Thermodynamics HW Solutions 362

Thermodynamics HW Solutions 362 - Chapter 4 Transient Heat...

This preview shows page 1. Sign up to view the full content.

Chapter 4 Transient Heat Conduction τ α == ×× => t L 2 6 115 10 0 025 6624 02 (. .. m / s)(60 min 60 s / min) m) 2 2 {} { } C 500 ° = ⎯→ = = ) , 0 , 0 , 0 ( 000186 . 0 ) 1016 . 1 ( ) 0580 . 1 ( 500 20 500 ) , 0 , 0 , 0 ( ) 624 . 6 ( ) 7910 . 0 ( 2 ) 624 . 6 ( ) 5932 . 0 ( 2 2 t T e e t T Note that τ > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction is applicable. Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r o = D /2 = 2.5 cm exposed to the hot gases with a heat transfer coefficient of h 40 W / m.C 2 and a plane wall of thickness 2 L = 5 cm exposed to the hot gases with h = ° 80 W / m.C 2 . After 10 minutes : The Biot number and the corresponding constants for the long cylinder are 400 . 0 ) C W/m. 5 . 2 ( ) m 025 . 0 )( C . W/m 40 ( 2 = ° ° = = k hr Bi o ⎯= = λ 11 08516 10931 and A To determine the center temperature, the product solution method can be written as [] [ ] C 370 ° = = = = = ) , 0 , 0 ( 271 . 0 ) 0931 . 1 ( ) 1016 . 1 ( 500
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

Ask a homework question - tutors are online