Answer to Question #234320 in Mechanical Engineering for nawab ul haq

Question

Consider a 55 wt% Sn–45 wt% Pb alloy at 160˚C,

a. Cite the phases present

b. Calculate the relative amount of each phase present in terms

of (a) mass fraction and (b) volume fraction.

Note: Locate the required points clearly on Lead-Tin phase diagram. Take

necessary values wherever required and mention the reference against

every value you take, e.g. any online source/book or research paper

Answer 1

mass fractions of α and liquid phases are both 0.5 for a 50 wt% Sn-45 wt% Pb alloy

The volume fraction of phase α, $V_\alpha$

$V_\alpha = \frac{V_\alpha}{V_\alpha+V_\beta}$

$V_\alpha = \frac{m_\alpha}{\rho_\alpha} = \frac{0.5}{7.31}= 0.06839$

$V_\beta = \frac{m_\beta}{\rho_\beta} = \frac{0.5}{11.29}= 0.0442$

$V_\alpha = \frac{V_\alpha}{V_\alpha+V_\beta} = \frac{0.06839}{0.11259} =0.6 cm^3$

$V_\beta = \frac{V_\beta}{V_\alpha+V_\beta} = \frac{0.0442}{0.11259} =0.4 cm^3$