{"id":71,"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-1-solution\/"},"modified":"2022-01-10T00:58:31","modified_gmt":"2022-01-10T00:58:31","slug":"exercise-1-solution","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/flux-equations-for-gas-diffusion-in-porous-media\/chapter\/exercise-1-solution\/","title":{"raw":"Exercise\u00a01 Solution","rendered":"Exercise\u00a01 Solution"},"content":{"raw":"
\r\n

Part a)<\/strong><\/p>\r\n

Arbitrarily set the origin of the coordinate, l<\/em>, at the left end of the diffusion chamber. The value of l<\/em> on the right-hand end of the chamber is L<\/em>, equal to 0.05\u00a0m. Equation\u00a023<\/a> is applicable because diffusion occurs in the molecular regime with no pressure gradient. At steady state, the total diffusive flux of each component is constant with respect to time and position along the diffusion path. Thus, Equation\u00a023<\/a> is integrated to obtain Equation\u00a0Exercise\u00a0Solution\u00a01-1.<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle N_{A}^{D}=\\frac{DC}{\\left ( 1-M_{AB}^{0.5} \\right )L}\\textup{ln}\\left\\{\\frac{1-\\left ( 1-M_{AB}^{0.5} \\right )x_{A}(L)}{1-\\left ( 1-M_{AB}^{0.5} \\right )x_{A}(0)} \\right\\}[\/latex]<\/td>\r\n(Exercise Solution 1-1)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

where:<\/p>\r\n\r\n\r\n\r\n\r\n\r\n
x<\/em>A<\/em><\/sub>(L<\/em>)<\/td>\r\n=<\/td>\r\nmole fraction of argon at l<\/em> = L<\/em><\/td>\r\n<\/tr>\r\n
x<\/em>A<\/em><\/sub>(0)<\/td>\r\n=<\/td>\r\nmole fraction of argon at l<\/em> = 0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

The molar concentration is computed from the ideal gas law, C<\/em>\u00a0=\u00a0p<\/em>\/RT<\/em>,<\/p>\r\n

with: p<\/em> = 1 \u00d7 105<\/sup> Pa, R<\/em> = 8.205 m3 <\/sup>Pa\/deg-mole and T<\/em> = 298 K.<\/p>\r\n

[latex]\\displaystyle C=\\frac{p}{RT}=\\frac{10^{5}\\ \\textup{Pa}}{8.205\\frac{\\textup{m}^{3}\\ \\textup{Pa}}{\\textup{deg\\ mole}}298\\ \\textup{K}}=40.9\\frac{\\textup{moles}}{\\textup{m}^{3}}[\/latex]<\/p>\r\n

Other parameter values are:<\/p>\r\n

x<\/em>A<\/em><\/sub>(L<\/em>) = 0, x<\/em>A<\/em><\/sub>(0) = 1, [latex]\\displaystyle M_{AB}^{0.5}[\/latex] = 3.158, L<\/em> = 0.05 m and D<\/em> = 2.37 \u00d7 10-5<\/sup> m2<\/sup>s-1<\/sup>.<\/p>\r\n

The total diffusive flux of argon is:<\/p>\r\n

[latex]\\displaystyle N_{A}^{D}=\\frac{(2.37\\times 10^{-5})(40.9){\\textup{ln}(3.158^{-1})}}{(1-3.158)(0.05)}=0.0103\\ \\frac{\\textup{moles}}{\\textup{m}^{2}\\ \\textup{s}}[\/latex]<\/p>\r\n

The flux of argon is positive because it is in the direction of increasing l<\/em>.<\/p>\r\n

The flux of helium can be calculated from Equation\u00a0Exercise\u00a0Solution\u00a01-1 after interchanging the subscripts. However, the flux of helium is computed more easily from Graham\u2019s law:<\/p>\r\n

[latex]\\displaystyle N_{B}^{D}=-M_{AB}^{0.5}N_{A}^{D}=-3.158(0.0103)=-0.0325\\ \\frac{\\textup{moles}}{\\textup{m}^{2}\\ \\textup{s}}[\/latex]<\/p>\r\n

The corresponding values of flux computed from Fick\u2019s law (Equation\u00a011<\/a>) are 0.019\u00a0moles\/(m2<\/sup>s) for species A<\/em> and -0.019\u00a0moles\/(m2<\/sup>s) for species B<\/em>. The Fick\u2019s law calculation results in a rather large error in this case as there is significant disparity between the molecular weights of the gas components.<\/p>\r\n

