Materials Science and Engineering: An Introduction

$c_{Li} = 2.38 wt%$
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: $c_{Li} = \frac{100p_{li}(p_{Al}-p_{ave})}{p_{ave}(p_{Al}-p_{Li})} = \frac{(100)(0.534 g/cm^{3})(2.71 g/cm^{3}- 2.47 g/cm^{3})}{(2.47 g/cm^{3})(2.71 g/cm^{3}-0.534 g/cm^{3})} = 2.38 wt$%