Question 10.44: According to Maxwell’s law, the partial pressure gradient in...
According to Maxwell’s law, the partial pressure gradient in a gas which is diffusing in a two-component mixture is proportional to the product of the molar concentrations of the two components multiplied by its mass transfer velocity relative to that of the second component. Show how this relationship can be adapted to apply to the absorption of a soluble gas from a multicomponent mixture in which the other gases are insoluble, and obtain an effective diffusivity for the multicomponent system in terms of the binary diffusion coefficients.
Carbon dioxide is absorbed in alkaline water from a mixture consisting of 30% CO₂ and 70% N₂, and the mass transfer rate is 0.1 kmol/s. The concentration of CO2 in the gas in contact with the water is effectively zero. The gas is then mixed with an equal molar quantity of a second gas stream of molar composition 20% CO₂, 50%, N₂ and 30% H₂. What will be the new mass transfer rate, if the surface area, temperature and pressure remain unchanged? It may be assumed that a steady-state film model is applicable and that the film thickness is unchanged.
Diffusivity of CO₂ in N₂ = 16 × 10^{-6} m²/s.
Diffusivity of CO₂ in H₂ = 35 × 10^{-6} m²/s.
For a binary system, Maxwell’s Law gives:
\frac{dC_A}{dy}=FC_A C_B(u_A-u_B) (equation 10.77)
For an ideal gas mixture:
-RT\frac{dC_A}{dy}=FC_A C_B(\frac{N^{\prime}_{A}}{C_A}-\frac{N^{\prime}_{B}}{C_B})
For a gas absorption process: N^{\prime}_{B}=0 if B is insoluble
or: -\frac{dC_A}{dy}=\frac{F C_B}{RT}N^{\prime}_{A} (i)
from Stefan’s law (equation 10.30):
N^{\prime}_{A}=-D\frac{C_T}{C_B}\frac{dC_A}{dy}
or: -\frac{dC_A}{dy}=N^{\prime}_{A}\frac{1}{D}\frac{C_B}{C_T} (equation 10.30) (ii)
=N^{\prime}_{A}\frac{1}{D}\frac{C_T-C_A}{C_T} (iii)
Comparing equations (i) and (ii):
\frac{F}{RT}=\frac{1}{DC_T}
or: F=\frac{RT}{DC_T} or D=\frac{RT}{FC_T}
Applying Maxwell’s law to a multicomponent system gives:
-\frac{dC_{A}}{dy} =F_{AB}C_{A}C_{B}(u_{A}-u_{B})+F_{AC}C_{A}C_{C}(u_{A}-u_{C})+…
For B, C … insoluble N_{B}^{\prime},N_{C}^{\prime}… = 0
Writing: F_{AB}=\frac{RT}{D_{AB}C_{T}} , and F_{AC}=\frac{RT}{D_{AC}C_{T}}
then: -RT\frac{dC_{A}}{dy} =\frac{RT}{D_{AC}C_{T}} N_{A}^{\prime}C_{B}+\frac{RT}{D_{AC}C_{T}} N_{A}^{\prime}C_{C}+…
∴ -\frac{dC_{A}}{dy} =\frac{N_{A}^{\prime}}{C_{T}} \left(\frac{C_{B}}{D_{AB}}+\frac{C_{C}}{D_{AC}}+… \right)
From equation (iii), using an effective diffusivity D′ for a multicomponent system gives:
-\frac{dC_{A}}{dy} =N_{A}^{\prime}\frac{1}{D^{\prime}} \frac{C_{T}-C_{A}}{C_{T}} (iv)
Comparing equations (iii) and (iv), then:
\frac{1}{D^{\prime}}=\frac{1}{C_{T}-C_{A}} \left\{\frac{C_{B}}{D_{AB}}+\frac{C_{C}}{D_{AC}}+… \right\}
=\frac{x_{B}^{\prime}}{D_{AB}} +\frac{x_{C}^{\prime}}{D_{AC}} +…
where x_{B}^{\prime},x^{\prime}_{C}…=\frac{C_{B}}{C_{T}-C_{A}} \gt \frac{C_{C}}{C_{T}-C_{A}} …
= mole fraction of B, C … in mixture of B, C
For absorption of CO₂ from mixture with N₂, the concentration driving force is:
\Delta C_{A}=C_{T}(0.3-0)=0.3C_{T}
where C_{T} is the total molar concentration
The mass transfer rate for an area A = AN_{A}^{\prime} =\Delta C_{A}\frac{D}{L} .\frac{C_{T}}{C_{Bm}} .A
where: C_{Bm} is the log mean of 1 ×C_{T} and 0.7C_{T}
or: C_{Bm}=\frac{C_{T}-0.7C_{T}}{\ln \frac{C_{T}}{0.7C_{T}} } =0.841C_{T} and: \frac{C_{T}}{C_{Bm}} = 1.189
Mass transfer rate, AN_{A}^{\prime}=0.3C_{T}\frac{16\times 10^{-6}}{L} 1.189 A=5.71\times 10^{-6}\frac{C_{T}A}{L} =0.10 kmol/s
or: \frac{C_{T}A}{L}=0.0175\times 10^{6} kmol/m².
For absorption of CO₂ from mixed stream, stream composition must be calculated.
100 moles stream 1 → 30 moles CO₂ 70 moles N₂
100 moles stream 2 → 20 moles CO₂ 50 moles N₂ 30 moles H₂
200 moles mixture → 50 moles CO₂ 120 moles N₂ 30 moles H₂
100 moles mixture → 25 moles CO₂ 60 moles N₂ 15 moles H₂
x^{\prime}_{N_{2}}=\frac{60}{75} =0.8 x^{\prime}_{H_{2}}=\frac{15}{75} =0.2
Diffusivity of CO₂ in mixture is given by:
\frac{1}{D^{\prime}} =\frac{0.8}{16\times 10^{-6}} +\frac{0.2}{35\times 10^{-6}}
\frac{10^{-6}}{D^{\prime}} =\frac{1}{20} +\frac{1}{175} = 0.0557 s/m²
D′ = 17.9 × 10^{-6} m^2/s
Thus: \Delta C_{A}=C_{T}(0.25-0)=0.25C_{T}
C_{Bm}=\frac{1-0.75}{\ln \frac{1}{0.75} } C_{T}=0.869C_{T}\frac{C_{T}}{C_{Bm}} =1.150
and the mass transfer rate:
AN_{A^{\prime}}=0.25C_{T}\frac{17.9\times 10^{-6}}{L} 1.150A=5.15\times 10^{-6}\frac{AC_{T}}{L} kmol/s = 0.090 kmol/s