Designing a heating system “by eye” can most likely lead to either an unjustified increase in the costs of its operation or to underheating of the home.
To prevent either one or the other from happening, it is necessary first of all to correctly calculate the heat loss of the house.
And only based on the results obtained, the power of the boiler and radiators is selected. Our conversation will focus on how these calculations are made and what needs to be taken into account.
The authors of many articles reduce the calculation of heat loss to one simple action: It is proposed to multiply the area of the heated room by 100 W. The only condition, which extends in this case, refers to the ceiling height - 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 to “taken from the air.” After all, the specific amount of heat loss is influenced by a number of factors: the material of the enclosing structures, outside temperature, area and type of glazing, air exchange rate, etc.
Heat loss at home
Moreover, even for houses with different heated areas, all other things being equal, its value will be different: in small house- more, in big - 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. This skill will not only allow you to select 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 its payback period.
The first thing the contractor needs to do is to decompose the total heat loss into three components:
- losses through enclosing structures;
- caused by 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.
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Heat loss calculation
Here's how to do the calculations:
Heat loss through building envelopes
For each material included in the building envelope, 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 enclosing structures, we determine the thermal resistance using the formula: R = S/Kt, where S is the thickness of this layer, m.
For multilayer structures, the resistances of all layers must be added together.
We determine heat loss for each structure using the formula Q = (A / R) *dT,
- A is the area of the enclosing structure, sq. m;
- dT - difference between external and internal temperatures.
- dT should be determined for the coldest five-day period.
Heat loss through ventilation
For this part of the calculation, it is necessary to know the air exchange rate.
In residential buildings built 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 renewed in an hour.
In houses built using European technology (DIN standard), in which the walls are lined with vapor barrier from the inside, the air exchange rate has to be increased to 2. That is, in an hour the air in the room should be renewed twice.
We determine heat loss through ventilation using the formula:
Qv = (V*Kv / 3600) * p * s * dT,
- V - volume of the room, cubic meters. m;
- Kv - air exchange rate;
- P - air density, taken equal to 1.2047 kg/cubic meter. m;
- C - specific heat capacity of air, taken equal to 1005 J/kg*C.
The above calculation allows us 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:
- lower the requirements for the level of comfort, that is, set the desired temperature in the coldest period at a minimum level, say, 18 degrees;
- during periods of severe cold, reduce the air exchange rate: minimum permissible capacity supply ventilation is 7 cu. m/h for each inhabitant of the house;
- provide for organization supply and exhaust ventilation with a recuperator.
Note that the recuperator is useful not only in winter, but also in summer: in hot weather it allows you to save the cold produced by the air conditioner, although at this time it does not work as efficiently as in cold weather.
When designing a house, it is best to perform zoning, that is, assign each room its own temperature based on the required comfort. For example, in a child’s room or an elderly person’s room, the temperature should be about 25 degrees, while for the living room 22 degrees will be enough. On a landing or in a room where residents rarely appear or where there are sources of heat generation, 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 amount of energy consumption during the cold season, the dT parameter must be calculated taking into account not the lowest, but the average temperature. Then you need to do the following:
W = ((Q + Qв) * 24 * N)/1000,
- W is the amount of energy required to replenish heat loss through building envelopes and ventilation, kWh;
- N is the number of days in the heating season.
However, this calculation will be incomplete if heat losses into the sewer system are not taken into account.
To perform hygiene procedures and wash dishes, residents of the house heat water and the heat produced goes into the sewer pipe.
But in this part of the calculation, it is necessary to take into account not only direct heating of water, but also indirect heating - heat is collected by water in the toilet tank and siphon, which is also discharged into the sewer.
Based on this, the average water heating temperature is assumed to be only 30 degrees. We calculate heat loss through the sewer using the following formula:
Qк = (Vв * T * р * с * 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 = 1000 kg/cubic. m;
- C is the heat capacity of water, we take c = 4183 J/kg*C;
- dT - temperature difference. Considering that the inlet water in winter has a temperature of about +7 degrees, and we agreed to consider the average temperature of the heated water to be equal to 30 degrees, we should take dT = 23 degrees.
- 3,600,000 - the number of joules (J) in 1 kWh.
An example of calculating heat loss at home
Let's calculate the heat loss of a 2-story building with a height of 7 m and dimensions of 10x10 m in plan.
