Calculation of heat loss at home
The house loses heat through the enclosing structures (walls, windows, roof, foundation), ventilation and sewerage. The main heat losses occur through the enclosing structures - 60-90% of all heat losses.
Calculation of heat loss at home is needed, at a minimum, to select the right boiler. You can also estimate how much money will be spent on heating in the planned house. Here is an example of a calculation for a gas boiler and an electric one. It is also possible, thanks to calculations, to analyze the financial efficiency of insulation, i.e. understand whether the costs of installing insulation will be recouped by fuel savings over the service life of the insulation.
Heat loss through building envelopes
I will give an example of calculation for external walls two-story house.| 1) We calculate the heat transfer resistance of the wall by dividing the thickness of the material by its thermal conductivity coefficient. For example, if a wall is built from warm ceramics 0.5 m thick with a thermal conductivity coefficient of 0.16 W/(m×°C), then divide 0.5 by 0.16: 0.5 m / 0.16 W/(m×°C) = 3.125 m 2 ×°C/W Thermal conductivity coefficients building materials can be taken . |
| 2) Calculate the total area of the external walls. Let me give you a simplified example of a square house: (10 m width × 7 m height × 4 sides) - (16 windows × 2.5 m 2) = 280 m 2 - 40 m 2 = 240 m 2 |
| 3) Divide the unit by the heat transfer resistance, thereby obtaining heat loss from one square meter of wall per degree temperature difference. 1 / 3.125 m 2 ×°C/W = 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 wall by the area of the walls and by the temperature difference between inside and outside the house. For example, if inside is +25°C and outside is -15°C, then the difference is 40°C. 0.32 W/m 2 ×°C × 240 m 2 × 40 °C = 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) It is more convenient to understand the meaning of heat loss in kilowatt-hours. In 1 hour, thermal energy is lost through our walls at a temperature difference of 40°C: 3072 W × 1 h = 3.072 kWh Energy lost in 24 hours: 3072 W × 24 h = 73.728 kWh |
For example, over 7 months of the heating period, the average difference in temperature indoors and outdoors 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 hours = 10838016 Wh = 10838 kWh
The number is quite “tangible”. For example, if the heating were 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 gas boiler. To do this, you need to know the cost of gas, the calorific value of 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 hours), 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 area of the walls, by these GSOP and by 24 hours, obtaining heat loss in kWh.
Similarly to walls, you need to calculate the heat loss values for windows, front door, roof, and foundation. Then sum everything up and get the value of heat loss through all enclosing structures. For windows, by the way, you will not need to find out the thickness and thermal conductivity; usually there is already a ready-made heat transfer resistance of the glass unit calculated by the manufacturer. For floor (in case slab foundation) the temperature difference will not be too great, 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 I don’t take furniture into account):10 m x 10 m x 7 m = 700 m 3
Air density at +20°C is 1.2047 kg/m3. The specific heat capacity of air is 1.005 kJ/(kg×°C). Air mass in the house:
700 m 3 × 1.2047 kg/m 3 = 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 indoor and outdoor temperatures of 28 °C over the entire heating period, the following thermal energy will be consumed on average per day to heat the incoming cold air:
5 × 28 °C × 843.29 kg × 1.005 kJ/(kg×°C) = 118650.903 kJ
118650.903 kJ = 32.96 kWh (1 kWh = 3600 kJ)
Those. During the heating season, with a fivefold replacement of air, the house through ventilation will lose an average of 32.96 kWh of thermal energy per day. Over 7 months of the heating period, energy losses will be:
7 × 30 × 32.96 kWh = 6921.6 kWh
Heat loss through sewerage
During the heating season, the water entering the house is quite cold, let’s say it has average temperature+7°C. Water heating is required when residents wash dishes and take baths. The water in the toilet cistern is also partially heated by the ambient air. Residents flush all the heat generated by water 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/m3. Let’s assume that on average the water entering the house heats up to +30°C, i.e. temperature difference 23°C.
Accordingly, per month heat loss through the sewerage system will be:
1000 kg/m 3 × 15 m 3 × 23°C × 4.183 kJ/(kg×°C) = 1443135 kJ
1443135 kJ = 400.87 kWh
During the 7 months of the heating period, residents pour into the sewer:
7 × 400.87 kWh = 2806.09 kWh
Conclusion
At the end, you need to add up the resulting numbers of heat loss through the building envelope, ventilation and sewerage. You will get the approximate total number of heat losses at home.It must be said that heat loss through ventilation and sewerage is quite stable and difficult to reduce. You won't shower less often or poorly ventilate your house. Although heat loss through ventilation can be partially 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 loss; additional coefficients are taken into account, but their influence is insignificant.
