Heat Conduction
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Transcript of Heat Conduction
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[Mechanical Engineering Laboratory 4]
SEGi University
EXPERIMENT 10: HEAT CONDUCTION
Candidate’s Name: Shaheer Qamar
Student ID: SCM – 027739
Lecturer/Supervisor: Dr. Vinod
Date of Experiment: 23/11/2015
Date of Submission: 30/11/2015
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FACULTY OF ENGINEERING & BUILT ENVIRONMENT
SUBJECT: EME3431 LABORATORY INVESTIGATIONS 4
EXPERIMENT 10: HEAT CONDUCTION
ABSTRACT:
In the experiment carried out, the conduction of heat along a composite bar and the
overall heat transfer coefficient. The experiment was successfully carried out with the
conductive compounds of steel and brass as sample. The experiment was prone to
some systematic and random errors which contributed to some deflected readings.
The experiment showed that different material have different value of thermal
conductivity and also that the overall heat transfer coefficient is dependent to the
input power.
OBJECTIVE:
To:
- Study the conduction of heat along a composite bar.
- Evaluate the overall heat transfer coefficient.
APPARATUS:
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PROCEDURES:
1. An intermediate position for the heater power control was selected.
2. Sufficient time was allowed for a steady state condition to be achieved before
recoding the temperature (T) at all nine sensor points and the input reading on the
wattmeter (W).
3. This procedure was repeated for other input powers.
4. After each increment of power, sufficient time was allowed to achieve steady state
conditions.
5. The results were tabulated in table 1
6. Step 1 was repeated with a second test sample.
7. The steps from 2-5 were then repeated and the results tabulated in table 2.
RESULTS:
Table 1: Temperature distribution along test sample 1
Test
No.
Wattmeter Q
(watts)
T1
(0C)
T2
(0C)
T3
(C)
T4
(C)
T5
(C)
T6
(0C)
T7
(0C)
T8
(0C)
T9
(0C)
1 4 41.7 39.9 38.6 33.5 30.3 31.5 28.5 28.2 27.9
2 8 42.6 40.8 39.5 34.1 30.7 31.9 28.6 28.3 28.0
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3 12 45.2 43.2 41.8 34.9 31.1 32.4 28.6 28.3 28.0
4 16 57.3 53.4 51.0 38.4 33.0 34.8 28.7 28.3 28.0
Distance (m) 0.005 0.015 0.025 0.035 0.045 0.055 0.065 0.075 0.085
𝑇@4 𝑊𝑎𝑡𝑡 = (T 1+T 2+T 3)
3 =
(41.7+39,9+38.6)3
= 40.07 0C
𝑇@8 𝑊𝑎𝑡𝑡 = (T 1+T 2+T 3)
3 =
(42.6+40.8+39.5 )3
= 40.97 0C
𝑇@12 𝑊𝑎𝑡𝑡 = (T 1+T 2+T 3)
3= (45.2+43.2+41,8)
3= 43.4 0C
𝑇@16 𝑊𝑎𝑡𝑡 = (T 1+T 2+T 3)
3= (57.3+53.4+51)
3= 53.9 0C
𝑇@4 𝑊𝑎𝑡𝑡 = (T 7+T 8+T 9)
3 =
(28.5+28.2+27.9)3
= 28.2 0C
𝑇@8 𝑊𝑎𝑡𝑡 = (T 7+T 8+T 9)
3 =
(28.6+28.3+28.0 )3
= 28.3 0C
𝑇@12 𝑊𝑎𝑡𝑡 = (T 7+T 8+T 9)
3=
(28.6+28.3+28.0 )3
= 28.3 0C
𝑇@16 𝑊𝑎𝑡𝑡 =(T 7+T 8+T 9)3
= (28.7+28.3+28.0 )
3= 28.33 0C
Watt (W) THS - Average (K) TCS - Average (K) THS - TCS (K)
4 313.07 301.2 11.87
7 313.97 301.3 12.67
12 316.4 301.3 15.1
16 326.9 301.33 25.57
𝑈 = Q
A (THS−TCS)
Where;
Q = heat transfer rate (W)
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U = Overall heat transfer coefficient (W/(m²·K))
A = Heat transfer surface area (m2)
T = Temperature (K)
- Diameter of the sample = 0.025 m
- Surface Area of sample = 𝜋𝑟2 = 0.00049 𝑚2
𝑈@4 𝑊𝑎𝑡𝑡 = 4
0.00049(11.87) = 687.722 W/(m²·K)
0.005 0.015 0.025 0.035 0.045 0.055 0.065 0.075 0.0850
10
20
30
40
50
60
70
f(x) = − 1.875 x + 42.7194444444444
Temperature vs length
4 watt Linear (4 watt) 8 watt12 watt 16 watt
Length
Tem
pera
ture
oC
Values to find thermal conductivity for Steel
Q = 4 Watt
dTdx = -1.875 Km-1
𝑘 = −Q
A (dTdx
)
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𝑘 = −4
0.00049(−1.875) = - 4353.74 W m∙ −1 K ∙ −1
Table 2: Temperature distribution along test sample 2
Test
No.
