Heat losses are determined for heated rooms 101, 102, 103, 201, 202 according to the floor plan.
Main heat losses, Q (W), are calculated using the formula:
Q = K × F × (t int - t ext) × n,
where: K – heat transfer coefficient of the enclosing structure;
F – area of enclosing structures;
n – coefficient taking into account the position of the enclosing structures in relation to the outside air, taken according to table. 6 “Coefficient that takes into account the dependence of the position of the enclosing structure in relation to the outside air” SNiP 02/23/2003 “Thermal protection of buildings”. For covering over cold basements and attic floors according to clause 2 n = 0.9.
General heat loss
According to clause 2a adj. 9 SNiP 2.04.05-91* additional heat loss is calculated depending on the orientation: walls, doors and windows facing north, east, northeast and northwest in the amount of 0.1, to the southeast and west - in the amount 0.05; in corner rooms additionally - 0.05 for each wall, door and window facing north, east, north-east and north-west.
According to paragraph 2d adj. 9 SNiP 2.04.05-91* additional heat loss for double doors with vestibules between them are taken equal to 0.27 H, where H is the height of the building.
Heat loss due to infiltration for residential premises, according to app. 10 SNiP 2.04.05-91* “Heating, ventilation and air conditioning”, adopted according to the formula
Q i = 0.28 × L × p × c × (t int - t ext) × k,
where: L is the consumption of exhaust air, not compensated by supply air: 1 m 3 / h per 1 m 2 of living space and kitchen area with a volume of more than 60 m 3;
c – specific heat capacity of air equal to 1 kJ / kg × °C;
p – density of outside air at t ext equal to 1.2 kg / m 3;
(t int - t ext) – difference between internal and external temperatures;
k – heat transfer coefficient – 0.7.
Q 101 = 0.28 × 108.3 m 3 × 1.2 kg / m 3 × 1 kJ / kg × °C × 57 × 0.7 = 1452,5 W,
Q 102 = 0.28 × 60.5 m 3 × 1.2 kg / m 3 × 1 kJ / kg × °C × 57 × 0.7 = 811,2 W,
Domestic heat gains are calculated at the rate of 10 W/m2 of the floor surface of residential premises.
Estimated heat loss of the room defined as Q calc = Q + Q i - Q life
Sheet for calculating heat loss in premises
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№ premises |
The name of a room |
Name of the enclosing structure |
Room orientation |
Fencing sizeF, m 2 |
Fencing area (F), m 2 |
Heat transfer coefficient, kW/m 2 ° C |
t vn - t nar , ° C |
Coefficient,n |
Main heat losses (Q basic ),W |
Additional heat loss % |
Additive factor |
Total heat loss, (Q generally ), W |
Heat consumption for infiltration, (Q i ), W |
Household heat input, W |
Estimated heat losses, (Q calc. ), W |
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|
For orientation |
other |
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Residential room |
Σ 1138,4 | |||||||||||||||
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Residential room |
Σ 474,3 | |||||||||||||||
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Residential room |
Σ 1161,4 | |||||||||||||||
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Residential room |
Σ 491,1 | |||||||||||||||
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staircase |
Σ 2225,2 | |||||||||||||||
NS – external wall, DO – double glazing, PL – floor, PT – ceiling, NDD – external double door with vestibule
The first step in organizing the heating of a private home is calculating heat loss. The purpose of this calculation is to find out how much heat escapes outward through walls, floors, roofing and windows (commonly known as building envelopes) during the most severe frosts in a given area. Knowing how to calculate heat loss according to the rules, you can get quite exact result and begin selecting a heat source based on power.
Basic formulas
To get a more or less accurate result, you need to perform calculations according to all the rules; a simplified method (100 W of heat per 1 m² of area) will not work here. The total heat loss of a building during the cold season consists of 2 parts:
- heat loss through enclosing structures;
- losses of energy used for heating ventilation air.
The basic formula for calculating the thermal energy consumption through external fences is as follows:
Q = 1/R x (t in - t n) x S x (1+ ∑β). Here:
- Q is the amount of heat lost by a structure of one type, W;
- R - thermal resistance of the construction material, m²°C / W;
- S—external fence area, m²;
- t in — internal air temperature, °C;
- t n - lowest temperature environment, °C;
- β - additional heat loss, depending on the orientation of the building.
The thermal resistance of the walls or roof of a building is determined based on the properties of the material from which they are made and the thickness of the structure. To do this, use the formula R = δ / λ, where:
- λ—reference value of the thermal conductivity of the wall material, W/(m°C);
- δ is the thickness of the layer of this material, m.
If a wall is built from 2 materials (for example, brick with mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summed up. Outdoor temperature is selected according to regulatory documents, and according to personal observations, internal - as necessary. Additional heat losses are coefficients determined by the standards:
- When a wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
- If the structure faces southeast or west, β = 0.05.
