How much electricity does an electric boiler consume: typical consumption

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The use of electricity as an energy source for heating a country house is attractive for many reasons: easy accessibility, prevalence, environmental friendliness. At the same time, the most important obstacle to the use of electric boilers remains fairly high rates.

Do you also think about the advisability of installing an electric boiler? Let's see together how much electric boiler consumes electricity. For what we will use the rules for performing calculations and formulas considered in our article.

Calculations will help to understand in detail how much kWh of electricity will have to be paid monthly in the case of using electric power boilers for heating a house or apartment. The obtained figures will make the final decision on the purchase / non-purchase of the boiler.

The content of the article:

  • Methods for calculating power boiler
  • The procedure for calculating the power of an electric boiler
    • Stage # 1 - collection of initial data for calculation
    • Stage # 2 - calculating the heat loss of the basement floor
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    • Stage # 3 - calculation of heat loss from the ceiling
    • Stage # 4 - calculation of total heat loss of the cottage
    • Stage # 5 - Calculate Electricity Costs
    • Stage # 6 - Calculate the seasonal heating costs.
  • Conclusions and useful video on the topic

Methods for calculating power boiler

There are two main methods for calculating the required power of an electric boiler. The first is based on the heated area, the second on the calculation of heat loss through the building envelope.

The calculation of the first option is very rough, based on a single indicator - power density. Specific power is given in reference books and depends on the region.

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Advantages of installing an electric boiler

Installation of electrical equipment for the heating system is distinguished by the lowest price and simple scheme

Strong advantages of operating an electrical unit

Electric boiler does not need to be heated, to provide fuel and to arrange a chimney. For the organization of heating with it does not need a boiler room

Disadvantages of heating systems with electric boiler

Weighty minus the use of electricity - inhuman tariffs for electricity and dependence on centralized networks

Selection of an electric boiler of sufficient power

Work requires good electrical power and uninterrupted power supply. Therefore, before buying you need to calculate everything, including expenses.

Advantages of installing an electric boiler

Advantages of installing an electric boiler

Strong advantages of operating an electrical unit

Strong advantages of operating an electrical unit

Disadvantages of heating systems with electric boiler

Disadvantages of heating systems with electric boiler

Selection of an electric boiler of sufficient power

Selection of an electric boiler of sufficient power

The calculation of the second option is more complicated, but takes into account the many individual indicators of a particular building. Full heat engineering calculation of the building is a rather complicated and painstaking task. Further, a simplified calculation will be considered, nevertheless possessing the necessary accuracy.

Regardless of the method of calculation, the quantity and quality of the collected source data directly affect the correct estimate of the required power of the electric boiler.

With low power, the equipment will constantly work with the maximum load, without providing the necessary comfort of living. With overpowering - unreasonably high electricity consumption is the high cost of heating equipment.

Electric meter

Unlike other types of fuel, electricity is an environmentally safe, fairly clean and simple option, but tied to the presence of an uninterrupted power network in the region

The procedure for calculating the power of an electric boiler

Further, we consider in detail how to calculate the required boiler capacity so that the equipment fully performs its task of heating the house.

Stage # 1 - collection of initial data for calculation

For the calculations will need the following information about the building:

  • S - the area of ​​the heated room.
  • Wud - power density.

The specific power indicator shows how much thermal energy is needed per 1 m2 at 1 o'clock

Depending on local environmental conditions, the following values ​​can be accepted:

  • for the central part of Russia: 120 - 150 W / m2;
  • for southern regions: 70-90 W / m2;
  • for northern regions: 150-200 W / m2.

Wud - a theoretical value, which is used mainly for very rough calculations, because it does not reflect the real heat loss of the building. Does not take into account the area of ​​the glazing, the number of doors, the material of the outer walls, the height of the ceilings.

Accurate thermal calculation is made using specialized programs, taking into account many factors. For our purposes, such a calculation is not needed; it is quite possible to dispense with the calculation of the heat loss of external enclosing structures.