The walls are 500 mm thick and are made of warm ceramics(Kt = 0.16 W/m*S), insulated from the outside mineral wool 50 mm thick (Kt = 0.04 W/m*S).
The house has 16 windows with an area of 2.5 square meters. m.
The outside temperature on the coldest five-day period is -25 degrees.
The average outside temperature during the heating period is (-5) degrees.
Inside the house it is required to ensure a temperature of +23 degrees.
Water consumption - 15 cubic meters. m/month
The duration of the heating period is 6 months.
We determine heat loss through the enclosing structures (for example, we will consider only the walls)
Thermal Resistance:
- base material: R1 = 0.5 / 0.16 = 3.125 sq. m*S/W;
- insulation: R2 = 0.05/0.04 = 1.25 sq. m*S/W.
The same for the wall as a whole: R = R1 + R2 = 3.125 + 1.25 = 4.375 sq. m*S/W.
We determine the area of the walls: A = 10 x 4 x 7 – 16 x 2.5 = 240 sq. m.
Heat loss through the walls will be:
Qс = (240 / 4.375) * (23 – (-25)) = 2633 W.
In a similar way, heat loss through the roof, floor, foundation, windows and entrance door is calculated, after which all the obtained values are summed up. Manufacturers usually indicate the thermal resistance of doors and windows in the product data sheet.
Please note that when calculating heat loss through the floor and foundation (if there is a basement), the temperature difference dT will be much smaller, since its calculation takes into account not the air temperature, but the soil temperature, which is much warmer in winter.
Heat loss through ventilation
We determine the volume of air in the room (to simplify the calculation, the thickness of the walls is not taken into account):
V = 10x10x7 = 700 cubic meters m.
Taking the air exchange rate Kv = 1, we determine the heat loss:
Qв = (700 * 1 / 3600) * 1.2047 * 1005 * (23 – (-25)) = 11300 W.
Ventilation in the house
Heat loss through sewerage
Taking into account the fact that residents consume 15 cubic meters. m of water per month, and billing period is 6 months, heat loss through the sewerage system will be:
Qk = (15 * 6 * 1000 * 4183 * 23) / 3,600,000 = 2405 kWh
If you don't live in country house In winter, in the off-season or in cold summer, it is still necessary to heat it. in this case it may be the most appropriate.
You can read about the reasons for the pressure drop in the heating system. Troubleshooting.
Estimation of the total amount of energy consumption
To estimate the entire volume of energy consumption during the heating period, it is necessary to recalculate heat loss through ventilation and enclosing structures, taking into account the average temperature, that is, dT will not be 48, but only 28 degrees.
Then the average power losses through the walls will be:
Qс = (240 / 4.375) * (23 – (-5)) = 1536 W.
Let's assume that an additional average of 800 W is lost through the roof, floor, windows and doors, then the total average power of heat loss through the building envelope will be Q = 1536 + 800 = 2336 W.
The average rate of heat loss through ventilation will be:
Qв = (700 * 1 / 3600) * 1.2047 * 1005 * (23 – (-5)) = 6592 W.
Then for the entire period you will have to spend on heating:
W = ((2336 + 6592)*24*183)/1000 = 39211 kWh.
To this value you need to add 2405 kWh of losses through the sewer system, so that the total 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 kWh of heat, then it will need 41616 / 9.45 = 4404 cubic meters. m.
Video on the topic
eeni2008
Let's look at how to calculate the heat loss of a house through the building envelope. The calculation is given using the example of a one-story residential building. This calculation can also be used to calculate the heat loss of a separate room, an entire house or an individual apartment.
An example of a technical specification for calculating heat loss
First, we draw up a simple house plan indicating the area of the premises, the size and location of windows and the front door. This is necessary to determine the surface area of the house through which heat loss occurs.
Formula for calculating heat loss
To calculate heat loss, we use the following formulas:
R=B/K- this is a formula for calculating the thermal resistance of the building envelope.
- R - thermal resistance, (m2*K)/W;
- K - coefficient of thermal conductivity of the material, W/(m*K);
- B - material thickness, m.
Q=S. dT/R- this is the formula for calculating heat loss.
- Q - heat loss, W;
- S - area of the building envelope, m2;
- dT - temperature difference between interior space and street, K;
- R - value thermal resistance structures, m2.K/W
To calculate the temperature inside the house, we take +21..+23°C - this mode is the most comfortable for a person. Minimum outside temperature To calculate heat loss, we took -30°C, since in winter period in the region: where the house was built (Yaroslavl region, Russia), such a temperature can last more than one week and it is recommended to include the lowest temperature indicator in the calculations, while the temperature difference is dT = 51..53, on average - 52 degrees.
