G. H. Raisoni College of Engineering, Nagpur. Department of Mechanical Engineering. Refrigeration and Air conditioning Lab Manual.

Transcription

1 G. H. Raisoni College of Engineering, Nagpur. Department of Mechanical Engineering Refrigeration and Air conditioning Lab Manual Semester- VIII

2 PRACTICAL: REFIRIGERATION AND AIR CONDITIONING Aim : Study of various tools and equipments used in Air Conditioning Introduction: 1. Controllers i) Temperature Controller ii) Humidity Controller 2. Cooling Coils and Heating Coils 3. Air Filter 4. Electric Filter 5. Air Duct 6. Grills and Diffusers as air outlet 7. Air Washers 8. Adiabatic Dehumidifier Conclusion

3 PRACTICAL: REFIRIGERATION AND AIR CONDITIONING Aim: Study of various types of compressor Introduction: Classification of Compressor 1. Hermetic Compressor Introduction Working Principle with neat sketch Application 2. Rotary Compressor Introduction Working Principle with neat sketch Application 3. Reciprocating Compressor Introduction Working Principle with neat sketch Application Conclusion

4 PRACTICAL: REFIRIGERATION AND AIR CONDITIONING Aim: Study of various condensers, evaporator and expansion devices used in Vapour Compression Refrigeration System. Introduction: CONDENSER Classification of condensers 1. Air cooler condenser Explain with working principle and neat sketch 2. Water cooled condenser Explain with working principle and neat sketch 3. Evaporative Condenser Explain with working principle and neat sketch EVAPORATOR Classification of Evaporator 1. According to types of construction Base tube coil evaporator Plate evaporator Fixed tube Shell and tube Shell and coil Tube in tube 2. According to manner in which liquid refrigerant is fed flooded evaporator Dry expansion evaporator 3. According to mode of heat transfer Actual natural convection Forced convection

5 4. According to operating condition Frosting evaporator Non frosting evaporator Departing evaporator EXPANSION DEVICES Types of Expansion device 1. Capillary tube 2. Hand operated expansion valve 3. Thermocoustic expansion valve 4. Low side float valve 5. High side float valve 6. Automatic or constant pressure expansion valve Conclusion :

6 Aim : Study and demonstration of various controls used in Refrigeration and air Conditioning Introduction : 1. Water level Indicator 2. High pressure control 3. Bulb and bellow element 4. Solenoid or electromagnetic element 5. Humidity sensitive element 6. Single phase induction motor 7. High and low pressure cutouts 8. Control room conditions at Partial load 9. On-Off Control 10. Volume Control Conclusion :

7 Aim : Introduction : Study of window air conditioning, packaged air conditioning and automotive air conditioning 1. Window Air conditioning Classification Working Principle Application 2. Packaged Air conditioning Working Principle Application 3. Automotive Air conditioning Working Principle Application Conclusion :

8 Aim : Introduction : Study of domestic Refrigerator Working Principle with neat sketch Various Components used Compressor Condenser Expansion device Evaporator Conclusion :

9 Aim : Introduction : Study of Testing and Charging of VCR system. 1. Methods for Testing and Charging Evaculation Dehydration Charging of refrigerant Testing for leaks - Soap bubble method - Sulphur candle method - Wet litmus paper - Halide torch method - Electric defector - Use of cold trap Conclusion :

10 DESERT COOLER ( EXPERIMENTAL UNIT ) INTRODUCTION : Evaporative coolers (also called air coolers or desert coolers) are cooling devices which uses simple evaporation of water in air. They differ from refrigeration or absorption air conditioning, which use the vapor-compression or absorption refrigeration cycles. small-scale evaporative coolers are called swamp coolers by some users due to the humid air conditions produced. Air washers and wet cooling towers utilize the same principles as evaporative coolers, but are optimized for purposes other than air cooling. Evaporative cooling is especially well suited for climates where the air is hot and humidity is low., if sufficient potable water is available. In dry climates, the installation and operating cost of an evaporative cooler can be much lower than air conditioning, often by 80% or so. But evaporative cooling and vapor-compression air conditioning are sometimes used is combination to yield the optimal performance. Some evaporative coolers may also serve as humidifiers in the heating season. In moderate humidity locations there are many cost-effective uses for evaporative cooling, in addition to their widespread use in dry climates. For example, industrial plants, commercial kitchens, laundries, dry cleaners, greenhouses, spot cooling (loading docks, warehouses, factories, construction sites, athletic events, workshops, garages, and kennels) and confinement farming (poultry ranches, hog, and dairy) all often employ evaporative cooling. In highly humid climates, evaporative cooling may have little thermal comfort benefit beyond the increased ventilation and air movement it provides. DESCRIPTION : Experinmental Desert cooler unit consists of water spray arrangement over the cooling pads at the centre and duct made of M.S. sheet on both sides. Inlet duct consistes of inlet orifice with inclind manometer, for air flow ( Q ) measurement. Inlet section having orifice at air entry and Air heater ( fin type, for simulation of summer condition ) after that and, DBT / WBT sensor at the centre. Water sprey with cooling pades after inlet section. Water sump is provided at the bottom. Pump sucks the water and discharges through nozzle on top of the cooling pades through water flow meter ( rotameter ). After water sprey section outlet section follows.

