Experiment #3: Linear Heat Conduction
Objective
To use Fourier’s heat conduction equation to 1) determine the conduction heat transfer rate 𝑸̇, 2)
the thermal conductivity of brass, and 3) the overall conduction heat transfer coefficient U for the
brass assembly.
Background
According to Fourier’s law of heat conduction, if a plane wall supports a
temperature difference ΔT then the heat transfer rate 𝑸̇ by conduction through
the wall is:
dT
ΔT
Q = −kA
 −kA
dX
ΔX
where k and A are the thermal conductivity and area of the wall, ΔT = Tb – Ta
and ΔX = Xb – Xa.
Since heat transfer is positive in the direction of temperature drop, there is a the negative sign in
the equation. For convenience the equation can be rearranged to avoid the negative sign simply by
redefining ΔT to be ΔT = Ta – Tb, which we will do in this experiment.
Experiment Apparatus
The Armfield linear heat conduction apparatus has a heating section and cooling section, which
can be clamped with interchangeable intermediate sections between them. The temperature
difference created by the application of heat to one end and cooling at the other end results in the
linear conduction of heat.
Thermocouple positions
T1 = Heated section high temperature
T2 = Heated section mid temperature
T3 = Heated section low temperature
T4 = Test section high temperature
T5 = Test section low temperature
T6 = Cooled section high temperature
T7 = Cooled section mid temperature
T8 = Cooled section low temperature
1
Experiment Information
Thermocouples are positioned along both the heated section and cooled sections at vertical intervals of
15 mm to measure the temperature gradient along the sections. A pressure regulator is incorporated to
minimize the effect of cooling water fluctuations in the supply pressure. A control valve allows the flow of
cooling water to be varied over the operating range of 0 to 1.5 liters/minute.
In this experiment you will use the 30 mm-long brass section with two imbedded thermocouples and the
same diameter as the heating and cooling sections. When this section is clamped between the heating
and cooling sections a long plane wall of uniform material and cross-section is created with temperatures
measured at eight positions
Physical Data
•
•
•
The vertical distance between each thermocouple is 0.015 m.
The diameter of each test section is 0.025 m, and its thickness is 30 mm.
The thermal conductivities of the test sections are approximately:
o brass is ≈120 W/mK,
o stainless steel is ≈ 25 W/mK,
o aluminum is ≈ 180 W/mK.
Experiment Procedure, Calculations, and Results
1) Apply a light coat of thermal paste to both sides of the intermediate brass
section containing thermocouples T4 and T5 and clamp it between the
heated and cooled sections of the apparatus.
2) Connect all eight thermocouples to the National Instruments NI-9219 data
acquisition unit.
3) Turn on the cooling water supply and set the heater voltage to 9 volts. When
the temperatures have stabilized record the data for the items in Table 1
below.
4) Set the heater voltage to 12 volts. When the temperatures have stabilized
record the data for the items in Table 1 below.
5) Set the heater voltage to 17 volts. When the temperatures have stabilized record the data for
the items in Table 1 below.
6) Plot a graph of temperature vs thermocouple spacing for each of
the heater power settings. Note: The distance between each
thermocouple is 0.015 m = 0.6 inch, the distance between T3 and
the top of the brass test section is 0.0075 m = 0.3 inch, and the
distance between T6 and the bottom of the brass test section is
also 0.0075 m = 0.3 inch.
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7) Compute the thermal conductivity of brass in the hot, mid, and cold sections as shown in
Table 2 below. Your results for brass should be in the range of 110 to 128 W/mK.
8) Finally, compute the overall conduction heat transfer coefficient U first using the thermal
resistance of the material, R
 X hot  Xmid  X cold
1
U=
R
where R =
k hot
+
k mid
+
k cold
and then using Fourier’s Law
U=
Q
A(T1 − T8)
and enter the values in Table 2 below.
Table 1: Thermocouple temperatures at three heater powers.
Item
Hot Section Heater Voltage V
Hot Section Heater current I
Hot Section Heater Power 𝑸̇ = VI
Thermocouple temperature T1
Thermocouple temperature T2
Thermocouple temperature T3
Thermocouple temperature T4
Thermocouple temperature T5
Thermocouple temperature T6
Thermocouple temperature T7
Thermocouple temperature T8
Units
Results
volts
amps
watts
⁰C
⁰C
⁰C
⁰C
⁰C
⁰C
⁰C
⁰C
9
12
17
Table 2: Thermal conductivities of the heated, test, and cooled sections and the overall
conduction heat transfer coefficient.
Item
Units
Hot Section Heater Voltage V
ΔThot = T1 – T3
ΔTmid = T4 – T5
ΔTcold = T6 – T8
A = πD2/4
ΔXhot = X3 – X1 = 0.030
ΔXmid = X5 – X4 = 0.015
ΔXcold = X8 – X6 = 0.030
khot = 𝑸̇Xhot/(ΔThotA)
kmid = 𝑸̇Xmid/(ΔTmidA)
kcold = 𝑸̇Xcold/(ΔTcoldA)
U = 1/R
U = 𝑸̇/([A(T1 – T8)]
volts
⁰C
⁰C
⁰C
m2
m
m
m
W/mK
W/mK
W/mK
W/m2K
W/m2K
9
4.91×10-4
Results
12
17
4.91×10-4
4.91×10-4
3
Questions to Be Answered in Your Report
1) Compute the percent difference kbrass = 120 W/mK and the thermal conductivity of the three
brass sections at the three-voltage setting in the table below.
Hot Section Heater Voltage V
Hot Section Heater Power 𝑸̇ = VI
9
12
17
 khot − kbrass

 kbrass

  100

 k mid − kbrass 

  100
kbrass


 kcold − kbrass 

  100
k
brass


2) Explain any differences in the overall conduction heat transfer coefficients U as determined by
using the thermal resistance R and Fourier’s Law.
4