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The choice of cooling system is of great importance. It determines the safety and shrinkage of cargo, energy consumption per unit of transported products, transportation safety, efficient use of cargo volume, etc.
Let's consider the main requirements that the ship's hold cooling system must meet:
Provide a uniform (homogeneous) temperature field at any point in the hold with minimal deviations from the optimal values for a given cargo;
Have a large storage capacity (inertia) in order to slow down the temperature increase in the hold during a temporary stop of the refrigeration machine;
Ensure the smallest possible temperature difference between the temperature of the cargo and the boiling point of the refrigerant. This will make it possible to obtain, at a given temperature of the chamber, the maximum value of the coefficient of performance of the machine and the lowest energy consumption for the transportation of goods.
Cooling devices and coolant sewage systems should be small in weight and dimensions. It is necessary to know that the small dimensions of the cooling surfaces can only be achieved by increasing the values of the heat transfer coefficients.
Ensure reliability, simplicity and convenience in operation, safety for people and fuzes, normal monitoring of the cooling regime, ease of its regulation, revision, repair, etc.
For provision chambers of a dry-cargo vessel, it is more economical to use an air cooling system with direct evaporation of the refrigerant in evaporative batteries. Because refrigerant systems are less economical than direct-cooling systems: heat transfer occurs twice - from air to brine and from brine to refrigerant. Therefore, ceteris paribus, the total temperature difference between the cargo and the evaporating refrigerant increases and amounts to 11 ... 12 ° C, which worsens the economic performance of the compressor and increases its size. In addition, the cost of driving the brine pumps is increasing.
Systems with an intermediate refrigerant also have a low refrigeration efficiency of the refrigerant, which predetermines the large weight and size indicators of brine systems.
The air cooling system has become widespread in transport and industrial refrigerators, especially when using freon refrigeration machines. This system is especially preferred for refrigerators carrying breathable goods (fruits, vegetables).
The air cooling system, served by refrigerating machines on freon-R-22, in the best way provides an increase in the technical and economic indicators of industrial and transport refrigerators.
The circulation of cooled air in the chambers is provided by fans that drive the air through direct-cooling air coolers.
Significantly smaller weight and dimensions of cooling devices significantly increase the usable volume of the chambers.
Air-cooled versus battery-cooled (“silent”) cooling system has a number of advantages and disadvantages, the mutual influence of which is taken into account in the technical and economic analysis of the compared systems. Advantages of the air system: significantly lower metal consumption, greater durability, more convenient operation, increased cargo capacity, all other things being equal. All these factors reduce depreciation charges, operating costs and improve the vessel's carrying capacity. In the presence of an air system, periodically conducted defrosts of the air coolers make it possible to use the performance of the refrigeration machine more efficiently, while with “silent” cooling, the layer of frost that grows over the entire period of the voyage significantly worsens the efficiency of the cooling batteries and leads to a decrease in the coefficient of performance of the mapgin with a corresponding increase energy consumption. The disadvantages of the air system include: increased cooling capacity of the installation, associated with the need to compensate for additional heat inflows equivalent to the power of the fans, and somewhat greater shrinkage of the product associated with more intense heat and mass transfer.
Feasibility studies of air cooling systems show the advantages of these systems over the battery cooling system, and therefore the air cooling system is considered the most progressive and promising.
Fig.2. Schematic diagram of the air cooling system with direct evaporation of ship refrigeration rooms.
4. Choice of insulating materials. Calculation of the insulating structure.
The main consumer of cold in refrigerated transportation is heat penetrating into the refrigerated premises from the outside through their enclosing structures. The reduction of external heat inflows contributes to the reduction of the ship's cold demand. This can be achieved by thermal insulation of enclosing surfaces. The lower the thermal conductivity of the insulating material and the greater its thickness, the less heat penetrates into the room. However, with an increase in the thickness of the insulation, the useful cargo volume of the insulated premises decreases, and the cost of the insulating material and its installation increase. On modern refrigerated ships, insulating structures reduce the volume of the hold by 15 ... 30%, which negatively affects the profitability of transportation. Therefore, materials with a low thermal conductivity coefficient are used for thermal insulation.
A number of other important requirements are imposed on insulating materials used in shipbuilding, which determine their high efficiency:
High heat-shielding properties (low coefficient of thermal conductivity λ [W/(m K)];
Low density ρ , kg / m 3;
High mechanical strength and elasticity, resisting vibration and deformation of the ship's hull;
Frost resistance (ability to resist the destruction of insulation under variable temperature loads);
Fire resistance and incombustibility;
Lack of odors and immunity to them;
Low moisture capacity and low hygroscopicity;
Minimum shrinkage of bulk insulation material;
Do not cause or contribute to corrosion of surfaces;
Do not affect people's health;
Sufficient resistance to putrefactive bacteria and fungi;
Cheapness, availability, ease of transportation, installation and operation, durability.
Existing insulating materials cannot adequately satisfy all of the above requirements at the same time. Therefore, when choosing them, they are guided by the fulfillment of only the basic requirements, depending on the purpose of the vessel, the navigation area, etc. In addition, the influence of a number of shortcomings can be eliminated or significantly reduced by the created rational insulating structure, which provides:
Protection of the insulation structure from moisture by installing a vapor-moisture protective coating and (or) installing drying layers for the day of drying the insulation during operation;
Protection of insulation against penetration of rodents by installing special metal meshes;
The continuity of the insulating layer and its thickness, contributing to the effectiveness of the heat-shielding properties of fences in a long operational period.
Materials consisting of small and closed pores have good insulating properties. In modern insulating materials, the number of closed pores contained in 1 cm 3 of the material reaches several thousand. Such materials do not require additional vapor barrier measures and do not need to be dried.
The most modern representatives of highly efficient heat-insulating materials are foam plastics. Recently, many different foams have been obtained that have high resistance to moisture, high strength, and low values of density and thermal conductivity.
Therefore, as a heat-insulating material for provision chambers, we will use plates made of polyvinyl chloride resin with an inorganic gas-forming agent PVC-1, which are a porous material, the cells of which are filled with air and isolated from each other by thin walls. PVC-1 does not rot, smolders in a flame, does not cause corrosion. Plates when heated allow you to create shaped parts in relation to the set of the vessel.
Thermophysical characteristics of the insulating material:
Density - ρ \u003d 90 ... 130 kg / m 3
λ and h = 0.058 W/(m K)
The insulating structures of the refrigerated spaces of ships are divided into three main types: hulls not cut through by a steel set; overlapping a set, or normal and bypassing a set.
X 
cooling chambers are located near the galley, therefore, we will use the first type of insulating structure to isolate smooth metal surfaces. Such structures do not cut through the steel set of the ship's hull, therefore they are made of materials with thermal conductivity coefficients that differ by no more than ten times. Structures of this kind are used to isolate the second bottom, decks, bulkheads and smooth sides of refrigerated spaces (Fig. 3.)
Fig.3. Bulkhead insulating structure.
