Appendix B
Derivation of the Taylor-Aris Dispersion

B.1 Derivation of the Taylor-Aris Dispersion

It is considered that the flow inside a straight cylindrical pipe is steady, driven by a constant pressure gradient (i.e, Poiseuille flow). The average velocity over the pipe cross-section can be given as:

u (r) = 2 \bar{u} (1 - \frac{r^{2}}{R^{2}}),

(B.1)

where:

\bar{u} = \frac{1}{π R^{2}} \int_{0}^{2 π} d 𝜃 \int_{0}^{R} r u d r .

(B.2)

In these equations $\bar{u}$ denotes the average quantity of the velocity flowing through the pipe. If it is assumed that an axisymmetric distribution of material $c (r, z, t)$ is released into the flow, the evolution of the propagation is described by the ADE in cylindrical form.

\frac{\partial c}{\partial t} + u (r) \frac{\partial c}{\partial z} = D_{m} (\frac{\partial^{2} c}{\partial z^{2}} + \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial c}{\partial r})) .

(B.3)

Since no particle can leave the system, the boundary conditions are must satisfy $\partial c ∕ \partial r = 0$ at $r = R$ . By separating $c$ using Reynold’s decomposition method, $c$ is separated into its cross-sectional average and $r$ variable parts.

c (r, z, t) = \bar{c} (z, t) + c^{'} (r, z, t),

(B.4)

where:

\bar{c} = \frac{2}{R^{2}} \int_{0}^{R} r c d r .

(B.5)

since the average of deviation is zero ( $\bar{c^{'}} = 0$ ) the equation can be written as:

\frac{\partial \bar{c}}{\partial t} + \frac{\partial c^{'}}{\partial t} + u (r) (\frac{\partial \bar{c}}{\partial z} + \frac{\partial c^{'}}{\partial z}) = D_{m} (\frac{\partial^{2} \bar{c}}{\partial z^{2}} + \frac{\partial^{2} c^{'}}{\partial z^{2}} + \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial c^{'}}{\partial r})) .

(B.6)

Taking the cross-sectional average of Eq. (B.6) yields the following simplification taking into account that $\partial c^{'} ∕ \partial t = 0$ on $r = R$ .

\frac{\partial \bar{c}}{\partial t} + \bar{u (r)} \frac{\partial \bar{c}}{\partial z} + \bar{u (r) \frac{\partial c^{'}}{\partial z}} = D_{m} \frac{\partial^{2} \bar{c}}{\partial z^{2}} .

(B.7)

The the mean concentration $\bar{c}$ depends on the average advection of the $r$ -varying part of $c$ (i.e., $c^{'} (r, z, t)$ ), which is calculated by subtracting Eq. (B.7) from Eq. (B.6) reveals the $r$ -varying component of Eq. (B.6),

\frac{\partial c^{'}}{\partial t} + (u (r) - \bar{u}) \frac{\partial \bar{c}}{\partial z} + u \frac{\partial c^{'}}{\partial z} - \bar{u \frac{\partial c^{'}}{\partial z}} = D_{m} \nabla^{2} c^{'} .

(B.8)

Based on this equation, an approximation is made whereby after a time of in the order $t = R^{2} ∕ D_{m}$ the radial diffusion to have almost smoothed out variation in the $r$ -axis. Thus for $t \sim O (R^{2} ∕ D_{m})$ , it is expected for $\bar{c} ≫ c^{'}$ . In addition, the gradients in the $r$ -direction are greater than those in the $z$ -direction. Therefore the primary balance is:

(u (r) - \bar{u}) \frac{\partial \bar{c}}{\partial z} ≃ \frac{D_{m}}{r} \frac{\partial}{\partial r} (r \frac{\partial c^{'}}{\partial r}),

(B.9)

Introducing Eq. (B.1) into (B.9), the following expression is derived.

\frac{\partial}{\partial r} (r \frac{\partial c^{'}}{\partial r}) = \frac{\bar{u}}{D_{m}} \frac{\partial \bar{c}}{\partial z} (r - \frac{2 r^{3}}{R^{2}}) .

(B.10)

As shown in the Reynold’s decomposition of $c$ in Eq. (B.4), $\bar{c}$ is independent from $r$ , so Eq. (B.10) can be integrated twice over,

c^{'} = \frac{\bar{u}}{D_{m}} \frac{\partial \bar{c}}{\partial z} (\frac{r^{2}}{4} - \frac{r^{4}}{8 R^{2}} + A + B l n r) .

(B.11)

Since $c^{'}$ is regular at $r = 0$ $B$ can be declared the value of 0. Furthermore, $c^{'}$ has zero average. This yields:

\int_{0}^{R} r u^{'} d r = 0,

(B.12)

This equation give $A$ the value of:

A = - \frac{R^{2}}{12},

(B.13)

c^{'} = \frac{\bar{u} R_{r}^{2}}{24 D_{m}} \frac{\partial \bar{c}}{\partial z} (6 R_{r}^{2} - 3 R_{r}^{4} - 2) w h e r e R_{r} = \frac{r}{R} .

(B.14)

Equation (4) requires the term $\bar{u (r) \partial c^{'} ∕ \partial z}$ , which is

\bar{u (r) \frac{\partial c^{'}}{\partial z}} = - \frac{R^{2} \bar{u}}{48 D_{m}} \frac{\partial^{2} \bar{c}}{\partial z^{2}} .

(B.15)

Substituting this result into (B.6), ADE for the mean concentration $\bar{c} (z, t)$ is derived.

\frac{\partial \bar{c}}{\partial t} + \bar{u} \frac{\partial \bar{c}}{\partial z} = (D_{m} + \frac{R^{2} {\bar{u}}^{2}}{48 D_{m}}) \frac{\partial^{2} \bar{c}}{\partial z^{2}} = D_{e f f} \frac{\partial^{2} \bar{c}}{\partial z^{2}} .

(B.16)

Appendix BDerivation of the Taylor-Aris Dispersion

B.1 Derivation of the Taylor-Aris Dispersion

Appendix B
Derivation of the Taylor-Aris Dispersion