Download ME224 Lab 5 - Thermal Diffusion

ME224 Lab 5
ME224 Lab 5 - Thermal Diffusion
(This lab is adapted from “IBM-PC in the laboratory” by B G Thomson & A F Kuckes,
Chapter 5)
1. Introduction
The experiments which you will be called upon to do in this lab give you a chance to
apply timing concepts and to review the use of the ADC while learning about the
phenomenon of diffusion. Specifically, you will be studying thermal diffusion but many
of the concepts encompass a variety of other phenomena.
In this lab you will learn to instrument and analyze experimental data to compute the
diffusion coefficient, thermal conductivity and heat capacity of copper.
2. Heat Flow Equation
In this section you will explore some of the physical and mathematical considerations of
one-dimensional heat diffusion. When heat is added to a material there are two
parameters that affect the distribution of temperatures: the specific heat (or heat capacity)
and the thermal conductivity. The specific heat indicates how much heat is added to a
mass of material for a specified temperature rise. The thermal conductivity indicates how
fast the thermal energy is transported through the material.
Figure 1. Copper rod schematic.
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ME224 Lab 5
Consider the flow of heat in a rod as shown in Figure 1. The specific heat C of a material
is the ratio of the amount of heat added dQ (Joules) to the resulting rise in temperature
dT (degrees Kelvin) per unit mass dm (kg); thus
⎛ dQ ⎞ 1
C =⎜
⎟
⎝ dT ⎠ dm
For a rod of cross-sectional area A, the volume dV=A⋅dz and dm=ρ⋅dV where ρ is the
density. So, the amount of heat added to the length dz of the rod is
dQ = C p dTdm
= C p dTρdV
(1)
= C v dTρAdz
where Cv = Cp is the volumetric heat capacity.
When one end of the rod is hotter than the other, there will be a net flow of energy from
the hot end to the cool end. The power P (watts) of this heat flow down the rod is the
heat energy per unit time flowing past a point on the rod P=dQ/dt. For one-dimensional
heat flow, P is proportional to the temperature gradient dT/dz, the thermal conductivity
κ (W/m·K) and the cross-sectional area:
⎛ dT ⎞
P = −κA⎜
⎟
⎝ dz ⎠
(2)
There is a minus sign because heat flows from higher to lower temperatures. In writing
this equation, it is assumed that the rod is insulated; no heat escapes from the rod by
conduction, convection or radiation. The net heat gain per unit time dQ/dt in the piece of
rod between z and z+dz is given by the difference in the power flowing in at z and the
power flowing out at z+dz, so
dQ
⎛ ∂P ⎞
= P( z ) − P( z + dz ) = −⎜ ⎟dz
dt
⎝ ∂z ⎠
(3)
Combining Equations (1), (2) and (3) gives the differential equation for heat flow in a rod
⎛ T⎞
⎛ ∂T ⎞
∂ ⎟
Cv ⎜
⎟ = κ⎜
⎜∂ 2⎟
⎝ ∂t ⎠
⎝ z ⎠
2
(4)
This equation has many solutions, depending on the initial and boundary conditions. For
a quantity of heat added to the rod quickly (a heat pulse) at z = 0, the solution can be
written as follows (B1 and B2 are constants).
T (t , z ) = B1 +
(
)
⎧ − z 2 Cv ⎫
B2
exp
⎨
⎬
t 1/ 2
⎩ 4κt ⎭
(5)
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ME224 Lab 5
Figure 2. Solution for the temperature distribution along the rod.
The solution (5) describes the temperature at any point in the rod as a function of time
after an impulse of heat has been added at z=0. Before proceeding further it is useful to
examine the graphs of temperature T vs. distance z of the solution at various times after
the impulse. These are shown in Figure 2. In this figure, T=T(t, z)-TS ( where TS –
initial temperature). At times near zero, the heat, and thus the excess temperature, is
concentrated near z=0. As time progresses the heat diffuses away from z = 0 to larger and
larger values of z with the peak temperature decreasing in time.
An important point is that since the solution is symmetric with respect to z, just as much
heat diffuses up as down the rod. Since there is no heat flow across the cross section at
z=0, “cutting” the rod at z=0 will not modify the form of the solution although now all
the heat added flows in one direction (of course: from the hot to the cold). This half-space
rod is the configuration that you will study experimentally.
