example2_00 - Examples Problem A large piece of glass with...

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1 Examples Problem A large piece of glass with a density of 100 g/cm 3 and a thickness of 1 cm containing 2.0 at % He is annealed at high temperatures for removing He. The surface concentrations of He for both surfaces are kept at 1.0 at % by a gas source. The diffusion coefficient of He in glass is 10 -10 cm 2 /s. (a) Write the differential equation for the evolution of concentration profile with appropriate initial and boundary conditions. (b) Derive the concentration profile of He in the glass after 10 4 s of annealing (c) Calculate the He flux per unit area at the surface at 10 4 s (d) Calculate the total amount of He per unit area which has diffused out of glass at 10 4 s (e) How long will it take to reduce the average concentration of He in glass to 1.5 at %? Solution: (a) Calculate Dt and compare with the sample size to see if we should use short-time or long- time solution. Dt = 10 - 10 × 10 4 = 10 - 3 cm << 1 cm which means we may apply the short-time solution. For constant surface concentration, the general error function solution can be used. Since the problem is symmetric, we can solve the concentration profile for one surface and obtain that for the other surface from the symmetry of the problem. Let's define one of the two surfaces at x = 0 as shown in Fig. 1. Since we view the sample as an infinite system, the appropriate initial and boundary conditions are given by C s = 0 C o x =0 Fig. 1 Cx ,0 (29 = C o = 2% × 100 gcm 3 = 2 3 C 0, t = C s = 1 3 C , t = C o = 2 3 The general error function solution is , t = A + Berf x 2 Dt Using the above boundary conditions, the undetermined constants, A and B, are given by
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2 A = 1 g/cm 3 ; B = C o - C s = 1g/cm 3 Therefore, the concentration profile at t = 10 4 s is given by Cx , t (29 = 1 + erf 500 x gcm 3 (b) The surface flux is given by J x = 0 =- D C x x = 0 DC o - C s 2 π e - x 2 4 Dt 1 2 Dt x = 0 D t 10 - 10 10 4 5.64 × 10 - 8 2 s (c) The total amount of He having diffused out of the glass at 10 4 s is given by m = 2 J x = 0 0 t o dt the factor 2 comes from two surfaces = 2 C o - C s D 1 t o t o dt = 4 C o - C s Dt = 4 10 - 10 × 10 4 = 2.26 × 10 - 3 2 (d) Since C - C s C o - C s < 0.8, we may apply the first term series solution, C - C s C o - C s = 8 2 e - 2 Dt h 2 Solve the above equation for t, we have t h 2 2 D ln C - C s 2 C o - C s 8 1 2 2 10 - 10 ln 0.5 2 8 = 4.9 × 10 8 s
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3 Problem A thick walled steel pressure vessel in an oil refinery contains high pressure hydrogen for a long time. To avoid hydrogen cracking on cooling, the vessel is to be held at temperature with no hydrogen inside until most of the H has diffused out. As initial and boundary conditions take for C(x,t): C ( x ,0) = C 1 + C 2 - C 1 (29 x / h C (0, t ) = C ( h , t ) = 0 (a) Derive C(x,t).
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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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example2_00 - Examples Problem A large piece of glass with...

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