Thermotechnical calculation of a building: specifics and formulas for performing calculations + practical examples
During the operation of the building, both overheating and freezing are undesirable. Determine the middle ground will allow thermal engineering calculation, which is no less important than the calculation of profitability, strength, resistance to fire, durability.
Based on heat engineering standards, climatic characteristics, vapor and moisture permeability, the choice of materials for the construction of enclosing structures is carried out. How to perform this calculation, we consider in the article.
The content of the article:
- The purpose of heat engineering calculation
- Parameters for performing calculations
- Formulas for calculating
- Conclusions and useful video on the topic
The purpose of heat engineering calculation
Much depends on the thermal characteristics of the capital fencing of the building. This is the humidity of the structural elements, and temperature indicators that affect the presence or absence of condensate on the interior partitions and ceilings.
The calculation will show whether stable temperature and humidity characteristics are maintained at plus and minus temperatures. The list of these characteristics also includes such an indicator as the amount of heat lost by the building envelope in the cold period.
You cannot start designing without having all this data. Based on them, choose the thickness of the walls and floors, the sequence of layers.
The heat engineering calculation aims to determine:
- Are the designs identical to the stated requirements in terms of thermal protection?
- Is the comfortable microclimate inside the building so fully ensured?
- Is optimal thermal protection of structures ensured?
The main principle is to maintain a balance of the difference in temperature indicators of the atmosphere of the internal structures of fences and rooms. If it is not observed, these surfaces will absorb heat, and inside the temperature will remain very low.
Changes in the heat flux should not significantly affect the internal temperature.This characteristic is called heat resistance.
By performing a thermal calculation, the optimal limits (minimum and maximum) of the dimensions of the walls and floors in thickness are determined. This is a guarantee of building operation over a long period, both without extreme freezing of structures and overheating.
Parameters for performing calculations
To perform heat calculation, you need the initial parameters.
They depend on a number of characteristics:
- Destination of the building and its type.
- Orientations of vertical building envelopes relative to the orientation to the cardinal points.
- The geographic parameters of the future home.
- The volume of the building, its number of floors, area.
- Types and dimensional data of door, window openings.
- Type of heating and its technical parameters.
- The number of permanent residents.
- Material of vertical and horizontal enclosing structures.
- Overlapping the top floor.
- Equipped with hot water.
- Type of ventilation.
Other structural features of the structure are taken into account in the calculation. The air permeability of building envelopes should not contribute to excessive cooling inside the house and reduce the heat-shielding characteristics of the elements.
Loss of heat causes and waterlogging of the walls, and in addition, this leads to dampness, adversely affecting the durability of the building.
In the calculation process, first of all, the thermal engineering data of building materials are determined, from which the building envelope is made. In addition, the reduced heat transfer resistance and conformity to its normative value is subject to determination.
Formulas for calculating
Leaks of heat lost by the house can be divided into two main parts: losses through building envelopes and losses caused by functioning ventilation system. In addition, heat is lost when warm water is discharged into the sewer system.
Losses through building envelopes
For the materials that make up the enclosing structures, it is necessary to find the value of the thermal conductivity index Kt (W / m x degree). They are in the relevant directories.
Now, knowing the thickness of the layers, according to the formula: R = S / CTcalculate the thermal resistance of each unit. If the design is multilayer, all the obtained values are added up.
Guided by this technique, take into account the moment that the materials that make up the structure have a different structure. It is also taken into account that the heat flux passing through them has different specifics.
For each individual design, heat loss is determined by the formula:
Q = (A / R) x dT
- A - area in m².
- R is the resistance of the heat transfer structure.
- dT is the temperature difference between the outside and the inside. It must be determined for the coldest 5-day period.
Performing the calculation in this way, you can get the result only for the coldest five-day period. The total heat loss for the entire cold season is determined by taking into account the parameter dT, taking into account the temperature, not the lowest, but the average.
Next, calculate the amount of energy needed to compensate for the loss of heat that has gone both through the building envelope, and through ventilation. It is indicated by W.
There is a formula for this:
W = ((Q + QB) x 24 x N) / 1000
In it N is the duration of the heating period in days.
The disadvantages of calculating the area
The calculation based on the area indicator is not very accurate. Here, such a parameter as climate, temperature indicators, both minimum and maximum, humidity, is not taken into account. Due to ignoring many important points, the calculation has significant errors.
Often trying to block them, the project provides for "stock".
