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Calculations for heat. Independent calculation of individual heating system

In a private house, you need to do everything yourself (specialist) "hands", including counting, designing, buying and installing a heating system. In order to start the organization of communications in the house, it is necessary to make a thermal calculation of the heating system. Below is explained how and why it is done.

Classic heat calculation heating system is a consolidated technical document that includes mandatory step-by-step standard methods of calculation.

But before studying these calculations of the main parameters it is necessary to determine the concept of the heating system itself.

Heating - a multicomponent system to ensure an approved temperature in the room / building. Is a separate part of the complex of communications of modern housing

The heating system is characterized by forced delivery and involuntary heat dissipation in the room. The main tasks of calculating and designing a heating system:

the most reliable is to determine the heat loss
determine the quantity and conditions of use of the coolant
as accurately as possible to choose elements of generation, movement and return of heat

When building a heating system, you must first collect a variety of data about the room / building where the heating system will be used. After calculating the thermal parameters of the system, analyze the results of arithmetic operations. Based on the data obtained, select the components of the heating system with the subsequent purchase, installation and commissioning.

It is noteworthy that this method thermal calculation allows us to accurately calculate a large number of quantities that specifically describe the future heating system. As a result of thermal calculation, the following information will be available:

number of heat losses, boiler output;
number and type of heat radiators for each room separately;
hydraulic characteristics of the pipeline;
volume, velocity of the heat carrier, power of the pump.

Thermal calculation is not theoretical sketches, but quite accurate and well-grounded results, which are recommended to be used in practice when selecting components of the heating system.

Room temperature conditions

Before carrying out any calculations of the parameters of the system, it is necessary, at a minimum, to know the order of the expected results, and also to have standardized characteristics of some tabulated values that must be substituted into formulas or oriented to them. After calculating the parameters with such constants, one can be sure of the reliability of the desired dynamic or constant parameter of the system.

For premises of various purposes, there are standard standards for temperature regimes for residential and non-residential premises. These norms are fixed in the so-called GOSTs

For a heating system, one of such global parameters is the room temperature, which should be constant regardless of the year and environmental conditions.

According to the regulations of sanitary regulations and rules there is a difference in temperature relative to the summer and winter period of the year. For the temperature regime of the room during the summer season, the air conditioning system is responsible, but the room temperature in the winter is provided by the heating system. I mean, we are interested in the temperature ranges and their tolerances for the winter season.

Most regulatory documents specify the following temperature ranges, which allow a person to be comfortable in the room. For non-residential premises of office type with an area of up to 100 m 2:

optimum air temperature 22-24 ° С
permissible fluctuation 1 ° С

For offices of office type with an area of more than 100 m 2, the temperature is 21-23 ° C. For non-residential premises of industrial type, the temperature ranges vary greatly depending on the intended use of the premises and the established occupational safety standards.

Comfortable room temperature for each person "own". Someone likes to be very warm in the room, someone is comfortable when the room is cool - it's all quite individual

As for living quarters: apartments, private houses, estates, etc., there are certain temperature ranges that can be adjusted depending on the wishes of the tenants. And yet for specific premises of an apartment and a house we have:

residential, including children's room, room 20-22 ° С, tolerance ± 2 ° С
kitchen, toilet 19-21 ° С, tolerance ± 2 ° С
bathroom, shower room, swimming pool 24-26 ° С, tolerance ± 1 ° С
corridors, hallways, staircases, storerooms 16-18 ° С, tolerance + 3 ° С

It is important to note that there are several basic parameters that influence the temperature in the room and which need to be oriented when calculating the heating system: humidity (40-60%), oxygen and carbon dioxide concentration in the air (250: 1), air movement speed masses (0.13-0.25 m / s), and so on.

Calculation of heat loss in the house

According to the second law of thermodynamics (school physics) there is no spontaneous transfer of energy from less heated to more heated mini- or macro-objects. A particular case of this law is the "striving" to create a temperature equilibrium between two thermodynamic systems.

