Relative Magnitudes of Electroosmotic and Pressure-Driven Flows Consider a circular tube of diameter d = 5 μm (r = 2.5 μm) and length L = 10 cm filled with an aqueous solution. If ζ = −100 mV, calculate the pressure required to give the same total flow that would be generated by the application of

Question

Relative Magnitudes of

Electroosmotic and Pressure-Driven Flows

Consider a circular tube of diameter [latex]d=5 \mu \mathrm{m}(r=2.5 \mu \mathrm{m})[/latex] and length [latex]L=10 \mathrm{~cm}[/latex] filled with an aqueous solution. If [latex]\zeta=-100 \mathrm{mV}[/latex], calculate the pressure required to give the same total flow that would be generated by the application of [latex]1 \mathrm{kV}[/latex].

Accepted Answer

Electroosmotic flow:

[latex]\begin{aligned}Q &=u A=-\frac{\varepsilon \zeta}{\mu} \frac{V}{L} A \&=-\frac{80 imes 8.85 imes 10^{-12} \mathrm{C} / \mathrm{V} \mathrm{m} imes(-0.1) \mathrm{V}}{10^{-3} \mathrm{~Pa} \mathrm{~s}} imes \frac{10^{3} \mathrm{~V}}{0.1 \mathrm{~m}} imes \pi\left(2.5 imes 10^{-6} \mathrm{~m} ight)^{2} \&=1.39 imes 10^{-14} \mathrm{~m}^{3} / \mathrm{s}=1.39 imes 10^{-11} \mathrm{l} / \mathrm{s}=13.9 \mathrm{p}(\mathrm{pico}) \mathrm{l} / \mathrm{s}.\end{aligned}[/latex]

Pressure-driven flow:

[latex]\begin{aligned}Q &=\frac{\pi r^{4}}{8 \mu}\left(-\frac{d p}{d x} ight)=\frac{\pi r^{4}}{8 \mu} \frac{(-\Delta p)}{L} ; \-\Delta p &=\frac{8 Q \mu L}{\pi r^{4}} \&=\frac{8 imes 1.39 imes 10^{-14} \mathrm{~m}^{3} / \mathrm{s} imes 10^{-3} \mathrm{~Pa} \mathrm{~s} imes 0.1 \mathrm{~m}}{\pi\left(2.5 imes 10^{-6} \mathrm{~m} ight)^{4}} \&=9.06 imes 10^{4} \mathrm{~Pa} ext { (about } 0.9 ext { atmospheres) } .\end{aligned}[/latex]

Question 12.2: Relative Magnitudes of Electroosmotic and Pressure-Driven Fl...

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