Heat pump power for heating a house. Do-it-yourself heat pump for heating a private house

Types of heat pump designs

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

The following varieties exist:

• TN “air-to-air”;
• HP "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 outdoors, and a unit with a condenser is installed inside the house. The latter is blown by a fan, due to which warm air mass is supplied to the room.

If such a system is equipped with a special heat exchanger with pipes, the result is an air-water heat exchanger. It is connected to a water heating system.

A HP evaporator of the “air-to-air” or “air-to-water” type can be placed not on the street, but in the exhaust ventilation duct (it must be forced). In this case, the efficiency of the HP will be increased several times.

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

Schematic diagram of the operation of a heat pump

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

Connecting a VT according to the “water-water” scheme is much less labor-intensive than “ground-water”, since there is no need for excavation work. The pipe is laid in a spiral at the bottom of the reservoir. Of course, only a reservoir that does not freeze to the bottom in winter is suitable for this scheme.

The time has come to substantively study foreign experience

Almost everyone now knows about heat pumps capable of extracting heat from the environment to heat buildings, and if until recently a potential customer, as a rule, asked the perplexed question “how is this possible?”, Now the question “how is this correct” 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 turn to the experience of specialists from those countries where heat pumps 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 about 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 transferring American methods to domestic soil is not an easy task. For Americans, everything is not the same as in Europe. Only they measure time in the same units as we do. All other units of measurement are purely American, or rather British. Americans are especially unlucky with heat flow, 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 adopted in the United States, to which one can get used to over time, but the absence in the mentioned book of a clear methodological basis for constructing a calculation algorithm. Too much space is devoted to routine and well-known calculation methods, while some important provisions remain completely undisclosed.

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

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

? = ?(t 2I,t 2 K). (1)

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 not difficult 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 coolant with a supply temperature of 50°C, then the maximum possible conversion coefficient of the heat pump will be about 3.5. In this case, the glycol temperature 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 using underfloor heating, coolant with a temperature of 35°C will flow from the heat pump condenser into the heating system. In this case, the heat pump can operate more efficiently, for example with a conversion factor of 4.3, if the temperature of the glycol cooled in the evaporator is around -2°C.

Using Excel spreadsheets, you can express 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 coefficient and a given value of the coolant temperature in a 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 presented as:

You can select the coolant temperature in the heating system at 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 identified these dependencies, we can now move directly to the American experience.

Calculation method for heat pumps

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

Heat loss through the building envelope (walls, ceilings, windows, doors) is determined. This can be done by applying the following relationship:

Qok = S*(tin – tout)* (1 + Σ β) *n / Rt(W)where

tout – outside air temperature (°C);

tin – internal air temperature (°C);

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

n – coefficient indicating the influence of the environment on the characteristics of the object. For rooms in direct contact with the outside 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 – coefficient of additional heat loss, which depends on the type of structure and its geographical location β can vary from 0.05 to 0.27;

Rt – thermal resistance, determined by the following expression:

Rt = 1/ α internal + Σ (δ i / λ i) + 1/ α external (m2*°C / W), where:

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

α nar – coefficient of thermal dissipation of external surfaces of enclosing structures (W/m2*оС);

α internal – coefficient of thermal absorption of internal surfaces of enclosing structures (W/m2*оС);

— The total heat loss of the structure is calculated using the formula:

Qt.pot = Qok + Qi – Qbp, where:

Qi is the energy consumption for heating the air entering the room through natural leaks;

Qbp ​​- heat generation 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.))(kW/hour per year.) where:

tout – outside air temperature;

tout.av – arithmetic mean value of the outside air temperature for the entire heating season;

d – number of days of the heating period.

Qgv = V * 17 (kW/hour per year) where:

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

Then the total consumption of thermal energy will be determined by the formula:

Q = Qgv + Qyear (kW/hour per year.)

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

Qтн=1.1*Q, where:

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

After calculating heat pumps, you can select 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 climate control system, one can note not only its functionality, but also its high aesthetic value.

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Types of Heat Pumps

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

• Air.
• Priming.
• Water - the source can be groundwater and surface water bodies.

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

“Air-to-water” is an air-type heat pump that heats a building by drawing air from outside through an external unit. It works on the principle of an air conditioner, only in reverse, converting air energy into heat. Such a heat pump does not require large installation costs; there is no need to allocate a plot of land for it and, especially, to drill a well. However, operating efficiency at low temperatures (-25ºС) is reduced and an additional source of thermal energy is required.

The “ground-water” device is a geothermal device and extracts heat from the ground using a collector placed at a depth below 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 location, you will need to drill a well.

“Water-water” is installed where there is a pond or groundwater nearby. In the first case, the collector is laid on the bottom of the reservoir, in the second, a well or several are drilled, 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 own advantages and disadvantages; if the building is located 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 in the cold season does not fall below -25º C.

