THERMODYNAMIC LAB: REFRIGERATION / AIR CONDITIONING CYCLE (T1) KM31401: LAB IV AZIRA BINTI MISMAN BK15110037 GROUP 3B FACULTY OF ENGINEERING, UMS
ABSTRACT
The objective of the refrigeration / air-conditioning cycle is to demonstrate vaporcompression refrigeration and to study the relationship of pressure and temperature in a refrigeration cycle. The vapor-compression cycle consists of four processes; isentropic compression, constant-pressure heat rejection, throttling in an expansion device, and constant-pressure heat absorption in an evaporator. A p-h diagram is used to determine the performance calculation of a refrigerating machine. There are t hree conditions tested on the air-conditioning unit, consisting of ambient, water heater (1 kW), and lower water heater (2 kW). The COPR obtained from the ambient, water heater and lower water heater are 3.18, 3.63 and 3.60, respectively, while the refrigeration capacities are 1507kJ/kg, 1885kJ/kg and 2385.8kJ/kg, respectively.
1.
INTRODUCTION INTRODUCTION / OBJECTIVE
The objective of this experiment is to demonstrate vapor compression refrigeration and to study the relationship of pressure and temperature in a refrigeration cycle. The A660 Air Conditioning Unit is used in this experiment. The temperature reading for ambient, water heater and lower water heater are recorded in a table. From the data obtained, P-h diagram is plotted at the chart to obtain the value of enthalpy. The value of enthalpy is required to calculate the COP of the air conditioning system and refrigerant capacity.
2.
PROCEDURES
a. The air conditioning conditioning unit is switched switched on and the air air flow is set to a convenient convenient value.
b. The orifice differential different ial pressure pressure is set to 4 mm H2O. c. When the condition is stabilized, stabilized, the following following observations observations are made, made, i. Evaporator outlet temperature ii. Condenser inlet temperature iii. Condenser outlet temperature iv. Evaporator outlet pressure (gauge) v. Condenser inlet pressure (gauge) vi. Condenser outlet pressure (gauge) vii. Orifice differential pressure, Z viii. Supply voltage, VL ix. Heater resistance d. The pre-heaters pre-heaters are switched switched on to give 1 kW kW (nominal) heating. heating. e. The procedures are repeated to give 2 kW heating, which is the lower lower water heater.
3.
RESULT AND DISCUSSION (~150 words)
TEST REF. Evaporator outlet temperature Condenser inlet temperature Condenser outlet temperature Supply volts Evaporator outlet pressure (gauge) Condenser inlet pressure (gauge) Condenser outlet pressure (gauge) Duct Differential Pressure Time interval R-134a mass flow rate
t13 t14 t15 VL p1 p2 p3 Z X ref
°C °C °C V AC -2 kNm kNm kNm mmH 2O s g/s
1 (Ambient) 12.9 78.4 47.3 225 275 1190 1175 4 180 11
2 (1 kW) 12.3 79.3 48.5 220 275 1160 1150 4 180 13
3 (2 kW) 16.9 81.2 50.4 225 325 1300 1250 4 180 15.8
The COPR and refrigerant capacity are calculated by the data obtained from the P-h diagram plotted on the psychometric chart. From the diagram plotted, the value for h 1, h2 and h 4 are obtained. Based on the data in the table, it can be said that when temperature increases, the pressure increases. However, the water heater data decreases slightly. It was supposed to increase as it goes from ambient to 1 kW and 2 kW. The errors might come from the air loss during experiment conducted since the glass on the outlet air is not closed properly, resulting in the pressure loss and drop in mass flow rate. Besides, the data display also kept on changing due to unstable conditions although we took 3 minutes interval before taking the reading. Such errors can be reduced by repeating the whole experiment a few times and also regularly maintaining the AC unit so that it will be in a good condition.
4.
CONCLUSION
As a conclusion, conclusion, the vapour compression cycle of the AC unit has been demonstrated and the relationship between temperature and pressure is defined. The COP R obtained from the ambient, water heater, and lower water heater are 3.18, 3.63 and 3.60, respectively, while the refrigeration capacities are 1507kJ/kg, 1885kJ/kg and 2385.8kJ/kg, respectively.
REFERENCES
Cengel, Y., Boles, M. A. (2001). ‘Thermodynamics ‘Thermodynamics : An Engineering Approach (Sixth Ed.) ’. Boston: McGraw-Hill Higher Education. JSRAE. (n.d.). ‘How to Draw a Refrigeration Cycle’. Retrieved on 8 th March 2018 from a website: http://www.jsrae.or.jp/jsrae/stady http://www.jsrae.or.jp/jsrae/stady/Eng%20saikuru /Eng%20saikuru.htm .htm
APPENDICES
a) Calculation Calculation of of COP for Ambient Ambient The enthalpies are obtained from the psychometric chart:-
= 407 kJ/kg, = 450 kJ/kg, = 270 kJ/kg, ̇ 11 COP for refrigeration cycle;
(407 – 270) 270) kJ/kg 137 kJ/kg (407 – ( )kJ/kg = 43 kJ/kg
Refrigeration capacity; ̇ ( ) ( RC ( ) kJ/kg = 1507 kJ/kg
b) Calculation Calculation of COP COP for water heater (1 kW) The enthalpies are obtained from the psychometric chart:-
= 415 kJ/kg, = 455 kJ/kg, = 270 kJ/kg, ̇ 13
COP for refrigeration cycle; (415 – 270) 270) kJ/kg 145 kJ/kg (415 – ( )kJ/kg = 40 kJ/kg
Refrigeration capacity; ̇ ( ) ( RC ( ) kJ/kg = 1885 kJ/kg
c) Calculation Calculation of COP COP for Lower water heater heater (2 kW) The enthalpies are obtained from the psychometric chart:-
= 421 kJ/kg, = 463 kJ/kg, = 270 kJ/kg, ̇ 15.8 COP for refrigeration cycle; (421 – (421 – 270) 270) kJ/kg 151 kJ/kg ( )kJ/kg = 42 kJ/kg
Refrigeration capacity; ̇ ( ) RC ( ) kJ/kg = 2385.8 kJ/kg
d) Psychometric chart:-
