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

Question #234320

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


1
Expert's answer
2021-09-07T23:53:02-0400

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, VαV_\alpha

Vα=VαVα+VβV_\alpha = \frac{V_\alpha}{V_\alpha+V_\beta}


Vα=mαρα=0.57.31=0.06839V_\alpha = \frac{m_\alpha}{\rho_\alpha} = \frac{0.5}{7.31}= 0.06839


Vβ=mβρβ=0.511.29=0.0442V_\beta = \frac{m_\beta}{\rho_\beta} = \frac{0.5}{11.29}= 0.0442


Vα=VαVα+Vβ=0.068390.11259=0.6cm3V_\alpha = \frac{V_\alpha}{V_\alpha+V_\beta} = \frac{0.06839}{0.11259} =0.6 cm^3


Vβ=VβVα+Vβ=0.04420.11259=0.4cm3V_\beta = \frac{V_\beta}{V_\alpha+V_\beta} = \frac{0.0442}{0.11259} =0.4 cm^3




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