Question

In your first job with a large U.S based steel company, you have been assigned to a team tasked with developing a new low carbon steel alloy. In the iron-carbon system, the kinetics of the austenite to pearlite transformation obeys the Avrami relationship. In initial experimental data shows that the transformation reaches 40% completion in 13.1 seconds and 60% completion in 16.2 seconds, Determine the time (in seconds) required for the transformation in this new steel to reach 95% completion. Assume ak value of 4.46 x 104.
a. 90

b. 36.2

c. 45

d. 28

e. 25

Answer 1

Answer:

option e

Explanation:

The avrami equation is given as follows

$y = 1 -e^{-kt^{n} }$

the above equation can be re-written as

$1-y = e^{-kt^{n} } \\\\\frac{1}{1-y} = e^{kt^{n}}\\\\ln(\frac{1}{1-y} )=kt^{n} \\\\hence\\\\t^{n} = \frac{ln(\frac{1}{1-y} )}{k}$

where y = the percentage

k = 4.46×10⁻⁴

t = time

n = constant

We determine n as follows

Using any of the cases, for instance case 1

when y = 40% or 0.4 , t = 13.1s

$t^{n} = \frac{ln(\frac{1}{1-y} )}{k}$

$13.1^{n} = \frac{ln(\frac{1}{1-0.4} )}{4.46\times 10^{-4} }$

solving the above equation, we have

$13.1^{n} = 1145.35\\\\ln(13.1^{n}) = ln(1145.35)\\\\n\times ln(13.1) = ln(1145.35)\\\\n = \frac{ln(1145.35}{ ln(13.1)}= 2.74$

The value of n is the same if we use the second case

that is, y = 60% or 0.6 and t = 16.2s

Now when y = 95% or 0.95 and using n = 2.74

$t^{2.74} = \frac{ln(\frac{1}{1-0.95} )}{4.46\times 10^{-4} }$

$t^{2.74}= 6716.88\\ \\ln(t^{2.74})=ln(6716.88)\\\\2.74 \times ln(t)=ln(6716.88)\\\\ln(t)=\frac{ln(6716.88)}{2.74}\\ \\ln(t)=3.2162\\\\t = e^{3.2162}\\ \\t = 24.93s$

Hence

t = 25s

Answer 2

Answer:

Option A

Explanation: