A Major Application Area Of Thermodynamics Engineering Essay

A Major Application Area Of Thermodynamics Engineering stressA major application atomic number 18a of thermodynamics is infrigidation, which is the transfer of vex from lower temperature region to a juicyer(prenominal) temperature mavin. The devices that produce refrigeration are c tot exclusivelyyed iceboxs, and the rounds on which they operate are called refrigeration takecycles. The most frequently use refrigeration cycle is a desiccation- contraction refrigeration cycle in which the cold is bollixified and compressed alternatively and is compressed in the vapor phase. There are number of refrigerants which sewer be used in here, but the most commonly used on a commercial scale is a R12 (used in this investigate as surface).The thermodynamics of ideal desiccation conglutination cycle clear be analyzed on a temperature versusentropy draw as depicted in Figure 1. At heighten 1 in the diagram, the circulating refrigerant en-ters the compressor as a saturated v apour. From prefigure 1 to point 2, the vapour is isentropicallycompressed (i.e., compressed at constant entropy) and exits the compressor as a super lovingness up va-pour. From point 2 to point 3, the super combusting trunked vapour travels through part of the sign upr which removes the super erupt by cooling the vapour. Between point 3 and point 4, the vapour travels through the remainder of the capacity and is condensed into a saturated unruffled. The condensation fulfil betides at essentially constant insistency.Between points 4 and 5, the saturated liquid refrigerant passes through the expansion valve( curb device) and chthoniangoes an abrupt decrease of twinge. This process results in the adia-batic flash evaporation and auto-refrigeration of a portion of the liquid (typically, less than halfof the liquid flashes). The adiabatic flash evaporation process is isenthalpic (i.e., occurs at con-stant henry).Figure 12 Temperature Entropy diagram1 www. wikipedia.org/wi ki/refrigeration2 http//upload.wikimedia.org/wikipedia/commons/f/f7/RefrigerationTS.pngUMAR DARAZ knave 3 of 22 Thermodynamics Lab 2Between points 5 and 1, the cold and partially vaporized refrigerant travels through the coil or tubes in the evaporator where it is totally vaporized by the warm air (from the space existence refrigerated) that a fan circulates across the coil or tubes in the evaporator. The evaporator operates at essentially constant contract. The resulting saturated refrigerant vapour returns to the compressor inlet at point 1 to complete the thermodynamic cycle.The area under the process curve on T-s diagram represents the heat transfer for internally reversible processes. The area under the process curve 5-1 represents the heat absorption in the evaporator, the area under the process 2- 4 represents the heat rejection in the capacitor. In the ideal vapour coalescence refrigeration cycle all the heat injuryes and disruptions are cosmos ignored, but in cer tain refrigeration cycle, we need to take these losses into consideration as they have been mentioned in this report later.The Hilton refrigeration laboratory exclusively R714 is capable of following entities Investigation of the variation in refrigerator duty or cooling ability for various condens-ing temperature and the heat delivered to the cooling water supply with variation in condensingtemperature. We hind end excessively investigate the variation in refrigeration coefficient of per-formance for the various condensing temperature. Investigation of the variation in coefficient of performance based on electrical, shaft andindicated power, determination of the overall heat transfer coefficient for the condensercooling coil and performance of the thermostatic expansion valve. Investigation of the heat delivered to the cooling water with variation in condensing tem-perature, coefficient of