The total heat loss of a house consists of the heat loss of all enclosing structures, therefore, using these formulas, we perform:
After the calculation we received the following data:
- Q walls - 0.49 kWh,
- Q ceiling - 0.49 kWh,
- Floor Q - 0.32 kWh,
- Q windows - 0.38 kWh.
- Q entrance door - 0.16 kWh.
Total: the total result of heat loss through the enclosing structures was 1.84 kWh.
To date heat saving is important parameter, which is taken into account when constructing a residential or office space. In accordance with SNiP 23-02-2003 “Thermal protection of buildings”, heat transfer resistance is calculated using one of two alternative approaches:
- Prescriptive;
- Consumer.
To calculate home heating systems, you can use the calculator for calculating heating and home heat loss.
Prescriptive Approach- these are the standards for individual elements of thermal protection of a building: external walls, floors above unheated spaces, coverings and attic floors, windows, entrance doors etc.
Consumer approach(heat transfer resistance can be reduced in relation to the prescribed level, provided that the design specific heat energy consumption for space heating is lower than the standard one).
Sanitary and hygienic requirements:
- The difference between indoor and outdoor air temperatures should not exceed certain permissible values. The maximum permissible temperature difference for an external wall is 4°C. for coating and attic floor 3°C and for covering over basements and crawl spaces 2°C.
- Temperature at inner surface fence must be above the dew point temperature.
For example: for Moscow and the Moscow region, the required thermal resistance of the wall according to the consumer approach is 1.97 °C m 2 /W, and according to the prescriptive approach:
- for home permanent residence 3.13 °C m 2 / W.
- for administrative and other public buildings, including structures seasonal residence 2.55 °C m 2 / W.
For this reason, when choosing a boiler or other heating devices solely according to those specified in their technical documentation parameters. You must ask yourself whether your house was built with strict regard to the requirements of SNiP 02/23/2003.
Therefore, for the right choice heating boiler power or heating devices, it is necessary to calculate real heat loss from your home. As a rule, a residential building loses heat through the walls, roof, windows, and ground; significant heat losses can also occur through ventilation.
Heat loss mainly depends on:
- temperature differences in the house and outside (the higher the difference, the higher the losses).
- heat-protective characteristics of walls, windows, ceilings, coatings.
Walls, windows, ceilings have a certain resistance to heat leakage, the heat-shielding properties of materials are assessed by a value called heat transfer resistance.
Heat transfer resistance will show how much heat will leak through square meter structures at a given temperature difference. This question can be formulated differently: what temperature difference will occur when a certain amount of heat passes through a square meter of fencing.
R = ΔT/q.
- q is the amount of heat that escapes through a square meter of wall or window surface. This amount of heat is measured in watts per square meter (W/m2);
- ΔT is the difference between the temperature outside and in the room (°C);
- R is the heat transfer resistance (°C/W/m2 or °C m2/W).
In cases where it concerns multilayer construction, then the resistance of the layers is simply summed up. For example, the resistance of a wall made of wood, which is lined with brick, is the sum of three resistances: brick and wooden wall and the air gap between them:
R(total)= R(wood) + R(air) + R(brick)
Temperature distribution and air boundary layers during heat transfer through a wall.
Heat loss calculation performed for the coldest period of the year, which is the frostiest and windiest week of the year. In construction literature, the thermal resistance of materials is often indicated based on this condition And climatic region(or outside temperature) where your home is located.
Heat transfer resistance table various materials
at ΔT = 50 °C (T external = -30 °C. T internal = 20 °C.)
|
Wall material and thickness |
Heat transfer resistance Rm.
|
|
Brick wall |
0.592 |
|
Log house Ø 25 |
0.550 |
|
Log house made of timber Thickness 20 centimeters |
0.806 |
|
Frame wall (board + |
|
|
Foam concrete wall 20 centimeters |
0.476 |
|
Plastering on brick, concrete. |
|
|
Ceiling (attic) floor |
|
|
Double wooden doors |
Window heat loss table various designs at ΔT = 50 °C (T external = -30 °C. T internal = 20 °C.)
|
Note
. Even numbers in symbol double glazed windows indicate air
gap in millimeters;
. The letters Ar mean that the gap is filled not with air, but with argon;
. The letter K means that the outer glass has a special transparent
heat-protective coating.
