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Simulation of heat sources Mathematical description of turboset Though this is not a topical question for the experts of TGSV, but the author found it necessary to include it because he devoted to it a certain time of his life. Wheelspace is divided into compartments – turbine sections with a constant flow rate. In the end of the compartment there is an output to the next compartment, steam bleedings on regenerative heaters (uncontrolled bleed), on production and central heating (controlled bleeds). In the end of a compartment there is a steam-out through seals. In terms of object-oriented simulation a compartment can be presented as an object consisting of the following ports: Port 1. Inlet steam from adjacent compartment. Port 2. Additional inlet steam into a compartment, for example, after the charging valve. Port 3. Steam outlet into adjacent compartment Port 4. Steam outlet through steals. Port 5. Steam outlet on breeding. Port 6. Mechanical power created by the steam. Compartment also contains the following channels-characteristics: o Compartment designation, o Compartment diameter, o Installation angle of blades, o Rotor speed, o Holdout flow rate of steam, o Velocity coefficient, o Degree of reaction of compartment, o Peripheral speed, o Fictitious speed, o Speed ratio, o Optimum speed ratio, o Maximal relative efficiency ratio, o Efficiency ratio, o Isotropic enthalpy, o Available enthalpy, o Mechanical efficiency. Vapour as the heating medium (port), consists of the following channels (parameters): o mass flow rate, o enthalpy, o temperature, o pressure, o dryness factor, o entropy, o specific volume, o number of area (zone) in I-S diagram (see section « Equations of water steam condition»). Value of each channel (except for channels with a text, such as name) is determined by corresponding formula, for example: Diameter of a compartment (d). At calculation it is necessary to use average value of diameter of a compartment. In reference literature we found out average diameters of the first (regulating) and the last stage. Other diameters were accepted by the results of turbine setup on nominal conditions. Angle of blade (1). At calculation angle of blade of 15° counted into the line target was used. Rotor speed (n). Standard speed is about 50 rotations/s Holdout flow rate of steam (Gо). Rated flow of steam through the compartment in adjusting mode was taken as holdout flow rate of steam. Velocity coefficient (). At calculation velocity coefficient was taken as 0,95. Degree of reaction of turbine (о). At calculations the degree of reaction of turbine was taken as 0,05. Peripheral speed (u) u=n*2*π*d/2. (10.1) Fictitious (virtual) speed (Cf) 1/2 Cf=(ho-hs) , (10.2) where ho- vapor enthalpy at compartment inlet, hs- isentropic enthalpy after steam expansion in compartment. Sped ratio (Y) Y= u/cf. (10.3) Optimum speed ratio (Yopt) 1/2 Yopt =*cos1/(2*(1-о) ). (10.4) Speed ratio (J) J= Y/ Yopt. (10.5) Maximal relative efficiency ratio of (ηoimax) For one-crown control stage [1] 1/2 ηoimax=0,83-(2E-4)*(P1/V1) /G1, (10.6) For two-crown control stage 1/2 ηoimax=0,8-(2E-4)*(P1/V1) /G1, (10.7) For other stages 1/2 ηoimax=0,925-(2E-4)*(P1/V1) /G1, (10.8) where P1- vapour pressure at compartment inlet, V1- specific volume, G1- vapour flow rate at compartment inlet. Efficiency ratio 2 3 ηoi = ηoimax*(2,1*J-1,19*J +0,09*J ), (10.9) Isentropic enthalpy (hs) hs=GetHbyPaS(P2;S1), (10.10) where GetHbyPaS – formula for enthalpy calculation by pressure and entropy (see appendix), P2- pressure at compartment outlet, S1- entropy at compartment inlet. Disposable enthalpy h=ho-hs, (10.12) where ho – vapour enthalpy at compartment inlet. Vapour pressure at compartment inlet (the same pressure will be at vapour outlet from adjacent compartment). Flugel formula [2] is used for determination of vapour pressure 2 2 2 2 1/2 P1=(P2 +(G1/G1op) *(P1op -P2op )*(T1/T1op)) , (10.13) where P1op – holdout vapour pressure at compartment inlet, P2(P2op) – vapour pressure (holdout pressure) at compartment outlet, G1 (G1op) - flow rate of vapour (holdout flow rate) at compartment inlet, T1(T1op) – vapour temperature (reference temperature) at compartment inlet. Vapour enthalpy at compartment outlet h2= hs+(1- ηoi)*h. (10.14) Dryness factor of