Chapter 5, Problem 5.44P

Interpretation Introduction

(a)

Interpretation:

The graph of diffusion coefficient (D) m²/s, v/s inverse of temperature (1/T) needs to be plotted using the given data.

Concept Introduction:

Diffusion coefficient (D) is given by the following relation:

$D=D_0\exp (\frac{-Q}{RT})$

Where,

$\begin{array}{l}Q\text{ = activation energy of atoms}\\ \text{R= gas constant}\\ \text{T= absolute temperature}\end{array}$

Answer to Problem 5.44P

The curves between diffusion coefficient and inverse temperature for Fe atom in FCC and BCC iron are as follows:

  

  

Explanation of Solution

Given Information:

Diffusion couple	Diffusion mechanism	Q(J/mol)	Do(m2/s)
C in FCC iron	Interstitial	$1.38\times {10}^5$	$2.3\times {10}^{-5}$
C in BCC iron	Interstitial	$8.74\times {10}^4$	$1.1\times {10}^{-6}$
Fe in FCC iron	vacancy	$2.79\times {10}^5$	$6.5\times {10}^{-5}$
Fe in BCC iron	vacancy	$2.46\times {10}^5$	$4.1\times {10}^{-4}$

Calculation:

The diffusion coefficient can be calculated as follows:

$D=D_0\exp (\frac{-Q}{RT})$ ---------------- (1)

Here,

$\begin{array}{l}Q\text{ = activation energy of atoms=1}\text{.38}\times {\text{10}}^{\text{5}}\text{ J/mol}\\ \text{R= gas constant=8}\text{.32}\text{J/mol}\\ \text{T= absolute temperature=2000 K}\\ {\text{D}}_o=\text{per}\text{expontial}\text{term}=2.3\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}\end{array}$

Equation (1) becomes

$\begin{array}{l}D=2.3\times {10}^{-5}\exp (\frac{-1.38\times {10}^5}{8.32\times 2000})\\ =5.75\times {10}^{-9}{\text{m}}^{\text{2}}\text{/s}\\ D\simeq 6\times {10}^{-9}{\text{m}}^{\text{2}}\text{/s}\end{array}$

Hence, coefficient of diffusion for C in FCC iron at $\text{2000 K}$ is ${\text{6×10}}^{\text{-9}}{\text{m}}^{\text{2}}\text{/s}$

Now, draw a table to calculate the coefficient of diffusion for C in FCC iron and for C in BCC iron diffusion temperature.

$\begin{array}{l}{\text{D}}_{\text{o}}\text{=2}{\text{.3×10}}^{\text{-5}}{\text{m}}^{\text{2}}\text{/s}\\ \text{Q=1}{\text{.38×10}}^{\text{5}}\text{J/mol}\\ {\text{D}}_{\text{o}}\text{=1}{\text{.1×10}}^{\text{-6}}{\text{m}}^{\text{2}}\text{/s}\\ \text{Q=8}{\text{.74×10}}^{\text{4}}\text{J/mol}\end{array}$

T(k)	1/T(k-1)	D for C in FCC iron ( $m^2/s$)	D for C in BCC iron ( $m^2/s$)
2000	0.0005	6E-09	6E-09
1750	0.0006	2E-09	3E-09
1500	0.0007	4E-10	1E-09
1250	0.0008	4E-11	2E-10
1200	0.0008	2E-11	2E-10
1150	0.0009	1E-11	1E-10
1100	0.0009	7E-12	8E-11
1000	0.0010	1E-12	3E-11
750	0.0013	6E-15	9E-13
500	0.0020	9E-20	8E-16

From the table, plot a curve between inverse temperature (1/T) and diffusion coefficient D.

  

Again draw a table to calculate the coefficient of diffusion for Fe in FCC iron and for Fe in BCC iron at different temperature.

$\begin{array}{l}{\text{D}}_{\text{o}}\text{=6}{\text{.5×10}}^{\text{-5}}{\text{m}}^{\text{2}}\text{/s}\\ \text{Q=2}{\text{.79×10}}^{\text{5}}\text{J/mol}\\ {\text{D}}_{\text{o}}\text{=4}{\text{.1×10}}^{\text{-4}}{\text{m}}^{\text{2}}\text{/s}\\ \text{Q=2}{\text{.46×10}}^{\text{5}}\text{J/mol}\end{array}$

T(k)	1/k(k-1)	D for C in FCC iron ( $m^2/s$)	D for C in BCC iron ( $m^2/s$)
2000	0.0005	3E-12	2E-10
1750	0.0006	3E-13	2E-11
1500	0.0007	1E-14	1E-12
1250	0.0008	1E-16	2E-14
1200	0.0008	5E-17	8E-15
1150	0.0009	1E-17	3E-15
1100	0.0009	4E-18	9E-16
1000	0.0010	2E-19	6E-17
750	0.0013	2E-24	3E-21
500	0.0020	5E-34	9E-30

Plotting the curve between inverse temperature (1/T) and diffusion coefficient (D) for Fe atom in FCC and BCC.

  

Interpretation Introduction

(b)

Interpretation:

The temperature range at which the iron-transition takes place needs to be determined from the graphs.

