This article describes options for home heating and hot water supply using a heat pump, a solar collector and a cavitation heat generator. An approximate method for calculating a heat pump and a heat generator is given. The approximate cost of heating a house with a heat pump is given.

Heat pump. home heating design

To understand its principle of operation, you can look at an ordinary household refrigerator or air conditioner.

Modern heat pumps use for their work low potential heat sources ground, groundwater, air. The same physical principle operates in both the refrigerator and the heat pump (physicists call this process the Carnot cycle). A heat pump is a device that "pumps out" the heat from the refrigerator compartment and throws it onto the radiator. The air conditioner "pumps out" the heat from the air of the room and throws it on the radiator, but located on the street. At the same time, to the heat "sucked" from the room, more heat is added, into which the electrical energy consumed by the air conditioner's electric motor has turned.

The number that expresses the ratio of the thermal energy produced by the heat pump (air conditioner or refrigerator) to the electrical energy consumed by it is called the “heating coefficient” by heat pump specialists. In the best heat pumps, the heating coefficient reaches 3-4. That is, for every kilowatt-hour of electricity consumed by the electric motor, 3-4 kilowatt-hours of thermal energy are generated. (One kilowatt-hour corresponds to 860 kilocalories.) This conversion factor (heating factor) directly depends on the temperature of the heat source, the higher the source temperature, the greater the conversion factor.

The air conditioner takes this heat energy from the outdoor air, and large heat pumps "pump out" this additional heat, usually from a reservoir/groundwater or ground.

Although the temperature of these sources is much lower than the air temperature in a heated house, but this low-temperature heat of the ground or water, the heat pump turns into high temperature needed to heat the house. Therefore, heat pumps are also called "heat transformers". (see transformation process below)

Note: Heat pumps not only warm houses, but also cool the water in the river, from which heat is pumped out. And in our time, when the rivers are too overheated by industrial and domestic wastewater, cooling the river is very useful for living organisms and fish to live in it. The lower the temperature of the water, the more oxygen can be dissolved in it, which is necessary for fish. In warm water, the fish suffocates, and in cold water it blisses. Therefore, heat pumps are very promising in saving environment from " thermal pollution".

But installing a heating system using heat pumps is still too expensive, because a lot of earthworks are required plus consumables, such as pipes to create a collector / heat exchanger.

It is also worth remembering that in heat pumps, as in conventional refrigerators, a compressor is used that compresses the working fluid - ammonia or freon. Heat pumps work better on freon, but freon has already been banned for use due to the fact that when it enters the atmosphere, it burns out ozone in its upper layers, which protects the Earth from the ultraviolet rays of the sun.

And yet, it seems to me that the future belongs to heat pumps. But they, no one yet mass-produces. Why? Not hard to guess.

If an alternative source of cheap energy appears, then where to put the produced gas, oil and coal, to whom to sell it. And what to write off the multibillion-dollar losses from explosions in mines and mines.

Schematic diagram of heating a house with a heat pump

How a heat pump works

The source of low-potential heat can be outdoor air with a temperature of -15 to +15°C, air discharged from the room with a temperature of 15-25°C, subsoil (4-10°C) and ground (more than 10°C) water , lake and river water (0-10°С), surface (0-10°С) and deep (more than 20 m) soil (10°С). In the Netherlands, for example, in the city of Heerlen, a flooded mine is used for this purpose. The water that fills the old mine at the level of 700 meters has a constant temperature of 32°C.

In the case of using atmospheric or ventilation air as a source of heat, the heating system operates according to the “air-to-water” scheme. The pump can be located indoors or outdoors. Air is supplied to its heat exchanger by means of a fan.

If groundwater is used as a heat source, then the system works according to the “water-water” scheme. Water is supplied from the well by means of a pump to the pump heat exchanger, and after heat is removed, it is discharged either into another well or into a reservoir. Antifreeze or antifreeze can be used as an intermediate coolant. If a reservoir acts as an energy source, a loop of a metal-plastic or plastic pipe is laid on its bottom. A solution of glycol (antifreeze) or antifreeze circulates through the pipeline, which transfers heat to freon through the heat pump heat exchanger.

When using the soil as a heat source, the system works according to the "soil-water" scheme. There are two options for the collector device - vertical and horizontal.

• With a horizontal collector, metal-plastic pipes are laid in trenches with a depth of 1.2-1.5 m or in the form of spirals in trenches with a depth of 2-4 m. This method of laying can significantly reduce the length of the trenches.

Diagram of a heat pump with a horizontal collector with spiral pipe laying

1 - heat pump; 2 - pipeline laid in the ground; 3 - indirect heating boiler; 4 - heating system "warm floor"; 5 - feed circuit hot water.

However, when laying in a spiral, the hydrodynamic resistance increases greatly, which leads to additional costs for pumping the coolant, and the resistance also increases as the length of the pipes increases.

• With a vertical arrangement of the collector, the pipes are laid in vertical wells to a depth of 20-100 m.

Schematic of the vertical probe

Photo of the probe in the bay

Installation of the probe in the well

Calculation of the horizontal collector of a heat pump

Calculation of the horizontal collector of a heat pump.

q - specific heat removal (from 1 m of the pipe).

• dry sand - 10 W/m,
• dry clay - 20 W/m,
• wet clay - 25 W/m,
• clay with high water content - 35 W/m.

Between the direct and return loops of the collector, a temperature difference of the coolant appears.

Usually for calculation it is taken equal to 3 ° C. The disadvantage of such a scheme is that it is not desirable to erect buildings on the site above the collector so that the heat of the earth is replenished due to solar radiation. The optimal distance between the pipes is considered to be 0.7-0.8 m. In this case, the length of one trench is selected from 30 to 120 m.

Heat pump calculation example

I will give an approximate calculation of a heat pump for our eco-house, described in the article.

It is believed that to heat a house with a ceiling height of 3 m, it is necessary to spend 1 kW. Thermal energy per 10 m2 area. With a house area of ​​​​10x10m \u003d 100 m2, 10 kW of thermal energy is needed.

When using a warm floor, the temperature of the heat carrier in the system must be 35°C, and the minimum temperature of the heat carrier - 0°C.

Table 1. Thermia Villa heat pump data.

To heat a building, choose a heat pump with a capacity of 15.6 kW (nearest larger size), which consumes 5 kW for the compressor. We select heat removal from the surface layer of soil according to the type of soil. For (wet clay) q is 25 W/m.

Calculate the power of the heat collector:

Qo=Qwp-P, where

Qo- thermal collector capacity, kW;

qwp- heat pump power, kW;

P- compressor electric power, kW.

The required heat output of the collector will be:

Qo=15.6-5=10.6 kW;

Now let's determine the total length of the pipes:

L=Qo/q, where q is specific heat removal (from 1 m. pipe), kW/m.

L \u003d 10.6 / 0.025 \u003d 424 m.

To organize such a collector, 5 contours with a length of 100 m each will be required. Based on this, we will determine the required area of ​​\u200b\u200bthe site for laying the contour.

A=Lxda, where da is the distance between the pipes (laying step), m.

With a laying step of 0.75 m, the required area of ​​\u200b\u200bthe site will be:

A \u003d 500x0.75 \u003d 375 m2.

Calculation of the vertical collector

When choosing a vertical collector, wells are drilled with a depth of 20 to 100 m. U-shaped metal-plastic or plastic pipes are immersed in them. To do this, two loops are inserted into one well, which are filled with cement mortar. Specific heat removal such a collector is 50 W/m.

For more accurate calculations, the following data is used:

• 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.

At depths of more than 15 m, the ground temperature is approximately +10°C. It must be taken into account that the distance between the wells must be more than 5 m. If there are underground currents in the soil, then the wells must be drilled perpendicular to the flow.

Example: L=Qo/q=10.6/0.05=212 m.

Thus, with a specific heat removal of a vertical collector of 50 W / m and a required power of 10.6 kW, the length of the pipe L should be 212 m.

To construct a collector, it is necessary to drill three wells with a depth of 75 m each. In each of them we place two loops from a metal-plastic pipe in total - 6 contours of 150 m each.

Heat pump operation when operating according to the "Soil-water" scheme

The pipeline is laid in the ground. When pumping a coolant through it, the latter heats up to the temperature of the soil. Further, according to the scheme, water enters the heat pump heat exchanger and gives off all the heat to the internal circuit of the heat pump.

Pressurized refrigerant has been pumped into the internal circuit of the heat pump. Freon or its substitutes are used as a refrigerant, since freon destroys the ozone layer of the atmosphere and is prohibited for use in new developments. The refrigerant has a low boiling point and therefore when the pressure drops sharply in the evaporator, it changes from a liquid state to a gas at a low temperature.

