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Chapter 4: Problem 24

Determine the approximate density of a Ti-6Al-4V titanium (Ti) alloy that has a composition of \(90 \mathrm{wt} \%\) Ti, \(6 \mathrm{wt} \% \mathrm{Al}\), and \(4 \mathrm{wt} \% \mathrm{~V}\)

Short Answer

Expert verified
Answer: The approximate density of the Ti-6Al-4V alloy is \(4.466 \frac{\text{g}}{\text{cm}^3}\).

Step by step solution

01

Identifying the densities of Ti, Al, and V

First, given the weights of the three elements, we will find the individual densities for Titanium (Ti), Aluminum (Al), and Vanadium (V). Consult a reference source to find these values, which are: Ti: \(4.51 \frac{\text{g}}{\text{cm}^3}\) Al: \(2.70 \frac{\text{g}}{\text{cm}^3}\) V: \(6.11 \frac{\text{g}}{\text{cm}^3}\)
02

Applying the rule of mixtures to the alloy's density (wt. %)

Next, apply the rule of mixtures to calculate the approximate density of the Ti-6Al-4V alloy. Assuming the volume fractions of Ti, Al, and V are constant, the density of the alloy can be calculated using the given weight percentages: \(Density_{alloy} = \rho_{Ti} * Weight \%_{Ti} + \rho_{Al} * Weight \%_{Al} + \rho_{V} * Weight \%_{V}\) Substitute the densities and weight percentages: \(Density_{alloy} = (4.51 \frac{\text{g}}{\text{cm}^3} * 0.90) + (2.70 \frac{\text{g}}{\text{cm}^3} * 0.06) + (6.11 \frac{\text{g}}{\text{cm}^3} * 0.04)\)
03

Calculating the alloy's density

Finally, perform the multiplication and addition operations to find the approximate density of the Ti-6Al-4V alloy: \(Density_{alloy} = (4.06 \frac{\text{g}}{\text{cm}^3}) + (0.162 \frac{\text{g}}{\text{cm}^3}) + (0.244 \frac{\text{g}}{\text{cm}^3}) = 4.466 \frac{\text{g}}{\text{cm}^3}\) The approximate density of the Ti-6Al-4V alloy is \(4.466 \frac{\text{g}}{\text{cm}^3}\).

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Most popular questions from this chapter

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?

Determine the ASTM grain-size number if 25 grains per square inch are measured at a magnification of \(75 \times\).

Gold (Au) forms a substitutional solid solution with silver (Ag). Compute the weight percent of gold that must be added to silver to yield an alloy that contains \(5.5 \times 10^{21}\) Au atoms per cubic centimeter. The densities of pure Au and \(\mathrm{Ag}\) are \(19.32\) and \(10.49 \mathrm{~g} / \mathrm{cm}^{3}\), respectively.

What is the composition, in atom percent, of an alloy that contains \(44.5 \mathrm{lb}_{\mathrm{m}}\) of \(\mathrm{Ag}, 83.7 \mathrm{lb}_{\mathrm{m}}\) of Au, and \(5.3 \mathrm{lb}_{\mathrm{m}}\) of \(\mathrm{Cu}\) ?

Following is a schematic micrograph that represents the microstructure of some hypothetical metal.Determine the following: (a) Mean intercept length (b) ASTM grain-size number, \(G\)
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