Question

In a gas-phase diffusion mass-transfer process, the steady-state flux of helium in a binary mixture of helium and neon is 3.8 x 10-9 kgmole/cm2 s, and the flux of neon is 0. At a particular point in the diffusion space, the concentration of helium is 0.023 kgmole/m3 and the concentration of neon is 0.045 kgmole/m3 . Estimate the individual net velocities of helium and neon along the direction of mass transfer, the average molar velocity, and the average mass velocity.''

Answer 1

Answer:

Explanation:

Let's assume that Helium is He and Neon is Ne

Then, the expression of the steady-state flux of Helium in a binary mixture of Helium and Neon is:

$N_{He} = c_{He}v_{He}$

where;

$c_{He}$ = concentration of Helium

$v _ {He}$ = net velocity of Helium

Making $v _ {He}$ the subject, we have:

$v_{He}=\dfrac{ N_{He} }{c_{He}}$

$v_{He}=\dfrac{ 3.8 \times 10^{-9} \ kg/mol /m^2.s}{0.023 \ kgmol/m^3}$

$v_{He}= 1.652 \times 10^{-7} \ m/s$

The expression for the steady-state flux of Neon

$N_{Ne} = c_{Ne} \ v_{Ne}$

Here;

$c_{Ne}$ = Concentration of neon

$v_{Ne}$ = net velocity of neon species

$v_{Ne} = \dfrac{N_{Ne} }{c_{Ne}}$

$v_{Ne} = \dfrac{ 0 \ kgmole/m^2 .s }{0.045 \ kgmole/m^3}$

$v_{Ne} = 0 \ m/s$

Thus, the net velocity of species Ne along the direction of mass transfer = 0 m/s

The average velocity V is:

$V _{avg }= \dfrac{1}{c}(c_{He}v_{He} + c_{Ne}v_{Ne})$

$= \dfrac{(N_{He} + N_{Ne})} {(C_{He} + C_{Ne})}$

$V _{avg}= \dfrac{(3.8 \times 10^{-9} + 0) \ kgmole /m^2.s} {(0.023 + 0.045) \ kgmole/m^3}$

$V _{avg}= 5.588 \times 10^{-8} \ m/s$

The average mass velocity is:

$V_{mass} = \dfrac{(0.023 \times 4 ) \times 1.652\times 10^{-7} +0}{(0.023 \times 4) + (0.045 \times 20) }$

$V_{mass} = 1.532 \times 10^{-8} \ m/s$