Method for calculating heat pump power

In addition to determining the optimal energy source, you will need to calculate the heat pump power required for heating. It depends on the amount of heat loss from 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 kW/h = 860 kcal/h;
• V is the volume of the house in m3 (the area is multiplied by the height of the ceilings);
• ∆T – the ratio of the minimum temperatures outside and inside the room 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 masonry in two layers k=1; for a well-insulated building k=0.6.

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

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

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

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

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

Taking into account all the above aspects, you can make the optimal 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.

Heat pumps have been used for a long time and very successfully in domestic and industrial refrigerators and air conditioners.

Today, these devices have begun to be used to perform the opposite function - heating a home during cold weather.

Let's look at how heat pumps are used to heat private homes and what you need to know in order to correctly calculate all its components.

Example of heat pump calculation

We will select a heating element for the heating system of a one-story house with a total area of ​​70 square meters. 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, it is necessary to spend 100 W of heat. Thus, to heat the entire house you will need:

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

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

Reservoir calculation

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

The collector power is determined by the formula:

Qk = W – N = 7.7 – 2.5 = 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 allocated for the collector will be 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 it will take for a person to get back his money invested in something, we mean how profitable the investment itself was. In the heating sector, 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 during use. 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, simply put, can be explained as follows:

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

And there is a more economical advanced technology - a heat pump, which 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, and also to estimate how much the annual costs of operating the selected types of equipment have decreased. And determine how many years it will take for more expensive equipment to pay for itself with the resulting savings. Ideally, several proposed design solutions are compared and the most cost-effective is selected.

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

Let's look at a specific example

• The house has 2 floors, is well insulated, with a total area of ​​150 sq. m.
• Heat distribution/heating system: circuit 1 – heated floor, circuit 2 – radiators (or fan coils).
• A gas boiler for heating and hot water supply (DHW) has been installed, for example 24 kW, double-circuit.
• Split air conditioning system 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, including the installation of the boiler room, design work, connection to the gas pipeline network and installation work - 6,100 euros.

1. The approximate cost of a Mycond heat pump with an additional fan coil system, installation work and connection to the electrical network is 6,650 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 home, apartment, office or commercial facility.

An intelligent control system with weekly or daily programming, automatic switching of seasonal settings, maintaining the temperature in the house, 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, and pumps allows for maximum energy savings.

Heat pump operation when working according to the ground-water scheme

The collector can be installed in the ground in three ways.

Horizontal option

The pipes are laid in trenches in a “snake” pattern to a depth exceeding the freezing depth of the soil (on average, from 1 to 1.5 m).

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

It should, however, be taken into account that constructing a heat exchanger manually is a rather labor-intensive process.

Vertical option

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

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

Combined option

This collector can be considered a type of horizontal, but its construction will require much less space.

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

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

Upon completion of installation work, the well is backfilled. As in the case of a horizontal heat exchanger, all the necessary work can be done with your own hands.

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

It should be taken into account that unlimited heat extraction from the ground, especially when the collector is located vertically, can lead to undesirable consequences for the geology and ecology of the site. Therefore, in the summer, it is highly desirable to operate HPs of the “soil-water” type in reverse mode - air conditioning.

A gas heating system has a lot of advantages and one of the main ones is the low cost of gas. The heating diagram for a private house with a gas boiler will tell you how to arrange heating of your home with gas. Let's consider the heating system design and replacement requirements.

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

Calculation of a horizontal heat pump collector

The efficiency of a horizontal collector depends on the temperature of the medium in which it is immersed, its thermal conductivity, and the area of ​​contact with the surface of the pipe. The calculation method is quite complex, so in most cases averaged data is used.

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

• 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 area of ​​land above the sewer is not developed, 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.

Operating principle of heat pumps

Any HP contains a working medium called 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 outlet pipe of the compressor, and the evaporator is connected to the inlet pipe. 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, and the condenser is in contact with the heating or hot water system.

Working principle of a heat pump

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

1. Ammonia, freon or other refrigerant, moving through the evaporator, is heated from the source medium, say, to a temperature of +5 degrees.
2. After passing through 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 is known, 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 also heats up.
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 exchange at the previous stage, its temperature drops below the original +5 degrees. Following the evaporator, it is heated again, then 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 every kWh of electricity spent, you can get from 3 to 5 kWh of heat.

Influence of initial data on the calculation result

Let us now use the mathematical model constructed during the calculations in order to monitor the influence of various initial data on the final result of the calculation. Let us note that calculations performed in Excel allow such analysis to be carried out very quickly.

First, let’s look at how the magnitude of the heat flow to the VGT from the soil is affected by its thermal conductivity.

A geothermal heat pump is the most economical way to heat and air condition a building. The cost of a heat pump is high, but continues to decrease as demand increases. This system is ideal for installing heated floors or heating radiators designed for low coolant temperatures. When designing it, the main thing is to choose the optimal power. In the last article we looked at self-assembly of a heat pump, however, for most people, more important information will be how to choose a heat pump, how much does it cost and what needs to be taken into account?