- β = 0 when the outer fence faces the south or southwest.
Calculation order
To take into account all the heat leaving the house, it is necessary to calculate the heat loss of the room, each separately. To do this, measurements are taken of all fences adjacent to the environment: walls, windows, roof, floor and doors.
Important point: measurements should be taken according to outside, capturing the corners of the building, otherwise the calculation of the heat loss of the house will give an underestimated heat consumption.
Windows and doors are measured by the opening they fill.
Based on the measurement results, the area of each structure is calculated and substituted into the first formula (S, m²). The value R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity coefficient building material. In the case of new windows made of metal-plastic, the R value will be told to you by a representative of the installer.
As an example, it is worth calculating heat loss through enclosing walls made of brick 25 cm thick, with an area of 5 m² at an ambient temperature of -25°C. It is assumed that the temperature inside will be +20°C, and the plane of the structure faces north (β = 0.1). First you need to take from reference books thermal conductivity coefficient of brick (λ), it is equal to 0.44 W/(m°C). Then, using the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:
R = 0.25 / 0.44 = 0.57 m²°C / W
To determine the heat loss of a room with this wall, all initial data must be substituted into the first formula:
Q = 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) = 434 W = 4.3 kW
If the room has a window, then after calculating its area, the heat loss through the translucent opening should be determined in the same way. The same actions are repeated regarding floors, roofing and front door. At the end, all the results are summed up, after which you can move on to the next room.
Heat metering for air heating
When calculating the heat loss of a building, it is important to take into account the amount of thermal energy consumed by the heating system to heat the ventilation air. The share of this energy reaches 30% of total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss of a house through the heat capacity of the air using a popular formula from a physics course:
Q air = cm (t in - t n). In it:
- Q air - heat consumed by the heating system to warm up the supply air, W;
- t in and t n - the same as in the first formula, °C;
- m is the mass flow of air entering the house from outside, kg;
- c is the heat capacity of the air mixture, equal to 0.28 W / (kg °C).
Here all quantities are known, except for the mass air flow rate during ventilation of premises. In order not to complicate your task, you should agree to the condition that the air environment in the entire house is renewed once an hour. Then the volumetric air flow rate can be easily calculated by adding the volumes of all rooms, and then you need to convert it into mass air flow through density. Since the density of the air mixture changes depending on its temperature, you need to take the appropriate value from the table:
m = 500 x 1.422 = 711 kg/h
Heating such a mass of air by 45°C will require the following amount of heat:
Q air = 0.28 x 711 x 45 = 8957 W, which is approximately equal to 9 kW.
At the end of the calculations, the results of heat losses through external fences are summed up with ventilation heat losses, which gives the total thermal load to the building's heating system.
The presented calculation methods can be simplified if the formulas are entered into Excel program in the form of tables with data, this will significantly speed up the calculation.
Below is a pretty simple one heat loss calculation buildings, which, however, will help to accurately determine the power required to heat your warehouse, shopping center or other similar building. This will make it possible, even at the design stage, to preliminarily estimate the cost of heating equipment and subsequent heating costs, and, if necessary, adjust the project.
Where does the heat go? Heat escapes through walls, floors, roofs and windows. In addition, heat is lost during ventilation of rooms. To calculate heat loss through building envelopes, use the formula:
Q – heat loss, W
S – structure area, m2
T – temperature difference between indoor and outdoor air, °C
R – value of thermal resistance of the structure, m2 °C/W
The calculation scheme is as follows: we calculate heat loss individual elements, sum up and add heat loss during ventilation. All.
Suppose we want to calculate the heat loss for the object shown in the figure. The height of the building is 5...6 m, width - 20 m, length - 40 m, and thirty windows measuring 1.5 x 1.4 meters. Room temperature 20 °C, external temperature -20 °C.
We calculate the areas of enclosing structures:
floor: 20 m * 40 m = 800 m2
roof: 20.2 m * 40 m = 808 m2
window: 1.5 m * 1.4 m * 30 pcs = 63 m2
walls:(20 m + 40 m + 20 m + 40 m) * 5 m = 600 m2 + 20 m2 (accounting pitched roof) = 620 m2 – 63 m2 (windows) = 557 m2
Now let's see thermal resistance materials used.
The value of thermal resistance can be taken from the table of thermal resistances or calculated based on the value of the thermal conductivity coefficient using the formula:
R – thermal resistance, (m2*K)/W
? – coefficient of thermal conductivity of the material, W/(m2*K)
d – material thickness, m
The value of thermal conductivity coefficients for different materials you can see .
floor: concrete screed 10 cm and mineral wool with a density of 150 kg/m3. 10 cm thick.
