TUGAS METALURGI FISIK II

Disusun Oleh :Abdullah Fikri

( 3334131750 )

Andri Subekti

( 3334131854 )

Putri Dewi Novianti

( 3334132213 )

Sukma Suci Fadarani

( 3334131140 )

Try Alif Shandy

( 3334130142 )

TEKNIK METALURGIFAKULTAS TEKNIK

UNIVERSITAS SULTAN AGENG TIRTAYASA

CILEGON BANTEN2015Contoh (1) :

A thick-walled pipe 3 cm in diameter contains a gas including 0.5 x 1020 N atoms per cm3 on one side of a 0.001 cm thick iron membrane. The gas is continuously introduced to the pipe. The gas on the other side of the membrane contains 1 x 1018 N atoms per cm3. Calculate the total number of nitrogen atoms passing through the iron membrane at 700 oC if the diffusion coefficient for nitrogen in iron is 4 x 10-7 cm2.s-1.Contoh (2) :Koefisien difusi untuk alumunium dalam paduan tembaga pada temperatur 200 oC dan 500 oC adalah masing-masing (0.8 dan 2,5) x 10-20 cm2.dt-1. Hitung energi aktivasi untuk difusi alumunium dalam paduan tembaga.

Jawab :

1. Dik :

C1 = 0.5 x 1020 N atoms per cm3 = 5 x 1019 N atom per cm3C2 = 1 x 1018 N atoms per cm3 C = C2 C1 = (1 x 1018 - 5 x 1019) N atom per cm3 = -4.9 x 1019 N atom per cm3 = 0.001 cm

D = 4 x 10-7 cm2/sDit : Total atom hydrogen/s

Jawab :

J = - D

= - (4 x 10-7 cm2/s) ()

= 196 x 108 N atom per cm3

= 1.96 x 1010 N atom per cm3

Total atom/s = J.A = J ( d2)

Total atom/s = (1.96 x 1010) () (32)

Total atom/s = 13.85 x 1010 atom/s

Total atom/s = 1.385 x 1011 atom/s2. Dik : T = 2000C = 473 K ( D = 0,8 x 10-20 cm2 dt-1

T = 5000 C = 773 K ( D = 2,5 x 10-20 cm2 dt-1Dit : G = ...?

Jawab :

Ln 3,125 = 0,000098677 Q

1,1394 = 0,000098677 Q

Q = 11547,111