Transcript Heat Flow in a Copper Rod

Heat Flow
in a Copper Rod
Alexander Williamson
Dr. Bruce Thompson
Ithaca College
A
p
p
a
r
a
t
u
s
Sample Data Set
1.6
z ~ 1"
z ~ 2"
1.4
Temperature change (K)
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
0
5
10
15
20
25
time (s)
30
35
40
45
50
T1
Analytical Model
q  CVT
Heat
Flow
z
z+dz
T2
dT
P  kA
dz
dq
dT
P
 C Adz 
dt
dt
1 T
TT
T2TT
Azs
kA
kA

C CCAz
k
2

A zt
tt
zz 
  z 2s 
Q

T  Ts 
exp 
A kst
 4kt 
Temperature vs Distance
Temperature vs Time
Data and Model Compared
The problem:
Heat is being lost to convection and radiation effects.
Convective and Radiative Heat Loss
Adds new term to partial differential equation.
h : transfer coefficient for free air
σ : Stefan-Boltzman constant
Ta: ambient room temperature
r: radius of the rod
T
 T 2
3
s
 k 2     h  4 Ta  T
t
z  r 
2
Heat Loss
z
T1
Heat
Flow
z+dz
T2
Solving The New PDE
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Analytical solution is impossible, so…
Now we turn to Matlab’s PDE solver!
Breaks up rod into n pieces along z and time
into m time steps
At first, very inconsistent: irrelevant parameters
changed function drastically
Realized amount of heat added was changing
Needed more detail near z=0 and t=0
Changed from linear to logarithmic steps
First solved original PDE with new method to confirm its accuracy
First solved original PDE with new method to confirm its accuracy
Next, added in heat loss factor and renormalized
method for calculating the peak time and amplitude
Looks good!
Almost perfect!
Next, added in heat loss factor and renormalized
method for calculating the peak time and amplitude
Looks good!
Almost perfect!
UH-OH!
Next steps
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Adding an “effective z”
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Heat conducting epoxy
around resistor conducts heat
~1000 times more slowly
Rough trials indicate more
like 3:2 ratio than 2:1 ratio
After Consistent Model
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Try other materials
Have gold rod to make similar
apparatus
The End
Thanks for listening.