R (concrete) = 0.1 / 1.75 = 0.057 (m2*K)/W
R (mineral wool) = 0.1 / 0.037 = 2.7 (m2*K)/W
R (floor) = R (concrete) + R (mineral wool) = 0.057 + 2.7 = 2.76 (m2*K)/W
roof:
R (roof) = 0.15 / 0.037 = 4.05 (m2*K)/W
window: The thermal resistance value of windows depends on the type of double-glazed window used
R (windows) = 0.40 (m2*K)/W for single-chamber glass 4–16–4 at? T = 40 °C
walls: panels from mineral wool 15 cm thick
R (walls) = 0.15 / 0.037 = 4.05 (m2*K)/W
Let's calculate the heat losses:
Q (floor) = 800 m2 * 20 °C / 2.76 (m2*K)/W = 5797 W = 5.8 kW
Q (roof) = 808 m2 * 40 °C / 4.05 (m2*K)/W = 7980 W = 8.0 kW
Q (windows) = 63 m2 * 40 °C / 0.40 (m2*K)/W = 6300 W = 6.3 kW
Q (walls) = 557 m2 * 40 °C / 4.05 (m2*K)/W = 5500 W = 5.5 kW
We find that the total heat loss through the enclosing structures will be:
Q (total) = 5.8 + 8.0 + 6.3 + 5.5 = 25.6 kW/h
Now about ventilation losses.
To heat 1 m3 of air from a temperature of – 20 °C to + 20 °C, 15.5 W will be required.
Q(1 m3 of air) = 1.4 * 1.0 * 40 / 3.6 = 15.5 W, here 1.4 is the air density (kg/m3), 1.0 is the specific heat capacity of air (kJ/( kg K)), 3.6 – conversion factor to watts.
It remains to determine the amount of air required. It is believed that during normal breathing a person needs 7 m3 of air per hour. If you use the building as a warehouse and 40 people work on it, then you need to heat 7 m3 * 40 people = 280 m3 of air per hour, this will require 280 m3 * 15.5 W = 4340 W = 4.3 kW. And if you have a supermarket and on average there are 400 people on the territory, then heating the air will require 43 kW.
Final result:
To heat the proposed building, a heating system of about 30 kW/h is required, and a ventilation system with a capacity of 3000 m3/h with a heater power of 45 kW/h.
Of course, the main sources of heat loss in a house are doors and windows, but when viewing the picture through a thermal imager screen, it is easy to see that these are not the only sources of leakage. Heat is also lost through poorly installed roofs, cold floors, and uninsulated walls. Heat loss at home today is calculated using a special calculator. This allows you to select best option heating and holding additional work for building insulation. It is interesting that for each type of building (made of timber, logs, the level of heat loss will be different. Let's talk about this in more detail.
Basics of calculating heat loss
Control of heat loss is systematically carried out only for rooms heated in accordance with the season. Premises not intended for seasonal living do not fall under the category of buildings amenable to thermal analysis. The home heat loss program in this case will have no practical significance.
To spend full analysis, calculate thermal insulation materials and choose a heating system with optimal power, you need to have knowledge of the real heat loss of your home. Walls, roof, windows and floors are not the only sources of energy leakage from a home. Most of heat leaves the room through improperly installed ventilation systems.
Factors influencing heat loss
The main factors influencing the level of heat loss are:
- High level of temperature difference between the internal microclimate of the room and the temperature outside.
- Character thermal insulation properties enclosing structures, which include walls, ceilings, windows, etc.
Heat loss measurement values
Enclosing structures perform a barrier function for heat and do not allow it to freely escape outside. This effect is explained by the thermal insulation properties of the products. The quantity used to measure thermal insulation properties is called heat transfer resistance. This indicator is responsible for reflecting the temperature difference when the nth amount of heat passes through a section of fencing structures with an area of 1 m2. So, let’s figure out how to calculate the heat loss of a house.
The main quantities necessary to calculate the heat loss of a house include:
- q is a value indicating the amount of heat leaving the room to the outside through 1 m 2 of the barrier structure. Measured in W/m2.
- ∆T is the difference between the temperature in the house and outside. It is measured in degrees (o C).
- R - heat transfer resistance. It is measured in °C/W/m² or °C·m²/W.
- S is the area of the building or surface (used as needed).
Formula for calculating heat loss
The home heat loss program is calculated using a special formula:
When making calculations, remember that for structures consisting of several layers, the resistance of each layer is summed up. So, how to calculate heat loss frame house lined with brick on the outside? The resistance to heat loss will be equal to the sum of the resistance of brick and wood, taking into account air gap between layers.
Important! Please note that the resistance calculation is carried out for the coldest time of the year, when the temperature difference reaches its peak. Reference books and manuals always indicate exactly this reference value, which is used for further calculations.
