How to calculate heating costs for a country house?

Calculations are made based on the following parameters:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cost TOTAL 324 thousand + 180 thousand = 504 thousand rubles

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

4.1. Operating principle of a heat pump

The use of alternative environmentally friendly energy sources can prevent the brewing energy crisis in Ukraine. Along with the search 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 realized by heat pumps (TH), which are essentially vapor-compression refrigeration machines (Fig. 4.1).

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

Heat pumps are used in various industries, residential and public sectors. 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 TN fleet is replenished by approximately 1 million units. Thus, in Stockholm, a heat pump 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 installed heat pump capacity in Europe was 4,531 MW, and the equivalent of 1.81 billion m 3 of natural gas was produced by heat pumps worldwide. Heat pumps using geothermal and groundwater are energy efficient. In the United States, federal legislation has approved requirements for the mandatory use of geothermal heat pumps (GHP) in the construction of new public buildings. In Sweden, 50% of all heating is provided by ground source heat pumps. By 2020, according to the World Energy Committee, the share of geothermal heat pumps will be 75%. The service life of the gas turbine unit is 25–50 years. The prospects for using heat pumps in Ukraine are shown in.

Heat pumps are divided according to the principle of operation (compressor, absorption) and 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, 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. Hydraulic heat pump diagram:

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

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

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

8 – high temperature heat exchanger

A heat pump with hydraulic piping (water pumps, heat exchangers, shut-off valves, etc.) is called a heat pump installation. 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 the 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. Air-water heat pump diagram

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 for indoor heating (Figure 4.3).

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

To supply water from heat pumps to heated rooms, heat exchangers are installed in them, called “fan coils” in the literature. Water is supplied to the fan coil units by a hydraulic system - a pumping station (Fig. 4.4).

Rice. 4.4. Pumping station diagram:

P – pressure gauges; RB – expansion tank; AB – storage tank; RP – flow switch; N – pump;

BC – balance valve; F – filter; OK – check valve; B – valve; T – thermometer;

PC – safety valve; TP – freon-liquid heat exchanger; ТХК – three-way valve; KPV – liquid replenishment valve; KPV – air make-up valve; KVV – air release valve

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

where is the cooling capacity of the HP, kW;

– volume of refrigerated premises, m 3 ;

– amount of water in the system, l;

Z – number of HP power stages.

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

To compensate for the thermal expansion of water, expansion tanks are installed in the hydraulic system. 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 – 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 only in cooling mode)

ΔT = t ambient – 4 °C; when operating in heat pump mode ΔT=60 °C – 4 °C = 56 °C);

P pre – setting the safety valve.

The pressure in the system (P syst) depends on the relative position of the pumping station and the end consumer (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-pumped with air to a pressure 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 – position of the membrane before installation (pre-pumping with air by 0.1–0.3 bar);

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

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

Expansion units are produced (Fig. 4.6) that maintain pressure on the water side in large-volume heating and air conditioning systems. The unit is equipped with a freely programmable processor and can be connected via an interface to a central control panel. This greatly simplifies control over the functioning of the system.

Specifications:

Volume, l 200–5,000;
Maximum overpressure, bar 10.0;
Maximum temperature, °C 120.

The flow switch (RP) turns off the refrigeration machine in the absence of liquid flow, which prevents freezing of the liquid in the heat exchanger (HE). A three-way valve mixes two fluid streams (A and B), maintaining a given fluid temperature. The three-way valve is controlled by a microcontroller.

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

The design of a three-way valve is 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. To move the shut-off cone over the entire stroke from one extreme position to the other, a control supply voltage is supplied to the electric drive in the range from 0 to 10 V. Electric motor power supply - 24 V.

Rice. 4.7. Three-way valve to regulate fluid flow

A control signal about the position of the shut-off cone is issued from the drive output. The full stroke time of the cone is 100–150 seconds. It is possible to manually move the cone using a hex key.

Liquid leakage when the channel is closed does not exceed 1% of the throughput. If the three-way valve and hydraulic system malfunction, after the three-way valve, fluid will circulate through the check valve (OK).

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

The expansion tank is connected to liquid feed valves (LPV) and air feed valves (APV). When installing the filter (F), you must pay attention to the direction of fluid flow through the filter. An automatic air release valve (VV) is installed at the highest point of the hydraulic circuit. The safety valve is adjusted to the maximum permissible pressure of the weakest element in the network plus 1 bar (7–10 bar).

If it is necessary to work according to a bivalent circuit, you can connect an electrically heated boiler in parallel with the HP according to the diagram shown in Fig. 4.8.

Rice. 4.8. Connection diagram of an electric boiler to a heat pump system

4.2.2. Heat pumps with water heat sources

Heat pumps with water heat sources (rivers, lakes, seas) use the accumulated energy of the Sun. 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 when extracting heat, water cannot be cooled below 0 °C, 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 necessary to increase the water consumption. For low-power HPs, it is not recommended to pump groundwater from a depth of more than 15 m. Otherwise, large costs will be required for pumps and their operation.

Rice. 4.9. Heat pump using groundwater heat

The heat extraction circuit from the reservoir can be open or closed. In the first case, water from the reservoir is pumped through a 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. Water intake and return must be carried out in the direction of groundwater flow to prevent “bypassing” of water. The intake line must have a check valve (4), located at the intake point or after the deep-well pump (5). The groundwater supply and drainage 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 outlet point in the return well must be below the groundwater level.