1 - metal sheathing; 2 - reinforcing wooden bars;
3 - insulating material; 4 - wooden lining of insulation.
Simple insulation structures for smooth bulkheads, decks made of materials with slightly different thermal conductivity coefficients are calculated according to the laws parallel to the heat flow.
Calculation of the insulating structure according to the method of parallel heat flows:
The main dimensions of the structure:
S= 800 mm
FROM= 60 mm
δ d= 60 mm
δ from=150 mm
Wooden lining and bars - pine along the fibers:
Density - ρ \u003d 500 kg / m 3
Coefficient of thermal conductivity - λ d= 0.4 W/(m K)
Heat capacity – c= 2.3 kJ/(kg K)
/(0.15+0.06)= 1.90W/(m K)
1/((0.15/0.058)+(0.06/)=0.37 W/(m K)
((1.90 0.06)+ 0.37(0.8-0.06))/0.8=0.48 W/(m K)
Calculation of the insulating structure by the circular flow method:
Spacing dimensions:
b=70 mm Fig.4. Normal insulation design
with longitudinal bars
the heat flow goes along the line of least resistance i.e. the greatest length of the arc of a quarter circle is equal to the height of the set profile:
(2 170)/π=0.108 m
The spacing is divided into 6 zones, the width of which is equal to:
II. 2h/π= 0.108 m
III. S-b-4h/π=(800-70-4 170/π)/1000=0.514 m
IV. H-e-a-h(1-2/π)=(300-150-60-170(1-2/π))/1000=0.028 m
V. h+e+a-H-c=(170+150+60-300-60)/1000=0.020 m
We calculate the heat flux of each zone:
m e \u003d λ from / λ d \u003d 0.058 / 0.4 \u003d 0.145 - thickness equivalent to a layer of wood 1 m thick;
I 
zone:
0,690 glad
The coefficient of thermal conductivity of the entire structure:
(0,0516+0,0425+0,1198+0,0072+0,00914+0,1311)/0,8=
Selection of a cooling system for REA of a given type. The cooling method largely determines the design of the REA, so even at an early design stage, i.e., at the stage of a technical proposal or draft design, it is necessary to select a REA cooling system. An unsuccessful solution to this problem can only be discovered at later stages of design (detailed study of the design, testing of a prototype, etc.), which can nullify the work of a large team, and the time for creating REA will increase significantly.
At the first stages of design, the designer has at his disposal a technical task (TOR), which usually contains the following very limited information:
The total power Ф of heat release in the block;
Range of possible temperature change environment
Ambient pressure limits -
Time of continuous operation of the device -
Admissible temperatures of elements -
Machine fill factor
(12.1)
Where Vi is the volume of the i-th CEA element; n is the number of elements; V is the volume occupied by REA. It is also required to set the horizontal (Li, L2) and vertical (L3) dimensions of the electronic equipment housing. These initial data are not sufficient for a detailed analysis of the CEA thermal regime, but they can be used for a preliminary assessment and selection of a cooling system. The latter is of a probabilistic nature, i.e., it makes it possible to assess the probability of providing the thermal mode of the REA specified according to the technical specifications for the selected cooling method. According to the results of processing statistical data for real structures, detailed thermal calculations and data from testing mock-ups, graphs were constructed (Fig. 12.1), characterizing the areas of appropriate application of various cooling methods. These graphs are built for the continuous operation of REA and link two main indicators: 
. First indicator 
overheating relative to the environment tc of the case of the least heat-resistant element, for which the allowable and given in the technical specification temperature has a minimum value.
Note that for free cooling, i.e., corresponds to the maximum ambient temperature according to the specification; for forced cooling, i.e., corresponds to the temperature of the air (liquid) at the inlet to the REA. The second indicator q is equal to the density of the heat flux passing through the conditional area of the heat exchange surface:
(12.2)
Figure 12.1 Appropriate areas for different cooling methods
Where F is the total power dissipated from this surface; coefficient taking into account air pressure (at atmospheric pressure, the filling factor determined by the formula (12.1).
On fig. 12.1 two types of areas are presented: in one, the use of any one cooling method can be recommended (not shaded: 1 - free air, 3 - forced air, 5 - forced evaporative); in another, it is possible to use two or three cooling methods (shaded: 2 - free and forced air, 4 - forced air and liquid, 6 - forced liquid and free evaporative, 7 - forced liquid, forced and free evaporative, 8 - free forced and free evaporative, 9-free and forced evaporative).
The upper curves in Fig. 2.1 is usually used to select the cooling of large elements - large lamps, magnets, chokes, etc. The lower curves are used to select the cooling system for blocks, racks, etc., performed on discrete microminiature elements.
If the CEA indicators fall within the shaded area (it is possible to use two or three cooling methods), then the task of choosing a cooling method becomes more complicated and more detailed calculations are required.
Let us give additional data that allow us to take into account air pressure; in formula (12.2), the latter is taken into account by the coefficient kp, which was found on the basis of calculations and experiments. With a decrease in air pressure, the temperature of the electronic equipment elements increases; let's designate the air pressure outside the unit p1 and inside - p2 for a sealed unit, the value of kp is given in the appendix (see Table A.11). The coefficient kp takes into account the deterioration of REA cooling at reduced pressure only under conditions of free air convection.
Note that the choice of a cooling system is not limited to determining the cooling area, it is also necessary to take into account the technical feasibility of implementing this method of cooling REA, i.e., mass, volume, power consumption. As experience shows, with rational design, it is possible to provide a given thermal regime of onboard REA at specific consumption air is not higher than 180-250 kg / (h * kW).
For stationary REA, where there are less stringent restrictions on dimensions, weight, power consumption, the air flow can be increased to 250-350 kg/(h-kW). For CEA cooled with air, the thermal regime has been studied most fully. In these cases, one can not only recommend one or another air cooling system, but also estimate the probability with which the selected cooling system will provide a given thermal regime.
RES heat exchangers.
A heat exchanger is a device in which the process of transferring heat from one coolant to another is carried out. Such devices are numerous and very diverse in terms of their technological purpose and design. According to the principle of operation, heat exchangers can be divided into recuperative, regenerative and mixing.
Recuperative devices are those in which heat from a hot coolant to a cold one is transferred through a wall separating them. Examples of such devices are steam generators, heaters, condensers, etc.
Regenerative devices are those in which the same heating surface is washed by either hot or cold coolant. When a hot liquid flows, heat is perceived by the walls of the apparatus and accumulates in them; when a cold liquid flows, this accumulated heat is perceived by it. An example of such devices are regenerators of open-hearth and glass-melting furnaces, air heaters of blast furnaces, etc.
In recuperative and regenerative apparatuses, the process of heat transfer is inevitably associated with the surface of a solid body. Therefore, such devices are also called surface.