vapour at compartment outlet X=SteamX3byPaH(P;h), (10.15) SteamX3byPaH(P;h) - formula for calculation of dryness factor by pressure and enthalpy (see appendix). Vapour entropy at compartment outlet S=GetSbyPaH(P;H), (10.15) GetSbyPaH(P;h)- formula for calculation of vapour entropy by pressure and enthalpy (see appendix). Specific volume of vapour at the outlet V=GetVbyPaH(P;h), (10.16) GetVbyPaH(P;h) - formula for calculation of specific volume of vapour by pressure and enthalpy (see appendix). Area of finding a point of vapour state at compartment outlet (see section «Equations of water steam condition») Z=GetZonePaT(P;T) (10.17) GetZonePaT(P;T) - formula for calculation of vapour state area by pressure and temperature (see appendix). Mechanical capacity of i compartment transferred by turbine shaft Nmi=G1*(h1-h2)* ηmec,+Nmi-1, (10.18) h1 – vapour enthalpy at compartment inlet, h2 - vapour enthalpy at compartment outlet, ηmec - mechanical efficiency of compartment, Nmi-1- mechanical capacity transferred by turbine shaft from (i-1) compartment. Validity of simulation results Validity of simulation results can be estimated only by comparison of the results received in the program MODEN with the results of experimental investigation. But at simulation it is necessary to repeat boundary conditions of experimental works. MODEN allows to fulfill computer (computing) experiments, the results of which can be compared to full-scale experiments (see [3,4,5]). Experiment 1. Comparison of thermal characteristics of turbine PT-60/70-130 in condensation modes Fig. 10.1. Comparison of experimental results Heat pump stations Refrigerators and recently heat pumps became an integral part of modern air conditioning systems. Vapour compression refrigerators are really popular. In vapour compression refrigerators, for example, freon with various additives is used as coolant. Simulation of such systems without sources of cold (heat) does not allow making correct conclusion about economy operation modes, variation consideration of consumer loads, time and values of climatic discomfort, etc. It is clear that it is mathematically complicated to describe not only system, but also a separate element of refrigerator. We will try to present such simplified model which can be quite sufficient for the experts of air conditioning systems. Vapour compression refrigerators consist of the following basic elements (see fig. 11.2): evaporator (03), condenser (01), compressor (02) and throttle (04). Evaporator is an equivalent to heat exchanger where evaporation of liquid coolant and cooling-down of external heating medium takes place. Condenser is a heat exchanger where condensation of coolant and heat of external heating medium takes place. Difficulty in description of refrigerators becomes more complicated because in the directories and catalogues machine data, received at standard operating conditions, is given. We cite a simple way of recalculation of parameters by conditions which occur at real time-use. Fig. 10.2. Function chart of heat pump Equation of heat exchange in evaporator Qi=(U*A) ev* (T1in-T1out)/(ln((T1in-Tev)/(T1out-Tev))) , (10.19) Equation of heat exchange in condenser Qcon=(U*A) con*(T2out-T2in)/(ln((Tcon-T2in)/(Tcon-T2out))), (10.20) where Qi – heat power of evaporator, W, Qi - heat power of condenser, W, T1in- temperature of heating medium evaporator inlet, °C, T1out- temperature of heating medium at evaporator outlet, °C, T2out- temperature of heating medium at condenser outlet, °C, T2in- temperature of heating medium at condenser inlet, °C, Tcon- temperature of condensation in condenser, °C, Ti – temperature of evaporation in evaporator, °C, (U*A) con - product U*A for condenser, (U*A) ev – product U*A for evaporator. Refrigerator is tested at standard conditions. If we assume that value (U*A) ev and (U*A) con do not vary, then the system of equations for heat exchange will be as follows Qsi=(U*A) ev* (T1in-T1out)*(ln((T1in-Tev)/(T1out-Tev)), (10.21) Qscon=(U*A) con*(T2out-T2in)/Ln((Tcon-T2in)/(Tcon-T2out)), (10.22) where Qsi – standard heat power of evaporator, W, Qsi – standard heat power of condenser, W, T1in- standard temperature of heating medium at evaporator inlet, °C, T1out- standard temperature of heating medium at evaporator outlet, °C, T2out- standard temperature of