heat conduction lab report

advertisement
EXPERIMENT C2: HEAT CONDUCTION STUDY BENCH
Abstract
The aim of the experiment is to study the factor which could affect linear
conduction heat transfer. The results demonstrate that linear heat transfer
obeys Fourier's law. Different materials (conductivity) and cross-sectional area
have different flux rates. Compared with stainless steel, Brass has better heat
conductivity and paper's is tiny. A large temperature drop occurs on between
contact surfaces of test section and cooler section
Introduction
1. Linear conduction heat transfer
Conduction is a mode of heat transfer, which is the heat transfer by microscopic
collisions of particles within a body.
According to Fourier’s law, the heat flux is proportional to the temperature
gradient and opposite to it in sign. For one-dimensional heat flow,
dq/dA=-k(dT/dx) . which states that:
Q  kA
dT
dx
(Eq.1)
where,
Q=heat flux(W)
K=thermal conductivity(W/(m•K)),
A=cross sectional area (m2)
dT/dx=temperature gradient(K/m),
Only in steady state, the temperature gradient could be replaced by
temperature difference over heat flux path


=
1 −2

(Eq.2)
For composite systems with multiply layers, it is often convenient to work with
an overall heat transfer coefficient U. Accordingly,
1
Q  UAΔT
(Eq.3)
Where,
ΔT=overall temperature difference(K)
And also,
1
=

1




1
2
2
=
+
+




1
2
(Eq.4)
2
Where,
x1, x2, x3=thickness of materials (m)
k1, k2, k3=thermal conductivity for materials (W/(m•K))
2. Heat conduction through different cross-sectional area
For the continuity heat flow rate and the same conductor, Q1=Q2 and k1=k2
and from Eq.1. Following equation can be derived:
ΔT2
A1
x
 2
A2 ΔT1
x1
(Eq.5)
Objective
To demonstrate the concept of Fourier’s law that relates the rate of onedimensional heat flow to heat resistance, area and temperature gradient.
Procedure
Before experiment start, the water supply was turned on. Next, the computer
and the power supply were switched on.
Firstly, control knob was turned to 5 W, and A brass conductor installed
sensors was inserted between the heater and the cooler sections. The heat
power was set to 5W and the auto recorder was start. For a moment, when
the reading per 5 min we got was closed, the experiment was paused and the
date was saved. The brass section was taken apart and another, stainless
steel section, was inserted in test area. The date saving was repeated as
2
above. Then the heat was switched off to make sure that temperature of linear
conduction module below than 100 Celsius. After it was cool down, the
stainless was replaced with brass section whose cross section is smaller. For
a sufficient time, the temperature was recorded as above. Next, the brass
section was removed, the thicker one (25mm diameter brass section) was put
in linear module and a little piece of paper was insert between cool section
and test section. The date saving was repeated as above. After that, paper
was taken put and conducting compound was applied on the side of brass
section adjant to cooler section. The date saving was repeated again as
above. After that, the compound applied on interface was wiped off.
In next three rounds, control knob was turned to 10W,15W,20W respectively
and record the temperature as above.
Last, the heater was switched off but the water supply was kept continuing
until the water cooled down.
Results
1. Fourier’s law study for linear conduction of heat along a homogeneous bar
All data shown below (the unit of temperature is Kelvin)
Power
T1
T2
T3
T4
T5
T6
T7
T8
T9
5W
307.25 307.15 306.85 302.05 304.85 304.75 301.65 302.95 303.95
10W
323.95 323.25 323.35 318.15 317.55 317.75 303.15 302.95 302.45
15W
335.35 333.35 334.45 326.75 326.25 326.65 303.05 302.75 302.35
20W
329.95 328.35 327.25 320.65 319.95 319.75 303.95 303.45 302.65
temperature/k
Table.1 Data of Experiment 1
345 y = -477,83x + 345,11
R² = 0,8381
340
335
330
y = -385,67x + 336,61
R² = 0,9014
325
y = -312,83x + 330,37
R² = 0,858
320
315
310
3
305
y = -55,833x + 307,4
R² = 0,5052
300
295
0
0,01
5W
0,02
0,03
0,04
10W
0,05
0,06
15W
0,07
0,08
20W
0,09
0,1
Figure.1 T-x graph of Experiment 1
dT
can obtained from the graph slope. Cross section A is known which is

π(
25×10−3 2
)
2

≈ 4.91 × 10−4 2 and from equation 1 we can get k=− .

when Q=5W, then dT/dx=-55.833,k1=−
5
=182.44
25×10−3 2
) ×(−55.833)
2
π(
when Q=10W, then dT/dx=-312.83,k2=−

10
=69.34
25×10−3 2
π(
) ×(−312.83)
2
when Q=15W, then dT/dx= -477.83,k3=−
when Q=20W, then dT/dx= -385.67,k4=−
15
=68.09
25×10−3 2
π(
) ×( −477.83)
2
20
=105.62
25×10−3 2
π(
) ×(−385.67)
2

so the average thermal conductivity kavg=(k1+k2+k3+k4)/4=106.37
·
2.Conduction of heat and overall heat transfer along a composite bar
All data shown below (the unit of temperature is Kelvin)
Power, Q
T1
T2
T3
T7
T8
T9
5W
314.55 315.05 316.85
301.75 301.95 302.05
10W
327.15 327.55 329.75
302.35 302.35 302.35
15W
339.45 342.65 345.25
302.65 302.65 302.15
20W
335.45 334.05 337.15
302.95 303.25 302.35
Table.2 Data of Experiment 2
360
350
y = -611,03x + 353,02
R² = 0,8974
340 y = -505,52x + 344,48
R² = 0,9211
330
y = -395,86x + 335,04
R² = 0,9067
320
4
Figure.2 T-x graph of Experiment 2
From Eq.3:
Q  UAΔT
From Eq.4:
1



=
+
+
   
Where A=4.91 × 10−4 2 ; kbrass=106.37

·
; L=0.03m
We can get thermal conductivity of the stainless steel
ksteel= 1

−
2
 
= ∆


−
2



When Q=5W, ∆T=T1-T9=12.50K, U=814.66
, k1=45.22
·
·


When Q=10W, ∆T=T1-T9=24.80K, U=821.23
, k2=45.90
·
·


When Q=15W, ∆T=T1-T9=37.30K, U=819.03
, k3=45.67
·
·


When Q=20W, ∆T=T1-T9=33.10K, U=1230.61
, k4=120.71
·
·

So kavg=(k1+k2+k3+k4)/4=64.37
·
From internet, we can get ideal thermal conductivity of brass and common
Steel

Kthoreticalbrass=109· ,

Kthoreticalsteel=50.2·
From eq.4,we can get
5
1
Uthoretical=

=871.03·



thoretical+ thoretical+ thoretical



brass

brass
1
Uexperimental=

=970.76·



experimental+ experimental+ experimental




brass
brass

Finally,Error
error=
U theoretical  U exp erimental 871.03−970.76
871.03  750.03

 13.9%
=|
|=11.4%
theoretical
871.03
871.03
U
3. The effect of a change in cross-sectional area on the temperature profile
along a thermal conductor
All data shown below (the unit of temperature is Kelvin)
Power, Q
T1
T2
T3
T7
T8
T9
5
316.05 316.55 318.15 301.85 301.85 301.85
10
325.65 327.25 328.35 302.65 302.65 302.85
15
332.05 333.55 335.85 302.75 302.85 302.35
20
341.95 341.75 345.85 302.55 302.85 302.85
Table.3 Data of Experiment 3
6
360
y = -620,17x + 353,98
R² = 0,9057
350
340
y = -477,76x + 342,12
R² = 0,904
330
320
y = -373,1x + 333,56
R² = 0,9028
y = -230,17x + 320,89
R² = 0,8961
310
300
290
0
0,01
0,02
5W
Линейная (5W)
0,03
10W
0,04
0,05
Линейная (10W)
0,06
15W
0,07
Линейная (15W)
0,08
20W
0,09
0,1
Distance/m
Линейная (20W)
Figure.3 T-x graph of Experiment 3
Cross section A is known which is π(
13×10−3 2
)
2
when Q=5W, then dT/dx=-230.17,k1=162.11
≈ 1.34 × 10−4 2

·

when Q=10W, then dT/dx=-373.1,k2=200.02
·

when Q=15W, then dT/dx= -477.76,k3=234.30
·

when Q=20W, then dT/dx= -620.17,k4=240.67
·

so the average thermal conductivity kavg=(k1+k2+k3+k4)/4=209.27
·
4. The influence of thermal insulation upon the conduction
All data shown below (the unit of temperature is Kelvin)
Power T1
T2
T3
T4
T5
T6
T7
T8
T9
5W
317.85 317.35 317.45 305.95 311.35 311.45 302.15 302.05 302.05
10W
324.55 323.55 324.15 318.65 318.15 318.35 302.65 302.35 302.25
15W
340.55 337.25 336.15 325.85 323.75 324.15 302.85 302.75 302.35
7
20W
349.85 345.75 345.25 329.65 329.15 329.05 303.05 302.85 302.75
temperature/k
Table.4 Data of Experiment 4
360
350
340
y = -670,17x + 359,88
R² = 0,9092
y = -541x + 348,79
R² = 0,9136
330
320
y = -326,83x + 331,3
R² = 0,846
y = -223,67x + 320,92
R² = 0,7922
310
300
290
0
0,01
5W
0,02
0,03
10W
Линейная (5W)
0,04
0,05
0,06
15W
Линейная (10W)
0,07
Линейная (15W)
0,08
20W
0,09
0,1
Thickness/m
Линейная (20W)
Figure.4 T-x graph of Experiment 4
From Eq.3: Q
 UAΔT
1



