Weak boiler power will ruin everything, including the standard of living. Due to excess power, the system will operate in impulses: fuel consumption will increase, and the service life of the equipment will be shortened. The boiler may boil. Or brains. How to calculate correctly, what parameters to take into account, what heat loss is and other customer questions about boiler power can be read here.
How to easily calculate the power of a heating boiler?
You can calculate the approximate power of a boiler for a home by area and volume .
1) A simplified version of calculations by area : 10 kW per 100 m² of a house (or heated area). And this figure will only show the minimum power, below which you cannot fall.
| Graph of boiler versus area |
To take into account climatic zones, coefficients have been developed that correct this formula:
- Central Russia is 1-1.5;
- Northern regions are 1.5-2;
- Southern territories – 0.7-0.9;
- Moscow and Moscow region – 1.2-1.5.
To get closer to a realistic figure, you also need to take into account possible heat loss. To do this, add 10-15% to the minimum value. If the ceilings are higher than 2.7 meters, then we divide the actual ceiling height by the standard height. We get another correction factor.
| Vacation home |
Example:
Let's calculate the power of a boiler for a house in the Moscow region. Ceilings – 3 meters, area – 150 m². A double-circuit boiler is needed - for heat and hot water supply.
The formula gives 15 kW - the minimum power value for the future boiler. Next, to the number 15 we add 10% of heat loss, multiply by the climate coefficient 1.2. The ceilings are higher than 2.7m, so we multiply the resulting figure by a factor of 1.1.
Boiler power = 15 kW (minimum) + 10% (heat loss) * 1.2 * 1.1 = 21.7, rounded to 22 kW.
2) Second formula based on volume : 1 m3 – 40 W. Plus the markups that were included in the first formula, except for the ceiling coefficient. Let's calculate the same house in the Moscow region using this formula.
Boiler power =((150 m²*3m)* 40 W + 10%) * 1.2 * 1.1 =23522 W ≈ 24 kW. The difference between the first and second calculations is 2 kW. The option of calculating boiler power by air volume is the most correct.
You could stop reading this article here. But the difference between an approximate and an exact calculation is in several nuances. What are the nuances, you ask. The answer to this question is contained in the following paragraphs.
How much electricity does an electric boiler consume?
Let us answer straightforwardly: an electric boiler consumes exactly as much as it produces in the form of heat. That is, if you need 10 kW/h for heating, then the boiler should consume exactly that much.
Let's calculate how much energy an electric heating boiler consumes per month for a house with an area of 100 sq.m.
Let’s make a reservation right away: in reality, the amount of heat that needs to be supplied depends on the heat loss of the house, and this depends on the material of the walls, the number of windows, doors and other nuances. To find out the exact amount of heat your room needs, as well as calculate how much the boiler will consume, you can order a thermal calculation of the room from us.
Here we will take the generally accepted norm: a rough calculation requires 1 kW/h per 10 square meters of room. This means that to heat a house with an area of 100 sq.m. you will need a 10 kW boiler.
Let's calculate its maximum consumption per day, month and winter season:
Per day:
24 (hours)*10 (kW) = 240 kW. Let’s convert it into costs in rubles, taking into account that on average, according to electricity tariffs in the Sverdlovsk region, 1 kW costs 4 rubles
240 (kW) * 4 (rub) = 960 rubles
Per month: 960 (rub per day) * 30 (days) = 28,800 rubles
In the winter season , we will calculate, taking for the heating period the time from October 15 to March 31 - 136 days): 136 (days) * 240 (kW consumption per day) = 32,640 kW - required for heating during the season.
Let's convert it into expenses in rubles at the same rate:
32,640 kW * 4 (rub) = 130,560 rubles - the maximum cost of electricity for the heating season for a house with an area of 100 square meters.
Let's summarize the data in a table:
What parameters, besides volume and area, influence the choice of boiler? And why is this important?
A simplified calculation formula often leads to the purchase of an unsuitable boiler. Each house is individual, and heat loss percentage cannot be equal for all houses. Before calculating power, consider the data of a specific house:
1) Measure the area of walls, windows, doors;
2) Specify the thickness of the walls, indicate the type of finishing and material, the height of the ceilings;
3) Observe the minimum temperature of the house in cold weather;
4) Determine the desired temperature as a result of installing the boiler;
5) Write down the thermal conductivity values for the materials from which the house was built.
| Wall material | Wall thickness and material thermal resistance | Required thickness for home |
| Brick (1600 kg/m³ - density) | 510 mm (if laying two bricks), R=0.73 °C m²/W | 1380 mm 2190 mm |
| Wooden beam | 150 mm, R=0.83 °С m²/W | 355 mm 565 mm |
| Expanded clay concrete (1200 kg/m³ - density.) | 300 mm, R=0.58 °С m²/W | 1025 mm 1630 mm |
| Wooden panel (inside filled with mineral wool + 25 mm layer of inner and outer cladding) | 150 mm, R=1.84 °С m²/W | 160 mm 235 mm |
| Arbolit | 0.80-0.17W/m² | — |
| Foam concrete | 0.14-0.38 W/m² | — |
| Aerated concrete | 0.18-0.28 W/m² | — |
Thermal resistance of materials
Why is this necessary? The key parameter influencing the choice of boiler is the heat loss of the house. Houses with the same area and volume, but different degrees of insulation, will require equipment of different power.