If you nevertheless chose this method for calculation, you need to consider the following nuances:
- With a height of vertical fences up to three meters and the presence of no more than two openings on one surface, the result is better to multiply by 100 watts.
- If the project has a balcony, two windows or a loggia are multiplied by an average of 125 watts.
- When the premises are industrial or warehouse, a 150W multiplier is used.
- If radiators are located near windows, their design capacity is increased by 25%.
The area formula is:
Q = S x 100 (150) W.
Here Q is a comfortable level of heat in the building, S is the area with heating in m². Numbers 100 or 150 - the specific amount of thermal energy spent for heating 1 m².
Losses through home ventilation
The key parameter in this case is the air exchange rate. Provided that the walls of the house are vapor permeable, this value is equal to unity.
It provides for a complete update of the air inside the building in one hour. Buildings constructed according to the DIN standard have walls with vapor barrier, therefore here the air exchange rate is taken to be two.
There is a formula by which heat loss through a ventilation system is determined:
Qw = (V x Qu: 3600) x P x C x dT
Here, the symbols indicate the following:
- Qв - heat loss.
- V is the volume of the room in mᶾ.
- P is the density of air. its value is taken equal to 1.2047 kg / mᶾ.
- Kv - the rate of air exchange.
- C is the specific heat. It is equal to 1005 J / kg x C.
Based on the results of this calculation, it is possible to determine the power of the heat generator of the heating system. If the power value is too high, the situation can become a way out. ventilation unit with recuperator. Let's look at a few examples for houses made of different materials.
An example of heat engineering calculation No. 1
We calculate a residential building located in 1 climatic region (Russia), subarea 1B. All data is taken from table 1 of SNiP 23-01-99. The coldest temperature observed for five days with a security of 0.92 - tn = -22⁰С.
In accordance with SNiP, the heating period (zop) lasts 148 days. The average temperature during the heating period with daily average temperature indices of air in the street is 8⁰ - tot = -2.3⁰. The outside temperature during the heating season is tht = -4.4⁰.
The condition is stipulated that the temperature of 22 дома should be provided in the rooms of the house. The house has two floors and walls with a thickness of 0.5 m. Its height is 7 m, dimensions in plan are 10 x 10 m. The material of the vertical walling is warm ceramics. For it, the thermal conductivity coefficient is 0.16 W / m x C.
Mineral wool was used as an external insulation, 5 cm thick. The value of CT for her is 0.04 W / m x C. The number of window openings in the house is 15 pcs. 2.5 m² each.
Heat loss through walls
First of all, it is necessary to determine the thermal resistance of both the ceramic wall and the insulation. In the first case, R1 = 0.5: 0.16 = 3.125 sq. mx C / W. In the second - R2 = 0.05: 0.04 = 1.25 sq. mx C / W. In general, for a vertical building envelope: R = R1 + R2 = 3.125 + 1.25 = 4.375 sq. mx C / W.
Since heat loss has a directly proportional relationship with the area of the building envelope, we calculate the area of the walls:
A = 10 x 4 x 7 - 15 x 2.5 = 242.5 m²
Now you can determine the heat loss through the walls:
Qc = (242.5: 4.375) x (22 - (-22)) = 2438.9 W.
Heat losses through horizontal walling are calculated in the same way. As a result, all the results are summarized.
If the basement under the floor of the first floor is heated, the floor can not be insulated.Basement walls are still better sheathed with insulation so that the heat does not go into the ground.
Determination of losses through ventilation
To simplify the calculation, do not take into account the thickness of the walls, but simply determine the volume of air inside:
V = 10х10х7 = 700 mᶾ.
With a multiplicity of air exchange Kv = 2, the heat loss will be:
Qw = (700 x 2): 3600) x 1.2047 x 1005 x (22 - (-22)) = 20 776 W.
If Kv = 1:
Qw = (700 x 1): 3600) x 1.2047 x 1005 x (22 - (-22)) = 10 358 W.
Effective ventilation of residential buildings is provided by rotary and plate recuperators. The efficiency of the former is higher, it reaches 90%.
An example of heat engineering calculation No. 2
It is required to calculate losses through a 51 cm thick brick wall. It is insulated with a 10-cm layer of mineral wool. Outside - 18⁰, inside - 22⁰. The dimensions of the wall are 2.7 m in height and 4 m in length. The only external wall of the room is oriented to the south, there are no external doors.
