The search for alternative sources that provide energy to many areas of human activity has recently become an urgent task. People are striving to more actively use the energy of the sun, wind, and water sources in order to reduce the costs of solving problems associated with the heat supply of buildings. At the same time, the issue of ecology is of no small importance, since reducing harmful emissions that pollute the atmosphere is more important than ever.

To create favorable and comfortable living conditions in the housing sector in last years began to use wind generators, solar collectors, and economical heat generators simultaneously with the implementation of measures that help increase the thermal insulation of a heat supply facility.

According to professionals working in this field, the use of geothermal sources of thermal energy - special pumps - is considered an effective and economical measure. Their fundamental design makes it possible to extract heat from environment, transform it and move it to the place of application (more details: " ").

The energy sources for heat pumps are water, air, and soil, and the heat generation process occurs due to the use of the physical properties of certain substances called refrigerants. They are able to boil even at low temperatures.

Calculation procedure for heat pumps

The result of the calculations depends mainly on the individual characteristics of the heated structure and consists of several stages:

1. First of all, heat losses occurring through the building envelope (these include windows, doors, walls, ceilings) are determined. To do this, use the following formula:
   
   Qok = Sx(tin – tout)x(1 + Σ β) x n / Rt (W), whereS – the sum of the areas of all enclosing structures (m²);
   tin – air temperature inside the building (°C);
   tout – outside air temperature (°C);
   
   n – coefficient reflecting the influence of the surrounding space on the characteristics of the structure. If the room is in direct contact with the outside environment through the ceiling, then this indicator is equal to 1. When the object has attic floors, n is equal to 0.9. If the object is located above the basement, the coefficient is 0.75 (more details: " ").
   β is the coefficient of additional heat loss, depending on the type of building and its geographical location. This indicator, when calculating a heat pump, is in the range from 0.05 to 0.27;Rt is an indicator of thermal resistance, which is determined by the following formula:Rt = 1/ α internal + Σ (δі / λі) + 1/ α external (m²x°C / W), where:α internal – coefficient characterizing the thermal absorption of the internal surfaces of fencing structures (W/m²x°C);
   δі / λі – is a calculated indicator of the thermal conductivity of materials used in construction;
   α nar – the value of thermal dissipation of the outer surfaces of fencing structures (W/m²x°C);
2. Next, to calculate heat pumps, use the formula to determinetotal heat loss of the building:
   
   Qt.pot = Qok + Qi – Qbp, where:
   
   Qi - the cost of heating the air that enters through natural leaky places;
   Qbp ​​- heat generation as a result of the operation of household appliances and human activity.
3. At this stage, the consumed thermal energy is calculated for each of the objects during the year:Qyear = 24x0.63xQt. pot.х((dх (tin - tout.medi.)/ (tin - tout.)) kW/hour), where:
   tout.av is the arithmetic mean value of the temperatures that are recorded in the outside air throughout the entire heating period;
   d – number of days in the heating season.
4. Then you need to determine the thermal power required to heat the water throughout the year, for which the expression is used:
   
   Qgv = V x17 (kW/hour per calendar year), where
   V x17 – daily volume of water heating up to 50 °C.
5. The total consumption of thermal energy is determined by the formula:
   
   Q = Qgv + Qyear (kW/hour for one year)

1.1 is a correction factor, since when critical temperatures occur, the load on the heat pump may increase.

When the necessary calculations have been made, it is easy to choose a heat pump suitable for a given room, which will provide a comfortable microclimate in it for the people in the room.

How to calculate heating costs for a country house?

Calculations are made based on the following parameters:

The first parameter is operating costs. To determine these costs, it is worth taking into account the cost of the fuel that will be used to generate heat. This item also includes maintenance costs. The most profitable in terms of this parameter will be heating, the energy carrier of which will be the supplied main gas. The next most efficient is the HEAT PUMP.

The second parameter is the cost of purchasing equipment and installing it. The most profitable and economical option at the procurement and installation stage would be to purchase an electric boiler. Maximum costs await if you decide to purchase boilers where the energy carriers are liquefied gas in gas tanks or diesel fuel. Here, too, a HEAT PUMP is optimal.

The third parameter should be considered convenience when using heating equipment. Solid fuel boilers in this case can be noted as the most demanding of attention. They require your presence and additional fuel loading, while electric ones and those powered by main gas supply operate independently. Therefore, gas and electric boilers are the most comfortable to use when heating country houses. And here the HEAT PUMP has an advantage. Climate control is the most comfortable characteristic of heat pumps.