To obtain a theoretical expression convenient for analyzing a quantitative experiment, it
is useful to relate the constant B2 in Equation (5) to the total heat Q added to the rod
(from z=0 to z=∞) by integrating Equation (1). Consider T as the excess temperature
above T(0), i.e., T=T(t, z)-TS = T′ -TS; integrating Equation (1) from temperature TS to
T′ gives
dQ = C v ρAdz (T '−Ts ) = C v ρAdzT
(6)
To integrate from z=0 to z=∞, use Equation (5) to describe the variation of temperature
at any z and t. Then
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ME224 Lab 5
C ⎞
⎛
B2 exp⎜ − z 2 v ⎟
4κt ⎠
⎝
Q = ∫ Cv A
dz
1/ 2
t
z =0
∞
=
Cv A
=
Cv A
t
1/ 2
t1/ 2
∞
C ⎞
⎛
B2 ∫ exp⎜ − z 2 v ⎟dz
4κt ⎠
⎝
0
B2
π 1 / 2 ⎛ 4κt ⎞
= B 2 (πκC v )
(7)
1/ 2
⎟
⎜
2 ⎜⎝ C v ⎟⎠
1/ 2
A
Solving for B2 and inserting into Equation (5) yields
T (t , z ) =
1
Q
A (πκC v )1 / 2
C ⎞
⎛
exp⎜ − z 2 v ⎟
4κt ⎠
⎝
+ Ts
1/ 2
t
(8)
As written Equation (8) is not an optimum form for displaying some of the important
features it contains. It is often very helpful, particularly for purposes of recognizing the
domain of behavior in a given physical situation, to relate the quantities in an equation to
physically significant parameters rather than simply measuring time in seconds,
temperature in degrees centigrade, etc. You saw this before in the equation for the
thermistor resistance as a function of temperature, the natural parameters there being R0
and T0. For displaying the change in temperature T as a function of time t at a fixed z,
Equation (8) can be written in terms of a characteristic time t1 and a characteristic
temperature T1 as
1/ 2
T ⎛ t1 ⎞
⎛ t ⎞
= ⎜ ⎟ exp⎜ − 1 ⎟
T1 ⎝ t ⎠
⎝ t ⎠
z2
t1 = C v
4κ
2Q
T1 =
AzC v π 1 / 2
(9)
T = T (t , z ) − Ts
where T is the excess temperature.
Equations (9) immediately show several important points. First, the variation of
temperature with time at a constant z can he related to just two parameters t1 and T1.
Second, the characteristic time scale t1 is proportional to z2; this is a general property of
diffusion phenomena.
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ME224 Lab 5
Activity 1: Graphing the Heat Diffusion Equation
1. The first equation in (9) is the normalized diffusion equation. Generate a graph of
T/T1 as a function of t/t1 from t/t1 = 0.1 to t/t1 = 10. Note that the independent
variable t/t1 is inverted both times it appears. Include this figure in your writeup.
2. Your plot should peak at t/t1 = 2 and T/T1 = 0.43. Derive these values analytically
from the first equation in (9) and include the proof in your writeup.
3. When you collect your data, it should have this shape, but the axes will not be
scaled properly (i.e., the peak will not occur at 2, nor will it have a value of 0.43).
You will need to normalize your data by finding scaling factors T1 and t1 to put
the peak in this location.
3. Numerical Integration of the Heat Flow Equation
General numerical integration of partial differential equations is a broad and difficult
subject. The following will be a simple procedure that works in this case but must be
used with care. It is really only meant to illustrate a general approach.
The basic equations for the flow in a rod are the static equation for the heat capacity,
Equation (1), and the dynamic equation with the thermal conductivity, Equation (2),
which can be combined to form the differential equation, Equation (4). However for
purposes of numerical integration, it is best to leave them separate and write them in this
form:
∆T =
∆Q
C v A∆z
∆Q = −κA∆T
(1’)
∆t
∆z
(2’)
where ∆z is assumed to approach zero.
Now break up the length of the rod (Figure 1) into Nz pieces of length ∆z each and
consider the ith piece; the heat flowing into this piece in the time ∆t will be:
Qin = κA(Ti −1 − Ti )
∆t
∆z
(10)
If the temperature in element i-1 is hotter than in the element i then Qin will be positive.
The heat flowing out of the piece will be:
Qout = κA(Ti − Ti +1 )
∆t
∆z
(11)
The difference of the two is the heat gained or lost in the element:
∆Qi = Qin − Qout
(12)
This heat changes the temperature of the element in proportion to its heat capacity:
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ME224 Lab 5
∆Ti =
∆Qi
C v A∆z
(13)
and so
∆Ti new = ∆Ti old + ∆Ti
(14)
4. Experimental Setup and Program Development
Enclosed in a
glass jar to reduce
convective heat
losses
Figure 3. Experimental apparatus.
