Question

S: (30 points). The Pb-Sn phase diagram is given below for your Pb-Sn solid couple (say, two O'C for t, - I and , 4 mallbas in physical comtacy is diffusion heat treasted at Liguid L α+L α+ β wt% Sn 3.1 Write down the solution of the Fick's second law of diffusion for a semi-infinite bar, where C, is initial solubility of Sn in pure lead (Pb) at t = 0 and C, is the solubility of Sn in Pb at 150°C for r>0. d. None of the above 2 mm and 1 hour. The 3.2 Calculate C(x.4) and C(x.1,) of Sn at a diffusion depth ofx diffusivity of Sn in Pb is D = 2.5x10-12 m2/s . The answer is a. C(r,t.)-296 Sn c, C(x,h)-4% Sn d. None of the above 3.3 Calculate C(r,1) and C(x,4) of Sn at a diffusion depth ofx=2mm and 48 hour. The diffusivity of Sn in Pb is D-25x10-12 m2/s·The answer is a. C(r,r,)-15% Sn c. C.(x4)=68% Sn b, cr,J-32% Sn d. None of the above

Answer 1

3.1) Answer - b

Solution:

According to Fick's Second Law for semi infinite rod

$\frac{C-C_{o}}{C_{s} - C_{o}}= 1-erf(\frac{x}{2*(Dt)^{0.5}})$

Since the initial solubility of Sn in Pb is zero, substituting $C_{o}=0$ in the above equation

$C= C_{o}*(1-erf(\frac{x}{2*(Dt)^{0.5}}))$

3.2) Ans: b

Using the equation from the solution in 3.1,

x = 2mm, D = $2.5\times 10^{-12} m^{2}/s$ , t = 1hr = 36000s

$erf(\frac{x}{2*(Dt)^{0.5}})= erf(14.9)= 1$

Therefore,

C = $C_{s}$ *0 = 0% Sn

3.3)

Ans: d

Using the equation from the solution in 3.1,

x = 2mm, D = $2.5\times 10^{-12} m^{2}/s$ , t = 48hr = 48 * 36000s

$erf(\frac{x}{2*(Dt)^{0.5}})= erf(1.52145)= 0.9685699$

Therefore,

% Sn

Therefore None of these is the answer