Wednesday, October 18, 2017

MEC554 UITM PULAU PINANG LABORATORY REPORT (HVAC)

Heating Ventilating Air Conditioning (HVAC)

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Common Report
Individual Report

Discussion
No process
In this section, the temperature in this system does not changed where temperature is 25.2°C. This is because no process occurs in this section. There is no factor contributes to change the air property changes in this process. We, conclude that all the value readings recorded at temperature in and out are all in the room temperature
Heating process
After the experiment is conducted, we clearly can take the reading temperature at fan inlet with 25.2°C (dry) and 21.3°C (wet) on test Ref. 1. After steady condition is achieved, the temperature was raised to 36°C (dry) and 29.5°C (wet). Based on this experiment, the temperature increase is not consistent because of the factor that contributes to heating process. The air temperature reached to maximum value of 36°C (dry) and 29.5°C (wet) but the relative humidity value slightly down compare to relative humidity at fan inlet. This shows that the moisture is decreasing upon the heating process. The air relative humidity values drop from 70% to 64% (dry).
Cooling process
Based on the experiment, at this section, the compressor of refrigerator system is switched on and the process called cooling process. By referring to the result, we can determined that the temperature of air decreasing from 36°C (dry) and 29.5°C (wet) to 27°C (dry) and 26.5°C (wet) respectively. The data proved that the cooling process decrease temperature value. The air relative humidity is increasing from 64% to 98%. This shows that the moisture is increasing upon cooling process.
Humidification process
For this section, the water vapor or moisture are added to the air without changing the temperature. The temperature still has changed after humidification process that decreased and increased from started to second section. Based on the theoretical, humidification process does not affect the temperature but in experiment, the temperature still changed. The value of air humidity in this section also has changed but in controlled value. Thus, the percentage error  between theoretical and the experiment data also in controlled value.




Dehumidification
Dehumidification is a psychometric process where the water is removes from the air as the air temperature falls below the dew point temperature. The air relative humidity decreased from 98% to 68%. This shows us that moisture percentage in the air drop in dehumidification process. In addition, the temperature of the air is increased from 27°C to 36°C. This prove that temperature different involves in dehumidification process.
Based on the result that we have calculated, the specific work done for test ref. 1 are 35kJ,14kJ and 2kJ for section B, section c, section D respectively. Next, for actual work done are 336.78kJ, 836.72kJ and 769.33kJ for section B, section C and D respectively. Power input for test ref. 1 is 187.2W and the efficiency is 97.73%.
Next, for the test ref. 2 the specific work done are 337.75kJ, 835.77kJ and 769.34kJ for section B, section C and section D. Besides, the actual work done is the same as test ref. 1 because we have used average value for test ref. 1 and test ref. 2. Lastly, the power input is 204.1W and the efficiency of the process is 97.63%.


Conclusion
As the conclusion, we can understand the air conditioning system. After the experiment was conducted, the most importance role is heat and moisture in the air-conditioning system. Air conditioning is the cooling of indoor air for thermal comfort. In a broader sense, the term can refer to any form of cooling, heating, ventilation, or disinfection that modifies the condition of air. An air conditioner is an appliance, system, or machine designed to stabilize the air temperature and humidity within an area (used for cooling as well as heating depending on the air properties at a given time), typically using a refrigeration cycle but sometimes using evaporation, commonly for comfort cooling in buildings and motor vehicles. (Ustudy.in)
There are some advantages of air-conditioning system; (infomechy)
  • Improved working efficiency, reduce stress and absenteeism
  • Increased sales, productivity, comfortless
  •  Less cleaning materials and furniture last longer
  • Suitable environment for plants growth, process, electronic and laboratory areas
  • Reduce internal and external pollution
  • Less risk of fire or damage due to static electricity caused by dry air
From the result of the experiment, the value of mass flow rate of air and vapor mixture is higher than the value of mass flow rate of the dry air. It shows that the flow rate of air and vapor mixture is much more used in the air conditioning unit. During the heating process, the highest temperature can be obtained is 32.53°C. It is because human cannot resist in a high temperature that more than around 33°C since the normal temperature of human body is 37°C. The temperature of cooling process, humidification process and dehumidification process that obtained during the experiment is in the range from 19°C to 27.6°C. If compared with the theoretical analysis, there was a difference that maybe occurred by an error during the experiment. Maybe the air conditioning unit was not being checked regularly by the schedule maintenance and the reading should be taking at least three times for a better or accurate result