For example, the first system is an environment with a temperature of -20 ° C, the second system is a building with an internal temperature of + 20 ° C. According to the law, these two systems will seek to balance themselves through the exchange of energy. This will occur with the help of heat losses from the second system and cooling in the first.

It can be unequivocally said that the ambient temperature depends on the latitude on which the a private house. And the temperature difference affects the amount of heat leakage from the building

By heat loss means the involuntary release of heat (energy) from some object (home, apartment). For an ordinary apartment, this process is not so "noticeable" in comparison with a private house, since the apartment is located inside the building and "neighbors" with other apartments. In a private house, through the external walls, the floor, the roof, the windows and the doors, heat "goes away" to some extent.

Knowing the amount of heat losses for the most unfavorable weather conditions and the characteristics of these conditions, it is possible to calculate the power of the heating system with high accuracy.

So, the amount of heat leakage from the building is calculated by the following formula:

Q = Q Floor + Q Wall + Q Window + Q Roof + Q Door + ... + Q i

where Qi is the volume of heat loss from a homogeneous shell shell. Each component of the formula is calculated by the formula:

Q = S * ΔT / R

where Q is the thermal leakage (Watts), S is the area of a particular type of construction (m 2), ΔT is the difference between ambient air temperature and indoor temperature (° C), R is the thermal resistance of a certain type of construction (m 2 * ° C / W).

The very value of thermal resistance for real materials is recommended to be taken from auxiliary tables. In addition, the thermal resistance can be obtained by the following relationship:

R = d / k

where R is the thermal resistance ((m 2 * K) / W), k is the thermal conductivity of the material (W / (m 2 * K)), d is the thickness of this material (m).

In old houses with a damp roof construction, heat leakage occurs through the upper part of the building, namely through the roof and attic. If you insulate the attic space and the roof, then the total heat loss from the house can be significantly reduced

In the house there are several more types of heat losses through the cracks in the structures, ventilation system, kitchen hood, opening windows and doors. But to take into account their volume does not make sense, since they make up no more than 5% of the total number of major heat losses.

Determination of boiler output

To maintain the temperature difference between the environment and the temperature inside the house, an autonomous heating system is needed, which maintains the desired temperature in each room of the private house.

The basis of the heating system is a boiler: liquid or solid fuel, electric or gas - at this stage it does not matter. The boiler is the central node of the heating system, which generates heat. The main characteristic of the boiler is its power, namely the rate of conversion of the amount of heat per unit of time.

After calculating the heat load for the heating, we obtain the required rated output of the boiler. For a typical multi-room apartment, the boiler output is calculated through the area and specific power:

Р boiler = (S rooms * Р specific) / 10

where S rooms - the total area of the heated room, R - specific power relative to climatic conditions. But this formula does not take into account the heat losses that are sufficient in a private house. There is another relation that takes into account this parameter:

P boiler = (Q losses * S) / 100

where P boiler - boiler power (W), Q losses - heat loss, S - heated area (m 2).

In most private house heating systems, it is recommended to use an expansion tank in which the coolant supply will be stored. Each private house needs hot water supply

In order to provide the reserve capacity of the boiler, taking into account the heating of the water for the kitchen and the bathroom, it is necessary to add the safety factor K to the last formula:

P of the boiler = (Q losses * S * K) / 100

where K - will be equal to 1.25, that is, the design capacity of the boiler will be increased by 25%. Thus, the boiler's capacity makes it possible to maintain the normative air temperature in the building's rooms, and also to have an initial and additional volume hot water in home.

Features of selection of radiators

The standard components for providing heat in the room are radiators, panels, "warm" floor systems, convectors, etc. The most common parts of the heating system are radiators.

The heat radiator is a special hollow construction of a modular type made of an alloy with a high heat output. It is made of steel, aluminum, cast iron, ceramics and other alloys. The principle of the radiator is reduced to the emission of energy from the coolant into the space of the room through the "petals".

Aluminum and bimetallic radiator has replaced massive cast-iron batteries. The simplicity of production, high heat transfer, successful design and design have made this product a popular and widespread tool for radiating heat in a room

There are several methods for calculating the number of radiator sections in the room. The following list of methods is sorted in order of increasing the accuracy of the calculation.

