DESIGN AND FABRICATION OF PLATE FREEZER full report
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DESIGN AND FABRICATION OF PLATE FREEZER
PROJECT REPORT

ABSTRACT
Techniques of freezing vary for each application, The type of refrigeration used for preserving fruits cannot suit the need of the fish industries. Later it has been discovered that the number of viable vegetative microorganisms in food are usually greatly reduced by freezing if quick freezing is employed. The conventional freezers could not cope with this higher rate of freezing. By the development of the plate freezer, the challenge of quick freezing has been met to enhance effective preservation.
Plate type evaporators may be used in single or in banks. The plates may be manifolds for parallel flow of the refrigerant or they may be connected for series flow. Plate evaporators are especially useful used for liquid cooling installation where unusual peak load conditions are encountered periodically.
In the present work, a plate freezer consisting of two plates connected in parallel and having a cooling load of 0.335 KW is to be designed. Installing formed tubing between two metal plates, which are brazed together at the edges, forms the plate surface evaporators.
The refrigerant used is R-134a a refrigerant which is now used as a replacement for R-12.


Submitted by:
LIJUSH PAUL
MIDHUN MOHAN
RAJESH K R
SUNIL K K


CHAPTER 1
INTRODUCTION
Conventional food preservation is done by keeping the food inside chambers having evaporator coils around it. This chamber is insulated from the surroundings by a casing. The vapour compression system is the most widely used system.
Fig 1.1 Conventional Evaporator
The heat transfer takes place from the food to the freezer (chamber) surface through the air gap. As air is a bad conductor of heat, the freezing rate is low and time consuming. The freezing rate was increased by the development of the freezer. Only compactable foods can utilize this method.
The Plate freezer under consideration is a multi plate freezer. In the plate freezer there are two plates through which the refrigerant expands. The food is placed between the plates and are brought closer so that the food gets pressed to a pre-determined pressure. As the plates are in direct contact with the food, there is better heat transfer and hence the freezing rate is increased.
The plate used is copper plate, which is having a high heat transfer coefficient. The two plates are brought closer manually.
Fig1.2 Flat plate Evaporator
Compared to the conventional method of freezing this method takes only a Quarter of the time required to bring 1 kg of meat from 30 ºC to -20 ºC, the freezing rate is increased about four times. This is the most important reason why plate freezers are replacing conventional freezing equipments in the recent past.
CHAPTER 2
CARNOT CYCLE
The Carnot cycle is the model of perfection for a refrigeration cycle. However, it is not practicable to realize this perfect thermodynamic cycle with hardware. Firstly, two compression machines, one realizing isentropic compression, the other realizing isothermal compression would be required. Secondly, the small amount of expansion work available does not support the idea of incorporating a work producing expander into the system. The Carnot cycle is shown for a refrigerator. Heat is delivered isothermally at TH and received isothermally at TL.
Carnot vapour compression cycle
Fig2.1 Carnot vapour compression cycle

Fig2.2 Thermodynamic model of Heat Pump
Carnot coefficient of performance is given by.