After the evaporator, the gaseous refrigerant enters the compressor and is compressed by the compressor. At the same time, it warms up, and its pressure rises. The hot refrigerant enters the condenser, where heat exchange takes place between it and the heat carrier from the return pipeline. Giving up its heat, the refrigerant cools and turns into a liquid state. The coolant enters the heating system and cools again, transfers its heat to the room. When the refrigerant passes through pressure reducing valve, its pressure drops, and it again passes into the liquid phase. After that, the cycle repeats.

In the cold season, the heat pump works as a heater, and in hot weather it can be used to cool the room (at the same time, the heat pump does not heat up, but cools the heat carrier - water. And the chilled water, in turn, can be used to cool the air in the room).

In general, a heat pump is a Carnot machine running in the opposite direction. The refrigerator pumps heat from the cooled volume into the surrounding air. If you place a refrigerator on the street, then, by extracting heat from the outside air and transferring it inside the house, you can, to some extent, heat the room in such a simple way.

However, as practice shows, only one heat pump to supply the house with heat and hot water not enough. I dare to offer the optimal, in my opinion, scheme for heating and hot water supply at home.

The proposed scheme for supplying the house with heat and hot water

1 - heat generator; 2 - solar collector; 3 - indirect heating boiler; 4 - heat pump; 5 - pipeline in the ground; 6 - circulation block of the solar system; 7 - heating radiator; 8 - hot water supply circuit; 9 - heating system "warm floor".

This scheme assumes the simultaneous use of three heat sources. The main role is played in it by the heat generator (1), the heat pump (4) and solar collector(2), which serve as auxiliary elements and help to reduce the cost of consumed electricity, as a result, and increase the heating efficiency. The simultaneous use of three heating sources almost completely eliminates the danger defrosting system.

After all, the probability of failure at the same time and the heat generator, and the heat pump, and the solar collector is negligible. The diagram shows two options for space heating: radiators (7) and "warm floor" (9). This does not mean that both options should be used, but only illustrates the possibility of using both one and the second.

The principle of operation of the heating circuit

The heat generator (1) supplies heated water to the boiler (3) and the circuit consisting of heating radiators (7). Also, the heated coolant from the heat pump (4) and the solar collector (2) enters the boiler. Part of the water heated by the heat pump is also supplied to the heat generator inlet. Mixing with the "return" of the heating circuit, it increases its temperature. This contributes to more efficient heating of water in the cavitator of the heat generator. The water heated and accumulated in the boiler is supplied to the “warm floor” system circuit (9) and the hot water supply circuit (8).

Of course, the effectiveness of this scheme will be different in different latitudes. After all, the solar collector will have the greatest efficiency in the summer and, of course, in sunny weather. In our latitudes, there is no need to heat residential premises in summer, so the heat generator can be turned off altogether. And since our summer is quite hot and we can hardly imagine our life without an air conditioner, the heat pump is supposed to be turned on for cooling mode. Naturally, the pipeline from the heat pump to the boiler will be blocked. Thus, it is supposed to solve the problem of hot water supply only with the help of a solar system. And only if the solar system does not cope with this task, use a heat generator.

As you can see, the scheme is quite complex and expensive. The general approximate costs depending on the chosen scheme are given below.

Costs for a vertical collector:

• Heat pump 6000 €;
• Drilling works 6000 €;
• Operating costs (electricity): approx. 400 € per year.

For a horizontal manifold:

• Heat pump 6000 €;
• Drilling works 3000 €;
• Operating costs (electricity): about 450 euros per year.

Of the large costs, it will be necessary to purchase pipes and pay workers.

Installing a flat solar collector (eg Vitosol 100-F and a 300 l water heater) will cost 3200 €.

So let's go from simple to complex. First, we will assemble a simple house heating scheme based on a heat generator, debug it, and gradually add new elements to it, which will increase the efficiency of the installation.

Let's assemble the heating system according to the scheme:

House heating scheme using a heat generator

1 - heat generator; 2 - indirect heating boiler; 3 - heating system "warm floor"; 4 - hot water supply circuit.

As a result, we got the simplest heat supply scheme for the house. I shared my thoughts in order to encourage initiative people to develop alternative energy sources. If someone has any ideas or objections about what was written above, let's share our thoughts, let's accumulate knowledge and experience in this matter, and we will save our environment and make life a little better.

As we see here, the main and only element that heats the coolant is the heat generator. Although the scheme provides only one source of heating, it provides for the possibility of further addition of additional heating devices. For this, it is assumed to use an indirect heating boiler with the possibility of adding or removing heat exchangers.

The use of heating radiators available in the circuit shown in the figure one above is not intended. As you know, the "warm floor" system copes more effectively with the task of heating the premises and saves energy.

Attention: Prices are valid for 2009.

4.1. How a heat pump works

The use of alternative environmentally friendly energy sources can prevent a brewing energy crisis in Ukraine. Along with the search for and development of traditional sources (gas, oil), a promising direction is the use of energy accumulated in reservoirs, soil, geothermal sources, technological emissions (air, water, wastewater, etc.). However, the temperature of these sources is quite low (0–25 °C), and for their effective use it is necessary to transfer this energy to a higher temperature level (50–90 °C). This transformation is implemented by heat pumps (TH), which, in fact, are vapor compression refrigeration machines (Fig. 4.1).

The low-temperature source (LTS) heats the evaporator (3), in which the refrigerant boils at a temperature of –10 °С…+5 °С. Further, the heat transferred to the refrigerant is transferred by the classical vapor-compression cycle to the condenser (4), from where it goes to the consumer (HTP) at a higher level.

Heat pumps are used in various industries industry, residential and public sector. Currently, more than 10 million heat pumps of various capacities are in operation in the world: from tens of kilowatts to megawatts. Every year, the HP fleet is replenished by about 1 million units. So, in Stockholm, a thermal pumping station with a capacity of 320 MW, using sea water with a temperature of +4 ° C in winter, provides heat to the entire city. In 2004, the capacity of heat pumps installed in Europe was 4,531 MW, and the equivalent of 1.81 billion m 3 of thermal energy was generated by heat pumps worldwide. natural gas. Energy efficient heat pumps using geothermal and groundwater. In the United States, federal legislation has mandated the use of geothermal heat pumps (GHPs) in the construction of new public buildings. In Sweden, 50% of all heating is provided by geothermal heat pumps. By 2020, according to the forecasts of the World Energy Committee, the share of geothermal heat pumps will be 75%. The service life of a gas turbine pump is 25–50 years. The prospects for the use of heat pumps in Ukraine are shown in.

Heat pumps are divided according to the principle of operation (compressor, absorption) and according to the type of heat transfer chain "source-consumer". The following types of heat pumps are distinguished: air-to-air, air-to-water, water-to-air, water-to-water, ground-to-air, ground-to-water, where the heat source is indicated first. If only a heat pump is used for heating, then the system is called monovalent. If, in addition to the heat pump, another heat source is connected, operating separately or in parallel with the heat pump, the system is called bivalent.

Rice. 4.1. Diagram of hydraulic heat pump:

1 - compressor; 2 – low-level heat source (LHL); 3 – heat pump evaporator;

4 - heat pump condenser; 5 - heat consumer high level(HTP);

6 - low-temperature heat exchanger; 7 - refrigerant flow regulator;

8 - high temperature heat exchanger

Heat pump with hydraulic piping (water pumps, heat exchangers, shutoff valves etc.) is called a heat pump unit. If the medium cooled in the evaporator is the same as the medium heated in the condenser (water-water, air-air), then by changing the flows of these media, it is possible to change the HP mode to reverse (cooling to heating and vice versa). If the media are gases, then such a change in regime is called a reversible pneumatic cycle, if liquids - a reversible hydraulic cycle (Fig. 4.2).

Rice. 4.3. Diagram of an air-to-water heat pump

Air-to-water heat pumps are widely used in air conditioning systems. Outside air is blown through the evaporator and the heat removed from the condenser heats the water used to heat the room in the room (Figure 4.3).

The advantage of such systems is the availability of a low-potential heat source (air). However, the air temperature varies in a wide range, reaching negative values. In this case, the efficiency of the heat pump is greatly reduced. Thus, a change in the outdoor air temperature from 7 °С to minus 10 °С leads to a decrease in the performance of the heat pump by 1.5–2 times.

To supply water from the HP to the heated premises, heat exchangers are installed in them, referred to in the literature as "fan coils". Water is supplied to the fan coils by a hydraulic system - a pumping station (Fig. 4.4).

Rice. 4.4. Pumping station scheme:

P - pressure gauges; RB - expansion tank; AB - storage tank; RP - flow switch; H - pump;

BK - balance valve; F - filter; OK - check valve; B - valve; T - thermometer;

PC - safety valve; TP – freon-liquid heat exchanger; THC - three-way valve; KPZh - liquid make-up valve; KPV - air supply valve; KVV - air release valve

To improve the accuracy of maintaining the temperature in the room and reduce the inertia, storage tanks are installed in the hydraulic system. The capacity of the storage tank can be determined by the formula:

where is the HP cooling capacity, kW;

- the volume of cooled rooms, m 3;

is the amount of water in the system, l;

Z is the number of HP power steps.

If V AB turns out to be negative, then the storage tank is not installed.

To compensate for the thermal expansion of water in the hydraulic system, expansion tanks are installed. Expansion tanks are installed on the suction side of the pump. The volume of the expansion tank is determined by the formula:

where V syst is the volume of the system, l;

k is the coefficient of volumetric expansion of the liquid (water 3.7 10 -4, antifreeze (4.0–5.5) 10 -4);

ΔT - liquid temperature difference (when operating in cooling mode only)

ΔT \u003d t env - 4 ° С; when operating in heat pump mode ΔT=60 °С – 4 °С = 56 °С);

R prev - setting the safety valve.

The pressure in the system (P syst) depends on the relative position of the pumping station and the end user (fan coil). If the pumping station is located below the end consumer, then the pressure (P syst) is determined as the maximum height difference (in bar) plus 0.3 bar. If the pumping station is located above all consumers, then P syst = 1.5 bar.

The expansion tank is pre-inflated with air to a pressure of 0.1–0.3 bar less than the calculated one, and after installation, the pressure is brought to normal.

The design of expansion tanks is shown in fig. 4.5.

Rice. 4.5. Expansion tank design:

1 - the position of the membrane before installation (preliminary pumping with air by 0.1–0.3 bar);

2 - the position of the membrane after connecting the tank to the network;

3 - the position of the membrane during the thermal expansion of the liquid.

Expansion units are available (Figure 4.6) to maintain pressure on the water side of large volume heating and air conditioning systems. The unit is equipped with a freely programmable processor and can be interfaced to a central control panel. This greatly simplifies the control over the functioning of the system.

Specifications:

1. Volume, l 200–5,000;
2. Maximum overpressure, bar 10.0;
3. Maximum temperature, °С 120.

The flow switch (RP) shuts down the chiller when there is no liquid flow, which prevents the liquid from freezing in the heat exchanger (TP). A three-way valve mixes two fluid streams (A and B) while maintaining the desired fluid temperature. The three-way valve is controlled by a microcontroller.

Rice. 4.6. Expansion unit for heating and air conditioning systems

Design three-way valve shown in fig. 4.7.

In the lower extreme position of the shut-off cone, the passage to flow B is closed, in the upper position of the cone, the passage to flow A is closed. 24 V.

Rice. 4.7. Three-way valve for fluid flow control

The output of the drive gives a control signal about the position of the shut-off cone. The full travel time of the cone is 100–150 seconds. It is possible to manually move the cone with a hex key.

Fluid leakage with a closed channel does not exceed 1% of the throughput. In the event of a malfunction of the three-way valve and hydraulic system after the three-way valve, the liquid will circulate through the check valve (OK).

To set the estimated flow rate in the system, a balancing valve is used, which is a high-precision manual or automatic control valve. The balancing valve has outlets for measuring fluid flow and pressure. Balancing valves are available, adjusted by the adjustment controller. To adjust the balancing valve, the calculated flow and pressure values ​​are entered into the adjustment controller, after which the controller automatically sets the balancing valve to the required position.

Liquid make-up valves (KPZh) and air make-up valves (KPV) are connected to the expansion tank. When installing the filter (F), pay attention to the direction of fluid flow through the filter. An automatic air release valve (VC) is installed at the highest point of the hydraulic circuit. Safety valve set according to the maximum allowable pressure of the weakest element in the network plus 1 bar (7–10 bar).

If it is necessary to work according to a bivalent scheme, it is possible to connect a boiler with electric heating in parallel with the HP according to the scheme shown in fig. 4.8.

Rice. 4.8. Wiring diagram electric boiler to the heat pump system

4.2.2. Heat pumps with water heat sources

Heat pumps with water heat sources (rivers, lakes, seas) use the accumulated solar energy. This energy is an ideal source for heat pumps, as it is supplied continuously, although it is less available than air. The water temperature in non-freezing reservoirs does not fall below 4 °C, and artesian water has an almost constant temperature of 10–12 °C. Considering that the water cannot be cooled below 0 °C during heat extraction, the temperature difference across the heat exchanger is several degrees. At the same time, to increase the selection of the required amount of heat, it is required to increase the water flow. For HP of small capacity, it is not recommended to pump ground water from a depth of more than 15 m. Otherwise, high costs for pumps and their operation will be required.

Rice. 4.9. Heat pump using heat ground water

The heat extraction circuit from the reservoir can be open or closed. In the first case, water from the reservoir is pumped through the cooler, cooled and returned to the reservoir (Fig. 4.9). Such a system requires filtration of the water supplied to the cooler and periodic cleaning of the heat exchanger. As a rule, an intermediate collapsible heat exchanger is installed. The intake and return of water must be carried out in the direction of the flow of groundwater in order to avoid "bypassing" the water. The intake line must be with check valve(4), located at the point of intake or after the deep pump (5). The groundwater supply and discharge to the heat pump must be protected from freezing and laid with an inclination towards the well.

The distance between the intake (2) and return (1) wells must be at least 5 m. The water exit point in the return well must be below the groundwater level.

The volume flow of water is determined from the cooling capacity of the heat pump

where L is the volumetric flow rate of water, m 3 / h

c p is the specific heat capacity of water, equal to 1.163 10 -3 kWh/kg K;

– density of water, 1000 kg/m3;

- temperature difference between intake and return water.

Where . (4)

If we take Q x \u003d 12 kW (determined according to the heat pump passport), a \u003d 4 K, then m 3 / h.

A closed circuit is laid on the bottom of the reservoir. The approximate value of thermal power per 1 m of a closed loop pipeline is about 30 W. That is, to obtain 10 kW of heat, the circuit must have a length of 300 m. In order for the circuit not to float, it is necessary to install a load of about 5 kg per 1 running meter.

4.2.3. Heat pumps with ground heat exchangers

Ground HP uses thermal energy accumulated in the ground due to its heating by the Sun or other sources. The heat stored in the ground is transformed by horizontally laid ground heat exchangers (also called ground collectors) or by vertically placed ground heat exchangers (ground probes).

Rice. 4.10. Ground source heat pump

As a rule, ground heat exchangers are made of polyethylene or metal-plastic pipes with a diameter of 25–40 mm.

With a horizontal version (Fig. 4.10), the pipeline in which the liquid circulates is buried in the ground to a depth below the soil freezing level (1.2–1.5 m). The minimum distance between pipes is 0.7–1.0 m. Depending on the pipe diameter, 1.4–2.0 m of pipes can be laid for each square meter of heat intake area. The length of each branch of the horizontal collector must not exceed 100 m, otherwise the pressure loss in the pipe and the required pump power will be too large.

The amount of transformed heat, and, consequently, the size of the required surface for the location of the ground collector significantly depends on the thermophysical properties of the soil and the climatic conditions of the area. Thermophysical properties, such as heat capacity and thermal conductivity, are very dependent on the composition and condition of the soil. In this regard, the determining factor is the proportion of water, the content of mineral components (quartz, feldspar), as well as the proportion and size of pores filled with air. The storage properties and thermal conductivity of the soil are the higher, the greater the proportion of water, mineral components and the lower the content of pores.

The average value of the specific thermal power of the soil is given in table 1.

Table 1. The average value of the specific thermal power of the soil

Soil type	Specific power of the soil collector, W / m 2	Specific power of the ground probe, W/m
sandy dry	10–15	20
sandy wet	15–20	40
Clay dry	20–25	60
Clay wet	25–30	80
Aquifer	30–35	80–100

The required area for the location of the collector is calculated by formulas (5) and (6)

where is the heat output of HP, W;

– HP power consumption from the network, W;

g - specific power of the soil collector, W / m 2.

So, if the cooling capacity of the HP is 10 kW, then in sandy moist soil (g \u003d 20 W / m 2) an area will be required to place the collector

To transform heat from such an area, it is necessary to lay in the ground polyethylene pipes with a diameter of 25 × 2.3 mm and a length of 500 × 1.4 = 700 m. (1.4 is the specific pipe consumption per square meter of area). Pipes must be laid in separate circuits of 100 m each, i.e. 7 circuits.

All distributors and collectors should be located in accessible places for inspection, for example, in separate distribution shafts outside the house or in the basement shaft of the house. Fittings must be made of corrosion resistant materials. All piping in the house and wall penetrations must be thermally insulated with diffusion impermeability to vapor to avoid condensation, as in the supply and return lines there is a cold (relative to the basement temperature) coolant.