Concept Introduction:

Iron transition is the period in which a subject change from one state to another takes place.

Answer to Problem 5.44P

The iron transition for BCC to FCC takes place in a range of temperature 1150 K-1200 K.

The iron transition for FCC to BCC takes place in the range of 1600 K-1700 K.

Explanation of Solution

Given Information:

Diffusion couple	Diffusion mechanism	Q(J/mol)	Do(m2/s)
C in FCC iron	Interstitial	$1.38\times {10}^5$	$2.3\times {10}^{-5}$
C in BCC iron	Interstitial	$8.74\times {10}^4$	$1.1\times {10}^{-6}$
Fe in FCC iron	vacancy	$2.79\times {10}^5$	$6.5\times {10}^{-5}$
Fe in BCC iron	vacancy	$2.46\times {10}^5$	$4.1\times {10}^{-4}$

The curves between diffusion coefficient and inverse temperature for Fe atom in FCC and BCC iron are as follows:

  

  

From the curve (1),

BCC iron to FCC iron transition takes place in a range of temperature 1150 K-1200 K.

From the curve (2),the iron transition from BCC to FCC iron take place in the temperature range 1600 K-1700 K.

Interpretation Introduction

(c)

Interpretation:

The reason for the higher activation energy for diffusion in FCC iron than carbon needs to be explained.

Concept Introduction:

Activation energy, D is related to the size of the atom.

Answer to Problem 5.44P

An atomic number of carbon (C) is 6 while the atomic number of iron (Fe) is 26.

As we know that activation is related to the size of atom hence more atomic number greater is the activation energy required.

Explanation of Solution

Diffusion atom is placed in a vacancy position.

But in some atoms, there is a defect due to more interstitial position and less vacancy. Diffusion atom takes less time to reach in interstitial position than vacancy. Activation energy co-relates with atomic number. Hence, more atomic number, maximum activation energy is required. The atomic number of carbon is 4 and that of iron is 26. Thus, the activation energy for diffusion of Fe is higher than that of C.

Interpretation Introduction

(d)

Interpretation:

The reason for the higher activation energy for the diffusion of FCC than BCC iron needs to be explained.

Concept Introduction:

Activation energy is directly proportional to the packing fraction. Packing fraction is a bond formed by atoms.

Answer to Problem 5.44P

Packing fraction of FCC structure is 0.71 while that for the BCC structure is 0.68.

Hence, the activation energy required for FCC is more than BCC.

Explanation of Solution

Packing fraction is the measurement of loss or gain of the total mass in a group of the nucleus which directly relates atomic bond. Packing fraction is correlated with activation energy. The packing fraction for FCC structure is 0.74 and that for the BCC structure is 0.68.

The activation energy for diffusion higher in FCC as compared to the BCC iron.

Interpretation Introduction

(e)

Interpretation:

Whether the diffusion rate in FCC is faster than in BCC or not needs to be determined.

Concept Introduction:

Diffusion coefficient (D) is given by the following relation:

$D=D_0\exp (\frac{-Q}{RT})$

Where,

$\begin{array}{l}Q\text{ = activation energy of atoms}\\ \text{R= gas constant}\\ \text{T= absolute temperature}\\ {\text{D}}_o=\text{per}\text{expontial}\text{term}\end{array}$

Answer to Problem 5.44P

Diffusion rate in FCC structure is slower than the BCC structure.

Explanation of Solution

Taking T= 1175 K from temperature range of BCC to FCC iron transition.

Since,

$D=D_0\exp (\frac{-Q}{RT})$ ----------------- (1)

Substitute,

$\begin{array}{l}{\text{D}}_{\text{o}}\text{=2}{\text{.3×10}}^{\text{-5}}{\text{m}}^{\text{2}}\text{/s}\\ \text{Q=1}{\text{.38×10}}^{\text{5}}\text{J/mol}\\ \text{R=8}\text{.32J/mol}\text{.K}\\ \text{T=1175 K}\end{array}$

Equation (1) becomes,

$\begin{array}{l}D=2.3\times {10}^{-5}\exp (\frac{-1.38\times {10}^5}{8.32\times 1175})\\ D=1.7027\times {10}^{-11}{m}^2/s\end{array}$ ---------------- (a)

Again to find out 'D' in BCC iron for carbon.

Put

$\begin{array}{l}{\text{D}}_{\text{o}}\text{=1}{\text{.1×10}}^{\text{-6}}{\text{m}}^{\text{2}}\text{/s}\\ \text{T=1175K}\\ \text{Q=8}{\text{.74×10}}^{\text{4}}\text{J/mol}\\ \text{R=8}\text{.32J/mol}\text{.K}\end{array}$

Equation (1) becomes,

$\begin{array}{l}D=1.1\times {10}^{-6}\exp \left(\frac{8.74\times {10}^4}{8.32\times 1175}\right)\\ D=1.44107\times {10}^{-10}{\text{m}}^{\text{2}}\text{/s}\end{array}$ ------------------- (b)

From (a) and (b), one can say that the diffusion coefficient is more in BCC structure than the FCC.

Hence, diffusion atoms move faster in BCC structure compared to FCC structure.