The volumetric flow rate of water is determined from the cooling capacity of the HP

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

c p – 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 = 12 kW (determined from the heat pump passport), a = 4 K, then m 3 / h.

The closed circuit is laid on the bottom of the reservoir. The approximate value of thermal power per 1 m of a closed circuit pipeline is about 30 W. That is, to produce 10 kW of heat, the circuit must have a length of 300 m. To prevent the circuit from floating, a load of about 5 kg must be installed per 1 linear meter.

4.2.3. Heat pumps with ground heat exchangers

Ground HP uses thermal energy accumulated in the soil due to its heating by the Sun or other sources. The heat accumulated in the soil is transformed using horizontally laid ground heat exchangers (also called ground collectors) or using vertically located 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 design (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 diameter of the pipe, 1.4–2.0 m of pipe can be laid for each square meter of heat intake area. The length of each horizontal collector branch should not exceed 100 m, otherwise the pressure loss in the pipe and the required pump power will be too high.

The amount of transformed heat, and, consequently, the size of the required surface for the location of a soil 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, very much depend on the composition and condition of the soil. In this regard, the determining factors are the proportion of water, the content of mineral components (quartz, feldspar), as well as the proportion and size of pores filled with air. The higher the proportion of water and mineral components and the lower the pore content, the higher the storage properties and thermal conductivity of soil.

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

Table 1. Average value of specific thermal power of soil

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

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

where is the heat output of the heat pump, W;

– power consumption of the transformer from the network, W;

g – specific power of the soil collector, W/m2.

So, if the cooling capacity of the HP is 10 kW, then in sandy wet soil (g = 20 W/m2) an area will be required to place the collector

To transform heat from such an area, it is necessary to lay polyethylene pipes with a diameter of 25 × 2.3 mm and a length of 500 × 1.4 = 700 m in the ground (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 manifolds should be located in accessible locations for inspection, such as separate distribution shafts outside the house or in a basement shaft of the house. Fittings must be made of corrosion-resistant materials. All pipelines in the house and entries through the wall must be thermally insulated to ensure diffusion impermeability to steam in order to avoid the appearance of condensation, because there is cold (relative to the basement temperature) coolant in the supply and return lines.

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


A	b

Rice. 4.11. Heat pump with ground probe

a – general diagram, b – soil probe diagram

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

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

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

Table 2. Specific internal volume of pipes

Pressure loss in pipes depends on the diameter of the pipes, density and flow rate of the coolant and is determined according to the data of the pipe manufacturer. 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. Hence 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 ground collector depends on the acidity of the soil: at normal acidity (pH = 5.0) – 50–75 years, at 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, equal to the ratio of the thermal output of the heat pump to the power consumed by the compressor. In cooling mode, the EER (energy efficiency ratio) is used to evaluate efficiency, 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 – condensation and boiling temperatures in the heat pump.

The temperature is determined by the condensation pressure 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 COP will be equal to 7.7. If heat is removed by water, then various refrigerants allow you to achieve the following temperatures: R717, R502, R22 - about +50 °C, R134a - +70 °C, R142 - +100 °C.

You should remember the basic rule that follows from (4): the smaller the temperature difference between the heat source and heat sink in the heat pump, the higher the conversion coefficient.

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

With an equipotential cycle =

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

In absorption heat pumps, the conversion coefficient is lower than in compression heat pumps due to large losses in the elements of the absorption circuit. Thus, when using groundwater with T0 = 281.16 K (8°C) and useful heat temperature = 323.16 K (50 °C), the absorption heat 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 should be at least 150 °C.

During the heating season (October–May), heating 100 m2 of living space with an electric boiler will require 37,440 kW of electricity, and with 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 Santekhnik LTD and Co LLC).

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

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

4.2. Selecting equipment for heat pumps

The choice of equipment begins with calculating the heat consumption of the building. Currently, there are a variety of programs for calculating heat consumption on a PC, which can be found on the Internet or obtained from equipment suppliers.

An approximate calculation can be made based on the heated area of the building and the amount of hot water consumed. Also, in case of periodic planned power outages, it is necessary to increase the thermal power 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/m2;
new building, good thermal insulation - 50 W/m2;
building with standard thermal insulation - 80 W/m2;
old buildings without special insulation - 120 W/m2.

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

Determine daily (24 hours) heat consumption

where is the heating capacity of the heating element, kW;

– time without electricity.

The calculation of additional thermal power for preparing hot water is based on the consumption of about 50 liters of water by one person 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 hot water consumption

Category	Water consumption, l/person		Specific heat consumption, Wh/person	Heat consumption for hot water, kW/person
Category	pace. water 60°С	pace. water 45°С	Specific heat consumption, Wh/person	Heat consumption for hot water, kW/person
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. Heating capacity 510 kW.

2. Low-temperature source – sea water with a temperature:

warm period of the year ≤20 °С,

cold period of the year 7 °C.

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

4. Minimum outdoor temperature – minus 10 °C (Crimea, Ukraine).

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

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

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

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

Using the CIAT equipment selection program, we will 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, the appearance of which is shown in Fig. 12.

