7/30/2019 Latihan Soal Difusi Padatan

1/3

Latihan Soal Difusi Padatan

1. A sheet of steel 1, 5 mm thick has nitrogen atmospheres on both sides at

1200C and is permitted to achieve a steady-state diffusion condition. The

diffusion 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 a linear concentration profile.

2. Determine the carburizing time necessary to achieve a carbon concentration of0.45 wt% at a position 2 mm into an ironcarbon alloy that initially contains

0.20 wt% C. The surface concentration is to be maintained at 1.30 wt% C, and

the treatment is to be conducted at 1000C. (D0 = 2,3 x 105 m2/s)

3. At what temperature will the diffusion coefficient for the diffusion of copper in

nickel have a value of 6.5 10-17 m2/s. Use the diffusion data (tabulasi data

difusi).4. The diffusion coefficients for iron in nickel are given at two temperatures:

T (K) D (m2/s)

1273 9.4 1016

1473 2.4 1014

a. Determine the values of D0 and the activation energy Qd.

b. What is the magnitude of D at 1100C (1373 K)?

.

7/30/2019 Latihan Soal Difusi Padatan

2/3

Jawaban Latihan Soal

J = DCA CB

xA xB

1. Gunakan rumus laju difusi steady state

xB = xA + DCA CB

J

xA = 0 (pada permukaan)

D = 6 x 1011

m2

/sCA = 4 kg/m

3

CB = 2 kg/m3

J = 1,2 x 107 kg/m2.s

Jadi, xB = 1 mm

2. Gunakan rumus laju difusi non steady state

Cx C0Cs C0

= 1 erfx

2 Dt

Cx = 0,45

C0 = 0,20

Cs = 1,30

x = 2 mm = 2 10-3 m

erfx

2 Dt

= 1 0.2273 = 0.7727

z = 0,854 (hasil interpolasi)

z = 0.854 =x

2 Dt

Jadi, t = 7.1 104 s = 19,7 h

7/30/2019 Latihan Soal Difusi Padatan

3/3

Jawaban Latihan Soal

3. Gunakan rumus hubungan koefisien difusi dengan temperatur

T = QdR(lnD lnD0)

D0 = 2,7 10-5 m2/sQd = 256000 J/mol

D =6.5 10-17 m2/s

Gunakan R = 8,31 J/mol.K

Jadi, T = 1152 K

4. Gunakan rumus hubungan koefisien difusi dengan temperatur

Qd = RlnD1 lnD2

1

T1

1

T2

D0 = D1 exp

Qd

RT1

Qd

= 252400 J/mol

D0 = 2,2 x 10-5 m2/s

D (saat 1100C) = 5,4 x 10-15 m2/s