With a vertical version of the soil probe, a well is drilled with a depth of 60–200 m, into which several U-shaped pipelines are lowered (Fig. 4.11).

A	b

Rice. 4.11. Ground source heat pump

a - general scheme, b - scheme of a soil probe

1 - return line, 2 - supply line, 3 - loop probe, 4 - protective cap

In clay moist soil with a heat pump cooling capacity of 10 kW, the probe length (well depth) should be

It is advisable to make 2 loops with a depth of 50 m with a diameter of D y \u003d 32 × 3 mm. The total length of the pipes will be 200 m. The well with pipes is filled with betonite, which conducts heat well. The amount of coolant is determined by the internal volume of the pipes of the collector (probe) and supply pipes. The diameter of the supply pipes is taken one size larger than the collector pipe. In our example, with a probe pipe D y = 32 × 3 mm and a supply pipe D y = 40 × 2.3 mm 10 m long, the internal volume (table 2), taking into account the supply line, will be 2 × 100 × 0.531 + 10 × 0.984 = 116 .04 l. The flow rate of the heat pump coolant is found according to the passport for the heat pump. Let's take 1600 l / h. Then the flow rate per loop will be 800 l / h.

Table 2. Specific internal volume of pipes

The pressure loss in the pipes depends on the diameter of the pipes, the density and flow rate of the coolant and is determined according to the pipe manufacturer's data. So, for HDPE pipes (high density polyethylene) 32 × 3 mm and a flow rate of 800 l / h is 154.78 Pa / m, and for pipes with a diameter of 40 × 2.3 - 520.61 Pa / m. From where the total pressure drop in the network will be 36161.1 Pa, which must be taken into account when choosing a pump.

The service life of a soil collector depends on the acidity of the soil: with normal acidity (pH = 5.0) - 50–75 years, with high acidity (pH > 5.0) - 25–30 years.

4.1. Heat pump efficiency

As the main indicator of the efficiency of a heat pump, the conversion coefficient or heating coefficient COP (coefficient of performance) is used, which is equal to the ratio of the heat output of the heat pump to the power consumed by the compressor. In cooling mode, the energy efficiency ratio (EER) is used to evaluate efficiency, which is equal to the ratio of the cooling capacity of the heat pump to the power consumed by the compressor.

where is the energy given off by the HTP;

- thermal energy taken from the INT;

- consumed electricity;

And are the condensing and boiling temperatures in the heat pump.

The temperature is determined by the pressure of condensation of the refrigerant in the HP, and - by the temperature of the HP. So, if we take = 281.16 K (8 ° C) and = 323.16 K (50 ° C), then the COP will be equal to 7.7. If heat is removed by water, then various refrigerants allow reaching the following temperatures: R717, R502, R22 - about +50 °C, R134a - +70 °C, R142 - +100 °C.

It should be remembered about the basic rule following from (4): the smaller the temperature difference between both the source and the heat receiver in the heat pump, the higher the conversion factor.

When heat pumps use heat and cold at the same time (for example, cooling cold rooms and heating office space), then

With an equipotential cycle =

At the above temperatures, the total conversion factor can reach 12.7, which characterizes the high energy efficiency of the heat pump. The real ROPs are somewhat lower and are on the order of 3–5.

In absorption heat pumps, the conversion coefficient is lower than in compression ones, due to large losses in the elements of the absorption circuit. So, when using groundwater with T 0 = 281.16 K (8 ° C) and useful heat temperature = 323.16 K (50 ° C), the absorption HP conversion coefficient will be only 1.45. The useful heat temperature in absorption heat pumps also depends on the heating temperature of the generator. At the temperatures indicated above, the heating of the generator must be at least 150 °C.

During the heating season (October-May), heating 100 m 2 of a living space with an electric boiler will require 37,440 kW of electricity, and a heat pump - 12,024 kW. At a tariff of 0.24 UAH per 1 kW of electricity, the savings will be 6100 UAH. (data from Santechnik LTD and Co., Ltd.).

According to http://www.aeroprof.by, the use of HP is 1.2–1.5 times more profitable than the most efficient gas boiler.

The cost of a heat pump can be approximately estimated at the rate of 750-1500 UAH per 1 kW of generated heat power. Payback period is 7–14 years.

4.2. Selection of equipment for heat pumps

The choice of equipment begins with the calculation of the heat demand of the building. There are currently a variety of PC-based heat calculation programs that can be found on the Internet or obtained from equipment suppliers.

An approximate calculation can be made based on the heated area of ​​\u200b\u200bthe building and the amount of hot water consumed. Also, in the event of periodic planned power outages, it is necessary to increase the heat output of the heat pump. If the power outage time does not exceed 2 hours, this factor can be ignored.

Specific heat consumption depends on the type of building:

• building with low consumption (modern materials, wall insulation, double-glazed windows) - 40 W / m 2;
• new building, good thermal insulation - 50 W / m 2;
• building with standard thermal insulation - 80 W / m 2;
• old buildings without special insulation - 120 W / m 2.

Accounting for additional thermal power to compensate for heat losses during planned power outages is carried out as follows.

Determine the daily (for 24 hours) heat consumption

where is the heating capacity of the HP, kW;

- time of power outage.

The calculation of the additional heat output for the preparation of hot water is based on the consumption by one person of about 50 liters of water at a temperature of 45 ° C, which corresponds to 0.25 kW / person. A more accurate calculation can be performed using the data in Table 3.

Table 3. Daily consumption of hot water

Category	Water consumption, l/person		Specific heat consumption, Wh/person	Heat consumption for hot water, kW / person
	pace. water 60°C	pace. water 45°C
Low consumption	10–20	15–30	600–1200	0,08–0,15
Standard consumption	20–40	30–60	1200–2400	0,15–0,3
Apartment occupying a floor	32	45	1800	0,225
Single-family residential building	35	50	2000	0,25

Let's consider an example of building a heat pump with a reversible hydraulic cycle, operating year-round in two modes (cooling or heating), depending on the period of the year, using equipment and software from CIAT (France).

Initial requirements:

1. Heat output 510 kW.

2. Low temperature source - sea ​​water with temperature:

warm period of the year ≤20 °С,

cold period of the year 7 °С.

3. High-temperature consumer - water with a temperature at the outlet of the heat exchanger 55 °C.

4. The minimum outdoor air temperature is minus 10 °С (Crimea, Ukraine).

We will solve this problem using a heat pump with a reversible hydraulic cycle, the scheme of which is shown in fig. 2.

Given that the outside air temperature is negative (minus 10 °C) in the heat pump, we use a two-circuit system. In the primary circuit, we use an ethylene glycol solution with a freezing point below -10 °C (a 20% mixture of ethylene glycol and water).

In accordance with the initial requirements, we choose the temperature difference in the high-temperature circuit Dt out = 5 °C (50/55 °C). Then the temperature of the coolant in the condenser circuit should be 55/60 °C, respectively. To obtain such temperatures in the heat pump, it is advisable to use the refrigerant R134a.

In accordance with the initial requirements, we set the temperature difference INT 7/4 °C, then in the evaporator circuit the temperature difference will be 5/2 °C, respectively.

Using the CIAT equipment selection program, we determine the type and parameters of the heat pump in heating and cooling operating modes. The program selected the heat pump HYDROCIAT 2500B X LW/LWP R134a with the parameters given in table. 4, appearance which is shown in Fig. 12.

Table 4. Specifications of the HYDROCIAT 2500B X LW/LWP R134a water chiller

Parameter	Heating mode	Cooling mode
Evaporator capacity, kW	326,0	395,9
coolant	MEG20%	MEG20%
Heat carrier temperature in the evaporator (inlet/outlet), °C	5,0/2,0	6,0/2,0
Coolant flow through the evaporator, m 3 / h	102,8	93,4
Capacitor capacity, kW	517,0	553,9
Coolant temperature in the condenser (inlet/outlet), °C	55/60	45,1/50
Coolant flow through the condenser, m 3 / h	93,4	102,1
Power consumption, kW	191	158,0

Rice. 4.12. Heat pump HYDROCIAT 2500B X LW/LWP R134a

1. Water temperature (outlet-inlet): 55/50 °С.
2. Temperature of 20% ethylene glycol solution in the primary circuit (outlet-inlet): 60/55 °C.
3. Consumption of a 20% ethylene glycol solution: 93.4 m 3 / h (see table. 1).

The CIAT program selects a PWB 30 11 plate heat exchanger with a capacity of 517 kW (Table 5).

