homework2solution

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 =× × = < < 11 0 1 0 0 0 0 1 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: cx c g cm ct ctc (,) / 01 1 0 00 0 53 0 ==× = ∞= The general error function solution is cxt A Ber f 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: cer f x Dt erf x Dt ( ) ( ) == × 0 5 2 0 2 b) The surface flux is given by JD c x c D t g cm s xx =− =− × × × × 0 5 6 10 2 0 0 100 564 10 ∂π π .( . ) c) The total amount of hydrogen diffused out of the nickel at 100s is given by mJ d t c D t dt c Dt x == = = ∫∫ 22 4 0 0 100 0 0 100 0 ππ × × 41 0 10 100 226 5 6 7 2 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 π exp( ) . Solve the above equation, we get t t h D s =− × × 2 2 2 26 2 5 8 1 10 80 212 10 ln( . ) ln . So, after s, the average concentration of hydrogen will reduce to 1 10 5 . × 10 63 × gc m / Problem 2 a) The failure of the detector means that the diffusion distance of elements reaches the magnitude of the sample size, i.e.
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homework2solution - Solutions to Homework # 2 Problem 1 a)...

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