References
1. (n.d.). Retrieved from Ustudy.in: http://www.ustudy.in/node/1206
2. (n.d.). Retrieved from infomechy: http://infomechy.blogspot.my/2012/11/lab-report-sand-testing-preparation-gfn_4.html
3. Thermal Engineering. (2013). Singapore: Mc Graw Hill Education.
4. R.S.Khurmi, J.K.Gupta. (2007). Refrigeration and Air Conditioning. Ram Nagar, New Delhi: Eurasia Publishing House.
5. Yunus A.Cengel, Michael A.Boles. (2011). Thermodynamics An Engineering Approach. 1221 Avenue of Americas, New York, NY10020 : McGraw-Hill.



MEC554 UITM PULAU PINANG LABORATORY REPORT (HEAT CONDUCTION)

HEAT CONDUCTION COMMON REPORT
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LIST OF CONTENTS

No.
Content
Pages
1.
Table of Content
1
2.
Introduction
2
3.
Objectives
3
4.
Theory
3
5.
Apparatus
5
6.
Procedure
7
7.
Result
8

















INTRODUCTION
Heat is defined as the form of energy that can be transferred from one system to another as a result of temperature difference. Transfer of energy as heat always from the higher temperature to the lower temperature.
Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Thermal conduction is the transfer of heat from hotter to cooler parts of a body resulting in equalizing of temperature. It is a result of a direct energy transfer between particles such as molecules, atoms, and electrons.
The rate of heat conduction through a medium depends on the geometry of the medium, its thickness, material of the medium and temperature difference across the medium. In solids, it is due to combination of vibrations of the molecules in a lattice and the energy transport by free electrons.
The basic law of thermal conduction is called Fourier’s Law of heat conduction after J. Fourier which states that the heat flux (a heat per unit area q/A) resulting from thermal conduction is proportional to the temperature gradient:

Where,
q = heat flow vector, (W)
k = thermal conductivity (W/m℃)
A = cross-sectional area of the conduction, (m2)
 = gradient of temperature (K/m.K)
    
The relation indicates that the rate of heat conduction in a given direction is proportional to the temperature gradient in that direction. Heat is conducted in the direction of decreasing temperature and the temperature gradient becomes negative when temperature decreases with increasing x.
OBJECTIVES
·       To investigate the thermal conductivity of brass in linear direction.
·       To investigate Fourier’s Law for linear conduction of heat along a simple bar.


THEORY
If a plane wall of thickness (∆X) and area (A), supports a temperature difference (∆T) then the heat transfer rate per unit time (Q) by conduction through the wall is found to be:
= kA
where,
 = heat flow rate, (W)
k = thermal conductivity of the material, (W/mK)
A = cross-sectional area of the conduction, (m2)
dT = changes of temperature between 2 points, (K)
dx = changes of displacement between 2 points, (m)

If the material of the wall is homogeneous and has a thermal conductivity (k) then,
 = A
Heat flow is positive in the direction of temperature fall.

Figure 1:Linear Temperature Distributions.

