The apparatus for these experiments is illustrated in Figure 3. In the top of the copper rod
(8.25 mm diameter) is set a resistor that is used as a heater. Current can be switched into
the heater under program control using the IRF 520 HEXFET in a manner similar to that
used in the Thermistor Experiment. After generating a short pulse of heat by momentarily
turning on the HEXFET, the computer will measure the increase in temperature at two
positions down the rod using two thermistors. The thermistor positions are as shown on
Figure 3 (z1 = 30 mm and z2 = 60 mm). A plot of the temperature vs. time at each of these
thermistors will yield values for the heat capacity and thermal conduction constants of
copper and also demonstrate the functional dependence of heat diffusion on time and
distance. Ideally, one thermistor would be sufficient to calculate these properties, but
experiments are rarely ideal. Here you will want to calculate the material properties
independently for each thermistor, then average them to get a more reliable result.
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ME224 Lab 5
5. Voltage Amplifier
The change in temperature of each thermistor from an initial temperature (T(t, z) – T0) is
the significant quantity to measure in this experiment. However the temperature
increments and thus the voltage changes are very small; if the ADC is connected directly
to the thermistor, the changes are close to the step size of digitization. To overcome this
problem an amplifier is used to boost the voltage change. On the protoboard attached to
the experimental apparatus is an amplifier using a CA3140 operational amplifier as
shown in Figure 4.
Figure 4. Schematic of a CA3140 operational amplifier.
It is not necessary to understand the details of this amplifier circuit except to note that the
relationship between the three voltages VA (output) (pin 6), V1 (pin 2) and VT (pin 3) is
given by
V A = G (VT − V1 )
(15)
For the circuit components used, the gain G is equal to 21. Equation (15) is the equation
corresponding to a differential amplifier circuit as discussed in class (see handouts)
The amplifier output (VA) is constrained by the characteristics of the CA3140 to be
between 0 V and +3 V. Since a rise in thermistor temperature will lead to a rise in the
output voltage of the amplifier, the potentiometer R1 should be set so that the output
voltage of the circuit starts near the lowest voltage before a heat pulse is applied. This
will allow the greatest voltage swing as the thermistor heats up without exceeding the 3V
limit. Use the oscilloscope or the multimeter to monitor the output voltage of each
amplifier. Set the potentiometers (one for each amplifier-thermistor combination) so that
the amplifier outputs are about 0.20 V before you start each run. When this is done each
potentiometer R1 has been adjusted to be essentially the same resistance as the thermistor
resistance RT before a temperature pulse is applied. Since the amplifier gain is 21, the
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ME224 Lab 5
change in the output voltage ∆VA will be 21 times greater than the change in the
thermistor voltage ∆VA.
Activity 2: Heat Impulse VI
Write a VI that will turn on the heater for a specified time. The VI should output a 0 to
the DAQ digital output until a button is pushed. Then it should output a 1 for the
specified time, then go back to 0 before the program ends so the DAQ output will not
float at 1.
Activity 3: Amplifier Check
Before developing a detailed program, write a simple program to see that the apparatus is
functioning following the outline shown in Figure 5.
Figure 5. Amplifier check algorithm
When you run this program you should see on the oscilloscope the voltage output rise
and then slowly fall. It should start above 0 V and should NOT exceed 3 V.
Do the same to check thermistor 2, the lower thermistor. You will need to let the
apparatus cool down and reset the potentiometer between heat pulses.
In order to measure both of the thermistor voltages at once, you will need to make a new
task in MAX. Choose Analog input and voltage, then choose both ai0 and ai1 as the
analog inputs from the DAQ. Give it a name, then set both Voltage0 and Voltage1 to
have the range from -10 to 10 and use 1 sample on-demand. Hook this task constant to
the start, read, and stop blocks as usual. Use the finger tool to change the drop box below
the read block. Use multiple channels, single sample, 1D DBL. The output of the read
block is now an array containing both the ai0 and ai1 values. They can be split easily by
using the “Array to Cluster” block followed by the “Unbundle” block, and wiring the top
two outputs of “Unbundle.”
Activity 4: Heat Flow Real-Time Plot
1. The next task is to make thermistor ADC measurements at specified times. To do
this, modify the previous two programs. Put in time delays as indicated by Figure
6. Note that a sample is taken before the heater is turned on. This records the
baseline ADC reading. The heating of the rod then changes the ADC reading from
this starting value.
