Chapter 8.15, Problem 53SEP

Based on the Al₂O₃−SiO₂ phase diagram in Figure 8.27, determine the wt% of phases present for Al₂O₃−55 wt% SiO₂ over the 1900°C to 1500°C temperature range (use 100°C increments).

To determine

Refer Figure 8.27, and determine weight percentage of phases present for ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{-55}\text{wt}\%{\text{SiO}}_{\text{2}}$ over the given temperature range.

Explanation of Solution

Refer Figure 8.27, for the composition of alloy ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{-55}\text{wt}\%{\text{SiO}}_{\text{2}}$:

At temperature of $\text{1900°C}$, the available phase is liquid and the composition is ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{-55\%SiO}}_{\text{2}}$.

Hence, the amount of phase is 100% liquid phase.

At temperature of $\text{1800°C}$, the available phases are $M\left(\text{Mullite}\right)+L\left(\text{Liquid}\right)$.

Upon drawing a horizontal line at temperature of $\text{1800°C}$, write the weight percentage as,

$\begin{array}{l}{w}_M=30\%{\text{SiO}}_{\text{2}}\\ w_l=60\%{\text{SiO}}_{\text{2}}\\ w_0=55\%{\text{SiO}}_{\text{2}}\end{array}$

Express the weight fraction of mullite phase.

$X_M=\frac{w_0-w_M}{w_l-w_M}$ (I)

Substitute $55\%$ for $w_0$, 60% for $w_l$ and 30% for $w_M$ in Equation (I).

$\begin{array}{c}{X}_M=\frac{55-30}{60-30}\\ =0.8333\\ =83.33\%{\text{SiO}}_{\text{2}}\end{array}$

Hence the weight fraction of mullite phase is $83.33\%{\text{SiO}}_{\text{2}}$.

Express the weight fraction of liquid phase.

$X_l=\frac{w_l-w_0}{w_l-w_M}$ (II)

Substitute $55\%$ for $w_0$, 60% for $w_l$ and 30% for $w_M$ in Equation (II).

$\begin{array}{c}{X}_l=\frac{60-55}{60-30}\\ =0.167\\ =16.67\%{\text{SiO}}_{\text{2}}\end{array}$

Hence the weight fraction of liquid phase is $16.67\%{\text{SiO}}_{\text{2}}$.

At temperature of $\text{1700°C}$, the available phases are $M\left(\text{Mullite}\right)+L\left(\text{Liquid}\right)$.

Upon drawing a horizontal line at temperature of $\text{1700°C}$, write the weight percentage as,

$\begin{array}{l}{w}_M=30\%{\text{SiO}}_{\text{2}}\\ w_l=79\%{\text{SiO}}_{\text{2}}\\ w_0=55\%{\text{SiO}}_{\text{2}}\end{array}$

Express the weight fraction of mullite phase.

$X_M=\frac{w_0-w_M}{w_l-w_M}$ (III)

Substitute $55\%$ for $w_0$, 79% for $w_l$ and 30% for $w_M$ in Equation (III).

$\begin{array}{c}{X}_M=\frac{55-30}{79-30}\\ =0.5102\\ =51.02\%{\text{SiO}}_{\text{2}}\end{array}$

Hence the weight fraction of mullite phase is $51.02\%{\text{SiO}}_{\text{2}}$.

Express the weight fraction of liquid phase.

$X_l=\frac{w_l-w_0}{w_l-w_M}$ (IV)

Substitute $55\%$ for $w_0$, 79% for $w_l$ and 30% for $w_M$ in Equation (IV).

$\begin{array}{c}{X}_l=\frac{79-55}{79-30}\\ =0.4898\\ =48.98\%{\text{SiO}}_{\text{2}}\end{array}$

Hence the weight fraction of liquid phase is $48.98\%{\text{SiO}}_{\text{2}}$.

At temperature of $\text{1600°C}$, the available phases are $M\left(\text{Mullite}\right)+L\left(\text{Liquid}\right)$.

Upon drawing a horizontal line at temperature of $\text{1600°C}$, write the weight percentage as,

$\begin{array}{l}{w}_M=30\%{\text{SiO}}_{\text{2}}\\ w_l=93\%{\text{SiO}}_{\text{2}}\\ w_0=55\%{\text{SiO}}_{\text{2}}\end{array}$

Express the weight fraction of mullite phase.

$X_M=\frac{w_0-w_M}{w_l-w_M}$ (V)

Substitute $55\%$ for $w_0$, 93% for $w_l$ and 30% for $w_M$ in Equation (V).

$\begin{array}{c}{X}_M=\frac{55-30}{93-30}\\ =0.3968\\ =39.68\%{\text{SiO}}_{\text{2}}\end{array}$

Hence the weight fraction of mullite phase is $39.68\%{\text{SiO}}_{\text{2}}$.

Express the weight fraction of liquid phase.

$X_l=\frac{w_l-w_0}{w_l-w_M}$ (VI)

Substitute $55\%$ for $w_0$, 93% for $w_l$ and 30% for $w_M$ in Equation (VI).

$\begin{array}{c}{X}_l=\frac{93-55}{93-30}\\ =0.6032\\ =60.32\%{\text{SiO}}_{\text{2}}\end{array}$

Hence the weight fraction of liquid phase is $60.32\%{\text{SiO}}_{\text{2}}$.