The values ​​that need to be used in the calculations:

R - heat transfer resistance or coefficient of heat resistance. This is the ratio of the temperature difference along the edges of the building envelope to the heat flux passing through this structure. Has dimension m2×⁰C / W.

In fact, everything is simple - R expresses the ability of the material to retain heat.

Q - value showing the amount of heat flux passing through 1 m2 surface at a temperature difference of 1⁰C for 1 hour. That is, it shows how much heat loss 1 m2 fencing per hour with a temperature difference of 1 degree. Has the dimension of W / m2×h

For the calculations presented here, there is no difference between Kelvin and degrees Celsius, since it is not the absolute temperature that is important, but only the difference.

Qgeneral- the amount of heat flux passing through the area S of the building envelope per hour. It has the dimension W / h.

P - power of the heating boiler. Calculated as the required maximum power of the heating equipment at the maximum temperature difference between the outdoor and indoor air. In other words, the boiler has enough power to heat the building during the coldest season. It has the dimension W / h.

Efficiency - The efficiency of the heating boiler, a dimensionless quantity showing the ratio of energy received to energy expended. The documentation for the equipment is usually given as a percentage of 100, for example 99%. In the calculations, the value from 1 is used. 0.99.

∆T - shows the temperature difference from two sides of the building envelope. To make it clearer how the difference is calculated correctly see an example. If outside: -30 °C, and inside +22 ° C, then ∆T = 22 - (-30) = 52 ° С

Or the same, but in kelvins: ∆T = 293 - 243 = 52K

That is, the difference will always be the same for degrees and kelvins, so reference data in kelvins can be used without corrections for calculations.

d - thickness of the building envelope in meters.

k - coefficient of thermal conductivity of the building envelope material, which is taken from reference books or SNiP II-3-79 "Building heat engineering" (SNiP - building codes and regulations). It has the dimension W / m × K or W / m × С.

The following list of formulas shows the relationship of values:

  • R = d / k
  • R = ∆T / Q
  • Q = ∆T / R
  • Qgeneral = Q × S
  • P = Qgeneral / Efficiency

For multilayer structures, the heat transfer resistance R is calculated for each structure separately and then summed.

Sometimes the calculation of multilayer structures can be too cumbersome, for example, when calculating the heat loss of a window glass unit.

What should be considered when calculating the heat transfer resistance for windows:

  • glass thickness;
  • the number of glasses and air gaps between them;
  • type of gas between the glasses: inert or air;
  • the presence of thermal insulation coating window glass.

However, you can find ready-made values ​​for the whole structure, either at the manufacturer or in the reference book, at the end of this article there is a table for the double-glazed windows of a common structure.

Stage # 2 - calculating the heat loss of the basement floor

Separately, it is necessary to stop the calculation of heat loss through the floor of the building, as the soil has a significant resistance to heat transfer.

When calculating the heat loss of the basement, it is necessary to take into account the penetration into the ground. If the house is at ground level, then the depth is assumed to be 0.

According to the generally accepted method, the floor area is divided into 4 zones.

  • 1 zone - recedes 2 m from the outer wall to the center of the floor around the perimeter. In the case of deepening of the building, it retreats from the ground level to the level of the floor along a vertical wall. If the wall is buried in the ground for 2 m, then zone 1 will be completely on the wall.
  • 2 zone - retreats 2 m along the perimeter to the center from the border of zone 1.
  • 3 zone - retreats 2 m along the perimeter to the center from the border of zone 2.
  • 4 zone - the remaining sex.

For each zone from the established practice, their own R are set:

  • R1 = 2.1 m2×° C / W;
  • R2 = 4.3 m2×° C / W;
  • R3 = 8.6 m2×° C / W;
  • R4 = 14.2 m2×° C / W

The R values ​​given are valid for uncoated floors. In the case of insulation, each R is increased by R insulation.