Table 5. Technical data of the heat exchanger PWB 30 11 with 43 plates (heat pump - consumer) in heating mode

The low-temperature heat exchanger "sea water-heat pump" in the heating mode is selected according to the following initial data:

1. Source of low-grade heat (primary circuit): sea water with an inlet/outlet temperature of 7/4 °С.
2. The temperature of the 20% ethylene glycol solution in the primary circuit is 5/2 °C.
3. The consumption of a 20% solution of ethylene glycol is 102.8 m 3 / h.

The CIAT program selects the PWB 45 11 plate heat exchanger.

Table 6. Technical data of the PWB 45 11 heat exchanger with 63 plates (sea-heat pump)

Let's perform a verification calculation of the previously calculated heat exchanger PWB 30 11 with 43 plates for the warm period of the year and determine the water temperature at the outlet / inlet to the consumer.

The CIAT program has shown that during the summer period the heat exchanger PWB 30 11 will have a capacity of 437 kW and coolant temperatures of (outlet/inlet) 7.5/12°C. (Table 7)

Table 7. Technical data of the PWB 30 11 heat exchanger with 43 plates (heat pump - consumer) in cooling mode

Thus, the selected HYDROCIAT 2500 XLW/LWP R134a heat pump provides:

• during the cold period of the year, the heating capacity is 517 kW with a power consumption of 191 kW;
• in the warm period of the year, the cooling capacity is 395.9 kW with a power consumption of 158 kW.

Below is a circuit diagram of a reversible hydraulic cycle heat pump calculated above.

Rice. 4.13. Schematic diagram of a heat pump with a reversible hydraulic cycle

The nomenclature of some CIAT heat pumps is given in Table. 8.

Table 8. Heat pumps from CIAT (France)

Heat pump type		Productivity, kW		Application area
		in the cold	by warmth	individual houses	apartment buildings	public buildings	production
AUREA 2		7…28	9…36	+
DYNACIAT LG/LGP/ILG		35…350	40…370		+	+
HYDROCIAT LW/LWP		275…1140	350…1420		+	+	+

Conclusion.

1. Heat pumps using renewable heat sources are the most energy efficient heating equipment.
2. Systems based on HP are reliable, safe and durable.
3. Receiving heat with a heat pump is an environmentally friendly technological process.
4. Modern climatic equipment (for example, CIAT, France) makes it possible to create HP with a capacity from tens of kW to MW.

Literature.

1. V. Maake, G.-Yu. Eckert, J.-L. Koshpen. Refrigeration textbook: Per. from French - M .: Publishing house of Moscow University, 1998. - 1142 p., ill.
2. Ray D., McMichael D. Heat pumps: Per. from English. - M.: Energoizdat, 1982. - 224 p., ill.
3. El Sadin Hasan. The choice of optimal parameters for the heating and cold supply system of a residential building // Kholodilnaya Tekhnika, 2003, No. 3, pp. 18–21.
4. Ovcharenko V.A. Ovcharenko A.V. Vikoristannya heat pumps / / Cold M + T, 2006, No. 2 p. 34–36.
5. Five steps towards getting rid of methane dependence//Heating Water supply Ventilation + air conditioners, 2006, No. 1, p. 30–41.
6. Bondar E.S., Kalugin P.V. Energy-saving air conditioning systems with cold storage//S.O.K., 2006, No. 3, p. 44–48.
7. Viesmann.Heat pump systems. Design instruction.5829 122-2 GUS 2/2000
8. Belova. Air conditioning systems with chillers and fan coil units

Types of designs of heat pumps

The type of HP is usually denoted by a phrase indicating the source medium and the heat carrier of the heating system.

There are the following varieties:

• TN "air - air";
• TN "air - water";
• TN "soil - water";
• TN "water - water".

The very first option is a conventional split system operating in heating mode. The evaporator is mounted on the street, and a block with a condenser is installed inside the house. The latter is blown by a fan, due to which a warm air mass is supplied to the room.

If such a system is equipped with a special heat exchanger with branch pipes, an air-to-water heat pump will be obtained. It is connected to the water heating system.

Air-to-air or air-to-water HP evaporator can be placed not on the street, but in the channel exhaust ventilation(it should be forced). In this case, the efficiency of HP will be increased several times.

Heat pumps of the "water - water" and "soil - water" types use the so-called external heat exchanger or, as it is also called, a collector to extract heat.

Schematic diagram of the heat pump

This is a long looped pipe, usually plastic, through which a liquid medium circulates, washing the evaporator. Both types of HP are the same device: in one case, the collector is immersed to the bottom of a surface reservoir, and in the second, to the ground. The condenser of such a HP is located in a heat exchanger connected to a water heating system.

Connecting a HP according to the "water - water" scheme is much less laborious than "soil - water", since there is no need for earthworks. At the bottom of the reservoir, the pipe is laid in the form of a spiral. Of course, only such a body of water is suitable for this scheme, which does not freeze to the bottom in winter.

It is time to study foreign experience in detail

Almost everyone already knows about heat pumps capable of extracting ambient heat for heating buildings, and if until recently a potential customer, as a rule, asked a bewildered question “how is this possible?”, Now the question “how is it right” is increasingly heard. do?".

It is not easy to answer this question.

In search of an answer to the numerous questions that inevitably arise when trying to design heating systems with heat pumps, it is advisable to refer to the experience of specialists from those countries where heat pumps based on ground heat exchangers have been used for a long time.

A visit* to the American exhibition AHR EXPO-2008, which was undertaken mainly to obtain information on the methods of engineering calculations of ground heat exchangers, did not bring direct results in this direction, but a book was sold at the ASHRAE exhibition stand, some of the provisions of which served as the basis for this publications.

It should be said right away that the transfer of American methods to domestic soil is not an easy task. Americans do not do things the way they do in Europe. Only they measure time in the same units as we do. All other units of measurement are purely American, or rather, British. The Americans were especially unlucky with the heat flux, which can be measured both in British thermal units per unit of time, and in tons of cooling, which were probably invented in America.

The main problem, however, was not the technical inconvenience of recalculating the units of measurement accepted in the United States, to which one can eventually get used, but the absence in the mentioned book of a clear methodological basis for constructing a calculation algorithm. Too much space is given to routine and well-known calculation methods, while some important provisions remain completely undisclosed.

In particular, such physically related initial data for the calculation of vertical ground heat exchangers, as the temperature of the liquid circulating in the heat exchanger and the heat pump conversion coefficient, cannot be set arbitrarily, and before proceeding with calculations related to unsteady heat transfer in the soil, it is necessary to determine the dependencies connecting these options.

The criterion for the efficiency of a heat pump is the conversion factor?, the value of which is determined by the ratio of its thermal power to the power of the compressor electric drive. This value is a function of the boiling temperatures in the evaporator t u and condensation t k , and in relation to heat pumps "water-water" we can talk about the temperatures of the liquid at the outlet of the evaporator t 2I and at the outlet of the condenser t 2 K:

? \u003d? (t 2I, t 2 K). (1)

An analysis of the catalog characteristics of serial refrigeration machines and water-to-water heat pumps made it possible to display this function in the form of a diagram (Fig. 1).

Using the diagram, it is easy to determine the parameters of the heat pump at the very initial stages of design. It is obvious, for example, that if the heating system connected to the heat pump is designed to supply a heating medium with a flow temperature of 50°C, then the maximum possible conversion factor of the heat pump will be about 3.5. At the same time, the temperature of the glycol at the outlet of the evaporator should not be lower than +3°C, which means that an expensive ground heat exchanger will be required.

At the same time, if the house is heated by underfloor heating, a coolant with a temperature of 35°C will enter the heating system from the heat pump condenser. In this case, the heat pump can work more efficiently, for example, with a conversion factor of 4.3, if the temperature of the cooled glycol in the evaporator is around -2°C.

Using Excel spreadsheets, you can express the function (1) as an equation:

0.1729 (41.5 + t 2I - 0.015t 2I t 2 K - 0.437 t 2 K (2)

If, with the desired conversion factor and a given value of the coolant temperature in the heating system powered by a heat pump, it is necessary to determine the temperature of the liquid cooled in the evaporator, then equation (2) can be represented as:

It is possible to choose the heat carrier temperature in the heating system for the given values ​​of the heat pump conversion coefficient and the liquid temperature at the outlet of the evaporator using the formula:

In formulas (2)…(4) temperatures are expressed in degrees Celsius.

Having determined these dependencies, we can now proceed directly to the American experience.