11 Outlet section contains DBT / WBT sensor for outlet air condition measurement. wheel type anemometer is placed at the centre of the duct for air velocity measurement. At the end of the outlet section axial fan is fitted to suck the air. All ducts ( inlet / outlet ) are having superlon insulation to reduce the heat losses. Front side of the duct is covered by acrylic sheet. Necessary controls and computer interface converters and switches are fitted over the main stand in panels. SPECIFICATION : 1. Axial fan : 250 watts. 2. Duct inlet : 250 mm x 250 mm 3. Outlet duct : 250 mm x 250 mm 4. Inlet orifice : 125 mm 5. Inlet / oulet air condition : DBT / WBT sensor 6. Air Heater : 250 watts x 2 nos with fins. 7. Water sprey chamber : 300 mm width. 8. Cooling pades : 3 Nos. 9. Sprey pipes : 3 nos with 10 no. each holes of 1.2 mm 10. Water sump tank : 8 ltrs 11. Water pump : ½ HP 12. Rotameter : LPH 13. Ball valve : 1 No. 14. Anemometer : wheel type ( 30 m/sec max ) 15. Temperature indicator : 6 channel, serial interface. 16. Heater voltage / current converter. Direct Evaporative Cooling ( open circuit ) :

12 DEC is used to lower the temperature of air by using latent heat of evaporation, changing water to vapor. In this process, the energy in the air does not change. Warm dry air is changed to cool moist air. Heat in the air, and the water too, is used to evaporate water. Typically, residential and industrial evaporative coolers use direct evaporation and can be described as an enclosed metal or plastic box with vented sides containing a centrifugal fan or 'blower', electric motor with pulleys (known as 'sheaves' in HVAC), and a water pump to wet the evaporative cooling pads. The units can be mounted on the roof (down draft, or downflow), or exterior walls or windows (side draft, or horizontal flow) of buildings. To cool, the fan draws ambient air through vents on the unit's sides and through the damp pads. Heat in the air evaporates water from the pads which are constantly re-dampened to continue the cooling process. Thus cooled, moist air is then delivered to the building via a vent in the roof or wall. Because the cooling air originates outside the building, one or more large vents must exist to allow air to move from inside to outside. Air should only be allowed to pass once through the system, or the cooling effect will decrease. This is due to the air reaching the saturation point. Often 15 or so air changes per hour (ACHs) occur in spaces served by evaporative coolers. Cooler pads Traditionally, evaporative cooler pads consist of excelsior (wood wool) (aspen wood fiber) inside a containment net, but more modern materials, such as some plastics and melamine paper, are entering use as cooler-pad media. Wood absorbs some of the water, which allows the wood fibers to cool passing air to a lower temperature than some synthetic materials. The thickness of the padding media plays a large part in cooling efficiency, allowing longer air contact. For example, an eight-inch-thick pad with its increased interface will be more efficient than a one-inch pad.

13 OBSERVATION TABLE : SR.No. Manometer reading ( h) mm Heater Volts Heater Amp. Water flow rate Inlet DBT (T 1 ) Inlet WBT (T 2 ) Outlet DBT (T 3 ) Outlet WBT (T 4 ) CALCULATION : 1) Amount of water added :- m w = m air x ( ω) m w = m air x (ω 1 ω 2 ) where, ω 2 = water content in air at inlet ω 1 = water content in air at outlet m air = C d x (π/ 4) d 2 x 2 x g x (ρ w / ρ g )x h 2) Desert Cooler Efficiency (η) η = T 1 - T 3 T 1 - T 2 Where, T 1 = Dry bulb temperature at inlet T 2 = Wet bulb temperature at inlet T 3 = Dry bulb temperature at outlet Performance Understanding evaporative cooling performance requires an understanding of psychrometrics. Evaporative cooling performance is dynamic due to changes in external temperature and humidity level. Under typical operating conditions, an evaporative cooler will nearly always deliver air cooler than 27 Celsius (80 Fahrenheit). A typical residential 'swamp cooler' in good working order should cool air to within 3 C - 4 C (6 F - 8 F) of the wet-bulb temperature.

14 Advantages Less expensive to install 1 Estimated cost for installation is 1/8 to 1/2 that of refrigerated air conditioning Less expensive to operate 1 Estimated cost of operation is 1/4 that of refrigerated air. 2 Power consumption is limited to the fan and water pump vs. compressors, pumps, and blowers. Ventilation air 1 The constant and high volumetric flow rate of air through the building reduces the age-of-air in the building drammatically. Disadvantages Performance 1 High temperature, high humidity outside conditions decrease the cooling capability of the evaporative cooler. 2 No dehumidification. Traditional air conditioners remove moisture from the air, which is usually a design requirement except in very dry locations. Evaporative cooling adds moisture, which, in dry climates, may improve thermal comfort. Comfort 1 The air supplied by the evaporative cooler is typically 80-90% relative humidity. 2 Very humid air reduces the evaporation rate of moisture from the skin, nose, lungs, and eyes. 3 High humidity in air accelerates corrosion. This can considerably shorten the life of electronic and other equipment. 4 High humidity in air may cause condensation. This can be a problem for some situations (e.g., electrical equipment, paper/books, old wood). Water 1 Evaporative coolers require a constant supply of water to wet the pads. 2 Water high in mineral content will leave mineral deposits on the pads and interior of the cooler. Water softeners, bleed-off, and refill systems may reduce this problem,

15 however. 3 The water supply line needs protection against freeze bursting during off-season, winter temperatures. The cooler itself needs to be drained too, as well as cleaned periodically and the pads replaced.

16 THERMOELECTRICTRIC COOLING In 1822 seebeck observed that, if a closed circuit were made of two dissimilar metals, an electrical current flows in the circuit when the two junctions were maintained at different temperatures. His investigations covered a wide range of elements and compounds. This resulted in publication of a series in which the investigated materials were arranged in the order of magnitude of their effect. However, he failed to realize the significance of his discovery. In Peltier observed the inverse effect, if an electrical current flows across the junction between two dissimilar material, heat was either absorbed or evolved. Peltier did not realize the significance of his discovery either, and moreover, failed to recognize the connection between his discovery and that of Seebeck. It is alleged that Lenz ended all conjecture surrounding these discoveries by freezing a small quantity of water placed in the vicinity of the junction between a bismuth and antimony rod through which a direct current was passed. These were two of the materials in which Seebeck had observed the most pronounced effect. A third effect was pointed out by Thomson (Lord Kelvin) in It related heat absorbed or evolved in a single conductor to the temperature gradient along it and the current flowing through it. This effect takes place in addition to the Joule (FR) heating. However, in thermoelectric cooling materials, Kelvin s is a second order effect when compared with Peltier s and Seebeck s, and will not be further considered. For many years. Practical application of these thermoelectric effects was almost exclusively restricted to thermocouples for temperature measurement, because metals exhibit a comparatively small Seebeck effect. However, the Seebeck effect in semiconductors can be considerably greater. The advent of the transistor and other semiconductor devices have stimulated research pertaining to properties of semiconductors in general, from this, materials have been developed in which thermoelectric effects are of sufficient magnitude so that fabrication of useful devices has become a reality.

