Chapter 3 Unidirectional transport - Chemical Engineering ...

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20 CHAPTER 3. UNIDIRECTIONAL TRANSPORTũ ∗ xThis equation is easily solved to obtain= 0 at z = L (3.80)ũ ∗ x (z∗ ) = C 1 exp ( √ ıω ∗ z ∗ ) + C 2 exp (− √ ıω ∗ z ∗ ) (3.81)where C 1 and C 2 are determined from the boundary conditions at z ∗ = 0and z ∗ = 1,ũ ∗ x (z∗ ) = exp (√ ıω ∗ z ∗ ) − exp ( √ ıω ∗ (2 − z ∗ ))1 − exp (2 √ ıω ∗ )(3.82)The physical velocity field, which is the real part of the product of ũ ∗ x andexp (ıω ∗ t ∗ ), is[ √exp (u ∗ x(z ∗ ıω∗ z ∗ ) − exp ( √ ıω) = Real∗ (2 − z ∗ ]))1 − exp (2 √ exp (ıω ∗ t ∗ ) (3.83)ıω ∗ )Before solving the above equation, it is useful to examine some limitingcases for the frequency of oscillation. In the limit ω ∗ → 0, the solution forthe fluid velocity, 3.83, isu ∗ x(z ∗ ) = (1 − z ∗ ) exp (ıω ∗ t ∗ ) (3.84)The physical reason for this result is as follows. The scaled frequency ofoscillation is obtained by dividing the frequency, ω, by the inverst of thetime required for the momentum to diffuse across the length of the channel,(L 2 /ν). If the scaled frequency ω ∗ is small, then the time period of variationof the velocity of the bottom plate is long compared to the time required formomentum to diffuse across the channel. In this case, it is expected, thatthe velocity profile at any time instant along the oscillation cycle is identicalto the velocity for a flow driven by a steady plate velocity which is equal tothe instanteneous value of the plate velocity at that instant in the cycle. Inthe limit ω ∗ ≫ 1, the fluid velocity field is given byu ∗ x(z ∗ ) = exp (− √ ıω ∗ z ∗ ) (3.85)In this case, the velocity field decreases over a distance z ∗ ∼ (1/ √ ω ∗ ) fromthe surface. This is because the frequency of oscillation is large comparedto the time required for diffusion of momentum across the channel, and themomentum diffuses only to a distance comparable to (L/ √ ω ∗ ). Beyond thisdistance, the momentum generated during the positive and negative parts ofa cycle cancel out, and the fluid velocity approaches zero.
3.2. EFFECT OF BULK FLOW AND REACTION IN MASS TRANSFER21z=LDry airzz=0¡¡¡¢¡¢¡¢¡¢¢¡¢¡¢¡¢¡ ¡ ¡Water¡¡¡¢¡¢¡¢¡¢¢¡¢¡¢¡¢¡ ¡ ¡¡¡¡¢¡¢¡¢¡¢¡ ¡ ¡x¢¡¢¡¢¡¢¡ ¡ ¡¢¡¢¡¢¡¢Figure 3.5: Diffusion with bulk flow.3.2 Effect of bulk flow and reaction in masstransferIn this section, the special effects of bulk flow and reactions on the solutionsfor unidirection mass transfer problems are examined.3.2.1 Diffusion in a stagnant filmWater evaporates from a container through a stagnant air film through aglass tube into dry air flowing at the top of the tube, as shown in figure 3.5.If the mole fraction at the surface of the liquid surface is the saturation molefraction x W s , and the dry air flowing past the tube does not contain anywater, what is the concentration profile of water in the glass tube?Though the air in the tube is stationary, the mean velocity across anyhorizontal surface in the tube is not zero, because of the flow of water vapour
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