Obtain an expression for the effective diffusivity of component A in a gaseous mixture of A, B and C in terms of the binary diffusion coefficients DAB for A in B, and DAC for A in C.The gas-phase mass transfer coefficient for the absorption of ammonia into water from a mixture of composition NH3 20

Question

Obtain an expression for the effective diffusivity of component A in a gaseous mixture
of A, B and C in terms of the binary diffusion coefficients [latex]D_{AB}[/latex] for A in B, and [latex]D_{AC}[/latex] for A in C. The gas-phase mass transfer coefficient for the absorption of ammonia into water from a mixture of composition [latex]NH_3 20\%, N_2 73\%. H_2 7\%[/latex] is found experimentally to be 0.030 m/s. What would you expect the transfer coefficient to be for a mixture of composition [latex]NH_3 5\%, N_2 60\%, H_2 35\%[/latex]? All compositions are given on a molar basis. The total pressure and temperature are the same in both cases. The transfer coefficients are based on a steady-state film model and the effective film thickness may be assumed constant.
Neglect the solubility of [latex]N_2[/latex] and [latex]H_₂[/latex] in water.
Diffusivity of [latex]NH_3 ext{ in }N_2 = 23 imes 10^{-6}[/latex] m²/s.
Diffusivity of [latex]NH_3 ext{ in }H_2 = 52 imes 10^{-6}[/latex] m²/s.

Accepted Answer

For case 1:
The effective diffusivity D′ is given by: [latex]\frac{1}{D^{\prime}} =\left(\frac{73/80}{23 imes 10^{-6}} ight) +\left(\frac{7/80}{52 imes 10^{-6}} ight) [/latex](from equation 10.90)

or: D′= 24.2 ×[latex] 10^{-6}[/latex] m²/s

The mass transfer coefficient is: [latex]\frac{D^{\prime}}{L} \left(\frac{1}{ ext{log mean 1 and 0.8}} ight) [/latex]

[latex]=\frac{24.2 imes 10^{-6}}{L} \left(\frac{0.2}{\ln \frac{1}{0.8} } ight) ^{-1}=\frac{1}{L} imes 27 imes 10^{-6}=\underline{\underline{0.030 ext{ m/s}}}[/latex]

and hence: L = 0.90 × [latex]10^{-3}[/latex] m
Case 2:

The effective diffusivity D′ is given by:[latex]\frac{1}{D^{\prime}}=\left(\frac{60/95}{23 imes 10^{-6}} ight) +\left(\frac{35/95}{52 imes 10^{-6}} ight) [/latex]

or: D′ = 28.9 × [latex] 10^{-6}[/latex] m²/s

The mean transfer coefficient is:[latex]\frac{D^{\prime}}{L}.\left(\frac{1}{ ext{ log mean 1 and 0.95}} ight) [/latex]

[latex]=\left(\frac{28.9 imes 10^{-6}}{0.9 imes 10^{-3}} ight) \left(\frac{0.05}{\ln \frac{1}{0.95} } ight) ^{-1}=\underline{\underline{0.033 ext{ m/s}}}[/latex] .

Question 10.41: Obtain an expression for the effective diffusivity of compon...

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