Question 10.41: Obtain an expression for the effective diffusivity of compon...
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 D_{AB} for A in B, and D_{AC} for A in C. The gas-phase mass transfer coefficient for the absorption of ammonia into water from a mixture of composition NH_3 20\%, N_2 73\%. H_2 7\% is found experimentally to be 0.030 m/s. What would you expect the transfer coefficient to be for a mixture of composition NH_3 5\%, N_2 60\%, H_2 35\%? 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 N_2 and H_₂ in water.
Diffusivity of NH_3 \text{ in }N_2 = 23 \times 10^{-6} m²/s.
Diffusivity of NH_3 \text{ in }H_2 = 52 \times 10^{-6} m²/s.
For case 1:
The effective diffusivity D′ is given by: \frac{1}{D^{\prime}} =\left(\frac{73/80}{23\times 10^{-6}} \right) +\left(\frac{7/80}{52\times 10^{-6}} \right) (from equation 10.90)
or: D′= 24.2 × 10^{-6} m²/s
The mass transfer coefficient is: \frac{D^{\prime}}{L} \left(\frac{1}{\text{log mean 1 and 0.8}} \right)
=\frac{24.2\times 10^{-6}}{L} \left(\frac{0.2}{\ln \frac{1}{0.8} } \right) ^{-1}=\frac{1}{L} \times 27\times 10^{-6}=\underline{\underline{0.030 \text{ m/s}}}
and hence: L = 0.90 × 10^{-3} m
Case 2:
The effective diffusivity D′ is given by:\frac{1}{D^{\prime}}=\left(\frac{60/95}{23\times 10^{-6}} \right) +\left(\frac{35/95}{52\times 10^{-6}} \right)
or: D′ = 28.9 × 10^{-6} m²/s
The mean transfer coefficient is:\frac{D^{\prime}}{L}.\left(\frac{1}{\text{ log mean 1 and 0.95}} \right)
=\left(\frac{28.9\times 10^{-6}}{0.9\times 10^{-3}} \right) \left(\frac{0.05}{\ln \frac{1}{0.95} } \right) ^{-1}=\underline{\underline{0.033\text{ m/s}}} .