1.0 TITLE
ANALYSIS OF AIR CONDITIONING PROCESSES
2.0 INTRODUCTION 3.0 OBJECTIVE 4.0 THEORY
1
5.0 APPARATUS
1) Computer Linked Air Conditioning Laboratory Unit (P.A Hilton)
Figure 5.1: AC575 Computer Linked Air Conditioning Unit
2) Computer and Printer
Figure 5.2: PC computer with printer
3)
Measuring container
Figure 5.3: Container and pipe
2
6.0 EXPERIMENTAL PROCEDURE
1) The unit was started by having suction fan running and the screen displaying the Master Menu. Programme 1 showed the process data displayed on a schematic layout on the system. While, programme 2 displayed the properties of the treated air on the psychrometric chart. 2) No Process – data and psychometric chart were printed to read the initial properties of the air as it entered the air-conditioning unit. 3) Sensible heatingi)
1 kW pre-heater was switched on and held for 5 minutes. Data and psychometric chart was printed out.
ii)
0.5 kW heater was then switched on and allowed for 5 minutes too. Data and psychometric chart was printed out.
iii)
For this process, temperature rise of the air was calculated at the exit.
4) Steam humidification – All water heaters were switched on to boil the water. When steam was produced, only 3 kW heater was switched on to maintain the steam and were allowed for 5 minutes. Data and psychometric chart were printed out. Then, the amount of steam produced, the change in relative humidity, and the corresponding rise of temperature were calculated. 5) Cooling and Dehumidification – Compressor and refrigeration system were switched on. The air inside were cooled until 18℃ to 20℃ (stable temperature). As soon as it reached the stable, it was then left for 5 minutes. Data and psychometric chart were then printed out. Heat rate and the amount of moisture removed from the air were also calculated. During the experiment, time and the rate of condensation (100ml) were measured from the beginning of the cooling process (stable temperature). It was then compared with analysis.
3
7.0 RESULTS AND DATA
Experiment No process Reading
Sensible Heating 1kW 1.5kW
Steam Humidification
Cooling and Dehumidification
(TA d), ℃
28.8
29.8
29.7
30.8
32.0
32.4
(TA w), ℃
24.9
25.4
25.3
26.1
26.3
26.7
(TB d), ℃
28.8
41.9
44.1
33.5
32.1
32.4
(TB w), ℃
25.3
30.4
31.2
31.3
27.1
27.5
(TC d), ℃
27.8
39.2
43.7
32.8
27.2
27.7
6 (TC w), ℃
24.6
28.2
29.4
30.3
25.2
25.6
(TD d), ℃
26.8
36.2
46.6
33.0
25.2
25.6
(TD w), ℃
27.3
29.0
31.1
31.1
29.0
29.4
9 (T1), ℃
X
X
X
28.9
29.1
(T2), ℃
X
X
X
86.9
90.8
(T3), ℃
X
X
X
36.8
37.3
(T4), ℃
X
X
X
-5.1
-4.0
, W
X
1013.1
1017.3
X
X
, W
X
0.0
541.0
X
X
, W
X
X
2814.8
X
X
X
X
94.8
104.2
X
X
X
832.1
859.4
54.4
211.2
211.0 0.0
(P1), (P3), ,
56.0
56.2
187.3
X
X
X
0.0
Time
X
X
X
16 min 32 sec
Drain Water,
X
X
X
100 ml
,
Figure 7.1: Table of data and result
4
8.0 SAMPLE CALCULATION a) No Process In
Out
TAdry = 28.8°C
TDdry = 28.8°C
TAwet = 24.9°C
TDwet = 25.3°C
b) Sensible Heating For 1kW pre-heater
State1
State 2
TAdry = 29.8 °C
TDdry = 36.2°C
TAwet = 25.4°C
TDwet = 29.0°C
Temperature rises at exit ∆Tdry = 36.2°C - 29.8 °C = 6.4°C ∆Twet = 29.0°C - 25.4°C = 3.6°C
For 1.5kW pre-heater
State 1
State 2
TAdry = 29.7 °C
TDdry = 46.6°C
TAwet = 25.3°C
TDwet= 31.1°C
Temperature rises at exit ∆TDry = 46.6°C - 29.7 °C = 16.9°C ∆Twet= 31.1°C - 25.3°C = 5.8°C
5
c) Steam Humidification
State 2
State 3
Tdry = 30.8°C
Tdry =33.5°C
Twet= 26.1°C
Twet= 31.3°C
ma= 0.0544 kg/s
ma= 0.0544 kg/s
From the psychometric chart: Ø2 = 72.5%
Ø3= 90%
ω 2=ω1= 18.0 x −kgv/kga
ω3=33.00 x 10−kgv/kga
Amount of steam
mw = ma(ω3-ω2) = 0.0915 kg/s (33.00 x 10−- 18.0 x 10−) kgv/kg =1.3725 x −kg/s
Change in relative humidity ∆Ø = Ø2- Ø1 = 0.90 – 0.725= 0.175 @ 17.5%