VAPOUR COMPRESSION CYCLE
All vapour compression refrigeration systems are designed and built around these basic thermodynamic principles.
¢ Fluids absorb heat while changing from a liquid phase to vapour phase and reject heat in changing from a vapour phase to a liquid phase.
¢ The temperature at which a change in phase occurs is constant during the change, but this temperature will vary with the pressure. At one fixed pressure vaporization takes place only at fixed corresponding temperature. However, the temperatures of vaporization at a particular pressure are different for different fluids.
¢ Heat will flow from a body at higher temperature to a body at lower temperature.
¢ In Selecting metallic parts of cooling and condensing units, metals are selected which have high heat conductivity.
¢ Heat energy and other forms of energy are mutually convertible with directional relationship imposed by second by second law of thermodynamics.
In order to approach the simple Carnot cycle, it is necessary to achieve the delivery and removal of heat under virtually isothermal conditions. To this end, a volatile liquid, known as working fluid or refrigerant is chosen which will change phase at useful temperatures and pressures. It will absorb heat by evaporation and it will deliver heat by condensation.
Fig2.3 The theoretical vapour compression cycle
The figure identifies main elements of a system working on the vapour compression cycle principle. They are: compressor, condenser, evaporator and throttling valve. A reciprocating hermetic Compressor is mostly used for vapour compression in small systems while open type is used in larger systems. Heat exchangers are usually in the form of staggered tubes with closely packed wavy fins (Spline-fins or bristle are some times used.) There are basically two types of expansion devices in application today. Constant flow area restrictors are distinctly prevailing in small capacity units such as household refrigerators, window type air conditioners and central residential air conditioners and heat pumps.
PRESSURE-ENTHALPY DIAGRAM
A refrigerator thermodynamic cycle can be explained considering the process that the refrigerant undergoes in the two heat exchangers, compressor and expansion device. The most convenient diagram for such explanation and performance analysis is that of a pressure vs. enthalpy coordinate system, as shown in fig: The compressor receives low pressure and low temperature refrigerant at state 1 and compresses it to a high pressure. This compression process is associated with an increase of refrigerant temperature. At state 2, the high pressure, high temperature vapour enters the condenser. The refrigerant passing through the condenser rejects heat to the high temperature reservoir and changes to a saturated liquid at state 3. Then the refrigerant flows through the expansion device undergoing a drop in pressure and temperature. Finally, the low pressure, low temperature and low quality refrigerant at state 4 enters the evaporator, where it picks up heat from the low temperature reservoir reaching a dry saturated vapour state at the evaporator exit.
PERFORMANCE OF THE STANDARD VAPOUR COMPRESSION CYCLE
With the help pressure-enthalpy diagram, the significant quantities of the standard vapour compression can be determined. These quantities are:
¢ The work of compression
¢ The heat rejection rate
¢ The refrigeration effect
¢ The coefficient of performance
¢ The volume rate of flow /kW of refrigeration
¢ The power/kW of refrigeration