Additionally, for floors laid on logs, R is multiplied by a factor of 1.18.

Floor zone layout

Zone 1 is 2 meters wide. If the house is buried, then you need to take the height of the walls in the ground, take from 2 meters, and transfer the rest to the floor

Stage # 3 - calculation of heat loss from the ceiling

Now you can start the calculations.

The formula, which can be used for a rough estimate of the power of an electric boiler:

W = Wud × S

Task: calculate the required boiler capacity in Moscow, heated area 150m².

When making calculations, we take into account that Moscow belongs to the central region, i.e. Wud can be taken as 130 W / m2.

Wud = 130 × 150 = 19500W / h or 19.5kW / h

This figure is so inaccurate that does not require consideration of the efficiency of heating equipment.

Now we determine the heat loss in 15m2 the area of ​​the ceiling, insulated with mineral wool. The thickness of the insulation layer is 150mm, the outdoor temperature is -30 ° C, inside the building is +22 ° C for 3 hours.

Solution: according to the table we find the coefficient of thermal conductivity of mineral wool, k = 0.036 W / m×° s Thickness d must be taken in meters.

The calculation procedure is as follows:

  • R = 0.15 / 0.036 = 4.167 m2×° C / W
  • ∆T = 22 - (-30) = 52 ° С
  • Q = 52 / 4.167 = 12.48 W / m2× h
  • Qgeneral = 12,48 × 15 = 187 W / h.

Calculated that the loss of heat through the ceiling in our example will be 187 * 3 = 561W.

For our purposes, it is quite possible to simplify the calculations, calculating the heat loss of only the external structures: walls and ceilings, without paying attention to the internal partitions and doors.

In addition, you can do without calculating the heat loss to the ventilation and sewage. We will not take into account the infiltration and wind load. Dependence of the location of the building on the cardinal points and the amount of received solar radiation.

From general considerations, one conclusion can be made. The greater the volume of the building, the less heat loss per 1 m2. This is easy to explain, since the area of ​​the walls increases quadratically and the volume in the cube. The ball has the least heat loss.

In enclosing structures only closed air layers are taken into account. If your house has a ventilated facade, then this air layer is not closed, it is not taken into account. Not taken all the layers that follow in front of an open air layer: facade tiles or cassettes.

Closed air layers, for example, in glass units are taken into account.


All walls of the house are external. The attic is not heated, the thermal resistance of roofing materials is not taken into account

Stage # 4 - calculation of total heat loss of the cottage

After the theoretical part, you can proceed to the practical.

For example, we calculate the house:

  • external wall dimensions: 9x10 m;
  • height: 3 m;
  • double-glazed window 1.5×1.5 m: 4 pcs;
  • oak door 2.1×0.9 m, thickness 50 mm;
  • pine floors 28 mm, on top of extruded polystyrene with a thickness of 30 mm, laid on logs;
  • ceiling GKL 9 mm, on top of mineral wool 150mm thick;
  • wall material: masonry 2 silicate bricks, mineral wool insulation 50 mm;
  • the coldest period is 30 ° С, the design temperature inside the building is 20 ° С.

We will make preliminary calculations of the required space. When calculating the zones on the floor, we take the zero depth of the walls. Board floor laid on the logs.

  • windows - 9 m2;
  • door - 1.9 m2;
  • walls, minus windows and doors - 103.1 m2;
  • ceiling - 90 m2;
  • area of ​​floor zones: S1 = 60 m2, S2 = 18 m2, S3 = 10 m2, S4 = 2 m2;
  • ΔT = 50 ° С.

Further, using reference books or tables given at the end of this chapter, we select the necessary values ​​of the thermal conductivity coefficient for each material. We recommend to get acquainted in more detail with thermal conductivity coefficient and its values ​​for the most popular building materials.

For pine boards, the coefficient of heat conductivity must be taken along the fibers.

