Steady State Molecular Diffusion In Fluids Under Stagnant And Laminar Flow Conditions | Mass Transfer - Chemical Engineering PDF Download

2.2 STEADY STATE MOLECULAR DIFFUSION IN FLUIDS UNDER STAGNANT AND LAMINAR FLOW CONDITIONS 
2.2.1 Steady state diffusion through a constant area 
Steady state diffusion through a stagnant gas film 
Assume steady state diffusion in the Z direction without any chemical reaction in a binary gaseous mixture of species A and B. For one dimensional diffusion of species A, the Equation of molar flux can be written as
                               (2.11)
Separating the variables in Equation (2.11), it can be expressed as
                                          (2.12)
For the gaseous mixture, at constant pressure and temperature C and DAB are constant, independent of position and composition. Also all the molar fluxes are constant in Equation (2.12). Therefore the Equation (2.12) can be integrated between two boundary conditions as follows:
at          Z = Z₁,             y_A = yA₁
at          Z = Z₂,            y_A = y_A2
where 1 indicates the start of the diffusion path and 2 indicates the end of the diffusion path. After integration with the above boundary conditions the Equation for diffusion for the said condition can be expressed as
         (2.13) 

For steady state one dimensional diffusion of A through non-diffusing B, N_B = 0 and N_A = constant. Therefore N_A /(N_A + N_B ) =1 . Hence Equation (2.13) becomes
                                             (2.14)

Since for an ideal gas  and for mixture of ideal gases    , the Equation (2.14) can be expressed in terms of partial pressures as
                                (2.15)
Where P is the total pressure and p_A1 and p_A2 are the partial pressures of A at point 1 and 2 respectively. For diffusion under turbulent conditions, the flux is usually calculated based on linear driving force. For this purpose the Equation (2.13) can be manipulated to rewrite it in terms of a linear driving force. Since for the binary gas mixture of total pressure P, P-_A2 = P_{B2 ;} P-P_{A1 =} P_B1 ; P_A1 - P_A2 = P_B2 - P_B1 Then the Equation (2.15) can be written as
                  (2.16)
Or
                        (2.17)
Where p_B,M is called logarithmic mean partial pressure of species B which is defined as
                                                       (2.18)
A schematic concentration profile for diffusion A through stagnant B is shown in Figure 2.1. The component A diffuses by concentration gradient,  . Here flux is inversely proportional to the distance through which diffusion occurs and the concentration of the stagnant gas ( p_B,M ) because with increase in Z and p_B,M , resistance increases and flux decreases. 

Figure 2.1: Partial pressure distribution of A in non-diffusing B

2.2.2 Steady state equimolar counter diffusion: 
This is the case for the diffusion of two ideal gases, where an equal number of moles of the gases diffusing counter-current to each other. In this case N_B = -N_A = constant and N_A+ N_B = 0. The molar flux Equation (Equation (2.11)) at steady state can then be written as
                                        (2.19)
Integrating the Equation (2.19) with the boundary conditions: at Z = Z₁, y_A = y_A1; at Z = Z₂ y_A = y_A2, the Equation of molar diffusion for steady-state equimolar counter diffusion can be represented as
                      (2.20)
It may be noted here also that molar latent heats of vaporization of A and B are equal. So,  where,   are molar latent heats of vaporization of A and B, respectively. The concentration profile in terms of partial pressure is shown in Figure 2.2.

Figure 2.2: Equimolar counter diffusion of A and B: Partial pressure distribution with position 

2.2.3 Non-equimolar counter diffusion
In some practical cases, A and B molecules diffuse in opposite directions at different molar velocities [1]. Let carbon monoxide is generated from the reaction between hot char and oxygen. The stoichiometry is as follows:
                                      (2.21)
When one mole oxygen molecule diffuses towards char, two moles carbon monoxide molecules diffuse in opposite direction. Here, N_A = -N_B / 2 and molar latent heats of vaporization are not equal. Hence,
                                          (2.22