# Choosing a heating boiler: how to calculate the boiler power

## Recommendation points

Private developers will probably be interested in how to correctly calculate the power of heating equipment. The article will acquaint readers with the basics of the methodology of heat engineering calculations, so that buying a boiler does not turn out to be unnecessarily expensive, and a house in winter is too cold.

## What determines the required power

Each building is unique due to the heat engineering processes taking place in it. When choosing a boiler power, the most interesting indicators are heat losses. Strictly speaking, the boiler output must be equivalent to the amount of heat that leaves the house due to the temperature difference between the two environments. There are a number of major areas with the highest leak rates:

1. Roof and ceiling slab – up to 30%.
2. Walls and floors – up to 50%.
3. Windows and doors – 15%.
4. Ventilation – 5-10%.

To set quantitative values ​​of heat loss, you can use all kinds of calculators. More accurate data are calculated by individual calculations, for which they are of fundamental importance:

1. Thickness and material of walls, floors.
2. Temperature difference between outdoor and indoor air.
3. The area of ​​the enclosing walls.
4. Number of doors and windows, type of glass units.
5. Ventilation capacity.
6. The depth of soil freezing, average wind loads on the house and other factors.

There are two ways to reach the nominal power required to maintain a comfortable temperature inside the building. The first consists in an almost unlimited increase in the capacity of heating equipment. The second option implies a detailed study of the thermal technical features of the house, the calculation of the value of heat losses for each enclosing plane and the feasible elimination of leaks. In the latter aspect, thermal imaging research of premises and the house as a whole is becoming more and more relevant..

## Calculation of building heat losses

The essence of the heat engineering calculation will be easiest to explain using the example of a rather primitive one-story building. The initial data are as follows:

1. Total area of ​​the house: 100 m2.
2. Ceiling height: 2.5 m.
3. Walls: aerated concrete 30 cm.
4. Overlap: wooden beams 25 cm thick with mineral wool filling with a density of 60 kg / m3.
5. Thermal insulation: external, expanded polystyrene 50 mm thick.
6. Ventilation: air exchange up to 40 m3 in hour.
7. Floor: monolithic concrete with 20 cm filling on the ground from crushed expanded clay.

Desired temperature inside the house: 23-25 ​​° С, in the region the average January temperature is -5 …- 6 ° С. Since the calculation is performed to determine the maximum output of the heating system, it is necessary to correlate this data with the lowest temperature that occurs during the year. Let’s say it is -25 ° С, the calculation for this value will be performed in parallel.

For convenience, we will divide the house into four zones of 25, 35 and 40 m2, using the contours of the internal partitions as a guide. This breakdown of calculations into small stages also makes the process easier..

In an area of ​​25 m2 there are two external walls, their total area is 26.9 m2. In addition, we have almost 25 m2 floor and the same ceiling. For each type of building envelope, we calculate individually using the formula:

• Qsweat = S x (? T) x (1 + Qext1 + Qext2 + … + QaddN) x Kposes / Rogre

Here: S is the area of ​​a uniform fence (m2).

Qext – the share of additional losses: through ventilation, opening doors or cold bridges, also with ceilings over 4 m. The value is conditional, the total value of the order of 1.5-2 will give a good “safety margin”.

TOposes – tabular value of the coefficient of position (position) of the design structure relative to the outside air. In rough calculations, it is taken equal to 1, depending on the orientation of the wall, it varies from 1.05 to 1.1.

Rogre – resistance to thermal transfer, each material is different (m2Kelvin / W).

Substituting the values ​​we know into the formula, we get:

1. For floor Qsweat = 25 x 14 x 1.22 x 1.06 / 0.853 = 0.53 kWh for average temperature (0.72 kWh for minimum).
2. For ceiling Qsweat = 25 x 27 x 1.95 x 1 / 1.3 = 1.01 kWh for average temperature (1.76 kWh for minimum).
3. For walls Qsweat = 26.9 x 30 x 1.85 x 1.05 / 1.12 = 1.44 kWh for average temperature (2.33 kWh for minimum).

Accordingly, the total heat loss of the first zone is 2.98 kWh. Avoiding additional calculations, we will proportionally increase the heat loss of the room by 25 m2 four times, having received 11.92 kWh average and 19.24 kWh peak heat loss for the whole house. The latter, in fact, is the desired power of the heating boiler (very close to reality), but not everything is so simple.

## Types of energy – what are the differences

If you are not ready or do not want to supply your home with geothermal heating, then there are few options for choosing an energy carrier: electricity, gas or solid fuel. Regarding the last two, one thing can be said unequivocally: the boiler equipment on them has the ability to convert only a certain part of the fuel into useful heat, the rest is wasted.

There are several reasons for this: incomplete combustion of fuel, loss of part of the heat with combustion products, automatic failures leading to inertial overheating. Thus, up to a third of the money spent on gas heating is wasted, solid fuels have even higher leakages.

Electric boilers are devoid of this drawback: how many kilowatts of power the input cable missed, almost the same amount remained inside the house, because energy conversion is performed with an efficiency of 99% for almost all types of heaters.

## What determines the efficiency of the boiler

First of all, from the correct download. For gas equipment, this is the burner performance, for solid fuel equipment – the mass of fuel in the furnace. Optimally select the amount of fuel in such a way that all the heat from its combustion can be absorbed by heat exchangers.

It is very important to understand that the closer the boiler’s rated power is to the peak, the more pronounced incomplete heat transfer and chemically defective combustion are. It is optimal to select the boiler power 20-25% higher than the expected peak, so that the equipment does not work in the mode of enhanced forcing.

Much also depends on the boiler design. Modern units provide protection against heat loss through the body, continuous environment of the furnace with a coolant and highly efficient automatic control schemes. Do not forget also about the importance of thrust control: most of the fuel does not burn out due to lack of oxygen.

## Automation and auxiliary equipment

It’s up to you: just to increase the boiler’s power or try to improve its efficiency. The latter can be achieved by installing automation, which will heat the coolant at optimal moments of high temperature difference, maintaining a constant air temperature. The efficiency of heating the water in the system can also be increased by forced circulation and overpressure..

There is a considerable arsenal of means for utilizing residual heat – all kinds of recuperators and economizers. We can confidently say that such boiler equipment works well at capacities over 40 kW, but even at a smaller scale, the equipment can be quite effective..

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