# 4.28: Untitled Page 83

## Chapter 4

(1)

in which v represents the mass average velocity defined by Eq. 4‐38. One can use this result, along with the definition of the mass average velocity, to prove that (2)

This means that the approximation, v

A

v B requires the restriction

(3)

Since 1   is always less than one, we can always satisfy this inequality A

whenever the mass diffusion velocity is small compared to the species velocity, i.e.,

u



A

v A

(4)

For the sulfur dioxide mass transfer process illustrated in Figure 4‐4, this means that the approximation

(5)

is valid whenever the mass diffusion velocity is restricted by (6)

In many practical cases, this restriction is satisfied and all species velocities can be approximated by the mass average velocity.

Next we direct our attention to the mass transfer process at the gas‐liquid interface illustrated in Figure 4‐5. If we assume that there is no mass transfer of air into or out of the liquid phase, we can prove that (7)

Under these circumstances, the mass diffusion velocity is never small compared to the species velocity for practical conditions. Thus the type of approximation indicated by Eq. 4‐46 is never valid for the component of the velocity normal to the gas‐liquid interface. As a simplification, we can treat the sulfur dioxide‐air

Multicomponent systems

148

system as a binary system with species A representing the sulfur dioxide and species B representing the air.

(a) Use the definition of the mass average velocity given by Eq. 4‐38

to prove Eq. 2.

(b) Use

in order to prove Eq. 7.

Section 4.4

4‐11. A three component liquid mixture flows in a pipe with a mass averaged velocity of v  0 . 9 m/s . The density of the mixture is  = 850 kg/m3. The components of the mixture and their mole fractions are: n‐pentane, x  0 . 2

P

,

benzene, x  0 3

. , and naphthalene,

B

x

0 . 5

N

. The diffusion fluxes of each

component in the streamwise direction are:

pentane,

6

2

 1.564 10 kg/m s

P P

u

benzene,

6

2

 1.563 10 kg/m s

B B

u

naphthalene,

6

2

 3 127 10 kg/m s

N

N

u

.

.

Determine the diffusion velocities and the species velocities of the three

components. Use this result to determine the molar averaged velocity, v . Note that you must use eight significant figures in your computation.

Section 4.5

4‐12. Sometimes heterogeneous chemical reactions take place at the walls of tubes in which reactive mixtures are flowing. If species A is being consumed at a tube wall because of a chemical reaction, the concentration profile may be of the form

2

o

c ( r)

c 1

 

A

A

r o r

(1)

Here r is the radial position and

is the tube radius. The parameter  depends

o

r

on the net rate of production of chemical species at the wall and the molecular diffusivity, and it is bounded by 0    1 . If  is zero, the concentration across the tube is uniform at the valu o

e cA . If the flow in the tube is laminar, the velocity profile is given by

2

v ( r)

2 v

1

 

z

z

r o r

(2)

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