Table 4. Technical characteristics of the water cooling machine HYDROCIAT 2500B X LW/LWP R134a

Parameter	Heating mode	Cooling mode
Evaporator capacity, kW	326,0	395,9
Coolant	MEG20%	MEG20%
Coolant temperature in the evaporator (input/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 (input/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

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

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

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

We select a low-temperature seawater-heat pump heat exchanger in heating mode according to the following initial data:

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

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

Table 6. Technical characteristics of the heat exchanger PWB 45 11 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 showed that in summer the performance of the PWB 30 11 heat exchanger will be 437 kW and the coolant temperatures will be (outlet/inlet) 7.5/12 ºС. (Table 7)

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

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

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

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

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

The range of some CIAT heat pumps is given in table. 8.

Table 8. Heat pumps from CIAT (France)

Heat pump type	Productivity, kW		Application area
Heat pump type	in the cold	by heat	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.

Heat pumps using renewable heat sources are the most energy efficient heating equipment.
Systems built on the basis of TN are reliable, safe and durable.
Producing heat through a heat pump is an environmentally friendly technological process.
Modern climate control equipment (for example, CIAT, France) makes it possible to create HP with a capacity of tens of kW to MW.

Literature.

W. Maake, G.-Y. Eckert, J.-L. Cochepin. Textbook on refrigeration technology: Transl. from French – M.: Moscow University Publishing House, 1998. – 1142 p., ill.
Ray D., McMichael D. Heat pumps: Transl. from English – M.: Energoizdat, 1982. – 224 p., ill.
El Sadeen Hassan. Selection of optimal parameters for a heat and cold supply system for a residential building // Refrigeration equipment, 2003, No. 3, pp. 18–21.
Ovcharenko V.A. Ovcharenko A.V. Vikoristannya of heat pumps//Holod M+T, 2006, No. 2 p. 34–36.
Five steps towards getting rid of methane addiction//Heating Water Supply Ventilation + Air Conditioners, 2006, No. 1, p. 30–41.
Bondar E.S., Kalugin P.V. Energy-saving air conditioning systems with cold accumulation//S.O.K., 2006, No. 3, p. 44–48.
Viesmann. Heat pump systems. Design instructions.5829 122-2 GUS 2/2000
Belova. Air conditioning systems with chillers and fan coils

Many owners of private houses decide to create an autonomous heating system in their home. While carrying out work to create it, they have to face a number of difficulties. Already at the very beginning they are forced to decide which energy carrier to use in the system.

If a main gas pipeline runs near the site, then in this case the choice is obvious. To supply gas to your home, you just need to submit documents for gasification, and after a while specialists will connect your home to natural gas. However, in our country, despite the high rate of gasification of regions and districts, many people do not have the opportunity to supply gas to their private home. Therefore, they have to use gas in cylinders.

What to do in such a situation? Using a conventional stove that runs on wood and coal for heating is a troublesome task. And if you install equipment that runs on electrical energy, it will be quite expensive, although in this case there will be less cold air flowing in. However there are new solutions, which recently appeared on the market. Installing equipment that uses alternative energy sources during operation is an opportunity to provide heat in the home at minimal cost. In the case of this heating option, heat is obtained from the earth, water and air.

It makes it possible to extract heat from the earth, water and air.

One of the new solutions that is available on the market is a heating system that uses a heat pump as the main working element. It is not necessary to buy this equipment if you decide to use it as part of your heating system. It is quite possible to make such a pump with your own hands. The main thing is to have a desire.

A heating system based on a heat pump includes, in addition to this equipment, devices for heat intake and distribution. If we talk about the composition of the internal circuit of such pumping equipment, we will highlight the following components:

Note that the basic principles of operation of this equipment were developed two centuries ago and known as the Carnot cycle. The heat pump works as follows:

Non-freezing liquid is used as a coolant, which is supplied to the collector. Antifreeze can be:
- water diluted with alcohol;
- brine;
- glycol mixture.
- These substances have the ability to absorb thermal energy and transport it to the pump.
Once in the evaporator, the heat is directed to the refrigerant. This substance has a low boiling point. Under the influence of thermal energy, the refrigerant boils. As a result, steam is formed.
A running compressor increases steam pressure, which causes an increase in air temperature.
Heat is transferred from water to the heating system through another element - a condenser. In order to extract additional heat, the refrigerant is cooled again, turned into a liquid, and then sent to the collector.
This process is then repeated in the same cycle.

In simple terms, a heat pump is a piece of equipment that works on almost the same principle as a refrigerator, only in reverse. If you take an ordinary refrigerator, then in it the refrigerant moving along the circuit receives heat from the food products placed for storage. At the end of the cycle, he brings it to the back wall. The same heat is used in the case of a heat pump, only it is used to heat the coolant, thanks to which air heating is provided.

A heat pump heating system, of course, consumes electrical energy. But, note that the amount required for operation is immeasurably less than for a conventional electric boiler. Thus, consuming 1 kW of electrical energy, a boiler that heats water produces 5 kW of thermal energy.

The costs that arise when purchasing this equipment and when installing a heat pump are quite high. They are higher than the costs of installing a heating boiler powered by electrical energy. Here, anyone who is thinking about creating their own autonomous heating system in the house may have a question: Is it profitable to set up such a system? In this regard, we can say the following: if the system is installed in a house with an area of 100 square meters, then the additional costs incurred for installing the equipment will pay off within 2 years. Then the owner of the home will only save on heating.

A heating system based on a heat pump has one important advantage: it can not only heat the room, but also cool the air, that is, work like an air conditioner. Therefore, in the summer, in order to get rid of unnecessary heat in the premises of the house, you can turn on a special operating mode of the heat pump.

How to calculate equipment?

