Suppose you wanted to pick up a boiler, radiators and pipes for the heating system of a private house yourself. Task number 1 - to calculate the heat load for heating, in other words, to determine the total heat consumption necessary for heating the building to a comfortable indoor temperature. We propose to study 3 calculation methods - different in complexity and accuracy of the results.
Methods for determining the load
First, we explain the meaning of the term. Thermal load is the total amount of heat consumed by the heating system to heat the premises to the standard temperature in the coldest period. The value is calculated in units of energy - kilowatts, kilocalories (less often - kilojoules) and is indicated in the formulas with the Latin letter Q.
Knowing the load on heating a private house in general and the need of each room in particular, it is not difficult to choose a boiler, heaters and batteries of the water system for power. How to calculate this parameter:
- If the ceiling height does not reach 3 m, an enlarged calculation is made according to the area of the heated rooms.
- With a ceiling height of 3 m or more, heat consumption is considered by the volume of the premises.
- Determination of heat loss through external fencing and the cost of heating ventilation air according to SNiP.
Note. In recent years, online calculators located on the pages of various Internet resources have gained wide popularity. With their help, the determination of the amount of thermal energy is quick and does not require additional instructions. Minus - the reliability of the results needs to be checked, because the programs are written by people who are not heat engineers.
The first two calculation methods are based on the application of specific thermal characteristics in relation to the heated area or the volume of the building. The algorithm is simple, it is used everywhere, but gives very approximate results and does not take into account the degree of insulation of the cottage.
According to SNiP, it is much more difficult to calculate thermal energy consumption according to SNiP. You will have to collect a lot of reference data and work on the calculations, but the final numbers will reflect the real picture with an accuracy of 95%. We will try to simplify the methodology and make the calculation of the heating load as accessible as possible to understanding.
For example, a project of a one-story house of 100 m²
To clearly explain all the methods for determining the amount of thermal energy, we suggest taking as an example a one-story house with a total area of 100 squares (according to the external measurement), shown in the drawing. We list the technical characteristics of the building:
- construction region - a strip of temperate climate (Minsk, Moscow);
- the thickness of the external fencing is 38 cm, the material is silicate brick;
- external wall insulation - polystyrene with a thickness of 100 mm, density - 25 kg / m³;
- floors - concrete on the ground, there is no basement;
- overlapping - reinforced concrete slabs insulated from the side of the cold attic with 10 cm polystyrene;
- windows - standard metal-plastic on 2 glasses, size - 1500 x 1570 mm (h);
- the front door is metal 100 x 200 cm, insulated with extruded polystyrene foam 20 mm from the inside.
The cottage has interior partitions in a half-brick (12 cm), the boiler room is located in a separate building. The area of the rooms is indicated on the drawing, the height of the ceilings will be taken depending on the calculation method explained - 2.8 or 3 m.
We calculate the heat consumption by quadrature
For an approximate estimate of the heating load, the simplest thermal calculation is usually used: the area of the building is taken according to the external measurement and multiplied by 100 watts. Accordingly, the heat consumption of a cottage of 100 m² will be 10,000 W or 10 kW.The result allows you to choose a boiler with a safety factor of 1.2-1.3, in this case, the unit power is taken equal to 12.5 kW.
We suggest performing more accurate calculations taking into account the location of the rooms, the number of windows and the region of development. So, with ceilings up to 3 m, it is recommended to use the following formula:
The calculation is carried out for each room separately, then the results are summed up and multiplied by the regional coefficient. Explanation of the notation of the formula:
- Q is the desired load, W;
- Spom - the quadrature of the room, m²;
- q is the indicator of the specific thermal characteristic related to the area of the room, W / m²;
- k - coefficient taking into account the climate in the area of residence.
For reference. If a private house is located in a temperate zone, the coefficient k is taken to be equal to one. In the southern regions, k = 0.7; in the northern regions, values of 1.5–2 are used.
An approximate calculation of the total quadrature indicator q = 100 W / m². This approach does not take into account the location of rooms and the different number of light openings. The corridor inside the cottage will lose much less heat than a corner bedroom with windows of the same area. We propose taking the value of the specific thermal characteristic q as follows:
- for rooms with one external wall and window (or door) q = 100 W / m²;
- corner rooms with one light opening - 120 W / m²;
- the same with two windows - 130 W / m².
How to choose the q value is clearly shown on the floor plan. For our example, the calculation looks like this:
Q = (15.75 x 130 + 21 x 120 + 5 x 100 + 7 x 100 + 6 x 100 + 15.75 x 130 + 21 x 120) x 1 = 10935 W ≈ 11 kW.
As you can see, the refined calculations gave a different result - in fact, for heating a particular house 100 m² will be spent on 1 kW of thermal energy more. The figure takes into account the heat consumption for heating the outdoor air penetrating into the home through openings and walls (infiltration).
Calculation of heat load by volume of rooms
When the distance between the floors and the ceiling reaches 3 m or more, the previous calculation option cannot be used - the result will be incorrect. In such cases, the heating load is considered to be according to specific aggregated indicators of heat consumption per 1 m³ of the volume of the room.