Where does the heat go:
| Surface | Heat loss in% |
| Roof and ventilation | 20-25% |
| Foundation, if it is adjacent to the ground | up to 15% |
| Walls, windows and doors | 10-15% |
| Ground floor and unheated rooms, basement, for example | up to 15% |
It also matters: how different the outside temperature is from the inside, the climatic region, the strength and direction of the wind, how the house stands relative to parts of the world.
| Heat loss |
Maximum electricity costs when heating a house of 100 m2
| Cost of 1 kW/h, rub | Heating cost per day, rub. | Heating cost per month, rub | Heating cost per season, rub |
| 4 | 960 | 28800 | 130560 |
Of course, the boiler is unlikely to be operated all season without stopping. But exactly how long the electric boiler will work - as we wrote above - depends on insulation, heat loss, maximum winter temperatures in your region , that is, on the conditions of a particular room.
The boiler operating period per day can be either 18 hours or 7. If you know approximately how many hours per day it takes to heat your home, you can calculate consumption and costs using the above example.
You can compare the heating costs of different types of fuel and more accurately calculate the consumption of different types of boilers by using our calculator:
How to calculate power taking into account heat loss?
What is heat loss? Let’s say it’s -20 degrees outside, but the average temperature at home is +20 degrees. These quantities are balanced through the exchange of energy. Heat losses occur. The amount of heat loss under severe weather conditions helps to calculate the boiler power with high accuracy.
Step 1
Heat loss is determined by the formula: Q = Q roof + Q walls + Q floor + Q doors + Q windows .
Where the extreme value of Q is the heat loss of each surface of the house.
Each Q value is calculated using the formula: Q = S * T / R
Where Q is heat loss in W, S is the area of a specific surface in m², T is the difference between street and room temperatures in degrees, R is reference data on thermal resistance by type of material.
Step 2
This formula additionally includes involuntary heat loss through cracks, ventilation, exhaust hoods, opening doors and ventilation through windows. For independent calculations without a program, an additional 5% of the total leakage figure is added.
Step 3
Next we move on to determining the boiler power. There are only two formulas to choose from:
Rkot. = ( S room * P beat ) / 10 , where Rkot. — boiler power, Sroom. - the total area of rooms in the house where heating is planned, Pud. — specific power according to climate conditions.
Rkot. = ( Q losses * S from area ) / 100, where Rkot. - boiler power, Qloss - heat loss, Sot. area – total area of heated rooms.
Step 4
For an electric and gas boiler, you can use the table to check:
| Option | House area, m² | Heating, kW | Recommended number of devices | How many people live | DHW boiler, l/kW | Warm floor, m² | Warm floor, kW | Total power | Boiler power | Standard range of boilers, Cat, Ns/A/Nd |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 2 | 150 | 19 | 10 | 4 | 100/28 | 16 | 0,75 | 48 | 28 | 28/27/28 |
| 3 | 200 | 22 | 11 | 4 | 100/28 | 20 | 1 | 51 | 28 | 28/27/28 |
| 4 | 250 | 25,5 | 17 | 4 | 160/33 | 20 | 1 | 60 | 33 | 32/35/36 |
| 5 | 300 | 27 | 20 | 6 | 160/33 | 30 | 1,5 | 62 | 34 | -/35/36 |
| 6 | 350 | 31 | 26 | 6 | 200/33 | 40 | 2 | 66 | 39 | -/44/44 |
| 7 | 400 | 4 | 30 | 6 | 200/36 | 50 | 2,5 | 70 | 43 | -/44/44 |
| 8 | 450 | 36 | 44 | 8 | 300/36 | 60 | 3 | 75 | 45 | -/53/52 |
Coolant speed
Then, using the obtained coolant flow values, it is necessary for each section of pipes in front of the radiators to calculate the speed of water movement in the pipes using the formula:
,
where V is the coolant movement speed, m/s;
m is the coolant flow through the pipe section, kg/s
ρ—density of water, kg/cub.m. can be taken equal to 1000 kg/cub.m.
f - cross-sectional area of the pipe, sq.m. can be calculated using the formula: π * r2, where r is the internal diameter divided by 2
Coolant velocity calculator
m = l/s; pipe mm by mm; V = m/s
Why count if you can buy a boiler with a power reserve?
