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Transcript
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 CVT Heat Flow z z+dz T2 dT P kA dz dq dT P C Adz dt dt 1 T T T2TT T Az kA kA C CCAz s k 2 A zt 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 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 Adding an “effective z” 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 Try other materials Have gold rod to make similar apparatus The End Thanks for listening.