performance as a heat pump for various condensing temperature,as well(p) as power arous al based on electrical, shaft and indicated power.The important aspect of this report is to raise the two laws of thermodynamics i.e.first and second law of thermodynamics. The first law is simply an expression of the conservation of competency ruler, and it asserts that energy is thermodynamic property.Qout = Wnet + Qin Equation (1)In this taste the Qin is provided by stimulant drug voltage, this input is used to do the net work done on the refrigerant by compressor and force, and the result of this produces the heat which is existence outback(a) by the condenser i.e. Qout.The second law of thermodynamics asserts that energy has quality and quantity, and actual processes occur in the commissioning of decreasing quality of energy.UMAR DARAZ Page 4 of 22 Thermodynamics Lab 2Aims and objectives The Hilton refrigeration laboratory unit R712 has been designed to allow students to respectabley investigate the performance of a vapour compression cycle under various conditions o f evaporator load and condenser pressure.The main objectives of this laboratory are listed below The demonstration of application of the premier and second law of thermodynamics. The introduction of to refrigeration go down and calculate the various coefficient of perform-ance. Investigation of system losses, this includes motor, compressor, evaporator and con-denser losses. These losses (friction, heat losses) occur only in working/commercial refrigerator, thither are no losses in ideal vapour compressor refrigerator.UMAR DARAZ Page 5 of 22 Thermodynamics Lab 2ApparatusThe bod shown below looks corresponding a refrigeration laboratory unit R712 ( non exactly it) and it consists of the following componentsFigure 23 Refrigeration laboratory unitPanel High quality glass reinforced plastic on which the following components are mounted. Refrigerant R12Digital Thermometer A device that measures temperature.Wattmeter Allows measurement of the power input to every evaporator or motor. Voltage Controller To vary evaporator load.Variable Area Flow meters Variable area types to power R12 and H2O hang up rates. rack Gauges To indicate R12 pressure in evaporator and condenser.Spring Balance and Tachometer These two to calculateher allow measurement of power required to drive the compressor.Expansion Valve Thermostatically masteryled type i.e. restrain device. Evaporator Electrically modify device i.e. heat exchangerCompressor (Internally mounted) Twin cylinder belt driven unit, on with take shape balance force system. electrical capacity A device or unit used to condense vapor into liquid. It is also called heat exchanger.Motor A machine that converts electricity into a mechanical motion.3 www.p-a-hilton.co.uk/English/Products/ Refrigeration__2_/refrigeration__2_.htmlUMAR DARAZ Page 6 of 22 Thermodynamics Lab 2Procedure4In prior performing an experiment the most important things to do are, to measure the atmos-pheric pressure, which would be added to the gauge pressure to get an absolute pressure for bothcondenser and evaporator, and to balance the two tips of the spring balance force, being appliedon the compressor. In failure to do these things would cause a sufficient inwardness of error in thefinal results.In this experiment the condenser pressure is being kept constant i.e. 900KPa.Step-1 Turn on the refrigeration plant apply one of the control breakers, and setting the evaporator voltage i.e. 40 vitamin C volts, at the same era balancing the two tips of compressor load and set the condenser pressure to 900KPa, using rota-meter.Step-2 Record the following set Evaporator Amps (1-2.42A), from wattmeter, compressor speed using tachometer, water and refrigerant flow rate using flow meter.Step-3 Record the spring balance force, reading directly from the scale.The hot water is in the tubes is indicated by red and cold water is indicated by blue sign in the refrigeration plant.Step-4 The flow rate is controlled by a throttling device (valve), the small changes in possibility and closing the valve, effect the condenser pressure.Step-5 The temperature valuates of the refrigerant at different branchs in the whole cycle at constant pressure is given by temperature dialler. straightway we had all the values we needed, now we changed evaporator Amps value, preserve rest of the values as mentioned sooner and repeated the whole experiment for leash to four times.The Refrigeration Laboratory Unit has ternary controls. Firstly a have miniature circuit breaker and switch turns on both the compressor motor and the supply to the electrically heated evaporator. A combined variable area water flow meter and valve allow control of the condenser pressure and a panel mounted voltage controller allows control of the evaporator load from zero to full power.Refrigerant R12 vapour is drawn into the compressor from the evaporator mounted on the frontof the panel. Work is done on the gas in the compressor and its pressure and t emperature areraised. This hot, high pressure gas discharges from the compressor and flows into the panelmounted water cooled condenser, where heat is removed from it. This liquid then flows througha thermostatic expansion valve. Here it passes through a controlled orifice, which allows itspressure to fall from that of the condenser to that of the evaporator. The refrigerant has a satu-rated vapour phase at this point. The voltage across the heater elements may be varied from zeroto that of the mains supply voltage by adjustment of a voltage controller situated on the frontpanel. Measurement of the power is carried out by a panel mounted digital wattmeter.4 http//www.p-a-hilton.co.uk/R714-Edition-2-GREY.pdfUMAR DARAZ Page 7 of 22 Thermodynamics Lab 2ResultsThe thoughtfulness table below shows all the values of different components in the refrigerationplant along with input indices and output indices, henry of the cycle and losses in the system.The calculations required to get thos e results (to complete the table) are also listed after this tablebelow.1 Condenser pressure (gauge) Pc KNm-2 900 900 9002 Evaporator pressure (gauge) Pe KNm-2 -20 20 403 Condenser pressure (Abs) Pc KNm-2 1001.663 1001.663 1001.6634 Evaporator pressure (Abs) Pe KNm-2 81.663 121.663 141.6635 Compressor suction t1 0 C -23.5 -22.6 -5.26 Compressor delivery t2 0 C 59.9 68.5 69.47 gas leaving condenser t3 0 C 31.6 34.8 33.88 Evaporator inlet t4 0 C -32 -23.6 -19.19 Water inlet t5 0 C 23.8 21.6 21.410 Water outlet t6 0 C 41.2 38.6 39.511 Water flow rate Mw g s-1 1.5 5.0 6.012 R 12 Flow rate Mr g s-1 0.7 1.5 1.913 Evaporator Volts Ve V 40 70 10014 Evaporator Amps Ie I 1 A 1.70 A 2.42 A15 Motor Volts Vm V 235 232 23216 Motor Amps Im A 3.6 3.6 3.617 Spring balance Force F N 5.5 7.5 8.218 Compressor speed nc rpm 477 474 473UMAR DARAZ Page 8 of 22 Thermodynamics Lab 219 Motor promote = 3.17 - nc Nm rpm 1512.09 1502.58 1449.7120 h1 KJ/Kg 340 345 36021 h2 KJ/Kg 385 400 42022 h3 KJ/Kg 225 240 2 5023 h4 KJ/Kg clx 170 18024 Qe,Elec = Ve - Ie W 40 119 24225 Qe, R 12 = Mr(h1 h4) W 126 262.50 34226 Wc = 0.0172-F-Nm W 143.043 193.832 204.46727 Power factor at shaft (power Wc) pf 0.43 0.48 0.5228 Wm = Vm. Im. pf W 363.78 400.89 434.3129 Wc = Mr (h2 h1) W 31.5 82.50 114.030 Q cond = Mr (h2 h3) W 112 240 32331 Qw = Mw - 4.18 (t6 t5) W 109.09 376.20 428.8732 catchnet = Qe, Elec / Wm 0.109 0.296 0.55733 nail down R12 = (h1 h4)/(h2 h1) 4.0 3.1818 3.0034 t41 can be found by (t1 t4) 0 C 8.5 1.00 13.935 CoP (te-t2) = t41 / (t2-t41) 0.165 0.015 0.25036 Motor loss = Wc Wm W -220.73 -207.06 -229.8437 Compressor loss = Wc-Wc W -111.54 -110.33 -90.4738 System loss = Qcond Qw W 2.91 -136.20 -105.8739 System loss = Qe, R12 Qe,Elec W 86 143.50 100.0UMAR DARAZ Page 9 of 22 Thermodynamics Lab 2Figure 