Experiment No 8 DP 2

Experiment No. 8

TEST OF A COMPRESSION REFRIGERATING PLANT: COEFFICIENT OF PERFORMANCE (COP)

Course Code: MEP510L2Program: MECHANICAL ENGINEERING

Course Title: ME LABORATORY 3Date Performed: AUGUST 18, 2015

Section: ME51FA1Date Submitted: SEPTEMBER 1, 2015

Leader: 1. ARCAY, MICHAEL JOHN ROLAND Instructor: ENGR. NELSON DELA PENA

Members: 2. ALUAN

3. VASQUEZ, RHINE STEPHEN M.

1. Objective:

The activity aims to demonstrate the operation of an air conditioning unit.

2. Intended Learning Outcomes (ILOs):

The students shall be able to:2.1 Explain fundamental principles of refrigeration system.2.2 Operate and test the air conditioning unit2.3 Collect and measure the required data for testing and calculations.2.4 Use the proper reference materials, tables and charts in making an engineering representations and diagrams.2.5 Develop professional work ethics, including precision, neatness, safety and ability to follow instruction.

3. Discussion:

Before any evaluation of the performance of a refrigeration system can be made, an effectiveness term must be defined. The index of performance is not called "efficiency", however, since that term is usually reserved for the ratio of output over the input. The ratio output to input would be misleading if applied to a refrigeration system since the output is usually wasted.The two most important results of a test of a refrigerating system are capacity and economy. Refrigerating capacity, or output, is expressed as "standard commercial tons of refrigeration produced". The economy is stated as "horsepower of compressor per standard commercial ton of refrigeration". This is based either on indicated horsepower of the compressor or on the brake horsepower required to drive it. usually both values are determined. The economy is sometimes stated in other terms also, depending largely on the type of the driving unit.

Other important test results are the volumetric efficiency, the mechanical efficiency and the compression efficiency of the compressor. The philosophy of the performance index of the refrigeration cycle is the same as efficiency, however, in that, it represents the ratio:Amount of desired commodity/Amount of expenditureThe performance term in the refrigeration cycle is called the "coefficient of performance". It is defined as: Useful Refrigeration Coefficient of Performance = Net Work of Compression

The two terms which make up the coefficient of performance must be in the same units, so that the coefficient of performance is dimensionless.

Air Conditioning Unit

4. Materials and Equipment:

Air Conditioning Unit Log sheets Personal protective equipment Sling Psychrometer

5. Procedure:

1. Locate and label the pressure gages and temperature gages of the air conditioning trainer for proper recording of test results.

2. Check the flow meters, pressure gages, and temperature gages.

3. Open all the valves then switch on the compressor motor.

4. Get the corresponding pressure and temperature readings at the high and the low sides respectively.

5. During the duration of the test, get the air properties inside the room and at the condenser area.

6. Evaluate the COP of the system.

7. After filling all the data sheets, switch of the motor then clean and dry the area.

6. Data and Results:

Form 8a. Data Log Sheet from the Air Conditioning Unit

Trial 1Mass Flow rate of refrigerantPressureTemperatureEnthalpy

Suction of Compressor0.004259 kg/sec508.00 kPa10.364C408 kJ/kg

Discharge of the Compressor0.004259 kg/sec1914.146kPa72.086C450 kJ/kg

Condenser Exit0.004259 kg/sec1914.146kPa49.362C262.379kJ/kg

Evaporator Entrance0.004259 kg/sec508.00 kPa0.634C262.379kJ/kg

Trial 2

Suction of Compressor0.00508 kg/sec473.539kPa8.486C407kJ/kg

Discharge of the Compressor0.00508 kg/sec1865.896kPa75.343C441kJ/kg


Evaporator Entrance0.00508 kg/sec473.539kPa-1.514C247.759kJ/kg

Trial 3

Suction of Compressor0.004736kg/sec445.97kPa6.6817C410 kJ/kg

Discharge of the Compressor0.004736kg/sec1528.46kPa72.2799C445kJ/kg

Condenser Exit0.004736kg/sec1528.46kPa39.85C280.57kJ/kg

Evaporator Entrance0.004736kg/sec445.97kPa-3.318C280.57kJ/kg

Trial 4

Suction of Compressor0.005075 kg/sec370.146kPa1.240C407 kJ/kg




Trial 5


Discharge of the Compressor0.004965kg/sec1779.814kPa77.85C447 kJ/kg



Trial 6Mass Flow rate of refrigerantPressureTemperatureEnthalpy





Trial 7

Suction of Compressor0.005061 kg/sec473.539kPa8.486C407kJ/kg




Trial 8


Discharge of the Compressor0.004736kg/sec1528.46kPa72.2799C445kJ/kg



Trial 9

Suction of Compressor0.005013 kg/sec370.146kPa1.240C407 kJ/kg




Trial 10





Average


Discharge of the Compressor0.004788kg/sec1779.814kPa77.85C447 kJ/kg



7. Computation, Analysis and Interpretation of Data:

COMPUTATION:TRIAL 1:Air Out = 13.33CAir In = 22.7CLS = 59psi + 14.7psi = 73.7 psiLS=508.00 kPaHS = 263psi + 14.7psi = 277.7psiHS = 1914.146kPaTwb = 16.1CTdb = 22.7CT1 = t @ Pin + 10CT1 = 10.364Ch1= by PH-diagramh1 = 408kJ/kgT2= 69C