By area. N = (S * 100) / C, where N is the number of sections, S is the area of the room (m 2), C is the heat transfer of one section of the radiator (W, taken from those passports or the product certificate), 100 W is the amount of heat flow , which is necessary for heating 1 m 2 (empirical value). The question arises: how to take into account the height of the ceiling of the room?
By volume. N = (S * H * 41) / C, where N, S, C are similar. H - room height, 41 W - the amount of heat flow, which is necessary for heating 1 m 3 (empirical value).
By the coefficients. N = (100 * S * k1 * k2 * k3 * k4 * k5 * k6 * k7) / C, where N, S, C and 100 are similar. k1 - the number of cells in the double-glazed windows of the room window, k2 - the thermal insulation of the walls, k3 - the ratio of the area of windows to the area of the room, k4 - the average minus temperature in the coldest week of winter, k5 - the number of exterior walls of the room (which "go" to the street) k6 - type of premise on top, k7 - ceiling height.

This is the most accurate version of the calculation of the number of sections. Naturally, rounding of fractional results of calculations is always made to the next integer.

Hydraulic calculation of water supply

Of course, the "picture" of calculating heat for heating can not be complete without calculating characteristics such as volume and velocity of the coolant. In most cases, ordinary water in a liquid or gaseous state is the coolant.

The actual volume of the coolant is recommended to be calculated by summing all the cavities in the heating system. When using a single-circuit boiler - this the best option. When using double-circuit boilers in a heating system, it is necessary to take into account the costs of hot water for hygienic and other household purposes

Calculation of the volume of water heated by a double-circuit boiler to provide occupants hot water and heating of the coolant, is made by summing the internal volume of the heating circuit and the actual needs of users in heated water.

The volume of hot water in the heating system is calculated by the formula:

W = k * P

where W is the volume of the heat carrier, P is the capacity of the heating boiler, k is the power factor (the number of liters per unit of power is 13.5, the range is from 10 to 15 liters). As a result, the final formula looks like this:

W = 13.5 * P

The velocity of the heat carrier is the final dynamic evaluation of the heating system, which characterizes the rate of circulation of the liquid in the system. This value helps to evaluate the type and diameter of the pipeline:

V = (0.86 * P * μ) / ΔT

where P is the boiler output, μ is the boiler efficiency, ΔT is the temperature difference between the water supplied and the water return circuit.

Summarizing the above methods of calculating the characteristics, in the end, real results of calculations will be available, which are the "foundation" of the future heating system.

Example of heat calculation

As an example of heat calculation, there is a usual one-storey house with four living rooms, a kitchen, a bathroom, a "winter garden" and utility rooms.

The foundation is made of a monolithic reinforced concrete slab (20 cm), the outer walls are concrete (25 cm) with plaster, the roof is covered with wooden beams, the roof is made of metal and mineral wool (10 cm)

Dimensions of the building. The height of the floor is 3 meters. Small window of the front and rear part of the building 1470 * 1420 mm, large facade window 2080 * 1420 mm, entrance doors 2000 * 900 mm, rear door (exit to the terrace) 2000 * 1400 (700 + 700) mm.

The total width of the building is 9.5 m 2, the length is 16 m 2. Only residential rooms (4 pcs.), A bathroom and a kitchen will be heated. To accurately calculate heat losses on walls from the area of external walls, you need to subtract the area of all windows and doors - this is a completely different type of material with its thermal resistance

We start with the calculation of areas of homogeneous materials:

floor area 152 m 2
roof area 180 m 2 (considering the height of the attic of 1.3 meters and the width of the run - 4 meters)
the area of windows 3 * 1.47 * 1.42 + 2.08 * 1.42 = 9.22 m 2
the area of the doors will be 2 * 0.9 + 2 * 2 * 1.4 = 7.4 m 2

The area of the outer walls will be 51 * 3-9.22-7.4 = 136.38 m 2. Let's proceed to the calculation of heat loss on each material:

Q floor = S * ΔT * k / d = 152 * 20 * 0.2 / 1.7 = 357.65 W
Q roof = 180 * 40 * 0.1 / 0.05 = 14400 W
Q window = 9.22 * 40 * 0.36 / 0.5 = 265.54 W
Q doors = 7.4 * 40 * 0.15 / 0.75 = 59.2 W

And also Q wall is equivalent to 136.38 * 40 * 0.25 / 0.3 = 4546. The sum of all heat losses will be 19628.4 W. As a result, we calculate the boiler output:

P boiler = Q loss * S heating room * K / 100 =
19628.4 * (10.4 + 10.4 + 13.5 + 27.9 + 14.1 + 7.4) * 1.25 / 100 = 19628.4 * 83.7 * 1.25 / 100 = 20536.2 = 21 kW

Calculate the number of radiator sections for one of the rooms. For all others, the calculations are similar. For example, the corner room (on the left, the bottom corner of the scheme) is 10.4 m2.

N = (100 * k1 * k2 * k3 * k4 * k5 * k6 * k7) / C = (100 * 10.4 * 1.0 * 1.0 * 0.9 * 1.3 * 1.2 * 1.0 * 1.05) /180=8.5176=9

For this room you need 9 sections of a radiator with 180 W heat dissipation. Let's pass to the calculation of the amount of coolant in the system:

W = 13.5 * P = 13.5 * 21 = 283.5 liters

The velocity of the coolant will be:

V = (0.86 * P * μ) / ΔT = (0.86 * 21000 * 0.9) /20 = 812.7 liters

As a result, the total turnover of the entire volume of the coolant in the system will be equivalent to 2.87 times in one hour.

Video materials for the calculation of heating

A simple calculation of the heating system for a private house is presented in the following overview:

All the subtleties and common methods for calculating the heat loss of the building are shown below:

Another option for calculating heat leakage in a typical private house:

This video tells about the features of the energy carrier circulation for heating the home:

Thermal calculation of the heating system is individual, it must be done correctly and accurately. The more accurately the calculations are made, the less the overpayers will have to the owners a country house in the process of operation.

The system of water heating has been increasingly popular recently as the main way for heating a private house. Water heating can be supplemented with devices such as heaters that work on electricity. Some devices and heating systems appeared on the domestic market quite recently, but already managed to gain popularity. These include infrared heaters, oil coolers, a warm floor system and others. To heat a local type, a device such as a fireplace is often used.

However, recently fireplaces perform more decorative function than heating. On how well the design and calculation of private house heating has been carried out, and the water heating system has been installed, its durability and efficiency during operation depend. During the operation of such a heating system, it is necessary to adhere to certain rules so that it works as efficiently and efficiently as possible.

The heating system of a private house is not only components such as a boiler or radiators. A water-type heating system includes such elements as:

Pumps;
Means of automation;
Pipeline;
Heat carrier;
Devices for adjustment.

To calculate the heating of a private house, you need to be guided by such parameters as the capacity of the boiler. For each of the rooms in the house, it is also necessary to calculate the power of the radiators.

Boiler selection

The boiler can be of several types:

A boiler operating on liquid fuel;
A gas boiler;
Solid fuel boiler;
Combined boiler.

The choice of a boiler that will use the heating scheme of a residential house should depend on which type of fuel is the most affordable and inexpensive.

In addition to the cost of fuel, it is required at least once a year to conduct a preventive inspection of the boiler. It is best for these purposes to call a specialist. Also, you need to perform preventive cleaning of filters. The most simple to operate are boilers that operate on gas. They are also quite cheap in maintenance and repair. The gas boiler will only fit in those houses that have access to the gas main.

Gas is a type of fuel that does not require individual transportation or storage space. In addition to this, many gas boilers modern type can boast a fairly high efficiency.

Boilers of this class are distinguished by a high degree of safety. Modern boilers are designed in such a way that they do not need to allocate a special room for the boiler house. Modern boilers are characterized by a beautiful appearance and are able to successfully fit into the interior of any kitchen.

