503hw2 - Solution to homework#2 1 Since C t C s 0.95Cs C s...

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Solution to homework #2 1. Since Ct C CC C s s ss s () . .. - - = - - =< 0 095 0 005 08 , the first term of the solution can be used: C s s - - 0 =0.05= 8 2 2 2 p p exp - & L ± ² l ³ Dt L therefore, L Dt 2 2 2 005 8 = - & L ± ² l ³ p p ln . L= - & L ± ² l ³ - p p 25 2 10 1 8 xx x ln . = 0.006cm 2. (a) For the detector to lose its sensitivity due to diffusion at different temperature, the diffusion distance is constant: DT t 11 2 2 = or D D DQ R T R T t t 1 2 01 02 2 1 = - - = exp[ / ( )] exp[ / ( )] therefore, Q RT T t t 21 2 1 - & L ± ² l ³= & L ± ² l ³ ln 2(a) and Q tt TT R x kJ mol = - = +- + = ln / // ln / /( ) ) ( / ) 250 55 1 273 110 1 273 162 8 314 40 3 . According to Eqn. 2(a) and using t 1 =55hr, T 1 =435K and T 3 =298K: ln( / ) . . t x 3 3 55 40 3 10 8 314 1 298 1 435 =- & L ± ² l ³ t 3 =9300 (hr) (b) Assume that the detector loses its sensitivity when the diffusion distance exceeds the spacing of the interlayer, i.e. Dt nm = 13 Therefore, D = (13x10 -7 cm) 2 /(55hr x 3600 s/hr)=8.5x10 -18 cm 2 s -1 3. (a) Since Dt cm s x s cm cm r = = << = -- 10 1 0 01 0 5 42 1 , the short time solution can be applied: C(x,t)=A+Berf x Dt 2 & L ± ² l ³ From C(0,t)=0 and C(x,0) = C 0 , we have A=0 and B=C 0 .
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Therefore, C(x,t)=C 0 erf x Dt 2 & L ± ² l ³ where C 0 =25x10 -6 x7.8gcm -3 =1.95x10 -4 gcm -3 . The flux of hydrogen from the surface is JD C x DC x erf x Dt DC Dt x Dt xx x x | | | exp | == = = =- & L ± ² l ³ Ø º Œ ø û œ - & L ± ² l ³ 00 0 0 0 2 0 2 21 2 4 p JC D t x x gcm s x |.
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This note was uploaded on 01/19/2010 for the course MAT SCI 503 taught by Professor Chen during the Spring '02 term at Penn State.

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503hw2 - Solution to homework#2 1 Since C t C s 0.95Cs C s...

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