Features of calculating heat loss of a wooden house
The calculation of heat loss in a house, the features of which must be taken into account when calculating, is carried out in several stages. The process requires special attention and concentration. You can calculate heat loss in a private house using a simple scheme like this:
- Determined through the walls.
- Calculated through window structures.
- Through doorways.
- Calculations are made through the floors.
- Calculate heat loss wooden house through the floor covering.
- Add the previously obtained values.
- Taking into account thermal resistance and energy loss through ventilation: from 10 to 360%.
For the results of points 1-5, the standard formula for calculating the heat loss of a house (made of timber, brick, wood) is used.
Important! Thermal resistance for window designs taken from SNIP II-3-79.
Construction reference books often contain information in a simplified form, that is, the results of calculating the heat loss of a house made of timber are given for different types walls and ceilings. For example, they calculate the resistance at a temperature difference for atypical rooms: corner and not corner rooms, single- and multi-storey buildings.
The need to calculate heat loss
Arrangement comfortable home requires strict process control at each stage of the work. Therefore, the organization of the heating system, which is preceded by the choice of the method of heating the room itself, should not be overlooked. When working on building a house, you will have to devote a lot of time not only project documentation, but also the calculation of heat loss at home. If in the future you are going to work in the field of design, then the engineering skills of calculating heat loss will definitely be useful to you. So why not practice doing this work through experience and make a detailed calculation of heat loss for your own home.
Important! The choice of method and power of the heating system directly depends on the calculations you have made. If you calculate the heat loss indicator incorrectly, you risk freezing in cold weather or sweltering from the heat due to excessive heating of the room. It is necessary not only to choose the right device, but also to determine the number of batteries or radiators that can heat one room.
Estimation of heat loss using a calculated example
If you do not need to study the calculation of heat loss at home in detail, we will focus on the evaluation analysis and determination of heat loss. Sometimes errors occur during the calculation process, so it is better to add minimum value to estimated power heating system. In order to begin calculations, you need to know the resistance indicator of the walls. It differs depending on the type of material from which the building is made.
Resistance (R) for houses made of ceramic bricks(with a masonry thickness of two bricks - 51 cm) is equal to 0.73 °C m²/W. The minimum thickness with this value should be 138 cm. When using expanded clay concrete as a base material (with a wall thickness of 30 cm), R is 0.58 °C m²/W with a minimum thickness of 102 cm. wooden house or a timber building with a wall thickness of 15 cm and a resistance level of 0.83 °C m²/W is required minimum thickness at 36 cm.
Building materials and their resistance to heat transfer
Based on these parameters, you can easily carry out calculations. You can find resistance values in the reference book. In construction, bricks, timber or log frames, foam concrete, wooden floor, ceilings.
Heat transfer resistance values for:
- brick wall (2 bricks thick) - 0.4;
- timber frame (200 mm thick) - 0.81;
- log house (diameter 200 mm) - 0.45;
- foam concrete (thickness 300 mm) - 0.71;
- wooden floor - 1.86;
- ceiling overlap - 1.44.
Based on the information provided above, we can conclude that for correct calculation heat loss requires only two values: the temperature difference and the level of heat transfer resistance. For example, a house is made of wood (logs) 200 mm thick. Then the resistance is 0.45 °C m²/W. Knowing this data, you can calculate the percentage of heat loss. To do this, a division operation is performed: 50/0.45 = 111.11 W/m².
Calculation of heat loss by area is performed as follows: heat loss is multiplied by 100 (111.11*100=11111 W). Taking into account the decoding of the value (1 W=3600), we multiply the resulting number by 3600 J/hour: 11111*3600=39.999 MJ/hour. By carrying out such simple mathematical operations, any owner can find out about the heat loss of his home in an hour.
Calculation of heat loss in a room online
There are many sites on the Internet offering the service of online calculation of heat loss of a building in real time. The calculator is a program with a special form to fill out, where you enter your data and after automatic calculation you will see the result - a figure that will indicate the amount of heat released from the living space.
A residential building is a building in which people live throughout the entire heating season. As a rule, country houses, where the heating system operates periodically and as needed, do not fall into the category of residential buildings. To retool and achieve optimal mode heat supply, you will have to carry out a number of works and, if necessary, increase the power of the heating system. Such re-equipment may take a long period. In general, the whole process depends on design features home and indicators of increasing the power of the heating system.
Many have not even heard of the existence of such a thing as “heat loss at home,” and subsequently, having made constructive correct installation heating system, suffer all their lives from a lack or excess of heat in the house, without even realizing the real reason. That is why it is so important to take into account every detail when designing a home, to personally control and build it in order to ultimately get a high-quality result. In any case, a home, regardless of what material it is built from, should be comfortable. And such an indicator as the heat loss of a residential building will help make staying at home even more pleasant.