As can be seen from the above table, modern double-glazed windows make it possible reduce heat loss windows almost doubled. For example, for 10 windows measuring 1.0 m x 1.6 m, savings can reach up to 720 kilowatt-hours per month.
To correctly select materials and wall thickness, apply this information to a specific example.
Two quantities are involved in calculating heat losses per m2:
- temperature difference ΔT.
- heat transfer resistance R.
Let's say the room temperature is 20 °C. and the outside temperature will be -30 °C. In this case, the temperature difference ΔT will be equal to 50 °C. The walls are made of timber 20 centimeters thick, then R = 0.806 °C m 2 / W.
Heat losses will be 50 / 0.806 = 62 (W/m2).
To simplify calculations of heat loss in construction reference books indicate heat loss various types walls, ceilings, etc. for some values of winter air temperature. Typically, different numbers are given for corner rooms (the turbulence of the air that swells the house influences this) and non-angular, and also takes into account the difference in temperatures for the rooms of the first and upper floors.
Table of specific heat loss of building enclosure elements (per 1 m2 along the internal contour of the walls) depending on the average temperature of the coldest week of the year.
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Note. In the case when there is an external unheated room behind the wall (canopy, glazed veranda etc.), then the heat loss through it will be 70% of the calculated value, and if there is another outdoor room behind this unheated room, then the heat loss will be 40% of the calculated value.
Table of specific heat loss of building enclosure elements (per 1 m2 along the internal contour) depending on the average temperature of the coldest week of the year.
Example 1.
Corner room(1st floor)
Room characteristics:
- 1st floor.
- room area - 16 m2 (5x3.2).
- ceiling height - 2.75 m.
- There are two external walls.
- material and thickness of the external walls - timber 18 centimeters thick, covered with plasterboard and covered with wallpaper.
- windows - two (height 1.6 m, width 1.0 m) with double glazing.
- floors - wooden insulated. basement below.
- above the attic floor.
- estimated outside temperature -30 °C.
- required room temperature +20 °C.
- Area of external walls minus windows: S walls (5+3.2)x2.7-2x1.0x1.6 = 18.94 m2.
- Window area: S windows = 2x1.0x1.6 = 3.2 m2
- Floor area: S floor = 5x3.2 = 16 m2
- Ceiling area: Ceiling S = 5x3.2 = 16 m2
The area of the internal partitions is not included in the calculation, since the temperature on both sides of the partition is the same, therefore heat does not escape through the partitions.
Now let's calculate the heat loss of each surface:
- Q walls = 18.94x89 = 1686 W.
- Q windows = 3.2x135 = 432 W.
- Floor Q = 16x26 = 416 W.
- Ceiling Q = 16x35 = 560 W.
The total heat loss of the room will be: Q total = 3094 W.
It should be borne in mind that much more heat escapes through walls than through windows, floors and ceilings.
Example 2
Room under the roof (attic)
Room characteristics:
- top floor.
- area 16 m2 (3.8x4.2).
- ceiling height 2.4 m.
- external walls; two roof slopes (slate, continuous sheathing, 10 centimeters of mineral wool, lining). pediments (beams 10 centimeters thick covered with clapboard) and side partitions ( frame wall with expanded clay filling 10 centimeters).
- windows - 4 (two on each gable), 1.6 m high and 1.0 m wide with double glazing.
- estimated outside temperature -30°C.
- required room temperature +20°C.
- Area of the end external walls minus windows: S end walls = 2x(2.4x3.8-0.9x0.6-2x1.6x0.8) = 12 m2
- Area of roof slopes bordering the room: S sloped walls = 2x1.0x4.2 = 8.4 m2
- Area of side partitions: S side partition = 2x1.5x4.2 = 12.6 m 2
- Window area: S windows = 4x1.6x1.0 = 6.4 m2
- Ceiling area: Ceiling S = 2.6x4.2 = 10.92 m2
Next, we will calculate the heat losses of these surfaces, while it is necessary to take into account that in this case the heat will not escape through the floor, since there is a warm room below. Heat loss for walls We calculate as for corner rooms, and for the ceiling and side partitions we enter a 70 percent coefficient, since unheated rooms are located behind them.
- Q end walls = 12x89 = 1068 W.
- Q pitched walls = 8.4x142 = 1193 W.