The work of compression is the change in enthalpy in process 1-2 of figure or h1-h2. This relation derives from the steady flow energy equation

where the changes in kinetic and potential energy are negligible. Because in the adiabatic compression, the heat transfer q is zero and work w equals h1-h2. The difference in enthalpy is a negative quantity indicating the work is done in the system. Even though the compressor may be of the reciprocating type, where the flow is intermittent rather than steady, process 1-2 still represents the action of the compressor. At a short distance in the pipe away from the compressor, the flow has smoothed out and approaches steady flow. Knowledge of the work of compression is important because it may be one of the main operating costs of the system.
The heat rejection is the heat transferred from the refrigerant in process 2-3 which is h3-h2. This knowledge also comes from the steady flow energy equation, in which the kinetic energy, potential energy and work terms drop out. The value of h3-h2 is negative, indicating the heat is transferred from the refrigerant. The value of the heat rejection is used in sizing the condenser and calculating the required flow quantities of the condenser cooling fluid.
The refrigeration effect is the heat transferred in process 4-1 or h1-h4. Knowledge of the magnitude of the necessary because performing this process is the ultimate purpose of the entire system.
The coefficient of performance of the standard vapour compression cycle is the ratio of the refrigeration effect to the work of compression
The volume flow rate is computed at the compressor inlet or state point 1. The volume flow rate is a rough indication of the physical size of the compressor.
The power/kW of refrigeration is the inverse of the coefficient of performance and an efficient refrigeration system has a low value power\kW of refrigeration but a high coefficient performance.
FLOW DIAGRAM OF PLATE FREEZER
Fig2.4 Flow diagram of plate freezer
In the plate freezer the evaporator plates are arranged in parallel as show in Fig: 2.5. The refrigerant after compression enters the condenser where the latent heat is rejected. The it is passed through a drier filter to remove moisture and dust particles from the liquid refrigerant. The expansion device used is capillary tube. The refrigerant enters the evaporator plates which are arranged in parallel. An accumulator is provided before the suction line to prevent any liquid refrigerant entering the compressor.
CHAPTER 3
SYSTEM DESIGN
Cooling Load Calculations
Chilling load
When a product enters a storage space at a temperature above the temperature of the space, the product will give off heat until it cools to the space temperature.
Chilling load, Q =
where W - weight of the product to be frozen
S - Specific heat above freezing
T 1 - space temperature
T 2 - initial temperature
For meat
S = 3.488 kJ/kgOC
T2 = 30°C
T1 = -5°C
W = 4kg
Chilling time = 2 hours
Chilling factor = 0.5
Q = 2x3.488x35 = 0.067 kW
2x3600x.5
Freezing Load
For meat latent heat = 252 kJ/kg
Q = 2x252 = 0.07 kW
2x3600
Cooling load below freezing
Specific heat below freezing = 1.918 kJ/ kg OC
Q = 2x1.918x20/2x3600 = 0.048 kW
Wall gain load
Volume of the space = 0.48 x 0.5 x 0.2 m3
= 0.048 m3
Total surface area = 2 x (0.48 x 0.5 + 0.48 x 0.2 + 0.5 x 0.2)
= 0.872 m2
where k- Thermal conductivity of insulation material
(PUF) = 0.033 W/mK
S - Surface area of the outer wall of the storage chamber
X - Thickness of insulating material
Temperature outside, T1 = 30°e.
Temperature inside, T2 = -23°C