17 Specifications : 1. Module assembly : KRYOTHERM, Russia make. Frost 74 HT 1.05 ohm resistance at 295 K T max > 74 0 C Height : 3.85 mm Parallel difference : mm 2. Control panel : 1. Separate controls to module and heater 3. Multi pole switch. 4. Two variable power supplies DC. 5. Two Ammeter 6. Two Voltmeter 7. Temperature indicator ( digital ) with serial interface. 8. Voltage and current module for computer interface. THERMOELECTRIC EFFECTS Seebeck Effect : For a small temperature difference between the two junction of materials A and B, the open circuit voltage developed is proportional to the temperature difference and is given by: Where, E = α AB T E = open circuit voltage developed α AB = relative Seebeck coefficient (the difference between absolute Seebeck Coefficients for materials A and B). T = temperature difference between junctions of materials A and B.

18 In practice, the absolute Seebeck coefficient, α, of a material is determined with respect to a material such as lead, in which the Seebeck coefficient is negligible. In metals, α does not exceed volt/deg C in semiconductors available for thermoelectric applications in 1966, α is typically to volt/deg C. Peltier Effect : The same circuit can be considered to be made up of materials A and B, into which a battery is introduced to provide a direct current, (I). At the junction between the two dissimilar materials, the heat evolved or absorbed in unit time is proportional to the current flowing and is given by : where Q = π AB I Q = heat evolved or absorbed in unit time, watts. π AB = relative Peltier coefficient for materials A and B. I = direct current flowing, amperes. Lord Kelvin, by performing a thermodynamic analysis of such a Thermoelectric circuit, showed a relationship exists between α and π Where π = αt T = the absolute temperature, Kelvin. Hence, heat absorbed or evolved per unit at the junction between two dissimilar materials is given by: Q = α AB IT Although thermodynamics do not accept Kelvin s derivation of his well-known relations because he considered the effects reversible, a more rigorous treatment, by application of irreversible thermodynamic, leads to the same results.

19 Thermodynamics and Refrigeration Cycles : The relative hot and cold surfaces are the result of the basic heat transport phenomenon called the Peltier effect, and are not necessary to make it occur. Investigations in solid-state physics have shown that thermoelectric refrigeration is means of heat pumping which utilizes the energy level changes in electrons for transporting thermal energy. Electrons flowing across a junction of two dissimilar thermoelectric materials, i.e., materials with different available electron energy levels, must undergo an energy change which results in either the evolutions or absorption of heat. The direction of current flow determines which will occur. The bibliography lists several sources of detailed information on solid state theory. PERFORMANCE OF THERMOELECTRIC COUPLES A couple circuit consisting of two dissimilar thermoelectric materials is referred to as a couple. The two thermoelectric materials are represented by n and p. The n-type has a negative Seebeck coefficient and an excess of electrons. The p-type has a positive Seebeck coefficient and a deficiency of electrons actually flow in the opposite direction. While there are a total of four connections between the thermoelectric junctions; the upper or cold one, at which heat is a absorbed; and the lower or hot one, at which heat is evolved. If the direction of current were reversed, the upper junction would evolve heat, the lower would absorb it. The temperature gradient between hot and cold junctions is not linear because of production of Joule heat within each leg. The conducted heat arriving at the cold junction is given by: C T + 0.5( I 2 R ) Where C = thermal conductance of couple legs, watts per degree Celsius. APPLICATIONS OF THERMOELECTRIC COOLING : Thermoelectric cooling is different from conventional compression Refrigeration in that there are no moving parts in the process of producing cold. However, as energy is being transferred as heat in the cooling process, heat must be removed from the hot side of the couples in order that cold can be produced. For improved efficiency on the cold side, heat transfer surfaces are frequency used. These heat transfer surfaces take the forms of fins and fans in gas systems, or with heat transfer to fluid systems using pumps for circulation.

20 Two different systems for cooling : 1) by using vapour compression system, 2) Thermoelectric module VAPOR COMPRESSION THERMOELECTRIC COMPRESSOR HOT COLD BATTERY HOT COLD SYSTEM ANALOGES System can be interchanged by reversing the polarity of the director current applied to it. Since there are no moving parts, there is nothing to generate noise. There is no refrigerant to contain, so the problem of handling a two-phase changeover is simplified. The pressure tight tubing is replaced by electrical wiring. Since the cooing capacity of a single thermoelectric couple is small, it is practical to make systems of low refrigerating capacity. In the cooling of infrared detectors, small thermoelectric assemblies can cool to temperatures as low as 145 K, using cascading. Thermoelectric systems have been built in capacities up to ten tons of air conditioning by using many couples. Capacity control in a thermoelectric system can be achieved by varying the voltage applied to the couples either by a variable voltage control or by switching series and parallel circuits. As the voltage drops, the temperature difference between hot and cold side reduces. Another advantage is that a thermoelectric system will operate under zero gravity, or many times the force of earth gravity, and will operate in any orientation. This has been an important consideration in selecting thermoelectric cooling application in the U.S. space program. A thermoelectric system is capable of operating satisfactorily at temperatures levels of C.

