homework2solution - Solutions to Homework 2 Problem 1 a...

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Solutions to Homework # 2 Problem 1 a) Calculate Dt and compare with the sample size to see if we should use short-time or long-time solution. Since Dt cm cm = × × = << 1 10 100 0 01 1 6 . we should apply the short-time solution. One of the two surfaces at x=0 is schematiclly shown in Figure. x C o C s = 0 c x =0 The initial and boundary conditions are given by: c x c g cm c t c t c ( , ) / ( , ) ( , ) 0 1 10 0 0 0 5 3 0 = = × = = The general error function solution is c x t A Berf x Dt ( , ) ( ) = + 2 Using the above boundary conditions, the constant A and B are given by: A = 0 B = c o Therefore, the concentration profile after 100s is: c x t c erf x Dt erf x Dt ( , ) ( ) ( ) = = × 0 5 2 1 10 2 b) The surface flux is given by J D c x c D t g cm s x x = = = − = − = − × × × = − × 0 0 0 5 6 10 2 1 10 1 10 100 564 10 π π . ( . ) c) The total amount of hydrogen diffused out of the nickel at 100s is given by m J dt c D t dt c Dt x = = = = 2 2 4 0 0 100 0 0 100 0 π π = × × × = × 4 10 10 100 2 26 10 5 6 7 2 π . ( g cm ) d) This is a long-time problem. Since 1
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c c = = < 0 6 5 10 10 01 08 . . we may apply the first term of series solution. c c Dt h = = 0 2 2 2 8 01 π π exp( ) . Solve the above equation, we get t t h D s = − × = − × = × 2 2 2 2 6 2 5 01 8 1 10 80 212 10 π π π π ln( . ) ln . So, after s, the average concentration of hydrogen will reduce to 1 212 10 5 . × 10 6 3 × g cm / Problem 2 a)
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