To date, semi-automatic boilers operating on solid fuel are very popular. True, there is one disadvantage of such boilers, which is that once a day you need to load fuel. Many manufacturers produce such boilers that are fully automated. In such boilers solid fuel is loaded in an autonomous mode.

Make a calculation of the heating system of a private house is possible in the case of a boiler operating on electricity.

However, such boilers are a bit more problematic. In addition to the main problem, which is that now electricity is quite expensive, they can still reboot the network. In small settlements, an average of 3 kW per hour is allocated per house, and this is not enough for a boiler, and it should be borne in mind that the network will be loaded not only by the operation of the boiler.

For the organization of the heating system of a private house, it is possible to install a liquid-fuel type boiler. The disadvantage of such boilers is that they can cause criticism in terms of ecology and safety.

Calculation of boiler power

Before calculating the heating in the house, it is necessary to do this with the calculation of the boiler output. The efficiency of the entire heating system will depend on the boiler's capacity, first of all. The main thing in this matter is not to overdo it, since a too powerful boiler will consume more fuel than necessary. And if the boiler is too weak, it will not be possible to heat the house properly, and this will negatively affect the comfort in the house. Therefore, the calculation of the heating system of a country house is important. To choose a boiler of the necessary capacity it is possible, if in parallel to calculate the specific heat losses of the building for the entire heating period. Calculation of home heating - specific heat loss can be the following method:

q house = Q year / F h

Qgod is the heat energy consumption for the entire heating period;

Fh is the area of the house that is heated;

In order to calculate the heating of a country house - the energy consumption that will leave the heating of a private house, you need to use the following formula and a tool such as a calculator:

Q year = β h *

As it is easy to see, we need an INNER diameter of 20 millimeters, which for polypropylene corresponds to an external diameter of 25 mm.

Conclusion

Some additional information on the calculation methods of heating systems can be found in the video attached to the article. Warm winters!

Heating a private house is a necessary element of comfortable housing. And the arrangement of the heating system should be approached carefully, because errors will not be cheap. Let us consider how the calculation of the heating system of a private house is performed to effectively fill the heat losses in the winter months.

The building loses its heat due to the difference in air temperatures inside and outside the house. The heat loss is the higher, the more significant the area of the enclosing structures of the building (windows, roof, walls, foundation).

Also, the loss of thermal energy is associated with the materials of the enclosing structures and their dimensions. For example, heat losses of thin walls are greater than thick ones.

The walls, the roof, the windows and the doors - everything passes the heat in winter outward. The thermal imager will clearly show the heat leakage

Effective calculation of heating for a private house necessarily takes into account the materials used in the construction of enclosing structures. For example, with an equal thickness of a wall made of wood and brick, heat is carried out with varying intensity - heat loss through wooden structures is slower. Some materials let the heat pass better (metal, brick, concrete), others worse (wood, mineral wool, expanded polystyrene).

Significantly reduce heat leakage, passing through the building structure, door / window openings can properly arranged system of thermal insulation (+)

The atmosphere inside the residential building is indirectly related to the external air environment. Walls, openings of windows and doors, roof and foundation in winter transmit heat from the house to the outside, supplying the cold in return. They account for 70-90% of the total heat loss of the cottage.

The constant leakage of heat energy for the heating season also occurs through ventilation and sewerage. When calculating the heat loss of the IZHS construction, these data are usually not taken into account. But the inclusion in the overall heat calculation of the house heat losses through the sewer and ventilation system - the solution is still the right one.

For a competent calculation of the heating system, the coefficient of thermal conductivity of common building materials will be required (+)

Calculate the autonomous heating circuit of a country house without an estimate of the heat loss of its enclosing structures is impossible. More precisely, it will not be possible to determine the capacity of the heating boiler, sufficient to heat the cottage in the most severe frosts.

The analysis of the real heat energy flow through the walls will allow to compare the costs of boiler equipment and fuel with the costs of thermal insulation of the enclosing structures. After all, the more energy-efficient the house is, i.e. The less heat energy it loses in the winter months, the lower the cost of purchasing fuel.