21 For many applications, the advantage of thermoelectric cooling outweigh its chief disadvantage of low coefficient of performance. The lack of moving parts and eliminations of cooling liquids are appealing features. For military applications, thermoelectric prove advantageous because of noiselessness, ruggedness, and lack of Thermodynamic Analysis of Thermoelectric Refrigeration System Fig shows an enlarged view of thermoelectric elements with control volume. In order to analyze the system to obtain refrigeration effect COP, etc., the following assumptions have been made: (i) Heat transfer takes place through semi conductors at the ends only. (ii) No energy exchange between the elements through space separating them, and (iii) Properties such as conductivity, etc., are invariant with temperature. Now cooling and heating due to the thermoelectric effect is given by (Peltier effect): q c = ab IT l q h = ab IT h ADVANTAGES OF THERMOLECTRIC REFRIGERATION SYSTEM It possesses several advantages as listed below: (i) Absence of moving parts eliminates vibration problem as well as regular attendance. Therefore, it can be best suited for system where vibration is undesirable. In addition there is no were due to rubbing as such the life is expected to be almost infinite compared with other systems. (ii) It is easy to overload if desired, just by increasing the current to a certain limit.

22 (iii) The load can be easily controlled by means of adjusting the current to meet the situation. (iv) Very compact in size since even the system boundary may be used as the cooling surface as exhibited in fig. Here one of the walls of the room forms the evaporator surface. Therefore, the cost of manufacture of that wall is taken care by the evaporator surface. (V) It can be operated in any position in contrast with vapour absorption, vapour-compression or steam - jet refrigeration systems. As for example, if a vertical compressor is used in a refrigeration system, it cannot be operated in any other position. (vi) It is lighter in weight for the same capacity of refrigeration. (vii) Since electric current passes through conductors, there is no problem of leakage which is most undesirable in other refrigeration systems. Further, the leakage of refrigeration from the system causes the drastic decrease in the capacity in addition to extra cost for the refrigeration and charging operations. Thus a thermoelectric refrigeration system operates at the same capacity for long and eliminates the cost of charging and extra materials. (viii) It is most easy to operate as a heat pump, just by reversing the terminals. Hence, a thermoelectric refrigeration system can be considered as an year round air conditioner*. (ix) Since no refrigerant is used, there is no question of toxication etc. and can be used directly for air conditioning. (x) Its design and manufacture are rather simpler than the other refrigeration systems.

23 (xi) It is most suitable for the production of cooling suit. Main disadvantage of thermoelectric refrigeration system is the unavailability of suitable material of high figure of merit. Presently the total cost of refrigeration system will be a few times higher that the vapour compression or other systems for a few ton capacity. In addition, the running cost is found t obe much higher compared to vapour compression system. That is why the vapour compression or other systems of refrigeration are most commonly employed. The overall COP of a thermoelectric refrigeration system is of the order of 01 to 0.2. Experimentation : 1. Evaluation of the peltier effect : DC Power is applied to the module at a variety of settings. When settled both temperatures are noted. Observe that at higher power inputs I 2 R and heat conduction factors tend to overwhelm the cooling effect. Module power voltage current Graph : Module Power v/s cold & hot side Temperature and temp. difference

24 2. Evaluation of Seebeck effect : By use of heater power create a temperature differential across the module. When settled note the open circuit voltage across the module voltmeter. Voltage and temperature can be seen to have a linear relationship. Heater power voltage current Graph : Seebeck open ckt voltage v/s Temperature gradient T 3. Evaluation of Generating effect : Create a temperature gradient across the module by applying heater power and switch in electrical load. When settled, note the generator voltage and amperage. Heater power Module Power Voltage Current Voltage Current Graph : module voltage / current / power v/s Temperature Gradient T 4. Coefficient of performance :

25 Power to the module is set and heater power adjusted to maintain ambient temperature on the cold side of the module. Repeat over the whole range. The curve shows that the coefficient is highest when module has to do the minimum work. Heater power Module Power Voltage Current Voltage Current Graph : Module power v/s COP, COP = W h /W m W h : Heater power W m : Module power

26 VAPOUR ABSORPTION REFRIGERATION SYSTEM INTRODUCTION & WORKING PRINCIPLE The vapor absorption Refrigeration System consists of a vapor absorption unit mounted on a cabinet fixed on a display frame of the set-up. The principle works on 3-fluid system. There is no solution circulation pump. Total pressure is same throughout the system. The third fluid remains mainly in the evaporator thus reducing partial pressure of refrigerant to enable it to evaporate at low pressure and hence low temperature. The schematic diagram of vapor absorption refrigerator working on NH 3 -H 2 O system with H 2 as the third fluid is shown in Fig 1. Liquid NH3 evaporates in the evaporator in the presence of H 2. Hydrogen is chosen as it is non-corrosive and insoluble in water A thermo siphon bubble pump is used to lift the weak aqua from generator to the separator. The discharge tube from the generator is extended down below the liquid level in the generator. The bubbles rise and carry slugs of weak NH 3 -H 2 O solution into the separator. Two U bends are provided as vapour locks to prevent H 2 from getting into the high side or solution circuit. Partial pressure of H 2 provides the pressure difference of NH 3 between the condenser and the evaporator. Accordingly, we have:.in condenser, Pure NH 3 vapour pressure = Total pressure In evaporator, NH3 vapour pressure = Total pressure Partial pressure of H 2 For example, consider the condenser temperature as 50 0 C, and evaporator temperature as 15 0 C. the corresponding vapour pressures of NH 3 are Condenser, P k = bar Evaporator outlet, Po 2 = 2.36 bar

27 The approximate pressures in various parts of the system, then will be as given in the table below: (Example for partial pressure effect) Section NH 3 H 2 O H 2 Total Condenser Evaporator Inlet Evaporator Exit Generator Top It has been assumed that vapours leaving generator top are in equilibrium with entering rich solution at 40 0 C, at which temperature saturation pressure of NH 3 is bar. It has also been assumed that the temperature at evaporator inlet is 25 0 C at which temperature saturation pressure of NH3 is bar. DESCRIPTION OF THE SET-UP The unit is charged with ammonia solution in water and hydrogen gas. The maximum pressure it can withstand is 30 bar and the ammonia charge is 0.245Kg NH 3 + H 2 O. The unit works on electric heater 90 watt capacity. A voltmeter / ammeter is provided to measure the power. The heater works with intermittent action of 18 sec on and 3 sec off to control the temperature rise. A temperature indicator is provided for measurement of temperature at 10 different locations as displayed on the mimic diagram presented on the board. A thermostatic control is provided for the sake of safety and it controls the heater.. An acrylic window in front of the system and open portion from behind gives the actual view of the system.