3 A graph represents the relationship amongst net CoP and evaporator temperatureFigure 4 A comparison of different losses of the system in one graph against Evaporator temperatureThe flu ctuation and entropy in the graphs is because of the poor calibration and lessnumber of repeated results (less tests provide less information), and most of the recorded results are based on guessed values.CalculationsTo hap absolute pressure, we need an atmospheric and gauge pressure of the component. instanter for two individual components, CondenserAs we know Patm = gh = 13600 kg/m3 - 9.81 m/s2 - 762 -10-3m= 101.663-103 Kg / ms2 = 101.663 KN/m2Hence Pgauge,cond = 900 KN/m2Pabs,cond = Patm + Pgauge,cond = 101.663 + 900 = 1001.663 KN/m2 EvaporatorAs Patm = gh = 13600 kg/m3 - 9.81 m/s2 - 762 -10-3m= 101.663-103 Kg / ms2 = 101.663 KN/m2i. Pgauge,Evap = -20 KN/m2Pabs,Evap = Patm + Pgauge,EvapTherefore = 101.663 + (-20)= 81.663 KN/m2ii. Pgauge,Evap = 20 KN/m2Pabs,Evap = Patm + Pgauge,Evap = 101.663 + (20)= 121.663 KN/m2iii. Pgauge,Evap = 40 KN/m2Pabs,Evap = Patm + Pgauge,EvapTherefore = 101.663 + 40 = 141.663 KN/m2To find Qw (Heat removal from condenser)As we repeated the experiment three times, so water flow rate have three different values, hence we need to find Qw at three points,Qw = Mw - 4.18 (t6 t5)When Mw = 1.5 gs-1, t6 = 41.2 0C, t5 = 23.8 0CQw = 1.5 -4.18 (41.2 23.8) = 109.098 WUMAR DARAZ Page 11 of 22 Thermodynamics Lab 2As Qw = Mw - 4.18 (t6 t5)When Mw = 5.0 gs-1, t6 = 39.6 0C, t5 = 21.6 0CSo Qw = 5.0 -4.18 (39.6 21.6) = 376.2 WQw = Mw - 4.18 (t6 t5)When Mw = 6.0 gs-1, t6 = 38.5 0C, t5 = 21.4 0CQw = 6.0 -4.18 (38.5 21.4) = 428.87 WTo find Wc (work done by the compressor or a shaft loss)The work done by the compressor depends on spring balance force and motor speed, hence to get more work done out of the compressor we need to increase any of the supra mentioned parameters. ThereforeWc = 0.0172-F-Nmi. Wc = 0.0172-5.5-1512.09 = 143.043 Wii. Wc = 0.0172-7.5-1502.58 = 193.832 Wiii. Wc = 0.0172-8.2-1449.71 = 204.467 WTo find Wm (work done by the motor on a shaft to rotate)The work done by the motor is a product of voltage provided, amount of topica l flowing the motor and power factor of the shaft, which gives us the following valuesWm = Vm - Im - pfi. Wm = 235 - 3.6 - 0.43 = 363.78ii. Wm = 232 - 3.6 - 0.48 = 400.89iii. Wm = 232 - 3.6 - 0.52 = 434.31UMAR DARAZ Page 12 of 22 Thermodynamics Lab 2To find CoPnet (Total coefficient of performance of refrigerant)CoPnet = Qe, Elec / WmBy substituting different values of electric input heat energy (artificial input energy) and the work done by the motor, we get net coefficient of performance of the cycle,i. CoPnet = 40 / 363.78 = 0.109 = 11%ii. CoPnet = 119 / 400.89 = 0.296 = 30%iii. CoPnet = 242 / 434.31 = 0.557 = 56%To find CoP (te-t2)This is the coefficient of performance of ratio of temperature values at point 1-4 and difference of it, to the temperature of the refrigerant after compression, so we get followingCoP (te-t2) = t41 / (t2-t41)i. CoP (te-t2) = 13.9 / (69.4 13.9) = 0.250 = 25%ii. CoP (te-t2) = 8.5 / (59.9 8.5) = 0.165 = 16%iii. CoP (te-t2) = 1.0 / (68.5 1.0) = 0.015 = 1.5%To find Qe, R 12(Heat removal from Evaporator) The given equation is Qe, R 12 = Mr (h1 h4)By substituting different values of atomic number 1, which we recorded from a pressure total heat diagram, so we geti. Qe, R 12 = 0.7 (340 160) = 126.0ii. Qe, R 12 = 1.5 (345 170) = 262.5iii. Qe, R 12 = 1.9 (360 180) = 342.0UMAR DARAZ Page 13 of 22 Thermodynamics Lab 2To find Wc (Input work done or compressor work loss)The input work done by the compressor can be calculated by finding flow rate of the refrigerant R12 and the difference of heat content of refrigerant before and after the compression.Wc = Mr (h2 h1)Substituting all three values of the above parameters (variables), we geti. Wc = 0.7 (385 340) = 31.5ii. Wc = 1.5 (400 345) = 