APPARATUS

Text Box: 4Text Box: 3Text Box: 1Text Box: 2

Text Box: 6Text Box: 5                                                                                                                            


1.     Water supply
2.     Heater power regulator
3.     Temperature indicator
4.     Heater power indicator
5.     Thermocouple connectors
6.     Brass sample

                                                                                                                                 
Figure 1 : Brass sample
















PROCEDURE
Procedure
Start-up
1. The main switch was determined off. An intermediate section was inserted into linear
module and clamped together.
2. The cooling water tubes were connected to water supply and the drain.
3. The heater supply lead for the linear module was connected to power supply at control
panel.
4. Nine sensor leads were connected to nine plugs on top off the linear conduction module
and another exits at control panel labeled TT1, TT2 until TT9.
5. The water supply was determined on to ensure the water flowing.
6. The heater power control knob was set to 0 Watts position by turning it anticlockwise.
7. The main switch was turned on and the equipment was ready for the experiment.
Experiment
1. The heater power control knob was set to 10 Watts. Temperature was taken after 5
minutes to warm up the heater.
2. After 10 seconds, the temperature selector was turned to No 2 and the temperature was
taken.
3. Repeat step No. 2 until temperature selector No. 9
4. Then, the heater power control knob was set to 20 Watts. Temperature was taken after 5
minutes to warm up the heater.
5. After 10 seconds, the temperature selector was turned to No 2 and the temperature was
taken.
6. Repeat step No. 2 until temperature selector No. 9
7. Lastly, the heater power control knob was set to 20 Watts. Temperature was taken after 5
minutes to warm up the heater.
8. After 10 seconds, the temperature selector was turned to No 2 and the temperature was
taken.
9. Repeat step No. 2 until temperature selector No. 9
Shut-down
1. The heater power control knob was turned to 0 Watts and water supply was flowing for 5
minutes.
2. Main switch was turned off and then the power supply cable was unplug

Start-up
1.     The main switch was determined off. An intermediate section was inserted into linear module and clamped together.
2.     The cooling water tubes were connected to water supply and the drain.
3.     The heater supply lead for the linear module was connected to power supply at control panel.
4.     Nine sensor leads were connected to nine plugs on top off the linear conduction module and another exits at control panel labeled TT1, TT2 until TT9.
5.     The water supply was determined on to ensure the water flowing.
6.     The heater power control knob was set to 0 Watts position by turning it anticlockwise.
7.     The main switch was turned on and the equipment was ready for the experiment.
Experiment
1.     The heater power control knob was set to 10 Watts. Temperature was taken after 5 minutes to warm up the heater.
2.     After 10 seconds, the temperature selector was turned to No 2 and the temperature was taken.
3.     Repeat step No. 2 until temperature selector No. 9
4.     Then, the heater power control knob was set to 20 Watts. Temperature was taken after 5 minutes to warm up the heater.
5.     After 10 seconds, the temperature selector was turned to No 2 and the temperature was taken.
6.     Repeat step No. 2 until temperature selector No. 9
7.     Lastly, the heater power control knob was set to 20 Watts. Temperature was taken after 5 minutes to warm up the heater.
8.     After 10 seconds, the temperature selector was turned to No 2 and the temperature was taken.
9.     Repeat step No. 2 until temperature selector No. 9
Shut-down
1.     The heater power control knob was turned to 0 Watts and water supply was flowing for 5 minutes.
2.     Main switch was turned off and then the power supply cable was unplug.




RESULT
Section
Heater
Heater
Heater
Brass
Brass
Brass
Cooler
Cooler
Cooler
X (mm)
0
10
20
30
40
50
60
70
80
X (m)
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Test
Q (w)
T1
 (°C )
T2
 (°C )
T3
 (°C )
T4
 (°C)
T5
 (°C )
T6
(°C )
T7
(°C )
T8
(°C )
T9
(°C )
A
10
36.5
36.0
35.6
32.7
32.4
32.1
27.6
27.5
27.1
B
20
55.2
54.3
52.9
46.8
45.4
44.2
31.1
30.3
29.3
C
30
78.5
77.0
74.8
64.0
61.3
59.1
35.7
34.1
32.3


Table 1 : Result on temperature vs heater power