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
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
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