1. EMM3812
MECHANICAL ENGINEERING
LABORATORY (IV)
Semester 1, 2018/2019
Title : Heat Conduction
Group : B1
No. Matric No. Name
1 187542 HAZLAN BIN HUSAING
2 189860 SITI KHAIRIYAH SULAIMAN
3 190185 MUHAMMAD SYUKRI BIN MOHD HAMIDI
4 190493 MUHAMMAD HAFIZ BIN MOHD YUNUS
2. 2
Introduction
Heat transfer is the study of thermal energy transport within a medium or among neighbouring media
by molecular interaction, fluid motion, and electro-magnetic waves, resulting from a spatial variation
in temperature. Basically, heat transfer is from a body of hot temperature to a body of lower temperature.
This variation in temperature is governed by the principle of energy conservation, which when applied
to a control volume or a control mass, states that the sum of the flow of energy and heat across the
system, the work done on the system, and the energy stored and converted within the system, is zero.
Heat transfer finds application in many important areas, namely design of thermal and nuclear power
plants including heat engines, steam generators, condensers and other heat exchange equipment,
catalytic convertors, heat shields for space vehicles, furnaces, electronic equipment etc, internal
combustion engines, refrigeration and air conditioning units, design of cooling systems for electric
motors generators and transformers, heating and cooling of fluids etc. in chemical operations,
construction of dams and structures, minimization of building heat losses using improved insulation
techniques, thermal control of space vehicles, heat treatment of metals, dispersion of atmospheric
pollutants.
Objective
i. To determine the unsteady state heat transfer by Lumped Capacitance Method.
ii. To define and compare the heat transfer coefficient with expected values
Theory
Heat transfer is a physical process of spontaneous, irreversible heat transport from hotter (i.e., having a
higher temperature) bodies or parts of bodies to cooler ones. It is also a field of science that is concerned
with the study and quantitative description of heat transfer laws.
Basically, heat transfer coefficient is determined by Lumped heat capacity model. A lumped
capacitance model, also called lumped system analysis, reduces a thermal system to a number of
discrete “lumps” and assumes that the temperature difference inside each lump is negligible. This
approximation is useful to simplify otherwise complex differential heat equations. It was developed as
a mathematical analogy of electrical capacitance, although it also includes thermal analogy of electrical
resistance as well.
Our initial assumption was that the temperature variations within the mass should be small, when
compared with the temperature difference from the surroundings. We said that in that case, we can
neglect such small variations and assume the entire mass to be at a single temperature. From our earlier
results on conduction, we know that for temperature variations to be small, the resistance to conduction
must be small. Thus, the assumption we made is good when Internal resistance over External resistance
is less than 1. Suppose the mass is a solid sphere of radius R and thermal conductivity s k . Then, the
length of the path for heat conduction within the sphere is R and the internal resistance can be
approximately represented by R kA /(s). The external resistance to heat transfer is the convective
resistance 1/(hA) .
You can verify that the group (hR ) / ks is dimensionless. It is called the Biot Number and abbreviated
as Bi=hR/ks for a sphere. If the solid has some other shape, a suitable characteristic length that represents
3. 3
the conduction path can be used instead of the radius of the sphere used here. We have specifically used
the subscript s for the thermal conductivity to specify that it is the thermal conductivity of the solid and
not that of the fluid. Thus, the condition for the lumped thermal capacity model to be a good
representation of the actual heat transfer system is Bi 1 .