- Q side burnout = 12.6x126x0.7 = 1111 W.
- Q windows = 6.4x135 = 864 W.
- Ceiling Q = 10.92x35x0.7 = 268 W.
The total heat loss of the room will be: Q total = 4504 W.
As we see, warm room 1st floor loses (or consumes) significantly less heat than attic room with thin walls and a large glazing area.
To make this room suitable for winter accommodation, it is necessary first of all to insulate the walls, side partitions and windows.
Any enclosing surface can be presented in the form of a multilayer wall, each layer of which has its own thermal resistance and its own resistance to air passage. By summing the thermal resistance of all layers, we get the thermal resistance of the entire wall. Also, if you sum up the resistance to the passage of air of all layers, you can understand how the wall breathes. The most best wall made of timber should be equivalent to a wall made of timber with a thickness of 15 - 20 centimeters. The table below will help with this.
Table of resistance to heat transfer and air passage of various materials ΔT = 40 ° C (T external = -20 ° C. T internal = 20 ° C.)
|
|
Thickness |
Resistance |
Resistance |
|
|
Equivalent |
||||
|
Brickwork from the usual 12 centimeters |
12 |
0.15 |
12 |
6 |
|
Masonry made of expanded clay concrete blocks 1000 kg/m3 |
1.0 |
75 |
17 |
|
|
Foam aerated concrete 30 cm thick 300 kg/m3 |
2.5 |
190 |
7 |
|
|
Thick timbered wall (pine) 10 centimeters |
10 |
0.6 |
45 |
10 |
To get a complete picture of the heat loss of the entire room, you need to take into account
- Heat loss through the contact of the foundation with frozen soil is usually assumed to be 15% of the heat loss through the walls of the first floor (taking into account the complexity of the calculation).
- Heat losses associated with ventilation. These losses are calculated taking into account building codes(SNiP). A residential building requires about one air change per hour, that is, during this time it is necessary to supply the same volume fresh air. Thus, the losses associated with ventilation will be slightly less than the amount of heat loss attributable to the enclosing structures. It turns out that heat loss through walls and glazing is only 40%, and heat loss for ventilation 50%. IN European standards ventilation and wall insulation, the heat loss ratio is 30% and 60%.
- If the wall “breathes”, like a wall made of timber or logs 15 - 20 centimeters thick, then heat returns. This allows you to reduce heat losses by 30%. therefore, the value of the thermal resistance of the wall obtained during the calculation must be multiplied by 1.3 (or, accordingly reduce heat loss).
By summing up all the heat loss at home, you can understand what power the boiler has and heating devices necessary for comfortable heating of the house on the coldest and windiest days. Also, such calculations will show where the “weak link” is and how to eliminate it using additional insulation.
You can also calculate heat consumption using aggregated indicators. So, in 1-2 storey houses that are not very insulated, at an outside temperature of -25 ° C, 213 W per 1 m 2 of total area is required, and at -30 ° C - 230 W. For well-insulated houses, this figure will be: at -25 °C - 173 W per m 2 of total area, and at -30 °C - 177 W.
During the cold period, when the indoor air temperature is much higher than the outdoor air temperature, heat flows (heat loss) occur through the building enclosure.
Heat loss in premises consists of two main components: transmission heat loss and heat consumption for heating the air infiltrated through leaks.
Transmission heat loss is the loss of heat through external enclosures due to heat transfer.
Transmission heat losses are found using the formulas:
where is heat loss, W;
Thermal resistance of the fence ()/W, determined thermotechnical calculation;
K - heat transfer coefficient of the fence W / (),
F is the surface area of the fence,
– design air temperature in the room, °C, table 2
Estimated outside air temperature equal to the average temperature of the coldest five-day period, °C, Table 3
N – correction factor to the calculated temperature difference;
Additional heat loss, W.
To calculate the surface areas F in formulas (1.24) and (1.25.), we are guided by the generally accepted methodology for determining the linear dimensions of the enclosing structure.
Rice. 2. Measurement of fences:
a – vertically; b – in plan; 1 – floor on the ground; 2- floor along joists; 3 – floor above the basement; O – windows; NS – external wall; Pl – floor; Fri – ceiling.
It is customary to determine the heat loss of the floor lying on the ground by zones. Each zone has its own thermal resistance.
