ENGR 321: Homework Set 4
WINTER 1997

Prof. W. H. Warnes
Office: Rogers Hall 308
Phone and Voice Mail: 737-7016

This page last updated February 17, 1997

email: warnesw@engr.orst.edu

HOMEWORK ASSIGNMENT 4
DUE IN CLASS FEBRUARY 17

 
PHASE DIAGRAMS
1) BNT 4-6
2) BNT 4-13
3) BNT 4-17
4) BNT 4-19
5) BNT 4-20

SOLUTIONS

1) Use the Gibbs Phase Rule: P + F = C + 1 (since we only consider T to be a variable, with pressure fixed at one atmosphere). In the problem, we know there are three components (C = 3).
Even if we have F = 0 (an invarient point), the most phases we can have in equilibrium together at one time is four, so the claim of five phases is not possible.

2) Refer to the binary phase diagram for Pb Sn (page128). The composition Pb 20wt%Sn appears on the phase diagram as 80wt% Pb. At 250 C, the alloy is on a two phase region, with an equilibrium between BETA phase solid, and LIQUID.

Using Gibbs Phase Rule, P + F = C + 1, we have 2 phases and two components, so that F = 1. This means that we can arbitrarily choose the temperature, but then the isotherm at that temperature will determine the composition of the phases in equilibrium. Alternatively, we can choose the composition of the liquid, and this will fix the temperature and the composition of the BETA phase. F = 1 means we are free to choose one variable (temperature or composition), but not both.

At 183 C, the alloy is at the eutectic temperature. Three phases (ALPHA, BETA and LIQUID) are in equilibrium, so that F = 0. The temperature of the eutectic point and the compositions of all three phases are uniquely determined, so there are zero degrees of freedom.

3) This problem has several parts. The first thing to do is to draw the vertical line at the composition we are interested in, 0.5wt% C (the red line in the figure below). At each of the four temperatures we get the following information:

Point on Plot

Discussion

Phases Present

Phase Compositions

Weight Fraction of Phases

A

At 1470C, the alloy is in the two phase region between GAMMA-Fe and liquid. We can use the tie line (isotherm) to determine the compositions of these two phases in equilibrium with each other. The tie line intersects the liquidus line at about 1wt%C (red spot on tie line), so this is the composition of the liquid phase. The tie line intersects the solidus line at about 0.35wt%C, which is the composition of the solid GAMMA

The lever rule can be used to find the weight fraction of the two phases: The amount of GAMMA phase is found from

(1.0-0.5)/(1.0-0.35)=77%.

GAMMA + LIQUID

0.35wt%C, 1.0wt%C.

77% by wt. 23% by wt.

B

At 1100C the alloy is in the single phase GAMMA region, so no tie line is needed.

GAMMA

0.5wt%C

100%

C

At 740C, just above the eutectic temperature, the alloy is in a two phase region with ALPHA and GAMMA iron. The tie line on the figure intersects the phase boundaries and gives the phase compositions. The lever rule gives the relative amounts of the phases.

GAMMA + ALPHA

0.02wt%C, 0.6wt%C

79% by wt. 21% by wt.

D

At 500C the alloy is in the two phase ALPHA and Fe3C (cementite, or iron-carbide) region. The tie line on the figure intersects the phase boundaries and gives the phase compositions. The lever rule gives the relative amounts of the phases.

ALPHA + Fe3C

0.01wt%C, 6.7wt%C

93% by wt. 7% by wt.
The largest carbon content in the austenite (GAMMA) iron phase occurs at point E on the figure below, at about 2.14 wt% C and a temperature of 1147C.

To answer the last question about the structure at 700C, we need to first look at a temperature just above the eutectoid temperature. At T=728C (point F on the figure), we have a balance between ALPHA-Fe and GAMMA-Fe. We can perform a lever rule at this temperature to find out how much ALPHA has formed before the eutectoid reaction. ALPHA has a composition of 0.022wt%C, and GAMMA of the eutectiod composition, 0.76wt%C. Using the lever rule gives us 35% by weight of ALPHA, and 65% GAMMA.

On cooling through the eutectoid temperature, there is no change in the ALPHA phase. Only the GAMMA phase changes, converting to ALPHA+Fe3C. This adds ALPHA phase to the alloy (see point D above), which appears in the eutectoid lamellar structure, and brings the total amount of ALPHA up to 93% of the alloy. The answer to the question is that only 35wt% of the alloy is ALPHA that formed BEFORE the eutectoid reaction.

4) In this problem we are asked to mix 50% SiO2 with 50% mullite by weight. From the phase diagram (see below) we can see that mullite is a mixture of SiO2 and Al2O3 with a composition of about 72wt% Al2O3. Thus a 50/50 mix of SiO2 and mullite will have a composition of 36wt%Al2O3 (the vertical red line below). The phases present on cooling from 2000C are summarized as follows:

Point on Plot

Phases Present

A

100% liquid

B

The first solid forms, and is mullite.

C

In this two phase region there is equilibrium between liquid and solid mullite.

D

The eutectic temperature, which has three phases in equilibrium, liquid, mullite and (essentially) pure SiO2.

E

The two phases region between SiO2 and mullite.

The ceramic alloy is completely solid at the eutectic temperature of 1595C.

5) By adding 10% by weight of mullite to the silica (which is about 7.2wt%Al2O3 overall composition), the engineer has made a lower melting point material because of the eutectic in the system. His/her bricks will melt completely at about 1620C, but will BEGIN to melt at the eutectic temperature of 1595C.

 

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