28 SEQUENCE OF OPERATION 1] Strong ammonia solution flows from the absorber vessel to the boiler thro the inner tube. 2] When the ammonia solution is heated in the boiler bubbles of ammonia gas rise from the pump. 3] The ammonia vapour passes in to the condenser. 4] Weak ammonia solution flows in to the annular tube. 5] Air circulating over the fins of the condenser cools down the ammonia vapour, {after 2} condensing it to liquid ammonia. 6] Liquid ammonia flows through the pipe to the evaporator after sub cooling & throttling. 7] A } The hydrogen in the evaporator lowers the ammonia vapour pressure and makes it to evaporate. B} This mixture of hydrogen and ammonia passes from the evaporator, to the Absorber. C} This process extracts heat from the evaporator, which in turn extracts heat from the storage space, thereby, the temperature inside the refrigerator is lowered. 8] As the weak solution is fed from the boiler system as it runs to the absorber vessel, it absorbs the ammonia from the ammonia hydrogen mixture and gets ready for another round of cycle in the boiler.

29 Principal Operating Line 1 The actual ammonia water system requires the separation of water vapour as much as practicable in order to avoid the high operating temperature of evaporator & the large amount of purging of water into the absorber. 2 To meet the above requirement an analyser & a rectifier are incorporated with the generator. deflegmator is not used.. 3 Thus the actual absorption system comprises an analyser, rectifier, a pre heater and a pre cooler in addition to the basic components. 4 The strong solution of ammonia is pumped from the absorber into generator thro A pre heater, a counter flow heat exchange.thus the hot and weak solution returning from the generator heats the aqua solution before it entered generator, causing reduction in heat input to generator. 5 The heat transfer to the generator brings about the separating of ammonia vapour accompanied by a small fraction of water vapour. 6 This mixture of ammonia and water vapour passes thro the analyser where a part of Water vapour is separated from the vapour mixture. 7 The mixture then enters the rectifier in a simple form and not in the form of deflegmeter.water particles are drained back to analyzer due to presence of contraction and bend. 8 The ammonia vapour then condenses in the condenser, which, in turn,is sub cooled by the vapour leaving the evaporator in a pre cooler,a counter flow heat exchanger. 9 Thus the refrigeration effect is increased. The sub cooled condensate is allowed to expand through the throttle valve down to the evaporator pressure. The energy transfer to evaporation causes vaporization of ammonia. The ammonia vapor are then reabsorbed. 10 The analyzer column is an open type counter flow heat exchanger having baffle plate effect provided with the dents to the analyzer column, increasing the area of contact between the vapour leaving the generator and strong aqua ammonia entering the analyzer at appropriate location.

30 11 The warm fluid from the pre heater enters the analyzer column where it comes in direct contact with the hot vapour. The aqua solution gets heated at the same time a part of water vapour is condensed since the ammonia vapour leaving the generator Contains small percentage of water vapour. Thus it acts as a column which purifies the ammonia vapour by reducing the amount of water vapour a kind of rectifying action. 12 As soon as strong aqua is heated in the generator, there is separation of vapour of ammonia leaving behind weak solution of ammonia, corresponding to the generator temperature and pressure. 13 The vapour leaving the generator with refrigerant concentration ξ v a in equilibrium L with the boiling poor solution having concentration ξ a enters the analyzer at 2 v As it travels upwards, counter flow to the entering rich solution at 1 with concentration ξ L r, the vapour encounters heat and mass exchange with the falling rich solution ultimately leaving the analyzer enriched in the refrigerant with vapour concentration ξ 5 In equilibrium with the rich solution having concentration ξ L r. SPECIFICATIONS Kg NH 3 + H 2 O Refrigerant 2. Heater capacity 90 watt,1 ph.ac Supply 3. Air cooled evaporator 4. Digital temperature indicator (10 channel, Cr-Al) 5. Digital voltmeter 6. Digital ammeter OPERATING INSTRUCTIONS 1. Start the electric connection-heater. 2. Keep the unit in working for about 4 hours to reach the equilibrium. 3. Take readings when steady state is reached 4. Find out the coefficient of performance COP as per the example illustrated with the help of Enthalpy-concentration chart and refrigerant properties of ammonia water. from The standard chart. 5. Keep the unit working for 24 hours. OBSERVATION TABLE

31 Temperature Readings (Thermocouple numbers as shown in diagram) Heater input T 0 1 C T 0 2 C T 0 3 C T 0 4 C T 0 5 C T 0 6 C T 0 7 C T 0 8 C T 0 9 C T 0 10 C V Vol t I amp Precautions :- 1] Do not start the switch when there is cover on the unit from behind. 2] Remove the cover and then start the unit. 3] Do not disturb the thermocouples placed at various locations. 4]Do not remove the cover wrapped over generator / boiler. 5] Do not operate the thermostatic switch at the right hand bottom side on the main panel. ACTUAL VAPOUR ABSORPTION CYCLE & ITS REPRESENTATION ON ENTHALPY-COMPOSITION DIAGRAM. Figure 2 shows the schematic arrangement of the actual vapor absorption cycle and figure 3 represents its thermodynamic cycle on the h - diagram. The system consists of generator G together with analyzer AN, and condenser C on the high-pressure side, and the evaporator E and Absorber A on the low-pressure side. Pump P, expansion valve VI, and pressure reducing valve VII separate the two sides. In addition, liquid-vapor heat exchanger HEI is also provided. The vapors at 5 distilled from the generator-analyzer, enter the refrigerant circuit. The vapors are condensed to 8 in the condenser, precooked to 9 in the liquid-vapor regenerative heat exchanger and throttled to 10 before entering the evaporator. The state 10 is at the same point as state 9 in the h - diagram, as both enthalpy and composition remain the same before and after throttling.