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ME224 Lab 5
Read initial value of
each thermistor.
Apply heat pulse
For 500 msec
For I = 1 to 40
Wait 0.5 sec
Read value of each
thermistor
Save data to a file
Figure 6. Thermistor data acquisition algorithm.
2. Combine the data gathering with plotting so that these unprocessed data are
plotted as they are gathered, i.e., in real-time.
3. Send the collected data to a file for later analysis.
4. Once the data has been collected, unscrew the glass jar. Turn the heater on and
quickly measure the resistance of the heater and the voltage across it.
5. Use these values to compute the power dissipated by the heater.
6. This value will allow you to calculate the heat input.
6. Data Analysis
Before proceeding to more data plots and analysis, here are some additional mathematical
considerations. We will assume that the temperature and voltage changes at the
thermistor are small enough so that their behaviors are adequately described by
differentials. Thus: (change in amplifier output voltage) = (gain) x (change in the input
voltage)
dV A = GdVT
(16)
The relationship between VT and RT is similar to the thermistor experiment, i.e.,
VT/V0 = R1/(R1 + RT) with V0 = 5 V. The relationship between dVT and thermistor
resistance changes dRT can be obtained by differentiation; the result (which you should
work out) is
dVT
dR
=− T
V0
R1
⎞
⎛
1
⎟⎟
⎜⎜
⎝ 1 + RT R1 ⎠
2
(17)
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ME224 Lab 5
Noting that R1 and RT are adjusted to be nearly equal at the outset gives
dVT
dR
=− T
V0
4 RT
(18)
The next task is to relate a change in the thermistor resistance to a change in temperature.
The relation between thermistor resistance and temperature is RT = R0 exp(T0/Ta) as
discussed in the Thermistor Experiment. Differentiation of RT with respect to temperature
Ta gives
T dT
dRT
=− 0 a
RT
Ta Ts
(19)
where Ta is the absolute temperature (K) (not the excess temperature, T(t, z) – TS) and
dTa is a small temperature change due to the heat pulse. Thus if dTa is small, it can be
approximated by the measured temperature change of the apparatus (i.e., the excess
temperature) and Ta can be approximated by room temperature. Appropriately combining
Equations (16), (18) and (19) gives the result
dTa = Ta
4 Ts dVa
G T0 V0
(20)
Where: Ta ≅ Ts is the room temperature.
As Equation (20) shows, the change in output voltage in volts is not important, only its
ratio with V0. This ratio dVA/V0 is equal to the ratio of the change in ADC units to the
ADC full-scale reading.
Activity 5: Thermal Conductivity and Specific Heat of Copper
1. We are only interested in the change of temperature, so we need to subtract the
initial value to get a change of the thermistor reading (dVa) versus time t.
2. Use equation (20) to get a change in temperature (dTa) versus time t (T0 = 3440 K
for the GB32J2 thermistor, V0 = 5 V).
3. Plot the curve of dTa versus time for the two thermistors (you will notice that the
curves you plotted look similar to the one you got in Exercise 1).
4. Normalize your curves to be the same as the one you got in Exercise 1. Find a
value of T1 so that the curve peaks at T/T1=0.43 and a value of t1 so that the peak
occurs at t/t1=2 for each curve.
5. Use these estimates (T1, t1) to draw curves on your graphs of the data and check
the fits. Then you may want to change your estimates and try other fits.
6. When you are satisfied with your values of T1 and t1, use them to calculate via
Equation (9) the thermal conductivity κ and volume heat capacity Cv. Calculate a
set of values using the data from each thermistor and average the values.
7. Calculate the diffusion constant D = κ/CV and the heat capacity C = CV/ρ, where ρ
is the density.
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ME224 Lab 5
8. Look up the values of D, κ, and CV in a RELIABLE source. Compare them to
your measured values.
9. Make an estimate of the error made in differential evaluation of the temperature
change. For doing this estimate, use the maximum change that can be measured
using the amplifier circuit employed.
Another consideration can be applied to the data analysis. In deriving Equation (5) we
assumed that the time during which the heater was on (t) was very small in relation to the
time the heat takes to diffuse down the rod (T1), i.e., it was an impulse of heat. In doing
your experiments this approximation is valid as long as you make t = 0 on your graph
correspond to the midpoint of the heating time and if the heating time is less than any t1.
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