ENGR 321: Homework Set 6
Fall 2004

Prof. W. H. Warnes
Office: Rogers Hall 308
Phone and Voice Mail: (541) 737-7016
FAX: (541) 737-2600

This page last updated December 4, 2004

DUE IN CLASS Friday, December 3, 2004

PHASE DIAGRAMS

1) Referring to the Fe-C phase diagram, describe the equilibrium cooling process for a steel with a composition of 1.0 wt% C as it slowly cools from the liquid state. Be sure to describe the phases present, their approximate compositions and the relative amounts of each, and the probable microstructure.

2) An Fe-C alloy is examined and is found to have a microstructure consisting of pro-eutectoid Ferrite surrounded by the eutectoid microstructure of Pearlite. At room temperature there is about 50% of the pro-eutectoid Ferrite phase. Use the lever rule to determine the overall alloy composition in terms of wt%C.

KINETICS

3) Refering to figure 10.9 for the transformation of austenite, briefly explain why it takes a long time for the transformation to begin at high temperatures (near the eutectoid temperature) and also at low temperatures (around 250 °C), but less time at intermediate temperatures.

4) A certain phase transition ( ALPHA phase --> BETA phase) is found to obey the Avrami relationship, with n = 1.75, and k = 0.003 [sec^-1.75] at a given temperature. Make a plot of the amount of transformed BETA phase versus the time to transform, using a LOGARITHMIC time scale. Determine the rate of the transformation process.

5) Refering to Figure 10.9, describe the microstructure of the final steel using the following cooling processes:

from 750 °C rapidly cool to 650 °C, hold for 3 seconds, and rapidly cool to room temperature;
from 750 °C rapidly cool to 550 °C, hold for 3 seconds, and rapidly cool to room temperature;
from 750 °C rapidly cool to 300 °C, hold for 1000 seconds, and rapidly cool to room temperature;
from 750 °C rapidly cool to 650 °C, hold for 10 seconds, rapidly cool to 400 °C, hold for 1000 seconds,and rapidly cool to room temperature.

SOLUTIONS

The composition of 1wt%C is shown as ther vertical RED line in the figure to the right. I have marked 7 key points along the cooling curve to describe the cooling process. (NOTE: temperatures and compositions are approximate, taken by eyeball off the phase diagram.)

a) Above point a (about T > 1470 °C), the system is in the single phase Liquid region:

100% L of composition 1wt%C
structure: 100% L

b) For temperatures between T_a = 1470 °C and T_c = 1350 °C, the system is in the two-phase region of GAMMA and Liquid. For the temperature shown, T_b = 1400 °C, the isotherm (in green) gives the following compositions of the phases, and the lever rule gives the relavtive amounts of each phase as:

72% L of comp. 1.66wt%C and
28% GAMMA of comp. 0.75wt%C
structure: L with islands of GAMMA solid.

c-d) For temperatures between T_c = 1350 °C and T_d = 800 °C, the system is in the single phase GAMMA region:

100% GAMMA of composition 1wt%C
structure: 100% GAMMA

e) For temperatures between T_d = 800 °C and T_f = 727 °C, the system is in the two-phase region of GAMMA and Fe₃C. For the temperature shown, T_e = 750 °C, the isotherm (in green) gives the following compositions of the phases, and the lever rule gives the relavtive amounts of each phase as:

2% Fe₃C of comp. 6.70wt%C and
98% GAMMA of comp. 0.9wt%C
structure: Grains of GAMMA solid
with precipitates of Fe₃C solid.

f) At T_f = 727 °C, the system goes through the EUTECTOID transformation, and the remaining GAMMA phase transforms into ALPHA + Fe₃C in the lamellar, eutectoid microstructure.

g) For temperatures below T_f = 727 °C, the system is in the two-phase region of ALPHA and Fe₃C. For the temperature shown, T_g = 500 °C, the isotherm (in green) gives the following compositions of the phases, and the lever rule gives the relavtive amounts of each phase as:

2% Fe₃C of comp. 6.70wt%C and
98% ALPHA of comp. 0.01wt%C
structure: PEARLITE (eutectoid mixture of ALPHA
and Fe₃C solids) with small
precipitates of Fe₃C solid.

