Degradation in parallel-disk rheometry

Abstract

We analyze quantitatively the oxidative degradation of a sample in a parallel-disk rheometer, as oxygen diffuses inward, radially, from the free boundary. We examine rheometer error mitigation by means of nitrogen blanketing, and also, of parallel-disk partitioning. We arrive at exact analytical expressions for the oxygen concentration and, thus, for the degradation rate. We then integrate this rate over time to get the amount of oxygen reacted as a function of radial position and time in the degrading sample. To illustrate the usefulness of our analytical expressions, we provide two worked examples investigating the effect of nitrogen blanketing and parallel-disk partitioning. We find that, though nitrogen blanketing always produces less degradation, its benefits are limited for short times. Additionally, parallel-disk partitioning provides a simpler solution and allows samples to be run for longer times without degradation compromising measurement, even in samples initially saturated with oxygen. We also consider the effect of antioxidants. We also consider an important special case, without chemical reaction, where the sample dries by evaporation from its free surface. We close by comparing the roles played by polymer degradation in parallel-disk flow versus cone-plate flow.

Appendices

Appendices

Solving Eq. (16)

Equation (16) has the form of the Emden-Fowler equation (Fowler 1914; Fowler 1931), which has general solution:

$$ \chi \equiv \frac{c_A}{c_{AR}}={C}_1{J}_0\left( i\kappa \alpha \right)+{C}_2{Y}_0\left(- i\kappa \alpha \right) $$

(56)

which is subject to Eqs. (6) and (7). In Eq. (56), J₀ is zeroth order Bessel function of the first kind, and Y₀, of the second kind. Differentiating Eq. (56), we find:

$$ \frac{d\chi}{d\alpha}= i\kappa {C}_2{Y}_1\left(- i\kappa \alpha \right)- i\kappa {C}_1{J}_1\left( i\kappa \alpha \right) $$

(57)

Subjecting Eq. (57) to Eq. (7) gives:

$$ 0={C}_2{Y}_1(0)-{C}_1{J}_1(0) $$

(58)

$$ 0={C}_2\left(\infty \right)-{C}_1(0) $$

(59)

and thus:

$$ {C}_2=0 $$

(60)

so that:

$$ \chi ={C}_1{J}_0\left( i\kappa \alpha \right) $$

(61)

which, subject to Eq. (6), gives:

$$ 1={C}_1{J}_0\left( i\kappa \right) $$

(62)

or:

$$ {C}_1=\frac{1}{J_0\left( i\kappa \right)} $$

(63)

so that:

$$ \chi =\frac{J_0\left( i\kappa \alpha \right)}{J_0\left( i\kappa \right)}=\frac{I_0\left(\kappa \alpha \right)}{I_0\left(\kappa \right)} $$

(64)

which is Eq. (17) above.

Solving Eq. (29)

Supposing that:

$$ X\left(\alpha, \tau \right)\equiv f\left(\tau \right)g\left(\alpha \right) $$

(65)

so that from Eq. (30):

$$ f\left(\tau \right)g(1)=0 $$

(66)

and thus, we have the rim condition:

$$ g(1)=0 $$

(67)

and from Eq. (32):

$$ \frac{\partial }{\partial \alpha}\left[f\left(\tau \right)g(0)\right]=0 $$

(68)

$$ {\left.\frac{\partial g}{\partial \alpha}\right|}_{\alpha =0}=0 $$

(69)

Substituting Eq. (65) into Eq. (29):

$$ \frac{1}{f}\frac{\partial f}{\partial \tau }=\frac{1}{g\alpha}\frac{\partial }{\partial \alpha}\left(\alpha \frac{\partial g}{\partial \alpha}\right)-{\kappa}^2 $$

(70)

$$ \frac{1}{f}\frac{\partial f}{\partial \tau }=C=\frac{1}{g\alpha}\frac{\partial }{\partial \alpha}\left(\alpha \frac{\partial g}{\partial \alpha}\right)-{\kappa}^2 $$

(71)

$$ \frac{1}{f}\frac{d f}{d\tau}=C $$

(72)

which has general solution:

$$ f={C}_1{e}^{C\tau} $$

(73)

and:

$$ \frac{d^2g}{d{\alpha}^2}+\frac{1}{\alpha}\frac{d g}{d\alpha}-{D}^2g=0 $$

(74)

in which:

$$ {D}^2\equiv C+{\kappa}^2 $$

(75)