In mixers, the heat transfer process occurs by direct contact and mixing of hot and cold coolants. In this case, the heat transfer proceeds simultaneously with the material exchange. An example of such heat exchangers are cooling towers (cooling towers), scrubbers, etc. The special names of heat exchangers are usually determined by their purpose, for example, steam generators, furnaces, water heaters, evaporators, superheaters, condensers, deaerators, etc. However, despite the wide variety of heat exchangers according to the type, device, principle of operation and working bodies, their purpose is ultimately the same, it is the transfer of heat from one, hot, liquid to another, cold. Therefore, the main provisions of thermal calculation for them remain common.
Heat exchangers differ in the characteristics of temperature distribution along the length of the channel:
where T 1 ’ and T 2 ’ are the temperatures at the inlet of the heat exchanger; T 1 "" and T 2 "" - at the output.
All heat exchangers are classified into two groups based on heat exchange conditions. Heat transfer from a hot coolant to a cold coolant can occur either through a solid wall or through a phase interface. Through a solid wall - a recuperative heat exchanger, through a phase boundary - a cooling tower.
The OST reference books contain the characteristics of heat exchangers manufactured by the industry for RES.
The main characteristic of heat exchangers is the specific area of the heat exchange surface:
; S beats ≈ 4500 and more.
Features of the operation of heat exchangers:
1. The mode of movement of the coolant. A turbulent regime must be implemented in the coolant. Gas - V ≈ 100 ÷ 150 m/s; liquid - V ≈ 2.5 ÷ 3 m/s. The modes that are implemented in the heat exchanger must be chosen in an optimal way.
2. Thermal design of heat exchangers is reduced to the implementation of design and verification calculations.
a) When performing a design calculation, the design of the apparatus is carried out, the purpose of the calculation is to determine the working surface area of the heat exchanger, if the mass flow rates of the hot and cold coolant, their inlet and outlet temperatures, as well as their specific heat capacities are given.
b) The verification calculation is carried out for a heat exchanger with a known surface area (for example, for a designed heat exchanger). The purpose of the calculation is to determine the temperatures of the coolant at the outlet of the heat exchanger and the flow F of heat transferred from the hot coolant to the cold one, that is, to set the operating mode of the apparatus.
Introduction
1 Selection of design parameters for outdoor and indoor air
1.1 Outside air design parameters
1.2 Design parameters of indoor air
2 Compilation of heat and humidity balances of the room
2.1 Calculation of heat gains
2.1.1 Calculation of heat gains from people
2.1.2 Calculation of heat gains from artificial lighting
2.1.3 Calculation of heat gains through external light openings
and coatings due to solar radiation
2.1.4 Calculation of heat gains through external enclosures
2.1.5 Calculation of heat gains through glazed openings due to
temperature difference between outdoor and indoor air
2.2 Moisture calculation
2.3 Determining the slope of the process beam in the room
3 Calculation of the air conditioning system
3.1 Selection and justification of the type of air conditioning systems
3.2 Selection of air distribution schemes. Definition of admissible and
operating temperature difference
3.3 Determining the capacity of air conditioning systems
3.4 Determining the amount of outside air
3.5 Mapping air conditioning processes
on Jd-diagram
3.5.1 Building an air conditioning process diagram for
warm period of the year
3.5.2 Building an air conditioning process diagram for
cold season
3.6 Determining the demand for heat and cold in systems
air conditioning
3.7 Selecting the brand of air conditioner and its layout
3.8 Calculations and selection of air conditioner elements
3.8.1 Calculation of the irrigation chamber
3.8.2 Calculation of air heaters
3.8.3 Selecting air filters
3.8.4 Calculation of the aerodynamic drag of air conditioning systems
3.9 Selecting an air conditioning fan
3.10 Selection of a pump for an irrigation chamber
3.11 Calculation and selection of the main equipment of the refrigeration system
4 UNIRS - Calculation of SCR on a computer
Appendix A - Jd-diagram. Warm period of the year
Appendix B -Jd-diagram. Cold period of the year
Annex D - Refrigeration scheme
Annex D - Specification
Annex E - Plan at the mark - 2.000
INTRODUCTION
Air conditioning is the automated maintenance of all or individual parameters air (temperature, relative humidity, purity and speed of air movement) in order to provide optimal conditions most favorable for the well-being of people, maintaining technological process, ensuring the preservation of cultural values.
Air conditioning is divided into three classes:
1. To ensure the meteorological conditions required for the technological process with allowable deviations outside the design parameters of the outside air. An average of 100 hours per year with 24-hour work or 70 hours per year with one-shift daytime work.
2. To ensure optimal, sanitary or technological standards with allowable deviations on average 250 hours per year with round-the-clock work or 125 hours per year with one-shift work in the daytime.
3. To maintain acceptable parameters if they cannot be provided with ventilation, an average of 450 hours per year for round-the-clock work or 315 hours per year for one-shift operation during the day.
Regulatory documents set the optimal and permissible air parameters.
Optimal air parameters ensure the preservation of normative and functional thermal state body, a sense of thermal comfort and prerequisites for high level performance.
Permissible air parameters are a combination of them in which there is no damage or a violation of the state of health, but uncomfortable heat sensations, deterioration in well-being and a decrease in efficiency can be observed.
Permissible conditions, as a rule, apply in buildings equipped with only a ventilation system.
Optimal conditions are provided by controlled air conditioning systems (SCR). Thus, SLE is used to create and maintain optimal conditions and clean indoor air all year round.
The purpose of this course work is to consolidate theoretical knowledge and acquire practical calculation skills, as well as the design of air conditioning systems (ACS).
In this term paper the air-conditioned room is the auditorium of the city club for 500 seats in the city of Odessa. The height of this room is 6.3 m, the floor area is 289 m 2, the attic floor area is 289 m 2, the volume of the room is 1820.7 m 3.
1 SELECTION OF DESIGN PARAMETERS FOR OUTDOOR AND INDOOR AIR
Estimated parameters of outdoor air.
The design parameters of the outdoor air are selected depending on the geographical location of the object.
Table 1 - Estimated parameters of outdoor air.
Estimated parameters of indoor air.
The design parameters of indoor air are selected depending on the purpose of the room and the time of year.
Table 2 - Calculated parameters of indoor air.
2 PREPARATION OF HEAT AND HUMIDITY BALANCES OF THE PREMISES
The purpose of compiling the heat and humidity balances of the room is to determine the heat and moisture surpluses in the room, as well as the angular coefficient of the process beam, which is used in the graphic-analytical method for calculating the SCR.
Heat and moisture balances are compiled separately for the warm and cold periods of the year.
The sources of heat emissions in the room can be people, artificial lighting, solar radiation, food, equipment, as well as heat gains through internal and external fences or through glazed openings due to the temperature difference between the external and internal air.
2.1 Calculation of heat gains
2.1.1 Calculation of heat gains from people
Heat dissipation in the room from people Q floor, W, is determined by the formula
Q floor = q floor n,(1)
where q floor is the amount of total heat generated by one person, W;
n is the number of people, pers.
Q rev = q rev n,(2)
where q av is the amount of sensible heat generated by one person, W;
n is the number of people, pers.