For brick, the thermal conductivity coefficient Kt = 0.58 W / m º C, for mineral wool - 0.04 W / m º C. Thermal resistance:
R1 = 0.51: 0.58 = 0.879 sq. mx C / W. R2 = 0.1: 0.04 = 2.5 sq. mx C / W. In general, for a vertical building envelope: R = R1 + R2 = 0.879 + 2.5 = 3.379 square meters. mx C / W.
External wall area A = 2.7 x 4 = 10.8 m²
Heat loss through the wall:
Qc = (10.8: 3.379) x (22 - (-18)) = 127.9 W.
To calculate losses through windows, the same formula is used, but their thermal resistance is usually indicated in the passport and it is not necessary to calculate it.
If the house’s windows with dimensions of 1.5 x 1.5 m² are energy-saving, oriented to the North, and the thermal resistance is 0.87 m2 ° C / W, then the losses will be:
Qo = (2.25: 0.87) x (22 - (-18)) = 103.4 t.
An example of heat engineering calculation No. 3
We perform a thermal calculation of a wooden log building with a facade erected from pine logs with a layer thickness of 0.22 m. The coefficient for this material is K = 0.15. In this situation, heat loss will amount to:
R = 0.22: 0.15 = 1.47 m² x ⁰C / W.
The lowest five-day temperature is -18⁰, for comfort in the house the temperature is set to 21⁰. The difference is 39⁰. If we proceed from an area of 120 m², we get the result:
Qc = 120 x 39: 1.47 = 3184 W.
For comparison, we determine the loss of a brick house. The coefficient for silicate brick is 0.72.
R = 0.22: 0.72 = 0.306 m² x ⁰C / W.
Qs = 120 x 39: 0.306 = 15,294 watts.
In the same conditions, a wooden house is more economical. Silicate brick for walling is not suitable here at all.
Builders and architects recommend doing heat consumption during heating for the competent selection of equipment and at the design stage of the house to select the appropriate insulation system.
Heat calculation example No. 4
The house will be built in the Moscow region. For the calculation, a wall created from foam blocks was taken. How insulation is applied extruded polystyrene foam. Finishing the structure - plaster on both sides. Its structure is calcareous and sandy.
Expanded polystyrene has a density of 24 kg / mᶾ.
Relative air humidity in the room is 55% at an average temperature of 20⁰. Layer thickness:
- plaster - 0.01 m;
- foam concrete - 0.2 m;
- polystyrene foam - 0.065 m.
The task is to find the necessary heat transfer resistance and the actual one. The necessary Rtr is determined by substituting the values in the expression:
Rtr = a x GSOP + b
where GOSP is the degree-day of the heating season, and a and b are the coefficients taken from table No. 3 of the Code of Rules 50.13330.2012. Since the building is residential, a is 0,00035, b = 1.4.
GSOP is calculated by the formula taken from the same SP:
GOSP = (tv - tot) x zot.
In this formula, tv = 20⁰, tf = -2.2⁰, zf - 205 - the heating period in days. Hence:
GSOP = (20 - (-2,2)) x 205 = 4551⁰ С x day .;
Rtr = 0.00035 x 4551 + 1.4 = 2.99 m2 x C / W.
Using table No. 2 SP50.13330.2012, determine the thermal conductivity for each layer of the wall:
- λb1 = 0.81 W / m ⁰С;
- λb2 = 0.26 W / m ⁰С;
- λb3 = 0.041 W / m ⁰С;
- λb4 = 0.81 W / m ⁰С.
The total conditional resistance to heat transfer Ro, equal to the sum of the resistances of all layers. Calculate it by the formula:
Substituting the values receive: = 2.54 m2 ° C / W. Rf is determined by multiplying Ro by a coefficient r equal to 0.9:
Rf = 2.54 x 0.9 = 2.3 m2 x ° C / W.
The result obliges to change the design of the enclosing element, since the actual thermal resistance is less than the calculated one.
There are many computer services that speed up and simplify calculations.
Thermal engineering calculations are directly related to the definition of dew point. You will learn what it is and how to find its value from the article we recommend.
Conclusions and useful video on the topic
Performing a heat engineering calculation using an online calculator:
The correct heat engineering calculation:
A competent heat engineering calculation will allow you to evaluate the effectiveness of insulation of the external elements of the house, to determine the power of the necessary heating equipment.
As a result, you can save on the purchase of materials and heating appliances. It is better to know in advance whether the equipment can handle the heating and conditioning of the building than to buy everything at random.
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