When calculating the power of a heat pump, you first need to focus on the level of heat loss in your home. Naturally, before installing such a heating system in your home, it is necessary carry out thermal insulation work Houses. It is necessary to insulate not only the walls and floor, but also the roof and windows.

It is optimal if such a heating system is installed still at the building design stage. This will make it possible to create a heating system that provides the most efficient heating of the building's premises in winter.

Practical experience shows that the best option for a heating system based on a heat pump is a water heated floor. When installing it, it is necessary to take into account the type of flooring. Ceramic tiles are an ideal material for flooring. But carpets, laminate and parquet have low thermal conductivity. Therefore, when using such a system, the water temperature should be above 8 degrees.

How to make a heat pump with your own hands?

The cost of a heat pump is quite high, even if you do not take into account the payment for the services of a specialist who will install it. Not everyone has sufficient financial resources to immediately pay for the installation of such equipment. In this regard, many are beginning to wonder whether it is possible to make a heat pump with your own hands from scrap materials? It is quite possible. In addition, during the work you can use not new, but used spare parts.

So, if you decide to create a heat pump with your own hands, then before starting work you need to:

check the condition of the wiring in your home;
make sure the electric meter is working and check that the power of this device is at least 40 amperes.

The first thing you need to do is buy a compressor. You can buy it from specialized companies or by contacting a refrigeration equipment repair shop. There you can purchase an air conditioning compressor. It is quite suitable for creating a heat pump. Next, it must be mounted on the wall using L-300 brackets.

Now you can move on to the next stage - manufacturing the capacitor. To do this, you need to find a stainless steel water tank with a volume of up to 120 liters. It is cut in half, and a coil is installed inside it. You can make it yourself using a copper tube from a refrigerator. Or you can create it from a small diameter copper pipe.

In order not to experience problems with making a coil, you need to take a regular gas cylinder and wrap copper wire around it. During this work, it is necessary to pay attention to the distance between the turns, which should be the same. To ensure that the tube is fixed in this position, you should use an aluminum corner with perforation, which is used to protect the corners of the putty. Using coils, the tubes should be positioned so that the coils of wire are opposite the holes in the corner. This will ensure the same pitch of turns, and in addition, the structure will be quite strong.

When the coil is installed, the two halves of the prepared tank are connected by welding. In this case, you need to take care of welding the threaded connections.

To create an evaporator, you can use plastic water containers with a total volume of 60 - 80 liters. A coil made of pipe with a diameter of ¾ inches is mounted into it. Regular water pipes can be used to deliver and drain water.

On the wall using an L-bracket of the required size The evaporator is being secured.

When all the work is completed, all that remains is to invite a refrigeration equipment specialist. He will assemble the system, weld copper pipes and pump in freon.

DIY heat pump installation

Now that the main part of the system is ready, all that remains is to connect it to the heat intake and distribution devices. This work can be done independently. There's nothing complicated about it. The process of installing a heat intake device can be different and largely depends on the type of pump that will be used as part of the heating system.

Vertical soil water pump

Here, too, certain costs will be required, since when installing such a pump, you simply cannot do without using a drilling rig. All work begins with the creation of a well, the depth of which should be 50-150 meters. Next, the geothermal probe is lowered, after which it is connected to the pump.

Horizontal soil water pump

When installing such a pump, it is necessary to use a manifold formed by a pipe system. It should be located below the soil freezing level. The accuracy and depth of placement of the collector largely depends on the climate zone. First, the soil layer is removed. Then the pipes are laid, and then they are backfilled with earth.

You can also use another method - laying individual pipes for water in a pre-dug trench. Having decided to use it, you first need to dig trenches, the depth of which should be below the freezing level.

Conclusion

If using an electric boiler to heat your home is an expensive pleasure, then you can opt for a heating system based on a heat pump. To save money, you can make a heat pump yourself. Its design is simple. You just need to allocate a little of your time to carry out this work and purchase the necessary parts and components. Having done it, you will receive a heating system that will create a warm atmosphere at minimal cost.

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Posted on http://www.allbest.ru/

MINISTRY OF EDUCATION AND SCIENCE OF RUSSIA

Federal State Budgetary Educational Institution of Higher Professional Education

St. Petersburg State Economic University

Institute of Service for Motor Transport, Utilities and Household Appliances

Department of "Machines and equipment for household and housing and communal services"

COURSE WORK

on the topic: Calculation of heat pumps

in the discipline: "Household machines and appliances"

The work was performed by: Melnik A.O.

The work was checked by: Lepesh G.V.

St. Petersburg - 2014

1. Heat sources. Geothermal heat pumps

2. Operating principle of a heat pump

3. Five advantages of heat pumps over traditional types of heating

4. Efficiency of heat pump application

5. Comparison of current heating costs for the population as of August 2008

6. Capital costs

7. Some background information

8. Calculation examples

1. Heat sources. Geothermal heat pumps

As is known, geothermal heat pumps use free and renewable energy sources: low-grade heat of air, soil, underground, waste and waste waters of 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, sources of low-grade heat can be external air with temperatures from -15 to +15 °C, air removed from the room (15-25 °C), subsoil (4-10 °C) and groundwater (more than 10 °C) , lake and river water (0-10 °C), surface (0-10 °C) and deep (more than 20 m) soil (10 °C).

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. Air is supplied to its heat exchanger using a fan.

