Answer:
46 and 20.67% wt A
Explanation:
Using.
W = (Cb - C)/(Cb - Ca)
Where
0.59 = (Cb - 60/(Cb - Ca) -- (1)
0.14 = (Cb - 30)/(Cb - Ca) ---- (2)
From (1)
(Cb - Ca)0.59 = Cb - 60
0.59Cb - 0.59Ca = Cb - 60
1.59Cb = 60 - 0.59Ca
From (2)
(Cb - Ca)0.14 = Cb - 30
0.14Cb - 0.14Ca = Cb - 30
1.14Cb = 30 - 0.14Ca
Cb = (30 - 0.14Ca)/1.14
Substitute (30 - 0.14Ca)/1.14 for Cb
1.59(30 - 0.14Ca)/1.14 = 60 - 0.59Ca
47.7 - 0.2226Ca = 1.14 * 60 - 1.14 * 0.59Ca
47.7 - 0.2226Ca = 68.4 - 0.6726Ca
47.7 - 68.4 =- 0.6726Ca + 0.2226Ca
-0.45Ca = -20.7
Ca = -20.7/-0.45
Ca = 46
Cb = (30 - 0.14Ca)/1.14
Cb = 30 - 0.14 * 46)/1.14
Cb = 20.67
Answer:
Alpha Boundary contains 87.33% of A and 12.66% of B
Beta Boundary contains 20.67% of A and 79.33% of B
Explanation:
For the two alloys with composition as given in the question the values of the Weightage is given as
[tex]W_a_1=\dfrac{C_{\beta}-C_o_1}{C_{\beta}-C_{\alpha}}[/tex]
Here
Wa1 os given as 0.59 as value of Alpha Phase for 1st Alloy
Co1 is given as 60% for the first Alloy so the equation becomes
[tex]W_a_1=\dfrac{C_{\beta}-C_o_1}{C_{\beta}-C_{\alpha}}\\0.59=\dfrac{C_{\beta}-60}{C_{\beta}-C_{\alpha}}\\0.59C_{\beta}-0.59C_{\alpha}=C_{\beta}-60\\C_{\beta}-60-0.59C_{\beta}+0.59C_{\alpha}=0\\0.41C_{\beta}+0.59C_{\alpha}-60=0[/tex]
Similarly for the 2nd alloy
[tex]W_a_2=\dfrac{C_{\beta}-C_o_2}{C_{\beta}-C_{\alpha}}[/tex]
Here
Wa2 os given as 0.14 as value of Alpha Phase for 1st Alloy
Co1 is given as 30% for the first Alloy so the equation becomes
[tex]\geq W_a_2=\dfrac{C_{\beta}-C_o_2}{C_{\beta}-C_{\alpha}}\\0.14=\dfrac{C_{\beta}-30}{C_{\beta}-C_{\alpha}}\\0.14C_{\beta}-0.14C_{\alpha}=C_{\beta}-30\\C_{\beta}-30-0.14C_{\beta}+0.14C_{\alpha}=0\\0.86C_{\beta}+0.14C_{\alpha}-30=0[/tex]
Solving the two equations as
[tex]0.41C_{\beta}+0.59C_{\alpha}-60=0\\0.86C_{\beta}+0.14C_{\alpha}-30=0\\C_{\beta}=20.67\%\\C_{\alpha}=87.33\%[/tex]
So the boundaries are as
Alpha Boundary contains 87.33% of A and 12.66% of B
Beta Boundary contains 20.67% of A and 79.33% of B