Addition.
Calculation of heat loss at home can also be done using SP 50.13330.2012 (updated edition of SNiP 02/23/2003). There is Appendix D “Calculation of the specific characteristics of thermal energy consumption for heating and ventilation of residential and public buildings“, the calculation itself will be much more complicated, more factors and coefficients are used.
Showing the 25 most recent comments. Show all comments (54).
|
Andrew Vladimirovich (11.01.2018 14:52)
In general, everything is fine for mere mortals. The only thing I would advise, for those who like to point out inaccuracies, is to indicate a more complete formula at the beginning of the article Q=S*(tin-tout)*(1+∑β)*n/Rо and explain that (1+∑β)*n, taking into account all coefficients, will differ slightly from 1 and cannot grossly distort the calculation of heat loss of the entire enclosing designs, i.e. We take as a basis the formula Q=S*(tin-tout)*1/Ro. I don’t agree with the calculation of ventilation heat loss, I think differently. I would calculate the total heat capacity of the entire volume, and then multiply it by the real factor. I would still take the specific heat of the air from frosty air (we will heat the street air), but it will be significantly higher. And it is better to take the heat capacity of the air mixture directly in W, equal to 0.28 W / (kg °C). |
Comfort is a fickle thing. Sub-zero temperatures arrive, you immediately feel chilly, and are uncontrollably drawn to home improvement. “Global warming” begins. And there is one “but” here - even after calculating the heat loss of the house and installing the heating “according to plan,” you can be left face to face with the quickly disappearing heat. The process is not visually noticeable, but is perfectly felt through woolen socks and large heating bills. The question remains: where did the “precious” heat go?
Natural heat loss is well hidden behind load-bearing structures or “well-made” insulation, where there should be no gaps by default. But is it? Let's look at the issue of heat leaks for different structural elements.
Up to 30% of all heat loss in a house occurs on the walls. IN modern construction They are multilayer structures made of materials of different thermal conductivity. Calculations for each wall can be carried out individually, but there are common errors for all, through which heat leaves the room and cold enters the house from outside.
The place where the insulating properties are weakened is called a “cold bridge”. For walls it is:
The optimal masonry seam is 3mm. It is achieved more often adhesive compositions fine texture. When the volume of mortar between the blocks increases, the thermal conductivity of the entire wall increases. Moreover, the temperature of the masonry seam can be 2-4 degrees colder than the base material (brick, block, etc.).
Masonry joints as a “thermal bridge”
Reinforced concrete has one of the highest thermal conductivity coefficients among building materials (1.28 - 1.61 W/(m*K)). This makes it a source of heat loss. The issue is not completely resolved by cellular or foam concrete lintels. Temperature difference reinforced concrete beam and the main wall is often close to 10 degrees.
You can insulate the lintel from the cold with continuous external insulation. And inside the house - by assembling a box from HA under the cornice. This creates an additional air layer for heat.
Connecting an air conditioner or TV antenna leaves gaps in the overall insulation. The through metal fasteners and the passage hole must be tightly sealed with insulation.
And if possible, do not withdraw metal fastenings outwards, fixing them inside the wall.
Installation of damaged material (with chips, compression, etc.) leaves vulnerable areas for heat leaks. This is clearly visible when examining a house with a thermal imager. Bright spots indicate gaps in the external insulation.
During operation, it is important to monitor the general condition of the insulation. An error in choosing an adhesive (not a special one for thermal insulation, but a tile one) can cause cracks in the structure within 2 years. Yes, and the main insulation materials also have their disadvantages. For example:
When working with both materials, it is important to ensure a precise fit of the locks of the insulation boards and the cross arrangement of the sheets.
Experience! Heat losses can increase during operation, because all materials have their own nuances. It is better to periodically assess the condition of the insulation and repair damage immediately. A crack on the surface is a “fast” road to destruction of the insulation inside.
Concrete is the predominant material in foundation construction. Its high thermal conductivity and direct contact with the ground result in up to 20% heat loss along the entire perimeter of the building. The foundation conducts heat particularly strongly from the basement and improperly installed heated floors on the first floor.
Heat loss is also increased by excess moisture that is not removed from the house. It destroys the foundation, creating openings for the cold. Many people are sensitive to humidity thermal insulation materials. For example, mineral wool, which often goes onto the foundation from general insulation. It is easily damaged by moisture and therefore requires a dense protective frame. Expanded clay also permanently loses its thermal insulation properties wet ground. Its structure creates an air cushion and well compensates for ground pressure during freezing, but the constant presence of moisture minimizes beneficial features expanded clay insulation. That is why the creation of working drainage is required condition long life of the foundation and heat retention.
