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2A4_Lecture02_2008

Analysis a l2 0015m 22510 4 m 2 2 4 pelec qconv

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Unformatted text preview: ll surface and a large enclosure, (3) Negligible heat transfer from sides of chip or from back of chip by conduction through the substrate. Analysis: A = L2 = ( 0.015m ) =2.25×10-4 m 2 2 4 Pelec = qconv + qrad = hA ( Ts − T∞ ) + ε Aσ ( Ts4 − Tsur ) (a) If heat transfer is by natural convection, 5/ 4 qconv = CA ( Ts − T∞ ) =4.2W/m 2 ⋅ K 5/4 2.25×10-4 m 2 qrad = 0.60 ( 2.25×10-4 m 2 ) 5.67×10-8 W/m 2 ⋅ K 4 Pelec = 0.158W+0.065W=0.223W (b) If heat transfer is by forced convection, ( ) ( 60K ) ( 358 -298 ) K 4 4 5/4 4 =0.158W =0.065W qconv = hA ( Ts − T∞ ) =250W/m 2 ⋅ K 4 ( 2.25×10-4 m 2 ) ( 60K ) =3.375W Pelec = 3.375W+0.065W=3.44W Announcements Announcements Assignment 1: Due January 16 at the end of class. at In-class Quiz: Monday January 21 In-class Monday Office hours & Mid-term dates: to be determined Tomorrow’s tutorial: Please, bring your textbook (if you have got it) you...
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