{"id":73,"date":"2021-10-02T23:21:37","date_gmt":"2021-10-02T23:21:37","guid":{"rendered":"https:\/\/books.gw-project.org\/flux-equations-for-gas-diffusion-in-porous-media\/chapter\/exercise-3-solution\/"},"modified":"2022-01-10T01:09:48","modified_gmt":"2022-01-10T01:09:48","slug":"exercise-3-solution","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/flux-equations-for-gas-diffusion-in-porous-media\/chapter\/exercise-3-solution\/","title":{"raw":"Exercise\u00a03 Solution","rendered":"Exercise\u00a03 Solution"},"content":{"raw":"
\r\n

From Equation 8c<\/a>, the condition N<\/em>A<\/em><\/sub> + N<\/em>B<\/em><\/sub> = <\/em>0 requires that the viscous flux cancel the non-equimolar flux: N<\/em>v<\/em><\/sup> = <\/em>-<\/em>N<\/em>D<\/em><\/sup>.<\/em><\/p>\r\n

The value of the non-equimolar flux (N<\/em>D<\/em><\/sup> = \u22120.0222 [latex]\\frac{\\textup{moles}}{\\textup{m}^{2}\\ \\textup{s}}[\/latex]) computed in Exercise 1<\/a> applies here because the diffusive fluxes are not appreciably affected by the pressure gradient. So N<\/em>v<\/em><\/sup> = \u22120.0222 [latex]\\frac{\\textup{moles}}{\\textup{m}^{2}\\ \\textup{s}}[\/latex] and, from Equation 2<\/a>, [latex]N^{v}=\\frac{k_{g}}{\\mu RT}pdp\/dl[\/latex] = 0.0222 [latex]\\frac{\\textup{moles}}{\\textup{m}^{2}\\ s}[\/latex].<\/p>\r\n

This is integrated to obtain the following expression<\/p>\r\n

[latex]\\displaystyle \\frac{k_{g}}{2\\mu RTL}\\left ( p_{0}^{2}-p_{L}^{2} \\right )=0.0222[\/latex]<\/p>\r\n

which can be used to directly calculate the desired pressure difference. However, computations are facilitated by rewriting the left side using:<\/p>\r\n

[latex]\\displaystyle \\frac{1}{2}\\left ( p_{0}^{2}-p_{L}^{2} \\right )=\\bar{p}\\left ( p_{0}-p_{L} \\right )[\/latex]<\/p>\r\n

where, [latex]\\bar{p}[\/latex] is the average pressure in the chamber.<\/p>\r\n

[latex]\\displaystyle p_{0}-p_{L}=0.0222\\frac{\\mu L}{Ck_{g}}=\\frac{(0.0222)(2.3\\times 10^{-5})(0.05)}{(1\\times 10^{-12})(40.9)}=624\\ \\textup{Pa}[\/latex]<\/p>\r\n

The pressure drop is quite small (about 6\u00a0cm\u00a0H2<\/sub>O) relative to the 1\u00a0\u00d7\u00a0105<\/sup>\u00a0Pa pressure maintained in the right-hand header, justifying the assumption that the mean pressure is closely approximated by the controlled pressure in the right-hand header.<\/p>\r\n

Return to E<\/span>xercise\u00a0<\/span>3<\/span><\/a><\/p>\r\n

<\/p>\r\n\r\n<\/div>","rendered":"

\n

From Equation 8c<\/a>, the condition N<\/em>A<\/em><\/sub> + N<\/em>B<\/em><\/sub> = <\/em>0 requires that the viscous flux cancel the non-equimolar flux: N<\/em>v<\/em><\/sup> = <\/em>–<\/em>N<\/em>D<\/em><\/sup>.<\/em><\/p>\n

The value of the non-equimolar flux (N<\/em>D<\/em><\/sup> = \u22120.0222 \"\frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}\") computed in Exercise 1<\/a> applies here because the diffusive fluxes are not appreciably affected by the pressure gradient. So N<\/em>v<\/em><\/sup> = \u22120.0222 \"\frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}\" and, from Equation 2<\/a>, \"N^{v}=\frac{k_{g}}{\mu RT}pdp/dl\" = 0.0222 \"\frac{\textup{moles}}{\textup{m}^{2}\ s}\".<\/p>\n

This is integrated to obtain the following expression<\/p>\n

\"\displaystyle \frac{k_{g}}{2\mu RTL}\left ( p_{0}^{2}-p_{L}^{2} \right )=0.0222\"<\/p>\n

which can be used to directly calculate the desired pressure difference. However, computations are facilitated by rewriting the left side using:<\/p>\n

\"\displaystyle \frac{1}{2}\left ( p_{0}^{2}-p_{L}^{2} \right )=\bar{p}\left ( p_{0}-p_{L} \right )\"<\/p>\n

where, \"\bar{p}\" is the average pressure in the chamber.<\/p>\n

\"\displaystyle p_{0}-p_{L}=0.0222\frac{\mu L}{Ck_{g}}=\frac{(0.0222)(2.3\times 10^{-5})(0.05)}{(1\times 10^{-12})(40.9)}=624\ \textup{Pa}\"<\/p>\n

The pressure drop is quite small (about 6\u00a0cm\u00a0H2<\/sub>O) relative to the 1\u00a0\u00d7\u00a0105<\/sup>\u00a0Pa pressure maintained in the right-hand header, justifying the assumption that the mean pressure is closely approximated by the controlled pressure in the right-hand header.<\/p>\n

Return to E<\/span>xercise\u00a0<\/span>3<\/span><\/a><\/p>\n

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