Part b)<\/strong><\/p>\r\n

The non-equimolar flux or net flux is given by Equation\u00a05<\/a>.<\/p>\r\n

[latex]\\displaystyle N^{D}=N_{A}^{D}+N_{B}^{D}=0.0103-0.0325=-0.0222\\ \\frac{\\textup{moles}}{\\textup{m}^{2}\\ \\textup{s}}[\/latex]<\/p>\r\n

This is the flux of the gas as a whole. It is engendered entirely by diffusion, occurs without loss of momentum by viscous shear, and is in the direction of diffusion of the lighter component (helium).<\/p>\r\n

Part c)<\/strong><\/p>\r\n

Integrate Equation\u00a023<\/a> subject to x<\/em>A<\/em> <\/sub>= x<\/em>A<\/em><\/sub>(l<\/em>) at position l<\/em> and x<\/em>A<\/em><\/sub>\u00a0=\u00a01 at l<\/em>\u00a0=\u00a00 or use Equation\u00a0Exercise\u00a0Solution\u00a01-1 to obtain Exercise\u00a0Solution\u00a01-2.<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle N_{A}^{D}=\\frac{DC}{\\left ( 1-M_{AB}^{0.5} \\right )l}\\textup{ln}\\left\\{\\frac{1-\\left ( 1-M_{AB}^{0.5} \\right )x_{A}(l)}{M_{AB}^{0.5}} \\right\\}[\/latex]<\/td>\r\n(Exercise Solution 1-2)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

However, x<\/em>A<\/em><\/sub>(L<\/em>)\u00a0=\u00a00, consequently, Equation\u00a0Exercise\u00a0Solution\u00a01-2 becomes Equation\u00a0Exercise\u00a0Solution\u00a01-3.<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle N_{A}^{D}=\\frac{DC}{\\left ( 1-M_{AB}^{0.5} \\right )L}\\textup{ln}\\left\\{\\frac{1}{M_{AB}^{0.5}} \\right\\}[\/latex]<\/td>\r\n(Exercise Solution 1-3)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

The desired result is obtained by combining these two equations and solving for x<\/em>A<\/em><\/sub>(l<\/em>) to obtain Exercise\u00a0Solution\u00a01-4.<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle x_{A}(l)=\\frac{1-\\textup{exp}\\left\\{(1-l\/L)\\textup{ln}(M_{AB}^{0.5}) \\right\\}}{1-M_{AB}^{0.5}}[\/latex]<\/td>\r\n(Exercise Solution 1-4)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Part d)<\/strong><\/p>\r\n

Equation\u00a010<\/a> is combined with Graham\u2019s law to obtain Exercise\u00a0Solution\u00a01-5.<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle J_{A}=N_{A}^{D}\\left\\{1-x_{A}(1-M_{AB}^{0.5}) \\right\\}[\/latex]<\/td>\r\n(Exercise Solution 1-5)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Then Equation\u00a0Exercise\u00a0Solution\u00a01-4 is substituted for x<\/em>A<\/em><\/sub> to arrive at Equation\u00a0Exercise\u00a0Solution\u00a01-6.<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\displaystyle J_{A}(l)=N_{A}^{D}\\textup{exp}\\left\\{(1-l\/L)\\textup{ln}(M_{AB}^{0.5}) \\right\\}[\/latex]<\/td>\r\n(Exercise Solution 1-6)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Equation\u00a0Exercise\u00a0Solution\u00a01-6 shows how the equimolar flux varies over the length of the diffusion chamber.<\/p>\r\n

At a point midway between the supply headers (l<\/em>\/L<\/em>\u00a0=\u00a00.5), we have<\/p>\r\n

J<\/em>A<\/em><\/sub> = (0.0103) exp{0.5 ln(3.158)} = 0.0183 [latex]\\frac{\\textup{moles}}{\\textup{m}^{2}\\ s}[\/latex]<\/p>\r\n

The corresponding helium flux follows from Equation\u00a0Exercise\u00a0Solution\u00a01-6 by interchanging the subscripts to obtain:<\/p>\r\n

J<\/em>B<\/em><\/sub> = (\u22120.0325) exp{0.5 ln(0.317)} = \u22120.0183 [latex]\\frac{\\textup{moles}}{\\textup{m}^{2}\\ s}[\/latex]<\/p>\r\n