Calculation of heat losses through walls

On the example of a conditional two-story cottage we calculate the heat loss through its wall structures. Initial data: a square "box" with facade walls 12 m wide and 7 m high; in the walls of 16 openings, the area of each 2.5 m 2; material of facade walls - full brick ceramic; the thickness of the wall is 2 bricks.

For accurate calculations, the thermal conductivity coefficient of the heat-insulating materials listed in the table used in construction (+)

Resistance to heat transfer. To find this figure for the front wall, you need to divide the thickness of the wall material by its thermal conductivity coefficient. For a number of structural materials, the data on the coefficient of thermal conductivity are presented in the images above and below.

Our conventional wall is built of ceramic full-bodied brick, the coefficient of thermal conductivity of which is 0.56 W / m С. С. C. Its thickness, taking into account the masonry on the CPR - 0,51 m. Dividing the thickness of the wall by the coefficient of thermal conductivity of the brick, we obtain resistance to the heat transfer of the wall:

0.51: 0.56 = 0.91 W / m 2 × o C

The result of dividing is rounded to two decimal places, there is no need for more accurate data on the resistance of heat transfer.

Area of external walls

Since an example is a square building, the area of its walls is determined by multiplying the width by the height of one wall, then by the number of external walls:

12 · 7 · 4 = 336 m 2

So, we know the area of the facade walls. But how do the openings of windows and doors occupying 40 m2 (2.5 × 16 = 40 m 2) of the front wall together, do they need to be taken into account? Indeed, how correctly to calculate the autonomous heating in wooden house without taking into account the resistance to heat transfer of window and door structures.

Coefficient of thermal conductivity of heat-insulating materials used for insulation bearing walls (+)

However, for low-rise buildings IZHS, built from traditional materials, door and window openings can be neglected. Those. Do not take their area out of the total area of the facade walls.

Total heat loss of walls

We find out the heat loss of the wall from its one square meter with the difference in the temperature of the air inside and outside the house of one degree. To do this, divide the unit by the resistance of the wall heat transfer, calculated earlier:

1: 0.91 = 1.09 W / m 2 · ° C

Knowing the heat loss per square meter of the perimeter of the outer walls, it is possible to determine the heat loss at certain street temperatures. For example, if the temperature of the cottage is +20 о С, and in the street -17 о С, the temperature difference will be 20 + 17 = 37 о С. In this situation, the total heat loss of the walls of our conditional house will be:

0,91 (resistance to heat transfer per square meter of the wall) · 336 (area of facade walls) · 37 (temperature difference between indoor and outdoor atmosphere) = 11313 W

Thermal conductivity coefficient of heat-insulating materials used for floor / wall insulation, for dry floor screeding and wall leveling (+)

We recalculate the received value of heat loss in kilowatt-hours, they are more convenient for the perception and subsequent calculations of the capacity of the heating system.

Heat losses of walls in kilowatt-hours

First, let's find out how much heat energy will go through the walls in one hour with a temperature difference of 37 ° C. We remind you that the calculation is for a house with design characteristics that are conditionally chosen for demonstration-demonstration calculations:

11313 (heat loss value obtained earlier) · 1 (hour): 1000 (number of watts in kilowatts) = 11,313 kWh.

Coefficient of thermal conductivity of building materials used for insulation of walls and ceilings (+)

To calculate the heat loss for a day, the heat loss value for an hour is multiplied by 24 hours:

11,313 · 24 = 271,512 kW · h

For clarity, let's find out the heat losses for the full heating season:

7 (number of months in the heating season) · 30 (number of days in the month) · 271,512 (daily heat loss of walls) = 57017,52 kW · h

So, the estimated heat losses at home with the characteristics of the enclosing structures selected above are 57017.52 kWh for the seven months of the heating season.