For the cold season
Q floor \u003d 120 285 \u003d 34200 W
Q av \u003d 90 285 \u003d 25650 W
For the warm period
Q floor \u003d 80 285 \u003d 22800 W
Q av \u003d 78 285 \u003d 22230 W
2.1.2 Calculation of heat gains from artificial lighting
Heat input from artificial lighting Q osv, W, is determined by the formula
Q sv \u003d q sv E F, (3)
where E - illumination, lx;
F - floor area of the room, m 2;
q sv - specific heat release, W / (m 2 lx).
Q osv \u003d 0.067 400 289 \u003d 7745.2 W
2.1.3 Calculation of heat gain due to solar radiation
Solar radiation Q p = 9400 W.
2.1.4 Calculation of heat gains through external enclosures
Heat gains through external fences, W, are determined by the formula
Q limit \u003d k st F st (t n - t c) + k cock F cb (t n - t c), (4)
where k i is the heat transfer coefficient through the fences, W / (m 2 K);
F i - surface area of the fence, m 2;
t n, t in - the temperature of the external and internal air, respectively, ° С.
Q limit \u003d 0.26 289 (26.6-22) \u003d 345.6 W
2.1.5 Calculation of heat gains through glazed openings
The calculation of heat gains into the room through glazed openings due to the temperature difference between the outdoor and indoor air is determined by the formula
Q r.p. = [(t n - t c) / R o ]F total, (5)
where R o is the thermal resistance of glazed openings, (m 2 K) / W, which is determined by the formula
R o = 1/k windows (6)
F total - the total area of glazed openings, m 2.
Q o.p = 0 W, since there are no glazed openings.
Table 3 - Heat balance of the premises in different periods of the year
2.2 Moisture calculation
Moisture enters the room from evaporation from the surface of the skin of people and from their breathing, from the free surface of the liquid, from the wet surfaces of materials and products, as well as from the drying of materials, chemical reactions, and the operation of technological equipment.
Moisture release from people W l, kg / h, depending on their state (rest, type of work they perform) and ambient temperature, is determined by the formula
W l \u003d w l n 10 -3, (7)
where w l - moisture release by one person, g / h;
n is the number of people, pers.
W l cold \u003d 40 285 10 -3 \u003d 11.4 kg / h
W l heat \u003d 44 285 10 -3 \u003d 12.54 kg / h
2.3 Determining the slope of the process beam in the room
Based on the calculation of heat and moisture balances, the angular coefficient of the beam of the process in the room is determined for warm ε t and cold ε x periods of the year, kJ / kg
ε t = (ΣQ t 3.6)/W t,(8)
ε x = (ΣQ x 3.6)/W x.(9)
Numerical values ε t and ε x characterize the tangent of the beam inclination angle of the process in the room.
ε t \u003d (40290.8 3.6) / 12.54 \u003d 11567
ε x \u003d (41945.2 3.6) / 11.4 \u003d 13246
3 CALCULATION OF THE AIR CONDITIONING SYSTEM
3.1 Selection and justification of the type of air conditioning systems
The choice and justification of the type of SCR is carried out on the basis of an analysis of the operating conditions of the air-conditioned object specified in the design task.
Based on the number of rooms, single or multi-zone air conditioning systems are envisaged, and then an assessment is made of the possibility of their use with exhaust air recirculation, which allows to reduce the consumption of heat and cold.
SCR with first and second recirculation is usually used for rooms that do not require high accuracy in temperature and relative humidity control.
The final decision on the choice of the concept of air treatment is made after determining the performance of the SCR and the flow rate of outdoor air.
3.2 Selection of air distribution schemes. Determination of permissible and operating temperature difference.
In terms of hygienic indicators and the uniformity of the distribution of parameters in the working area, for most air-conditioned rooms, the most acceptable supply of supply air with an inclination into the working area at a level of 4 ... 6 m and with the removal of the general exchange hood in the upper zone.
1. Determine the allowable temperature difference
Δt add \u003d 2 ° С.
2. Determine the supply air temperature
t p \u003d t in - Δt add (10)
t p heat \u003d 22 - 2 \u003d 20 ° С,
t p cold \u003d 20 - 2 \u003d 18 ° С.
3. Determine the temperature of the outgoing air
t y \u003d t in + grad t (H - h), (11)
where gradt is the temperature gradient along the height of the room above the working area, °С;
H is the height of the room, m;
h is the height of the working area, m.
The temperature gradient along the height of the room is determined depending on the specific excess of sensible heat in the room q I, W
q i = ΣQ / V pom = (ΣQ p -Q p + Q i) / V pom (12)
q i warm \u003d (40290.8 - 22800 + 22230) / 1820.7 \u003d 21.8 W
q I cold \u003d (41945.2 - 34200 + 25650) / 1820.7 \u003d 18.3 W
t at heat \u003d 22 + 1.2 (6.3 - 1.5) \u003d 27.76 ° С;
t at cold \u003d 20 + 0.3 (6.3 - 1.5) \u003d 21.44 ° С.
4. Determine the operating temperature difference
Δt p \u003d t y - t p (13)
Δt p heat \u003d 27.76 - 20 \u003d 7.76 ° С;
Δt p cold \u003d 21.44 - 18 \u003d 3.44 ° С.
3.3 Determining the capacity of air conditioning systems
For air conditioning systems, a distinction is made between the total capacity G, taking into account the loss of air due to leakage in supply air duct networks, kg / h, and the useful performance G p used in air-conditioned rooms, kg / h.
The useful performance of the SCR is determined by the formula
G p \u003d ΣQ t / [(J y - J p) 0.278], (14)
where ΣQ t is the total heat surplus in the room during the warm period of the year, W;
J y, J p - specific enthalpy of outgoing and supply air in the warm period of the year, kJ / kg.
G p \u003d 40290.8 / [(51 - 40)) 0.278] \u003d 13176 kg / h.
The total productivity is calculated by the formula
G = K p G p, (15)
where K p is a coefficient that takes into account the amount of losses in the air ducts.
G \u003d 1.1 13176 \u003d 14493.6 kg / h.
The volumetric productivity of air conditioning systems L, m 3 / h, is found by the formula
where ρ is the density of supply air, kg / m 3
ρ = 353/(273+t p)(17)
ρ \u003d 353 / (273 + 20) \u003d 1.2 kg / m 3;
L \u003d 14493.6 / 1.2 \u003d 12078 m 3 / h.
3.4 Determining the amount of outside air
The amount of outside air used in SCR affects the costs of heat and cold during heat and moisture treatment, as well as the consumption of electricity for dust removal. In this regard, one should always strive for a possible reduction in its quantity.
Minimum allowable amount outdoor air in air conditioning systems is determined based on the requirements:
Ensuring the required sanitary norm of air supply per person, m 3 / h
L n ΄ = l n,(18)
where l is the normalized consumption of outdoor air supplied per person, m 3 / h;
n is the number of people in the room, pers.