This also includes in importance the waterproofing protection of the base, as well as a multi-layer blind area, not wide less than a meter. At columnar foundation or heaving soil, the blind area around the perimeter is insulated to protect the soil at the base of the house from freezing. The blind area is insulated with expanded clay, sheets of expanded polystyrene or polystyrene.
It is better to choose sheet materials for foundation insulation with a groove connection, and treat it with a special silicone compound. The tightness of the locks blocks access to the cold and guarantees continuous protection of the foundation. In this matter, seamless spraying of polyurethane foam has an undeniable advantage. In addition, the material is elastic and does not crack when the soil heaves.
For all types of foundations, you can use the developed insulation schemes. An exception may be a foundation on piles due to its design. Here, when processing the grillage, it is important to take into account the heaving of the soil and choose a technology that does not destroy the piles. This is a complex calculation. Practice shows that a house on stilts is protected from the cold by a properly insulated floor on the first floor.
Attention! If the house has a basement and it often floods, then this must be taken into account when insulating the foundation. Since the insulation/insulator is in this case will clog moisture in the foundation and destroy it. Accordingly, heat will be lost even more. The first thing that needs to be resolved is the flooding issue.
An uninsulated ceiling transfers a significant portion of the heat to the foundation and walls. This is especially noticeable if the heated floor is installed incorrectly - the heating element cools down faster, increasing the cost of heating the room.
To ensure that the heat from the floor goes into the room and not outside, you need to make sure that the installation follows all the rules. The main ones are:
Serious insulation is important for any floor, and not necessarily with heating. Poor thermal insulation turns the floor into a large “radiator” for the ground. Is it worth heating it in winter?!
Important! Cold floors and dampness appear in the house when the ventilation of the underground space is not working or not done (vents are not organized). No heating system can compensate for such a deficiency.
The compounds disrupt the integrity of the materials. Therefore, corners, joints and abutments are so vulnerable to cold and moisture. The joints of concrete panels become damp first, and fungus and mold appear there. The temperature difference between the corner of the room (the junction of the structures) and the main wall can range from 5-6 degrees, to sub-zero temperatures and condensation inside the corner.
Clue! At the sites of such connections, craftsmen recommend making an increased layer of insulation on the outside.
Heat often escapes through interfloor covering, when the slab is laid over the entire thickness of the wall and its edges face the street. Here the heat loss of both the first and second floors increases. Drafts form. Again, if there is a heated floor on the second floor, the external insulation should be designed for this.
Heat is removed from the room through equipped ventilation ducts, ensuring healthy air exchange. Ventilation that works “in reverse” draws in the cold from the street. This happens when there is a shortage of air in the room. For example, when the fan in the hood is turned on, it takes too much air from the room, due to which it begins to be drawn in from the street through other exhaust ducts (without filters and heating).
Questions about how not to withdraw a large number of heat out, and how not to let it in cold air into the house, have long had their own professional solutions:
Comfort costs good ventilation. With normal air exchange, mold does not form and a healthy microclimate for living is created. That is why a well-insulated house with a combination of insulating materials must have working ventilation.
Bottom line! To reduce heat loss through ventilation ducts It is necessary to eliminate errors in air redistribution in the room. In properly functioning ventilation only warm air leaves the house, some of the heat from which can be returned back.
A house loses up to 25% of heat through door and window openings. The weak points for doors are a leaky seal, which can be easily replaced with a new one, and thermal insulation that has become loose inside. It can be replaced by removing the casing.
Vulnerable spots for wooden and plastic doors similar to “cold bridges” in similar window designs. Therefore, we will consider the general process using their example.
What indicates “window” heat loss:
The sashes may not fit tightly when the window is not adjusted and the rubber bands around the perimeter are worn out. The position of the valves can be adjusted independently, as well as the seal can be changed. It is better to completely replace it once every 2-3 years, and preferably with a “native” made seal. Seasonal cleaning and lubrication of rubber bands maintains their elasticity during temperature changes. Then the seal does not let the cold in for a long time.
The gaps in the frame itself (relevant for wooden windows) are filled silicone sealant, better transparent. When it hits the glass it is not so noticeable.
The joints of the slopes and the window profile are also sealed with sealant or liquid plastic. IN difficult situation, you can use self-adhesive polyethylene foam - “insulating” tape for windows.