Methodology for calculating heat pumps

Of course, the process of selecting and calculating a heat pump is a technically very complex operation and depends on individual features object, but roughly it can be reduced to the following steps:

Heat losses through the building envelope (walls, ceilings, windows, doors) are determined. This can be done using the following ratio:

Qok \u003d S * (tin - tout) * (1 + Σ β) * n / Rt (W) where

tout - outside air temperature (°С);

tin – internal air temperature (°С);

S is the total area of ​​all enclosing structures (m2);

n - coefficient indicating the influence of the environment on the characteristics of the object. For premises in direct contact with the external environment through ceilings n=1; for objects with attic floors n=0.9; if the object is located above the basement n = 0.75;

β is the coefficient of additional heat loss, which depends on the type of building and its geographical location; β can vary from 0.05 to 0.27;

Rt - thermal resistance, is determined by the following expression:

Rt \u003d 1 / α int + Σ (δ i / λ i) + 1 / α out (m2 * ° С / W), where:

δ і / λі - calculated indicator of thermal conductivity of materials used in construction.

α nar - coefficient of thermal dissipation of the outer surfaces of enclosing structures (W / m2 * ° C);

α int - coefficient of thermal absorption of the internal surfaces of enclosing structures (W / m2 * ° C);

- The total heat loss of the structure is calculated according to the formula:

Qt.pot \u003d Qok + Qi - Qbp, where:

Qi - energy costs for heating the air entering the room through natural leaks;

Qbp ​​- heat release due to the functioning of household appliances and human activities.

2. Based on the data obtained, the annual consumption of thermal energy is calculated for each individual object:

Qyear = 24*0.63*Qt. sweat.*((d*(tin — tout.av.)/ (tin — tout.)) (kWh per year) where:

tout - outside air temperature;

tout.average - the arithmetic mean of the outdoor air temperature for the entire heating season;

d is the number of days of the heating period.

Qhv \u003d V * 17 (kW / h per year.) Where:

V is the volume of daily heating of water up to 50 °C.

Then the total consumption of thermal energy is determined by the formula:

Q \u003d Qgw + Qyear (kW / h per year.)

Taking into account the obtained data, it will not be difficult to choose the most suitable heat pump for heating and hot water supply. Moreover, the calculated power is determined as. Qtn=1.1*Q, where:

Qtn=1.1*Q, where:

1.1 - correction factor indicating the possibility of increasing the load on the heat pump during the occurrence of critical temperatures.

After performing the calculation of heat pumps, you can choose the most suitable heat pump that can provide the required microclimate parameters in rooms with any technical characteristics. And given the possibility of integrating this system with a heated floor air conditioner, it can be noted not only its functionality, but also its high aesthetic value.

If you liked the material, I will be grateful if you recommend it to friends or leave a useful comment.

Types of heat pumps

Heat pumps are divided into three main types according to the source of low-grade energy:

• Air.
• Priming.
• Water - the source can be groundwater and reservoirs on the surface.

For water heating systems, which are more common, the following types of heat pumps are used:

"Air-water" - air type a heat pump that heats the building by taking in air from outside through an outdoor unit. It works on the principle of an air conditioner, only in reverse, converting the energy of the air into heat. Such a heat pump does not require large installation costs, it does not need to allocate a piece of land for it and, moreover, drill a well. However, the efficiency of operation at low temperatures (-25ºС) decreases and an additional source of thermal energy is required.

The "ground-water" device refers to geothermal and produces heat from the ground using a collector laid to a depth below the freezing of the soil. There is also a dependence on the area of ​​​​the site and the landscape, if the collector is located horizontally. For a vertical arrangement, a well will need to be drilled.

"Water-water" is installed where there is a reservoir or groundwater nearby. In the first case, the collector is laid on the bottom of the reservoir, in the second, a well is drilled or several, if the area of ​​​​the site allows. Sometimes the depth of groundwater is too great, so the cost of installing such a heat pump can be very high.

Each type of heat pump has its advantages and disadvantages, if the building is far from a body of water or the groundwater is too deep, then water-to-water will not work. "Air-water" will be relevant only in relatively warm regions, where the air temperature during the cold season does not fall below -25º C.

Method for calculating the power of a heat pump

In addition to determining the optimal energy source, it will be necessary to calculate the power of the heat pump required for heating. It depends on the amount of heat loss of the building. Let's calculate the power of a heat pump for heating a house using a specific example.

To do this, we use the formula Q=k*V*∆T, where

• Q is heat loss (kcal/hour). 1 kWh = 860 kcal/h;
• V is the volume of the house in m3 (we multiply the area by the height of the ceilings);
• ∆Т is the ratio of the minimum temperatures outside and inside the premises during the coldest period of the year, °C. From the internal tº we subtract the external one;
• k is the generalized heat transfer coefficient of the building. For a brick building with two layers of masonry k=1; for a well-insulated building k=0.6.

Thus, the calculation of the power of a heat pump for heating a brick house of 100 sq.m and a ceiling height of 2.5 m, with a difference in ttº from -30º outside to +20º inside, will be as follows:

Q \u003d (100x2.5) x (20- (-30)) x 1 \u003d 12500 kcal / hour

12500/860= 14.53 kW. That is, for a standard brick house with an area of ​​100 m2, you will need a 14-kilowatt device.

The consumer accepts the choice of the type and power of the heat pump based on a number of conditions:

• geographical features of the area (proximity of water bodies, the presence of groundwater, a free area for a collector);
• climate features (temperature);
• type and internal volume of the room;
• financial opportunities.

Considering all the above aspects, you will be able to make the best choice of equipment. For a more efficient and correct selection of a heat pump, it is better to contact specialists, they will be able to make more detailed calculations and provide the economic feasibility of installing the equipment.

For a long time and very successfully, heat pumps have been used in household and industrial refrigerators and air conditioners.

Today, these devices began to be used to perform the function of the opposite nature - heating the home during the cold season.

Let's see how heat pumps are used for heating private houses and what you need to know in order to correctly calculate all its components.

Heat pump calculation example

We will select a heat pump for the heating system of a one-story house with a total area of ​​70 sq. m with a standard ceiling height (2.5 m), rational architecture and thermal insulation of enclosing structures that meet the requirements of modern building codes. For heating the 1st sq. m of such an object, according to generally accepted standards, you have to spend 100 W of heat. Thus, for heating the whole house you will need:

Q \u003d 70 x 100 \u003d 7000 W \u003d 7 kW of thermal energy.

We choose a heat pump brand "TeploDarom" (model L-024-WLC) with a heat output of W = 7.7 kW. The compressor of the unit consumes N = 2.5 kW of electricity.

Collector calculation

The soil in the area allotted for the construction of the collector is clayey, the groundwater level is high (we take the calorific value p = 35 W/m).

Collector power is determined by the formula:

Qk \u003d W - N \u003d 7.7 - 2.5 \u003d 5.2 kW.

L = 5200 / 35 = 148.5 m (approx.).

Based on the fact that laying a circuit longer than 100 m is irrational due to excessively high hydraulic resistance, we accept the following: the heat pump collector will consist of two circuits - 100 m and 50 m long.

The area of ​​​​the site that will need to be taken under the collector is determined by the formula:

Where A is the step between adjacent sections of the contour. We accept: A = 0.8 m.

Then S = 150 x 0.8 = 120 sq. m.

Payback of a heat pump

When it comes to how long a person will be able to return his money invested in something, it means how profitable the investment itself was. In the field of heating, everything is quite difficult, since we provide ourselves with comfort and warmth, and all systems are expensive, but in this case, you can look for an option that would return the money spent by reducing costs when using. And when you start looking for a suitable solution, you compare everything: a gas boiler, a heat pump or an electric boiler. We will analyze which system will pay off faster and more efficiently.

The concept of payback, in this case, the introduction of a heat pump to modernize the existing heat supply system, if simply, can be explained as follows:

There is one system - an individual gas boiler, which provides heating system and DHW. There is a split-system type air conditioner that provides cold to one room. Installed 3 split systems in different rooms.

And there is a more economical advanced technology - a heat pump that will heat / cool houses and heat water in the right quantities for a house or apartment. It is necessary to determine how much the total cost of equipment and initial costs has changed, as well as to assess how much the annual costs of operating the selected types of equipment have decreased. And to determine how many years more expensive equipment will pay off with the resulting savings. Ideally, several proposed design solutions are compared and the most cost-effective one is selected.

We will carry out the calculation and find out what is the payback period of a heat pump in Ukraine

Consider a specific example

• House on 2 floors, well insulated, with a total area of ​​150 sq. m.
• Heat / heating distribution system: circuit 1 - underfloor heating, circuit 2 - radiators (or fan coil units).
• A gas boiler for heating and hot water supply (DHW), for example, 24kW, double-circuit, is installed.
• Air conditioning system from split systems for 3 rooms of the house.

Annual heating and water heating costs

1. The approximate cost of a boiler room with a 24 kW gas boiler (boiler, piping, wiring, tank, meter, installation) is about 1000 Euros. An air conditioning system (one split system) for such a house will cost about 800 euros. In total, with the arrangement of the boiler room, design work, connection to the gas pipeline network and installation work - 6100 euros.