These molar fluxes are equal in magnitude and opposite in direction so they satisfy Equation\u00a07<\/a>. This numerical computation applies to a particular point, and we leave it to the reader to demonstrate that Equation\u00a0Exercise\u00a0Solution\u00a01-6 and its counterpart for component B<\/em> sum to zero at all points. Of course, this must be the case because Equation\u00a07<\/a> is at the heart of the analysis leading to the starting point for this example, Equation\u00a023<\/a>.<\/p>\r\n

Return to E<\/span>xercise\u00a0<\/span>1<\/span><\/a><\/p>\r\n

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

\n

Part a)<\/strong><\/p>\n

Arbitrarily set the origin of the coordinate, l<\/em>, at the left end of the diffusion chamber. The value of l<\/em> on the right-hand end of the chamber is L<\/em>, equal to 0.05\u00a0m. Equation\u00a023<\/a> is applicable because diffusion occurs in the molecular regime with no pressure gradient. At steady state, the total diffusive flux of each component is constant with respect to time and position along the diffusion path. Thus, Equation\u00a023<\/a> is integrated to obtain Equation\u00a0Exercise\u00a0Solution\u00a01-1.<\/p>\n\n\n\n
\"\displaystyle N_{A}^{D}=\frac{DC}{\left ( 1-M_{AB}^{0.5} \right )L}\textup{ln}\left\{\frac{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}(L)}{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}(0)} \right\}\"<\/td>\n(Exercise Solution 1-1)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

where:<\/p>\n\n\n\n\n
x<\/em>A<\/em><\/sub>(L<\/em>)<\/td>\n=<\/td>\nmole fraction of argon at l<\/em> = L<\/em><\/td>\n<\/tr>\n
x<\/em>A<\/em><\/sub>(0)<\/td>\n=<\/td>\nmole fraction of argon at l<\/em> = 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The molar concentration is computed from the ideal gas law, C<\/em>\u00a0=\u00a0p<\/em>\/RT<\/em>,<\/p>\n

with: p<\/em> = 1 \u00d7 105<\/sup> Pa, R<\/em> = 8.205 m3 <\/sup>Pa\/deg-mole and T<\/em> = 298 K.<\/p>\n

\"\displaystyle C=\frac{p}{RT}=\frac{10^{5}\ \textup{Pa}}{8.205\frac{\textup{m}^{3}\ \textup{Pa}}{\textup{deg\ mole}}298\ \textup{K}}=40.9\frac{\textup{moles}}{\textup{m}^{3}}\"<\/p>\n

Other parameter values are:<\/p>\n

x<\/em>A<\/em><\/sub>(L<\/em>) = 0, x<\/em>A<\/em><\/sub>(0) = 1, \"\displaystyle M_{AB}^{0.5}\" = 3.158, L<\/em> = 0.05 m and D<\/em> = 2.37 \u00d7 10-5<\/sup> m2<\/sup>s-1<\/sup>.<\/p>\n

The total diffusive flux of argon is:<\/p>\n

\"\displaystyle N_{A}^{D}=\frac{(2.37\times 10^{-5})(40.9){\textup{ln}(3.158^{-1})}}{(1-3.158)(0.05)}=0.0103\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}\"<\/p>\n

The flux of argon is positive because it is in the direction of increasing l<\/em>.<\/p>\n

The flux of helium can be calculated from Equation\u00a0Exercise\u00a0Solution\u00a01-1 after interchanging the subscripts. However, the flux of helium is computed more easily from Graham\u2019s law:<\/p>\n

\"\displaystyle N_{B}^{D}=-M_{AB}^{0.5}N_{A}^{D}=-3.158(0.0103)=-0.0325\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}\"<\/p>\n

The corresponding values of flux computed from Fick\u2019s law (Equation\u00a011<\/a>) are 0.019\u00a0moles\/(m2<\/sup>s) for species A<\/em> and -0.019\u00a0moles\/(m2<\/sup>s) for species B<\/em>. The Fick\u2019s law calculation results in a rather large error in this case as there is significant disparity between the molecular weights of the gas components.<\/p>\n

Part b)<\/strong><\/p>\n

The non-equimolar flux or net flux is given by Equation\u00a05<\/a>.<\/p>\n

\"\displaystyle N^{D}=N_{A}^{D}+N_{B}^{D}=0.0103-0.0325=-0.0222\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}\"<\/p>\n

