3.205 Recitation 3 Notes

Massachusetts Institute of Technology Department of Materials Science and Engineering 3.205 Thermodynamics and Kinetics of Materials—Fall 2003 November 7, 2003 We are looking at the incorporation of Al2 O3 into MgO. Let’s first start by writing the incorporation reaction as follows: M gO × •

Al2 O3 → 2AlM g + 3OO + VM g Note that this is just one possible reaction. There are many other ways you could incorporate alumina into magnesia, creating different types of defects in each case. Now we need some real data to figure out which type of defect is most common in MgO. According to Physical Ceramics by Chiang, et. al. the dominant defects in MgO are Schottky defects. The reaction for the formation of Schottky defects is: �� null → VM g + VO•• with a corresponding equilibrium relationship:  �� Keq = VM g 
[VO•• ] = exp −GfS kT  We now have to account for charge neutrality, taking into consideration both the extrinsic defects (namely the impurities) and the intrinsic Schottky defects.  ��   •  2 VM g = 2 [VO•• ] + AlM g Here is where I made the error. The more interesting case is to look at how the cation vacancy concentration depends on the impurity levels and temperature. However, in class I continued on from this point looking at the anion vacancies. My analysis was incorrect and will now present both solutions correctly. 1 Cation Vacancy  ��  For the cation vacancy concentration, namely VM g , we can use the neutrality condition to get:  �� Keq = VM g  2   �� VM g   1 •  AlM g − 2 1  ��   •  V AlM g − Keq = 0 2 Mg  ��  We can now solve this quadratic equation for VM g yielding: �� VM g −   1  1 �� • AlM VM g = g + 4 4  • AlM g 2 + 16Keq  ��  This relationship gives us VM g but the term Keq can be rewritten to give us a better physical understanding of the solution. Since we know in PURE MgO (with only Schottky defects), the following is true:  ��  Keq = VM g [VO•• ] And from charge neutrality in the pure crystal:   �� VM g = [VO•• ] Thus,  �� 2 Keq = VM g pure Now we have a relation that relates Keq to something more physical (the cation vacancy concentration in the pure crystal). We now go back to our original problem and get:   1  1 �� • VM g = AlM g + 4 4  • AlM g 2  2 �� + 16 VM g pure  ��  From this we can look at the different temperature regimes to understand variation in VM g .  High Temperature Low Temperature  �� VM g �� VM g  pure pure  •  >> AlM g  •  << AlM g    ��  �� VM g ≈ VM g  ��   • pure  1 VM g ≈ 2 AlM g So at high temperatures the cation vacancy concentration is dominated by the intrinsic Schottky defects and at low temperatures its dominated by the vacancy creation from the impurity addition. 2 Anion Vacancy Now let us consider the case of the anion vacancies which I attempted to do in recitation. We can start with the same two tools, the Schottky equilibrium statement and charge neutrality.  ��  Keq = VM g [VO•• ]  ��   •  2 VM g = 2 [VO•• ] + AlM g This time we will solve for [VO•• ] .  Keq = [VO•• ] +  1 •  AlM g [VO•• ] 2 2 [VO•• ] + 1  •  •• AlM g [VO ] − Keq = 0 2 Solving this equation for [VO•• ] yields (this is where I made the error): [VO•• ] = − 1 •  1 AlM g + 4 4  • AlM g 2 + 16Keq 2 Using the same a similar substitution as in the previous case, we can eliminate Keq (Keq = [VO•• ]pure . [VO•• ] = − 1 •  1 AlM g + 4 4  • AlM g 2 2 + 16 [VO•• ]pure Again, let’s look at the limits at high and low temperature: High Temperature Low Temerature  •  2 [VO•• ]pure >> AlM  • g [VO•• ]pure << AlM g [VO•• ] ≈ [VO•• ]pure [VO•• ] ≈ 