Chapter 27, Problem 27.1P

Interpretation Introduction

Interpretation:

The equation for average concentration, ${\overline{c}}_A$ within the control volume at a given time is to be derived and the dimensionless concentration profile $\left({\overline{c}}_A-c_{As}\right)/\left(c_{A0}-c_{As}\right)$ is to be plotted as the function of the dimensionless time ratio, $X_D$ .

Concept introduction:

The equation for average concentration is derived using the concentration profile resulting from transient one-dimensional diffusion in a slab under conditions of negligible surface resistance.

Explanation of Solution

The concentration profile resulting from transient one-dimensional diffusion in a slab under conditions of negligible surface resistance is as follows:

  $\frac{c_A-c_{As}}{c_{A0}-c_{As}}=\frac{4}{\pi }{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}\sin \left(\frac{n\pi z}{L}\right)e^{-{\left(n\pi /2\right)}^2{X}_D}},n=1,3,5\mathrm{.....}$ …………… (1)

Where,

• $c_{A0}$ represents the initial concentration
• $c_{As}$ represents the surface concentration
• $c_A$ concentration at a thickness of $z$
• $L$ represent the total thickness of the slab
• $X_D$ represents the relative time ratio

The concentration varies along the position; therefore, the average concentration can be written as follows:

  $\overline{c}=\frac{{\displaystyle \underset{0}{\overset{L}{\int }}{c}_{A}dz}}{{\displaystyle \underset{0}{\overset{L}{\int }}dz}}$ ……………. (2)

Rearrange equation (1)

  $c_A=c_{As}+\left(c_{A0}-c_{As}\right)\left(\frac{4}{\pi }{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}\sin \left(\frac{n\pi z}{L}\right)e^{-{\left(n\pi /2\right)}^2{X}_D}}\right),n=1,3,5\mathrm{.....}$

Therefore, the average concentration becomes,

  $\begin{array}{l}{\overline{c}}_A=\frac{{\displaystyle \underset{0}{\overset{L}{\int }}{c}_{As}dz}}{L}+\frac{\frac{4}{\pi }\left(c_{A0}-c_{As}\right)\left[{\displaystyle \underset{0}{\overset{L}{\int }}\frac{1}{n}\sin \left(\frac{n\pi z}{L}\right)e^{-{\left(n\pi /2\right)}^2{X}_D}dz}\right]}{L},n=1,3,5\mathrm{.....}\\ {\overline{c}}_A=\frac{{\left(c_{As}z\right)}_0^L}{L}-\frac{4}{\pi L}\left(c_{A0}-c_{As}\right){\left[\frac{1}{n}{e}^{-{\left(n\pi /2\right)}^2{X}_D}\cos \left(\frac{n\pi z}{L}\right)\times \frac{L}{n\pi }\right]}_0^L,n=1,3,5\mathrm{.....}\end{array}$

  ${\overline{c}}_A=c_{As}-\frac{4}{{\pi }^2}\left(c_{A0}-c_{As}\right){\left[\frac{1}{n^2}{e}^{-{\left(n\pi /2\right)}^2{X}_D}\cos \left(\frac{n\pi z}{L}\right)\right]}_0^L,n=1,3,5\mathrm{.....}$

  $\begin{array}{l}{\overline{c}}_A=c_{As}-\frac{4}{{\pi }^2}\left(c_{A0}-c_{As}\right)\left[{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}{e}^{-{\left(n\pi /2\right)}^2{X}_D}}\left[-1-\left(1\right)\right]\right],n=1,3,5\mathrm{.....}\\ {\overline{c}}_A-c_{As}=\frac{8}{{\pi }^2}\left(c_{A0}-c_{As}\right)\left[{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}{e}^{-{\left(n\pi /2\right)}^2{X}_D}}\right],n=1,3,5\mathrm{.....}\end{array}$

  $\frac{{\overline{c}}_A-c_{As}}{c_{A0}-c_{As}}=\frac{8}{{\pi }^2}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}{e}^{-{\left(n\pi /2\right)}^2{X}_D}},n=1,3,5\mathrm{.....}$ ………. (3)

Equation $\left(3\right)$ represents the desired expression.

Replace the concentration ratio with a dimensionless variable, $Y$

  $Y=\frac{{\overline{c}}_A-c_{As}}{c_{A0}-c_{As}}$

Therefore, equitation $\left(3\right)$ becomes,

  $Y=\frac{8}{{\pi }^2}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}{e}^{-{\left(n\pi /2\right)}^2{X}_D},n=1,3,5\mathrm{.....}}$

Expand the expression up to $n=5$

  $Y=\frac{8}{{\pi }^2}\left[\frac{1}{1^2}{e}^{-{\left(1\times \pi /2\right)}^2{X}_D}+\frac{1}{3^2}{e}^{-{\left(3\times \pi /2\right)}^2{X}_D}+\frac{1}{5^2}{e}^{-{\left(5\times \pi /2\right)}^2{X}_D}+\mathrm{.......}\right]$

  $Y=\frac{8}{{\pi }^2}\left[e^{-{\left(\pi /2\right)}^2{X}_D}+\frac{1}{9}{e}^{-{\left(3\pi /2\right)}^2{X}_D}+\frac{1}{25}{e}^{-{\left(5\pi /2\right)}^2{X}_D}+\mathrm{.......}\right]$ …………. (4)

Vary the values of the relative time ratio from $0to0.5$ and calculate the corresponding value of $Y$ and then plot the data.

The spreadsheet is as follows: