Solution6

Solution6 - HW 6 Solution 10-19(a L 49 wt W at W= 49 183.85...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
HW 6 Solution 10-19 (a) L: 49 wt% W, at% W= % 7 . 32 % 100 847 . 71 / 51 85 . 183 / 49 85 . 183 / 49 = × + α: 70 wt% W, at% W= % 1 . 54 % 100 847 . 71 / 30 85 . 183 / 70 85 . 183 / 70 = × + (b) wt% L = % 6 . 47 % 100 49 70 60 70 = × wt% α= 52.4% The original composition , in wt% Nb, is: % 1 . 43 % 100 847 . 71 / 40 85 . 183 / 60 85 . 183 / 60 = × + at% L = % 4 . 51 % 100 1 . 32 1 . 54 1 . 43 1 . 54 = × at%, α =48.6% (c) Vol % L= % 2 . 51 % 100 05 . 16 / 4 . 52 91 . 13 / 6 . 47 91 . 13 / 6 . 47 = × + , α =48.8% 10-21 At 1300C°, the composition of the two phases in equilibrium are
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/09/2009 for the course MAE 614106 taught by Professor Marcmeyers during the Spring '09 term at UCSD.

Page1 / 2

Solution6 - HW 6 Solution 10-19(a L 49 wt W at W= 49 183.85...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online