Important! It is worth making sure that in the finishing of external slopes the insulation (foam plastic, etc.) completely covers the seam polyurethane foam and the distance to the middle of the window frame.
Modern ways to reduce heat loss through glass:
Healthy! Reduce heat loss through glass - organized air curtains above the windows (possibly in the form of warm baseboards) or protective shutters at night. Especially relevant when panoramic glazing and severe sub-zero temperatures.
Heat loss also applies to heating, where heat leaks often occur for two reasons.
Following simple rules reduces heat loss and prevents the heating system from running idle:
The note! When refilling, it is better to add anti-corrosion inhibitors to the water. This will support metal elements systems.
Heat initially tends to the top of the house, making the roof one of the most vulnerable elements. It accounts for up to 25% of all heat loss.
Cold attic space or a residential attic are insulated equally tightly. The main heat losses occur at the junctions of materials, it does not matter whether it is insulation or structural elements. Thus, an often overlooked bridge of cold is the boundary of the walls with the transition to the roof. It is advisable to treat this area together with the Mauerlat.
Basic insulation also has its own nuances, related more to the materials used. For example:
Practice! IN upper structures any gap can drain a lot of expensive heat. Here it is important to place emphasis on dense and continuous insulation.
It is useful to know the places of heat loss not only for arranging a house and living in comfortable conditions, but also so as not to overpay for heating. Proper insulation in practice pays for itself in 5 years. The term is long. But we’re not building a house for two years.
Today, many families choose for themselves Vacation home like a place permanent residence or year-round holiday. However, its content, and especially payment utilities, - are quite expensive, while most homeowners are not oligarchs at all. One of the most significant expenses for any homeowner is heating costs. To minimize them, it is necessary to think about energy saving even at the stage of building a cottage. Let's consider this issue in more detail.
« The problems of energy efficiency of housing are usually remembered from the perspective of urban housing and communal services, but owners individual houses this topic is sometimes much closer,- thinks Sergey Yakubov , Deputy Director for Sales and Marketing, leading manufacturer of roofing and facade systems in Russia. - The cost of heating a house can be much more than half the cost of maintaining it during the cold season and sometimes reach tens of thousands of rubles. However, with a competent approach to thermal insulation of a residential building, this amount can be significantly reduced.».
Actually, you need to heat the house in order to constantly maintain it comfortable temperature, no matter what is happening on the street. In this case, it is necessary to take into account heat loss both through the enclosing structures and through ventilation, because heat leaves along with heated air, which is replaced by cooled air, as well as the fact that a certain amount of heat is emitted by people in the house, Appliances, incandescent lamps, etc.
To understand how much heat we should get from our heating system and how much money we will have to spend on it, let’s try to evaluate the contribution of each of the other factors to the heat balance using the example of a two-story brick house located in the Moscow region with with total area premises 150 m2 (to simplify the calculations, we assumed that the dimensions of the cottage in plan are approximately 8.7x8.7 m and it has 2 floors 2.5 m high).
Heat loss through enclosing structures (roof, walls, floor)
The intensity of heat loss is determined by two factors: the difference in temperatures inside and outside the house and the resistance of its enclosing structures to heat transfer. By dividing the temperature difference Δt by the heat transfer resistance coefficient Ro of walls, roofs, floors, windows and doors and multiplying by their surface area S, you can calculate the heat loss rate Q:
Q = (Δt/R o)*S
The temperature difference Δt is not a constant value, it changes from season to season, during the day, depending on the weather, etc. However, our task is simplified by the fact that we need to estimate the total heat demand for the year. Therefore, for an approximate calculation, we can easily use such an indicator as the average annual air temperature for the selected area. For the Moscow region it is +5.8°C. If we take +23°C as a comfortable temperature in the house, then our average difference will be
Δt = 23°C - 5.8°C = 17.2°C
Walls. The area of the walls of our house (2 square floors 8.7x8.7 m, height 2.5 m) will be approximately equal to
S = 8.7 * 8.7 * 2.5 * 2 = 175 m2
However, from this we need to subtract the area of windows and doors, for which we will calculate heat loss separately. Suppose we have one front door, standard size 900x2000 mm, i.e. area
Door S = 0.9 * 2 = 1.8 m2,
and there are 16 windows (2 on each side of the house on both floors) measuring 1500x1500 mm, the total area of which will be
S windows = 1.5 * 1.5 * 16 = 36 m2.
Total - 37.8 m2. Remaining area brick walls -
S walls = 175 - 37.8 = 137.2 m2.
