Embed Size (px)
Citation preview
EKC 291
CHEMICAL ENGINEERING LABORATORY I
SEM II, 2010 - 2011
EXPERIMENT NO. 9
EXPERIMENT TITLE: HEAT CONDUCTION STUDY BENCH
GROUP NO: GROUP 4
GROUP MEMBERS: KAN CHEE KIT (104710)
: NURUL ASYIKIN BINTI SHUBANRIO (104753)
: NURUL DARSANI BINTI AMAT DARBIS (104754)
LECTURE-IN-CHARGE: Dr. SUZYLAWATI ISMAIL
DATE OF EXPERIMENT: 18 JANUARY 2011
DATE OF SUBMISSION: 1 FEBRUARY 2011
1
TABLE OF CONTENTS
TITLE PAGES
FRONT PAGE 1
CONTENTS 2
OBJECTIVE 3
THEORY 3
PROCEDURE 5
DATA, CALCULATION and
GRAPH
6
DISCUSSION 14
CONCLUSION 15
RELEVANCE to INDUSTRY 16
MODIFICATIONS
SUGGESTED
16
APPENDIX 17
REFERENCES 19
OBJECTIVE
2
1. Objective of EXPERIMENT 1 is to:
Determine the heat conduction along a homogenous bar.
2. Objectives of EXPERIMENT 2 are to:
Determine the heat conduction along a composite bar or wall.
Evaluating the overall heat transfer coefficient.
THEORY
Experiment 1
When a temperature gradient exists in a body, there is an energy transfer
from high temperature region to low temperature region.
We can observed that the heat energy is transferred by conduction, Qis
directly proportional to area A, butvaries inversely with ∆x. Q is also
directly proportional to temperature gradient, ∆T, the collective effect is
then
Q∝ A dTdx
In changing the material, we would find that the above proportionality
remains valid. This suggests that the proportionality may be converted to
equality by introducing a coefficient that is a measure of the material
behavior. Hence we write
QA
=k dTdx
k=QA
1
( dTdx )where k = conduction heat transfer coefficient or thermal conductivity of
material (W/m.K)
Q/A = heat transfer rate per solid cross section area (W/m2)
dT/dx= temperature gradient in the direction of the heat flow
3
Heat transfer rate, Q=−kaverage A ( dTdx )
Experiment 2
The rate of heat transfer per unit area flows through successive slabs is
the same. Hence from Fourier’s Law,
QA
=kHT HS−T HIX H
=kST HI−T CIX S
=kCT CI−T CSXC
From which it follows that
1U
=XHkH
+XSkS
+XCkC
Where U = overall heat transfer coefficient (W/m2K)
X = length of each bar (m)
k = heat transfer coefficient of each bar (W/mK)
QA
=U (T HS−T CS )or
QA
=Uaverage (T 1−T 2)
Where T = temperature (K)
Therefore U is overall heat transfer coefficient for the composite wall and
1/U = Σ Resistances.
4
This figure shows the schematic diagram of composite wall of material, H,
S, and C in order to determine the overall heat transfer coefficient U for
the wall.
PROCEDURE
Experiment 1
An intermediate section that contains the homogeneous bar with silicon applied
to the surface of both sides is inserted into the linear module and clamped together.One
of the water tubes is connected to the water supply and the other to drain. The heater
supply lead for the linear conduction module is connected into the power supply socket
on the control panel. The 9 sensors leads are connected to the 9 plugs on top of the
linear conduction module. The left-hand sensor lead is connected from the module to
the place marked TT1 on the control panel. This procedure is repeated for the
remaining 8 sensor leads, connecting them from left to right on the module and in
numeral order on the control panel. The AUTO/OFF switch on the electrical console is
checked to be in the OFF position. The water supply is turned on and then water is
ensured flowing from the free end of the water pipe to drain. The heater power control
knob in the control panel is rotated to the fully anticlockwise position.The main
AUTO/OFF switch is set to the AUTO position when the digital reading outs are
illuminated. The D-shell 9 pin connector is connected from the control panel to the
computer’s com 2.The computer is switched on and loading the heat conduction
experiment software. When assembling the sample between the heater and the cooler,
care is taken to match the shallow shoulders in the housings. The temperature
measurement points are aligned along the longitudinal axis of the unit.