When groundwater is used as a heat source, it is supplied from a well using a pump into the heat exchanger of a pump operating according to the “water-to-water” scheme, and is either pumped into another well or discharged into a reservoir.

2. Operating principle of a heat pump

A heat pump, the operating principle of which is based on the Carnot cycle, is essentially a heat engine, which, unlike the traditional combustion process, allows for heat supply to an object using environmental heat or return (waste) heat from technological processes. An important factor is the extremely low energy consumption of the heat pump for its operation - by spending 1 kW of electricity, the heat pump is capable of generating 4 kW of heat. For some types of heat pumps this figure may be higher. In other words, the principle of operation of a heat pump is based on the transfer of thermal energy from a low-potential source (water, air, earth) to the consumer (coolant) due to the expenditure of energy on the transformation of the working fluid. Schematically, a heat pump can be represented by four main elements: an evaporator, a compressor, a condenser and a relief valve. Two more circuits are connected to the working circuit of the heat pump itself: the primary (external) circuit, in which the working medium (water, antifreeze or air) circulates, removing heat from the environment (earth, air, water), and the secondary circuit - water in heating and hot water systems. water supply

The operating principle of heat pumps is based on the ability of the working fluid, which is a liquid that can boil and evaporate even at sub-zero temperatures (for example, freon). The temperature of the low-potential energy source perceived by the evaporator is higher than the boiling point of freon at the corresponding pressure. Due to heat transfer, freon boils and goes into a gaseous state. Freon vapor enters the compressor, in which it is compressed. At the same time, its pressure and temperature increase. Then the hot and compressed freon is sent to the condenser, cooled by the coolant. On the cooled surfaces of the condenser, freon vapor condenses, turning into a liquid state, and its heat is transferred to the coolant, which is subsequently used in heating and hot water supply systems. Liquid freon is directed to the relief valve, passing through which it reduces the pressure and temperature and returns to the evaporator. The cycle is then completed and will automatically repeat as long as the compressor is running.

3. Fiveadvantages of heat pumps over traditional types of heating

Economical - high power factor - 1 kW of electricity is used to produce 4 kW of thermal energy, i.e. Three of the kilowatts received will cost the consumer free of charge - this is heat taken from the environment by the pump. In practice, this means annual savings in operating costs.

Versatility - with the help of a heat pump you can solve not only the problem of heating, but also cooling.

Independence from the presence of a heat source.

Exceptional durability - the only element that is subject to mechanical wear is the compressor

Fire and environmental safety - heat generation is not accompanied by a combustion process.

Heat sources for heat pumps

In heat supply systems for objects of any functional purpose, natural, continuously renewable resources of the Earth can be used as sources of low-potential thermal energy:

Atmospheric air

Surface waters and groundwater

The soil is below the freezing depth.

The following can act as artificial, technogenic sources of low-grade heat:

Exhaust ventilation air

Sewage system wastewater

Industrial discharges of process waters

Types of heat pumps

The type of heat pump is determined by the type of heat source it uses as the primary heat source. Let us recall that the primary source of heat can be either of natural origin (soil, water, air) or industrial (air removed by ventilation, process and purified wastewater).

Air-to-water heat pumps

Ambient atmospheric air is especially attractive for use as a heat source; it is available everywhere and unlimitedly. Air source heat pumps require neither horizontal collectors nor vertical probes. The compact outdoor unit effectively removes heat from the air and fits seamlessly into any interior. Air-to-water heat pumps can operate all year round, both in winter and summer. However, at temperatures below -15C, the heating system must be supplemented with a second heating device, for example, a gas or solid fuel boiler. The advantage is reduced investment costs compared to other types of heat pumps due to the absence of auxiliary excavation work, simplicity of design for use for both heating and cooling purposes. The disadvantage is the temperature limit of the primary heat source. Power factor - 1.5-2.

Water-to-water heat pumps

Groundwater is a good accumulator of solar thermal energy. Even in winter, they maintain a constant positive temperature (for example, for the North-West region this indicator is at +5+7°C). However, in our opinion, heat pumps operating on the heat of waste and process water have the best application prospects. The continuous water flow and its high temperature level guarantee a constantly high power factor. For industrial enterprises, investing in a heat transfer unit immediately from the moment of launch will provide savings on heating costs and reduce dependence on centralized heating networks. In this case, the heat discharged into wastewater is, in fact, a source of additional income, which would not be possible without the use of a heat pump. The advantage is stability. Disadvantage - for stable operation a constant flow of water of satisfactory quality is required. Power factor - 4-6.

Ground-to-water heat pumps

The thermal energy of the Sun is perceived by the soil either directly in the form of radiation, or indirectly in the form of heat received from rain or from the air. The heat accumulated in the soil is collected either by vertical soil probes or by horizontally laid soil collectors. This type of pump is also called geothermal heat pump. The advantage is stability of operation and the highest heat removal among all types of heat pumps. The disadvantage is the relatively high cost of drilling work in the case of a geothermal heat pump and the large area for placing horizontal ground collectors (with a heat demand of about 10 kW and dry clay soil, the collector area must be at least 450 sq. m). Power factor 3-5.

geothermal heat pump heating

4 . Heat pump efficiency

It is possible to reduce the total gas consumption by more than half, or, if alternative sources of electricity are available, to abandon it altogether, then for specific objects currently a lot depends on the tariff policy of the state, location, thermal insulation properties of the object, etc.