82.5iii. Wc = 1.9 (420 360) = 114To find Q cond (Heat loss by the condenser)Similarly heat loss by the condenser is a product of refrigerant flow rate to the difference of enthalpy values of it, before entering and leaving the condenser, we getQ cond = Mr (h2 h3)Now, using above stated equationi. Q cond = 0.7 (385 225) = 112ii. Q cond = 1.5 (400 240) = 240iii. Q cond = 1.9 (420 250) = 323To find CoPR12 (Coefficient of performance of refrigerant) CoP R12 = (h1 h4)/(h2 h1)Coefficient of performance of refrigerant is a ratio of all the enthalpy values in the cycle, here note that for ideal vapour compression refrigeration cycle h3 = h4Hence we geti. CoP R12 = (340 160) / (385 340) = 4.00ii. CoP R12 = (345 170) / (400 345) = 3.1818iii. CoP R12 = (360 180) / (420 360) = 3.00UMAR DARAZ Page 14 of 22 Thermodynamics Lab 2Systems lossesMotor loss = Wc Wm= 143.043 363.78 = -220.75= 193.832 400.89 = -207.06= 204.467 434.31 = -229.84Compressor loss = Wc-Wc= 31.5 143.043 = -111.54= 82.5 -193.832 = -110.33= 114 204.467 = -90.47System loss = Qcond Qw= 112 109.09 = 2.91= 240 376.20 = -136.20= 323 428.87 = -105.87System loss = Qe, R12 Qe,Elec= 126 40 = 86.00= 262.5 119 = 143.50= 342 242 = 100.00UMAR DARAZ Page 15 of 22 Thermodynamics Lab 2Discussion of ResultsThe observation table of results has been listed on page 8 9, and it is followed by all the calculations required to complete the table or to get the results.The experiment has been repeated three times, so all the results (values have been listed threetimes. In the calculation section the system losses and heat energy are shown as negative val-ues, its because the work is done on the system and heat is being removed from that particu-lar system, in this case its condenser. The positive values of system loss and heat energyshows that heat is being add in the system and work is done by the system, and in this case itsevaporator. The condenser pressure i.e.900 KPa, was not exactly 900 KPa. As we were set-ting the pressure manually, so in the whole experiment the pressure was 900 KPa 10%, itwas because of the fluctuation in the gauge needle, so we assumed the considered pressure.The compressor pressure applied by spring balance force, a ffected the work done of the com-pressor on the refrigerant R12, because to get an immaculate compressor work done, the two tipsof the spring balance should be in balance (level), but during an experiment we were acquiringrandom values (results), so then I realised that something is wrong, so I looked at all thecomponents of the refrigeration plant, and I found that the two tips of the spring were not bal-ance. Hence to get right results we had to reconstruct the experiment. The throttling device or valvehas a huge impact on condenser pressure, because by opening or closing i.e. changing a flowrate make a considerable amount of difference on condenser pressure and evaporator tem-perature.Motor loss refers to the consumption of electrical energy not converted to useful mechanicalenergy output, but in this case energy loss mover less input energy to the compressor, whichmeans a refrigerant would be less compressed by a compressor, so less heat would be re-moved by the condenser, an d even after extremely through the valve the refrigerant would stillhave a high temperature and pressure, hence less refrigeration would occur in a vapour com-pression cycle. Therefore we need to take into account power losses in the electric motor.In order to study this process more closely, refrigeration engineers use this pressure en-thalpy diagram shown in Figure 5. This diagram is a way of describing the liquid and gasphase of a substance. hydrogen can be thought of as the quantity of heat in a given quantity,or mass of substance. The curved line is called the saturation curve and it defines theboundary of pure liquid and pure gas, or vapour. In the region marked vapour, its pure va-pour. In the