Procedure
i. Around 3 litres of raw water or tap water has been prepared.
ii. Both valves DV1 are being ensured to fully close.
iii. The carrier with specimen located on top of the water bath has been removed by pulling up.
iv. The raw water has been poured into the tank through the flow duct chamber until the water
reached half of the flow duct hole.
v. The power supply that located on at the control panel has been turned on.
vi. The heater has been turned on and water bath temperature controller has been set at 65 𝑜
C.
vii. Flow control valve V1 has been fully opened and the water pump has been started in order for
water to circulate.
viii. Waited until the temperature has been stabilized.
ix. The brass slab specimen’s PT-100 wire has been connected into the connector that located at
the control box. The reading of the initial temperature of the specimen that is located on the
control box has been taken.
x. The brass slab specimen has been plunged into the duct by lifting the carrier that is attached to
the specimen. Once the temperature of brass slab specimen same as the temperature of water
bath, the carrier has been pulled up and leaved it to cold down in the air. All the temperatures
data has been taken at 30 seconds interval.
xi. The experiment for other materials and types of specimens has been repeated.
xii. Once all the specimens have been conducted, the pump has been shut down and the main power
at the control panel has been turned. If it is necessary to drain water inside the water bath tank,
valve DV1 has been connected to a hose to a proper drainage system and then the valve DV1
has been opened.
Apparatus
Figures below show the setup of the experiments and the materials used in the experiment.
Figure 1: LS-17020 unsteady state heat transfer apparatus.
4. 4
i. Water bath tank
ii. Flow duct
iii. Specimen carrier
iv. Geometric center
v. Flow meter
vi. Flow control valve
vii. Water bath
viii. Heater
ix. Water pump
x. Control panel
xi. Low level switch
xii. Drain valve
Figure 2: Specimen types of (a) slab, (b) cylinder and (c) sphere.
Results
Table 1 shows the temperature drop in interval of 30 seconds for stainless steel specimen.
Initial Geometric Centre Temperature : 65 (°C)
Geometric Centre Temperature (°C) 𝑻𝑻 𝟐
Recorded
Time (s)
Air Temperature
(°C) 𝑻𝑻 𝟏
Cylinder
Specimen (°C)
Sphere
Specimen (°C)
Slab
Specimen (°C)
0 27 65.0 65.0 65.0
30 27 64.1 63.2 64.2
60 27 62.9 62.3 62.7
90 27 62.3 62.0 61.5
120 27 61.5 61.0 60.5
150 27 61.0 60.6 59.6
180 27 60.9 60.0 58.8
210 27 60.5 59.4 58.0
240 27 60.3 58.8 57.3
270 27 59.7 58.3 56.8
300 27 59.5 57.8 56.2
6. 6
450 27 55.1 54.2 51.0
480 27 54.6 53.7 50.6
510 27 54.2 53.3 50.1
540 27 53.8 52.8 49.7
570 27 53.3 52.4 49.2
600 27 52.9 52.0 48.8
630 27 52.6 51.5 48.4
660 27 52.0 51.0 48.0
690 27 51.6 50.6 47.6
720 27 51.3 50.3 47.2
750 27 50.9 49.8 46.8
780 27 50.6 49.5 46.4
810 27 50.2 49.0 46.0
840 27 49.9 48.7 45.7
870 27 49.5 48.4 45.3
900 27 49.2 48.0 45.0
Table 2: Temperature Drop for Brass Specimen (Cylinder, Sphere, Slab)
Graph 1 and 2 shows temperature against time during cooling for stainless steel and brass specimen.
Graph 1: Temperature against Time for Stainless Steel Specimen During Cooling
0
10
20
30
40
50
60
70
0 60 120 180 240 300 360 420 480 540 600 660 720 780 840
Temperature(°C)
Recorded Time (s)
Temperature againts Recorded Time for Stainless Steel
Specimen During Cooling
Cylinder Specimen (°C)
Sphere Specimen (°C)
Slab Specimen (°C)
7. 7
Graph 2: Temperature against Time for Brass Specimen During Cooling
Graphs below show temperature against time for sphere, cylinder and slab specimens.