Heat pump power calculation

When choosing equipment, it is necessary to take into account the heat loss of the house. But this is not always possible or very expensive, and purchasing a heat pump with a large power reserve is very expensive. Therefore, it is necessary to have a backup heat source in case of severe frosts (for example, a wood boiler). This will allow you to choose a heat pump with a power one third less than that required to compensate for heat loss in the coldest weather. This equipment can operate in any of three modes:monoelectric, monovalent and bivalent . The choice of mode depends on the level of consumption.

How to calculate heat consumption depending on area

It is necessary to take measures to insulate the building and reduce heat loss to 40-80 W/m². Then for further calculation we will accept the following data.

1. A house without thermal insulation requires 120 W/m² for heating.
2. The same for a building with normal thermal insulation – 80 W/m².
3. New building with good thermal insulation - about 50 W/m².
4. House with energy-saving technologies – 40 W/m².
5. With passive energy consumption – 10 W/m².

We give an approximate calculation of a heat pump, with which you can determine how to choose a heat pump. Let’s assume that the total area of ​​all heated rooms in the house is 180 m². Thermal insulation is good and heat consumption is around 9 kW. Then the heat loss will be: 180 × 50 = 9000 W. A temporary power outage is taken into account as 3 × 2 = 6 hours, but we will not take into account 2 hours, since the building is inert. We get the final figure: 9000 W × 24 hours = 216 kW hour. Then 216 kW hour / (18 hour + 2 hour) = 10.8 kW.
Thus, to heat this house, it is necessary to install a 10.8 kW heat pump. To simplify the calculation, you need to add 20% to the heat loss value (that is, increase 9000 W by 20%). But this does not take into account the cost of heating water to meet domestic needs.

Accounting for energy consumption for water heating

To determine the full power of the pump, we add the energy consumption for heating water (up to t = 45 ˚C) at the rate of 50 liters per day per person. So for four people this would be equal to 0.35 x 4 = 1.4 kW. Hence the total power: 10.8 kW + 1.4 kW = 12.4 kW.

Dependence of power on operating mode

Thermal deposition calculation must be carried out taking into account the operating mode.

1. Monovalent The mode involves the use of this equipment without auxiliary equipment (as the only one). To determine the total heat load, you should take into account the costs of compensating for emergency power outages (maximum - for 2 hours, 3 times a day).
2. Monoenergeticmode: it uses a second heat generator, the operation of which uses the same type of energy (electricity). It is connected to the system if it is necessary to increase the temperature of the coolant. This can be done automatically (installation of a heat pump also includes the installation of temperature-monitoring sensors and control equipment) or manually. But even in severe winter conditions there are not so many cold days and the additional heat generator does not have to be activated often. But such an organization of heating allows you to save on equipment: a 30% less powerful heat pump is cheaper, but it will be enough to provide heat for 90% of the heating period.
3. With bivalent mode, the heat pump is assisted by a gas boiler or one running on liquid fuel. The process is controlled by a processor that receives information from temperature sensors. Such equipment can be installed as additional (during the reconstruction of the building) to the existing one.

Heat pump market overview

There are various types of equipment available on the market today. It is worth noting the geothermal heat pumps of the Austrian company OCHSNER : they have been improved by the manufacturer for 35 years. Well established brand Waterkotte : boilers with an external coating of this brand have the highest productivity. Among Russian equipment, one can highlight those produced under the brand name “ HENK."
To make it easier to imagine the upcoming expenses, we will indicate the cost of the main equipment and its installation.

1. Heat pump with earth probe:

• drilling work – 6 thousand euros;
• heat pump price – 6 thousand euros;
• electricity costs (per year) – 400 euros.

2. With horizontal manifold:

• the cost of the pump itself is about 6 thousand euros;
• drilling work will require 3 thousand euros;
• electricity costs – 450 euros for the heating period.

3. Air heat pump:

• pump price – 8 thousand euros;
• installation work – 500 euros;
• electricity – 600 euros.

4. Water-to-water pump:

• the pump can be purchased for 6 thousand euros;
• well drilling – 4 thousand euros;
• electricity costs (per year) – 360 euros.

These are approximate data for equipment with a power of about 6 - 8 kW. Ultimately, everything depends on many factors (installation prices, drilling depth, pump of required power, etc.) and costs can increase several times. But by choosing heating using a heat pump, the customer has the opportunity to gain independence from rising prices for traditional coolants and refuse the services of heat and power companies.

An overview of using a heat pump-based system can be seen in this video

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.

Fig. 2. Diagram of a water-to-water heat pump

There it is compressed, leading to an increase in the temperature of the gas, and further supply to the condenser. In the condenser, 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, and 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.

Figure 4. Assembly and appearance of a heat exchanger made of copper and plastic pipes

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.

Rice. 5 Basic equipment for heat pump

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.

Rice. 6 Appearance of a self-assembled heat pump

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 ability to obtain large amounts of 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 losses on 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 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 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.