Chapter 9, Problem 9.37P

Interpretation Introduction

(a)

Interpretation:

Thedendritic fraction of silver nucleates at ${10}^0\text{C}$ undercooling needs to be determined.

Concept Introduction:

Dendritic solidification is the process in which a liquid having lack of nucleating properties are undercooled to solidify the liquid.

A dendritic fraction is given by the following relation:

  $F=\frac{C\Delta T}{\Delta Hf}$

Answer to Problem 9.37P

Dendritic fraction of silver nucleates at ${10}^0\text{C}$ is 0.0337.

Explanation of Solution

Given Information:

Undercooling temperature ${10}^0\text{C}$ and specific heat $3.2{\text{5J/cm}}^{\text{3}\text{0}}\text{C}$

Calculation:

A generic fraction is given by the following relation:

  $F=\frac{C\Delta T}{\Delta Hf}$ ---------------- (1)

Here,

  $\begin{array}{l}\begin{array}{l}F=\text{Dendritic fraction}\hfill \\ C=\text{Specific heat}\hfill \\ \Delta T=\text{Undercooling temperature}\hfill \end{array}\\ \Delta H_f=\text{Latent heat of fusion}\end{array}$

Latent heat of fusion for silver metal is as follows:

  $\Delta H_F=96{\text{5 J/cm}}^{\text{3}}$ (Fromthetable of latent heat of fusion)

  $\begin{array}{l}C=3.{\text{25 J/cm}}^{\text{3}}{}^{\text{o}}\text{C}\\ \Delta T={10}^0\text{C}\end{array}$

Equation (1) becomes

  $\begin{array}{l}F=\frac{3.25\times 10}{965}\\ F=0.0337\end{array}$

Therefore, dendritic fraction of silver metal at ${10}^0\text{C}$ is 0.0337

Interpretation Introduction

(b)

Interpretation:

Thedendritic fraction of silver metal at undercooling of ${100}^0\text{C}$ needs to be determined.

Concept Introduction:

Lack of nucleation growth in liquid causes it difficult to solidifies hence liquid is undercooling for this solidify which produces small solid bulge called dendrites and process of formation of solid called dendritic solidification.

Answer to Problem 9.37P

Dendritic fraction of silver metal at ${100}^0\text{C}$ undercooling is 0.337.

Explanation of Solution

Given Information:

The undercooling temperature of silver metal is ${100}^0\text{C}$

Specific heat of metal $\text{3}{\text{.25J/cm}}^{\text{3}}{}^{\circ }\text{C}$

Calculation:

Equation dendritic fraction is given by

  $F=\frac{C\Delta T}{\Delta Hf}$ --------------------- (1)

Where,

  $\begin{array}{l}\begin{array}{l}F=\text{Dendritic fraction}\hfill \\ C=\text{Specific heat}\hfill \\ \Delta T=\text{Undercooling temperature}\hfill \end{array}\\ \Delta H_f=\text{Latent heat of fusion}\end{array}$

But,

  $\begin{array}{l}\Delta T={100}^0\text{C}\\ C=3.25{\text{J/cm}}^{\text{3}\text{0}}\text{C}\\ \Delta Hf=96{\text{5 J/cm}}^{\text{3}}\end{array}$

(Form latent heat table)

Equation (1) becomes,

  $\begin{array}{l}F=\frac{3.25\times 100}{965}\\ F=0.337\end{array}$

Therefore dendritic fraction at ${100}^0\text{C}$ undercooling for silver metal is 0.337.

Interpretation Introduction

(c)

Interpretation:

The fraction of solidification of silver nucleates at homogeneity needs to be determined.

Concept Introduction:

Dendritic solidification is the process in which a liquid having lack of nucleating properties are undercooled to solidify the liquid.

A dendritic fraction is given by the following relation:

  $F=\frac{C\Delta T}{\Delta Hf}$

Answer to Problem 9.37P

The dendritic fraction of solid silver metal when it is homogeneous is 0.842.

Explanation of Solution

Given Information:

Thehomogeneous undercooling temperature is ${250}^0\text{C}$

Specific heat $\text{3}{\text{.25J/cm}}^{\text{3}\text{0}}\text{C}$

Calculation:

The dendritic fraction of silver metal is given by the following relation:

  $F=\frac{C\Delta T}{\Delta Hf}$ --------------------- (1)

Here,

At the homogeneous state of silver metal $\Delta T={250}^0\text{C}$

  $\Delta H_f={\text{965 J/cm}}^{\text{3}}$ (From latent heat table)

  $C=3.2{\text{5J/cm}}^{\text{3}\text{0}}\text{C}$

Equation (1) becomes,

  $\begin{array}{l}F=\frac{3.25\times 250}{965}\\ F=0.842\end{array}$

The dendritic fraction of silver metal at homogeneous is 0.842.