Today, the following price situation has developed in the Moscow region... Connecting gas to private houses costs about 600 thousand rubles. Also required design work and the corresponding approvals, which sometimes last for years and also cost money. Add here the cost of the equipment and the relatively short period of its wear and tear (which is why gas companies offer more powerful gas boilers so that the wear and tear of the boiler takes longer). Heating with heat pumps is already comparable to the above price, but does not require any approvals. A heat pump is a common electric household appliance that consumes 4 times less electricity than a conventional electric boiler and is also a climate control device, i.e. an air conditioner. The motor life of modern heat pumps, and especially high-quality ones (premium class), allows them to operate for more than 20 years.

We give examples of calculating heat pumps for various types and sizes of houses.

First, you need to determine the heat loss of your building, depending on the region of its location. Read more in "Full news"

First of all, you need to decide on the power of the heat pump or boiler, since this is one of the decisive technical characteristics. It is selected based on the amount of heat loss of the building. Calculation of the heat balance of a house, taking into account the features of its design, should be carried out by a specialist, however, for a rough estimate of this parameter, if the house construction is designed taking into account building standards, you can use the following formula:
Q = k V ΔT
1 kW/h = 860 kcal/h
Where
Q - heat loss, (kcal/h)
V is the volume of the room (length × width × height), m3;
ΔT - maximum difference between the air temperature outside and inside the room in winter, °C;
k is the generalized heat transfer coefficient of the building;
k = 3…4 - building made of boards;
k = 2…3 - brick walls in one layer;
k min-max = 1…2 - standard masonry (brick in two layers);

k = 0.6...1 - well-insulated building;

An example of calculating the power of a gas boiler for your home:

For a building with volume V = 10m × 10m × 3m = 300 m3;

Heat loss of a brick building (k max= 2) will be:
Q = 2 ×300 × 50 = 30000 kcal/hour = 30000 / 860 = 35 kW
This will be the required minimum boiler power, calculated to the maximum...

Typically, a 1.5-fold power reserve is selected, however, factors such as constantly running ventilation of the room, open windows and doors, large glazing area, etc. should be taken into account. If you plan to use a double-circuit boiler (heating the room and supplying hot water), then its power should be further increased by 10 - 40%. The additive depends on the amount of hot water flow.

An example of calculating the power of a heat pump for your home:

At ΔT = (Tvn - Tnar) = 20 - (-30) = 50°C;
Heat loss of a brick building (k min= 1) will be:
Q = 1 ×300 × 50 = 15000 kcal/hour = 30000 / 860 = 17 kW
This will be the required minimum power of the boiler, calculated to the minimum, since there is no burnout in the heat pump and the resource depends on its motor life and cycling during the day... To reduce the number of on/off cycles of the heat pump, heat accumulator tanks are used.

So: You need the heat pump to cycle 3-5 times per hour.
those. 17 kW/hour -3 cycles

You will need a buffer tank - 3 cycles - 30 l/kW; 5 strokes - 20 l/kW.

17 kW*30l=500l storage capacity!!! The calculations are approximate, here a large battery is good, but in practice they use 200 liters.

Now let’s calculate the cost of a heat pump and its installation for your home:

The volume of the building is the same V = 10m × 10m × 3m = 300 m3;
We calculated the approximate power to be -17 kW. Different manufacturers have different power lines, so choose a heat pump based on quality and cost together with our consultants. For example, Waterkotte has an 18 kW heat pump, but you can also install a 15 kW heat pump, since if there is insufficient power, there is a 6 kW peak closer in each heat pump. Peak reheating occurs relatively quickly and therefore there is no need to overpay for a heat pump. Therefore, you can choose 15 kW, since in the short term 15+6=21 kW is higher than your heat needs.

Let's stop at 18 kW. Check the cost of the heat pump with consultants, since today delivery conditions are “to put it mildly” unpredictable. Therefore, the factory version is presented on the site.

If you are in the southern regions, then the heat loss of your home based on the above calculations will be less, since ΔT = (Tvn - Tnar) = 20 - (-10) = 30°C. or even ΔT = (Tvn - Tnar) = 20 - (-0) = 20°C. You can choose a heat pump of lower power and also on the air-to-water operating principle. Our air source heat pumps operate efficiently down to -25 degrees and therefore do not require drilling work.

