{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture note 16 _10-27-2011__1_

# Lecture note 16 _10-27-2011__1_ - MAE 581 Advanced...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAE 581 Advanced Materials Science Lecture note 16 Thermodynamics of Alloys The equilibrium conditions (Eqs. 1 and 2) are given by 0 Example 1 In a binary system of A and B atoms with the phases α and β in equilibrium at constant T, V, each change in F can be described as the sum of the changes in both phases. Thus in equilibrium 0 , , We can simplify the above equation as follows 0 Because , we can further simplify the above equation as follows 0 Here and 1 We then can get the equilibrium condition for this binary system and Example 2 The following figure is the phase diagram for the Cu-Ag system. The region above AEB is the liquid phase, to the right of ACF is the α-solid solution phase, and to the left of BDG is the βsolid solution phase. The triple point E is called the eutectic point. K 1,083 B liquid phase A Temperature β α D 200 E 779 G Cu 960.5 C H F weight fraction Ag (1) How will a solution behave if it is cooled down from the liquid state K in the figure? (2) How will it behave if an α-solid solution is heated up from the point H in the figure? 2 Solution of Example 2 (1) The temperature merely drops in the liquid phase along the vertical line passing through K until it reaches the intersection K′, which can be seen in the figure below, with curve BE. At the temperature of K′ the β phase of the solid solution starts to solidify the concentration of which is given by K′′. Since the solid solution which solidifies has a smaller concentration of Ag than the liquid phase, the Ag concentration of the remaining liquid becomes greater. So, as the temperature decreases further, the β phase solid solution freezes out with increasing Ag concentration. Thus, the state of the liquid solution travels along K′E and accordingly the β phase solid solution along K′′D. When the liquid reaches the point E, the α phase solid solution starts to freeze out and thereafter, the concentration of the liquid phase remain constant while the α phase (C) and β phase (D) solids are freezing out until all is solidified. (When the temperature is further decreased to, say, K1 and a sufficiently long time has passed, then the solid solutions corresponding to K α and K β should exist as stable solid phases. The ratio for the numbers of / atoms is equal to : . However this process takes a very long time because it is a reaction between solid phases.) K 1,083 B liquid phase A K′′ K′ Temperature β α D K 200 960.5 E 779 H K K G Cu C F weight fraction 3 Ag (2) Similarly, in the figure shown below, when the temperature is raised from H, the α phase remains unchanged until it reaches concentration corresponding to , when the liquid phase starts to appear with a . The Ag concentration is less than that of the solid. If a sufficiently long time is allowed, the α phase moves to that with higher concentrations of Ag, namely to the right in the figure. Correspondingly, the liquid phase also becomes richer in Ag, so that both and move to the right. Finally and all becomes liquid. arrives at in the figure when arrives at must have the same concentration as the original point H, because all is liquid. This condition determines the position of . After this stage, heating merely increases the temperature of the melt. If, however, the melted solution is constantly separated, the solid phase continues past and will first become pure solid silver at A and then become pure liquid Ag. K 1,083 B liquid phase A H H Temperature β D E 779 H 960.5 H C H α 200 G Cu F weight fraction 4 Ag Example 3 Consider the phase equilibrium between a binary solution and its vapor. The figure shows that relationship between the mole fraction in the liquid or the gas phase and the temperature under constant pressure. The gas phase is above the solid curve and liquid phase lies below the dotted curve. Suppose now that the gas mixture at the point P in the figure is cooled down gradually. At point A, the liquid phase appears and at point C, the gas phase disappears completely. (1) Find the mole fractions of the solute at A (2) Find the mole fractions in the gas at C (3) Find the mole fractions in the liquid and the gas at an intermediate stage represented by a point B in the figure. Show that the ratio of the mass in the gas and the liquid phases at this point is given by / . P T 1 2 5 Solution of Example 3 of the solute in the liquid phase which appears at point A is given by the (1) The mole fraction intersection A of the horizontal line drawn from the point A and the dotted line (liquid line). in the gas phase at point C is given by point (2) Also the mole fraction . (3) Generally at an intermediate temperature corresponding to point B, the mole fraction of the solute in the liquid phase and that in the gas phase are given by the and that of the point coordinate of the point respectively. Let the mole number in the gas phase be and that of the . Then one has liquid phase be , where and are the mole fractions in the gas phase and in the liquid phase, and are the coordinates of the point and the point , respectively. If this is divided by the total mole number N, one obtains and therefore ⁄ ⁄ ⁄1 This shows the required results. 6 Example 4 Consider a binary solution, which contains moles of the constituent 1 and be some extensive quantity of this solution ⁄ moles of 2. Let J changes with the mole fraction , for instance, as shown by the solid line in the figure shown below. The broken line QPR shows the tangent to this curve at P. Prove the following relationships: (1) QO J (2) RO J J ,T, ,T, J R J n B n P Q A O O 0 1 7 Solution of Example 4 By assumption, J is extensive as a function of and , Differentiate both sides by ; that is , (1) 1, to get and put , (2) and 1 Therefore, 1 (3) (Here we have used the Gibbs-Duhem relation which will be derived as follows: . Hence From Eq. (2), we have, 1 0 . Now the equation of the tangent QPR is given by 1 where is the coordinate of the point P. 8 At the point Q: 0, OQ J At the point R: 1, RQ J 9 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern