Hot extrusion of nanocrystalline yttria-stabilized tetragonal zirconia polycrystals

Abstract

Hot indirect extrusion behaviors of fully dense 3 mol% Y₂O₃-stabilized ZrO₂ polycrystals with a grain size of 80 nm were investigated by using a conical graphite die. During early extrusion at 1843 K and a constant compression stress of 60–80 MPa, the relative moving billet velocity (v₀) shows a sharp maximum, followed by a continuous decrease until it reaches a steady-state v₀ value. Visual observations of the material flow by using graphite foil markers indicate that the deformation zone at the steady-state v₀ is well represented by a spherical velocity field. The decrease in v₀ corresponds to the transition from heterogeneous to homogeneous deformation. The stress exponent (n) of 2.07 was obtained from the log–log plots of the steady-state v₀ versus the external applied stress that is compensated by threshold and sliding friction stresses. This n value suggests plastic deformation through grain-boundary sliding in indirect extrusion.

APPENDIX

A. Internal power of deformation

When materials flow toward the apex in the deformation zone between S_I and S_II (spherical velocity field10) in Fig. A, the relative moving velocity of the billets is expressed as follows:

$${\dot U_{\rm{r}}} = v = - {v_{\rm{f}}}\cos \,\theta {\left( {{{{r_{\rm{f}}}} \over r}} \right)^2},\>\>{\dot U_\theta } = {\dot U_\varphi } = 0\quad .$$

(A1)

FIG. A

Schematic of spherical velocity field proposed by Avitzur.10

From Eq. (A1), the strain-rates as a function of velocity components are

$${\dot \varepsilon _{rr}} = {{\partial {{\dot U}_r}} \over {\partial r}} = 2{v_{\rm{f}}}r_{\rm{f}}^2{{\cos \theta } \over {{r^3}}}\quad ,$$

(A2a)

$${\dot \varepsilon _{\theta \theta }} = {{{{\dot U}_r}} \over r} + {{\partial {{\dot U}_\theta }} \over {r\partial \theta }} = - {v_{\rm{f}}}r_{\rm{f}}^2{{\cos \theta } \over {{r^3}}}\quad ,$$

(A2b)

$${\dot \varepsilon _{\varphi \varphi }} = {{{{\dot U}_r}} \over r} + {{\partial {{\dot U}_\varphi }} \over {r\partial \theta }} = - {v_{\rm{f}}}r_{\rm{f}}^2{{\cos \theta } \over {{r^3}}}\quad ,$$

(A2c)

$${\dot \varepsilon _{rr}} = - 2{\dot \varepsilon _{\theta \theta }} = - 2{\dot \varepsilon _{\varphi \varphi }}\quad ,$$

(A2d)

$${\dot \varepsilon _{r\theta }} = {1 \over 2}{\rm{\cdot}}{1 \over r}{\rm{\cdot}}{{\partial {{\dot U}_r}} \over {\partial \theta }} = {1 \over 2}{v_{\rm{f}}}r_{\rm{f}}^2{{\sin \,\theta } \over {{r^3}}}\quad ,$$

(A2e)

$${\dot \varepsilon _{\theta \varphi }} = {\dot \varepsilon _{\varphi r}} = 0\quad .$$

(A2f)

The equivalent strain-rate is defined as follows:

$$\eqalign{ & \dot \varepsilon _{\rm{e}}^2 = \>{2 \over 9}\left[ {{{\left( {{{\dot \varepsilon }_{rr}} - {{\dot \varepsilon }_{\theta \theta }}} \right)}^2} + {{\left( {{{\dot \varepsilon }_{\theta \theta }} - {{\dot \varepsilon }_{\varphi \varphi }}} \right)}^2} + {{\left( {{{\dot \varepsilon }_{\varphi \varphi }} - {{\dot \varepsilon }_{rr}}} \right)}^2}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\, + {4 \over 3}\left( {\dot \varepsilon _{r\theta }^2 + \dot \varepsilon _{\theta \varphi }^2 + \dot \varepsilon _{\varphi r}^2} \right) \cr} $$

$$ = \dot \varepsilon _{rr}^2 + {4 \over 3}\dot \varepsilon _{r\theta }^2\quad .$$

(A3)