32 The refrigerant entering the evaporator at 10, leaving the evaporator at 11, and the liquid vapor heat exchanger at 12, comprise a liquid plus vapor mixture. The refrigerant is finally absorbed by the poor solution at 2 returning from the generator and after being cooled in the liquid-liquid heat exchanger to 3 and throttled to 3a whereas the rich solution from the absorber at 4 is pumped to 4 a and heated to 1a before entering the analyzer. The state points 1,2,3 & 4 can be located on the h - diagram according to their temperatures and compositions as the enthalpy of liquid is independent of pressure. Also point 3a lies at 3 only (isenthalpic process) and point 4a lies approximately at 4 itself as the pump work s very small. State point 5 of the vapor is along the isothermal tie line drawn from 1. Point 8 is the saturation state at P ike. Point 9 after sub cooling of the liquid can be plotted according to the temperature and composition and point 10 is at 9 itself (isenthalpic process). Point 11 is on the tie line corresponding to the evaporator leaving temperature t 02 = t 2 and pressure P 0.the composition is same at 7,8,9,10,11 and 12. Point 12 can be similarly located by knowing the temperature from the energy balance of the heat exchanger. The absorber pressure in the absorber system is equal to the evaporator pressure Po & the generator pressure is similarly equal to condenser pressure Pk. Hence at a given generator pressure, the poor solution concentration is determined by the heating temperature t h and at a given absorber pressure, the rich solution concentration is determined by the cooling temp. t A the absorber temperature. STEPS FROM START TO FINAL RESULT OBSERVATIONS I) 1) Generator Temperature

33 II) 2) Temperature after Generator outlet. 3) Temperature before condensation 4) Temperature after condensation 5) Temperature after sub cooling 6) Evaporator Inlet Temperature 7) Evaporator outlet Temperature 8) Absorber Inlet Temperature 9) Temperature of Inlet to absorber vessel 10) Temperature of Outlet of Absorber Refer to the Properties of Ammonia - Table (Pk & Po) 1) Note Temperature after condensation t h = t c. With this as saturation temperature refer to the Properties of Ammonia & find out the condenser pressure = P k = (Psat)t h = t c 2) Note temperature at the inlet of Evaporator = t 6 = t 0, with this as saturation temperature refer to the properties of Ammonia & find out the Evaporator pressure p 0 = (Psat) t 6 = t 0 III) 1) Refer to the Enthalpy Concentration chart of Ammonia Note the temperature of generator = t 1 =t h at condenser pressure P k & call this point as 2. The poor solution concentration.a is obtained at this point on x axis. 2) Evaporator pressure and absorber temperature, t 8 = t A call this point as 4, The rich solution concentrations r is obtained at this point.points 1 & 3 are obtained as illustrated on the Evlhalpy chart by drawing r & a Lines & where they intersect the pressure pk & P0 are the points resp, 1 &3. 3) Note temperature after condensation & condenser pressure Pk on the chart and this is the point 8. The concentrations of vapour leaving the analyzer, in equilibrium with the entering rich solution is r = v r This point can also be traced as given below Draw vertical line from point 4-1, where it meets the auxiliary line, from that point draw a horizontal line and where it meets the condenser pressure line that point is 5 v & r = v r= 5v. Is the same as described above. f ξ ξ = 5 a ξ r ξ a 4) Now specific rich solution rate Specific poor solution rate = (f-1)

34 5) Temperature after sub cooling = t 9 Note this point on the line 5v 6) The vapour state 10 v at v 10 = 1 and then join the point 10 v to 9 and extend the line to in tersect the saturated liquid line for evaporator pressure p0 at 10 L which gives the temperature after expansion and the temperature at the in let to Evaporator as t 01 = t 10 & t 02 = t 11 7) Draw the isothermal tie line for t 02 = t 11 & p 0 The intersection of this line with the 11 is the point 11 8) Refrigerating Effect = q 0 = h 11 - h 10 9) Heat transfer in the liquid Heat Exchanger HE II q = (f-1) (h 2 h 3 ) = f(h 1 a h 4 ) From this h 1a is known as h 2, h 3, h 4 are known from the Enthalpy Concentration chart. 10) Heat added in the generator qh = h 5 +h 2 (f-1) fh 1a qo COP = qh 11) Coefficient of Performance 12) h 12 = h 11 + (h 8 h 9 ) 13) qa = Heat rejected in Absorber qa = h 12 h 3 +f (h 3 h 4 ) qa + qc = Heat rejected (Total) 14) Heat rejected in the condenser qo + qh = total heat given 15) Energy balance qa + qc = qo +qh Heat rejected in the condenser + Heat rejected in the Absorber = Heat added in generator +Refrigerating Effect. Sea that both the LHS & RHS match well showing that balance is achieved well. SET No. 1 I = Observations Vapour Absorption Refrigeration system