Heat losses during ventilation of a private house

Calculation of the ventilation losses of heat in the heating season as an example will be carried out for a conditional cottage with a square shape, with a wall of 12 m width and 7 m height. Without furniture and internal walls, the internal volume of the atmosphere in this building will be:

12 · 12 · 7 = 1008 m 3

At an air temperature of +20 o C (the norm during the heating season), its density is 1.2047 kg / m 3, and the specific heat is 1.005 kJ / (kg · о С). We calculate the mass of the atmosphere in the house:

1008 (volume of home atmosphere) · 1,2047 (air density at t +20 о С) = 1214.34 kg

Table with the value of the coefficient of thermal conductivity of materials, which may be required for accurate calculations (+)

Suppose a five-fold change in the air volume in the premises of the house. Note that the exact demand for fresh air is dependent on the number of residents of the cottage. With an average temperature difference between the house and the street in the heating season equal to 27 ° C (20 ° C domestic, -7 ° C external atmosphere) a day to heat the fresh air supply, you need thermal energy:

5 (number of indoor air changes) · 27 (difference in room and outdoor atmosphere temperatures) · 1214.34 (air density at t + 20 ° C) · 1,005 (specific heat of air) = 164755.58 kJ

Let's translate kilojoules into kilowatt-hours:

164755,58: 3600 (number of kilojoules in one kilowatt hour) = 45.76 kWh

Having found out the cost of heat energy for heating the air in the house with a five-fold replacement through intake ventilation, you can calculate the "air" heat loss for a seven-month heating season:

7 (number of "heated" months) · 30 (average number of days in a month) · 45.76 (daily heat consumption for heating of supply air) = 9609.6 kWh

Ventilation (infiltration) energy costs are inevitable, since renewal of air in the cottage's rooms is vital. The heating needs of a changing air atmosphere in the house are to be calculated, summed up with heat losses through the enclosing structures and taken into account when choosing a heating boiler. There is another type of thermal energy consumption, the latter - sewage heat loss.

Heat losses for hot water

If in warmer months a crock enters the cottage cold water, then in the heating season it is icy, with a temperature not exceeding +5 ° C. Bathing, washing dishes and washing are impossible without heating the water. The water that is put into the toilet bowl contacts the walls with a home atmosphere, taking a little heat. What happens to water heated by burning non-free fuel and spent on domestic needs? It is poured into the sewer.

Double-circuit boiler with boiler indirect heating, used both for heating the coolant, and for the supply of hot water in the contour constructed for it

Consider the example. A family of three, say, spends 17 m 3 of water every month. 1000 kg / m 3 is the density of water, and 4,183 kJ / kg · о С is its specific heat. The average temperature of heating water intended for domestic needs, let it be +40 o C. Accordingly, the difference in the average temperature between those entering the house cold water (+5 о С) and heated in a boiler (+30 о С) it turns out 25 о С.

For the calculation of sewage heat loss we consider:

17 (monthly water flow rate) · 1000 (water density) · 25 (difference in temperatures of cold and heated water) · 4.183 (specific heat of water) = 1777775 kJ

To recalculate kilojoules into more understandable kilowatt-hours:

1777775: 3600 = 493.82 kWh

Thus, during the seven-month period of the heating season, the sewage system leaves thermal energy in volume:

493.82 · 7 = 3456.74 kWh

The consumption of heat energy for heating water for hygiene needs is low, in comparison with heat losses through walls and ventilation. But this is also energy consumption, which loads a boiler or boiler and causes fuel consumption.

Calculation of boiler output

The boiler in the heating system is designed to compensate for the heat loss of the building. And also, in the case of a two-circuit system or when the boiler is equipped with an indirect heating boiler, to warm the water to hygienic needs.

Calculating the daily heat losses and the flow of warm water "to the sewer", you can accurately determine the required boiler output for a cottage of a certain area and the characteristics of the enclosing structures.

A single-circuit boiler produces only heating water for the heating system

To determine the capacity of the heating boiler, it is necessary to calculate the cost of heat energy at home through the facade walls and the heating of the changing air atmosphere of the interior. Data on heat loss in kilowatt-hours per day are required - in the case of a conditional house, counted as an example, this is:

271,512 (daily heat losses by external walls) + 45.76 (daily heat loss for heating of supply air) = 317.272 kWh

Accordingly, the required heating capacity of the boiler will be:

317,272: 24 (hours) = 13.22 kW

However, such a boiler will be under constant high load, reducing its service life. And in especially frosty days, the estimated capacity of the boiler will not be enough, because with a high temperature difference between room and street atmosphere, the heat loss of the building will increase dramatically.