In central Russia and Siberia, geothermal heat pumps operating on the “water-to-water” principle are much more effective.

Drilling for a geothermal field will cost differently depending on the region. In the Moscow region, the cost calculation is as follows:

We take the power of our heat pump -18 kW. The electrical consumption of such a geothermal heat pump is approximately 18/4 = 4.5 kW/hour from an outlet. Waterkotte has even less (this characteristic is called COP. Waterkotte heat pumps have a COP of 5 or more). According to the law of conservation of power, electrical power is transferred to the system, converted into thermal power. We obtain the missing power from a geothermal source, that is, from probes that need to be drilled. 18-4.5 = 13.5 kW from the Earth for example (since the source in this case can be a horizontal collector, a pond, etc.).

The heat transfer of soils in different places, even in the Moscow region, is different. On average, from 30 to 60 W per 1 m.p., depending on soil moisture.

13.5 kW or 13500 W divided by heat transfer. on average it is 50W so 13500/50=270 meters. Drilling work costs an average of 1200 rubles/m.p. We get 270*1200=324000 rubles. turnkey with entry into the heating station.

The cost of an economy class heat pump is 6-7 thousand dollars. those. 180-200 thousand rubles

Cost TOTAL 324 thousand + 180 thousand = 504 thousand rubles

Add the cost of installation and the cost of a heat accumulator and you will get a little more than 600 thousand rubles, which is comparable to the cost of supplying main gas. Q.E.D.

The use of low-grade ambient heat for water heating and heating becomes economically beneficial with long-term use of the system. An obstacle to the widespread use of such devices is the high initial cost of the equipment and its installation. Therefore, complete or partial installation of a heat pump with your own hands is always relevant, allowing you to save significant money.

Rice. 1 Water-to-water heat pump in the house

When creating heat pumps for heating, natural low-grade heat of air masses, soil and water is used. Aquatic species absorb thermal energy from wells, wells, ponds and other open bodies of water. A heat pump works like a refrigerator, which takes heat from the refrigerator compartment and releases it outside through an external radiator.

During installation, the primary heat exchanger with a circulating coolant is placed in a container of water, from which heat is taken. Water is sucked in by a water pump, passes through a pipe system and then enters the evaporator - in the device, when the liquid is heated, it evaporates. In the evaporator, the coolant transfers heat to freon, for which a small positive temperature of 6 - 8 C is the boiling point, and the gaseous refrigerant enters the compressor.

There it is compressed, leading to an increase in the temperature of the gas, and further supply to the condenser. In a capacitor thermal energy from gas with a temperature of 40 - 70 C is transferred to water in the heating system, the cooled gas condenses and enters the pressure reducing valve (throttle). Its pressure decreases - this leads to greater cooling of the gas to a liquid state, in which it is again supplied to the evaporator. The system operates in a circular closed cyclic mode.

Heat pump calculation

To design a system with your own hands, you first need to perform a calculation taking into account the needs for thermal energy (pumps can additionally be used to provide hot water supply to the house) and possible losses. The calculation algorithm consists of the following operations.

1. The area of ​​the heated room is calculated.
2. Based on the obtained values, the total amount of energy required for heating is determined based on the calculation of 70 - 100 watts per square meter. The parameter depends on the height of the ceilings, the material of manufacture and the degree of thermal conductivity of the house.
3. When providing hot water supply, the obtained value is increased by 15 - 20%.
4. Based on the received power, a compressor is selected, the main components of the system are calculated and designed: pipeline, evaporator, condenser, electric pump and other components.

Components for a heating system with a heat pump when manufactured independently

It is quite difficult for an ordinary homeowner to compete with industrial heat pumps from domestic and foreign manufacturers, however, its installation and production of individual components is not an impossible task. The main task when installing a heat pump remains the correctness of the calculations, because if there is an error, the system may have low efficiency and become ineffective.

Compressor

For installation you will need a new or used one. the compressor is in working condition with an unexpired resource of suitable power. The usual compressor power should be 20 - 30% of the calculated one; you can use standard factory units for refrigerators or spiral air conditioners, which have a higher efficiency compared to piston devices.