1. Approximate cost of a Mycond heat pump with additional fan coil system, installation work and electrical connection is 6650 euros.

1. The growth of capital investments is: K2-K1 = 6650 - 6100 = 550 euros (or about 16500 UAH)
2. The reduction in operating costs is: C1-C2 = 27252 - 7644 = 19608 UAH.
3. Payback period Tokup. = 16500 / 19608 = 0.84 years!

Ease of use of the heat pump

Heat pumps are the most versatile, multifunctional and energy efficient equipment for heating a house, apartment, office or commercial facility.

An intelligent control system with weekly or daily programming, automatic switching of seasonal settings, maintaining the temperature in the home, economical modes, control of a slave boiler, boiler, circulation pumps, temperature control in two heating circuits, is the most advanced and advanced. Inverter control of the compressor, fan, pumps, allows maximum energy savings.

Heat pump operation during ground-water operation

Laying the collector in the ground can be done in three ways.

Horizontal option

Pipes are laid in trenches "snake" to a depth exceeding the depth of soil freezing (on average - from 1 to 1.5 m).

Such a collector will require a plot of land of a sufficiently large area, but any homeowner can build it - no skills other than the ability to work with a shovel will be needed.

It should, however, be taken into account that the construction of a heat exchanger by hand is a rather laborious process.

Vertical option

Collector pipes in the form of loops, having the shape of the letter “U”, are immersed in wells with a depth of 20 to 100 m. If necessary, several such wells can be built. After the pipes are installed, the wells are filled with cement mortar.

The advantage of a vertical collector is that a very small area is needed for its construction. However, there is no way to drill wells with a depth of more than 20 m on your own - you will have to hire a team of drillers.

Combined variant

This collector can be considered a variation of the horizontal one, but it will require much less space to build.

A round well is dug on the site with a depth of 2 m.

The heat exchanger pipes are laid in a spiral, so that the circuit is like a vertically mounted spring.

Upon completion of the installation work, the well falls asleep. As in the case of a horizontal heat exchanger, all the necessary amount of work can be done by hand.

The collector is filled with antifreeze - antifreeze or ethylene glycol solution. To ensure its circulation, a special pump crashes into the circuit. Having absorbed the heat of the soil, the antifreeze enters the evaporator, where heat exchange takes place between it and the refrigerant.

It should be taken into account that the unlimited extraction of heat from the soil, especially with a vertical collector, can lead to undesirable consequences for the geology and ecology of the site. Therefore, in the summer period, it is highly desirable to operate the HP of the "soil - water" type in the reverse mode - air conditioning.

The gas heating system has a lot of advantages and one of the main ones is the low cost of gas. How to equip home heating with gas, you will be prompted by the heating scheme of a private house with a gas boiler. Consider the design of the heating system and the requirements for replacement.

Read about the features of choosing solar panels for home heating in this topic.

Calculation of the horizontal collector of a heat pump

The efficiency of a horizontal collector depends on the temperature of the medium in which it is immersed, its thermal conductivity, as well as the area of ​​contact with the pipe surface. The calculation method is rather complicated, therefore, in most cases, averaged data are used.

It is believed that each meter of the heat exchanger provides the HP with the following heat output:

• 10 W - when buried in dry sandy or rocky soil;
• 20 W - in dry clay soil;
• 25 W - in wet clay soil;
• 35 W - in very damp clay soil.

Thus, to calculate the length of the collector (L), the required thermal power (Q) should be divided by the calorific value of the soil (p):

• The land above the collector is not built up, shaded, or planted with trees or bushes.
• The distance between adjacent turns of the spiral or sections of the "snake" is at least 0.7 m.

How heat pumps work

In any HP there is a working medium called a refrigerant. Usually freon acts in this capacity, less often - ammonia. The device itself consists of only three components:

The evaporator and condenser are two reservoirs that look like long curved tubes - coils. The condenser is connected at one end to the compressor outlet, and the evaporator to the inlet. The ends of the coils are joined and a pressure reducing valve is installed at the junction between them. The evaporator is in contact - directly or indirectly - with the source medium, while the condenser is in contact with the heating or DHW system.

How a heat pump works

The operation of the HP is based on the interdependence of the volume, pressure and temperature of the gas. Here is what happens inside the aggregate:

1. Ammonia, freon or other refrigerant, moving through the evaporator, heats up from the source medium, for example, to a temperature of +5 degrees.
2. After passing the evaporator, the gas reaches the compressor, which pumps it into the condenser.
3. The refrigerant pumped by the compressor is held in the condenser by a pressure reducing valve, so its pressure here is higher than in the evaporator. As you know, with increasing pressure, the temperature of any gas increases. This is exactly what happens to the refrigerant - it heats up to 60 - 70 degrees. Since the condenser is washed by the coolant circulating in the heating system, the latter is also heated.
4. Through the pressure reducing valve, the refrigerant is discharged in small portions into the evaporator, where its pressure drops again. The gas expands and cools, and since part of the internal energy was lost by it as a result of heat transfer at the previous stage, its temperature drops below the initial +5 degrees. Following the evaporator, it heats up again, then it is pumped into the condenser by the compressor - and so on in a circle. Scientifically, this process is called the Carnot cycle.

But HP still remains very profitable: for each kWh of electricity spent, it is possible to get from 3 to 5 kWh of heat.

Influence of initial data on the calculation result

Let us now use the mathematical model built in the course of calculations in order to trace the influence of various initial data on the final result of the calculation. It should be noted that the calculations performed on Excel allow such an analysis to be carried out very quickly.

To begin with, let's see how its thermal conductivity affects the value of the heat flux to the WGT from the ground.

As you know, heat pumps use free and renewable energy sources: low-grade heat of air, soil, underground, waste and waste water from technological processes, open non-freezing reservoirs. Electricity is spent on this, but the ratio of the amount of thermal energy received to the amount of electrical energy consumed is about 3-7. More precisely, the sources of low-grade heat can be outdoor air with a temperature of -15 to +15°C, exhaust air (15-25°C), subsoil (4-10°C) and groundwater (more than 10°C) water , lake and river water (0-10°С), surface (0-10°С) and deep (more than 20 m) soil (10°С).

If atmospheric or ventilation air is selected as the heat source, heat pumps operating according to the "air-to-water" scheme are used. The pump can be located indoors or outdoors. Air is supplied to its heat exchanger by means of a fan.

When groundwater is used as a source of heat, it is pumped from a well by a pump into a heat exchanger of a water-to-water pump and either pumped into another well or discharged into a reservoir.
If the source is a reservoir, a loop of a metal-plastic or plastic pipe is laid on its bottom. A glycol solution (antifreeze) circulates through the pipeline, which transfers heat to freon through the heat pump heat exchanger.

There are two options for obtaining low-grade heat from the soil: laying metal-plastic pipes in trenches 1.2-1.5 m deep or in vertical wells 20-100 m deep. Sometimes pipes are laid in the form of spirals in trenches 2-4 m deep. This significantly reduces the total length of the trenches. The maximum heat transfer of the surface soil is 50-70 kWh/m 2 per year. According to foreign companies, the service life of trenches and wells is more than 100 years.

Calculation of the horizontal collector of a heat pump

The removal of heat from each meter of pipe depends on many parameters: laying depth, availability of groundwater, soil quality, etc. Tentatively, it can be considered that for horizontal collectors it is 20 W / m. More precisely: dry sand - 10, dry clay - 20, wet clay - 25, clay with a high water content - 35 W/m. The difference in the temperature of the coolant in the direct and return lines of the loop in the calculations is usually assumed to be 3 °C. Buildings should not be erected on the site above the collector so that the heat of the earth is replenished due to solar radiation.

The minimum distance between the laid pipes should be 0.7-0.8 m. The length of one trench is usually from 30 to 120 m. It is recommended to use a 25% glycol solution as the primary circuit coolant. In calculations, it should be taken into account that its heat capacity at a temperature of 0 ° C is 3.7 kJ / (kg.K), density - 1.05 g / cm 3. When using antifreeze, the pressure loss in the pipes is 1.5 times greater than when water is circulating. To calculate the parameters of the primary circuit of a heat pump installation, it will be necessary to determine the antifreeze consumption:

Vs = Qo.3600 / (1.05.3.7..t),

Where.t is the temperature difference between the supply and return lines, which is often assumed to be 3 K, and Qo is the thermal power received from a low-potential source (soil). The latter value is calculated as the difference between the total power of the heat pump Qwp and the electric power spent on heating the freon P:

Qo = Qwp - P, kW.