The heat transfer resistance coefficient of a wall of 2 bricks is 0.405 m2°C/W. For simplicity, we will neglect the heat transfer resistance of the layer of plaster covering the walls of the house from the inside. Thus, the heat release from all walls of the house will be:
Q walls = (17.2°C / 0.405m 2°C/W) * 137.2 m 2 = 5.83 kW
Roof. For simplicity of calculations, we will assume that the heat transfer resistance roofing pie equal to the heat transfer resistance of the insulation layer. For lightweight mineral wool thermal insulation with a thickness of 50-100 mm, most often used for roof insulation, it is approximately equal to 1.7 m 2 °C/W. Heat transfer resistance attic floor Let's neglect: let's assume that the house has an attic, which communicates with other rooms and heat is distributed evenly between all of them.
The area of a gable roof with a slope of 30° will be
Roof S = 2 * 8.7 * 8.7 / Cos30° = 87 m 2.
Thus, its heat release will be:
Q roof = (17.2°C / 1.7m 2 °C/W) * 87 m 2 = 0.88 kW
Floor. The heat transfer resistance of a wooden floor is approximately 1.85 m2°C/W. Having made similar calculations, we obtain the heat release:
Q floor = (17.2°C / 1.85m 2 °C/W) * 75 2 = 0.7 kW
Doors and windows. Their heat transfer resistance is approximately equal to 0.21 m 2 °C/W (double wooden door) and 0.5 m 2 °C/W (typical double-glazed window, without additional energy-efficient “bells and whistles”). As a result, we get the heat release:
Q door = (17.2°C / 0.21W/m2°C) * 1.8m2 = 0.15 kW
Q window = (17.2°C / 0.5m 2 °C/W) * 36m 2 = 1.25 kW
Ventilation. According to building codes, the air exchange coefficient for a residential premises must be at least 0.5, and better - 1, i.e. Within an hour, the air in the room should be completely renewed. Thus, with a ceiling height of 2.5 m, this is approximately 2.5 m 3 of air per hour per square meter area. This air must be heated from outside temperature(+5.8°C) to room temperature (+23°C).
The specific heat capacity of air is the amount of heat required to increase the temperature of 1 kg of a substance by 1°C - equal to approximately 1.01 kJ/kg°C. In this case, the air density in the temperature range of interest to us is approximately 1.25 kg/m 3, i.e. the mass of 1 cubic meter is 1.25 kg. Thus, to heat the air by 23-5.8 = 17.2°C for each square meter of area you will need:
1.01 kJ/kg°C * 1.25 kg/m 3 * 2.5 m 3 /hour * 17.2°C = 54.3 kJ/hour
For a house with an area of 150 m2 it will be:
54.3 * 150 = 8145 kJ/hour = 2.26 kW
| Heat loss through | Temperature difference, °C | Area, m2 |
Heat transfer resistance, m2°C/W |
Heat loss, kW |
| Walls |
17,2 |
175 |
0,41 |
5,83 |
| Roof |
17,2 |
87 |
1,7 |
0,88 |
| Floor |
17,2 |
75 |
1,85 |
0,7 |
| Doors |
17,2 |
1,8 |
0,21 |
0,15 |
| Window |
17,2 |
36 |
0,5 |
0,24 |
| Ventilation |
17,2 |
- |
- |
2,26 |
|
Total: |
|
|
|
11,06 |
Now let's breathe!
Let's assume that a family of two adults with two children lives in a house. The nutritional norm for an adult is 2600-3000 calories per day, which is equivalent to a heat output power of 126 W. We will estimate the heat release of a child to be half the heat release of an adult. If everyone living at home is in it 2/3 of the time, then we get:
(2*126 + 2*126/2)*2/3 = 252 W
Let’s assume that there are 5 rooms in the house, illuminated by ordinary 60 W incandescent lamps (not energy-saving), 3 per room, which are turned on for an average of 6 hours a day (i.e. 1/4 of the total time). Approximately 85% of the power consumed by the lamp turns into heat. In total we get:
5*60*3*0.85*1/4 = 191 W
Refrigerator - very efficient heating device. Its heat dissipation is 30% of the maximum power consumption, i.e. 750 W.
Other household appliances (let them be washing and dishwasher) releases about 30% of the maximum power consumption as heat. The average power of these devices is 2.5 kW, they operate approximately 2 hours a day. In total we get 125 W.
A standard electric stove with oven has a power of approximately 11 kW, but the built-in limiter regulates the operation of the heating elements so that their simultaneous consumption does not exceed 6 kW. However, it is unlikely that we ever use more than half of the burners at the same time or all the oven heating elements at once. Therefore, we will assume that the average operating power of the stove is approximately 3 kW. If it works for 3 hours a day, we get 375 W of heat.