This is the flux of the gas as a whole. It is engendered entirely by diffusion, occurs without loss of momentum by viscous shear, and is in the direction of diffusion of the lighter component (helium).<\/p>\n

Part c)<\/strong><\/p>\n

Integrate Equation\u00a023<\/a> subject to x<\/em>A<\/em> <\/sub>= x<\/em>A<\/em><\/sub>(l<\/em>) at position l<\/em> and x<\/em>A<\/em><\/sub>\u00a0=\u00a01 at l<\/em>\u00a0=\u00a00 or use Equation\u00a0Exercise\u00a0Solution\u00a01-1 to obtain Exercise\u00a0Solution\u00a01-2.<\/p>\n\n\n\n
\"\displaystyle N_{A}^{D}=\frac{DC}{\left ( 1-M_{AB}^{0.5} \right )l}\textup{ln}\left\{\frac{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}(l)}{M_{AB}^{0.5}} \right\}\"<\/td>\n(Exercise Solution 1-2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

However, x<\/em>A<\/em><\/sub>(L<\/em>)\u00a0=\u00a00, consequently, Equation\u00a0Exercise\u00a0Solution\u00a01-2 becomes Equation\u00a0Exercise\u00a0Solution\u00a01-3.<\/p>\n\n\n\n
\"\displaystyle N_{A}^{D}=\frac{DC}{\left ( 1-M_{AB}^{0.5} \right )L}\textup{ln}\left\{\frac{1}{M_{AB}^{0.5}} \right\}\"<\/td>\n(Exercise Solution 1-3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The desired result is obtained by combining these two equations and solving for x<\/em>A<\/em><\/sub>(l<\/em>) to obtain Exercise\u00a0Solution\u00a01-4.<\/p>\n\n\n\n
\"\displaystyle x_{A}(l)=\frac{1-\textup{exp}\left\{(1-l/L)\textup{ln}(M_{AB}^{0.5}) \right\}}{1-M_{AB}^{0.5}}\"<\/td>\n(Exercise Solution 1-4)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Part d)<\/strong><\/p>\n

Equation\u00a010<\/a> is combined with Graham\u2019s law to obtain Exercise\u00a0Solution\u00a01-5.<\/p>\n\n\n\n
\"\displaystyle J_{A}=N_{A}^{D}\left\{1-x_{A}(1-M_{AB}^{0.5}) \right\}\"<\/td>\n(Exercise Solution 1-5)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Then Equation\u00a0Exercise\u00a0Solution\u00a01-4 is substituted for x<\/em>A<\/em><\/sub> to arrive at Equation\u00a0Exercise\u00a0Solution\u00a01-6.<\/p>\n\n\n\n
\"\displaystyle J_{A}(l)=N_{A}^{D}\textup{exp}\left\{(1-l/L)\textup{ln}(M_{AB}^{0.5}) \right\}\"<\/td>\n(Exercise Solution 1-6)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Equation\u00a0Exercise\u00a0Solution\u00a01-6 shows how the equimolar flux varies over the length of the diffusion chamber.<\/p>\n

At a point midway between the supply headers (l<\/em>\/L<\/em>\u00a0=\u00a00.5), we have<\/p>\n

J<\/em>A<\/em><\/sub> = (0.0103) exp{0.5 ln(3.158)} = 0.0183 \"\frac{\textup{moles}}{\textup{m}^{2}\ s}\"<\/p>\n

The corresponding helium flux follows from Equation\u00a0Exercise\u00a0Solution\u00a01-6 by interchanging the subscripts to obtain:<\/p>\n

J<\/em>B<\/em><\/sub> = (\u22120.0325) exp{0.5 ln(0.317)} = \u22120.0183 \"\frac{\textup{moles}}{\textup{m}^{2}\ s}\"<\/p>\n

These molar fluxes are equal in magnitude and opposite in direction so they satisfy Equation\u00a07<\/a>. This numerical computation applies to a particular point, and we leave it to the reader to demonstrate that Equation\u00a0Exercise\u00a0Solution\u00a01-6 and its counterpart for component B<\/em> sum to zero at all points. Of course, this must be the case because Equation\u00a07<\/a> is at the heart of the analysis leading to the starting point for this example, Equation\u00a023<\/a>.<\/p>\n

Return to E<\/span>xercise\u00a0<\/span>1<\/span><\/a><\/p>\n

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