Respuesta :
Solution :
Avrami relationship
[tex]$1-y=exp(-kt^n)$[/tex]
[tex]$\ln(1-y)=-kt^n$[/tex]
[tex]$-\ln(1-y)=kt^n$[/tex]
[tex]$\ln\left[\ln\left(\frac{1}{1-y}\right)\right]=\ln k + n \ln t$[/tex]
The fraction transformed is 0.2 at 12.6 s,
[tex]$\ln\left[\ln\left(\frac{1}{1-0.2}\right)\right]=\ln k + n \ln (12.6)$[/tex]
[tex]$-1.5 = \ln k + 2.533 n$[/tex] .........(i)
The fraction transformed is 0.7 at 25.7 s,
[tex]$\ln\left[\ln\left(\frac{1}{1-0.7}\right)\right]=\ln k + n \ln (25.7)$[/tex]
[tex]$0.5 = \ln k + 3.24 n$[/tex] ..............(ii)
Subtract (ii) from (i),
-2 = -0.71 n
n = 2.81
Therefore, from (i),
[tex]$-1.5 = \ln k + 2.533 (2.81)$[/tex]
k = 0.000181
Now if the fraction transformed is 0.92, then
[tex]$kt^n=-\ln (1-y)$[/tex]
[tex]$t=\left[\frac{-\ln (1-y)}{k}\right]^{\frac{1}{n}}$[/tex]
[tex]$t=\left[\frac{-\ln (1-0.92)}{0.000181}\right]^{\frac{1}{2.81}}$[/tex]
t = 29.8528 s