Because of volume constancy (${v_0}r_0^2 = {v_{\rm{f}}}r_{\rm{f}}^2$), the equivalent strain rate (Eq. A3) with Eqs. (A2d) and (A2e) can be rewritten as follows:

$${\dot \varepsilon _{\rm{e}}} = 2{v_{\rm{f}}}r_{\rm{f}}^2{1 \over {{r^3}}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } = 2{v_0}r_0^2{1 \over {{r^3}}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \quad .$$

(A4)

The constitutive equation for steady-state plastic deformation at high temperature is represented as follows3:

$${\dot \varepsilon _{\rm{e}}} = C\sigma _{\rm{e}}^n\quad .$$

(A5)

Because ${{{{\dot \varepsilon }_{ij}}} \over {{\sigma _{ij}}}} = {3 \over 2}{{{{\dot \varepsilon }_{\rm{e}}}} \over {{\sigma _{\rm{e}}}}}$, the stress component is rewritten by using Eq. (A5) as follows:

$${\sigma _{ij}} = {2 \over 3}{\left( {{{{{\dot \varepsilon }_{\rm{e}}}} \over C}} \right)^{{1 \over n}}}{{{{\dot \varepsilon }_{ij}}} \over {{{\dot \varepsilon }_{\rm{e}}}}}\quad .$$

(A6)

Thus, by using Eq. (A6), the internal power per unit volume within the deformation zone can be represented as follows:

$${\dot w_1} = {\sigma _{ij}}{\dot \varepsilon _{ij}} = {1 \over {\root n \of C }}{\left( {{{\dot \varepsilon }_{\rm{e}}}} \right)^{{{n + 1} \over n}}}\quad .$$

(A7)

After integration of Eq. (A7),

$${\dot W_1} = \int_V {{\sigma _{ij}}{{\dot \varepsilon }_{ij}}{\rm{d}}V} = {1 \over {\root n \of C }}\int_V {{{\left( {{{\dot \varepsilon }_{\rm{e}}}} \right)}^{{{n + 1} \over n}}}\,{\rm{d}}V} \quad ,$$

(A8a)

where ${\rm{d}}V = 2\pi r\sin \,\theta r\,{\rm{d}}\theta \,{\rm{d}}r$. By using Eq. (A4), Eq. (A8a) is rewritten as follows:

$${\dot W_1} = {1 \over {\root n \of C }}\int_0^\theta {{{\int_{{r_f}}^{{r_0}} {\left[ {{{2{v_0}r_0^2} \over {{r^3}}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } } \right]} }^{{{n + 1} \over n}}}2\pi r\sin \,\theta r\,{\rm{d}}\theta \,{\rm{d}}r} $$

$$ = 2\pi {{{{\left( {2{v_0}r_0^2} \right)}^{{{n + 1} \over n}}}} \over {\root n \of C }}\int_0^\phi {\left[ {{{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{n + 1} \over {2n}}}}\sin \theta \int_{{r_{\rm{f}}}}^{{r_0}} {{{\left( {{1 \over {{r^3}}}} \right)}^{{{n + 1} \over n}}}{r^2}\,{\rm{d}}r} } \right]} \,{\rm{d}}\theta $$

(A8b)

Because $\int_{{r_{\rm{f}}}}^{{r_0}} {{{\left( {{1 \over {{r^3}}}} \right)}^{{{n + 1} \over n}}}{r^2}\,{\rm{d}}r = {n \over 3}{{\left( {{1 \over {{r_0}}}} \right)}^{{3 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right]}$, and r₀ sin ϕ = D₀/2, Eq. (A8b) is rewritten as follows:

$$\eqalign{ & {{\dot W}_1} = \>{{n\pi } \over 3}{v_0}D_0^2{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right] \cr & \times \,\,{\sin ^{{1 \over n} - 2}}\phi \int_0^\phi {{{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{n + 1} \over {2n}}}}\sin \theta \>{\rm{d}}\theta } \quad . \cr} $$

(A8c)

An approximation of $1 - {{11} \over {12}}{\sin ^2}\,\theta \simeq 1 - {\sin ^2}\,\theta$ leads to

$$\eqalign{ & {{\dot W}_1} = {{n\pi } \over 3}{v_0}\,D_0^2{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right] \cr & \,\,\,\,\,\,\,\,\,\,\, \times {\sin ^{{1 \over n} - 2}}\phi \left[ {{{1 - {{\cos }^{2 + {1 \over n}}}\phi } \over {\left( {2 + 1/n} \right){{\sin }^{2 - {1 \over n}}}\phi }}} \right]\quad . \cr} $$

(A8d)

B. Power loss because of velocity discontinuities

The stress component along S_I is represented by Eq. (A6) as follows:

$${\sigma _{r\theta }} = {2 \over 3}{\left( {{{{{\dot \varepsilon }_{\rm{e}}}} \over C}} \right)^{{1 \over n}}}{{{{\dot \varepsilon }_{r\theta }}} \over {{{\dot \varepsilon }_{\rm{e}}}}}\quad .$$

(B1)

The strain-rate component along S_I is represented by Eq. (A2e) as follows:

$${\dot \varepsilon _{r\theta }}\left( {{S_{\rm{I}}}} \right) = {1 \over 2}{v_0}r_0^2{{\sin \theta } \over {r_0^3}} = {{{v_0}} \over {{D_0}}}\sin \phi \sin \theta \quad .$$

(B2)

The equivalent strain-rate component along S_I is represented by Eq. (A4) as follows:

$$\eqalign{ & {{\dot \varepsilon }_{\rm{e}}}\left( {{S_{\rm{I}}}} \right) = 2{v_0}{r_0}{1 \over {r_0^3}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4{{{v_0}} \over {{D_0}}}\sin \phi \sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \quad . \cr} $$

(B3)

Modification of Eq. (B3) leads to

$${v_0} = {{{D_0}} \over {4\sin \phi }}{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)^{ - {1 \over 2}}}{\dot \varepsilon _{\rm{e}}}\quad .$$

(B4)

Because ${\rm{d}}S = 2\pi {r^2}\sin \theta \,{\rm{d}}\theta$, power loss of velocity discontinuity along S_I is obtained through Eqs. (B1), (B2), (B3), and (B4) as follows:

$${\dot W_2} = \int_{{S_{\rm{I}}}} {\left| {\Delta {v_{{S_{\rm{I}}}}}} \right|{\sigma _{r\theta }}{\rm{d}}S} = \int_{{S_{\rm{I}}}} {\left| {{v_0}\sin \theta } \right|{\sigma _{r\theta }}{\rm{d}}S} $$

$$\eqalign{ & = - \int_0^\phi {\left[ {{{{D_0}} \over {4\,\sin \phi }}{{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{ - {1 \over 2}}}} \right.} \cr & \,\,\,\,\,\,\, \times \,\,\sin \theta {2 \over 3}{\left( {4{{{v_0}} \over {C{D_0}}}\sin \phi \sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } } \right)^{{1 \over n}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {{{{v_0}} \over {{D_0}}}\sin \phi \sin \theta } \right)2\pi r_0^2\sin \theta \,{\rm{d}}\theta \cr} $$

$$ = - {{\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi \int_0^\phi {{{\sin }^3}\,\theta {{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{1 - n} \over {2n}}}}{\rm{d}}\theta } \quad .$$

(B5a)

An approximation by $1 - {{11} \over {12}}{\sin ^2}\,\theta \simeq 1 - {\sin ^2}\,\theta$ leads to

$${\dot W_2} = - {{n\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\,\phi \left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\,\phi {{2n} \over {2n + 1}}} \right)\quad .$$

(B5b)

Power loss of velocity discontinuity along S_II (${\dot W_3}$) is obtained in the same manner as ${\dot W_2}$. The strain-rate component along S_II is represented by Eq. (A2e) as follows:

$${\dot \varepsilon _{r\theta }}\left( {{S_{{\rm{II}}}}} \right) = {1 \over 2}{v_{\rm{f}}}r_{\rm{f}}^2{{\sin \theta } \over {r_{\rm{f}}^3}} = {{{v_{\rm{f}}}} \over {{D_{\rm{f}}}}}\sin \phi \sin \theta \quad .$$

(B6)

The equivalent strain-rate component along S_II is represented by Eq. (A4) as follows:

$$\eqalign{ & {{\dot \varepsilon }_{\rm{e}}}\left( {{S_{{\rm{II}}}}} \right) = 2{v_{\rm{f}}}{r_{\rm{f}}}{1 \over {r_{\rm{f}}^3}}\sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4{{{v_{\rm{f}}}} \over {{D_{\rm{f}}}}}\sin \phi \sqrt {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \quad . \cr} $$

(B7)