heating medium at condenser outlet, °C, T2in- standard temperature of heating medium at condenser inlet, °C, Tcon- standard temperature of condensation in condenser, °C, Ti – standard temperature of evaporation in evaporator, °C. From the equations for standard conditions we can find the values of (U*A) ev and (U*A)con which then are substituted into the equations for current values. Standard rating conditions of heat pumps are prescribed, for example, by Eurovent. Some standard conditions are presented in the tables 10.1 and 10.2 Table 10.1. Standard parameters of heat pump "water-water" testing Parameters T1in T1out T2in T2out Values 12 7 40 45 Table 10.2. Standard parameters of heat pump "air-water" testing Parameters T1in φ1in T1out φ1out T2in T2out Values 12 60 7 87 40 45 Transformation coefficient Simulation has shown that transformation coefficient (COP) of heat pump does not fully depend on the external temperature. It is connected with the fact that at all periods of the year we maintain constant temperature in hydraulic mixer. Some fluctuations of COP are connected with temperature variation of water in the container of coarse filter. Average values of COP from the start of heat pump to stationary conditions are presented in the figure 10.3. As we can see from the figure, its value verges towards value 3.4. Therefore, in the following, we will consider that average transformation coefficient HP is CОР=3,4. Achievement of the maximal values of transformation coefficient is one of the main problems at the selection of heat pumps and schemes of their installation. Therefore at simulation we tried to select such operation modes of TH work where COP was maximal. It is possible to sum up requirements which lead to the growth of COP (for specified HP): Temperature decrease of water inlet and outlet condenser, Temperature increase of condensation (in condenser), Temperature increase of evaporation (in evaporator), Temperature increase of water inlet (outlet) in evaporator. Fig. 10.3. Chart of COP variation from the start of heat pump to the stationary conditions (Tout=-20°C) Operating mode of a heat pump Heat pump has three basic operating modes: Normal mode, Standby mode, Emergency switching mode. There are only two modes in the model (the second and the third are incorporated). These modes are determined by the value of number n which can take only two values 0 and 1. If HP is at normal mode, then n=1 if it does not work n=0. The diagram of operating mode of a heat pump is presented in the figure 10.4. It is clear that operating mode of a heat pump depends on the external temperature. The higher external temperature, the more time HP will be working. So, at Тн =-20°C HP does not pass through shutdown condition, since its power is not enough for water heating in hydraulic mixer to the upper limit (55°C). Fig. 10.4. Operating mode of a heat pumpТн =-10°C T-i diagram of a heat pump work Basis for the analysis of HP work is T-i (T-s) work diagram. Fig. 10.5. Т-I diagram of heat pump Evaporator work of heat pump As it was said earlier if you want to raise COP it is necessary to raise temperature of water inlet (outlet) in evaporator. Temperature increase is possible only up to a certain limit (not higher than temperature of water in the container of coarse filter). The maximal approximant to this temperature causes the growth of a tube surface that should be put into the container (to raise a surface of heat exchange). The decrease of this temperature leads to the decrease of COP and to danger of break by the minimal temperature of water outlet from the evaporator. Fig. 10.6. Diagram of temperature variation of heating mediums in evaporator. Condenser work of heat pump Fig. 10.7. Diagram of temperature variation of heating mediums in condenser Water temperature in hydraulic mixer The assumption that all water volume of heating systems is concentrated in hydraulic mixer was made 3 during the calculation. This volume is known, it is 13 m . Graph of temperature variation in the mixer from start of a heat pump (Tout =-10°C) is presented in the figure 10.8. At start of HP this volume starts to warm up gradually and in about 8 hours reaches 55°C (upper limit). HP stops. But heat analysis from mixer proceeds and it cools down. Achieving 53°C (lower limit) HP activates. The