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Transport Phenomena: Transfer

Procedure

1) Mass balance over a thin shell

2) Obtain 1st ODE

3) Insert the relation between mass and concentration gradient

4) Result in 2nd ODE

5) Integration

6) Apply BCs to determine the constants (from integration)

7) Obtain concentration distribution

The law of conservation of mass of species A in a binary system is written over the volume of the shell in the form

Rate of Rate of rate of production of

Mass of - Mass of - mass of A by =0

A in A out homogeneous reaction

Mass flux = the number of moles of A that go through a unit area in unit time.

Combined molecular convective

flux flux flux (Diffusive flow) (Due to Bulk motion of A)

Boundary conditions

1) The concentration at surface can be specified; for example .

2) The mass flux at surface can be specified; for example, .

3) Solid surface substance A is lost to a surrounding stream

4) Rate of chemical reaction at the surface can be specified; for example,

Example 1 Diffusivity through a stagnant gas (gas-air)

Air flow

B

S: Cross sectional area

z2

z1 A Loop A z

0

Figure 1 Steady-state of A through stagnant B with the liquid vapor interface maintained at a fixed position.

Assumption

1) B does not dissolve in liquid A

2) A does not react to B

3) and is constant

Mole A balance around shell

Divided by and taking limit we obtain

Mass diffusion in binary system

From Fick’s law;

* is the molar flux of component A in the z direction due to .

* is the molecular diffusivity of the A in B

Mass in binary system

From Binary system:

From molar relative fixed co-ordinate:

Set equation (3) = equation (4) we get

Since B is stagnant,

We obtain the molar flux of A with

Substitute equation (5) into equation (1) we get

Divided through the equation by we obtain

Integration with respect to z gives

Second integration then gives

If we replace by and by , the above equation becomes

and may then be determined from the boundary conditions

BC1:

BC2:

Apply BC1,

Apply BC2,

Divided equation (7) by equation (8) gives

Substitute into equation (7) we obtain

Substitute and into the equation (6) we get

The profile of gas A are

The profile of gas B are obtained by using