From Eq.4: =
+


+


Where A=4.91 × 10−4 2 ; kbrass=106.37
+



·

; L=0.03m ;  =64.37·;
x paper  1mm  0.001m
We can get thermal conductivity of the paper
Kpaper= ∆


2

−
 
−

When Q=5W, ∆T=T1-T9=15.8K, k1=1.92
·

When Q=5W, ∆T=T1-T9=22.3K, k1=15.4
·

When Q=15W, ∆T=T1-T9=38.2K, k1=4.54
·

When Q=5W, ∆T=T1-T9=47.1K, k1=7.93
·
8

So kavg=(k1+k2+k3+k4)/4=7.45
·
5.The effect of surface contact on thermal conduction between adjacent slabs
of material
All data shown below (the unit of temperature is Kelvin)
Q
T1
T2
T3
T4
T5
T6
T7
T8
T9
5W
318.55
317.95 318.35 304.35 313.75 313.85 302.35 302.15 302.05
10W
320.85
320.15 320.75 305.25 315.65 315.85 302.45 302.25 302.15
15W
344.85
342.05 341.25 331.05 330.25 330.75 303.15 302.85 302.45
20W
343.05
342.25 340.95 329.25 327.05 327.05 303.95 303.15 302.45
temperature/k
Table.5 Data of Experiment 5
360 y = -606,17x + 355,71
R² = 0,8745
350
340
330
320
310
y = -593,17x + 354,01
R² = 0,9063
y = -257,5x + 324,58
R² = 0,6889
y = -226,5x + 321,7
R² = 0,6857
300
290
0
0,01
5W
0,02
Линейная (5W)
0,03
10W
0,04
0,05
15W
0,06
0,07
0,08
20W
0,09
Линейная (10W)
Линейная (15W)
Линейная (20W)
0,1
Thickness/m
Figure.5 T-x graph of Experiment 5
Sharp temperature drop occurred between the contact surfaces of three
sections. But compared with contact surface of heater section and test
section, the temperature drop in contact surface where the conducting
compound was applied is sharper.
Discussion

From experiment 1,the thermal conductivity of brass is 106.37·. From wiki
9

we get, the theoretical value is 109·.. The error equal 2.41%. The
experiment measured value agree well with theoretical value.

Also, from experiment 2,the thermal conductivity of stainless steel is 64.37·.

The error is 28.2% compared with theoretical value 50.2·. And the error in
overall heat transfer coefficient is 11.4%. The reason cause the error may be
the contact between test section is poor and the contact resistance could not
be neglected. And if we calculate the conductivity as the date from experiment

5, the value is 54.7· which more close to theoretical value. That supports
what we suppose above that the contact resistance could not be neglected in
experiment 2.
In experiment 3, the slope of trend line was steeper than experiment 1. It is
consistent with the Eq.1. But the thermal conduction we got is far larger than
theoretical value. From the graph, we could see the temperature increase in
heater section. That means the date we got is not under steady state and the
date is null.
In experiment 4, the thermal conductivity of paper we gained is 7.45

·
,
which is not conform to theoretical value (0.05 watts). But once consider that
the thickness should be thinned than 1mm, the error is possible.
In experiment 5, temperature drop between the contact surfaces of test and
heater sections is unreasonably sharp. I guess that the linear module didn’t
achieve steady state.
Conclusion
The experiment requires us to explore the effect of material and cross-sectional
area to the heat transfer and the influence of thermal insulation and conducting
compounds on the thermal conduction.


The thermal conductivity of brass is 106.37· and stainless steel is 64.37·.
From experiment we could see the contact resistance couldn’t be ignored
10
most of time In the reality. And Recording the data until the steady state is
achieved are necessary.
Reference
1.FOURIER'S LAW, available from:
http://www.thermopedia.com/content/781/
2. Thermal conductivity of paper
https://translate.google.com/translate?hl=zhCN&sl=en&u=https://sciencing.com/thermal-properties-paper6893512.html&prev=search
3. Properties: Stainless Steel - Grade 304 (UNS S30400) - AZoM
https://www.azom.com/properties.aspx?ArticleID=965
11
Download
Related flashcards
Create Flashcards