Latihan Soal Difusi Padatan

Latihan Soal Difusi Padatan 1. A sheet of steel 1, 5 mm thick thick has has nitrogen nitrogen atmosph atmospheres eres on on both sides at at 1200 °C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient coefficient for nitrogen in steel at this temperature is 6 x 10 -11 m2/s, and the diffusion flux is found to be 1,2 x 10 -7 kg/m2s. Also, it is known that the concentration of nitrogen in the steel at the high pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2,0 kg/m3? Assume Assume a linear concentration profile. 2. Determine Determine the the carburizing carburizing time necessar necessary y to achieve achieve a carbon carbon concentra concentration tion of  0.45 wt% at a position 2 mm into an iron – carbon carbon alloy that initially contains 0.20 wt% C. The surface concentration is to be maintained at 1.30 1.30 wt% C, and . the treatment is to be conducted at 1000 C. (D0 = 2,3 x 10 5 m2/s) 3. At what what temperature temperature will will the diffusio diffusion n coefficient coefficient for the diff diffusion usion of of copper copper in nickel have a value of 6.5  10-17 m2/s. Use the diffusion data (tabulasi data difusi). 4. The diffu diffusion sion coef coefficient ficientss for iron in in nickel nickel are given given at at two temperatu temperatures: res: T (K) D (m2/s) 16 1273 9.4 × 10 16 14 1473 2.4 × 10 14 a. Determine the the values values of D0 and the activation energy Q d.  b. What is the magnitude of D at 1100ºC 1100ºC (1373 K)?  – 

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Jawaban Latihan Soal 1. Gunakan rumus laju difusi steady state xA = 0 (pada permukaan) C A  C B =   J   D D = 6 x 10 11 m2/s  xA   xB 3 CA = 4 kg/m C A  C B  CB = 2 kg/m3  xB =  xA +  D   J = 1,2 x 10 7 kg/m2.s  J    Jadi, xB = 1 mm  – 

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2. Gunakan rumus laju difusi non steady state Cx = 0,45 C0 = 0,20 Cs = 1,30 x = 2 mm = 2

C  x  C 0

  x   = 1  erf       2 C  C   Dt   s 0



10-3 m

 x erf   = 1  0.2273 = 0.7727 2  Dt  

z = 0,854 (hasil interpolasi)  x  z  = 0.854 = 2  Dt  Jadi, t = 7.1  104 s = 19,7 h

Jawaban Latihan Soal 3. Gunakan rumus hubungan koefisien difusi dengan temperatur  D0 = 2,7  10-5 m2/s Qd = 256000 J/mol D =6.5  10-17 m2/s Gunakan R = 8,31 J/mol.K 

T  = 

Qd 

Jadi, T = 1152 K 

 R (ln D  ln D0 )

4. Gunakan rumus hubungan koefisien difusi dengan temperatur 

Qd  =   R

ln D1  ln D2 1 T 1



 Qd    D0 =  D1 exp   RT     1 

1 T 2

Qd = 252400 J/mol D0 = 2,2 x 10-5 m2/s D (saat 1100°C) = 5,4 x 10 -15 m2/s