L n ΄ \u003d 25 285 \u003d 7125 m 3 / h;
Compensation for local exhaust and the creation of excess pressure in the room
L n ΄΄ = L mo + V pom K΄΄ , (19)
where L mo is the volume of local extract, m 3 / h;
V pom - the volume of the room, m 3;
К΄΄-frequency rate of air exchange.
L n ΄΄ \u003d 0 + 1820.7 2 \u003d 3641.4 m 3 / h.
We choose a larger value from L n ΄ and L n ΄΄ and take for further calculations L n ΄ \u003d 7125 m 3 / h.
We determine the flow rate of outdoor air according to the formula
G n = L n ρ n, (20)
where ρ n is the density of the outside air, kg / m 3.
G n \u003d 7125 1.18 \u003d 8407.5 kg / h.
We check SLE for recirculation:
14493.6 kg/h >8407.5 kg/h, the condition is met.
2. J< J н
51kJ/kg< 60 кДж/кг, условие выполняется.
3. The air must not contain toxic substances.
Note: all conditions are met, so we apply the SCR scheme with recirculation.
The accepted flow rate of outdoor L n must be at least 10% of the total amount of supply air, that is, the condition must be met
8407.5kg/h ≥ 0.1 14493.6
8407.5 kg/h ≥ 1449.36 kg/h, the condition is met.
3.5 Building a scheme of air conditioning processes on J - d diagram
3.5.1 Building a scheme of air conditioning processes for the warm period of the year
Scheme of air conditioning processes on j-d chart for the warm period of the year is given in Appendix A.
Consider the procedure for constructing a SCR scheme with the first recirculation.
a) finding on the J-d diagram the position of points H and B, characterizing the state of the outdoor and indoor air, according to the parameters that are given in tables 1 and 2;
b) carrying out through t. In the beam of the process, taking into account the magnitude of the slope ε t;
c) determining the position of other points:
T. P (that is, the state of the supply air), which lies at the intersection of the isotherm t p with the beam of the process;
T. P΄ (that is, the state of the supply air at the outlet of the second air heater VN2), for which a segment of 1 ° C is laid vertically down from the point P (the segment PP΄ characterizes the heating of the supply air in the air ducts and the fan);
T. O (that is, the state of the air at the outlet of the irrigation chamber), for which a line is drawn from t. П΄ down the line d \u003d const to the intersection with the segment φ \u003d 90% (the segment OP΄ characterizes the heating of air in the second air heater VN2) ;
T. Y (that is, the state of the air leaving the room) lying at the intersection of the isotherm t y with the process beam (the segment of the PVU characterizes the assimilation of heat and moisture by the air in the room);
T. U΄ (that is, the state of recirculation air before mixing it with outside air), for which from t. U along the line d \u003d const
set aside a segment of 0.5 ° C (the segment YU΄ characterizes the heating of the outgoing air in the fan);
T. C (i.e. the state of the air after mixing the recirculated air with the outside air).
The points U΄ and H are connected by a straight line. Segment U΄N characterizes the process of mixing recirculation and outside air. Point C is on the straight line У΄Н (at the intersection with J c).
The specific enthalpy J s, kJ/kg, point C is calculated by the formula
J c = (G n J n + G 1p J y΄)/ G, (21)
where J n - specific enthalpy of outdoor air, kJ / kg;
J c - specific enthalpy of the air formed after mixing the outside and recirculation air, kJ / kg;
G 1r - air consumption of the first recirculation, kg / h
G 1p \u003d G - G n (22)
G 1r \u003d 14493.6– 8407.5 \u003d 6086.1 kg / h
J c \u003d (8407.5 60 + 6086.1 51) / 14493.6 \u003d 56.4 kJ / kg
Points C and O are connected by a straight line. The resulting segment of CO characterizes the polytropic process of heat and moisture treatment of air in the irrigation chamber. This completes the construction of the SCR process. The parameters of the base points are entered according to the form in Table 4.
3.5.2 Building a scheme of air conditioning processes for the cold season
The scheme of air conditioning processes on the J-d diagram for the cold period of the year is given in Appendix B.
Consider the procedure for constructing a circuit with the first air recirculation on the J-d diagram.
a) finding on the J-d diagram the position of the base points B and H, characterizing the state of the outdoor and indoor air, according to the parameters that are given in table. 12;
b) carrying out through t. In the beam of the process, taking into account the magnitude of the slope ε x;
c) determination of the position of points P, U, O:
T. U, located at the intersection of the isotherm t y (for the cold period) with the beam of the process;
T. P, located at the intersection of the isoenthalpe J p with the beam of the process; the numerical value of the specific enthalpy J p of the supply air for the cold period of the year is calculated previously from the equation
J p \u003d J y - [ΣQ x / (0.278 G)], (23)
where J y is the specific enthalpy of air leaving the room during the cold season, kJ / kg;
Q x - total total heat surpluses in the room during the cold season, W;
G is the productivity of SCR in the warm period of the year, kg/h.
J p \u003d 47 - \u003d 38.6 kJ / kg
The PVU section characterizes the change in the parameters of the air in the room.
T. O (that is, the state of the air at the outlet of the irrigation chamber), located at the intersection of the line d p with the line φ \u003d 90%; segment OP characterizes air heating in the second air heater VN2;
T. C (that is, the state of the air after mixing the outside air, which has been heated in the first air heater BH1, with the air leaving the room), located at the intersection of the isoenthalpe J about with the line d c; the numerical value is calculated by the formula
d c \u003d (G n d n + G 1p d y) / G (24)
d c \u003d (8407.5 0.8 + 6086.1 10) / 14493.6 \u003d 4.7 g / kg.
T. K, characterizing the state of the air at the outlet of the first air heater VN1 and located at the intersection of d n (moisture content of the outside air) with the continuation of the straight line US.
The air parameters for the base points are entered according to the form in Table 5.
Table 5 - Air parameters at base points during the cold season
| Air parameters | |||||
temperature t, |
Specific enthalpy J, kJ/kg |
Moisture content d, g/kg | Relative humidity φ, % |
||
| P | 13,8 | 38,6 | 9,2 | 85 | |
| AT | 20 | 45 | 9,8 | 68 | |
| At | 21,44 | 47 | 10 | 62 | |
| O | 14,2 | 37 | 9,2 | 90 | |
| FROM | 25 | 37 | 4,8 | 25 | |
| H | -18 | -16,3 | 0,8 | ||
| To | 28 | 30 | 0,8 | 4 | |
3.6 Determining the demand for heat and cold in air conditioning systems
During the warm period of the year, heat consumption in the second air heater, W
Q t VH2 \u003d G (J p΄ - J o) 0.278, (25)
where J p΄ - specific enthalpy of air at the outlet of the second heater, kJ/kg;
J o - specific enthalpy of air at the inlet to the second heater, kJ/kg.
Q t VH2 \u003d 14493.6 (38 - 32.2) 0.278 \u003d 23369.5 W
Cold consumption for the implementation of the cooling and drying process, W, is determined by the formula
Q cool \u003d G (J c - J o) 0.278, (26)
where J c is the specific enthalpy of air at the inlet to the irrigation chamber, kJ/kg;
J o - specific enthalpy of air at the outlet of the irrigation chamber, kJ / kg.