5 . Comparison of current heating costs for the population as of August 2008

Tariffs: 1000 cubic meters gas -- 300 US dollars

1 kWh electricity -- 0.1 US dollars

For a conventional cast iron floor-standing boiler with efficiency = 0.82 out of 1000 cubic meters. gas we get:

1000 * 9.1 kWh. m.cub. * 0.82 = 7462 kWh. heat

For a state-of-the-art condensing boiler with efficiency = 1.05 - 9555 kWh. heat.

To obtain the same amount of heat using a medium-efficient universal heat pump, you need in the first case:

7462 / 4.5 = 1658 kWh. electricity costing $166.

in the second:

9555 / 4.5 = 2123 kWh, costing $212.

Cost reduction compared to the cost of gas ($300), respectively:

(300 - 166) / 300 -- 45%

(300 - 212) / 300 -- 29%

USA (Vermont)

1000 cubic meters -- $350

1 kWh electricity -- $0.12

Savings 27-43%.

Belarus

1000 cubic meters -- 141,600 rub. = $66

1 kWh electricity -- 74.7 rub. = $0.0349

This is if we use time-differentiated tariffs approved in 2007 in many countries, i.e. turn off the HP during periods of maximum load on the power system from 8.00 to 11.00 and from 19.00 to 22.00, which is feasible using heat accumulators. Savings compared to a conventional gas boiler - only up to 12%. But that's today. The situation when gas is sold at $200-230 cannot last long. Probably something similar will be introduced in Moldova.

6 . Capital expenditures

The cost of the heat pump itself is significantly higher than the cost of a gas boiler, which, however, will not greatly change the overall estimate for the new construction of a decent cottage. Prices are practically comparable if it is necessary to build 200--300 m of gas pipeline. If you are not building a temporary plywood house, but a permanent structure for children and grandchildren, it will be unsightly to leave them a legacy of dependence on the pressure in the gas pipe. Surely there will always be electricity in the country. But problems with gas may arise in the near future. The famous monopolist Gazprom, which has tens of billions of dollars in debt, is not doing well and is rapidly increasing gas prices not only for its closest allies, but also for domestic Russian consumers. There is simply no money to explore and develop new fields, or to patch pipelines built during the USSR. Especially when his main income from gas exports to Europe through Ukraine quietly floats away in an unknown direction through the Swiss founders of the exporting company UkrGazenergo and no one in Moldova cares. We don’t have any other suppliers and don’t expect to.

7 . Some background information

Reference data.

1. Natural gas price forecast:

2. Approximate dependence of the required heat output of a heat pump on the area of a house with good thermal insulation properties:

In each specific case, an individual calculation is made for the heat loss of the building. To reduce capital costs, HPs are often used in bivalent mode. In parallel with it, an additional peak heater is installed, or during reconstruction, it is left on any type of fuel, which is put into operation on the coldest days, of which we don’t have many. According to the Hydrometeorological Center, the average temperature in Molodova for January is 4.8°C, for the period December - February - 4.0°C. In the coldest year in the entire history of observations (2006), it was - 8.6 ... - 5.7 ° C during the same periods.

With this connection, the HP can either be turned off if it becomes ineffective (for example, “air-to-water” at high negative outdoor temperatures), or work

If the source is a reservoir, a loop of metal-plastic or plastic pipe is laid at its bottom. A glycol solution (antifreeze) circulates through the pipeline, which transfers heat to freon through the heat exchanger of the heat pump.

There are two options for obtaining low-grade heat from the ground: 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 surface soil is 50-70 kWh/m2 per year. According to foreign companies, the service life of trenches and wells is more than 100 years.

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. As a rough guide, it can be assumed 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 coolant temperature in the forward and return lines of the loop in calculations is usually taken to be 3 °C. No buildings should be erected in the area above the collector so that the heat of the earth is replenished by solar radiation.

The minimum distance between 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 coolant. In the calculations, it should be taken into account that its heat capacity at a temperature of 0 °C is 3.7 kJ/(kg K), density is 1.05 g/cm3. When using antifreeze, 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 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 taken to be 3 K, and 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 freon 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).

Example of heat pump calculation

Initial conditions: heat demand of a cottage with an area of 120-240 m2 (depending on thermal insulation) - 12 kW; the water temperature in the heating system should be 35 °C; minimum coolant temperature - 0 °C. To heat the building, a heat pump with a power of 14.5 kW (the nearest larger standard size) was selected, which consumes 3.22 kW of freon to heat. Heat removal from the surface layer of soil (dry clay) q is equal to 20 W/m. In accordance with the formulas shown above, we calculate:

1) the 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. 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 consumption of glycol solution Vs = 11.28 3600/ (1.05 3.7 3) = 3.51 m3/h, flow rate per circuit is 0.58 m3/h.

To install the collector, we select a high-density polyethylene (HDPE) pipe of size 32. The pressure loss in it will be 45 Pa/m; 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 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 filled 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 soil temperature 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 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 loops of metal-plastic pipe of standard size 26×3; 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% 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), an expansion tank is required in the primary circuit of the heat pump installation.

It is also recommended to install a storage tank on the return line: the heat pump compressor operates in an “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 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 a mixing valve, the drive of which is controlled by a heat pump or a general automation system.

