Lecture 13 - Ch 5 pt 3

# Lecture 13 - Ch 5 pt 3 - C h a p t e r 5- 3 5 N o n s t e a...

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Unformatted text preview: C h a p t e r 5- 3 5 N o n s t e a d y- S t a t e D i f f u s i o n E x a m p l e P r o b l e m : A n F C C i r o n- c a r b o n a l l o y i n i t i a l l y c o n t a i n i n g . 2 w t . % C i s c a r b u r i z e d a t a n e l e v a t e d t e m p e r a t u r e a n d i n a n a t m o s p h e r e t h a t g i v e s a s u r f a c e c a r b o n c o n c e n t r a t i o n c o n s t a n t a t 1 . w t . % . I f a f t e r 4 9 . 5 h t h e c o n c e n t r a t i o n o f c a r b o n i s . 3 5 w t . % a t a p o s i t i o n 4 . m m b e l o w t h e s u r f a c e , d e t e r m i n e t h e t e m p e r a t u r e a t w h i c h t h e t r e a t m e n t w a s c a r r i e d o u t . S o l u t i o n : u s e E q n . 5 . 5 = D t x C C C t x C o s o 2 e r f 1 ) , ( C h a p t e r 5- 3 6 S o l u t i o n : e r f ( z ) v a l u e s a r e g i v e n i n T a b l e 5 . 1 f o r v a r i o u s x / 2 D t v a l u e s C S C o C ( x , t ) ( ) = D t x C C C t , x C o s o 2 e r f 1 d y e y z 2 2 = C ( x , t ) = C o n c . a t p o i n t x a t t i m e t e r f ( z ) = G a u s s i a n e r r o r f u n c t i o n : C h a p t e r 5- 3 7 S o l u t i o n ( C o n t . ) : B o u n d a r y C o n d i t i o n s : t = 4 9 . 5 h x = 4 x 1- 3 m C x = . 3 5 w t . % C s = 1 . w t . % C o = . 2 w t . % = D t x C C C ) t , x ( C o s o 2 e r f 1 ) ( e r...
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## This note was uploaded on 07/27/2011 for the course ENGR 134 taught by Professor Marks during the Spring '11 term at Drexel.

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Lecture 13 - Ch 5 pt 3 - C h a p t e r 5- 3 5 N o n s t e a...

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