How much electricity does an electric boiler consume: how to calculate before purchasing

The use of electricity as an energy source for heating a country house is attractive for many reasons: easy availability, prevalence, and environmental friendliness.At the same time, the main obstacle to the use of electric boilers remains rather high tariffs.

Have you also thought about the feasibility of installing an electric boiler? Let's figure out together how much electricity an electric boiler consumes. For which we will use the calculation rules and formulas discussed in our article.

Calculations will help you understand in detail how many kW of electricity you will have to pay monthly if you use an electric boiler to heat a house or apartment. The obtained figures will allow you to make a final decision regarding the purchase/not purchase of the boiler.

Methods for calculating the power of an electric 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 according to the first option is very rough, based on a single indicator - specific power. Specific power is given in reference books and depends on the region.

The calculation for the second option is more complicated, but takes into account many individual indicators of a particular building. A complete thermal engineering calculation of a building is a rather complex and painstaking task. Next, a simplified calculation will be considered, which nevertheless has the necessary accuracy.

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

With reduced power, the equipment will constantly operate at maximum load, not providing the necessary living comfort. With overestimated power, there is an unreasonably large consumption of electricity and a high cost of heating equipment.

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

The procedure for calculating the power of an electric boiler

Next, we will consider in detail how to calculate the required boiler power so that the equipment fully fulfills its task of heating the house.

Stage #1 - collection of initial data for calculation

To carry out calculations, you will need the following information about the building:

• S – area of ​​the heated room.
• W_beat – specific power.

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

Depends on local natural conditions, the following values ​​can be taken:

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

W_beat - 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 glazing area, the number of doors, the material of the external walls, or the height of the ceilings.

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

Quantities that need to be used in the calculations:

R – heat transfer resistance or thermal resistance coefficient. This is the ratio of the temperature difference at the edges of the enclosing structure to the heat flow passing through this structure. Has dimension m²×⁰С/W.

It's actually simple - R expresses the ability of a material to retain heat.

Q – a value indicating the amount of heat flow passing through 1 m² surfaces with a temperature difference of 1⁰C for 1 hour. That is, it shows how much thermal energy 1 m loses² building envelope per hour with a temperature difference of 1 degree. Has a dimension W/m²×h.

For the calculations given here, there is no difference between kelvins and degrees Celsius, since it is not the absolute temperature that matters, only the difference.

Q_generally – the amount of heat flow passing through the area S of the enclosing structure per hour. Has the dimension W/h.

P – heating boiler power.It is calculated as the required maximum power of the heating equipment at the maximum difference in temperature of the external and internal air. In other words, sufficient boiler power to heat the building in the coldest season. Has the dimension W/h.

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

∆T – shows the temperature difference on two sides of the enclosing structure. To make it clearer how the difference is calculated correctly, look at the example. If outside: -30 °C, and inside +22 ° C, then ∆T = 22 - (-30) = 52 °C

Or the same, but in Kelvin: ∆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 for calculations without corrections.

d – thickness of the enclosing structure in meters.

k – thermal conductivity coefficient of the building envelope material, which is taken from reference books or SNiP II-3-79 “Building Heat Engineering” (SNiP - building codes and regulations). Has the dimension W/m×K or W/m×⁰С.

The following list of formulas shows the relationship between quantities:

• R=d/k
• R= ∆T / Q
• Q = ∆T/R
• Q_generally = Q × S
• P = Q_generally / 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 double-glazed window.

What needs to be taken into account 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;
• presence of thermal insulation coating of window glass.

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

Stage #2 - calculation of heat loss from the basement floor

Separately, it is necessary to dwell on the calculation of heat loss through the floor of the building, since the soil has significant resistance to heat transfer.

When calculating the heat loss of the basement floor, 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 - retreat 2 m from the outer wall to the center of the floor along the perimeter. In case of deepening of the building, it is retreated from the ground level to the floor level along a vertical wall. If the wall is buried 2 m into the ground, 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 floor.

Based on established practice, each zone has its own R:

• 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 increases by R of the insulation.

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

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

Stage #3 - calculation of ceiling heat loss

Now you can start making calculations.

A formula that can serve to roughly estimate the power of an electric boiler:

W=W_beat × S

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

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

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

This figure is so inaccurate that it does not require taking into account the efficiency of heating equipment.

Now let’s determine the heat loss after 15m² ceiling area insulated with mineral wool. The thickness of the thermal insulation layer is 150 mm, the outside air temperature is -30 ° C, inside the building +22 ° C in 3 hours.

Solution: using the table we find the thermal conductivity coefficient of mineral wool, k=0.036 W/m×°C. 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
• Q_generally = 12,48 × 15 = 187 W/h.

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

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

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

From general considerations, one conclusion can be drawn. The larger 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 increases in a 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 such an air layer is considered not closed and is not taken into account. All layers that come before the open air layer are not taken: facade tiles or cassettes.

Closed air layers, for example, in double-glazed windows 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 the total heat loss of the cottage

After the theoretical part, you can begin the practical part.

For example, let's calculate a house:

• dimensions of external walls: 9x10 m;
• height: 3 m;
• window with double glazing 1.5×1.5 m: 4 pcs;
• oak door 2.1×0.9 m, thickness 50 mm;
• 28 mm pine floors, on top of 30 mm thick extruded foam, laid on joists;
• gypsum board ceiling 9 mm, on top of mineral wool 150 mm thick;
• wall material: masonry of 2 silicate bricks, insulation with 50 mm mineral wool;
• the coldest period is 30 °C, the estimated temperature inside the building is 20 °C.

We will make preparatory calculations of the required areas. When calculating zones on the floor, we assume zero wall depth. The floor board is laid on joists.