Equation (74) has the form of the Emden-Fowler equation (Fowler 1914; Fowler 1931), so that:

$$ g={C}_3{J}_0\left( iD\alpha \right)+{C}_4{Y}_0\left(- iD\alpha \right) $$

(76)

Subjecting Eq. (76) to the rim condition, Eq. (67) yields:

$$ 0={C}_3{J}_0(iD)+{C}_4{Y}_0\left(- iD\right) $$

(77)

Differentiating Eq. (76) yields:

$$ \frac{d g}{d\alpha}= i\kappa {C}_4{Y}_1\left(- i\kappa \alpha \right)- i\kappa {C}_3{J}_1\left( i\kappa \alpha \right) $$

(78)

and subjecting this to Eq. (69):

$$ 0={C}_4 $$

(79)

and thus:

$$ g={C}_3{J}_0\left( iD\alpha \right) $$

(80)

and our rim condition, Eq. (67), gives:

$$ 0={C}_3{J}_0(iD) $$

(81)

so that:

$$ {D}_n=\pm i\;{j}_{0,n};n\ge 1 $$

(82)

where j_{0, n} is the nth zero of J₀, so that:

$$ {C}_n=-{j}_{0,n}^2-{\kappa}^2 $$

(83)

Substituting Eq. (82) into Eq. (80) yields:

$$ {g}_n={C}_{3,n}{J}_0\left(i\;{j}_{0,n}\alpha \right) $$

(84)

Substituting Eqs. (73), (83), and (84) into Eq. (65) yields:

$$ X\left(\alpha, \tau \right)=\sum \limits_{n=1}^{\infty }{A}_n{e}^{-\left({j}_{0,n}^2+{\kappa}^2\right)\tau }{J}_0\left(-{j}_{0,n}\alpha \right) $$

(85)

Applying Eq. (31) to Eq. (85):

$$ X\left(\alpha, 0\right)=1=\sum \limits_{n=1}^{\infty }{A}_n{J}_0\left(-{j}_{0,n}\alpha \right)=\sum \limits_{n=1}^{\infty }{A}_n{J}_0\left({j}_{0,n}\alpha \right) $$

(86)

since J₀(−j_{0, n}α) = J₀(j_{0, n}α). For such infinitely summed series, we multiply through by αJ₀(j_{0, m}α), and then integrate over the entire range of α to get:

$$ {\int}_0^1\alpha {J}_0\left({j}_{0,m}\alpha \right) d\alpha =\sum \limits_{n=1}^{\infty }{A}_n{\int}_0^1\alpha {J}_0\left({j}_{0,n}\alpha \right){J}_0\left({j}_{0,m}\alpha \right) d\alpha $$

(87)

Mindful of the following orthogonal properties (Eqs. (5.16) and (5.17) of Crank 1975):

$$ {\int}_0^1\alpha {J}_0\left({j}_{0,n}\alpha \right){J}_0\left({j}_{0,m}\alpha \right) d\alpha =\Big\{{\displaystyle \begin{array}{cc}0;& m\ne n\\ {}\frac{1}{2}{J}_1^2\left({j}_{0,n}\right);& m=n\end{array}}\kern1em $$

(88)

Equation (87) then becomes:

$$ {\int}_0^1\alpha {J}_0\left({j}_{0,m}\alpha \right) d\alpha ={A}_m\frac{1}{2}{J}_1^2\left({j}_{0,m}\right) $$

(89)

Reversing m back to n gives:

$$ {\int}_0^1\alpha {J}_0\left({j}_{0,n}\alpha \right) d\alpha ={A}_n\frac{1}{2}{J}_1^2\left({j}_{0,n}\right) $$

(90)

so that:

$$ {A}_n=\frac{2\int_0^1\alpha {J}_0\left({j}_{0,n}\alpha \right) d\alpha}{J_1^2\left({j}_{0,n}\right)}=\frac{2{J}_1\left({j}_{0,n}\right)}{j_{0,n}\;{J}_1^2\left({j}_{0,n}\right)}=\frac{2}{j_{0,n}\;{J}_1\left({j}_{0,n}\right)} $$

(91)

Substituting this into Eq. (85) gives:

$$ X\left(\alpha, \tau \right)=2\sum \limits_{n=1}^{\infty}\frac{J_0\left({j}_{0,n}\alpha \right)}{j_{0,n}\;{J}_1\left({j}_{0,n}\right)}{e}^{-\left({j}_{0,n}^2+{\kappa}^2\right)\tau } $$

(92)

which is Eq. (33) above.