Each computer (and there are 2 of them in the house) produces approximately 300 W of heat and operates 4 hours a day. Total - 100 W.
TV is 200 W and 6 hours a day, i.e. per circle - 50 W.
In total we get: 1.84 kW.
Now let's calculate the required thermal power heating systems:
Heating Q = 11.06 - 1.84 = 9.22 kW
Heating costs
Actually, above we calculated the power that will be needed to heat the coolant. And we will heat it, naturally, using a boiler. Thus, heating costs are the fuel costs for this boiler. Since we are considering the most general case, we will make a calculation for the most universal liquid (diesel) fuel, because Gas mains are not available everywhere (and the cost of connecting them is a figure with 6 zeros), and solid fuel you need, firstly, to bring it somehow, and secondly, to throw it into the boiler firebox every 2-3 hours.
To find out what volume V of diesel fuel per hour we will have to burn to heat the house, we need its specific heat of combustion q (the amount of heat released when burning a unit of mass or volume of fuel, for diesel fuel - approximately 13.95 kW*h/l) multiplied by Boiler efficiency η (approximately 0.93 for diesel engines) and then divide the required heating system power Qheating (9.22 kW) by the resulting figure:
V = Q heating /(q*η) = 9.22 kW / (13.95 kW*h/l) * 0.93) = 0.71 l/h
With the average cost of diesel fuel for the Moscow region being 30 rubles/l per year, it will take us
0.71 * 30 rub. * 24 hours * 365 days = 187 thousand rubles. (rounded) .
How to save money?
The natural desire of any homeowner is to reduce heating costs even at the construction stage. Where does it make sense to invest money?
First of all, you should think about insulating the facade, which, as we saw earlier, accounts for the bulk of all heat loss in the house. In general, external or internal additional insulation can be used for this. However internal insulation much less effective: when installing thermal insulation from the inside, the interface between the warm and cold areas “moves” inside the house, i.e. Moisture will condense in the thickness of the walls.
There are two ways to insulate facades: “wet” (plaster) and by installing a suspended ventilated facade. Practice shows that due to the need for constant repairs, “wet” insulation, taking into account operating costs, ends up being almost twice as expensive as a ventilated facade. The main disadvantage of a plaster facade is high price its maintenance and upkeep. " The initial costs for arranging such a facade are lower than for a ventilated curtain wall, by only 20-25%, maximum by 30%,- explains Sergey Yakubov (“Metal Profile”). - However, taking into account the costs of current repairs, which need to be done at least once every 5 years, after the first five-year period, the cost of a plaster facade will be equal to that of a ventilated façade, and over 50 years (the service life of a ventilated façade) it will be 4-5 times more expensive.».
What is a hinged ventilated facade? This is an external “screen” attached to the lung metal frame, which is attached to the wall with special brackets. Between the wall of the house and the screen, a lightweight insulation is placed (for example, Isover “VentFacade Bottom” with a thickness of 50 to 200 mm), as well as a wind and waterproof membrane (for example, Tyvek Housewrap). As external cladding can be used various materials, but in individual construction Steel siding is most often used. " The use of modern high-tech materials in the production of siding, such as steel coated with Colorcoat Prisma™, allows you to choose almost any design solution, - says Sergey Yakubov. - This material has excellent resistance to both corrosion and mechanical stress. The warranty period for it is 20 years with a real service life of 50 years or more. Those. provided that steel siding is used, all façade design will last 50 years without repair».
Additional layer facade insulation made of mineral wool has a heat transfer resistance of approximately 1.7 m2°C/W (see above). In construction, to calculate the heat transfer resistance of a multilayer wall, add up the corresponding values for each layer. As we remember, our main bearing wall 2 bricks have a heat transfer resistance of 0.405 m2°C/W. Therefore, for a wall with a ventilated facade we get:
0.405 + 1.7 = 2.105 m 2 °C/W
Thus, after insulation, the heat release of our walls will be
Q facade = (17.2°C / 2.105m 2 °C/W) * 137.2 m 2 = 1.12 kW,
which is 5.2 times less than the same indicator for a non-insulated facade. Impressive, isn't it?
Let us again calculate the required thermal power of the heating system:
Heating Q-1 = 6.35 - 1.84 = 4.51 kW
Diesel fuel consumption:
V 1 = 4.51 kW / (13.95 kW*h/l) * 0.93) = 0.35 l/h
Heating amount:
0.35 * 30 rub. * 24 hours * 365 days = 92 thousand rubles.
