Water flows at 0.50 m/s through a 20 mm tube lined with β-naphthol. What is the mass transfer coefficient if the Schmidt number is 2330?

Question

Accepted Answer

Reynolds number: Re = (0.020 x 0.50 x 1000/1 x [latex]10^{-3}[/latex]) = 10,000
From equation 12.144:

[latex]\frac{h_{D}}{u}\,=\,\frac{0.032R e^{-1/4}}{1+2.0R e^{-1/8}(S c\ -1)}[/latex] (12.144)

[latex]\frac{h_{D}}{u}\,=\,0.032R e^{-1/4}[1+2.0(S c-1)R e^{-1/8}]^{-1}[/latex]

∴ [latex]h_{D}=0.032 imes0.50 imes0.1[1+2 imes2329 imes0.316]^{-1}[/latex]

= [latex]\underline{\underline{1.085 imes 10^{-6}\ kmol/m^{2}\ s\ (kmol/m^{3})}}[/latex] = [latex]\underline{\underline{1.085 imes 10^{-6}\ m/s}}[/latex]

Question 12.2: Water flows at 0.50 m/s through a 20 mm tube lined with β-na......

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