In the event 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 - toff), where toff 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 Moldova, 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, the most appropriate is to operate the heat pump in bivalent mode, which involves turning on an additional heat generator during periods when the air temperature drops below a certain level: -5 °C - in the southern regions of Moldova, -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 Moldova, for a rough estimate when selecting a heat pump operating in bivalent mode, you can focus on the 70/30 ratio: 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 m2 for 4 people with a heat loss 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 ALTAL GWHP19 with a power of 19 kW, consuming 5.3 kW of electricity or a newer, higher conversion coefficient, heat pump with a multi-compressor system, GWHP16C (Copeland compressors, Carel controller, improved new generation heat exchangers, redundancy system, soft start, etc.).

In the case of using a bivalent system with an additional electric heater and a set temperature of -10 ° C, taking into account the need to obtain 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.

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.

8. Primeryfor calculation

So, having learned enough information to select a heat pump, all we have to do is calculate the minimum heating power required for our specific room.

A lot depends:

What heat sources can be used (sewage, exhaust, well...)?

The flow rate and depth of the water surface of the well, if there is one on the site?

Is the site located on the shore of a body of water?

What is the geology of the soil on the site (meaning: sand, clay, peat...)?

Levels of groundwater and groundwater in the area?

What are the heat losses at home?

Calculation of required thermal power

Accepted notations.

V - Volume of the heated room (width, length, height) - Mi

T - Difference between the outdoor air temperature and the required indoor temperature - °C

K - Dissipation coefficient (depending on the type of structure and insulation of the room)

K = 3.0 - 4.0 - Simplified wooden or corrugated sheet metal structure. No thermal insulation.

K = 2.0 - 2.9 - Simplified building design, single brickwork, simplified window and roof design. Little thermal insulation.

K = 1.0 - 1.9 - Standard construction, double brickwork, few windows, standard roof. Average thermal insulation.

K = 0.6 - 0.9 - Improved construction, brick walls with double thermal insulation, a small number of windows with double frames, thick floor base, roofing made of high-quality thermal insulation material. High thermal insulation.

Example of thermal power calculation

V = width 4 m, length 12 m, height 3 m = Volume of the heated room = 144 m. (V=144)

T = Outdoor temperature -5° C, + required indoor temperature +18° C, = difference between indoor and outdoor temperatures 23° C. (T = 23)

K - This coefficient depends on the type of construction and insulation of the room (see above)

Required thermal power

Now you can start choosing a heat pump model

Note. The units of measurement of power (performance) used in climate control equipment are interconnected by the following ratios:

Table of heating power required for various rooms

Thermal power kW	Volume of premises in the new building	Volume of premises in an old building	Greenhouse area made of insulated glass and double foil	Greenhouse area made of ordinary glass with foil
TEMPERATURE DIFFERENCE 30°C




	1050 - 1300 m
	1350 - 1600 m

	2100 - 2500 m	1400 - 1650 m
	2600 - 3300 m	1700 - 2200 m
	3400 - 4100 m	2300 - 2700 m
	4200 - 5000 m	2800 - 3300 m
	5000 - 6500 m	3400 - 4400 m

conclusions

1) Disadvantages: Versatility - using a heat pump you can solve not only the problem of heating, but also cooling.

2) Independence from the presence of a heat source.

3) Exceptional durability - the only element that is subject to mechanical wear is the compressor

4) Fire and environmental safety - heat generation is not accompanied by a combustion process.

5)Low payback period. Approximately 3-5 years.

6) Energy is the main source of heat. The most important thing is that it will definitely not end soon.

Flaws:

1) High cost of initial costs.

Posted on Allbest.ru

Heat pump. House heating design

To understand its operating principle, you can look at a regular household refrigerator or air conditioner.

Modern heat pumps are used 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” heat from the refrigerator compartment and throws it onto the radiator. The air conditioner “pumps” heat from the air in the room and throws it onto the radiator, but located outside. At the same time, to the heat “sucked” from the room, more heat is added, into which the electrical energy consumed by the electric motor of the air conditioner has been converted.

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

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

Although the temperature of these sources is much lower than the air temperature in the heated house, the heat pump converts this low-temperature heat from the soil or water 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 rivers are too overheated by industrial and domestic wastewater, cooling the river is very useful for the life of living organisms and fish in it. The lower the water temperature, the more oxygen needed for fish can dissolve in it. In warm water, fish suffocate, and in cold water they are blissful. Therefore, heat pumps are very promising in saving the environment from " thermal pollution".

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

It is also worth remembering that heat pumps, like conventional refrigerators, use a compressor that compresses the working fluid - ammonia or freon. Heat pumps work better with freon, but freon is already prohibited 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.

Still, it seems to me that the future belongs to heat pumps. But no one is producing them en masse yet. Why? Not hard to guess.

If an alternative source of cheap energy appears, then where to put the extracted gas, oil and coal, and who to sell it to. And what should we write off multi-billion dollar losses from explosions in mines and mines?

Schematic diagram of heating a house using a heat pump

Operating principle of a heat pump

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

If atmospheric or ventilation air is used as a heat source, the heating system operates according to the air-water scheme. The pump can be located indoors or outdoors. Air is supplied to its heat exchanger using a fan.

If groundwater is used as a heat source, the system operates according to the “water-to-water” scheme. Water is supplied from the well using a pump to the pump’s 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 body of water acts as a source of energy, a loop of metal-plastic or plastic pipe is placed at its bottom. A solution of glycol (antifreeze) or antifreeze circulates through the pipeline, which transfers heat to freon through the heat exchanger of the heat pump.