Evaporator and condenser

To cool and heat liquids, they are usually passed through copper pipes placed in a container with a heat exchanger. To increase the cooling area, the copper pipe is arranged in the form of a spiral; the required length is calculated using the formula for calculating the area divided by the cross-section. The volume of the heat exchange tank is calculated based on the implementation of effective heat exchange, the usual average value is about 120 liters. For a heat pump, it is rational to use pipes for air conditioners, which initially have a spiral shape and are sold in coils.

Rice. 3 Copper pipe and tank for heat exchanger

Many manufacturers of heat pumps have replaced this method of designing heat exchangers with a more compact one, using heat exchange on the “pipe-in-pipe” principle. The standard diameter of the plastic pipe for the evaporator is 32 mm, a copper pipe with a diameter of 19 mm is placed in it, the evaporator is thermally insulated, the total length of the heat exchanger is about 10 - 12 m. For the condenser, you can use 25 mm. metal-plastic pipe and 12.7 mm. copper.

To increase the area and efficiency of the heat exchanger, some craftsmen twist a braid of several small-diameter copper pipes, cover them with thin wire and place the structure in plastic. This allows you to obtain a heat exchange area of ​​about 1 cubic meter over a 10-meter segment.

Thermostatic valve

A properly selected device regulates the degree of filling of the evaporator and is largely responsible for the performance of the entire system. For example, if the supply of refrigerant is too large, it will not have time to completely evaporate, and drops of liquid will enter the compressor, leading to disruption of its operation and a decrease in the outlet gas temperature. Too little freon in the evaporator after increasing the temperature in the compressor will not be enough to warm up the required volume of water.

Sensors

For ease of use, monitoring operation, detecting faults and setting up the system, it is necessary to have built-in temperature sensors. Information is important at all stages of the system’s operation; only with its help, using formulas, can one establish the most important parameter of the installed equipment for water heat pumps - the COP efficiency indicator.

Pump equipment

When heat pumps operate, water is taken and supplied from a well, well or open reservoir using water pumps. Submersible or surface types can be used, usually their power is low, 100 - 200 W is enough to supply water. To control the operation and protect the pumps and the system, filters, a pressure gauge, water meters and simple automation are additionally installed.

Assembling heat pump equipment with your own hands does not present any great difficulties if you know how to handle a special tool for welding and soldering copper. The completed work will help save significant funds - the cost of components will be about 600 USD. That is, the purchase of industrial equipment will cost 10 times more (about 6000 USD). A self-assembled structure, if properly calculated and configured, has an efficiency (COP) of about 4, which corresponds to industrial designs.

As you know, heat pumps use free, renewable energy sources: low-grade heat from air, soil, underground, open, non-freezing reservoirs, waste and waste water and air, as well as waste heat from technological enterprises. In order to collect this, electricity is expended, but the ratio of the amount of thermal energy received to the amount of electrical energy consumed is about 3-7 times.

If we talk only about the sources of low-grade heat around us for use for heating purposes, this is; outside air with a temperature of -3 to +15 °C, air exhausted from the room (15-25 °C), subsoil (4-10 °C) and groundwater (about 10 °C), lake and river water (5-10 °C), ground surface (below the freezing point) (3-9°C) and ground deep (more than 6 m - 8 o C).

Heat extraction from the environment (inner district).

A liquid working medium, refrigerant, is pumped into the evaporator at low pressure. The thermal level of temperatures surrounding the evaporator is higher than the corresponding boiling point of the working medium (the refrigerant is selected such that it can boil even at sub-zero temperatures). Due to this temperature difference, heat is transferred to the environment, the working environment, which at these temperatures boils and evaporates (turns into steam). The heat required for this is taken from any of the above low-potential heat sources.

Learn more about renewable energy sources

If atmospheric or ventilation air is chosen as a heat source, heat pumps operating according to the air-water scheme are used. The pump can be located indoors or outdoors, with a built-in or remote condenser. Air is blown through the heat exchanger (evaporator) using a fan.

Groundwater with a relatively low temperature or soil from the surface layers of the earth can be used as a source of low-potential thermal energy. The heat content of the soil mass is generally higher. The thermal regime of the soil in the surface layers of the earth is formed under the influence of two main factors - solar radiation incident on the surface and the flow of radiogenic heat from the earth's interior. Seasonal and daily changes in the intensity of solar radiation and outside air temperature cause fluctuations in the temperature of the upper layers of the soil. The penetration depth of daily fluctuations in outside air temperature and the intensity of incident solar radiation, depending on specific soil and climatic conditions, ranges from several tens of centimeters to one and a half meters. The depth of penetration of seasonal fluctuations in outside air temperature and the intensity of incident solar radiation does not, as a rule, exceed 15-20 m.