Total Cooling Load = 0.067+0.07+0.0133+0.028 = 0.178 kW
DESIGN AND SELECTION OF PLATE FREEZER COMPONENTS
The four major equipments of the plate freezer, viz, the compressor, condenser, evaporator and expansion device, each have their individual characteristics which are functions of evaporator and condenser temperatures. This various design aspects of components used in the existing system are explained.
REFRIGERANT SELECTION
The objective is to present the criteria required of a refrigerant, discuss the reasons why CFCs were originally investigated as refrigerants (reasons which also make them the most promising in the search for alternatives) and finally to demonstrate that the inevitable tradeoffs among the various alternatives can be treated in a systematic way.
Requirements of a Refrigerant
The working fluid in a vapour compression refrigeration system must satisfy a number of requirements as summarized in Table 8.1. The most essential characteristic is chemical stability within the refrigeration system all the other properties would be meaningless if the material decomposed or reacted to form something else. Stability can be a double edged sword; once emitted to the atmosphere a refrigerant should not be so stable that it persists indefinitely. The ideal refrigerant would be totally stable in use but decompose within a few years in the atmosphere due to conditions (such as ultraviolet reactive chemical species) not present in the sealed system.
The next most important characteristics related to health and safety. As specified in the ASHRAE Safety Code for Mechanical Refrigeration, in residential and most commercial applications a refrigerant must be non-flammable and of a very low order of toxicity. These can be compromised in some industrial applications as evidenced by the use of hydrocarbons and ammonia. The new environmental requirement must now be added to the traditional health and safety criteria. A refrigerant should not contribute to ozone depletion, low level smog formation or greenhouse warming.
The thermodynamic and transport properties determine the performance of a refrigeration system. We will demonstrate below that the critical or boiling point temperatures and the heat capacity of the vapour are the most significant thermodynamic criteria. These two fundamental criteria account for all the desirable properties usually presented, such as high latent heat of vaporization, positive evaporator pressure, etc. By considering only the most fundamental thermodynamic criteria it is possible to establish a page link between bulk properties and molecular structure; this will yield insight into the type of molecule most likely to be a good refrigerant.
A number of other more practical criteria are also necessary or, at least, desirable. High oil solubility and high vapour dielectric strength are most important for hermetic compressors. A freezing point below the lowest expected system temperature is necessary. Finally, compatibility with common materials of construction, easy leak detection and low cost are obviously desirable.
Finding a new refrigerant is thus seen to be no small task. For a refrigerant to be used as a direct substitute in existing equipment virtually all of the above criteria must be satisfied. At most, some compromise in thermal properties could be tolerated at the expense of performance. For an alternate refrigerant for a newly designed system the situation is not as critical since equipment could be adapted for the different pressures and capacity that would accompany a change to a refrigerant with a different boiling point; heat exchangers could be adjusted for different transport properties; and the system could be constructed of different materials. Even in this case, however, it seems unlikely that compromises could be tolerated in the areas of chemical stability and health and safety.
TABLE 1
Refrigerant Criteria
Chemical:
Stable and inert
Health, Safety and Environmental:
Non-toxic
Non-flammable
Does not degrade the atmosphere
Thermal (Thermodynamic and Transport :
Critical point and boiling point temperatures appropriate for the application
Low vapour heat capacity Low viscosity
High thermal conductivity
Miscellaneous:
Satisfactory oil solubility
High dielectric strength of vapour Low freezing point
Reasonable containment materials Easy leak detection
Low cost
Refrigerant 134a (R134a)
The refrigerant used is R-134a. R-134a does not contain any chlorine or bromine atoms and therefore is widely accepted by scientists as unlikely to cause any destruction of stratospheric ozone. Furthermore, R-134a also possesses boiling point and thermodynamic properties that are very close to those of R-12. R-134a has the boiling point as that of R-12 (-29.8°C). R-134a is not a drop in placement for R-12 because the refrigerating effect is slightly different. It does not seem to be compatible with conventional lubricants or motor winding insulation. It gives higher benefit than R-12 in using in conventional air conditioning and refrigeration plants where reasonable condensing temperatures can be specified. The atmospheric lifetimes of hydrogen containing compounds are lower than that of halogenated compounds, due to the reaction of hydrogen in the molecule with compounds present in the lower atmosphere. Thus R-134a has lower life time compared to R-12 and it is the key for its environmental acceptability
These would appear to be no non-flammable and non-toxic substitute for R-12 at extreme pressure ratios apart from R -13 4a.
R-134a is presently under extensive investigation for oil-miscibility (as it has very low solubility in mineral oil lubricants) and winding cooling characteristics. ICI research workers are examining the potential of R-134a for use only as retrofit refrigerant in commercial and retail refrigeration and air-conditioning systems because of the following reasons.
(1) To enable equipment users to utilize existing refrigeration equipments of R-12 and to minimize the equipment modification to change over from R-12 to R-134a.
(2) Due to limited future availability of R-12, the top up costs of R-12 are likely to increase dramatically.
(3) Most systems of R-12 leak to a smaller or greater extent; therefore, top-up quantities of R12 will be required on a long term basis.
INSULATION
The Insulation used is polyurethane foam insulation (PUF). It is formed by mixing two chemicals (liquids) in equal proportion and is allowed to expand. The insulation got is very effective. The thermal conductivity is only about 33 mW/m K.
DESIGN OF COMPRESSOR
Fig 3.1 Compressor
The compression of suction vapour from the evaporator to the condenser pressure can be achieved by the use of compressor .It should take least power, within a wide range of operating conditions. It must run trouble free for a long time.
Design procedure
Enthalpy of refrigerant at compressor suction, h1 = 383 kJ/kg
Enthalpy of refrigerant at compressor discharge, h2 = 438 kJ / kg
Mass flow rate of refrigerant, mr = 3.3 x 10-3 kg/s
Compressor selection
Based on design, a R134a compressor of 1/3 hp was selected
SELECTED COMPRESSOR SPECIFICATIONS
Compressor model = CAJ35G (Kirloskar make)
Compressor power = 1/3 hp
Motor specifications = Power- 300W, Single Phase, 50 Hz, 230V, 2.7 A

