Answer
$p_{ave} = 4.38 g/cm^{3}$
Work Step by Step
Given:
Ti-6Al-4V titanium alloy with 90 wt% Titanium, 6 wt% Al, and 4 wt% V
Required:
approximate density
Solution:
Using a modified version of Equation 4.10a:
$p_{ave} = \frac{100}{\frac{c_{Ti}}{p_{Ti}}+\frac{c_{Al}}{p_{Al}}+\frac{c_{V}}{p_{V}}}$
where:
$p_{Ti} = 4.51 g/cm^{3}$
$p_{Al} = 2.71 g/cm^{3}$
$p_{V} = 6.1 g/cm^{3}$
Substituting the given values:
$p_{ave} = \frac{100}{\frac{(90 wt)}{(4.51 g/cm^{3})}+\frac{6 wt}{(2.71 g/cm^{3})}+\frac{4 wt}{(6.1 g/cm^{3})}} = 4.38 g/cm^{3}$