Types of horizontal heat exchangers:

1. heat exchanger made of series-connected pipes;
2. heat exchanger made of parallel connected pipes;
3. horizontal collector laid in a trench;
4. loop-shaped heat exchanger;
5. a heat exchanger in the shape of a spiral located horizontally (the so-called “slinky” collector);
6. heat exchanger in the form of a spiral located vertically.

Water accumulates solar heat well. Even in the cold winter period, groundwater has a constant temperature of +7 to +12°C. This is the advantage of this heat source. Due to the constant temperature level, this heat source has a high conversion rate through the heat pump throughout the year. Unfortunately, groundwater is not available in sufficient quantities everywhere. When using groundwater as a source, supply is carried out from a well using a submersible pump to the entrance to the heat exchanger (evaporator) of a heat pump operating according to the “water-to-water/open system” scheme; from the outlet of the heat exchanger, water is either pumped into another well, or dumped into a reservoir. The advantage of open systems is the possibility of obtaining large quantity thermal energy at relatively low costs. However, wells require maintenance. In addition, the use of such systems is not possible in all areas. The main requirements for soil and groundwater are as follows:
- sufficient permeability of the soil, allowing water reserves to be replenished;
- good chemical composition of groundwater (for example, low iron content), which avoids problems associated with the formation of deposits on pipe walls and corrosion.

Open systems are more often used to supply heating or cooling to large buildings. The world's largest geothermal heat transfer system uses groundwater as a source of low-grade thermal energy. This system is located in the USA in Louisville, Kentucky. The system is used for heat and cold supply of a hotel and office complex; its power is approximately 10 MW.

Let's take another source - a reservoir; loops of plastic pipe can be laid on its bottom, a “water-water/closed system” scheme. An ethylene glycol solution (antifreeze) circulates through the pipeline, which transfers heat to the refrigerant through the heat exchanger (evaporator) of the heat pump.
The soil has the ability to accumulate solar energy over a long period of time, which ensures a relatively uniform temperature of the heat source throughout the year and, thus, a high conversion coefficient of the heat pump. The temperature in the upper layers of the soil varies depending on the season. Below the freezing line, these temperature fluctuations are significantly reduced. The heat accumulated in the ground is extracted through horizontally laid sealed heat exchangers, also called ground collectors, or through vertically laid heat exchangers, so-called geothermal probes. The ambient heat is transferred by a mixture of water and ethylene glycol (brine or medium), the freezing point of which should be approximately -13°C (take into account the manufacturer's data). Thanks to this, the brine does not freeze during operation.
This means that there are two possible options for obtaining low-grade heat from the ground. Horizontal laying of plastic pipes in trenches with a depth of 1.3-1.7 m, depending on the climatic conditions of the area, or vertical wells with a depth of 20-100 m. Laying of pipes in trenches can also be done in the form of spirals, but with a laying depth of 2- 4 m, this will significantly reduce the total length of the trenches. The maximum heat transfer of surface soil is from 7 to 25 W per m.p., from geothermal 20-50 W per m.p. According to manufacturing companies, the service life of trenches and wells is more than 100 years.

A little more about vertical ground heat exchangers.

Since 1986, research has been carried out on a system with vertical ground heat exchangers in Switzerland, near Zurich. A vertical coaxial type ground heat exchanger with a depth of 105 m was installed in the soil mass. This heat exchanger was used as a source of low-grade thermal energy for a heat transfer system installed in a single-apartment residential building. The vertical ground heat exchanger provided a peak power of approximately 70 W per meter of length, creating a significant thermal load on the surrounding ground mass. The annual production of thermal energy is about 13 MWh.
At a distance of 0.5 and 1 m from the main well, two additional wells were drilled, in which temperature sensors were installed at a depth of 1, 2, 5, 10, 20, 35, 50, 65, 85 and 105 m, after which the wells were filled clay-cement mixture. Temperatures were measured every thirty minutes. In addition to the ground temperature, other parameters were recorded: the speed of movement of the coolant, energy consumption by the compressor drive, air temperature, etc.

