Download presentation

Presentation is loading. Please wait.

Published byKarley Bazley Modified about 1 year ago

1
Temperature of a 100 W Light Bulb Filament Joran Booth and Peter Hyatt ME 340 Winter 2009

2
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

3
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 3321.054910082.00E-051.91E-052.87E-022.72E-050.7033.01E-03 Ø fil =0.06149 σ =5.67E-08h r1 =481.0875139R 1 =-0.0434 Ø inner =0.0595Tin =368h r2 =8.111165792R 2 =7.00E-01 Ø outer =0.06T S =368h =6.56E+00R 3 =0.00028 k N =2T ∞ =300Nu =13.71767276R 4 =53.901 k glass =81q =96 WRa =8.34E+05R 5 =43.6038 A fil =0.00283 ε glass =0.95R total =24.8045without R1 A glass =2.97E-03 ε W =0.38T filament =2681K Ts =2681

4
Results

5
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

6
Appendix http://answers.yahoo.com/question/index? qid=20080508134327AAlv437. Fundamentals of Heat and Mass Transfer, Incropera, 6 th Edition. http://members.misty.com/don/bulb1.html

7
Additional Graphs Both graphs based on assumption of pure nitrogen inside bulb with no convection

8
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 =2933.96 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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google