; 4.3()/W;
The amount of heat loss through the i-th zone is found by the formula:
where is the resistance of the i-th zone, ()/ W;
– area of the i-th zone (area of a 2 m wide ring strip along the building’s contour). The area of zone I in the corners of the building is multiplied by 2.
Rice. 3. Heat flows from floors along the ground and buried walls:
a – through the floor; b – through a recessed wall; c – division of the floor into zones 1,2,3,4; d – division of the recessed shade and floor into zones 1,2,3,4.
Heat loss through the floors is obtained by summing the heat loss by zone
If the floors are laid on joists or on insulating material (have an air gap) and the thermal resistance of these additional elements The calculation method is preserved (in this case, the resistance of each zone increases by the amount of resistance of the underlying layers.)
The same technique is used to calculate heat loss through the walls of a building buried in the ground (heated basements).
The division into zones begins from the ground surface outside the building, the floors are considered as a continuation of the walls.
Additional heat loss is determined as follows:
1. Additions for orientation to the cardinal points are made to all vertical fences or vertical projections of inclined fences as follows:
N, N-W, N-E, E-10%; W, SE – 5%; S, S-W – 0%.
2. For the rush of cold air through external doors when they are opened briefly at a building height of N, m:
Double doors with vestibules – 27% of H;
The same without a vestibule - 34% of H;
Single doors – 22% of N.
3. For the floors of the first floor above the cold basements of buildings in areas with an estimated outside air temperature (five days) of minus 40 ° C and below, it is assumed to be equal to 5%.
By summing the transmission heat losses across all enclosures, we find the heat losses of the entire room.
CHAPTER 3. HEAT BALANCE OF ROOMS AND HEAT CONSUMES FOR HEATING BUILDINGS
Design power of heating systems
Thermal conditions can be constant or variable.
Constant - supported around the clock in residential, industrial buildings with continuous operation, children's and medical institutions, hotels, sanatoriums.
Variable - in industrial buildings with one- and two-shift work, administrative, commercial, educational buildings, service enterprises. During non-working hours, use the existing heating system, or standby heating - low temperature.
The heat balance is summarized in a form (Table 3.1).
Table 3.1. Heat balance form
If heat loss is greater than heat release, then heating is required.
Estimated thermal power of the heating system:
Q с,о = ∑Q sweat - ∑Q post, (3.1)
If in industrial building ∑Q post >∑Q sweat, then supply ventilation is arranged.
Heat loss through building envelopes
To determine heat loss you must have:
Floor plans with all building dimensions;
Copy from the general plan with the designation of the cardinal points and the wind rose;
The purpose of each room;
Geographical location of the building's construction;
Designs of all external fencing.
All rooms on the plans indicate:
Numbered from left to right, staircases are designated by letters or Roman numerals regardless of the floor and are considered as one room.
Heat loss in premises through enclosing structures, rounded up to 10 W:
Q limit = (F/R o)(t in – t n B)(1 + ∑β)n = kF(t in – t n B)(1 - ∑β)n,(3.2)
Where F, k, R o- design area, heat transfer coefficient, heat transfer resistance of the enclosing structure, m2, W/(m2 oC), (m2 oC)/W; t in- estimated room air temperature, o C; t n B- estimated outside air temperature (B) or air temperature in a colder room; n- coefficient taking into account the position of the outer surface of the enclosing structures in relation to the outside air (Table 2.4); β - additional heat losses in fractions of the main losses.
Heat exchange through fences between adjacent heated rooms is taken into account if the temperature difference in them is more than 3°C.
Square F, m2, fences (external walls (NS), windows (O), doors (D), lanterns (F), ceiling (Pt), floor (P)) are measured according to plans and sections of the building (Fig. 3.1).
1. Height of the walls of the first floor: if the floor is on the ground, between the floor levels of the first and second floors ( h 1); if the floor is on joists - from the external level of preparation of the floor on joists to the floor level of the second floor ( h 1 1); for an unheated basement or underground - from the level bottom surface floor structure of the first floor to the level of the finished floor of the second floor ( h 1 11), and in one-story buildings with an attic floor, the height is measured from the floor to the top of the insulating layer of the floor.
2. The height of the walls of the intermediate floor is between the levels of the finished floors of this and the overlying floors ( h 2), and the upper floor - from the level of its finished floor to the top of the insulating layer of the attic floor ( h 3) or roofless roofing.
3. The length of external walls in corner rooms - from the edge of the outer corner to the axes interior walls (l 1 And l 2l 3).