region its marked liquid, it is a pure liquid. If the pressure rises so that we areconsidering a region above the top of the curve, there is no distinction between liquid and va-pour. Above this pressure the gas cannot be liquefied. This is called the Critical Pressure. Inthe region underneath the curve, there is a mixture of liquid and vapour.UMAR DARAZ Page 16 of 22 Thermodynamics Lab 23 24 1Figure 65 Pressure Enthalpy diagramEvaporator Pressure line Condenser pressure linestage (Not a straight line) Isobar Condensation stagesion valve R12 Evaporation process5 http//www.mvsengineering.com/chapter18.pdfUMAR DARAZ Page 17 of 22Isentropic CompressionR12 passing through Expan-Thermodynamics Lab 2At the inlet of the compressor the temperature (t1) is the same as temperature of refrigerant R12 at the outlet of the evaporator. So we go straight from that temperature of go away side of the doom (saturated liquid) to the right side of the doom (saturated vapour line), and then following the temperature gradient line, we go down and record the enthalpy value at that temperature and pressure. Similarly for the stage 2, we find h2 on x-axis.When the refrigerant leaves the condenser, it obtains a saturated liquid phase (left side of thedoom), so taking the reference of condenser pressure line (red line), we take a straight lineparallel to the y-axis, and wherever it meets the x-axis gives a value of enthalpy (h3) at stagethree. In actual refrigerant plant, enthalpy at stage 3 and stage 4 is not same, but for the sakeof calculation we assume that its an ideal condition and enthalpy at these two points is same.Test 1As Compressor suction = t1 = -23.5 0C and condenser Pressure (Abs) = Pc =1001.663 KNm-2Hence the enthalpy h1 = 340 KJ/KgCompressor delivery = t2 = 59.9 0C and Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h2 = 385 KJ/KgHere Liquid leaving condenser = t3 = 31.6 0CAnd Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h3 = 225 KJ/KgAs mentioned earlier that h3 = h4 (Ideal condition) Hence the enthalpy h4 = 225 KJ/KgBut using temperature at evaporator inlet, t4 = -32 0C, we get Actual enthalpy value, h4 = 160 KJ/KgTest 2As Compressor suction = t1 = -22.6 0C and condenser Pressure (Abs) = Pc =1001.663 KNm-2 Hence the enthalpy h1 = 345 KJ/Kg (from above p-h diagram)Compressor delivery = t2 = 68.5 0C and Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence using Figure 4, we get enthalpy h2 = 400 KJ/KgHere Liquid leaving condenser = t3 = 34.8 0CAnd Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h3 = 240 KJ/KgUMAR DARAZ Page 18 of 22 Thermodynamics Lab 2As mentioned earlier that h3 = h4 (Ideal condition) Hence the enthalpy h4 = 240 KJ/KgBut using temperature at evaporator inlet, t4 = -23.6 0C, we get Actual enthalpy value using figure 4, h4 = 170 KJ/KgTest 3As Compressor suction = t1 = -5.2 0C and condenser Pressure (Abs) = Pc =1001.663 KNm-2Hence the enthalpy h1 = 360 KJ/KgCompressor delivery = t2 = 69.4 0C and Condenser Pressure (Abs) = Pc = 1001.663 KNm-2 Hence the enthalpy h2 = 420 KJ/KgHere Liquid leaving condenser = t3 = 33.8 0CAnd Condenser Pressure (Abs) = Pc = 1001.663 KNm-2, Evaporator Pressure = 40 KPa Hence the enthalpy h3 = 250 KJ/KgAs mentioned earlier tha t h3 = h4 (Ideal condition) Hence the enthalpy h4 = 250 KJ/KgBut using temperature at evaporator inlet, t4 = -19.1 0C, we get Actual enthalpy value at this stage, h4 = 180 KJ/Kg6However the expansion of the high pressure liquid, process 5 1 above is non reversible.Notice that Expansion is a constant enthalpy process. It is drawn as a vertical line on the P-hdiagram. No heat is absorbed or rejected during this expansion, the liquid just passes througha valve, like water coming out of a tap. The difference is that because the liquid is saturated atthe start of expansion by the end of the process it is partially vapour. Point 1 is inside the curveand not on the curve