temperature of water in the tank starts to grow. Fig. 10.8.Water temperature in a mixer Fig. 10.9.Temperature in all inlet ports of hydraulic mixer Horizontal ground heat exchangers From the point of view of heating experts, ground is ever-living source of thermal energy. It is possible to select geothermal heat (ground heat) only by means of heat pumps. Heat pump is a device which allows transferring heat from cold (low-temperature) source to warm (high-temperature) consumer. Heat pumps that are used for selection of ground heat are called sometimes as ground. This notion is rather conventional since the same heat pump can be used both for selection of ground heat and for selection of water and air heat. At selection of the Earth heat top layers, placed at a depth of about 100 meters from the surface, are used. From the point of view of heat exchange this layer is under influence of radiant energy of the Sun, radiogenic heat from deep layers of the Earth, convection heat exchange with atmospheric air and heat transfer at the expense of various mass transfer processes (rain, snow melting, ground water, etc.). There are various classifications of the ground in foreign literature. Most of all we are interested in classification of the ground by heat conduction. Data of famous American directory ASHRAE [6] is given in the table 10.3. Table 10.3 Ground classification by [6] Ground mode λ, Ground type W/(m*°C) Very low heat <1 Light clay (15% moisture conductivity content) Low heat conductivity <1,5 Heavy clay (5% moisture content) Normal heat <2 Heavy clay (15% moisture conductivity content) Light sand (15% moisture content) High moisture content <2,5 Heavy sand (5% moisture content) Very high heat >2.5 Heavy sand (15% moisture conductivity content) We found in Russian sources tables 3 SNiP 2.02.04-88 [7] on the basis of which it is possible to create a table 10.4 by determination of heat conduction of a thawed ground - λth. . Table 10.4. Ground classification by [7] Ground mode λ, Ground type W/(m*°C) Very low heat <1 Peaty soil and peat conductivity Low heat conductivity <1,5 Loamy soil and clay, dust loamy sand, light dust loamy sand Normal heat conductivity <2 Heavy dust loamy sand Light sand High moisture content <2,5 Heavy sand (5% moisture content) Very high heat >2.5 Heavy sand (15% moisture content) conductivity From comparison of the tables 10.3 and 10.4 is clear that data of American and Russian directories is rather adequate. For exact determination of heat conduction of the grounds, it is necessary to conduct experimental researches of heat conduction in a place of prospecting construction activities. Let us note that heat conduction of a ground is not a constant value within a year. It depends on humidity, aggregative state of moisture in a ground and temperatures. And most of all humidity varies at a ground freezing. Data [7] shows that heat conduction of frozen grounds λ f is λf=1.05…2.1* λth. (10.23) The author has only foreign sources (see fig.10.10 [8]) of data about ground temperature at various depths. From these data it is possible to draw a conclusion, that at a depth of more than 8 meters the temperature is practically constant within a year (only 1/20 changes on a surface). There is such notion abroad as ground temperature. The directory ASHRAE [6] offers to determine ground temperature by the temperature of ground waters in the relevant locality. If we take into account temperature of ground waters, then it ranges from 8-10°C for conditions of Belarus. Value of quantity of radiogenic heat is (for Central Europe) 0,05-0,12 W/m2 [8]. If it is unknown, then usually 0,1 Вт/m2 is taken. Fig. 10.10Temperature distribution of a ground by depth. There are two basic ways of geothermal heat selection - by means of open and closed circuits. By open circuit we mean the use of heat of ground waters that specify delivery of these waters onto the surface, the use of their heat and returning into the formation. By closed circuit we mean the use of heat of a ground by means of intermediate heat exchangers and heating mediums. In turn systems with closed circuits differ in the type of heat exchangers - horizontal (fig. 10.11а) and vertical (fig.10.11в). The device of closed circuits with vertical heat exchangers is more expensive, than that one with horizontal heat