Calculation and determination of heat load for heating: calculation methods, conclusion

Heat load means the amount of thermal energy, necessary to maintain a comfortable temperature in the house, apartment or separate room. The maximum hourly load on the heating means the amount of heat, necessary to maintain normalized values ​​for an hour in the most adverse conditions.

Factors, affecting the heat load

• Material and thickness of walls. Example, brick wall in 25 centimeters and aerated concrete wall 15 centimeters are able to pass different amounts of heat.
• Roof material and structure. Example, heat losses of a flat roof from reinforced concrete plates considerably differ from heat losses of the warmed attic.
• Ventilation. The loss of heat energy from the exhaust air depends on the performance of the ventilation system, presence or absence of heat recovery system.
• Glazing area. Windows lose more heat energy compared to solid walls.
• The level of insolation in different regions. Determined by the degree of absorption of solar heat by external coatings and the orientation of the planes of buildings in relation to the world.
• The difference in temperature between the street and the room. It is determined by the heat flux through the enclosing structures under the condition of constant heat transfer resistance.

Heat load distribution

With water heating, the maximum heat output of the boiler must be equal to the sum of the heat output of all heating appliances in the house. On the distribution of heating devices are influenced by the following factors:

• Room area and ceiling height;
• Location inside the house. Corner and end rooms lose more heat, than premises, located in the middle of the building;
• Distance from the heat source;
• Desirable temperature in the rooms.

SNiP recommends the following values:

• Living rooms in the middle of the house - 20 degrees;
• Corner and end living rooms - 22 degrees. At the same time due to the higher temperature the walls do not freeze;
• Kitchen - 18 degrees, because it has its own heat sources - gas or electric stoves, etc..
• Bathroom - 25 degrees.

With air heating heat flow, which enters a separate room, depends on the capacity of the air hose. Often the simplest way to adjust it is to adjust the position of the ventilation grilles with temperature control manually.

With a heating system, where a distributive heat source is used (convector, underfloor heating, electric heaters, etc.), the required temperature is set on the thermostat.

calculation methods

There are several ways to determine the heat load, having different complexity of calculation and reliability of the received results. The following are the three simplest methods of calculating the heat load.

method no. 1

According to the current SNiP, there is a simple method of calculating the heat load. On 10 square meters are taken 1 kilowatts of thermal power. Then the obtained data are multiplied by the regional coefficient:

• The southern regions have a coefficient 0,7-0,9;
• For temperate and cold climates (Moscow and Leningrad regions) the coefficient is equal to 1,2-1,3;
• The Far East and the Far North: for Novosibirsk from 1,5; for Oymyakon to 2,0.

Calculation by example:

1. Building area (10 * 10) is equal to 100 square meters.
2. Baseline heat load 100/10 = 10 kilowatt.
3. This value is multiplied by the regional coefficient, equal to 1,3, the result is 13 kW of thermal power, which are needed to maintain a comfortable temperature in the house.

pay attention! If you use this technique to determine the heat load, it is still necessary to take into account the power reserve in 20 percent, to compensate for errors and extreme cold.

method no. 2

The first method of determining the heat load has many errors:

• Different buildings have different ceiling heights. Considering that, heated area is not, and volume, this parameter is very important.
• More heat passes through doors and windows, than through the walls.
• You can not compare a city apartment with a private house, where below, above and behind the walls are not apartments, and the street.

Method adjustment:

• The basic heat load is equal to 40 watts on 1 cubic meter of room volume.
• Every door, leading to the street, adds to the baseline heat load 200 watts, each window - 100 watts.
• Corner and end apartments of an apartment building have a coefficient 1,2-1,3, which is affected by the thickness and material of the walls. A private house has a coefficient 1,5.
• Regional coefficients are equal: for the Central regions and the European part of Russia - 0,1-0,15; for the Northern regions - 0,15-0,2; for the southern regions - 0,07-0,09 kw / sq.m.

