a building
Any material has such a property as thermal conductivity, only the level of thermal resistance differs, that is throughput... From any house, even with thermal insulation arranged according to all the rules, heat leaves through windows, doors, walls, floor, ceiling (roof), as well as through ventilation. When there is a difference between external and internal temperatures, a so-called "dew point" necessarily occurs, with an average value. And only the microclimate in the premises, the material and thickness of the walls, as well as the characteristics of thermal insulation, determines where this point will be: inside, outside or directly in the wall, as well as what temperature it will be.
If you take a responsible approach to the task and calculate the heat loss of a building according to all the rules, it will take you many hours and you will have to draw up a lot of formulas, the calculations will take a whole notebook. Therefore, we will determine the indicators of interest to us by a simplified method, or by turning to SNiP and GOST for help. And, since it was decided to do the calculations not too deeply, we will leave aside the determination of the average annual temperature and humidity for the coldest five-day period in several years, as required by SNiP 23-01-99. Let's just mark the coldest day for the last winter season, let's say it will be -30 o C. Also, we will not take into account the average seasonal wind speed, humidity in the region and the duration of the heating period.
Building heat loss calculator
Specify the sizes and types of walls.| On the street average temperature for the day |
Select the value -40 ° C -30 ° C -20 ° C -15 ° C -10 ° C -5 ° C 0 ° C + 5C + 10C |
| Inside average temperature for the day |
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| Walls Outgoing only to the street walls! |
Add street-facing walls and specify which layers the wall is composed of |
| Rooms |
Add all used rooms, even corridors, and specify which layers the floors are made of |
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Heat loss: |
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However, what constitutes the microclimate in the living room? Comfortable conditions for residents depend on the air temperature t in, its humidity φ in and movement v in, arising in the presence of ventilation. And one more factor affects the level of heat - radiation of heat or cold t p characteristic of naturally heated (cooled) objects and surfaces in the setting. It determines the resulting temperature t п, using the formula [ t n \u003d ( t p + t at 2]. All these indicators for different premises can be considered in the table below.
Optimal parameters of the microclimate of residential buildings in accordance with GOST 30494-96
| Period of the year | Premises |
Indoor air temperature t in, ° С |
Resulting temperature t n, ° C |
Relates. indoor air humidity φ in, % |
Air speed v in, m / s |
| Cold | Living room | ||||
| The same, in areas with t 5 from -31 ° C | |||||
| Kitchen | |||||
| Restroom | |||||
| Bathroom, combined bathroom | |||||
| Recreation and study area | |||||
| Interroom corridor | |||||
| Lobby, staircase | |||||
| Pantry | |||||
| Warm | Living room |
The letters NN indicate non-standardized parameters.
2 We make a heat engineering calculation of the wall, taking into account all layers
As already said, each material has a characteristic resistance to heat transfer, and the thicker the wall or floor, the higher this value... However, do not forget about thermal insulation, in the presence of which the surfaces enclosing the room become multi-layer and much better prevent heat leakage. Each layer has its own resistance to the passage of heat, and the sum of all these values \u200b\u200bis denoted in the formulas as Σ R i (here the letter i defines the layer number).
Since materials with different properties constituting the enclosure of premises have some disturbance of the temperature regime in their structure, the total resistance to heat transfer is calculated. His formula is as follows: where R in and R n correspond to the resistance on the inner and outer surfaces of the fence, be it a wall or a ceiling. However, heaters make adjustments to the thermal engineering calculation of the wall, which are based on the coefficient of thermal uniformity rdefined by the formula.
Indicators with digital indices are, respectively, the coefficients of internal fasteners and the connection of the design fence with any other. First, that is r 1, is responsible just for fixing the heaters. If the thermal conductivity coefficient of the latter λ \u003d 0.08 W / (m ° C), the value r 1 will be large, if the thermal conductivity of the thermal insulation is estimated as λ \u003d 0.03 W / (m · ° С), then less.
The value of the coefficient of internal fasteners decreases as the thickness of the insulation layer increases.
In general, the picture is as follows. Let's say the thermal insulation is mounted by direct anchoring on a three-layer cellular concrete wall, facing with brick on the outside. Then, with a layer of insulation of 100 millimeters r 1 corresponds to 0.78-0.91, a thickness of 150 millimeters gives the coefficient of internal fasteners 0.77-0.90, the same indicator, but 200 mm, determines r 1 as 0.75-0.88. If the inner layer is also brick, then r 1 \u003d 0.78-0.92, and if the walls of the room are reinforced concrete, then the coefficient shifts to 0.79-0.93. But window slopes and ventilation make a difference r 2 \u003d 0.90-0.95. All these data should be taken into account in the future.
3 Some information on how to calculate the thickness of the insulation
In order to start calculating thermal insulation, we need, first of all, to calculate R o, then find out the required thermal resistance R req from the following table (shorthand).
Required values \u200b\u200bof resistance to heat transfer of enclosing structures
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Building/ premises |
Degree-day of the heating period D d, ° С day |
Reduced resistance to heat transfer of fences R req, m 2 ° С / W |
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walls |
coverings |
attic floor and ceiling over cold basements |
windows and balcony doors, showcases and stained glass |
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| 1. Residential, medical and prophylactic and children's institution, school, boarding school | |||||
| and | |||||
| b | |||||
| 2. Public, administrative, domestic and other premises with wet or wet conditions | |||||
| and | |||||
| b | |||||
Odds a and b are necessary for those cases where the value D d, ° С day differs from that given in the table, then R req, m 2 ° С / W is calculated by the formula R req \u003d a D d + b... For column 6 of the first group of buildings, there are amendments: if the value of the degree-day is less than 6000 ° С day, a \u003d 0.000075, and b \u003d 0.15, if the same indicator is in the range 6000-8000 ° С day, then a = 0,00005, b \u003d 0.3, if more than 8000 ° С day, then a \u003d 0.000025, a b \u003d 0.5. When all the data is collected, we proceed to the calculation of thermal insulation.
Now let's find out how to calculate the thickness of the insulation. You will have to turn to math here, so be prepared to work with formulas. Here is the first of them, according to it we determine the required conditional resistance to heat transfer R o conv. tr \u003d R req / r. We need this parameter to determine the required heat transfer resistance of the insulation R ut tr \u003d R o conv. tr - (R in + Σ R t. Iz + R n), here Σ R t. iz is the sum of the thermal resistance of the layers of the fence, excluding thermal insulation. We find the thickness of the insulation δ ut \u003d R ut tr λ ut (m), and λ is taken from table E.1 SP 23-101-2004, and round the result up to a constructive value, taking into account the manufacturer's nomenclature.