Experiment 2
The procedures are the same as Experiment 1, but the six sensor leads are
connected to the plugs on the top of the linear conduction module (TT1, TT2, TT3, TT7,
TT8 and TT9). The left hand sensor lead from the module is connected to the place
marked as TT1 on the control panel. This procedure is repeated for the other sensor
leads by connecting them from left to right on the module and in numeral order on the
control panel.
5
Safety for this experiment:
Turn OFF the power supply to the control panel before connecting the
heater power plug.
Never splash water to the control panel this will cause body injury and
damage to the equipment.
Never use bare hand to test the AC power supply. It may cause
hazardous injury.
Do not operate heat conduction bench where explosive vapors or
flammable material exist.
DATA,CALCULATION and GRAPH
Experiment 1
Experiment with Homogenous Bar (Brass) with diameter 25mm
Table 1(i): Data from experiment
Tes
t
no.
Wattmeter
, Q (watts) TT1⁰C TT2⁰C TT3⁰C TT4⁰C TT5⁰C TT6⁰C TT7⁰C TT8⁰C TT9⁰C
A 5 43.9 44.3 44.4 29 29 28.9 28.8 28.6 28.6
B 10 62.7 63.2 63.3 29.7 29.7 29.5 29.1 28.9 28.8
C 15 87.5 88.1 88.4 30.9 30.7 30.5 29.5 29.2 29.1
D 20 110.9 111.2 111.4 32 31.8 31 30 29.8 29.3
Length of
Homogenous bar
(m) 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
6
Graph 1: Graph of Temperature versus Length of Homogenous Bar (Brass
d=25mm)
0 2 4 6 8 10 120
2
4
6
8
10
12
f(x) = NaN x + NaNR² = 0f(x) = NaN x + NaNR² = 0f(x) = NaN x + NaNR² = 0f(x) = NaN x + NaNR² = 0 Graph of Temperature versus Length of Homogenous Bar (Brass
d=25mm)
Test A= 5 WattLinear (Test A= 5 Watt)Test B= 10 WattLinear (Test B= 10 Watt)Test C= 15 WattLinear (Test C= 15 Watt)Test D= 20 WattLinear (Test D= 20 Watt)
Length of Homogenous Bar (cm)
Te
mp
era
ture
(C
)⁰
Table 1(ii): Data for calculation Thermal conductivity, k and Heat transfer rate, Q
Test
no.
Wattmeter, Q
(watts)
dT/dx,
(K/m)
Thermal
conductivity, k
(W/mK)
Heat Transfer Rate, Q
(Watt)
A 5 -232.6 43.79145546 4.326642324
B 10 -511.8 39.80419124 9.520101211
C 15 -880.8 34.69309448 16.38394909
D 20 -1224 33.28723052 22.76788566
Average Thermal Conductivity, k 37.89399292
7
Experiment 2
Experiment with Composite Bar (Brass) with diameter 13mm
Table 2A (i): Data from experiment
Tes
t
no.
Wattmeter,
Q (watts)
TT
1⁰C TT2⁰C TT3⁰C
TT
4⁰C
TT
5⁰C
TT
6⁰C
TT
7⁰C
TT
8⁰C
TT
9⁰C
A 5 100.2 100 99.9 30.1 29.8 29.7
B 10 102.5 102.3 101.7 30.3 29.9 29.8
C 15 106.7 106.7 106.8 30.4 30.1 29.8
D 20 122 122.3 122.4 30.7 30.3 30
Length of
Composite Bar (m) 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
Graph 2A: Graph of Temperature versus Length of Composite Bar (Brass
d=13mm)
0 2 4 6 8 10 120
2
4
6
8
10
12
Graph of Temperature versus Length of Composite Bar (Brass d=13mm)
Test A= 5 Watt
Test B= 10 Watt
Test C= 15 Watt
Test D= 20 Watt
Length of Composite Bar (cm)
Te
mp
era
ture
(C
)⁰
8
Table 2A (ii): Data for calculation Thermal conductivity, kS
Tes
t
no.