Calculation by example:

1. The volume of the building 300 square meters (10 * 10 * 3 = 300).
2. Baseline heat load 12000 watts (300 * 40).
3. With eight windows and two doors, the heat output is equal to 13200 watts (12000+ (8 * 100) + (2 * 200)).
4. For a private house, the heat load increases by a regional factor and leaves 19800 watts (13200 * 1,5).
5. 19800 * 1,3 = 25740 watts (taking into account the regional coefficient for the Northern regions). so, a 28-kilowatt boiler will be required for heating.

method no. 3

Do not be tempted - the second method of calculating the heat load is also very imperfect. It conditionally takes into account the thermal resistance of the ceiling and walls; the temperature difference between the outside air and the air inside.

It is worth noting, this amount of heat energy is needed to maintain a constant temperature inside the house, which will be equal to all losses through the ventilation system and fencing devices. However, and in this method the calculations are simplified, as it is impossible to systematize and measure all factors.

of heat loss affects the material of the walls — 20-30 percentage of heat loss. It goes through the ventilation 30-40 percent, through the roof - 10-25 percent, through the windows - 15-25 percent, through the floor on the ground - 3-6 percent.

To simplify heat load calculations, heat losses through enclosing devices are calculated, and then this value is simply multiplied by 1,4. Temperature delta is easily measured, but data on thermal resistance can only be found in reference books. Here are some popular ones the value of thermal resistance:

• The thermal resistance of the wall in three bricks is the same 0,592 m2 * WITH / W.
• The walls of 2,5 brick is 0, 502.
• The walls of 2 bricks one 0,405.
• Walls in one brick (thickness 25 cm) is equal to 0,187.
• Wooden log house, where the diameter of the deck 25 cm - 0,550.
• Wooden log house, where the diameter of the deck 20 centimeters - 0,440.
• Log house, where the thickness of the log 20 cm - 0,806.
• Log house, where the thickness 10 cm - 0,353.
• Frame wall, the thickness of which 20 cm, insulated with mineral wool - 0,703.
• Aerated concrete walls, the thickness of which 20 cm - 0,476.
• Aerated concrete walls, the thickness of which 30 cm - 0,709.
• Plasters, the thickness of which 3 cm - 0,035.
• Ceiling or attic floor - 1,43.
• Wooden floor - 1,85.
• Double Wooden Doors - 0,21.

Calculation by example:

1. The temperature delta during the peak of frosts is equal to 50 degrees: inside the house plus 20 degrees, externally - minus 30 degrees.
2. Heat loss through one square meter 50 / 1,85 (indicator of thermal resistance of a floor from a tree) is approximately 27 watts. The whole floor will have 27 * 100 = 2700 watts.
3. Heat losses through the ceiling are (50 / 1,43) * 100 and in the same way, 3500 watts.
4. Wall area (10 * 3) * 4 and is equal to 120 square meters. Example, the walls are made of timber with a thickness 20 cm, thermal resistance = 0,806. so, heat loss will be (50 / 0,806) * 120 = 7444 W.
5. All the obtained values ​​of heat loss are added, and the value is obtained 13644 watts. This is the amount of heat the house will lose through the walls, floor and ceiling.
6. Then the value obtained is multiplied by a factor 1,4 (losses to the ventilation system) and it turns out 19101 watts. so, a 20-kilowatt boiler will be needed to heat such a house.

conclusion

As can be seen from the calculations, ways to determine the heat load have significant errors. Fortunately, excess boiler power will not hurt:

• Operation of the gas boiler at reduced power is carried out without a drop in efficiency, and the operation of condensing devices at partial load is carried out in economy mode.
• The same can be said for solar boilers.
• The efficiency of electric heating equipment is equal to 100 percent.

pay attention! Operation of solid fuel boilers at power less than the nominal value of power is contraindicated.

The calculation of the heat load for heating is an important factor, which must be calculated before creating a heating system. In case of approach to process with reason and competent performance of all works faultless work of heating is guaranteed, as well as significantly save money on extra costs.