The formula and calculation algorithm remain the same, only the area parameter S changes to the volume - V:
Accordingly, another indicator of specific consumption q is taken, referred to the cubic capacity of each room:
- a room inside the building or with one external wall and a window - 35 W / m³;
- corner room with one window - 40 W / m³;
- the same with two light openings - 45 W / m³.
Note. Increasing and decreasing regional coefficients k are applied in the formula without changes.
Now, for example, we determine the load on the heating of our cottage, taking the ceiling height equal to 3 m:
Q = (47.25 x 45 + 63 x 40 + 15 x 35 + 21 x 35 + 18 x 35 + 47.25 x 45 + 63 x 40) x 1 = 11182 W ≈ 11.2 kW.
It is noticeable that the required thermal power of the heating system increased by 200 watts compared to the previous calculation. If we take the height of the rooms 2.7–2.8 m and calculate the energy expenditure through cubic capacity, then the figures will be approximately the same. That is, the method is quite applicable for the integrated calculation of heat loss in rooms of any height.
Calculation algorithm according to SNiP
This method is the most accurate of all existing. If you use our instructions and correctly perform the calculation, you can be 100% sure of the result and calmly select heating equipment. The procedure looks like this:
- Measure the quadrature of the external walls, floors, and ceilings separately in each room. Determine the area of windows and front doors.
- Calculate heat loss across all outdoor fences.
- Find out the consumption of thermal energy used to heat the ventilation (infiltration) air.
- Summarize the results and get a real indicator of the heat load.
An important point. In a two-story cottage, internal ceilings are not taken into account, since they do not border the environment.
The essence of calculating heat loss is relatively simple: you need to find out how much energy each type of building structure loses, because windows, walls and floors are made of different materials. When determining the quadrature of the outer walls, subtract the area of the glazed openings - the latter allow a greater heat flux and therefore are considered separately.
When measuring the width of the rooms, add half the thickness of the inner partition to it and capture the outer corner, as shown in the diagram. The goal is to take into account the full quadrature of the external fence, losing heat over the entire surface.
We determine the heat loss of walls and roofs
The formula for calculating the heat flux passing through a structure of the same type (for example, a wall) is as follows:
Decipher the notation:
- the amount of heat loss through one fence we designated Qi, W;
- A - the squaring of the wall within the same room, m²;
- tv - comfortable temperature inside the room, usually +22 ° С;
- tн - the minimum temperature of street air that lasts for the 5 coldest winter days (take the real value for your area);
- R is the resistance of the thickness of the outer fence to heat transfer, m² ° C / W.
One uncertain parameter remains in the above list - R. Its value depends on the material of the wall structure and the thickness of the fence. To calculate the heat transfer resistance, proceed as follows:
- Determine the thickness of the supporting part of the external wall and, separately, the insulation layer. The letter in the formulas - δ, is considered in meters.
- Find out the thermal conductivity coefficients of structural materials λ from the reference tables, and the unit of measure is W / (mºС).
- Substitute the found values in the formula one by one:
- Define R for each wall layer separately, add the results, then use in the first formula.
Repeat the calculations separately for windows, walls and ceilings within the same room, then move on to the next room. Heat losses through floors are considered separately, as described below.
Tip. The correct thermal conductivity coefficients of various materials are indicated in the regulatory documentation. For Russia, this is the Code of Rules SP 50.13330.2012, for Ukraine - DBN V.2.6–31 ~ 2006. Attention! In the calculations, use the value of λ prescribed in column "B" for operating conditions.
Calculation Example for the living room of our one-story house (ceiling height 3 m):
- The area of the external walls together with the windows: (5.04 + 4.04) x 3 = 27.24 m². The square of the windows is 1.5 x 1.57 x 2 = 4.71 m². Net fencing area: 27.24 - 4.71 = 22.53 m².
- The thermal conductivity λ for masonry of silicate brick is 0.87 W / (m º C), foam 25 kg / m ³ - 0.044 W / (m º C). Thickness is 0.38 and 0.1 m, respectively, we consider the heat transfer resistance: R = 0.38 / 0.87 + 0.1 / 0.044 = 2.71 m² ° C / W.
- Outside temperature - minus 25 ° С, inside the living room - plus 22 ° С. The difference is 25 + 22 = 47 ° C.
- We determine the heat loss through the walls of the living room: Q = 1 / 2.71 x 47 x 22.53 = 391 watts.
Similarly, heat flow through windows and floors is considered. The thermal resistance of translucent structures is usually indicated by the manufacturer, the characteristics of reinforced concrete floors with a thickness of 22 cm are found in the normative or reference literature:
- R of insulated floor = 0.22 / 2.04 + 0.1 / 0.044 = 2.38 m² ° C / W, heat loss through the roof is 1 / 2.38 x 47 x 5.04 x 4.04 = 402 W.
- Losses through window openings: Q = 0.32 x 47 x71 = 70.8 W.
