Answer
$c_{Li} = 2.38 wt%$
Work Step by Step
Given:
Aluminum-lithium (Al-Li) alloy
desired density -$ 2.47 g/cm^{3}$
Required:
wt% of Li that is required
Solution:
Using Equation 4.10a, we find:
$p_{ave} = \frac{100}{\frac{c_{Li}}{p_{Li}} + \frac{100 c_{Li}}{p_{Al}}}$
Considering that $c_{Al}+c_{Li} =100$, and $p_{Al} = 2.71 g/cm^{3}$ and $p_{Li} = 0.534 g/cm^{3}$, it follows:
$c_{Li} = \frac{100p_{li}(p_{Al}-p_{ave})}{p_{ave}(p_{Al}-p_{Li})} = \frac{(100)(0.534 g/cm^{3})(2.71 g/cm^{3}- 2.47 g/cm^{3})}{(2.47 g/cm^{3})(2.71 g/cm^{3}-0.534 g/cm^{3})} = 2.38 wt$%