35 1) Generator temperature = c 2) Top of Generator temperature = c 3) Condenser in let temperature = c 4) Condenser out let temperature = c 5) Subcooled temperature = c 6).In let of E vapur = c 7) Out let of E vapur = c 8) In let of absorber = c 9) out let of absorber = c 10) out let of absorber vessel = c I I = Observations Referring NH 3 properties from Table 1) PK = Condenser pressure = (p sat) at c pk = kg / cm 2 2) PO = Evaporation pressure = (P sat) = 17 0 c po = 2.15 kg /cm 2 I I I = Observations Points on enthalpy conc. diagram POINT NO 2 Generator temperature Generator pressure kg / cm 2 ξa = c POINT NO 4 Evaporator pressure po = 2.15 kg / gm 2 ξ r = 0.4 Absorber Temp = c POINT NO 8 8 = 0.96 h 8 = 460 POINT NO 5 5 = 0.96 h 5 = 1850 POINT NO 1- r = 0.4 h 1 = 300 POINT NO 3- a = 0.14 h 3 = 280 POINT NO 9-9 = 0.96 h 9 = 280 POINT NO = 0.96 h 10 = 280 POINT NO = 0.96 h 11 = 1400 qo = h 11 h 10 = = f = = = 3.15

36 f = 3.15, (f-1) = 2.15 q = (f-1) (h2 h3) = 2.15 x ( ) = 2.15 x 320 = 688 = f(h 1a h 4 ) = 688 = 3.15 (h 1a 20) h1a = 238 qh = h5 +h2(f-1)-fh1a = x 2.15 = = qh = 2390 qo + qh = 3110 = Heat given = 3510 h 12 = h 11 + h 8 h 9 = = h 12 = 1580 qa = h 12 h 3 + f(h 3 h 4 ) = (280 20) = x 260 = x 260 = = 2119 qa = 2119 KJ/Kg qc = h5 h8 = qc = 1390 qa +qc = Heat Rejected = = 3509 qa + qc = 3509 qo +qh = 3510 Heat Balance is checked Well. Vapour Absorption Refrigeration system SET No. 2 I = Observations 1)Generator temperature = c 2)Top of Generator temperature = c

37 3)Condenser in let temperature = c 4)Condenser out let temperature = c 5) subcooled temperature = c 6) In let of E vapur = c 7) Out let of E vapur = c 8) In let of absorber = c 9) out let of absorber = c 10) out let of absorber vessel = c I I = Observations Referring NH 3 properties from Table 1) pk = (Psat) at ccondenser outlet pk = kg / cm 2 2) PO = (P sat)at c Evaporation Inlet Temperature po = 1.19 kg / cm 2 I I I = Observations POINT NO 2 Generator temperature c Condenser pressure = pk = kg / cm 2 h 2 = 600 KJ/kg POINT NO 3 a = 0.14 & po = 1.96 Kg/cm 2 h 3 = 280 POINT NO 4 po = Ev. Pressure & abs. Temperature Po = 1.96 Kg/cm 2 ta = c r = 0.4 h4 = f = = = POINT NO 1 r = 0.4 pk = kg/cm 2 H 1 = 300 f-1 = 2.15 POINT NO 5 h 5 = = 0.96 POINT NO 8 pk = kg/cm 2 T Temperature after condenser = = 0.96 h 8 = 470 POINT NO 9 h 9 = = 0.96 t subcooled = 7.5

38 POINT NO 10 h 10 = 260 POINT NO 11 h 11 = 1380 Draw 10 v to po= 1.96 kg/cm 2 & temperature at Ev. Outlet = c. This line intersects 5 at 11 h 12 = h 11 + h 8 h 9 = = h 12 = 1590 Kj/Kg q = (f-1) (h2 h3) = 2.15 ( ) q = 2.15 (320) q = 688 q = 688 = f (h 1a h 4 ) = 3.15 (h 1a 30) h 1a = 248 qo = Ref. Effect = h 11 h 10 = qo = 1120 qh = Heat added in gen. cop = cop = = h 5 + h 2 (f-1) fh 1a = x x 248 = = qh = 2359 qo qh qa +qh = Heat given = qo + qh = 3479 qc = heat rejected in condenser = h 5 h 8 = qc = 1380

39 qa = Heat rejected in Absorber = h 12 h 3 + f(h 3 h 4 ) = (280 30) = x 250 = qa = qa + qc = Total heat rejected = qa + qc = qo + qh = Total Heat given = qo +qh = 3479 KJ/Kg This qa +qc qo +qh Heat Balance matches well. ACTUAL VAPOUR ABSORPTION CYCLE & ITS REPRESENTION ON ENTHALPY COMPOSITION DIAGRAM. Figure 2 shows the schematic arrangement of the actual vapour absorption cycle and figure 3 represents its thermodynamic cycle on the h - diagram. The system consists of generator G together with analyzer AN, and condenser C on the high pressure side, and the evaporator E and Absorber A on the low pressure side. Pump P, extension valve VI, and

40 pressure reducing valve VII separate the two sides. In addition, liquid-vapour heat exchanger HEI is also provided. The vapour s at 5 distilled from the generator analyzer, enter the refrigerant circuit. The vapour s are condensed to 8 in the condenser, pre cooled to 9 in the liquid vapour regenerative heat exchanger and throttled to 10 before entering the evaporator. The state 10 is at the same point as state 9 in the h- diagram as both enthalpy and composition remain the same before and after throttling. The refrigerant entering the evaporator at 10, leaving the evaporator at 11, and the liquid vapour heat exchanger at 12, comprise a liquid plus vapour mixture. The refrigerant as finally absorbed by the poor solution at 2 returning from the generator and after being cooled in the liquid liquid heat exchanger to 3 and throttled to 3a whereas the rich solution from the absorber at 4 is pumped to 4 a and heated to 1a before entering the analyzer. PLOTTING STATE POINTS The state points 1,2,3&4 can be located on the h- diagram according to their temperatures and compositions as the enthalpy of liquid is independent of pressure. Also point 3a lies at 3 only (isenthalpic process) and point 4a lies approximately at 4 itself as the pump works very small. State point 5 of the vapour is along the isothermal tie line drawn from 1. point 8 is the saturation state at P k. Point 9 after sub cooling of the liquid can be plotted according to the temperature and composition and point 10 is at 9 itself (isenthalpic process). Point 11 is on the tie line corresponding to the evaporator leaving temperature t 02 = t 2 and pressure P 0.the composition is same at 7,8,9,10,11 and 12. Point 12 can be similarly located by knowing the temperature from the energy balance of heat exchanger. The absorber pressure in the absorber system is equal to the evaporator pressure Po & the generator pressure is similarly equal to condenser pressure Pk. Hence at a given generator pressure, the poor solution concentration is determined by the heating temperature t h and at a given absorber pressure, the rich solution concentration is determined by the cooling temp. ta the absorber temperate Sequence of procedure to calculate the performance of vapour absorption (3 fluid) refrigerator 1) To find the coefficient of performance c.o. p. = Heat absorbed Heat supplied 2) To find the enthalpy balance in the process HEAT ABSORBED + HEAT SUPPLIED HEAT REJECTED IN