Wattmeter,Q
(watts)
(y₂-y₁),
K
(x₂-x₁),
m
(dT/dx)S,
(K/m)
Thermal
conductivity,
kS (W/K)
A 5 -69.8 0.04 -1745 21.58723262
B 10 -71.4 0.04 -1785 42.20697023
C 15 -76.4 0.04 -1910 59.16710093
D 20 -91.7 0.04 -2292.5 65.72688493
Average Thermal Conductivity, kS 47.17204718
Average (x₂-x₁) 0.04
Table 2A (iii): Data for calculation Thermal conductivity, kH
Test
no.
Wattmeter,Q
(watts)
(y₂-
y₁), K
(x₂-x₁),
m
(dT/dx)H,
(K/m)
Thermal
conductivity,
kH (W/K)
A 5 -0.3 0.02 -15 2511.314729
B 10 -0.8 0.02 -40 1883.486046
C 15 0.1 0.02 5 -22601.83256
D 20 0.4 0.02 20 -7533.944186
Average Thermal Conductivity, kH -6435.243992
Average (x₂-x₁) 0.02
9
Table 2A (iv): Data for calculation Thermal conductivity, kC
Test
no.
Wattmeter,Q
(watts)
(y₂-
y₁), K
(x₂-x₁),
m
(dT/dx)C,
(K/m)
Thermal
conductivity,
kC (W/K)
A 5 -0.4 0.02 -20 1883.486046
B 10 -0.5 0.02 -25 3013.577674
C 15 -0.6 0.02 -30 3766.972093
D 20 -0.7 0.02 -35 4305.110963
Average Thermal Conductivity, kC 3242.286694
Average (x₂-x₁) 0.02
Table 2A (v): Data for calculation Overall Heat Transfer Coefficient, U
XS/kS XH/kH XC/kC 1/U
Overall Heat
Transfer
Coefficient, U
(W/m2K)
0.0008479
6 -3.10789×10-06 6.16849×10-06
0.0008510
2 1175.059953
Table 2A (vi): Data for calculation Heat Transfer Rate, Q
Test no. Wattmeter, Q (Watt) Heat Transfer Rate, Q (Watt)
A 5 10.99579777
B 10 11.33892905
C 15 11.99399786
D 20 14.34912617
10
Experiment with Composite Bar (Stainless Steel) with diameter 25mm
Table 2B (i): Data from experiment
Tes
t
no.
Wattmete
r, Q
(watts)
TT
1⁰C
TT
2⁰C
TT
3⁰C
TT
4⁰C
TT
5⁰C
TT
6⁰C
TT
7⁰C
TT
8⁰C
TT
9⁰C
A 5 83.3 83.1 82.9 30.1 29.8 29.7
B 10 86.7 86.6 86.6 30.3 29.9 29.9
C 15 101.2 101.3 101.5 30.7 30.2 29.8
D 20 123 123.1 123.1 31.2 30.6 30.2
Length of
Composite Bar
(m) 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
Graph 2B: Graph of Temperature versus Length of Composite Bar (Stainless steel
d=25mm)
11
0 2 4 6 8 10 120
2
4
6
8
10
12
Graph of Temperature versus Length of Composite Bar (Stain-less Steel d=25mm)
Test A= 5 Watt
Test B= 10 Watt
Test C= 15 Watt
Test D= 20 Watt
Length of Composite Bar (cm)
Te
mp
era
ture
(C
)⁰
Table 2B (ii): Data for calculation Thermal conductivity, kS
Tes
t
no.
Wattmeter,
Q (watts)
(y₂-y₁),
K
(x₂-
x₁), m
(dT/
dx)S,
(K/m)
Thermal
conductivity,
kS (W/K)
A 5 -52.8 0.04 -1320 7.716585257
B 10 -56.3 0.04 -1407.5 14.47373718
C 15 -70.8 0.04 -1770 17.26422464
D 20 -91.9 0.04 -2297.5 17.73387167
Average Thermal Conductivity, kS 14.29710469
Average (x₂-x₁) 0.04
Table 2B (iii): Data for calculation Thermal conductivity, kH
Tes
t
no.