DESIGN OF CONDENSER

Fig3.3 Condenser with fan
Condenser is a heat exchanger in which de-superheating of high temperature vapour, phase change (from vapour to in to liquid) and sub cooling of condensate occur. However, in some cases condensation may not be accomplished with de-superheating or sub cooling of condensate. It should be inexpensive, compact, strong, and light .The pressure drop should be minimum.
Design procedure
The desired temperature and pressure of condenser is 55°C and 14 bar respectively
Condenser capacity required, Qc = mr x (h2 - h3)
where mr is the mass flow rate of refrigerant
Tube selection
The following factors are considered while selecting the tube diameters:
a. The pressure drop should be quite low.
b. There should be as small contraction or expansion of tubes as possible. At the same time expansion and contraction should be very close to the matching parts in view of least distortion in the system.
c. Workability should be easily shaped and brazed or soldered if necessary. Copper tube is the best in this respect.
After considering the above factors, the pipe diameter and material is selected based on recommended practical value.
Thus copper tube with an inside diameter of 8.05 x 10-3 m and an outside diameter of 9.525 x 10-3 m is selected.
Refrigerant-side heat transfer coefficient
Assume 7°C drop in temperature of Refrigerant while flowing through the condenser.
Mean temperature of refrigerant (R-134a) in the condenser Tmr= 55 - (7/2)
= 51.5 ºC
Take the properties of R-134a at Tmr.
Thermal conductivity, kr = 0.0172 W/mK
Density, Pr = 1095.215 Kg/m3
Absolute viscosity, =1.5 X 10-4 Kg/ms
Specific heat, Cpr = 1.59 KJ/kgK
Inside area of condenser tube, Ai =
= 9.8528 X 10-5 m2
Reynolds number, Re =
= 1807.14
Prandtl number ,Pr =
= 13.8
Nusselt number, Nu = 0.026 x (Pr) 1/3 (Re) 0.8
= 0.026 x (13.8) 1/3 (1807.14) 0.8
= 25.12
But Nu =
Therefore, refrigerant-side heat transfer coefficient, hi = Nu x kr d
= 53.67 W /mK
Air side heat transfer coefficient
Face area, Af = Qa/ Va
= 0.21175/3
= 0.072 m2
Equivalent diameter, De = (0.072 x 4/1r )05
=0.3 m
Mean temperature of air while flowing through the condenser, Tma = 42.5 ºC
Properties of air at Tma
Thermal conductivity, ka = 0.0274 W/mK
Density, a = 1.1193Kg/m3
Absolute viscosity, = 2.008x 10-5 kg/ms
Prandtl Number, Pr = 0.705 kJ/kg K
Reynolds Number Re =

Since the air side heat transfer coefficient (ho) is small compared with the refrigerant side heat transfer coefficient (hi), it is essential to provide fins on the air side.
Overall heat transfer coefficient based on total fin side surface area
Since the tube wall thickness is very small, tube wall can be neglected.

Log Mean Temperature Difference
Temperature of the refrigerant in the condenser = 55 ºC
Temperature of the air reaching the condenser = 30 ºC
Temperature of the air leaving the condenser = 42.5 ºC
LMTD = (55-30)-(55-42.5)/ln(25/12.5)
Extended Surface area, At =

Therefore, bare surface area, A0 =
Length of tube required = 0.173/3.14x9.5x10-3 m
Designed Condenser Specifications
Heating capacity = 0.527 kW
Inside diameter of tube = 8.05 x 10-3 m
Outside diameter of tube = 9.525 x 10-3 m
Ratio of fin area to bare area = 15
Mass flow rate of refrigerant = 3.3 x10-3 kg/s
Length of tube = 5.7 m
Face area = 0.072 m2
Fin Specifications = rectangular (0.3 m x 0.10 m)