The whole calculation is quite simple:

Step # 1: The calculation of heat loss through bearing wall structures includes three steps.

Calculate the coefficient of heat loss of the walls of the brickwork: RCyrus = d / k = 0.51 / 0.7 = 0.73 m2×° C / W.

The same for wall insulation: Rut = d / k = 0.05 / 0.043 = 1.16 m2×° C / W.

Heat loss 1 m2 external walls: Q = ΔT / (RCyrus + Rut) = 50 / (0,73 + 1,16) = 26,46 m2×° C / W.

As a result, the total heat loss of the walls will be: Qst = Q × S = 26.46 × 103.1 = 2728 W / h.

Step 2: Calculate heat loss through windows: Qthe window = 9 × 50 / 0.32 = 1406W / h.

Step number 3: Calculation of heat leakage through the oak door: Qtwo = 1.9 × 50 / 0.23 = 413W / h.

Step 4: Heat loss through the upper ceiling - the ceiling: Qsweat = 90 × 50 / (0.06 + 4.17) = 1064W / h.

Step 5: Calculate Rut for the floor as well in several actions.

First we find the heat loss coefficient of insulation: Rut= 0,16 + 0,83 = 0,99 m2×° C / W.

Then add Rut to each zone:

  • R1 = 3.09 m2×° C / W; R2 = 5.29 m2×° C / W;
  • R3 = 9.59 m2×° C / W; R4 = 15.19 m2×° C / W.

Step 6: Since the floor is laid on logs multiplied by a factor of 1.18:

R1 = 3.64 m2×° C / W; R2 = 6.24 m2×° C / W;

R3 = 11.32 m2×° C / W; R4 = 17.92 m2×° C / W.

Step number 7: Calculate Q for each zone:

Q1 = 60 × 50 / 3.64 = 824W / h;

Q2 = 18 × 50 / 6.24 = 144W / h;

Q3 = 10 × 50 / 11.32 = 44W / h;

Q4 = 2 × 50 / 17.92 = 6W / h.

Step 8: Now you can calculate Q for the entire floor: Qfloor = 824 + 144 + 44 + 6 = 1018W / h.

Step 9: As a result of our calculations, we can denote the sum of the total heat loss:

Qgeneral = 2728 + 1406 + 413 + 1064 + 1018 = 6629W / h.

The calculation does not include heat losses associated with sewage and ventilation. In order not to complicate beyond measure, we simply add 5% to the listed leaks.

Of course, a margin of at least 10% is needed.

Thus, the final figure of heat loss given as an example at home will be:

Qgeneral = 6629 × 1.15 = 7623W / h.

Qgeneral shows the maximum heat loss at home when the difference between the temperature of the outdoor and indoor air is 50 ° C.

If you count on the first simplified version through Wud then:

Wud = 130 × 90 = 11700W / h.

It is clear that the second version of the calculation, albeit much more complicated, but gives a more realistic figure for buildings with insulation. The first option allows to obtain a generalized value of heat loss for buildings with a low degree of thermal insulation, or even without it.

In the first case, the boiler will have every hour to fully renew the heat loss occurring through the openings, floors, walls without insulation.

In the second case, it is necessary to heat only one time before reaching a comfortable temperature. Then the boiler will only need to recover heat losses, the value of which is significantly lower than the first option.

Table 1. Thermal conductivity of various building materials.

Heat conductivity table

The table shows the coefficients of thermal conductivity for common building materials.

Table 2. The thickness of the cement joint with different types of masonry.

Brick thickness

When calculating the thickness of the masonry, a thickness of 10 mm is taken into account. Due to the cement joints, the thermal conductivity of the masonry is somewhat higher than a single brick

Table 3. Thermal conductivity of various types of mineral wool slabs.

Thermal conductivity of insulation

The table shows the values ​​of thermal conductivity for various mineral wool plates. For the insulation of facades used hard plate

Table 4. Heat loss windows of various designs.