Reinforced concrete Concrete on gravel or crushed stone made from natural stone Dense silicate concrete Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=1800 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=1600 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=1400 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=1200 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=1000 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=800 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=600 Expanded clay concrete on expanded clay. sand and expanded clay foam concrete P=500 Expanded clay concrete on quartz sand with porous P=1200 Expanded clay concrete on quartz sand with porous P=1000 Expanded clay concrete on quartz sand with porous P=800 Perlite concrete P=1200 Perlite concrete P=1000 Perlite concrete P=800 Perlite concrete P=600 Agloporite concrete and concretes on fuel slag P=1800 Agloporite concrete and concrete on fuel slag P=1600 Agloporite concrete and concrete on fuel slag P=1400 Agloporite concrete and concrete on fuel slag P=1200 Agloporite concrete and concrete on fuel slag P=1000 Concrete on ash gravel P= 1400 Concrete on ash gravel P=1200 Concrete on ash gravel P=1000 Polystyrene concrete P=600 Polystyrene concrete P=500 Gas and foam concrete. gas and foam silicate P=1000 Gas and foam concrete. gas and foam silicate P=900 Gas and foam concrete. gas and foam silicate P=800 Gas and foam concrete. gas and foam silicate P=700 Gas and foam concrete. gas and foam silicate P=600 Gas and foam concrete. gas and foam silicate P=500 Gas and foam concrete. gas and foam silicate P=400 Gas and foam concrete. gas and foam silicate P=300 Gas and foam ash concrete P=1200 Gas and foam ash concrete P=100 Gas and foam ash concrete P=800 Cement-sand mortar Complex (sand, lime, cement) mortar Lime-sand mortar Cement-slag mortar P =1400 Cement-slag mortar P=1200 Cement-perlite mortar P=1000 Cement-perlite mortar P=800 Gypsum perlite mortar Porous gypsum perlite mortar P=500 Porous gypsum perlite mortar P=400 Gypsum slabs P=1200 Gypsum slabs P=1000 Sheets gypsum cladding (dry plaster) Clay ordinary brick Sand-lime brick P=2000 Sand-lime brick P=1900 Sand-lime brick P=1800 Sand-lime brick P=1700 Sand-lime brick P=1600 Ceramic brick P=1600 Ceramic brick P=1400 Ceramic stone P=1700 Thickened silicate brick P=1600 Thickened silicate brick P=1400 Silicate stone P=1400 Silicate stone P=1300 Granite. gneiss and basalt Marble Limestone P=2000 Limestone P=1800 Limestone P=1600 Limestone P=1400 Tuff P=2000 Tuff P=1800 Tuff P=1600 Tuff P=1400 Tuff P=1200 Tuff P=1000 Pine and spruce across the grain Pine and spruce along the grain Oak along the grain Oak along the grain Glued plywood Facing cardboard Multilayer construction cardboard Wood fiber boards. and wood chips, wood fibers. P=1000 Wood fiber boards. and wood chips, wood fibers. P=800 Wood fiber boards. and wood chips, wood fibers. P=400 Wood fiber boards. and wood chips, wood fibers. P=200 Fiberboard and wood concrete slabs on Portland cement P=800 Fiberboard and wood concrete slabs on Portland cement P=600 Fiberboard slabs and wood concrete on Portland cement P=400 Fiberboard and wood concrete slabs on Portland cement P=300 Fiber thermal insulation slabs made from artificial fur waste P=175 Plates fiber heat-insulating slabs made from artificial fur waste P=150 Fiber heat-insulating slabs made from faux fur waste P=125 Flax insulating slabs Peat heat-insulating slabs P=300 Peat heat-insulating slabs P=200 Tow Stitched mineral wool mats P=125 Stitched mineral wool mats P=100 Mineral wool mats new pierced Р=75 Stitched mineral wool mats Р=50 Mineral wool slabs with synthetic binder Р=250 Mineral wool slabs with synthetic binder Р=200 Mineral wool slabs with synthetic binder Р=175 Mineral wool slabs with synthetic binder Р=125 Mineral wool slabs with synthetic binder Р=75 Slabs polystyrene foam P=50 Polystyrene foam slabs P=35 Polystyrene foam slabs P=25 Polystyrene foam slabs P=15 Polyurethane foam P=80 Polyurethane foam P=60 Polyurethane foam P=40 Resol-phenol-formaldehyde foam slabs P=100 Resol-phenol-formaldehyde foam slabs P= 75 Plates made of resol-phenol-formaldehyde foam P =50 Resol-phenol-formaldehyde foam slabs P=40 Polystyrene concrete thermal insulation slabs P=300 Polystyrene concrete thermal insulation slabs P=260 Polystyrene concrete thermal insulation slabs P=230 Expanded clay gravel P=800 Expanded clay gravel P=600 Expanded clay gravel P=400 Grav expanded clay P=300 Expanded clay gravel P =200 Crushed stone and sand from expanded perlite P=600 Crushed stone and sand from expanded perlite P=400 Crushed stone and sand from expanded perlite P=200 Sand for construction work Foam glass and gas glass P=200 Foam glass and gas