The first observation period lasted from 1986 to 1991. Measurements have shown that the influence of the heat of the outside air and solar radiation is observed in the surface layer of soil at a depth of up to 15 m. Below this level, the thermal regime of the soil is formed mainly due to the heat of the earth's interior. During the first 2-3 years of operation, the temperature of the soil mass surrounding the vertical heat exchanger dropped sharply, but every year the temperature drop decreased, and after a few years the system reached a regime close to constant, when the temperature of the soil mass around the heat exchanger became 1 lower than the initial one. -2 °C.

In the fall of 1996, ten years after the system began operating, measurements were resumed. These measurements showed that the ground temperature did not change significantly. In subsequent years, slight fluctuations in ground temperature within 0.5 °C were recorded depending on the annual heating load. Thus, the system reached a quasi-stationary mode after the first few years of operation.

Based on experimental data, mathematical models of the processes occurring in the soil mass were constructed, which made it possible to make a long-term forecast of changes in the temperature of the soil mass.

Mathematical modeling showed that the annual decrease in temperature will gradually decrease, and the volume of the soil mass around the heat exchanger, subject to a decrease in temperature, will increase every year. At the end of the operating period, the regeneration process begins: the soil temperature begins to rise. The nature of the regeneration process is similar to the nature of the heat “selection” process: in the first years of operation there is a sharp increase in soil temperature, and in subsequent years the rate of temperature increase decreases. The length of the "regeneration" period depends on the length of the operating period. These two periods are approximately the same. In the case under consideration, the period of operation of the ground heat exchanger was thirty years, and the “regeneration” period is also estimated at thirty years

Thus, heating and cooling systems for buildings using low-grade heat from the earth represent a reliable source of energy that can be used everywhere. This source can be used for quite a long time and can be renewed at the end of the operating period.

Calculation of a horizontal heat pump collector

The heat removal from each meter of pipe depends on many parameters: laying depth, presence of groundwater, soil quality, etc. Approximately we can assume that for horizontal collectors it is 20 W.m.p. More precisely: dry sand - 10, dry clay - 20, wet clay - 25, clay with a high water content - 35 W.m.p. The difference in coolant temperature in the forward and return lines of the loop in calculations is usually taken to be 3 °C. On the collector site, no buildings should be erected so that the heat of the earth, i.e. our energy source was replenished with energy from solar radiation.

The minimum distance between laid pipes must be at least 0.7-0.8 m. The length of one trench can vary from 30 to 150 m, it is important that the lengths of the connected circuits are approximately the same. It is recommended to use an ethylene glycol solution (medium) with a freezing point of approximately -13 o C as the primary coolant. In calculations, it should be taken into account that the heat capacity of the solution at a temperature of 0 ° C is 3.7 kJ/(kg K), and the density is 1 .05 g/cm 3 . When using a medium, the pressure loss in the pipes is 1.5 times greater than when circulating water. To calculate the parameters of the primary circuit of a heat pump installation, you will need to determine the flow rate of the medium:

Vs = Qo 3600 / (1.05 3.7 .t),
where t is the temperature difference between the supply and return lines, which is often taken equal to 3 o K. Then Qo is the thermal power received from a low-potential source (ground). The latter value is calculated as the difference between the total power of the heat pump Qwp and the electrical power spent on heating the refrigerant P:

Qo = Qwp - P, kW.

The total length of the collector pipes L and the total area of ​​the area A for it are calculated using the formulas:

Here q is the specific (from 1 m of pipe) heat removal; da - distance between pipes (laying pitch).

Calculation example. Heat Pump.
Initial conditions: heat demand of a cottage with an area of ​​120-240 m2 (based on heat losses taking into account infiltration) - 13 kW; The water temperature in the heating system is taken to be 35 °C (underfloor heating); The minimum coolant temperature at the outlet to the evaporator is 0 °C. To heat the building, a heat pump with a capacity of 14.5 kW was selected from the existing technical range of equipment, taking into account the losses due to the viscosity of the medium, when selecting and transferring thermal energy from the ground, amounting to 3.22 kW. Heat removal from the surface layer of soil (dry clay), q equals 20 W/m.p. In accordance with the formulas we calculate:

1) required thermal power of the collector Qo = 14.5 - 3.22 = 11.28 kW;
2) total pipe length L = Qo/q = 11.28/0.020 = 564 m.p. To organize such a collector, 6 circuits 100 m long will be required;
3) with a laying step of 0.75 m, the required area of ​​the site is A = 600 x 0.75 = 450 m2;
4) total filling of ethylene glycol solution Vs = 11.28 3600/ (1.05 3.7 3) = 3.51 m3, in one circuit is 0.58 m3.