4. The length of the internal walls - from the internal surfaces of the external walls to the axes of the internal walls ( m 1) or between the axes of internal walls (T).
5. Areas of windows, doors and lanterns - according to the smallest dimensions of building openings in the light ( A And b).
6. The areas of ceilings and floors above basements and underground spaces in corner rooms - from the inner surface of the external walls to the axes of opposite walls ( m 1 And n), and in non-corner ones - between the axes of the internal walls ( T) and from the inner surface of the outer wall to the axis of the opposite wall ( n).
The error of linear dimensions is ±0.1 m, area error is ±0.1 m2.
Rice. 3.1. Measuring diagram for heat transfer fencing
Figure 3.2. Scheme for determining heat loss through floors and walls buried below ground level
1 - first zone; 2 – second zone; 3 – third zone; 4 – fourth zone (last).
Heat loss through the floors is determined by zone-strips 2 m wide, parallel to the external walls (Fig. 5.2).
Reduced heat transfer resistance R n.p., m 2 K/W, areas of uninsulated floors on the ground and walls below ground level, with thermal conductivity λ > 1.2 W/(m o C): for the 1st zone - 2.1; for zone 2 - 4.3; for the 3rd zone - 8.6; for the 4th zone (remaining floor area) - 14.2.
Formula (3.2) when calculating heat losses Q pl, W, through the floor located on the ground, takes the form:
Q pl = (F 1 / R 1n.p +F 2 / R 2n.p +F 3 / R 3n.p +F 4 / R 4n.p)(t in – t n B)(1 + ∑β) n,(3.3)
Where F 1 - F 4- area of 1 - 4 zone-strips, m2; R 1, n.p. - R 4, n.p.- heat transfer resistance of floor zones, m 2 K/W; n =1.
Heat transfer resistance of insulated floors on the ground and walls below ground level (λ< 1,2 Вт/(м· оС)) R y .п, m 2 o C/W, also determined for zones using the formula
R u.p = R n.p +∑(δ u.s. /λ u.s.),(3.4)
Where R n.a.- heat transfer resistance of non-insulated floor zones (Fig. 3.2), m 2 o C/W; sum of fraction- sum thermal resistances insulating layers, m 2 o C/W; δ у.с- thickness of the insulating layer, m.
Heat transfer resistance of floors on joists R l, m 2 o C/W:
R l.p = 1.18 (R n.p +∑(δ u.s. /λ u.s.)),(3.5)
The insulating layers are an air layer and a plank floor on joists.
When calculating heat losses, floor areas in the corners of external walls (in the first two-meter zone) are entered into the calculation twice in the direction of the walls.
Heat loss through the underground part of the external walls and the floors of the heated basement is also calculated in zones 2 m wide, counting them from ground level (see Fig. 3.2). Then the floors (when counting zones) are considered as a continuation of the underground part of the external walls. Heat transfer resistance is determined in the same way as for uninsulated or insulated floors.
Additional heat loss through fences. In (3.2) the term (1+∑β) takes into account additional heat losses as a fraction of the main heat losses:
1. On orientation in relation to the cardinal points. β external vertical and inclined (vertical projection) walls, windows and doors.
Rice. 3.3. Addition to the main heat loss depending on the orientation of the fences in relation to the cardinal points
2. For ventilation of rooms with two or more external walls. IN standard projects through walls, doors and windows facing all countries of the world β = 0.08 with one external wall and 0.13 for corner rooms and in all residential premises.
3. At the design temperature of the outside air. For unheated floors of the first floor above the cold undergrounds of buildings in areas with t n B minus 40°C and below - β = 0,05.
4. To heat the rushing cold air. For external doors, without air or air-thermal curtains, at building height N, m:
- β = 0,2N- for triple doors with two vestibules between them;
- β = 0,27 N - For double doors with a vestibule between them;
- β = 0,34 N - for double doors without vestibule;
- β = 0,22 N - for single doors.
For external non-equipped gates β =3 without vestibule and β = 1 - with a vestibule at the gate. For summer and emergency external doors and gates β = 0.
Heat losses through the building envelopes are entered in the form (Table 3.2).
Table 3.2. Form (form) for calculating heat loss
The area of the walls in the calculation is measured with the area of the windows, thus the area of the windows is taken into account twice, therefore in column 10 the coefficient k windows is taken as the difference between its values for windows and walls.
Heat loss calculations are carried out by room, floor, building.