Wattmeter
, Q (watts)
(y₂-y₁),
K
(x₂-x₁),
m
(dT/dx)H,
(K/m)
Thermal
conductivity,
kH (W/K)
A 5 -0.4 0.02 -20 509.2946269
12
B 10 -0.1 0.02 -5 4074.357016
C 15 0.3 0.02 15 -2037.178508
D 20 0.1 0.02 5 -8148.714031
Average Thermal Conductivity, kH -1400.560224
Average (x₂-x₁) 0.02
13
Table 2B (iv): Data for calculation Thermal conductivity, kC
Test
no.
Wattmeter,
Q (watts)
(y₂-y₁),
K
(x₂-x₁),
m
(dT/dx)C,
(K/m)
Thermal
conductivity, kC
(W/K)
A 5 -0.4 0.02 -20 509.2946269
B 10 -0.4 0.02 -20 1018.589254
C 15 -0.9 0.02 -45 679.0595026
D 20 -1 0.02 -50 814.8714031
Average Thermal Conductivity, kC 755.4536966
Average (x₂-x₁) 0.02
Table 2B (v): Data for calculation Overall Heat Transfer Coefficient, U
XS/kS XH/kH XC/kC 1/U
Overall Heat
Transfer
Coefficient, U
(W/m2K)
0.00279776
9 -1.428×10-05 2.64742×10-05
0.0028
1 355.8765192
Table 2B (vi): Data for calculation Heat Transfer Rate, Q
Test no. Wattmeter, Q (Watt) Heat Transfer Rate, Q (Watt)
A 5 9.363431508
B 10 9.922442344
C 15 12.47292929
D 20 16.21131425
14
DISCUSSION
This experiment contains two parts which is experiment 1 and experiment 2.
Experiment 1 is an experiment using homogenous bar which is brass with 25mm in
diameter while Experiment 2 using composite bar which is brass with 13mm in diameter
and stainless steel with 25mm in diameter.
From experiment 1, the temperature gradient increases as the rate of heat
transfer increases, however the heat transfer coefficient decreases. According to
Fourier’s Law, heat transfer coefficient is inversely proportional to the temperature
gradient. Averagethermal conductivity obtained from this experiment is37.89399292
W/mK that can be seen in the Table 1(i). That value is very low compared to the
theoretical value which is 111 W/mK. Besides that, heat transfer rate obtained from this
experiment is slightly different from the wattmeter reading that have been set before
start the experiment. These values can be seen in Table 1(ii). This is due to some heat
loss to surrounding.
From experiment 2 which is using brass of 13 mm diameter, the average thermal
conductivity in different sections are kS=47.17204718 W/mK, kH=-6435.243992W/mK
and kC=3242.286694W/mK in Table 2A (ii),(iii) and (iv). Overall heat transfer coefficient
from Table 2A (v) is U=1175.059953W/m2K and from Table 2A (vi) the values of heat
transfer rate obtained is largely different from the wattmeter reading. The overall of heat
transfer coefficient for brass composite of this experiment is higher than that in
experiment 1. It is shown that smaller diameter of brass has higher value of heat
transfer coefficient. According to Fourier’s law, heat transfer coefficient is inversely
proportional to the surface area.
Whereas for the stainless steel, the average thermal conductivity are
kS=14.29710469W/mK, kH=-1400.560224W/mK and kC=755.4536966W/mK from Table
2B (ii), (iii) and (iv) respectively. From Table 2B (v), it can be seen the value of overall
heat transfer coefficient, U=355.8765192W/m2K and from Table 2B (vi) the values of
heat transfer rate obtained having difference from the wattmeter reading.The overall
15
heat transfer coefficient for stainless steel is lower than that for brass. In other word, the
brass has better conductivity than stainless steel.
From the general equation of Fourier’s Law, QA
=−k dTdx
, cross sectional area is
proportional to surface area and inversely proportional to the temperature gradient.
There are some error occur while carrying out the experiment and causing some
effect to the result obtained from these experiment. The errors are:
The loss of heat from the gap between sections of the bar, caused by the loose
grip of the clamp.
The nine sensors are not properly connected that will affect the reading of the
temperature.
The glass sealant applied to minimize heat loss and eliminate air gap that might
causes some resistance to the heat transfer.