exchangers. At the same time circuits with horizontal heat exchangers occupy large territories that can be, in some cases, a rather critical condition. A. В. C. Fig. 10.11. Pipe distribution of horizontal (А), vertical (B) ground heat exchangers and variations of horizontal heat exchanger placement into the pipe trench (C) [9] Pipes of horizontal heat exchangers are placed into the pipe trenches. Placement of the pipes into the trenches is usually carried out by two basic ways: straight line and spiral pipes. In reality there are also other, sometimes rather exotic, ways, for example, pipelines are covered on top by copperplates (copper fins) - probably for improvement of heat exchange. In this article we could not take into consideration such question, heaving of heat pumps. Let us consider some important characteristics of heat pumps: heat conduction, COP – transformation coefficient, temperature of the heating medium at condenser inlet (or condensation temperature, which is above 10°C) and antifreeze temperature at evaporator outlet. For standard conditions all these values are given by the manufacturer of heat pump. We will choose well-known French firm CIAT. Data from the directory for heat pump LGN-100Z of this firm is presented in the table. Heat pump LGN-100Z uses R407c as coolant. We can easily get the value of COP from this table and it is COP=Qh/N. (10.24) Diagrams of transformation coefficient dependence on the temperature of water (antifreeze) at evaporator outlet are given in the figure 10.12. Table 10.5. Characteristics of heat pump LGN 100Z (at condensation temperature of 55°C) Water temperature at Qc, kW N, kW Qh, kW COP the output from evaporator, °C Water- -8 13,8 9 22,8 2,53 glycolic -4 16 9,1 25,1 2,76 mixture 2 20,4 9,4 29,9 3,18 Water 5 23,2 9,5 32,7 3,44 7 25 9,5 34,6 3,64 12 30 9,6 39,6 4,16 The choice of heat pump at design stage is not a simple problem. The reason is that heat pump is never selected for total peak heating load. If it happens, then capital expenditure will be so large, that payback of your decision will not ever happen. It is clear, that if you choose heat pump not for peak load, it will be necessary to find special peak terminal unit. Electric boilers are usually applied as peak terminal units. And again a question arises: « What load should be used for heat pumps? ». Fig.10.12. Characteristic chart of the heat pump - LGP. Work simulation of a heat pump together with a ground heat exchanger can be held by means of the simulation program - MODEN, developed by Energovent [10]. On the basis of numerous computer experiments and comparisons of the received data with known foreign methods, and also taking into consideration the experience of the heat pump device with ground heat exchangers on water abstraction Muhavetskiy (Brest), the recommendations on designing such systems were prepared in Energovent. In this article we will take only some data from these recommendations. For demonstration of this fact let us look at the behaviour of horizontal heat exchanger that consists of two pipes Dнар=32 mm placed in a trench 100m in length. Initial temperature of a ground will be 10°C and antifreeze temperature will be minus 10°C and we select heat of a ground (heat conduction of a ground is 3 /(m*°C) One more question. What is the power of a ground as a heat exchanger during heat pump work? For answering this question we will conduct a computer experiment. In the figure 10.13 there are results of the computer experiment executed within the limits of the program st st MODEN (version 2.1). The experiment begins on the 1 of November, and ends on the 1 March. Fig. 10.13.Results of computer experiment in MODEN Let us analyse the graph. At the first stage it is possible to select more than 4000 W, but this period lasts not long, then there is a fall up to 1500 W, and then a small growth starts again. Falling of a heat extraction is connected with a ground cooling (heat exchanger work and pulldown of temperature of external air), and the increase is caused by the increase of solar radiation and some increase of temperatures of external air. Average heat extraction for settlement period will be QCр=2232 W. Actually we have this heat extraction. At some period it can be increased, and vice verse, but average heat extraction should not exceed