NOTE: You only need to show ONE of the lever rule calculations/estimates, along with the rest of the single phase information, for credit on this problem.

The problem tells you that 1/2 of the microstructure is PEARLITE, and the other half is ALPHA-Fe that formed above the eutectoid temperature (PRO-EUTECTOID Ferrite). The pearlite comes from the decomposition of GAMMA phase of the eutectoid composition (0.76wt%C) into the equilibrium phases of ALPHA and Fe₃C. The only temperature at which eutectoid GAMMA and ALPHA are in equilibrium is just above the eutectoid temperature. The microstructure must have formed at this temperature. (Looking at the phase diagram, there are essentially no changes in the system as it cools down from the eutectoid temperature to room temperature.)

Use the lever rule at a temperature of about T = 728 °C. We know the composition of the GAMMA phase at this temperature (0.76wt%C) and the composition of the ALPHA-Fe is 0.022wt%C. These are the two ends of the lever. The alloy composition we are trying to find is the fulcrum of the lever. Since the mass fraction of ALPHA is 50%, the fulcrum must be at the mid-point between the two end compositions. This gives the alloy composition as:

C_alloy = (0.76 - 0.022)/2 = 0.37wt%C.

At high temperatures, there is lots of diffusion to allow growth of the new phases, but the undercooling is too small to permit nucleation to occur. With no nucleii, the phase tansformation cannot take place, and the time required is very long. This is the NUCLEATION LIMITED region.

At low temperatures, there is lots of undercooling, so the nucleation rate is high. However, the temperature is low enough that the diffusion is very slow. The nucleii of the new phases form, but they cannot grow to complete the transformation. This is the DIFFUSION LIMITED region.

At intermediate temperatures there is both enough undercooling to develop nucleii, and high enough thermal energy to allow diffusion, and the transformation takes place more rapidly.

Using the Avrami equation below, I made the following plot in EXCEL.

The RATE is defined as the inverse of the time needed for 50% transformation. Using the AVRAMI equation, the time for 50% transformation is:

5) Refering to Figure 10.9, describe the microstructure of the final steel using the following cooling processes:

(a) from 750 °C rapidly cool to 650 °C, hold for 3 seconds, and rapidly cool to room temperature;

The RED lines to the right show that, after three seconds at 650 °C, the austenite (A) has still not begun to transform to the equilibrium phases of ferrite and cementite (in the pearlite (P) microstructure). Rapidly cooling to room temperature "freezes in" the metastable austenite phase:

Ans: 100% austentite.

(b) from 750 °C rapidly cool to 550 °C, hold for 3 seconds, and rapidly cool to room temperature;

The GREEN lines show that, after three seconds at 550 °C, about 75% of the austenite has transformed into the pearlitic microstructure:

Ans: 75% pearlite, 25% austenite.

(c) from 750 °C rapidly cool to 300 °C, hold for 1000 seconds, and rapidly cool to room temperature;

The RED lines show that, after 1000 seconds at 300 °C, about 75% of the austenite has transformed into the equilibrium phases of ferrite and cementite, in the bainitic microstructure:

Ans: 75% bainite, 25% austenite.

(d) from 750 °C rapidly cool to 650 °C, hold for 10 seconds, rapidly cool to 400 °C, hold for 1000 seconds,and rapidly cool to room temperature.

The BLUE lines show that, after ten seconds at 650 °C, about 50% of the austenite has transformed into the pearlitic microstructure, so that we have 50% pearlite, 50% austenite. Dropping the temperature to 400 °C effectively resets the clock on the transformation of the remaining austenite (the pearlite does not change, since it is already the equilibrium phases). After an additional 1000 seconds at 400 °C the remaining austenite transforms fully to bainite:

Ans: 50% pearlite, 50% bainite.

End of File.