Q cool \u003d 14493.6 (56.7 - 32.2) 0.278 \u003d 47216 W
The amount of moisture condensed in the air, kg/h
W K \u003d G (d c - d o) 10 -3, (27)
where d c is the moisture content of the air at the inlet to the irrigation chamber, g/kg;
d o - moisture content of air at the outlet of the irrigation chamber, g/kg.
W K \u003d 14493.6 (11.5 - 8) 10 -3 \u003d 50.7 kg / h
During the cold period of the year, the heat consumption in the first air heater, W
Q x VH1 \u003d G (J k - J n) 0.278,
where J c - specific enthalpy of air at the outlet of the first air heater, kJ / kg;
J n - specific enthalpy of air at the inlet to the first air heater, kJ/kg.
Q x VH1 \u003d 14493.6 (30- (-16.3)) 0.278 \u003d 18655.3 W
Heat consumption in the cold season in the second air heater, W
Q x BH2 \u003d G (J p - J o) 0.278, (28)
where J p - specific enthalpy of air at the outlet of the second air heater in the cold season, kJ / kg;
J o - specific enthalpy of air at the inlet to the second air heater in the cold season, kJ / kg.
Q x VH2 \u003d 14493.6 (38.6 - 37) 0.278 \u003d 6447 W
Water consumption for air humidification in the irrigation chamber (for feeding the irrigation chamber), kg/h
W P \u003d G (d o - d s) 10 -3 (29)
W P \u003d 14493.6 (9.2 - 4.8) 10 -3 \u003d 63.8 kg / h.
3.7 Selecting the brand of air conditioner and its layout
KTZZ brand air conditioners can operate in two modes of air performance:
In nominal capacity mode
In maximum performance mode
Air conditioners of the KTCZ brand are manufactured only according to the basic equipment layout schemes or with their modifications formed by completing necessary equipment, replacement of one equipment by another or exclusion certain types equipment.
The KTZZ brand air conditioner index is determined taking into account the full volumetric performance.
L 1.25 \u003d 12078 1.25 \u003d 15097.5 m 3 / h
We choose the KTCZ-20 brand air conditioner.
3.8 Calculations and selection of air conditioner elements
3.8.1 Calculation of the irrigation chamber
The calculation of OKFZ is carried out according to the method of VNIIKonditsioner.
a) warm weather
Determine the volumetric performance of SCR
L \u003d 12078m 3 / h
version 1, total number of nozzles n f = 18 pcs.
We determine the coefficient of adiabatic efficiency of the process, taking into account the characteristics of the beam of the chamber process according to the formula
E a \u003d (J 1 - J 2) / (J 1 - J pr), (30)
where J 1 , J 2 - enthalpy of air at the inlet, at the outlet of the chamber, respectively,
J pr - enthalpy of the limiting state of air on the J-d diagram,
E a \u003d (56.7 - 32.2) / (56.7 - 21) \u003d 0.686
Determine the relative air temperature difference
Θ = 0.33 s w μ (1/ Е p – 1/ Е а) (31)
Θ = 0.33 4.19 1.22 (1/ 0.42 - 1/ 0.686) = 1.586
We calculate the initial temperature of the water in the chamber
t w 1 \u003d t in pr -Θ (J 1 - J 2) / w μ, (32)
where t in pr - limit temperature air, °C.
t w 1 \u003d 6.5-1.586 (56.7 - 32.2) / 4.19 1.22 \u003d 3.32 ° С
We calculate the final temperature of the water (at the outlet of the chamber) according to the formula
t w 2 \u003d t w 1 + (J 1 - J 2) / with w μ (33)
t w 2 \u003d 1.32 + (56.7 - 32.2) / 4.19 1.22 \u003d 9.11 ° С
Determination of the flow rate of sprayed water
G w = μ G(34)
G w \u003d 1.22 14493.6 \u003d 17682.2 kg / h (~ 17.7 m 3 / h)
We calculate the water flow through the nozzle (nozzle performance)
g f = G w /n f (35)
g f \u003d 17682.2 / 42 \u003d 421 kg / h
The required water pressure in front of the nozzle is determined by the formula
ΔР f = (g f /93.4) 1/0.49 (36)
ΔР f = (421/93.4) 1/0.49 = 21.6 kPa
The stable operation of the injectors corresponds to 20 kPa ≤ ΔР f ≤ 300 kPa. The condition is met.
The flow rate of cold water from the refrigeration station is determined by the formula
G w x \u003d Q cold / c w (t w 1 - t w 2) (37)
G w x \u003d 47216 / 4.19 (9.11 - 3.32) \u003d 4935.8 kg / h (~ 4.9 m 3 / h).
b) cold period
During this period of the year, OKFZ operates in the mode of adiabatic air humidification.
We determine the coefficient of heat transfer efficiency by the formula
E a \u003d (t 1 - t 2) / (t 1 - t m1) (38)
E a \u003d (25 - 14.2) / (25 -13.1) \u003d 0.908
Irrigation coefficient is determined from the graphic dependence E a =f(μ).
Also graphically by the value of μ we find the numerical value of the coefficient
reduced enthalpy efficiency factor E p.
We calculate the flow rate of sprayed water using the formula (34)
G w \u003d 1.85 14493.6 \u003d 26813.2 kg / h (~ 26.8 m 3 / h)
We determine the performance of the nozzle according to the formula (35)
g f \u003d 26813.2 / 42 \u003d 638 kg / h
We determine the required water pressure in front of the nozzles according to the formula (36)
ΔР f = (638/93.4) 1/0.49 = 50.4 kPa
We calculate the flow rate of evaporating water in the chamber according to the formula
G w isp \u003d G (d o - d s) 10 -3 (39)
G w isp \u003d 14493.6 (9.2– 4.8) 10 -3 \u003d 63.8 kg / h
As can be seen from the calculation, the highest water flow (26.8 m 3 /h) and the highest water pressure in front of the nozzles (50.4 kPa) correspond to the cold season. These parameters are taken as calculated when selecting a pump.
3.8.2 Calculation of air heaters
Calculation of air heaters is carried out for two periods of the year: first, they calculate for the cold period, then for the warm period of the year.
Also separately calculate the air heaters of the first and second heating.
The purpose of the calculation of air heaters is to determine the required and available heat transfer surfaces and their mode of operation.
When checking the calculation, they are set by the type and number of base air heaters, based on the brand of the central air conditioner, that is, at first they accept the standard layout, and refine it by calculation.
cold period
When calculating, calculate:
The heat required to heat the air, W
Q woz \u003d 18655.3 W;
Hot water consumption, kg/h:
G w = 3.6Q woz /4.19(t w n - t w k) = 0.859Q woz / (t w n - t w k) (40)
G w \u003d 0.859 18655.3 / (150 - 70) \u003d 200.3 kg / h;
Depending on the brand of the air conditioner, the number and type of base heat exchangers are selected, for which the mass air velocity in the free section of the air heater is calculated, kg / (m 2 s):
ρv = G woz /3600 f woz,(41)
where f woz is the open area for the passage of air in the air heater, m 2
Speed of hot water movement through heat exchanger pipes, m/s
w = G w /(ρ w f w 3600), (42)
where ρ w is the density of water at its average temperature, kg/m3;
f w - cross-sectional area for the passage of water, m 2.
w \u003d 200.3 / (1000 0.00148 3600) \u003d 0.038 m / s.