When using soil as a heat source, the system operates according to the “ground-water” scheme. There are two possible options for the collector design - vertical and horizontal.

When the collector is located horizontally, metal-plastic pipes are laid in trenches 1.2-1.5 m deep or in the form of spirals in trenches 2-4 m deep. 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 - “warm floor” heating system; 5 - hot water supply circuit.

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

When the collector is located vertically, the pipes are laid in vertical wells to a depth of 20-100 m.

Vertical probe diagram

Photo of the probe in the bay

Installing a probe into a well

Calculation of a horizontal heat pump collector

Calculation of a horizontal heat pump collector.

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

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

A coolant temperature difference appears between the direct and return loops of the collector.

Usually for calculation it is taken equal to 3°C. The disadvantage of this scheme is that it is not advisable to erect buildings on the area above the collector so that the heat of the earth is replenished due to solar radiation. The optimal distance between 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.

Example of heat pump calculation

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 consume 1 kW. Thermal energy per 10 m2 area. With a house area of 10x10m = 100 m2, 10 kW of thermal energy is required.

When using heated floors, the coolant temperature in the system should be 35°C, and the minimum coolant temperature should be 0°C.

Table 1. Thermia Villa heat pump data.

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

Let's calculate the power of the thermal collector:

Qo=Qwp-P, where

Qo- thermal collector power, kW;

Qwp- heat pump power, kW;

P- compressor electrical power, kW.

The required thermal power 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 the specific (from 1 m of running pipe) heat removal, kW/m.

L=10.6/0.025 = 424 m.

To organize such a collector, you will need 5 circuits each 100 m long. Based on this, we will determine the required area of the site for laying the circuit.

A=Lхda, where da is the distance between pipes (laying pitch), m.

With a laying step of 0.75 m, the required area of the site will be:

A=500x0.75=375 m2.

Vertical collector calculation

When choosing a vertical collector, wells are drilled to 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 and filled with cement mortar. Specific heat removal of such a collector is 50 W/m.

For more accurate calculations, use the following data:

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 flows 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 pipe length L should be 212 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 loops of metal-plastic pipe in total - 6 loops of 150 m each.

Heat pump operation when operating according to the “Ground-water” scheme

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

Refrigerant under pressure is 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 in the evaporator drops sharply, it changes from a liquid to a gas at a low temperature.

After the evaporator, the refrigerant gas enters the compressor and is compressed by the compressor. At the same time, it heats up and its pressure increases. The hot refrigerant enters the condenser, where heat exchange occurs between it and the coolant from the return pipeline. Giving up its heat, the refrigerant cools and turns into a liquid state. The coolant enters the heating system and, when cooled again, transfers its heat to the room. When the refrigerant passes through pressure reducing valve,its pressure drops and it goes back into the liquid phase. After this, 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 (in this case, the heat pump does not heat, but cools the coolant - 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 working in reverse. The refrigerator pumps heat from the cooled volume into the surrounding air. If you place a refrigerator outside, then by extracting heat from the outside air and transferring it inside the house, you can, to some extent, heat the room in this simple way.

However, as practice shows, a heat pump alone is not enough to supply a home with heat and hot water. I dare to propose what, in my opinion, is the optimal heating and hot water supply scheme for the house.

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 - solar system circulation unit; 7 - heating radiator; 8 - hot water supply circuit; 9 - “warm floor” heating system.

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

After all, the probability of failure of both the heat generator, the heat pump, and the solar collector at the same time is negligible. The diagram shows two options for heating rooms: radiators (7) and “warm floor” (9). This does not mean that you need to use both options, but only illustrates the possibility of using both one and the other.

Operating principle of the heating circuit

The heat generator (1) supplies heated water to the boiler (3) and a circuit consisting of heating radiators (7). The boiler also receives heated coolant from the heat pump (4) and the solar collector (2). Part of the water heated by the heat pump is also supplied to the input of the heat generator. 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 heated and accumulated water 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 at different latitudes. After all, the solar collector will be most efficient in the summer and, of course, in sunny weather. In our latitudes, there is no need to heat living quarters in the summer, so the heat generator can be turned off altogether. And since our summers are quite hot and we can hardly imagine our life without air conditioning, the heat pump is supposed to be turned on in cooling mode. Naturally, the pipeline going from the heat pump to the boiler will be blocked. Thus, the problem of hot water supply is supposed to be solved only with the help of a solar system. And only if the solar system cannot cope with this task, use a heat generator.

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

Costs for a vertical collector:

Heat pump 6000 €;
Drilling work 6000 €;
Operating costs (electricity): about 400 € per year.

For a horizontal collector:

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

Large expenses will include the purchase of pipes and the payment of workers.

Installation of a flat-plate solar collector (for example, 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 circuit based on a heat generator, debug it, and gradually add new elements to it that will allow us to increase the efficiency of the installation.

Let's assemble the heating system according to the following diagram:

Scheme of heating a house using a heat generator

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

As a result, we received the simplest scheme for heating a house. I shared my thoughts in order to encourage proactive people to develop alternative energy sources. If anyone has any ideas or objections to what is 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 circuit provides only one heating source, it allows for the possibility of further adding additional heating devices. To do this, it is assumed that an indirect heating boiler is used with the possibility of adding or removing heat exchangers.

The use of heating radiators available in the diagram shown in the figure one above is not intended. As you know, the “warm floor” system more effectively copes with the task of heating rooms and allows you to save energy.

Please note: Prices are valid for 2009.