Fig 3.4 Disassembled view of the condenser
Condenser Selection
Based on design, a condenser with two row 3/8 outer diameter copper tube and having a face area of 0.5m2 is selected.
Inside diameter of tube = 8.05 x 10 -3 m
Outside diameter of tube =9.5 x 10 -3 m
Total number of fins = 74
Face area = 0.05 m2
DESIGN OF EXPANSION DEVICE (CAPILLARY TUBE)
Fig 3.5 Capillary tube
An expansion device is essentially a restriction offering resistance to the flow so that the pressure drops as in a throttling process. Capillary tube is the expansion device used in the existing system.
The design of the capillary tube with the diameter selected as 0.000914 m (0.0316) is as follows:
Mass flow rate, m = 3.3 x 10-3 kg/s
Area of cross section =
= 6.362 x 10-7 m2
Let,
X = m/A = 5187.27 kg/sm2
= u/v
Y = X/2D = 2881817.9 kg/m3s
Z = DX = 4.6685 kg/ m s
Friction Factor Calculations:
Point 1 (55 ºC)
Viscosities
= 0.000142 cP
=
Re1 = Z/ = 4.6685 / (0.000142) =32876.76
f1 = 0.32 /(Re)0.25
= 0.32/(32876.76) 0.25
= 0.0238
= at -23 ºC
= 0.00365 cP
Re (at -23 ºC) = 4.6685/ 0.00365 = 1279.04
f2 = 0.0535
f = (f+ f2)/2 = 0.03866
Length Calculations:
Total pressure drop
= 14.76 “ 1.172
= 13.588 x105 N/m2
Acceleration Pressure drop
= X = 5187.27 x (4.8143 “ 3.801)
= 0.052563 N/m2
Friction Pressure Drop
= -
= 13.588 x105- 0.052563
= 13.5354 x 105 N/m2
Mean Friction Factor
f = 0.03866
Mean Velocity
u = (4.8143 + 3.801) = 4.30765 m/s
Incremental Length
=
= (13.5354 x 105)/ 2881817.9 x 0.03866 x 4.30765
= 2.825 m
Designed Capillary Tube Specifications
Diameter = 0.09144 m
Length = 2. 825 m
Selected Capillary Tube
A capillary tube of length 3.48 m and bore 9.144 x 10-4 m was selected based on the design.
EVAPORATOR DESIGN


Fig 3.6 Evaporator Plate
Temperature of the refrigerant = (-23) +5/2
= -20.5 ºC
Outside diameter of the tube taken = 9.525 x 10-3 m
Inside diameter of the tube taken = 8.025 x 10-3 m
Taking the properties of R-134 a at -20.5 ºC
Density, = 1351.67 kg/m2
Viscosity, = 350 x 10-6 Pa S
Specific Heat, Cpr = 1.25 x 103 J/kgK
Thermal conductivity, kr = 102 x 10-3 W/mK
Reynolds Number, Re =

Prandtl Number, Pr =

Nusselt Number, Nu =

Refrigerant side heat transfer coefficient, hi = Nu x kr/Di
= (8.57 x 102 x 10-3)/ 8.025 x 10-3
= 37.768 W/m2 K
Outside heat transfer coefficient
Taking mean temperature of glycol as 27 ºC
Thermal conductivity of glycol, ka = 248.9 W/mK
Log mean temperature difference =

Overall heat transfer coefficient,
where U is overall heat transfer coefficient
hi is refrigerant side heat transfer coefficient
x1 is average thickness in which glycol is filled
x2 is thickness of copper plates
kc is the thermal conductivity of copper
kg is the thermal conductivity of glycol