this figures. Let us take rather simple technique to find out the value of the required heat pump capacity that is very widespread in the USA. We determine the number of hours when the maximum of thermal capacity is used max=Qгод/Qmax. (10.25) Charge coefficient Kзагр=max /y. (10.26) Rated capacity of the heat pump in foreign literature is offered to determine by the formula QTout=2*Qmax* Kзагр. (10.27) Coefficient 2 in this formula takes into account that duration of the heating period is, approximately, 50 % of all the year. Now we will not prejudice this formula, though, undoubtedly, it requires specification. The capacity of ground heat exchanger is determined by simple formula Qгт=QTout*(COP-1)/CОР (10.28) As it was before, pipelines of horizontal heat exchanger are placed into the trenches. The number of pipes in the trench can be different (1, 2, 4, 6, 10, etc.), as well as the distance between trenches. What if we put all the pipes into one trench? Allure of such assumption is erroneous because both between close pipes, and between trenches interference can happen, i.e. imposition of temperature fields, that leads to essential decrease of a heat flow from the ground to pipes. Therefore, if we assume that the number of pipes placed into a single trench with one pipe is Lтр1, then the number of pipes in reality is Lp=Lтр1*Kтр*Kтран, (10.29) where Kтр – coefficient of correction that takes into account the number of pipes in a trench (see tab. 10.6), Kтран – coefficient of correction that takes into account the distance between trenches. Calculations show that if the distance between trenches is more than 2 m, then Kтран=1. . In general Lp=F(Kтр, Kтран, Tгр, Тж, λth, Cгр, Dнар, λCт….) . (10.30) It is not always possible to take into consideration all the factors by introduction of coefficients (Kтр and Kтран). Direct calculation for certain conditions is the most preferable. Nowadays such direct checking calculation can be executed by means of simulation programs. Making a number of such calculations, we want to show the reader of the article how separate parameters influence the value of heat extraction by heat exchanger. Influence of a ground temperature, heat conduction of a ground and antifreeze temperature (heating medium circulating through the evaporator of heat pump) can be estimated by means of the table 10.7. We can see from the table that the temperature of heating medium influences the value of heat extraction strongly then goes heat conduction of a ground and its temperature which influences this value most of all. Though this data is received by means of numerical calculation, the author does not understand exactly why the influence of thermal conductance coefficient is so insignificant? I thought that it will be a direct proportion, as it was in case of a stationary problem of heat conduction. In foreign sources that I read this dependence is not discussed at a numerical level. Table 10.6. Correction coefficient by the number of pipes in one trench - Ктр Base Numbers of tubes in a trench 2 4 6 Calculation in MODEN 1,45 1,97 2,34 [6] 1,43 1,73 2,16 Table 10.7. Heat transfer of horizontal ground heat exchanger from 2 pipes and the trench length of 100 m Тгр, °C λth, W/(m*°C) λth, W/(m*°C) λth, W/(m*°C) at Тж=-10 °C at Тж=-6 °C at Тж=-2 °C 1,5 3 1,5 3 1,5 3 8 1839 2287 1307 1561 794 833 10 1951 2391 1428 1674 897 958 Example. Calculate horizontal ground heat exchanger and choose heat pump for a building heating located in Minsk. Design load on the heating system is 80 kW, heat conduction of a ground is W/(m*°C) 1. For Minsk at heating system work at external temperature below +8°C, the value of maximum hours is 2320 (it is calculated in the program MODEN), thus charge coefficient is 0,264. We determine required capacity of heat pumps by the formula (10.27) QTout=120*0,264*2=63,3 kW. 2. So as to accept the most effective variant of installation, we make a number of calculations on various values of the number of pipes in a trench (1, 2 and 4) and antifreeze temperature (-10,-6 and-2°C). We show how to make one of the calculations: one pipe in a trench and antifreeze temperature -10°C 3. We install a heat pump of the firm CIAT and of the LGN mark. According to the graph in the fig. 10.12 for such heat pumps we choose COP that is equal 2.35. According