Sometimes boilers have a capacity reserve. This is good when the reserve is no more than 25%. Especially when the family plans to develop the area: complete a swimming pool, sauna, or other heated area. When the required power is exceeded significantly, the owner spends extra money, and the equipment operates in abnormal mode:
| Boiler repair |
- Breaks or malfunctions;
- The efficiency of the system decreases;
- A higher power boiler costs more;
- More fuel is consumed than is required to heat the house;
- A more powerful and expensive pump will be needed;
- The house will be very hot;
- Automatic regulation becomes difficult, the boiler may boil;
- The boiler begins to cycle - it turns on and off in a short period of time, equipment components wear out;
- Condensation appears in the chimney. When burned, the condensate reacts with the emissions and produces acid. It destroys the chimney and sometimes the boiler.
Conclusion: frequent switching on and off wastes more fuel than continuous operation. Buying a boiler with excess power not only makes no sense, but is also harmful to the budget and equipment.
How to solve the problem of high power and weak demand?
In an ideal situation, the boiler operates at a constant, rated output. At the same time, the outside temperature is constantly changing, and there are even abnormal jumps. What to do? Four-way mixing valves in the hydraulic system will help. Or an option with thermo-hydraulic distribution. These devices solve the problem not by adjusting the boiler power, but by adjusting the control valve. Or the speed of the circulation pump changes. The temperature of the coolant in the batteries becomes comfortable without disturbing the optimal conditions of the boiler. This solution has a minus - a high price.
| Four way mixing valve |
For gas and liquid fuel boilers, multi-stage burners solve this situation. A lower stage reduces the boiler power if necessary. Advanced models have a smooth adjustment of burner power - modulation. It is cheaper and not as troublesome as the first option.
| Multi-stage gas burner |
Solid fuel boilers also have built-in power settings and automatic fuel supply. This helps solve the issue of excess power when the external temperature changes.
| Automatic fuel supply device for liquid-heat boiler |
Determining the number of radiators for single-pipe systems
There is another very important point: all of the above is true for a two-pipe heating system, when a coolant with the same temperature enters the input of each radiator. A single-pipe system is considered much more complex: there, increasingly colder water flows to each subsequent heating device. And if you want to calculate the number of radiators for a one-pipe system, you need to recalculate the temperature every time, and this is difficult and time-consuming. Which exit? One of the possibilities is to determine the power of the radiators as for a two-pipe system, and then, in proportion to the drop in thermal power, add sections to increase the heat transfer of the battery as a whole.
In a single-pipe system, increasingly colder water flows to each radiator
Let's explain with an example. The diagram shows a single-pipe heating system with six radiators. The number of batteries was determined for two-pipe wiring. Now we need to make an adjustment. For the first heating device everything remains the same. The second one receives coolant with a lower temperature. We determine the % drop in power and increase the number of sections by the corresponding value. In the picture it turns out like this: 15kW-3kW=12kW. We find the percentage: the temperature drop is 20%. Accordingly, to compensate, we increase the number of radiators: if 8 pieces were needed, there will be 20% more - 9 or 10 pieces. This is where knowing the room will come in handy: if it’s a bedroom or a children’s room, round up, if it’s a living room or other similar room, round down
You also take into account the location relative to the cardinal points: in the north you round up, in the south you round down.
In single-pipe systems, it is necessary to add sections to radiators located further along the branch
This method is clearly not ideal: after all, it turns out that the last battery in the branch will have to be simply enormous in size: judging by the diagram, a coolant with a specific heat capacity equal to its power is supplied to its input, and in practice it is unrealistic to remove all 100%. Therefore, usually when determining the power of a boiler for single-pipe systems, they take a certain reserve, install shut-off valves and connect radiators through a bypass so that the heat transfer can be adjusted and thus compensate for the drop in coolant temperature. One thing follows from all this: the number and/or size of radiators in a single-pipe system must be increased, and more and more sections must be installed as you move away from the beginning of the branch.
What happens if you buy a boiler of lower power?
When the owner made a mistake with the power on the smaller side, this is just as bad as an oversupply. The system is working at its limit. Service life is reduced. The house is not heated sufficiently; the system may freeze during abnormal frosts.
| Freezing of the heating system |
Why contact a specialist?
We found out that buying a boiler from the square is wrong. It is important to consider the heat loss of the building. For example, a house of 300 m² can be heated by a 15 kW boiler if all surfaces are thoroughly insulated. And a house of 150 m² may require 30 kW equipment with thin walls and uninsulated roof and ventilation.
There are hundreds of standards and regulations on this topic, there are dozens of formulas. Sometimes one thing contradicts the other, or standards change and it is difficult for a non-professional to understand whether these requirements are relevant. You can calculate all this armed with a stack of reference books. Or contact specialists who will make accurate calculations in the program and explain all the details. They will help you save money, time, and model an efficient home heating system.