as described in the Evaporation process. At point 4 it starts to condenseand this continues until point 5 when all the vapour has dark into liquid. Point 5 is saturated liquid. If more heat is removed, the liquid cools. It is then called sub-cooled liquid. Hence h4 is on a saturated liquid line (left side of the doom), a nd does not appear in a vapour compression cycle, and this is the case in all three tests.6 http//www.alephzero.co.uk/ref/vapcom.htmphUMAR DARAZ Page 19 of 22 Thermodynamics Lab 2As there is no moving part in the whole refrigeration plant apart from motor shaft of a compressor, so work done by them is zero, i.e. w = 0So using steady state energy equation, we getW Q = h2 h1 Equation (2)As W =0, so equation (1) becomes Q = h2 h1Or Q = h1 h2 Equation (3)The coefficient of performance or COP (sometimes CP), of a heat pump (i.e. refrigerator) isthe ratio of the change in heat at the output (the heat reservoir of interest) to the suppliedwork.To find Cop value of refrigeration plant as well as for the refrigerant is a total practice, because this will illustrate that how much efficient of these two are.7It takes a lot of heat to evaporate liquid. In other words a small amount of liquid circulatingin a refrigerator can perform a large amount of cooling. This is one reason why the vapo urcompression cycle is widely used. The refrigeration system can be small and compact. Alsofrom a practical point of view heat exchange is much better when using change of state evaporation and condensation. However the expansion of the high pressure liquid, process 5 1 above is non reversible. And so the capability of this cycle can never even approach Carnotefficiency.7 http//www.alephzero.co.uk/ref/practcop.htmUMAR DARAZ Page 20 of 22 Thermodynamics Lab 2Conclusion8The vapour-compression cycle is used in most household refrigerators as well as in manylarge commercial and industrial refrigeration systems but the efficiency of this cycle cannever even approach Carnot efficiency, because of its low coefficient of performance.In the refrigeration plant the operating parameters can be varied by adjustment of condensercooling water flow and electrically heated evaporator supply voltage. Components have a lowthermal mass resulting in immediate response to control variations and rapid s tabilisation.Instrumentation includes all relevant temperatures, condenser pressure, evaporator pressure,refrigerant and cooling water flow rates, evaporator and motor power, motor torque and com-pressor speed.The most of components of refrigeration plant used in this experiment (R712) are manually calibrated scales (not digital), and based on this poor calibration all the recorded results are being guessed on the base of individual judgment, which is wrong most of the time. Anyway a small amount of liquid circulating in a refrigerator can perform a large amount of cooling. This is one reason why the vapour compression cycle is widely used.The enthalpy values which are being recorded directly from enthalpy pressure diagram (Figure 4), and based on how unclear that diagram is, I would say it is not a great source of information, but still we use this to find enthalpy. The system (refrigeration plant) has some losses, which have described earlier in this report, this includes motor l oss, condenser and evaporator loss.In conclusion, I would like to say that by doing this experiment I learnt a great amount ofknowledge, about refrigeration plant, and how it works, what mental of cycle more often use forthis, how much efficient is this and how to calculate the different losses in this system. I wouldsay by understanding the operation of this small scale refrigeration plant, I forecast I would beable to operate on an industrial scale refrigeration plant, because the basic principle is same.8 http//www.alephzero.co.uk/ref/vapcomcyc.htmUMAR DARAZ Page 21 of 22 Thermodynamics Lab 2