Solving χ _t(α, τ) in Eq. (37)

To get the transient part, we revise the atmospheric boundary condition, Eq. (21), to:

$$ {c}_A\left(R,t\right)={c}_{AR} $$

(93)

so that Eq. (30) becomes:

$$ \chi \left(1,\tau \right)={\chi}_s(1)+{\chi}_t\left(1,\tau \right)=1 $$

(94)

and from Eq. (17):

$$ {\chi}_s(1)=1 $$

(95)

we rearrange Eq. (94) to:

$$ {\chi}_t\left(1,\tau \right)=0 $$

(96)

The initial condition must also be revised to:

$$ \chi \left(\alpha, 0\right)={\chi}_s\left(\alpha \right)+{\chi}_t\left(\alpha, 0\right)=\frac{c_{A0}}{c_{AR}}\equiv {\chi}_0 $$

(97)

so that:

$$ {\chi}_t\left(\alpha, 0\right)={\chi}_0-\frac{J_0\left( i\kappa \alpha \right)}{J_0\left( i\kappa \right)} $$

(98)

Because the initial condition is the only difference from the nitrogen blanket, we can start by following Eq. (85):

$$ {\chi}_t=\sum \limits_{n=1}^{\infty }{B}_n{e}^{-\left({j}_{0,n}^2+{\kappa}^2\right)\tau }{J}_0\left(-{j}_{0,n}\alpha \right) $$

(99)

which we subject to Eq. (98) to get:

$$ {\chi}_0-\frac{J_0\left( i\kappa \alpha \right)}{J_0\left( i\kappa \right)}=\sum \limits_{n=1}^{\infty }{B}_n{J}_0\left({j}_{0,n}\alpha \right) $$

(100)

Following Eq. (90):

$$ {\chi}_0{\int}_0^1\alpha {J}_0\left({j}_{0,m}\alpha \right) d\alpha -{\int}_0^1\frac{J_0\left( i\kappa \alpha \right)}{J_0\left( i\kappa \right)}\alpha {J}_0\left({j}_{0,m}\alpha \right) d\alpha ={\int}_0^1\sum \limits_{n=1}^{\infty }{B}_n\alpha {J}_0\left({j}_{0,n}\alpha \right){J}_0\left({j}_{0,m}\alpha \right) d\alpha $$

(101)

which is only nonzero when m = n:

$$ {\chi}_0\frac{J_1\left({j}_{0,n}\right)}{j_{0,n}} - \frac{\kappa {J}_0\left({j}_{0,n}\right){I}_1\left(\kappa \right)+{j}_{0,n}{J}_1\left({j}_{0,n}\right){I}_0\left(\kappa \right)}{J_0\left( i\kappa \right)\left({j}_{0,n}^2+{\kappa}^2\right)}={B}_n\frac{1}{2}{J}_1^2\left({j}_{0,n}\right) $$

(102)

Solving for B_n, we get:

$$ {B}_n=\frac{2}{J_1\left({j}_{0,n}\right)}\left[\frac{\chi_0}{j_{0,n}} - \frac{j_{0,n}}{j_{0,n}^2+{\kappa}^2}\right] $$

(103)

Substituting this into Eq. (99) gives us the transient part:

$$ {\chi}_t=2\sum \limits_{n=1}^{\infty}\frac{1}{J_1\left({j}_{0,n}\right)}\left[\frac{\chi_0}{j_{0,n}} - \frac{j_{0,n}}{j_{0,n}^2+{\kappa}^2}\right]{e}^{-\left({j}_{0,n}^2+{\kappa}^2\right)\tau }{J}_0\left({j}_{0,n}\alpha \right) $$

(104)

which is Eq. (38) above.