Wattmeter Q
(watts)
T1
(0C)
T2
(0C)
T3
(0C)
T7
(0C)
T8
(0C)
T9
(0C)
1 4 54.3 51.8 50.2 28.5 28.2 28.0
2 8 53.7 51.3 49.6 28.5 28.2 28.0
3 12 54.0 51.4 49.8 28.5 28.2 28.0
4 16 55.7 52.8 50.9 28.5 28.2 28.0
0.005 0.015 0.025 0.065 0.075 0.0850
10
20
30
40
50
60
f(x) = − 6.4 x + 62.5666666666667
Temperature vs length
4 watt Linear (4 watt) 8 watt12 watt 16 watt
Length
Tem
pera
ture
oC
Values to find thermal conductivity for Brass
Q = 4 Watt
dTdx = -6.4 Km-1
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𝑘 = −Q
A (dTdx
)
𝑘 = −4
0.00049(−6.4 ) = - 1275.5 W m∙ −1 K ∙ −1
DISCUSSION:
For different materials, the thermal conductivity varies. It is because for each and
every material the atomic configuring varies and thus the heat transfer. For example
Heat moves very quickly through a metal spoon, leaving one end of a spoon in boiling
water will make the entire spoon hot very quickly. The entire spoon becomes hot, not
just the spot in the boiling water. On the other hand, heat moves very slowly through
the insulation in your house. This fact can also be seen from the above calculations
since the brass and steel have different thermal conductivities.
The experiment carried out was prone to both systematic and random errors. In the
experiment carried out, the percentage errors for the overall heat transfer coefficient
‘U’ lies beyond acceptable limit. However, the heat conductivity ‘k’ for brass and
steel were quite accurate. The reason for deflected reason may have been;
The shallow shoulder in the nylon housing were not perfectly matched.
The surface of the conductive compounds in contact were not very smooth thus
causing contact resistance to add in.
The adhesive used to attach the conductive compound too may have added to the
contact resistance.
The wattmeter scale was very sensitive and was thus difficult to set it at the
desired exact power input
The temperatures from the conductive compounds fluctuated too much.
The following improvements could be made to the experimental procedure to
improve the overall accuracy and the precision of the experiment.
Make sure that the shallow shoulder in the nylon housing is perfectly matched
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Use a conductive compound that is smooth and less adhesive to reduce the
contact resistance.
The instrument should have a selector to set the sensitivity of the device as
desired.
Perform the experiment thrice and take the average of the temperatures.
REFERENCES:
i) Adam Zofka. (2011). Methods of Heat Transfer. Available:
http://www.physicsclassroom.com/class/thermalP/Lesson-1/Methods-of
HeatTransfer.
ii) Baumeister, Theodore; Marks, Lionel S., eds. (1967), Standard Handbook for
Mechanical Engineers (Seventh ed.), McGraw-Hill.
iii) Charlie Dalton. (2012). CONDUCTION ALONG A SIMPLE BAR. Available:
http://www.engr.iupui.edu/~mrnalim/me314lab/lab01.html.