to the formula (10.28) the capacity of a ground heat exchanger will be Qгт= 63,3*(2,35-1)/2,35=36,36 kW. 4. According to the table 10.7 there is no data on heat transfer in the trench with a single pipe. Therefore we take a similar trench, but with 2 pipes. Average value of the selected heat from the trench of 100 m for the heating period is 2391 W. On 100 m of the pipe heat removal will be 2391/2=1196 W/100m. 5. If there is only one pipe in a trench instead of two pipes, then heat removal subject to Ктр will be 1196*1,45=1730 Вт/100m. 6. Total length of pipes and trenches will be Lтр=Lтран=36360*100/1730=2098 m. 7. Flow rate of electric energy is determined by the formula Nгод= Ny*2320*2 8. The analysis of calculation results shows (calculations are not presented) that the most economical, according to the capital expenditure, are variants with antifreeze temperature of -10°C. These are the variants with a minimum number of pipes and a greater standard size of the heat pump. Such projects are more popular in the Central Europe that is connected with a lack of space. In spite of great capital expenditure, the variant with antifreeze temperature of -2°C has smaller operating costs that is connected with a smaller size of the heat pump. A great number of pipes require a lot of space. Such projects are more popular in the USA and Canada. It is impossible to say what variant has real advantages. Deviations between the expenses are within the bounds of computing experiment. Summary 1. In the article we considered major factors influencing the work of horizontal ground heat exchangers: heat conduction of a ground, interaction of pipes in a trench and interaction of tranches among themselves, antifreeze temperature in a circuit of heat exchangers. 2. In the article we presented fundamentals of design technique of heat pump systems with horizontal ground heat exchangers. This technique is made on the basis of computer experiments executed by means of the program MODEN (version 2.1). 3. There is an example of heat extraction of heat pump system using the received results. Conventions COP- coefficient of heat pump transformation, 3 Сгр – specific heat of a ground, J/(m *°C), Dнар – external diameter of pipeline of ground heat exchanger, m, Kзагр – charge coefficient of heat pump, Kтран – coefficient that takes into account distance between trenches, Kтр – coefficient that takes into account the number of pipes in a trench, Lтран – length of a trench for allocation of horizontal ground heat exchanger, m, Lтр – length of the pipelines of horizontal ground heat exchanger, m, N – electric power of heat pump drive, W, Тгр – ground temperature, °C, Тж – average antifreeze temperature (of heat exchanger passing through evaporator),°С, Qгод - total annual expenditure of thermal energy, J, Qmax - maximum heat load, W, Qh – heat output of HP, W, Qc - cooling efficiency of HP, W, Qтн – required heat output of HP, W, Qгт – heat power of ground heat exchanger, W, max - number of hours used for maximum of heat load, s (hour), год – year duration, (hour), λ – heat conduction, W/(m*°C), λth - heat conduction of thawed ground, W/(m*°C), λf - heat conduction of frozen ground, W/(m*°C), λст - heat conduction of pipe wall, W/(m*°C). Literature 1. Тепловые и атомные электрические станции. Кн. 3: Справочник. – М.: Энергоатомиздат, 1989. – 608 с. 2. Вульман Ф.А. и др. Математическое моделирование тепловых схем паротурбинных установок на ЭВМ. – М.: Машиностроение, 1985. – 112 с. 3. Типовая нормативная характеристика турбоагрегата ПТ-60-130/13 ЛМЗ. – ОРГРЭС, 1975. – 34 с. 4. Типовая нормативная характеристика турбоагрегата ПТ-60-130/13 ЛМЗ. – ОРГРЭС, 1975. – 34 с. 5. Рубинштейн Я.М. и др. Тепловые характеристики теплофикационной турбины Т-100-130 ТМЗ. – Теплоэнергетика, 11, 1965, с. 12-20. 6. ASHRAE Handbook. 1999 HVAC Application. Chapter 31. Energy resources. 7. СНиП 2.02.04-88. Основания и фундаменты вечномерзлых грунтов. 8. Sanner B. DESCRIPTION OF GROUND SOURCE TYPES FOR THE HEAT PUMP.– www.geothermie.de/ueb_seiten/ub_sanner.htm 9. Closed Loop Ground-Coupled Heat Pumps - HPC-IFS2, January 2002. 10. Волов Г.Я. Внедрение имитационного моделирования в инженерную практику (программа МОДЭН, версия 2,0) – Энергия и менеджмент, 2001, NN2, с. 30-33.

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