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

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

Heat Pump Calculation Example

Initial conditions: heat demand of a cottage with an area of ​​120-240 m 2 (depending on thermal insulation) - 12 kW; the water temperature in the heating system should be 35 ° C; the minimum temperature of the heat carrier is 0 °С. To heat the building, a heat pump WPS 140 l (Buderus) with a capacity of 14.5 kW (the nearest larger standard size) was selected, which consumes 3.22 kW of freon for heating. Heat removal from the surface layer of soil (dry clay) q is 20 W/m. In accordance with the formulas shown above, we calculate:

1. the required heat output of the collector Qo = 14.5 - 3.22 = 11.28 kW;
2. the total length of the pipes L = Qo / q = 11.28 / 0.020 = 564 m. To organize such a collector, 6 circuits 100 m long are required;
3. with a laying step of 0.75 m, the required area of ​​\u200b\u200bthe site A \u003d 600 × 0.75 \u003d 450 m 2;
4. total consumption of glycol solution Vs = 11.28.3600 / (1.05.3.7.3) = 3.51 m 3 / h, the flow rate per circuit is 0.58 m 3 / h.

For the collector device, we select a metal-plastic pipe of size 32Ch3 (for example, Henco). The pressure loss in it will be 45 Pa / m; the resistance of one circuit is approximately 7 kPa; coolant flow rate - 0.3 m/s.

Probe calculation

When using vertical wells with a depth of 20 to 100 m, U-shaped metal-plastic or plastic (with diameters above 32 mm) pipes are immersed in them. As a rule, two loops are inserted into one well, after which it is poured with cement mortar. On average, the specific heat removal of such a probe can be taken equal to 50 W/m. 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 temperature of the soil at a depth of more than 15 m is constant and is approximately +10 °C. The distance between the wells should be more than 5 m. In the presence of underground currents, the wells should be located on a line perpendicular to the flow.

The selection of pipe diameters is carried out on the basis of pressure losses for the required coolant flow rate. Calculation of the liquid flow rate can be carried out for .t = 5 °C.

Calculation example: The initial data are the same as in the horizontal collector calculation above. With a specific heat removal of the probe of 50 W/m and a required power of 11.28 kW, the probe length L should be 225 m.

To construct a collector, it is necessary to drill three wells with a depth of 75 m. In each of them we place two loops from a metal-plastic pipe of size 26Ch3; in total - 6 contours of 150 m each.

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

Equipment selection

Since the antifreeze temperature can vary (from -5 to +20 °C), an expansion tank is required in the primary circuit of the heat pump unit.

It is also recommended to install a storage tank on the return line: the heat pump compressor operates in on/off mode. Too frequent starts can lead to accelerated wear of its parts. The tank is also useful as an energy accumulator - in case of a power outage. Its minimum volume is taken at the rate of 10-20 liters per 1 kW of heat pump power.

When using a second energy source (electric, gas, liquid or solid fuel boiler), it is connected to the circuit through mixing valve whose drive is controlled by a heat pump or common system automation.

In case of possible power outages, it is necessary to 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 outage.

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 temperature is a small part of the heating season. For example, for the Central region of 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 January temperature is approximately -10 °C. Therefore, the most expedient is the operation of the heat pump in a bivalent mode, which provides for the inclusion of an additional heat generator during periods when the air temperature drops below a certain one: -5 ° C - in the southern regions of Russia, -10 ° C - in the central ones. This makes it possible to reduce the cost of the heat pump and, especially, the installation of the primary circuit (laying trenches, drilling wells, etc.), which greatly increases with an increase in the capacity of the installation.

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

For a cottage with an area of ​​200 m 2 for 4 people with a heat loss of 70 W / m 2 (calculated at -28 ° C outside air temperature), the heat demand will be 14 kW. Add 700 W for domestic hot water to this value. As a result, the required power of the heat pump 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 heat pump power should be 17.6 kW (multiplier factor - 1.2). In the case of monovalent operation, you can choose a logafix WPS 160 L (Buderus) ground-water heat pump with a capacity of 17.1 kW, consuming 5.5 kW of electricity.

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

How to calculate the cost of heating a country house?

Calculations are made on the basis of the following parameters:

The first parameter is operating costs. To determine these costs, it is worth considering the cost of the fuel that will be used to generate heat. This item also includes maintenance costs. The most profitable in 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 its installation. The most profitable and economical at the stage of purchase and installation will be the purchase of an electric boiler. The maximum costs are expected if you decide to purchase boilers, where the energy carriers are liquefied gas in gas tanks or diesel fuel. Here, too, the HEAT PUMP is optimal.

The third parameter should be considered the convenience when using heating equipment. Solid fuel boilers in this case, it can be noted as the most demanding for attention. They require your presence and reloading of fuel, while electric and gas powered ones work independently. Because gas and electric boilers the most comfortable to use when heating country houses. And here the HEAT PUMP has an advantage. Climate control is the most comfortable feature of heat pumps.

As of today, the price situation in the Moscow Region is as follows... Connecting gas to private houses costs about 600,000 rubles. Also required design work and the corresponding approvals, which sometimes stretch for years and also cost money. Add here the cost of equipment and the relatively short period of wear (which is why gas companies offer more powerful gas boilers so that the wear-burnout of the boiler takes longer). Heating on heat pumps is already comparable to the above price, but does not require any approvals. A heat pump is an ordinary electrical 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 resource of modern heat pumps, and even more high-quality ones (premium class), allows them to work for more than 20 years.

We give examples of the calculation of heat pumps for various types and house sizes.

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

First of all, it is necessary to determine the power of the heat pump or boiler, as this is one of the decisive specifications. It is selected based on the magnitude of the heat loss of the building. The calculation of the heat balance of the house, taking into account the features of its design, should be carried out by a specialist, however, for an approximate assessment of this parameter, if the house building is designed taking into account building codes, you can use the following formula:
Q = k V ∆T
1 kWh = 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 from outside and inside the room in winter, °С;
k is the generalized heat transfer coefficient of the building;
k \u003d 3 ... 4 - building from boards;
k \u003d 2 ... 3 - brick walls in one layer;
k min-max \u003d 1 ... 2 - standard masonry (brick in two layers);

k \u003d 0.6 ... 1 - well-insulated building;

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

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

The heat loss of a brick building (k max \u003d 2) will be:
Q \u003d 2 × 300 × 50 \u003d 30000 kcal / h \u003d 30000 / 860 \u003d 35 kW
This will be the required minimum boiler power, calculated to the maximum ...

Usually 1.5 times the power margin is selected, however, factors such as constantly running ventilation of the room, open vents and doors big square glazing, etc. If it is planned to use a double-circuit boiler (space heating and hot water supply), then its capacity should be further increased by 10 - 40%. The additive depends on the amount of hot water consumption.

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

At ΔT = (Tvn - Tnar) = 20 - (-30) = 50°С;
The heat loss of a brick building (k min \u003d 1) will be:
Q \u003d 1 × 300 × 50 \u003d 15000 kcal / h \u003d 30000 / 860 \u003d 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 resource and cycling during the day ... To reduce the number of on / off cycles of the heat pump, heat storage tanks are used.

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

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

17 kW*30l=500l storage capacity!!! The calculations are approximate, here a large battery is good, but in practice they put 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;
Approximate power we calculated -17kW. Different manufacturers have a different range of capacities, so choose a heat pump according to quality and cost together with our consultants. For example, Waterkotte has a 18kW heat pump, but you can also supply 15kW, since in case of insufficient power there is a 6kW peak closer in each heat pump. Peak reheating takes a relatively short time and therefore there is no need to overpay for a heat pump. Therefore, you can also choose 15 kW, because in the short term 15+6=21kW is higher than your heat needs.

Let's stop at 18kW. Specify the cost of a heat pump with consultants, since today the delivery conditions are "to put it mildly" unpredictable. Therefore, the site presents the factory.

If you are in the southern regions, then the heat loss of your house based on the above calculations will be less, since ΔT \u003d (Tvn - Tnar) \u003d 20 - (-10) \u003d 30 ° С. and then ΔT \u003d (Tvn - Tnar) \u003d 20 - (-0) \u003d 20 ° С. You can choose a heat pump with a smaller power and, moreover, according to the principle of operation "air-water". Our air source heat pumps operate efficiently down to -25 degrees and therefore no drilling is required.

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

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

We take the power of our heat pump -18kW. The electrical consumption of such a ground source heat pump is approximately 18/4=4.5 kWh from the socket. 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, being converted into thermal power. We receive the missing power from a geothermal source, i.e. 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 60W per 1 rpm, 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 works cost an average of 1200 rubles / m.p. We get 270 * 1200 \u003d 324000 rubles. turnkey with input to the heat point.

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

The cost of 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.