To install the collector, we select a plastic pipe of size 32x3. The pressure loss in it will be 45 Pa/m.p.; resistance of one circuit is approximately 7 kPa; coolant flow speed - 0.3 m/s.

Probe calculation

When using vertical wells with a depth of 20 to 100 m, U-shaped plastic pipes (with diameters from 32 mm) are immersed in them. As a rule, two loops are inserted into one well, filled with a suspension solution. On average, the specific heat removal of such a probe can be taken equal to 50 W/m.p. You can also focus on the following data on heat removal:

• dry sedimentary rocks - 20 W/m;
• rocky soil and water-saturated sedimentary rocks - 50 W/m;
• rocks with high thermal conductivity - 70 W/m;
• groundwater - 80 W/m.

The soil temperature at a depth of more than 15 m is constant and is approximately +9 °C. The distance between the wells should be more than 5 m. In the presence of underground flows, the wells should be located on a line perpendicular to the flow.
The selection of pipe diameters is carried out based on pressure losses for the required coolant flow. Calculation of liquid flow can be carried out for t = 5 °C.

Calculation example.

The initial data are the same as in the above calculation of a horizontal reservoir. With a probe specific heat removal of 50 W/m and a required power of 11.28 kW, probe length L should be 225 m.
To install a collector, it is necessary to drill three wells with a depth of 75 m. In each of them we place two pipe loops of standard size 32x3; in total - 6 circuits of 150 m each.

The total coolant flow rate at t = 5 °C will be 2.1 m3/h; flow rate through one circuit is 0.35 m3/h. The circuits will have the following hydraulic characteristics: pressure loss in the pipe - 96 Pa/m (coolant - 25% ethylene glycol solution); circuit resistance - 14.4 kPa; flow speed - 0.3 m/s.

Equipment selection

Since the temperature of the antifreeze can vary (from -5 to +20 °C), a hydraulic expansion tank is required in the primary circuit of the heat pump installation.
It is also recommended to install a storage tank on the heating (condenser) line of the heat pump: the heat pump compressor operates in the “on-off” mode. Too frequent starts can lead to accelerated wear of its parts. The tank is also useful as an energy storage device in case of a power outage. Its minimum volume is taken at the rate of 20-30 liters per 1 kW of heat pump power.

When using a bivalence, a second energy source (electric, gas, liquid or solid fuel boiler), it is connected to the circuit through a battery tank, which is also a thermal hydraulic distributor; the boiler’s activation is controlled by a heat pump or an upper-level automation system.
In case of possible power outages, you can increase the power of the installed heat pump by a factor calculated by the formula: f = 24/(24 - t off), where t off is the duration of the power supply interruption.

In the event of a possible power outage for 4 hours, this coefficient will be equal to 1.2.
The power of the heat pump can be selected based on the monovalent or bivalent mode of its operation. In the first case, it is assumed that the heat pump is used as the only generator of thermal energy.

It should be taken into account: even in our country, the duration of periods with low air temperatures is a small part of the heating season. For example, for Central region In Russia, the time when the temperature drops below -10 °C is only 900 hours (38 days), while the duration of the season itself is 5112 hours, and the average temperature in January is approximately -10 °C. Therefore, it is most appropriate to operate the heat pump in bivalent mode, which involves turning on an additional source during periods when the air temperature drops below a certain level: -5 °C in the southern regions of Russia, -10 °C in the central regions. This allows you to reduce the cost of the heat pump and, especially, the installation of the primary circuit (laying trenches, drilling wells, etc.), which increases greatly with increasing installation power.

In the conditions of the Central region of Russia, for a rough estimate when selecting a heat pump operating in bivalent mode, you can focus on the ratio of 70/30: 70% of the heat demand is covered by a heat pump, and the remaining 30% by an electric or other source of thermal energy. In the southern regions, you can be guided by the ratio of the power of the heat pump and the additional heat source, often used in Western Europe: 50 to 50.