Graph 3 : Temperature against Time for Cylinder Specimen
0
10
20
30
40
50
60
70
0 60 120 180 240 300 360 420 480 540 600 660 720 780 840 900
Temperature(°C)
Recorded Time (s)
Temperature against Recorded Time for Brass Specimen
During Cooling
Cylinder Specimen (°C)
Sphere Specimen (°C)
Slab Specimen (°C)
0
10
20
30
40
50
60
70
0 60 120 180 240 300 360 420 480 540 600 660 720 780 840 900
Temperature(°C)
Recorded Time (s)
Temperature against Recorded Time for Cylinder Specimen During
Cooling
Stainless Steel Specimen (°C)
Brass Specimen (°C)
8. 8
Graph 4: Temperature against Time for Sphere Specimen
Graph 5: Temperature against Time for Slab Specimen
0
10
20
30
40
50
60
70
0 60 120 180 240 300 360 420 480 540 600 660 720 780 840 900
Temperature(°C)
Recorded Time (s)
Temperature against Recorded Time for Sphere Specimen During Cooling
Stainless Steel Specimen (°C)
Brass Specimen (°C)
0
10
20
30
40
50
60
70
0 60 120 180 240 300 360 420 480 540 600 660 720 780 840 900
Temperature(°C)
Recorded Time (s)
Temperature against Recorded Time for Slab Specimen During Cooling
Stainless Steel Specimen (°C)
Brass Specimen (°C)
9. 9
To determine the dimensionless temperature,
𝛉 𝟎 = (𝑻 𝟐–𝑻 𝟏) / (Ti–𝑻 𝟏)
where
𝑻 𝟐 : Geometric Center Temperature (°C)
𝑻 𝟏 : Temperature of Air (°C)
Ti : Initial Temperature of the Specimen (°C)
Sphere Stainless Steel Specimen at first 150 seconds,
θ 0 = (60.6 – 27) / (66 – 27)
= 33.6 / 39
= 0.86
To determine the Fourier Number,
𝑭 𝟎 = (kt) / (cp𝒓 𝟐
)
where
k : thermal conductivity (W/mK)
t : Current time (s)
c : Specific heat (J/kg)
p : Density (kg/𝑚3
)
𝒓 𝟐
: Radius of sphere (m), 0.05/2 = 0.025m
Sphere Stainless Steel Specimen at first 150 seconds
𝐹0 = (16.36 x 150) / (460 x 8500 x 0.0252)
= 2454 / 2443.75
= 1.0042
Biot number can be determined from the chart below using value of θ 0 and 𝐹0 calculated.
10. 10
From the chart, Biot number, Bi < 0.1. Therefore, The Lumped Capacitance Method can be applied.
The Lumped Capacitance Method
𝛉 𝟎 = exp [- (hAt / cpV)]
where
θ 0 : Dimensionless Temperature
h : Heat Transfer Coefficient (W/𝑚2
K)
A : Area of Sphere (𝑚2
)
t : Time (s)
c : Specific heat (J/kg)
p : Density (kg/𝑚3
)
V : Volume of Sphere (𝑚3
)
11. 11
Sphere Stainless Steel Specimen
- ln (0.86) = [h x (7.854 x 10−3
) x 150 ] / [460 x 8500 x (6.545 x 10−5
)]
0.15 = 1.1781 h / 255.91
h = 32.58 W/𝑚2
K
Discussion
Based on the results obtained and above graph temperature versus time illustrating the cooling for 3
different type of materials and shapes, it can be observed that all the specimens cools down with almost
linearly.
As observed, there are not much difference on cooling rates between all the test specimens.
Theoretically, brass materials has higher thermal conductivity compared to stainless steel which means
brass response faster than stainless steel when it is cooled down in normal room temperature from the
hot water bath.
Conclusion
From this experiment, students can determine the unsteady state heat transfer by Lumped Capacitance
Method. Besides, students can define and compare the heat transfer coefficient with expected values.
References
Book :
[1] Bejan, A., Kraus, A.D. (2003). Heat Transfer Handbook. John Wiley & Sons, Inc.
Electronic Sources :
[2] Kurganov,V.A. (2011). Heat Transfer, Retrieved from http://www.thermopedia.com/content/840/
[3] R. Shankar (n.d.), Unsteady Heat Transfer: Lumped Thermal Capacity Model, Retrieved from
http://web2.clarkson.edu/projects/subramanian/ch330/