where, is the LMTD
A = Q/U
= 0.388 m2
Total length of the pipe = A / ( x Do) = 0.338 / x 9.5 x 10-3= 11.3 m
DESIGNED EVAPORATOR SPECIFICATIONS
Evaporator capacity = 0.178 kW
Inside diameter of the tube = 8.025 x 10-3 m
Outside diameter of the tube = 9.525 x 10-3 m
Length of the tube in each evaporator = 5.65 m
SELECTION
Evaporator
Material Selected Copper
Area of the plates 46x48 cm2
Thickness of the plates 1 mm
Length of the tube in each plate 5.2 m
Fabrication of Evaporator Plates
As designed there are two evaporator connected in parallel. Each evaporator consists of copper tube diameter (3/8" OD) & length 5.4 m installed between two plates and the plates are brazed together along the edges so as to form a rectangular box plate of dimension (48 x 46 x l cm3). The space between the plates is filled with ethylene glycol which is a eutectic solution. Ethylene glycol acts as an anti freezing agent. Since the bottom plate is fixed and the top plate is moving, two flexible hoses are used to connect the top and bottom plate respectively
Drier filter
Fig 3.7 Drier Filter
A drier filter is used to remove moisture and dust particles from the liquid refrigerant. Dust particles in the refrigerant may cause blocking at the refrigerant control.
\
Accumulator
Fig 3.8 Accumulator
An accumulator is used in the suction line to prevent any non vaporised liquid from entering the compressor.
CHAPTER 4
TEMPERATURE MEASUREMENT
The most common electrical method of temperature measurement uses the thermo-electric sensor, also known as thermocouple. In the present plate freezer set up; measurement of temperature is to be required at a number of locations that are inaccessible from outside like the refrigerant line. A thermocouple connected to a temperature scanner is useful in determining surface temperatures and temperature of inaccessible locations and the scanner can position at a place of our convenience.
In our project under consideration we use thermometer for temperature measurement.
CHAPTER 5
PRESSURE MEASUREMENT
Fig 5.1 Pressure Gauge
Bourden tube gauges are used for measurement of pressure. The gauge is connected to the measuring point via capillary tube. The locations where the refrigerant measure is required are given below in the table.
Table 3
Sl. No. Measuring Location Bourden gauge reading(psi)
1 Refrigerant line before compressor -30 to 300
2 Line after compressor 0 to 300
CHAPTER 6
EXPERIMENTAL ANALYSIS OF THE SYSTEM
Readings were taken for a load of 200W at a time interval of ten minutes. The lowest possible temperature obtained is -15 ºC after one hour. The readings are tabulated below.
Table 4
Sl. No. Temperature at different points on evaporator plates Compressor discharge temperature
(ºC) Condenser discharge temperature
(ºC) Compressor suction temperature
(ºC) Compressor discharge pressure
(psi) Condenser discharge pressure
(psi)
Plates
1 2 3 4 5
1 -2 6 3 0 8 41 27 24 140 0.8
2 -7 1 0 -4 4 45 27 22 140 0.8
3 -10 -2 -3 -7 0 47 27 18 140 0.8
4 -11 -4 -4 -8 -1 47 28 18 145 1
5 -13 -5 -5 -9 -3 48 28 20 150 1
6 -13 -6 -6 -9 -4 48 28 20 150 1
7 -14 -7 -8 -10 -7 48 28 18 150 1
8 -15 -8 -8 -10 -6 49 29 15 150 1
9 -15 -10 -9 -11 -9 50 29 12 150 0.2
10 -15 -11 -11 -11 -10 53 29 16 150 0.2
SCOPE OF FUTURE WORK
The valves can be connected as individual expansion valve on each tube connecting the evaporator plates. After installation the test can be carried out so that, temperature shown by each thermocouple connected to each of the evaporator plates be the same, by adjusting the flow rate through each expansion valve. Then experiment can be conducted by applying different loads to the plates freezer and evaluate the performance of the modified system.
CONCLUSION
The experiments conducted of the existing system shows the shortage of the flow rate of the refrigerant through the top plate. We can get the temperature inside the plate freezer to exactly the same required temperature. Also we can use this system at different temperatures depending on our need.
REFERENCES
1. Roy J Dossat, Principles of Refrigeration
2. C P Arora, Refrigeration and Air Conditioning , Tata McGraw Hill Publications
3. W F Stoecker, Refrigeration and Air Conditioning Tata McGraw Hill Publications
4. Domkunwar S, A Course in Heat and Mass Transfer
5. ASHRAE, ASHRAE Handbook and Product directory
6. Manohar Prasad, Refrigeration and Air Conditioning Wiley Eastern Limited
7. P L Ballaney, Refrigeration and Air Conditioning, Khanna Publishers
8. Ananthraman, Basic Refrigeration and Air Conditioning, Tata McGraw Hill Publications
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thanks man. can you send me the ppt file please
adilshahzad2001[at]yahoo.com
adilshahzad2001[at]gmail.com
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give the project report on given topic
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How can i get access?
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