We accept the speed equal to 0.1 m / s
Heat transfer coefficient, W / (m 2 K)
K = a(ρv) q w r ,(43)
where a, q, r are coefficients
The average temperature difference between the coolants:
Δt cf = (t w n + t w k) / 2 - (t n + t k) / 2 (44)
Δtav = (150 + 70)/2 - (-18 +28)/2 = 35°C
Required heat exchange area, m 2
F tr \u003d Q woz / (K Δt cf) (45)
F tr \u003d 18655.3 / (27.8 35) \u003d 19.2 m 2
[(F r - F tr)/ F tr ] 100≤15%(46)
[(36.8 - 19.2)/ 19.2] 100 = 92%
The condition is not met, we accept the VH1 air heater with a margin.
a) cold weather
Q woz \u003d 6447 W;
Hot water consumption, kg / h, according to the formula (40)
G w \u003d 0.859 6447 / (150 - 70) \u003d 69.2 kg / h;
Depending on the brand of the air conditioner, the number and type of base heat exchangers are selected, for which the mass air velocity in the open section of the air heater is calculated, kg / (m 2 s), according to the formula (41) ρv \u003d 14493.6 / 3600 2.070 \u003d 1, 94 kg / (m 2 s);
The speed of movement of hot water through the pipes of the heat exchanger, m / s, according to the formula (42)
w \u003d 69.2 / (1000 0.00148 3600) \u003d 0.013 m / s.
We accept the speed equal to 0.1 m/s.
Heat transfer coefficient, W / (m 2 K), according to the formula (43)
K \u003d 28 (1.94) 0.448 0.1 0.129 \u003d 27.8 W / (m 2 K);
The average temperature difference between the coolants, according to the formula (44)
Δtav = (150 + 70)/2 - (13.8 +14.2)/2 = 26°C
The required heat exchange area, m 2, according to the formula (45)
F tr \u003d 6447 / (27.8 26) \u003d 8.9 m 2
We check the condition by the formula (46)
[(36.8 - 8.9)/ 8.9] 100 = 313%
b) warm period
According to the above proposed formulas (40) - (46), we recalculate for the warm period
Q woz \u003d 23369.5 W;
G w \u003d 0.859 23369.5 / (70 - 30) \u003d 501.8 kg / h
ρv \u003d 14493.6 / 3600 2.070 \u003d 1.94 kg / (m 2 s);
w \u003d 501.8 / (1000 0.00148 3600) \u003d 0.094 m / s.
For further calculations, we take the speed equal to 0.1 m/s.
K \u003d 28 (1.94) 0.448 0.1 0.129 \u003d 27.88 W / (m 2 K);
Δtav = (30 + 70)/2 - (12 +19)/2 = 34.5 °С
F tr \u003d 23369.5 / (27.88 34.5) \u003d 24.3 m 2
In this case, the following condition must be met: between the available surface F p (preliminarily selected air heater) and the required surface F tr, the reserve of the heat exchange surface should not exceed 15%
[(36.8 - 24.3)/ 24.3] 100 = 51%
The condition is not met, we accept the VH2 air heater with a margin.
3.8.3 Selecting air filters
To clean the air from dust in the SLE, filters are included, the design solution of which is determined by the nature of this dust and the required air purity.
The choice of the air filter is carried out according to [2, kn.2].
Based on the available data, we select the filter FR1-3.
3.8.4 Calculation of the aerodynamic drag of air conditioning systems
The total aerodynamic drag of the SCR is found by the formula
R s = ΔR pc + ΔR f + ΔR in1 + ΔR ok + ΔR in2 + ΔR pr + ΔR in.v. , (47)
whereΔР pc is the resistance of the receiving unit, Pa
ΔР pc = Δh pc (L/L c) 1.95 (48)
(here L is the calculated volumetric productivity of the SCW, m 3 /h;
L to - volumetric performance of the air conditioner, m 3 / h;
Δh pc - block resistance at the nominal capacity of the air conditioner (Δh pc = 24 Pa), Pa);
ΔР pc \u003d 24 (12078 / 20000) 1.95 \u003d 8.98 Pa;
ΔР f – aerodynamic resistance of the filter (at the maximum dust content of the filter ΔР f = 300 Pa), Pa;
ΔР в1 – aerodynamic resistance of the first air heater, Pa;
ΔР в1 = 6.82 (ρv) 1.97 R
ΔР в1 \u003d 6.82 (1.94) 1.97 0.99 \u003d 24.9 W.
ΔР в2 – aerodynamic resistance of the second air heater, Pa
ΔР в2 \u003d 10.64 (υρ) 1.15 R, (49)
(here R is a coefficient depending on the arithmetic mean air temperature in the air heater);
ΔР в2 \u003d 10.64 (1.94) 1.15 1.01 \u003d 23.03 Pa;
ΔР ok - aerodynamic resistance of the irrigation chamber, Pa
ΔР ok \u003d 35 υ ok 2, (50)
(here υ ok is the air velocity in the irrigation chamber, m/s);
ΔР ok \u003d 35 2.5 2 \u003d 218.75 Pa;
ΔР pr - aerodynamic resistance of the connecting section, Pa
ΔР pr = Δh pr (L/L c) 2 , (51)
(here Δh pr – section resistance at nominal capacity (Δh pr = 50 Pa), Pa);
ΔР pr \u003d 50 (12078/20000) 2 \u003d 18.2 Pa;
ΔР w.v - aerodynamic resistance in air ducts and air distributors (ΔР w.v = 200 Pa), Pa.
P c \u003d 8.98 + 300 + 24.9 + 218.75 + 23.03 + 18.2 + 200 \u003d 793.86 Pa.
3.9 Selecting an air conditioning fan
The initial data for fan selection are:
Fan performance L, m 3 /h;
Nominal pressure developed by the fan P y, Pa, and specified by the formula
P y \u003d P s [(273 + t p) / 293] P n / P b, (52)
where t p is the temperature of the supply air in the warm period of the year, °С;
P n - air pressure under normal conditions (P n \u003d 101320 Pa), Pa;
P b - barometric pressure at the fan installation site, Pa.
P y \u003d 793.86 [(273 + 20) / 293] 101230 / 101000 \u003d 796 Pa.
Based on the data obtained, we select the fan V.Ts4-75 version E8.095-1.
n in = 950 rpm
N y \u003d 4 kW
3.10 Selection of a pump for an irrigation chamber
The selection of the pump is carried out taking into account the flow rate of the liquid and the required
ora. The fluid flow must correspond to the maximum volume
consumption of circulating water in the irrigation chamber, m 3 / h
L w = G w max /ρ,(53)
where G w max is the maximum mass flow rate of water in the OCF, kg/h;
ρ is the density of water entering the OCF, kg/m 3 .