glass P=180 Foam glass and gas glass P=160 Flat asbestos-cement sheets P=1800 Flat asbestos-cement sheets P=1600 Petroleum construction and roofing bitumens P=1400 Petroleum construction and roofing bitumens P=1200 Petroleum construction and roofing bitumens roofing P=1000 Asphalt concrete Products made of expanded perlite on a bitumen binder P=400 Products made of expanded perlite on a bitumen binder P=300 Ruberoid. glassine roofing felt Multilayer polyvinyl chloride linoleum P=1800 Multilayer polyvinyl chloride linoleum P=1600 Fabric backed polyvinyl chloride linoleum P=1800 Fabric backed polyvinyl chloride linoleum P=1600 Fabric backed polyvinyl chloride linoleum P=1400 Reinforcing rod steel Cast iron Aluminum Copper Window glassConventionally, heat loss in a private home can be divided into two groups:
Below we present to your attention 15 examples of such “leaks”. These are real problems that are most often encountered in private homes. You will see what problems may be present in your home and what you should pay attention to.
Insulation does not work as effectively as it could. The thermogram shows that the temperature on the wall surface is distributed unevenly. That is, some areas of the wall heat up more than others (the brighter the color, the higher the temperature). This means that the heat loss is no greater, which is not correct for an insulated wall.
In this case, the bright areas are an example of ineffective insulation. It is likely that the foam in these places is damaged, poorly installed or missing altogether. Therefore, after insulating a building, it is important to make sure that the work is done efficiently and that the insulation works effectively.
Joint between wooden beam And mineral wool not compacted enough. This causes the insulation to not work effectively and causes additional heat loss through the roof that could be avoided.
One of the reasons why the house is cold is that some sections of the radiator do not heat up. This can be caused by several reasons: construction debris, air accumulation or manufacturing defects. But the result is the same - the radiator works at half its capacity heating power and does not warm the room enough.
Another example of inefficient radiator operation.
There is a radiator installed inside the room, which heats up the wall very much. As a result, part of the heat it generates goes outside. In fact, the heat is used to warm the street.
The underfloor heating pipe is laid close to external wall. The coolant in the system is cooled more intensively and has to be heated more often. The result is an increase in heating costs.
There are often cracks in windows that appear due to:
Cold air constantly enters the room through the cracks, causing drafts that are harmful to health and increasing heat loss in the building.
Also, cracks appear in balconies and entrance doors.
“Cold bridges” are areas of a building with lower thermal resistance compared to other areas. That is, they transmit more heat. For example, these are corners, concrete lintels above windows, junction points building structures and so on.
Why are cold bridges harmful?
Ventilation works in reverse. Instead of removing air from the room to the outside, cold street air is drawn into the room from the street. This, as in the example with windows, provides drafts and cools the room. In the example given, the temperature of the air that enters the room is -2.5 degrees, at a room temperature of ~20-22 degrees.
And in this case, the cold enters the room through the hatch into the attic.
Cold flow into the room through the air conditioner mounting hole.
The thermogram shows “heat bridges” associated with the use of materials with weaker resistance to heat transfer during the construction of the wall.
Often when insulating the wall of a building, they forget about another important area - the foundation. Heat loss also occurs through the foundation of the building, especially if the building has basement or a heated floor is installed inside.
Masonry joints between bricks are numerous cold bridges and increase heat loss through the walls. The example above shows that the difference between minimum temperature(masonry joint) and maximum (brick) is almost 2 degrees. Thermal resistance the walls have been lowered.
Cold bridge and air leak under the ceiling. It occurs due to insufficient sealing and insulation of the joints between the roof, wall and floor slab. As a result, the room is additionally cooled and drafts appear.
All this typical mistakes, which are found in most private homes. Many of them can be easily eliminated and can significantly improve the energy status of the building.
Let's list them again:
15 hidden heat leaks in a private home that you didn't know about