Modification of Eq. (B7) leads to

$${v_{\rm{f}}} = {{{D_{\rm{f}}}} \over {4\sin \phi }}{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)^{ - {1 \over 2}}}{\dot \varepsilon _{\rm{e}}}\quad .$$

(B8)

Thus, power loss because of velocity discontinuity along S_II is

$${\dot W_3} = \int_{{S_{{\rm{II}}}}} {\left| {\Delta {v_3}} \right|{\sigma _{r\theta }}{\rm{d}}S} = \int_{{S_{{\rm{II}}}}} {\left| {{v_{\rm{f}}}\sin \theta } \right|{\sigma _{r\theta }}{\rm{d}}S} $$

$$ = {{\pi {v_{\rm{f}}}D_{\rm{f}}^2} \over {{\rm{12}}}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi \int_0^\phi {{{\sin }^3}\,\theta {{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{1 - n} \over {2n}}}}{\rm{d}}\theta } $$

(B9a)

Because ${v_{\rm{f}}}r_{\rm{f}}^2 = {v_0}r_0^2$ and D_f = D₀(r_f/r₀), ${{{v_{\rm{f}}}} \over {{D_{\rm{f}}}}} = {{{v_0}} \over {{D_0}}}{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)^3}$, Eq. (B9a) can be rewritten as follows:

$${\dot W_3} = {{\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi {\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)^{{3 \over n}}}\int_0^\phi {{{\sin }^3}\,\theta {{\left( {1 - {{11} \over {12}}{{\sin }^2}\,\theta } \right)}^{{{1 - n} \over {2n}}}}{\rm{d}}\theta } \quad .$$

(B9b)

Approximation by $1 - {{11} \over {12}}{\sin ^2}\,\theta \simeq 1 - {\sin ^2}\,\theta$ leads to

$${\dot W_3} = {{n\pi {v_0}D_0^2} \over {12}}{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}{\sin ^{{1 \over n} - 2}}\phi {\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)^{{3 \over n}}}\left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\phi + {{2n} \over {2n + 1}}} \right)\quad .$$

(B9c)

C. Power loss because of sliding friction

Power loss because of sliding friction along the conical portion of S_III is represented as follows:

$${\dot W_4} = \int_{{S_{{\rm{III}}}}} {{\tau _{{\rm{sf}}}}\left| {\Delta {v_{{\rm{III}}}}} \right|{\rm{d}}S} = \int_{{r_{\rm{f}}}}^{{r_0}} {{\tau _{{\rm{sf}}}}{v_0}\cos \,\emptyset {{\left( {{{{r_0}} \over r}} \right)}^2}2\pi r\sin \,\emptyset \,{\rm{d}}r} $$

$$ = \left[ {{\tau _{{\rm{sf}}}}2\cot \,\emptyset \,{\rm{ln}}\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)} \right]\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0}\quad .$$

(C1)

For indirect extrusion, there is no need to consider power loss because of sliding friction along the cylindrical portion of S_IV (${\dot W_5}$). However, for direct extrusion, ${\dot W_5}$ is an important factor that affects extrusion behavior. ${\dot W_5}$ is represented as follows:

$$\eqalign{ & {{\dot W}_5} = \int_{{S_{{\rm{IV}}}}} {{\tau _{{\rm{sf}}}}\left| {\Delta {v_{{\rm{IV}}}}} \right|{\rm{d}}S} = \int_0^L {{\tau _{{\rm{sf}}}}{v_0}\pi {D_0}\,{\rm{d}}L} \cr & \,\,\,\,\,\, = {\tau _{{\rm{sf}}}}{v_0}\pi {D_0}L = \left[ {{\tau _{{\rm{sf}}}}{{4L} \over {{D_0}}}} \right]\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0}\quad . \cr} $$

(C2)

D. Applied power

When the applied power equals to the upper bound energies,10 the applied stress (σ₀) is represented as follows:

$${\sigma _0}\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0} = {\dot W_1} + {\dot W_2} + {\dot W_3} + {\dot W_4} + {\dot W_5}\quad .$$

(D1)

