what phases are present in a slow cooled 1090 stell and how much of each of these phases are present?
so the phases are cementite and pearlite and it's hypereutectoid.
The formula is hypereutectoid=cementite(6.67%)+pearlite(0.8%)
I know in a 1090 steel, there is .9%C.
I know the amount of cementite and pearlite is 100%
so if i find pearlite, then amount of cementite=100-pearlite.
the equation now becomes
.9%C=(100-pearlite)(6.67%)+pearlite(.8%)
from here, it's basic algebra and i just have to solve for pearlite.
apparently, the answer is 98.3% pearlite and cementite is 1.7%.
My question is how do I solve this equation because I keep getting 111% for pearlite.
The reason you did not get the right answer was because you used percentages on the right hand side and not fractions.
Let x = fraction of pearlite (1.0=100%)
and (1-x) = fraction of cementite,
using your equation:
0.9% = (1-x)*6.67% + x*0.8%
Solving gives
x=0.983 (pearlite) and
1-x=0.017 (cementite)
To solve the equation and find the amount of pearlite in the slow cooled 1090 steel, we can use algebraic calculations.
Let's start with the given equation:
0.9%C = (100 - pearlite)(6.67%) + pearlite(0.8%)
To simplify the equation, we can convert percentages to decimal form:
0.9C = (100 - pearlite)(0.0667) + pearlite(0.008)
Now, distribute the terms to express the equation more explicitly:
0.9C = 6.667 - 0.0667 * pearlite + 0.008 * pearlite
Rearrange the equation to isolate the term with pearlite:
0.0667 * pearlite - 0.008 * pearlite = 6.667 - 0.9C
Combine like terms:
0.0587 * pearlite = 6.667 - 0.9C
Divide both sides of the equation by 0.0587 to solve for pearlite:
pearlite = (6.667 - 0.9C) / 0.0587
Now, substitute the value of C (0.9) into the equation:
pearlite = (6.667 - 0.9 * 0.9) / 0.0587
Calculating this expression gives us:
pearlite = 98.319%
To find the amount of cementite, subtract the percentage of pearlite from 100%:
cementite = 100% - pearlite
cementite = 100% - 98.319%
cementite = 1.681%
Therefore, the slow cooled 1090 steel consists of approximately 98.319% pearlite and 1.681% cementite.