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Temperature of a 100 W Light Bulb Filament Joran Booth and Peter Hyatt ME 340 Winter 2009

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Heat Transfer Problem o Measured the electrical power dissipated o Measured the bulb surface temperature and the ambient temperature o Measured the diameter of the light bulb o Measured the thickness of the glass A SSUMPTIONS : o Near vacuum inside the light bulb o Geometry is a sphere o The radiation is a point source o Filament treated as a sphere o All electrical work converted to heat

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Solution Online sources show: T = 2550 ºC T s = 95 ºC T filament = ? k nitrogen = 81 W/m ·K A s, ε q in = 96 W 60 mm Ø 0.5 mm thick Glass Vacuum T avg ρcPcP μνkαPrbeta E E E E E-03 Ø fil = σ =5.67E-08h r1 = R 1 = Ø inner =0.0595Tin =368h r2 = R 2 =7.00E-01 Ø outer =0.06T S =368h =6.56E+00R 3 = k N =2T ∞ =300Nu = R 4 = k glass =81q =96 WRa =8.34E+05R 5 = A fil = ε glass =0.95R total = without R1 A glass =2.97E-03 ε W =0.38T filament =2681K Ts =2681

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Results

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Conclusions and Recommendations The temperature is between 4500 to 2500˚C. Main contributors to T filament : ◦ Surface Area of Filament ◦ Vacuum v. Conduction thru gas Measure different filament styles Verify/modify assumptions with other measurements

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Appendix qid= AAlv437. Fundamentals of Heat and Mass Transfer, Incropera, 6 th Edition.

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Additional Graphs Both graphs based on assumption of pure nitrogen inside bulb with no convection

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Calcuations A1 =2.97E-03 ε 1 =0.38Ts =2681KA1 =2.97E-05 ε 1 =0.38Ts =4316KA1 =2.97E-07 ε 1 =0.38Ts =11709K Ts =2681Ts =4320Ts =11690 A1 =2.97E-05 ε 1 =0.38Ts =4316KA1 =2.97E-05 ε 1 =0.27Ts =5010KA1 =2.97E-05 ε 1 =0.18Ts =4983K Ts =4320Ts =4320Ts =4984 A1 =2.97E-03 ε 1 =0.38Ts =2615KA1 =2.97E-05 ε 1 =0.38Ts =2643KA1 =2.70E-07 ε 1 =0.38Ts =2934K Ts =2615Ts =2643Ts = A1 =2.97E-05 ε 1 =0.38Ts =2643KA1 =2.97E-05 ε 1 =0.27Ts =2643KA1 =2.70E-05 ε 1 =0.18Ts =2644K Ts =2643Ts =2643Ts =2644 Assuming a vacuum Assuming nitrogen filled

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