This is related to the configuration of atoms generating the tetrahedron and octahedron packing. The octahedron voids are more in austenite compared to that in ferrite.
Dear Prof. Ojha,
I think you wrote the answer by yourself: "The octahedron voids are more in austenite compared to that in ferrite."
I usually use following pictures to explain such a huge difference in carbon solubility in alfa and gamma Fe:
http://brykov.com/Ojha/
Hope this will help.
Best Regards,
Michael
Dear Dr. Michael,
Thanks for providing very good illustrations. To know the quantitative value, one needs to know the exact ratio of two configuration in both ferrite and austenite.
Its good question, yes solubility of C in ferrite only 0.025% whereas in Austenite this is 2% even the atomic packing factor in austenite (fcc) (4 atom per unite cell) is higher than ferrite (bcc) (2 atoms per unit cell. The calculation of the amount of empty space in unfilled space in f.c.c is 25% and in b.c.c lattice is 32%, so it may seem strange that the solubility of C in austenite is much greater than in ferrite. The reason is
The largest hole in b.c.c ferrite is halfway between the center of the face and space between the two corner atoms. Two of the four possible positions for a Carbon atom on the front face of a body-centered cube. The largest interstitial sphere that would just fit has a radius of 0.36(10)-8 cm. The largest hole in f.c.c austenite is midway along the edge between two corner atoms. The largest interstitial sphere that would just fit has a radius of 0.52(10)-8 cm. therefore, austenite will have a greater solubility for carbon than ferrite.
FCC has 8 and 4 octahedral and tertrahedral voids per unit cell respectively. whereas BCC has 12 and 6 respectively. octahedral in FCC is bigger than tertrahedral. C occupies octahedral with less distortion. Octahedral in BCC is smaller than tertrahedral. even in that case C occupies octahedral void due to lesser distortion (only top atom and bottom need to be distorted). No. of voids in BCC are more than FCC whereas size of voids in BCC are significantly smaller than FCC. For this reason austenite have higher solubility of C than ferrite.
Carbon occupies interstitial sites of iron mainly the octahedral sites because in FCC(austenite) , these octahedral sites are larger than in BCC(ferrite) , irrespective of the more number the octahedral sites in bcc (ferrite) than fcc (austenite), so mainly it zeros on the size of the octahaderal voids.
FCC interstitial radius(0.41R) is larger than BCC (0.291R). R is the radius of atom.so more carbon atoms are soluble in FCC then BCC.
i have follow up question, FCC is what we call compact structure, so how come the free space in FCC is more important than BCC
The size of octahedral void is larger in austenite and other factor is that the number of octahedral void is more in ferrite but the size of octahdral site is very less. r/R in octahedral in ferrie is 0.155 and r/R in octahderal void in austenite is 0.414.
Dear S.N.Ojha,
Carbon is generally more soluble in austentie than ferrite. Reason is a carbon atom occupies octahedral interstitial sites in these crystals. Octahedral interstitial site of BCC ferrite is much smaller for carbon than that of FCC austenite. This is why solubility of carbon in ferrite is much smaller than solubility of carbon in austenite. Possible maximum carbon content in austenite may be 2%. Maximum solubility of carbon in iron is about bout 0.02 wt% at 727 °C (1,341 °F) and 0.001% carbon at 0 °C (32 °F). When it dissolves in iron, carbon atoms occupy interstitial "holes".
Ashish
Dear Ashok,
Can you explain how BCC has 12 octahedral sites and FCC has 8?
As far as I know BCC has 6 and FCC has 4 octahedral site per unit cell. Ashok Kamaraj
- Badges
- Science topic
- Similar topics
- Engineering
- Mechanical Engineering