• windows – 9 m²;
• door – 1.9 m²;
• walls, minus windows and doors - 103.1 m²;
• ceiling - 90 m²;
• floor areas: S1 = 60 m², S2 = 18 m², S3 = 10 m², S4 = 2 m²;
• ΔT = 50 °C.

Next, using reference books or tables given at the end of this chapter, we select the required values ​​of the thermal conductivity coefficient for each material. We recommend that you read more about thermal conductivity coefficient and its values ​​for the most popular building materials.

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

The whole calculation is quite simple:

Step #1: Calculation of heat loss through load-bearing wall structures includes three steps.

We calculate the heat loss coefficient of brick walls: R_Cyrus = d / k = 0.51 / 0.7 = 0.73 m²×°C/W.

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

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

As a result, the total heat loss from the walls will be: Q_st = Q×S = 26.46 × 103.1 = 2728 Wh.

Step #2: Calculation of thermal energy losses through windows: Q_windows = 9 × 50 / 0.32 = 1406 W/h.

Step #3: Calculation of thermal energy leaks through an oak door: Q_dv = 1.9 × 50 / 0.23 = 413 W/h.

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

Step #5: Calculating R_ut for the floor also in several steps.

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

Then we add R_ut 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, we multiply 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 #7: Let's calculate Q for each zone:

Q1 = 60 × 50 / 3.64 = 824 W/h;

Q2 = 18 × 50 / 6.24 = 144 W/h;

Q3 = 10 × 50 / 11.32 = 44 W/h;

Q4 = 2 × 50 / 17.92 = 6W/h.

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

Step #9: As a result of our calculations, we can indicate the amount of total heat loss:

Q_generally = 2728 + 1406 + 413 + 1064 + 1018 = 6629Wh.

The calculation did not include heat losses associated with sewerage and ventilation. In order not to complicate things beyond measure, let’s simply add 5% to the listed leaks.

Of course, a reserve is required, at least 10%.

Thus, the final figure for heat loss of the house given as an example will be:

Q_generally = 6629 × 1.15 = 7623 W/h.

Q_generally shows the maximum heat loss of a house when the temperature difference between the external and internal air is 50 °C.

If we calculate according to the first simplified version using Wsp then:

W_beat = 130 × 90 = 11700 W/h.

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

In the first case, the boiler will have to completely renew the loss of thermal energy occurring through openings, ceilings, and walls without insulation every hour.

In the second case, it is necessary to heat until a comfortable temperature is reached only once. Then the boiler will only need to restore heat loss, the value of which is significantly lower than the first option.

Table 1. Thermal conductivity of various building materials.

The table shows thermal conductivity coefficients for common building materials

Table 2. Cement joint thickness for various types of masonry.

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

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

The table shows the values ​​of the thermal conductivity coefficient for various mineral wool slabs. A rigid slab is used to insulate facades

Table 4.Heat loss from windows of various designs.

Designations in the table: Ar – filling of double-glazed windows with inert gas, K – outer glass has a heat-protective coating, glass thickness 4 mm, the remaining numbers indicate the gap between the glasses

7.6 kW/h is the estimated required maximum power that is spent on heating a well-insulated building. However, electric boilers also need some charge to power themselves to operate.

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

We have articles on our website on insulation methods and rules for choosing thermal insulation materials. We invite you to familiarize yourself with them:

• Insulation of a private house from the outside: popular technologies + review of materials
• Floor insulation using joists: materials for thermal insulation + insulation schemes
• Insulation of an attic roof: detailed instructions on installing thermal insulation in the attic of a low-rise building
• Types of insulation for the walls of a house from the inside: materials for insulation and their characteristics
• Insulation for the ceiling in a private house: types of materials used + how to choose the right one
• Do-it-yourself insulation of a balcony: popular options and technologies for insulating a balcony from the inside

Stage #5 - calculating energy costs

If we simplify the technical essence of a heating boiler, then we can call it a conventional converter of electrical energy into its thermal analogue. While performing the conversion work, it also consumes a certain amount of 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 for the electricity consumption of the electric heating boiler under study, its power (nominal in the first case and calculated in the second) must be divided by the efficiency value declared by the manufacturer.

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

7.6 / 0.98 = 7.8 kW/h.

All that remains is to multiply the value by the local tariff. Then calculate the total cost of electric heating and start looking for ways to reduce them.

For example, buy a two-tariff meter, which allows you to partially pay at lower “night” rates. Why do you need to replace the old electric meter with a new model? The procedure and rules for performing replacement in detail reviewed here.

Another way to reduce costs after replacing the meter is to include a thermal accumulator in the heating circuit to store cheap energy at night and use it during the day.

Stage #6 - calculating seasonal heating costs

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

According to SNiP 23-01-99 “Building climatology” in columns 13 and 14 we find for Moscow the duration of the period 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 Q_generally for ΔT=22.2 °C, it is not necessary to perform the entire calculation again.

It is enough to output Q_generally by 1 °C:

Q_generally = 7623 / 50 = 152.46 W/h

Accordingly, for ΔT= 22.2 °C:

Q_generally = 152.46 × 22.2 = 3385Wh

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

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

The above calculation is also interesting because it allows us to analyze the entire structure of the house from the point of view of the effectiveness of insulation.

We considered a simplified version of the calculations. We also recommend that you read the full thermal engineering 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 limited by the capabilities of electrical networks and the cost of electricity.

However, as an additional, for example to solid fuel boiler, can be very effective and useful. They can significantly reduce the time it takes to warm up 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 used to calculate the required power for your home. Or maybe you just want to buy an electric boiler and have questions? Ask them in the comments to the article - we will try to help you.