L w \u003d 26813.2 / 1000 \u003d 26.8 m 3 / h
Required pump head H tr, m water. Art., determined by the formula
Н tr = 0.1Р f + ΔН, (54)
where Р f is the water pressure in front of the nozzles, kPa;
ΔH - pressure loss in pipelines, taking into account the height of the rise to the collector (for irrigation chambers ΔH = 8 m w.c.), m w.c. st..
H tr \u003d 0.1 50.4 + 8 \u003d 13.04 m of water. Art.
According to the data obtained, we select the pump and the electric motor to it.
Parameters of the selected pump:
Name: KK45/30A;
Liquid consumption 35 m 3 /h;
Total head 22.5 m w.c. Art.;
Parameters of the selected electric motor:
Type A02-42-2;
Weight 57.6 kg;
Power 3.1 kW.
3.11 Calculation and selection of the main equipment of the refrigeration system
The purpose of calculating the main equipment of the refrigeration system is:
Calculation of the required cooling capacity and selection of the type of refrigeration machine;
Finding the operating parameters of the refrigeration machine and, on their basis, performing a verification calculation of the main elements of the refrigeration unit-evaporator and condenser.
The calculation is carried out in the following sequence:
a) find the required cooling capacity of the refrigeration machine, W
Q x \u003d 1.15 Q cool, (55)
where Q cool - cold consumption, W.
Q x \u003d 1.15 47216 \u003d 59623.4 W
b) taking into account the value of Q x, we select the type of the MKT40-2-1 refrigerating machine.
c) determine the mode of operation of the refrigeration machine, for which we calculate:
Refrigerant evaporation temperature, °С
t and \u003d (t w k + t x) / 2 - (4 ... 6), (56)
where t w k is the temperature of the liquid leaving the irrigation chamber and entering the evaporator, °С;
t x is the temperature of the liquid leaving the evaporator and entering the irrigation chamber, °C.
Refrigerant condensation temperature, °С
t k \u003d t w k2 +Δt, (57)
where t w k2 is the temperature of the water leaving the condenser, °С
t w k2 =t w k1 +Δt (58)
(here t w k1 is the temperature of the water entering the condenser, ° С (Δt \u003d 4 ... 5 ° С); while t k should not exceed + 36 ° С.)
t w k1 \u003d t mn + (3 ... 4), (59)
where t mn is the outdoor air temperature according to a wet bulb in the warm period of the year, °С.
t and \u003d (3.32 + 9.11) / 2 - 4 \u003d 2.215 ° С
t mn \u003d 10.5 ° С
t w k1 \u003d 10.5 + 4 \u003d 10.9 ° С
t w k2 \u003d 10.9 + 5 \u003d 15.9 ° С
t k \u003d 15.9 + 5 \u003d 20.9 ° С
Subcooling temperature of the liquid refrigerant in front of the control valve, °С
t lane \u003d t w k1 + (1 ... 2)
t lane \u003d 10.9 + 2 \u003d 12.9 ° С
Suction temperature of refrigerant vapor into the compressor cylinder, °С
t sun \u003d t and + (15 ... 30), (60)
where t and is the evaporation temperature of the refrigerant, °С
t sun \u003d 0.715 + 25 \u003d 25.715 ° С
d) make a verification calculation of equipment, for which they calculate:
Evaporator surface according to the formula
F and \u003d Q cool /K and Δt cf.i, (61)
where K and - heat transfer coefficient of a shell-and-tube evaporator operating on freon 12 (K and = (350 ... 530) W / m 2 K);
Δt av.i - the average temperature difference between the heat carriers in the evaporator, determined by the formula
Δt cf.i = (Δt b - Δt m) / 2.3lg Δt b / Δt m (62)
Δt b \u003d Δt w 2 - t and (63)
Δt b \u003d 9.11 - 2.215 \u003d 6.895 ° С (64)
Δt m \u003d 3.32 - 2.215 \u003d 1.105 ° С
Δt av.i \u003d (6.895– 1.105) / 2.3lg6.895 / 1.105 \u003d 3.72 ° С
F and \u003d 47216 / 530 3.72 \u003d 23.8 m 2
The calculated surface F and compare with the surface of the evaporator F and `, given in technical specification refrigeration machine; in this case, the condition
F and ≤ F and `
23.8 m2< 24 м 2 – условие выполняется
The surface of the capacitor according to the formula
F k \u003d Q k / K k Δt sr.k, (65)
Q k \u003d Q x + N k.in, (66)
(here N k.in is the consumed indicator power of the compressor; with a certain margin, the indicator power can be taken equal to the compressor power consumption, W);
K k - heat transfer coefficient of a shell-and-tube condenser operating on freon 12 (K k \u003d (400 ... 650) W / m 2 K);
Δtav.k - the average temperature difference between the heat carriers in the condenser, determined by the formula, ° С
Δt cf. = (Δt b – Δt m)/2.3lg Δt b / Δt m (67)
Δt b = t k - t w k1 (68)
Δt b \u003d 20.9 - 3.32 \u003d 17.58 ° С
Δt m = t to - t w to2 (69)
Δt m \u003d 20.9 - 9.11 \u003d 11.79 ° С
Δt av.c = (17.58 - 11.79) / 2.3lg17.58 / 11.79 = 14 ° С
Q k \u003d 59623.4 + 19800 \u003d 79423.4 W
F k \u003d 79423.4 / 400 14 \u003d 14.2 m 2
The calculated surface of the capacitor F to compare with the surface of the capacitor F to `, the numerical value of which is given in the technical characteristics of the refrigeration machine, while the condition must be met
F to ≤ F to `
14.2 m 2 ≤ 16.4 m 2 - the condition is met.
Water consumption in the condenser, kg / s, is calculated by the formula
W \u003d (1,1 Q c) / c w (t w c2 - t w c1), (70)
where c w is the specific heat capacity of water (c w = 4190 J/(kg K))
W \u003d (1.1 79423.4) / 4190 (9.11 - 1.32) \u003d 2.6 kg / s.
List of sources used
1. SNiP 2.04.05-91. Heating, ventilation and air conditioning. – M.: Stroyizdat, 1991.
2. Internal sanitary devices: Ventilation and air conditioning /B.V. Barkalov, N.N. Pavlov, S.S. Amirjanov and others; Ed. N.N. Pavlova Yu.I. Schiller: In 2 books. – 4th ed., revised. and additional - M .: Stroyizdat, 1992. Book. 1, 2. Part 3.
3. Averkin A. G. Examples and tasks for the course "Air conditioning and refrigeration": Textbook. allowance. - 2nd ed., Rev. and additional - M.: DIA Publishing House, 2003.
4. Averkin A. G. Air conditioning and refrigeration: Guidelines to course work. – Penza: PISI, 1995.