Equation (D1) is rewritten through Eq. (A8d), (B5b), (B9c), (C1), and (C2) as follows:

$$\left[ {{\sigma _0} - {\tau _{{\rm{sf}}}}2\,\cot \emptyset \ln \left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right) - {\tau _{{\rm{sf}}}}{{4L} \over {{D_0}}}} \right]\pi {\left( {{{{D_0}} \over 2}} \right)^2}{v_0}$$

$$\eqalign{ & = \pi n{v_0}D_0^2{\left( {{{4{v_0}} \over {C{D_0}}}} \right)^{{1 \over n}}}\left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right]{\sin ^{{1 \over n} - 2}}\phi \left\{ {{1 \over 3}\left[ {{{1 - {{\cos }^{2 + {1 \over n}}}\phi } \over {\left( {2 + 1/n} \right){{\sin }^{2 - {1 \over n}}}\phi }}} \right]} \right. \cr & \left. {\,\, + {1 \over {12}}\left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\phi + {{2n} \over {2n + 1}}} \right)} \right\}\quad . \cr} $$

(D2a)

Equation (D2) is rewritten as follows:

$${{{v_0}} \over {{D_0}}} = {C \over {4{{\left( {4nK} \right)}^n}}}\sigma _{{\rm{eff}}}^n\quad ,$$

(D2b)

where

$$\eqalign{ & K = \left[ {{{\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)}^{{3 \over n}}} - 1} \right]{\sin ^{{1 \over n} - 2}}\phi \left\{ {{1 \over 3}\left[ {{{1 - {{\cos }^{2 + {1 \over n}}}\phi } \over {\left( {2 + 1{\rm{/}}n} \right){{\sin }^{2 - {1 \over n}}}\phi }}} \right]} \right. \cr & \,\,\,\,\,\,\,\,\left. {\, + {1 \over {12}}\left( {{{{{\cos }^{{{2n + 1} \over n}}}\phi } \over {2n + 1}} - {{\cos }^{{1 \over n}}}\phi + {{2n} \over {2n + 1}}} \right)} \right\} \cr} $$

(D3)

and

$${\sigma _{{\rm{eff}}}} = {\sigma _0} - {\sigma _{{\rm{sf}}}}\quad .$$

(D4)

The external stresses required to overcome the interfacial sliding friction stress (σ_sf) in Eq. (D4) are given as follows:

For direct extrusion,

$${\sigma _{{\rm{sf}}}} = {\tau _{{\rm{sf}}}}2\,\cot \,\emptyset \,\ln \left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right) - {\tau _{{\rm{sf}}}}{{4L} \over {{D_0}}}\quad .$$

(D5a)

For indirect extrusion,

$${\sigma _{{\rm{sf}}}} = {\tau _{{\rm{sf}}}}2\cot \,\emptyset \,{\rm{ln}}\left( {{{{r_0}} \over {{r_{\rm{f}}}}}} \right)\quad .$$

(D5b)

E. Uniaxial creep deformation

The constitutive equation for steady-state uniaxial creep deformation obtained experimentally4 is

$${\dot \varepsilon _{rr}} = A{{Gb} \over {kT}}{\left( {{b \over d}} \right)^p}\left[ {{D^*}\exp \left( { - {Q \over {RT}}} \right)} \right]{\left( {{{{\sigma _{rr}}} \over G}} \right)^n}\quad .$$

(E1)

When the stress exponent (n) and grain size exponent (p) are constant during uniaxial creep deformation, Eq. (E1) can be rewritten as follows:

$${\dot \varepsilon _{rr}} = {{{A_1}} \over {T{d^p}}}\left[ {\exp \left( { - {Q \over {RT}}} \right)} \right]\sigma _{rr}^{\,\,\,n}\quad ,$$

(E2)

where

$${A_1} = A{{Gb} \over k}{\rm{\cdot}}{{{b^p}} \over {{G^n}}}{D^*}\quad .$$

(E3)

In uniaxial creep deformation, ${\dot \varepsilon _{rr}}$ and ${\sigma _{rr}}$ equal ${\dot \varepsilon _{\rm{e}}}$ and ${\sigma _{\rm{e}}}$, respectively, and Eq. (E2) can be rewritten as follows:

$${\dot \varepsilon _{\rm{e}}} = {{{A_1}} \over {T{d^p}}}\left[ {\exp \left( { - {Q \over {RT}}} \right)} \right]\sigma _{\rm{e}}^{\,\,\,n}\quad .$$

(E4)

A comparison of Eqs. (A5) and (E4) yields

$$C = {{{A_1}} \over {T{d^p}}}\exp \left( { - {Q \over {RT}}} \right)\quad .$$

(E5)