Chapter 12, Problem 12.17P

Interpretation Introduction

Interpretation:

The amount of Aluminum required to be added and the liquid mixture cooled

diagram needs to be determined by referring to the given phase.

Concept introduction:

The aluminum alloy consists of aluminum as the parent material or the main constituent of the alloy. Aluminum alloy develops a white surface which acts as a protective layer of aluminum oxide.

Aluminum alloy has its major applications in the field of Aerospace industry as they are lighter, less flammable and have more strength.

Phase diagram is used to represent the thermodynamic conditions for existence of distinct phases of a material. A point at which the free energy becomes non-analytic and there is discontinuous change in coordinates of the derivatives takes place are represented by the curves on the phase diagrams.

Answer to Problem 12.17P

The amount of aluminum to be added is calculated as $\text{1}\text{.01911 kg}$

Explanation of Solution

Given Information:

Amount of aluminum ${\text{W}}_{\text{Al}}\text{=45}\text{.7 kg}$

Weight percent of Cu-Al alloy $\text{W\%=35\%}$

Calculation:

The composition of the Cu-Al alloy is Cu- $\text{35\%}$ Al. The temperature is halfway between the lower and upper boundaries.

To calculate the temperature of the alloy, following expression is used,

  ${\text{T}}_{\text{alloy}}\text{=}\frac{{\text{T}}_{\text{upper}}{\text{+T}}_{\text{lower}}}{\text{2}}$

Where, the alloy temperature= ${\text{T}}_{\text{alloy}}$

The upper boundary temperature= ${\text{T}}_{\text{upper}}$

The lower boundary temperature= ${\text{T}}_{\text{lower}}$

Now, substituting the values as ${\text{T}}_{\text{lower}}\text{=591°C}$ and ${\text{T}}_{\text{upper}}\text{=626°C}$,

  ${\text{T}}_{\text{alloy}}\text{=}\frac{\text{591+626}}{\text{2}}$

  ${\text{T}}_{\text{alloy}}\text{=}\frac{\text{1217}}{\text{2}}$

  ${\text{T}}_{\text{alloy}}\text{=608}\text{.5°C}$

Fig. $2$ Phase diagram of Cu-Al

In the above-mentioned phase diagram, the Cu- $\text{35\%}$ Al is represented at temperature $\text{608}\text{.5°C}$

To calculate the weight percent of the alloy at temperature $\text{608}\text{.5°C}$, the formula used is as follows:

  $\text{\%wt η=}\frac{{\text{C}}_{\text{o}}{\text{-C}}_{\text{1}}}{{\text{C}}_{\text{o}}{\text{-C}}_{\text{2}}}$

Where, ${\text{C}}_{\text{o}}$ = composition of the aluminum in the liquid phase

  ${\text{C}}_{\text{1}}$ = composition of aluminum in the alloy

  ${\text{C}}_{\text{2}}$ = composition of the aluminum at the end

Substituting the values as ${\text{C}}_{\text{o}}=\left(\text{100-55}\text{.5}\right)\%$, ${\text{C}}_{\text{1}}=35\%$ and ${\text{C}}_{\text{2}}=\left(\text{100-70}\right)\%$,

  $\text{\%wt η=}\frac{\left(\text{100-55}\text{.5}\right)\text{-35}}{\left(\text{100-55}\text{.5}\right)\text{-}\left(\text{100-70}\right)}\text{×100}$

  $\begin{array}{l}\text{\%wt η=}\frac{\text{44}\text{.5-35}}{\text{44}\text{.5-30}}\text{×100}\\ \text{\%wt η=0}\text{.6651×100}\end{array}$

  $\text{\%wt η=65}\text{.51 wt\%}$ of Al

To calculate the weight percentage of the mentioned alloy at the eutectic temperature $\text{591°C}$, following expression is used,

  ${\text{\%wt η}}_{\text{eutectic}}\text{=}\frac{{\text{C}}_{\text{o}}^{\text{'}}{\text{-C}}_{\text{1}}}{{\text{C}}_{\text{o}}{\text{-C}}_{\text{2}}}$

Where ${\text{C}}_{\text{o}}^{\text{'}}$ = composition of the aluminum in the liquid phase at the eutectic temperature.

Substituting the values as ${\text{C}}_{\text{0}}\text{=}\left(\text{100-54}\text{.5}\right)\text{\%}$, ${\text{C}}_{\text{1}}\text{=35\%}$ and ${\text{C}}_{\text{2}}$ = ${\text{C}}_2=\left(\text{100-70}\right)\%$, ,

  ${\text{\%wt η}}_{\text{eutectic}}\text{=}\frac{\left(\text{100-54}\text{.5}\right)\text{-35}}{\left(\text{100-54}\text{.5}\right)\text{-}\left(\text{100-70}\right)}\text{×100}$

  ${\text{\%wt η}}_{\text{eutectic}}\text{=}\frac{\text{45}\text{.5-35}}{\text{45}\text{.5-30}}\text{×100}$

  ${\text{\%wt η}}_{\text{eutectic}}\text{=0}\text{.6774×100}$

  ${\text{\%wtη}}_{\text{eutectic}}\text{=67}\text{.74 wt\%}$ of Al

Now, to calculate the amount of the aluminum to be added more, the following formula is used:

  ${\left(\text{Amount}\right)}_{\text{Al}}\text{=}\left[\left({\text{\%wt η}}_{\text{eutectic}}\right)\text{-}\left(\text{\%wtη}\right)\right]{\text{m}}_{\text{alloy}}$

Where ${\text{m}}_{\text{alloy}}$ is termed as the mass of the mentioned alloy.

Substituting the values as,

  ${\text{\%wtη}}_{\text{eutectic}}\text{=67}\text{.74 wt\%}$ of Al,

  $\text{\%wtη= 65}\text{.51wt\%}$

  ${\text{m}}_{\text{alloy}}\text{= 45}\text{.7kg}$

Putting the values,

  ${\left(\text{Amount}\right)}_{\text{Al}}\text{=}\left[\frac{\text{67}\text{.74-65}\text{.51}}{\text{100}}\right]\left(\text{45}\text{.7}\right)$

  ${\left(\text{Amount}\right)}_{\text{Al}}\text{=1}\text{.01911 kg}$

To calculate the temperature change,

  ${\text{ΔT=T}}_{\text{alloy}}{\text{-T}}_{\text{eutectic}}$

Where the values substituted are ${\text{T}}_{\text{eutectic}}\text{=608}\text{.5°C}$ and ${\text{T}}_{\text{eutectic}}\text{=591°C}$,

  $\text{ΔT=608}\text{.5-591}$

  $\text{ΔT=17}\text{.5°C}$

Conclusion

Thus, based on the phase diagram of Cu- $\text{35\%}$ Al alloy, the temperature change and the amount of aluminum to be added is determined.