Temperature rises at exit ∆Tdry = 33.5°C - 30.8°C = 2.7°C ∆Twet= 31.3°C - 26.1°C =5.2°C
6
d) Cooling and Dehumidification State 1
State 2
Tdry = 32.1°C
Tdry = 27.2°C
Twet= 27.1°C
Twet = 25.2 °C
ma1= 0.0211 kg/s
ma2= 0.0211 kg/s
From the psychometric chart: Ø1= 74%
Ø2= 96.47%
ω 1= 17 x10−) kgv/kga
ω2= 11.95 x 10−) kgv/kga
h1=95.0kJ/kg
h2= 44.5kJ/kg
Amount of moisture
mw= ma(ω1-ω2) = 0.0211kg/s (17 x 10− – 1 1.95 x 10−) kgv/kga =1.0655 10−) kg/s
Amount of heat rate
hw= hf @ T2 = 17.2 from table A-4, therefore hw= ? Hf (kJ/kG) 104.83
Temperature (°C) 25 27.2 30
Hw 125.74 ℎ 104.83
27.2 25 = 125.74 104.83 3025
hw=114.02 kJ/kg Qout= ma(h – h2) - mwhw = 0.0211 (95.0 – 44.5) - (1.0655 x 10−) (114.02) = 10.643 KJ/s Rate of condensation: Rate of condensation = 0.1 liter / 512 sec = 1.95 × 10−ℓ/s 7
9.0 DISCUSSION 1. Answer all the problems in the experimental and discuss the results obtained by explaining the factors that contributes to the air property changes for each process problems. No process:
Based on the data for the first process, which is no process, we can see that there is no changes in the temperature because it does not involve with any variable yet. Even though, there is still a value for the first data but the value is just a reading taken from the surrounding condition. So, at this moment we cannot make any assumption yet and we still need more data from other variable to see clearly the relation between all the variables. Sensible Heating:
During running the sensible heating experiment we discovered that the temperature reading for both wet bulb and dry increased compare to no process experiment. The different between these two experiments is because in sensible heating experiment we use the pre-heat and re-heat to heat the air while in no process experiment there is nothing particular changes had been made. Theoretically, the pre-heat and reheat both of them is actually increasing the temperature of the air since both of them is provide heat to the air. Unfortunately, there is still some temperature drop during the air flow to the outlet. In order to solve this problem the air must be reheat to a certain temperature so that we can get the temperature that we want. Steam humidification:
Based on the data we can see that after we switch the water heater to boil the water the output temperature started to increase for both dry and wet temperature. Is obvious that the cause of the temperature to increase is because of the steam that had been produced from the water that had been boiled. It seem that the steam make the wet bulb temperature to increase more than the other processes. By using these reactions, we can take advantages from it which is by adding the air we can increase the humidity of the air whereas for the steam we can use it to increase the temperature of the air.
8
Cooling and dehumidification:
For the last experiment which is cooling and dehumidification, we switch on the compressor of the refrigeration system. From the data obtained we can see that the reading temperatures for both dry and wet bulb are the lowest from the previous experiment. This explains the idea where from the refrigeration system it will lower the temperature because there is cooling process occur at that time. At the same time the refrigeration system will produce the water vapor which we have to remove it by using the dehumidifier. So that is why the temperature drop drastic all because the process of cooling and dehumidification happens at the same time at the same places.
2. Find the schematic of a modern air-conditioning system with advanced air treatment processes and explain the function of the main devices.