Unsteady and inefficient power supply and also unsteady cooling the running
water may cause unsteady temperature distribution.
Some heat loss to the surroundingthrough convection and radiation while
carrying out the experiment.
The temperature reading has been taken every 15 minutes but maybe the
temperature does not reached steady state.
CONCLUSION
Experiment 1
Thermal conductivity of the brass with diameter 25mm is lower than the actual value.
When increasing the heat transfer rate, the heat transfer coefficient decreases as
the temperature gradient increases.
Experiment 2
Smaller diameter(surface area) of same material has higher heat transfer coefficient,
which means better thermal conductivity.
16
Overall heat transfer coefficient for stainless steel is lowers than brass. Brass is
better in conductivity for heat conduction.
The heat transfer rate for both bar having a quite large different to the wattmeter
reading.
RELEVANCE to INDUSTRY
In chemical industry, examples of the steady state heat flow through composite
systems: a furnace wall made up of a layer of refractory bricks followed by a steel sheet,
insulated spherical tanks for storage of liquefied gases and the walls of refrigerator.
In the case of furnace, a furnace wall normally uses various type fire bricks with
different thermal conductivities followed by the steel casting to provide high mechanical
strength at elevated temperatures.
MODIFICATION SUGGESTED
1. For more accurate reading of temperature, take the reading when the temperature
is truly under steady state.
2. Cover the air gap between the conduction modules thoroughly to prevent any heat
loss to surrounding.
3. Make sure the conduction bar in the module is in full contact with each other to
allow the maximum conduction.
4. Stabilize the flow rate of cooling water.
5. Close the fans nearby to avoid any convection of heat
17
APPENDIX
Sample of calculation
Experiment 1: Homogenous bar (brass, d=25mm) Test A= 5 Watt
Thermal conductivity, k (W/mK)
o Cross sectional area
d=25mm
A=π r2
A=π × ¿
A=4.90875×10−4m2
o Temperature gradient from excel
dTdx
=−232.6
o From Fourier’s Law
Q∝ A dTdx
Q=−kA dTdx
k=−QA
1dTdx
k=−( 5
4.90875×10−4 )( 1−232.6
)
k=43 .79145546 WmK
Heat transfer rate, Q (Watt)
k ave=k A+k B+kC+kD
4
k ave=43.79145546+39.80419124+34.69309448+33.28723052
4
k ave=43.79145546WmK
Q=−kave AdTdx
18
Q=− (43.79145546 ) (4.90875×10−4 )(−232.6)
Q=4 .326642324Watt
Experiment 2: Composite bar (brass, d=13mm) Test A= 5 Watt
Thermal conductivity, kS
o Cross sectional area
d=13mm
A=π r2
A=π × ¿
A=1.327326×10−4m2
o Temperature gradient
¿
¿
¿
Same goes to ¿
o From Fourier’s Law
Q∝ A dTdx
Q=−kA dTdx
k S=−QA
1dTdx
k S=−( 5
1.327326×10−4 )( 1−1745
)
k S=21 .58723262WmK
Same goes to kH and kC
k S , ave=k A+k B+kC+kD
4
19
k S , ave=21.58723262+42.20697023+59.16710093+65.72688493
4
k S , ave=47.17204718WmK
Same goes to kH , ave∧kC ,ave
Overall heat transfer coefficient, U
1U
=x SkS
+xHkH
+xCkC
1U
= 0.0447.17204718
+ 0.02−6435.243992
+ 0.023242.286694
1U
=0.00085102
U=1175 .059953 W
m2K
Heat transfer rate, Q (Watt)
Q=UA ∆T
Q= (1175.059953 ) (1.327326×10−4 )(100.2−29.7)Q=10 .99579777Watt
REFERENCE
Holman, J.P., Heat Transfer, McGraw-Hill Book Company, 2010.
Incropera, F.P., and Dewitt, D.P., Fundamentals of heat and mass
transfer, J. Wiley and sons, New York, 1996.
Kern, D.Q., and A.D. Kraus, Extended surface heat transfer, McGraw-Hill
book Co., New York, 1972.
EKC 291 Chemical Engineering Laboratory I Lab Manual, Semester 2,
2010/2011
20