41 CONDENCER + HEAT REJECTED IN ABSORBER I) To ascertain these two aspects of performance following data is essential 1) Generator temperature 4) Temperature after condensation 5) Temperature after sub cooling 6) Temperature at evaporator inlet 8) absorber inlet temperature II) properties of NH 3 ( R 717 refrigerant ) 1) pressure saturation temperature.table III) ENTHALPY - H / CONCERTRATION CHART ( h = K - cal / Kg ) (ξ = 0.1 to 1.0) Procedure to plot the process recurrence on the h ξ chart And obtain the values of h 1, h 2, h 3, h 4, h 8, h5 h 10 v,h 11 h 9, h 10, h 12, h ia I) obtain the partial pressure of NH 3 in condenser and boiler pressure P k as the pressure at saturation Temperature = condenser outlet temperature (T 4 ) ii) obtain the partial pressure of NH 3 in Evaporator & absorber Pressure P o as the pressure at saturation Temperature = absorber inlet Temperature iii) Mark the point 2 at P k = NH 3 pressure in condenser and boiler (generator Temperature T 1 ) note ξ a h 2 iv) Mark point 4 at P o = NH 3 pressure in evaporator Absorber and Absorber inlet temp. T 8 and note ξ of h 4 v) Mark point 3 at P o and ξa & note h 3 vi) Mark point 1 at P k and ξa & note h 1 vii) Mark point 8 at P k and temp. after condensation T 4 and note h 8, ξ 8 viii) mark point 9 at ξ 8 and temp. after sub cooling T 5 and Note h 9 iv) Mark point 10 at point 9 and note h 10 x) Mark 10 v at ξ = 1 and P k xi) join 10 v to 9 and draw a line to meet pressure P o and get the temp.. at inlet to evaporator and mark 10 L XII) Join 10 v to 11 L at P o and temp at inlet to Evaporator XIII) Where this line 10 v 10 L interserts the line ξ 8 8 is the point 11 and make h 11 XIV) Find the specific rich solution rate = f

42 f ξ ξ = 8 a ξ r ξ a XVI) Find the specific poor solution rate = (f 1) XVII) Mark point 5 on P k at E 8 and mark h 5 XVIII) Refere the Schematic diagram and find h 1a from equation qh = h5 + h2 ( f 1) fh1 a q = ( f i)( h2 h3 ) = f ( h1 a h4 ) XIX) Heat added in the generator = qh XX) Heat absorbed in Evaporator = q o q o = h 11 h 10 XX) Total heat supplied = q o + qh XXI) h 12 = h 11 + (h 8 h 9 ) qa = h h + f h ) 12 3 ( 3 h4 XXII) XXIII) Heat rejected in absorber = qa XXIV) Heat rejected in condenser = q C q C = h 5 h 8 XII) XIII) Total heat rejected = qa + q C Compare the total heat supplied = qo +qh with the total heat rejected = qa +q C and These should be almost the equality of qo +qh = qa + q C showing the energy +balance

43 VORTEX TUBE APPARATUS INTRODUCTION : George J. Ranque a French metallurgist discovered in 1928, that the temperature of central and peripharal layers of air in a cyclone dust separator are different, the core having low temperature. This led Ranque to develop the vortex tube which is often called the Ranque tube, vortex means fluid under rotation. Further development work was done by Hilsch from 1933 onwards. In 1946 he published a remarkable means of refrigeration. The Vortex Tube consists of a) Nozzle cum chamber b) Hot end tube fitted with c) Diaphragm and Cold end. OPERATION OF VORTEX TUBE :- Compressed air at about 7 atmospheres of pressure and at room temperature is admitted into vortex tube. The air attains a higher velocity while passing through the nozzle and its temperature drops. This air then enters the chamber where vortex is formed. The core has a higher static temperature than the outer layer and hence heat flows core to periphery. The outer layer gets heated due to heat transfer as well as due to viscous and wall friction effects. The length of the tube has a optimum value at which maximum cooling of the core results. The cold air at the core comes out from the cold end through the diaphragm while hot air escapes through the valve side. The cold end temperature decreases as the inlet pressure increases. The temperature drop between hot and cold sides is controlled by the valve position for a given inlet air pressure. It has been observed that for favorable cooling effect the diaphragm should be as near to the nozzle as possible.

44 APPARATUS The vortex tube test apparatus consists of a vortex tube with air inlet with pressure gauge and pressure transmitter. Air inlet is provided with pressure regulator. Hot air outlet is provided with balance valve for cooling effect. Two separate rotameters are provided to measure air flowrate of cold and hot air.differential pressure transmitter are placed in the air steam for air flow measurement. system. Temperature scanner with computer interface are provided at diffetent locations of the SPECIFICATIONS 1) Vortex Tube : Nominal capacity : 0.25 m 3 /min 2) Inlet Pressure : 0 10 Kgf/cm 2, Gauge and pressure transmitter. 3) Heat exchanger : Concentric tube, counter flow, heat transfer area 0.03 m 2 4) Pressure regulator : controls inlet pressure 5) Flowmeters : To measure cold and hot air mass flow rates. 6) Valves : 1) For isolating and pressure control. 2) For varying the mass ratio. 7) Digital temperature indicator with serial interface.