T2=72.086Ch2 = by PH-diagramh2 = 450kJ/kgt3 = tsat @ 1914.146kPat3 = 49.362Ct4 = tsat @ 508.00kPat4 = 0.634Ch3 = h4 = 262.379 kJ/kg

AREA:

TRIAL 2:Air Out = 9.44CAir In = 21.67CLS = 35psi + 14.7psi = 49.7 psiLS = 342.575kPaHS = 195psi + 14.7psi = 209.7psiHS = 1445.432kPaTwb = 15.56CTdb = 21.67CT1 = t @ Pin + 10CT1 = 0.96Ch1= by PH-diagramh1 = 411 kJ/kgT2= 78C

T2=82.05473Ch2 = by PH-diagramh2 = 453kJ/kgt3 = tsat@ 1445.432kPat3 = 37.5903Ct4 = tsat @ 342.575kPat4 = -10.9607C

AREA:

TRIAL 3:Air Out = 13.33CAir In = 22.7CLS = 50psi + 14.7psi = 64.7psiLS=445.96kPaHS = 207psi + 14.7psi = 221.7psiHS = 1528.146425kPaTwb = 16.11CTdb = 22.7CT1 = t @ Pin + 10CT1 = 6.6817Ch1= by PH-diagramh1 = 410kJ/kgT2= 69C

T2=72.2799Ch2 = by PH-diagramh2 = 445kJ/kgt3 = tsat @ 1528.146kPat3 = 39.8553Ct4 = tsat @ 445.97kPat4 = -3.3183C

AREA:


T2=82.04Ch2 = by PH-diagramh2 = 448kJ/kgt3 = tsat@ 1810.754kPat3 = 46.968Ct4 = tsat @ 342.575kPat4 = -8.759Ch3 = h4 = 259.079 kJ/kg

AREA:

TRIAL 5:Air Out = 13.33CAir In = 22.7CLS = 59psi + 14.7psi = 73.7 psiLS=508.00 kPaHS = 263psi + 14.7psi = 277.7psiHS = 1914.146kPaTwb = 16.11CTdb = 22.7CT1 = t @ Pin + 10CT1 = 10.364Ch1= by PH-diagramh1 = 408kJ/kgT2= 69C

T2=72.086Ch2 = by PH-diagramh2 = 450kJ/kgt3 = tsat @ 1914.146kPat3 = 49.362Ct4 = tsat @ 508.00kPat4 = 0.634Ch3 = h4 = 262.379 kJ/kg

AREA:


T2=82.05473Ch2 = by PH-diagramh2 = 453kJ/kgt3 = tsat@ 1445.432kPat3 = 37.5903Ct4 = tsat @ 342.575kPat4 = -10.9607C

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AVERAGE COEFFICIENT OF PERFORMANCE (COP):

Average P-H Diagram

8. Conclusion and Recommendation:

Conclusion: In operating an air conditioning unit, the fan exhausts the air from the indoor environment and conveys it through the air conditioning system. A chemical refrigerant is being compressed in the compressor from low pressure, low temperature to high pressure, high temperature then the compressor pumps it through the condenser coil. The refrigerant now passes a series of piping, and the cooler outside air passes across the coil. The air absorbs heat from the refrigerant which causes the refrigerant to condensate from a gas to liquid state then passes through the expansion valve. When the refrigerant enters the evaporator it evaporates, and absorbs heat from the surrounding air and produces cooled air.The results shown at the previous tables shows the thermodynamic properties of air based on the trials made by the group. The computed coefficient of performance (COP) is determined to represent the performance of the air-conditioning system.

Recommendation: Make sure the instrument to be used is calibrated and in good condition before proceeding with the experiment. Take your time reading the pressure and temperature at the high and low sides.

PICTURES:

Taking the Suction and discharge Getting the wet and dry bulb temperature. temperature of the compressor

Measuring the air velocity

9. Assessment Rubric:

T I P - V P A A 0 5 4 D Revision Status/Date:0/2009 September 09TECHNOLOGICAL INSTITUTE OF THE PHILIPPINESRUBRIC FOR LABORATORY PERFORMANCECRITERIABEGINNER1ACCEPTABLE2PROFICIENT3SCORE

Laboratory Skills

Manipulative SkillsMembers do not demonstrate needed skills.Members occasionally demonstrate needed skills.Members always demonstrate needed skills.

Experimental Set-upMembers are unable to set-up the materials.Members are able to set-up the materials with supervision.Members are able to set-up the material with minimum supervision.

Process SkillsMembers do not demonstrate targeted process skills.Members occasionally demonstrate targeted process skills.Members always demonstrate targeted process skills.

Safety PrecautionsMembers do not follow safety precautions.Members follow safety precautions most of the time.Members follow safety precautions at all times.

Work Habits

Time Management/Conduct of ExperimentMembers do not finish on time with incomplete data. Members finish on time with incomplete data.Members finish ahead of time with complete data and time to revise data.

Cooperative and Teamwork Members do not know their tasks and have no defined responsibilities. Group conflicts have to be settled by the teacher.Members have defined responsibilities most of the time. Group conflicts are cooperatively managed most of the time.Members are on tasks and have responsibilities at all times. Group conflicts are cooperatively managed at all times.

Neatness and OrderlinessMessy workplace during and after the experiment.Clean and orderly workplace with occasional mess during and after the experiment.Clean and orderly workplace at all times during and after the experiment.

Ability to do independent workMembers require supervision by the teacher.Members require occasional supervision by the teacher.Members do not need to be supervised by the teacher.

Other Comments/Observations:TOTAL SCORE

RATING= x 100%

Documents

Experiment No 8 DP 2