Total heat loss in the living room (excluding floors) will be 391 + 402 + 70.8 = 863.8 watts. Similar calculations are carried out for the remaining rooms, the results are summarized.
Please note: the corridor inside the building does not come into contact with the outer shell and loses heat only through the roof and floors. What fences should be taken into account in the calculation methodology, see the video.
Division of the floor into zones
To find out the amount of heat lost by floors on the ground, the building is divided into 2 m wide zones in the plan, as shown in the diagram. The first strip starts from the outer surface of the building structure.
The calculation algorithm is as follows:
- Outline the cottage plan, divide it into strips 2 m wide. The maximum number of zones is 4.
- Calculate the area of the floor that falls separately in each zone, neglecting the interior partitions. Please note: the quadrature at the corners is counted twice (shaded in the drawing).
- Using the calculation formula (for convenience we give it again), determine the heat loss in all areas, summarize the figures.
- The heat transfer resistance R for zone I is taken to be 2.1 m² ° C / W, II - 4.3, III - 8.6, the rest of the floor - 14.2 m² ° C / W.
Note. If we are talking about a heated basement, the first strip is located on the underground part of the wall, starting from the ground level.
Floors insulated with mineral wool or polystyrene foam are calculated identically, only the thermal resistance of the insulation layer, determined by the formula δ / λ, is added to fixed R values.
Calculation example in the living room of a country house:
- The quadrature of zone I is (5.04 + 4.04) x 2 = 18.16 m², plot II - 3.04 x 2 = 6.08 m². The remaining zones do not enter the living room.
- The energy consumption for the 1st zone will be 1 / 2.1 x 47 x 18.16 = 406.4 W, for the second - 1 / 4.3 x 47 x 6.08 = 66.5 W.
- The amount of heat flow through the living room floors is 406.4 + 66.5 = 473 watts.
Now it is easy to knock down the total heat loss in the room in question: 863.8 + 473 = 1336.8 W, rounded - 1.34 kW.
Ventilation air heating
The vast majority of private houses and apartments have natural ventilation. Street air penetrates through the narthexes of windows and doors, as well as supply air openings. The incoming cold mass is heated by the heating system, spending additional energy. How to find out the amount of these losses:
- Since the calculation of infiltration is too complicated, regulatory documents allow the allocation of 3 m³ of air per hour for each square meter of housing. The total supply air supply L is considered simple: the room quadrature is multiplied by 3.
- L is the volume, and the mass m of the air flow is needed. Find out by multiplying by the density of the gas taken from the table.
- The air mass m is substituted in the formula of the school physics course, which allows determining the amount of energy expended.
We calculate the required amount of heat on the example of a long-suffering living room with an area of 15.75 m². The inflow volume L = 15.75 x 3 = 47.25 m³ / h, mass - 47.25 x 1.422 = 67.2 kg / h. Assuming the heat capacity of air (indicated by the letter C) equal to 0.28 W / (kg ºС), we find the energy consumption: Qvent = 0.28 x 67.2 x 47 = 884 W. As you can see, the figure is quite impressive, which is why heating the air masses must be taken into account.
The final calculation of the heat loss of the building plus the heat consumption for ventilation is determined by summing all the previously obtained results. In particular, the load on the heating of the living room will result in the figure 0.88 + 1.34 = 2.22 kW. Similarly, all the rooms of the cottage are calculated, at the end of the energy costs add up to one digit.
Final settlement
If your brain has not yet begun to boil due to the abundance of formulas 😊, then it is certainly interesting to see the result throughout the one-story house. In the previous examples, we did the main work, it remains only to go through other rooms and find out the heat loss of the entire outer shell of the building. Found raw data:
- thermal resistance of walls - 2.71, windows - 0.32, floors - 2.38 m² ° С / W;
- ceiling height - 3 m;
- R for the front door insulated with extruded polystyrene foam is 0.65 m² ° C / W;
- internal temperature - 22, external - minus 25 ° С.
To simplify the calculations, we propose to create a table in Exel, then we will put in there the intermediate and final results.
At the end of the calculations and filling out the table, the following values of thermal energy consumption for the premises were obtained:
- living room - 2.22 kW;
- kitchen - 2.536 kW;
- hallway - 745 W;
- corridor - 586 W;
- bathroom - 676 W;
- bedroom - 2.22 kW;
- children's - 2.536 kW.
The total load on the heating system of a private house of 100 m² was 11.518 W, rounded - 11.6 kW.It is noteworthy that the result differs from approximate calculation methods by literally 5%.
How to use the results of calculations
Knowing the building’s heat demand, the homeowner can:
- clearly select the power of heat-power equipment for heating the cottage;
- dial the desired number of sections of radiators;
- determine the required thickness of the insulation and perform thermal insulation of the building;
- find out the flow rate of the coolant on any part of the system and, if necessary, perform hydraulic calculation of pipelines;
- Find out the average daily and monthly heat consumption.
The last paragraph is of particular interest. We found the value of the heat load in 1 hour, but it can be recalculated for a longer period and calculate the estimated fuel consumption - gas, firewood or pellets.