Thermal conductivity of glass

The designations in the table: Ar - filling of glass with inert gas, K - outer glass has a heat-shielding coating, glass thickness is 4mm, the remaining figures indicate the gap between the glasses

7.6 kW / h is the calculated required maximum power that is used for heating a well-insulated building. However, electric boilers also need some charge for their own power supply.

As you have noticed a poorly insulated house or apartment will require large amounts of electricity for heating. And this is true for any type of boiler. Proper insulation of the floor, ceiling and walls can significantly reduce costs.

We have articles on the methods of insulation and the rules for the choice of insulation material on our website. We invite you to familiarize yourself with them:

  • Insulation of a private house outside: popular technologies + materials review
  • Floor insulation by logs: materials for thermal insulation + insulation schemes
  • Insulation of the attic roof: detailed instructions on the insulation in the attic of a low-rise building
  • Types of insulation for the walls of the house from the inside: materials for insulation and their characteristics
  • Insulation for the ceiling in a private house: the types of materials used + how to choose
  • Warming the balcony with your own hands: popular options and technologies for warming the balcony from the inside

Stage # 5 - Calculate Electricity Costs

If you simplify the technical nature of the heating boiler, then you can call it a conventional converter of electrical energy into its thermal counterpart. While doing the conversion work, he also consumes some energy. Those. the boiler receives a full unit of electricity, and only 0.98 of it is supplied for heating.

To obtain an accurate figure of the power consumption of the investigated electric heating boiler it is necessary power (nominal in the first case and calculated in the second) divided by the manufacturer efficiency value.

On average, the efficiency of such equipment is 98%. As a result, the amount of energy consumption will be, for example, for the design variant:

7.6 / 0.98 = 7.8 kW / h.

It remains to multiply the value at the local rate. Then calculate the total cost of electric heating and look for ways to reduce them.

For example, buy a dvuhtarifny counter that allows partial payment at lower “night” tariffs. What will need to replace the old electric meter with a new model. The procedure and rules for the replacement of detailed reviewed here.

Another way to reduce costs after replacing the meter is to include a thermal accumulator in the heating circuit in order to stock up on cheap energy at night and spend it during the day.

Stage # 6 - Calculate the seasonal heating costs.

Now that you have mastered the method of calculating future heat losses, you can easily estimate the cost of heating during the entire heating period.

According to SNiP 23-01-99 "Building climatology" in columns 13 and 14 we find the duration of the period for Moscow with an average temperature below 10 ° C.

For Moscow, this period lasts 231 days and has an average temperature of -2.2 ° C. To calculate Qgeneral for ΔT = 22.2 ° C, it is not necessary to perform the entire calculation again.

It suffices to derive Qgeneral at 1 ° C:

Qgeneral = 7623/50 = 152.46 W / h

Accordingly, for ΔT = 22.2 ° С:

Qgeneral = 152.46 × 22.2 = 3385W / h

To find the electricity consumed, multiply by the heating period:

Q = 3385 × 231 × 24 × 1.05 = 18766440W = 18766kW

The above calculation is also interesting in that it allows an analysis of the entire structure of the house in terms of the effectiveness of the use of insulation.

We considered a simplified version of the calculations. We recommend that you also read the full thermal calculation of the building.

Conclusions and useful video on the topic

How to avoid heat loss through the foundation:

How to calculate heat loss online:

The use of electric boilers as the main heating equipment is very strongly limited by the capacity of the power grids and the cost of electricity.

However, as an additional, for example to solid fuel boilercan be very effective and helpful. Can significantly reduce the time for heating the heating system or be used as the main boiler at not very low temperatures.

Do you use an electric boiler for heating? Tell us what method you calculated the required power for your home. Or maybe you just want to buy an electric boiler and you have any questions? Ask them in the comments to the article - we will try to help you.

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