Therefore, the boiler, chosen by the average calculation of the cost of thermal energy, with strong frosts can not cope. It will be rational to increase the required capacity of boiler equipment by 20%:

13.22 · 0.2 + 13.22 = 15.86 kW

To calculate the required capacity of the second circuit of a boiler that heats water for washing dishes, bathing, etc., it is necessary to divide the monthly heat consumption of "sewer" heat losses by the number of days in a month and by 24 hours:

493.82: 30: 24 = 0.68 kW

As a result of calculations, the optimum boiler output for the example cottage is 15.86 kW for the heating circuit and 0.68 kW for the heating circuit.

Selection of radiators

Traditionally, the capacity of the heating radiator is recommended to be chosen according to the area of the heated room, and with a 15-20% overstatement of power requirements, just in case. On an example we will consider, how much the method of a choice of a radiator «10 m2 of the area - 1,2 kW» is correct.

The thermal power of the radiators depends on the way they are connected, which must be taken into account when calculating the heating system

Initial data: corner room on the first level two-story house IZhS; external wall of double-row masonry of ceramic bricks; width of the room 3 m, length 4 m, ceiling height 3 m. According to the simplified scheme of choice, it is proposed to calculate the area of the room, we consider:

3 (width) · 4 (length) = 12 m 2

Those. The required heat radiator power with a 20% extra charge is 14.4 kW. And now we calculate the power parameters of the radiator based on the heat loss of the room.

In fact, the area of the room affects the loss of thermal energy less than the area of its walls, leaving one side outside the building (facade). Therefore, we will consider exactly the area of "street" walls in the room:

3 (width) · 3 (height) + 4 (length) · 3 (height) = 21 m 2

Knowing the area of the walls transferring heat "to the street", we calculate the heat loss at a difference of room and street temperatures of 30 ° (in the house +18 о С, outside -12 о С), and immediately in kilowatt-hours:

0,91 (resistance to heat transfer in m2 of room walls facing the street) · 21 (area of "street" walls) · 30 (temperature difference inside and outside the house): 1000 (number of watts per kilowatt) = 0.57 kW

According to construction standards, heating appliances are located in places of maximum heat loss. For example, radiators are installed under window openings, heat guns - above the entrance to the house. In the corner rooms, the batteries are installed on blind walls, subject to maximum winds

It turns out that to compensate for heat losses through the facade walls of this design, at 30 ° C the temperature difference in the house and on the street is sufficient for heating with a capacity of 0.57 kWh. Increase the required power by 20, even by 30% - we get 0.74 kWh.

Thus, the real power heating requirements can be significantly lower than the trade scheme "1,2 kW per square meter of the room area." Moreover, the correct calculation of the required capacities of heating radiators will reduce the volume of the coolant in the heating system, which will reduce the load on the boiler and fuel costs.

Video calculations of the heating system

The preservation of heat in the premises of the house is the main task of the heating system in the winter months. However, the heat is constantly in short supply. Where the heat leaves home - answers are provided by a visual video:

In the video, the procedure for calculating heat losses at home through enclosing structures is considered. Knowing the heat loss, it will be possible to accurately calculate the capacity of the heating system:

The choice of capacity of the boiler depends on the condition of the house and the quality of insulation of its enclosing structures. The principle of "kilowatt per 10 squares of area" works in a cottage of average condition of facades, roof and foundation. A detailed video on the principles of selecting the power characteristics of a heating boiler see below:

The production of heat annually becomes more expensive - the prices for fuel are rising. You can not treat the energy costs of the cottage indifferently, it is completely unprofitable. On the one hand, each new heating season costs the homeowner more expensive and more expensive. On the other hand, the warming of the walls, the foundation and the roof of a suburban one costs good money. However, the less heat leaves the building, the cheaper it will be to heat it.