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Unformatted text preview: lt/ft. The resistivity ρ is
calculated as ρ = RA/L = (0.0001)(π)(0.03125)2/(1) = 3.06796 x 10-7 Ω-ft.
A conversion factor 3.415 (Btu/hr)/W must be included in the resistivity ρ so that the Joule heat units match the
thermal units ρ/3.415 = 8.983782 x 10-8. Current printout is divided by 3.415 to get electrical (amp) units. The
steady-state convergence procedures are used.
For the thermal-electric axisymmetric elements (SOLID69), nodes 1 through 16 are assumed to be ground nodes
for reference. The solution is based on a unit radian model. Since the problem is axisymmetric, only a one-element
sector is needed. A small angle Θ = 10° is used for approximating the circular boundary with a straight-sided
POST1 is used to extract results from the solution phase. Total heat dissipation is computed parametrically at
the outer surface as HRATE using q = h.area.(Ts-Ta). Results Comparison
°F 419.9 420.0 1.00 417.9 417.8 1.00 q, Btu/hr 341.5 341.5 1.00 Centerline Temperature,
°F 419.9 418.6 0.997...
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This note was uploaded on 12/09/2010 for the course DEPARTMENT E301 taught by Professor Kulasinghe during the Spring '09 term at University of Peradeniya.
- Spring '09
- The Land