0 This makes sense since at low temperatures there are very few intrinsic defects and the only contributions to the concentration of anion vacancies can come from the incorporation of the impurity. However, no oxygen vacancies are created when alumina is added. Please come see me if this explanation is not clear. Again I would like to thank those people who brought this error to my attention and I am very sorry for any confusion my mistakes might have caused. Continued on next page... 5.2 Find an expression for the steady-state concentration profile during the radial diffusion of a diffusant through a cylindrical shell of thickness, ∆R, and inner radius, Rin , in which the diffusivity is a function of radius D(r). The boundary conditions are c(r = Rin ) = cin and c(r = Rin + ∆R) = cout . Solution. The gradient operator in cylindrical coordinates is ∇= ∂ 1 ∂ ∂ eˆθ + eˆz eˆr + ∂r r ∂θ ∂z (5.82) The divergence of a flux J in cylindrical coordinates is 1 ∂(rJr ) ∂Jz 1 ∂Jθ ∇ · J = + + r ∂r r ∂θ ∂z (5.83) Therefore, the steady-state, radially symmetric, diffusion equation becomes 0= Continued on next page... ∂c ∂ rD(r) ∂r ∂r (5.84) which can be integrated twice to give r in c(r) = c + a1 Rin dρ ρD(ρ) (5.85) The integration constant a1 is determined by the boundary condition at Rin +∆R: a1 = cout − cin Rin +∆R Rin (5.86) dρ ρD(ρ) 5.3 Find the steady-state concentration profile during the radial diffusion of a dif­ fusant through a bilayer cylindrical shell of inner radius, Rin , where each layer has thickness ∆R/2 and the constant diffusivities in the inner and outer lay­ ers are Din and Dout . The boundary conditions are c(r = Rin ) = cin and c(r = Rin + ∆R) = cout . Will the total diffusion current through the cylinder be the same if the materials which make up the inner and outer shells are ex­ changed? Assume that the concentration of the diffusant is the same in the inner and outer layers at the bilayer interface. Solution. The concentration profile at the bilayer interface will not have continu­ ous derivatives. Break the problem up into separate diffusion problems in each layer and then impose the continuity of flux at the interface. Let the concentration at the bilayer interface be ci/o . Inner region: Rin ≤ r ≤ Rin + ∆R 2 Using Eq. 5.86, cin (r) = ci/o − cin ln Rin +∆R/2 ln Rin r + cin in R (5.87) The flux at the bilayer interface is J i/o = −Din Outer region: Rin + ∆R 2 ci/o − cin ln Rin +∆R/2 Rin 1 + ∆R/2 (5.88) r + ci/o ; + ∆R/2 (5.89) 1 + ∆R/2 (5.90) Rin ≤ r ≤ Rin + ∆R cout (r) = cout − ci/o ln Rin +∆R Rin +∆R/2 ln Rin The flux at the bilayer interface is J i/o = −Dout cout − ci/o ln Rin +∆R Rin +∆R/2 Rin Setting the fluxes at the interfaces equal and solving for ci/o : ci/o = αout cout + αin cin αout + αin (5.91) where D out αout ≡ ln Rin +∆R Rin +∆R/2 αin ≡ ; D in ln Rin +∆R/2 Rin (5.92) Putting Eq. 5.91 into Eqs. 5.87 and 5.89 yields the concentration profile of the whole cylinder. The total current diffusing through the cylinder (per unit length) is I = 2π Rin + ∆R J i/o = − 2 ln Using Eq. 5.91,  ci/o − cin =  2πDin ci/o − cin Rin +∆R/2 Rin αout cout − cin  (5.93)  (5.94) αout + αin If everything is kept constant except Din and Dout , use of Eq. 5.94 in Eq. 5.93 shows that Din Dout I∝ (5.95) α1 Dout + α2 Din where α1 and α2 are constants. Clearly, I will be different if the materials making up the inner and outer shells are exchanged and the values of Dout and Din are therefore exchanged. This contrasts with the result for the two adjoining flat slabs in Exercise 5.1.