9
The function of main device
The three main parts of an air conditioner unit are the compressor, the condenser and the evaporator. The compressor and condenser are located on the outside of the air conditioner and the evaporator is located on the inside. The basic functioning of the air conditioner is based on the principle of successive heating and cooling of a highly volatile liquid, such as a Freon. The liquid first will enters the compressor, where its function is to compress into a gas. This reaction will releases heat and makes the liquid cooler. After that, the dissipated heat is radiated outwards with the help of a fan. The liquid then enters the condenser, where its function is to absorb heat from the surroundings to reconvert into a gas. Hence, the surroundings temperature will become cool. The entire process continues and eventually causes of the cooling of the room Besides the three main air conditioner parts, an air conditioner also has a hot coil on the outside to dissipate heat, a cool coil on the inside to absorb heat, two fans (one outside and one inside) and a control circuit to modify the temperature. This is done b y changing the rotation speeds of the fans using a potentiometer. 3. Explain with the suitable diagrams the operation and arrangements in an automotive airconditioning unit.
10
Components of Automotive Air Conditioning
In automotive air conditioning we have several main parts that have to be focuses which are a compressor, a condenser, an evaporator, refrigeration lines and a couple of sensors. We will go through it part by part.
Compressor: Compressor is like the heart of your a/c system where it takes the refrigerant
(the gas) and pressurizes it so it will cool the air surrounding. It's run by an engine belt. The compressor also has an electrically operated clutch that turns the compressor on and off as you demands more cool air
Condenser : The condenser is like a miniature radiator, usually mounted at the front of the
car right next to the big radiator. Sometimes the condenser will have its own electric cooling fan, too. The hot, compressed air passes through the condenser and gets lots cooler. As it cools, it becomes a liquid.
Evaporator : The evaporator is another little radiator that does just the opposite task as the
condenser. When the super-cool liquid is passed through its tubes, air is forced through and gets really cold, right before it hits your face. As it warms up again, the refrigerant starts turning back into a gas.
Thermal Expansion Valve : In order to prevent our toes from freeze off, we can control it
with thermal expansion valve so that we can control the flow of super cool refrigerant to the evaporator. With this we can regulate how cold the air blowing on us. Even though, there are a few types of valves in use these days, but they all do the same thing.
Drier or Accumulator: The drier or accumulator, also known as the receiver-drier, is sort
of the safety catch for our system. The compressor is only supposed to compress the gas form of our refrigerant. But, there's always a chance that some liquid could make it back that far. The drier catches this liquid before it can damage your compressor. Since even the tiniest leak or careless installation can introduce water moisture to the system, the drier absorbs this chemically, using what's called a desiccant. The drier also has a filter that catches any gunk that might be in there.
11
10.0 CONCLUSION
Throughout of the experiment, we can conclude that the experiment is success since we manage to achieve our goal which is to observe and understand the changes in air properties as it is treated in a basic air-conditioning. We knew that the air properties are changes based on what types of process that we running it. Each types of process which is the simple heating, steam humidification, simple cooling and dehumidification have their own result and reason why it is occur like that. With that data and result we can study throughout about the air-conditioning. But still an error must be avoided in order to get the precise data.
12
11.0 REFERENCES
1. Yunus A. Cengel, (2007), Heat and Mass Transfer: A Practical Approach, Third Edition, Published by McGraw Hill. 2. Yunus A. Cengel, Michael A. Boles, (2007) Thermodynamics: An Engineering Approach, Sixth Edition (SI Units), Published by McGraw Hill. 3. Bill Whitman, Bill Johnson, John Tomczyk, Eugene Silberstein, (2007), Refrigeration and Air Conditioning Technology, Eighth Edition. 4. Thermal Physics (2nd Edition) by Charles Kittel 5. Bruce R. Munson, Ted H. Okishi, Wade W. Huebsch, Alric P. Rothmayer (2010), Fluid Mechanics, Seventh Edition, Published by John Wiley & Sons. 6. Fundamentals of Engineering Thermodynamics by Michael Moran 7. Lab sheet 3, Heat Transfer Laboratory Sheet, Provided by UITM Shah Alam.
13
12.0 APPENDIXS
14