For a cottage with an area of ​​200 m2 for 4 people with heat losses of 70 W/m2 (calculated at -28 °C outside air temperature), the heat requirement will be 14 kW. To this value should be added 700 W for the preparation of sanitary hot water. As a result, the required heat pump power will be 14.7 kW.

If there is a possibility of a temporary power outage, you need to increase this number by the appropriate factor. Let's say the daily shutdown time is 4 hours, then the power of the heat pump should be 17.6 kW (increasing factor - 1.2). In the case of monovalent mode, you can choose a ground-water heat pump with a power of 17.1 kW, consuming 6.0 kW of electricity.

For a bivalent system with an additional electric heater and a cold water supply temperature of 10 ° C for the need for hot water and safety factor, the power of the heat pump should be 11.4 W, and the electric boiler - 6.2 kW (total - 17.6) . The peak electrical power consumed by the system will be 9.7 kW.

The approximate cost of electricity consumed per season when the heat pump is operating in monovalent mode will be 500 rubles, and in bivalent mode at temperatures below (-10C) - 12,500. The cost of energy when using only the appropriate boiler will be: electricity - 42,000, diesel fuel - 25,000, and gas - about 8,000 rubles. (in the presence of a supplied pipe and low gas prices existing in Russia). Currently, for our conditions, in terms of operating efficiency, a heat pump can only be compared with a gas boiler of new series, and in terms of operating costs, durability, safety (no boiler room required) and environmental friendliness, it surpasses all other types of thermal energy production.

Note that when installing heat pumps, first of all, you should take care of insulating the building and installing double-glazed windows with low thermal conductivity, which will reduce the heat losses of the building, and therefore the cost of work and equipment.

Taking into account the fact that a heat pump is equipment that requires quite significant costs for acquisition and installation, the issue of its selection should be treated especially carefully. The first thing a potential buyer needs to do is to make at least an approximate calculation of the power of the equipment that is suitable for efficient operation in specific conditions. Of course, you can turn to specialists to draw up a heat pump design, but in order to estimate approximate costs, you can do some initial calculations yourself.

A heat pump, the design of which is a rather complex undertaking, is chosen depending on the area of ​​the house, the degree of its insulation, and average temperature values ​​in the cold season. In addition to calculating the required power, the complete project involves determining the parameters of the earthen reservoir for the geothermal pump, calculating the number and diameter of pipes for the well in the case of a water-water system. Correct calculation of a heat pump involves taking into account many factors: from the characteristics of the soil on the site to the material from which the house is built.

Development of a heating system based on a heat pump

If you are seriously interested in such a progressive method of home heating as heat pumps, then it is best to prefer the services of specialists with specialized education and extensive experience working with such equipment. This is because the correct development of a heat pump and the entire heating system for the home will allow you to forget about heat problems for many years, enjoying stable efficient work equipment.

First of all, it is worth deciding on the source of heat that will be converted into energy for the coolant in the heating system. Whether it is soil, water or air determines both the production of heat pumps (or rather, the manufacturing technology), as well as the productivity and price of the equipment itself and installation work. One of the most effective systems is water-to-water, but it requires the presence of a reservoir near the house or a sufficient amount of groundwater on the site.

It is worth considering that the heat pump is more used for low-temperature heat sources, ideally combined with a “warm floor” system, but combination with traditional generators is also possible. When choosing heat pumps, their thermal calculation is carried out in such a way as to take into account whether it is capable of independently heating the room even in the most extreme cold, or whether it is necessary to provide an additional heat source in the system, for example, an electric boiler. The thermodynamic calculation takes into account the minimum temperatures that can be reached in winter.

It is also necessary to take into account the need for hot water supply at home; if such functionality is required, then an additional 20% is included in the required power.

Example of heat pump calculation

So, we have a two-story building with an area of ​​250 sq.m. with a ceiling height of 2.7 m. Let us assume that the temperature in the room is +20°C and outside -26°C. Next, we calculate the power of the heat pump for heating the house:

0.434*250*2.7*(20-(-26)) = 13475.7 kW - maximum required heating power in accordance with SP 50.13330-2012

This calculation does not imply large losses. Losses in this case may be even less than 13475.7 kW.

More accurate thermal calculations can be made individually. It will take into account all materials of walls, windows, ceilings, etc.

Calculation of the heat pump circuit, which will be used for heating and cooling the room, is more complex and is carried out by specialists.