Chap 1(a) molecular-diffusion_in_gas(2)

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BKF 2432: MASS TRANSFER
FKKSA, UMP
Principles of Mass Transfer
CHAPTER 1CHAPTER 1
Molecular Diffusion in GasesMolecular Diffusion in Gases
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BKF 2432: MASS TRANSFER FKKSA, UMP
Topic Outcomes
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 It is expected that student will be able to:
 Apply the diffusivity coefficient of molecular
diffusion in gases.
 Solve mathematical solution of molecular diffusion
in gases.

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CONTENTS
Mass Transfer
Molecular Diffusion Convective Mass Transfer
Gases Liquid Solid
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Introduction
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Equimolar Counter diffussion
in Gases
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Molecular Diffusion in Gases
 Equimolar Counter diffussion in Gases
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 For a binary gas mixture of A and B, the diffusivity coefficient
DAB=DBA
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Problem 6.1-1 (pg 452) Diffusion of Methane
Through Helium
A gas of CH4 and He is contained in a tube at 101.32 kPa pressure and 298
K. At one point the partial pressure of methane is pA1 = 60.79 kPa, and at a
point 0.02 m distance away, pA2 = 20.26 kPa. If the total pressure is
constant throughout the tube, calculate the flux of CH4(methane) at
steady state for equimolar counter diffusion.
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Example 6.2-1 (pg 415) Equimolar Counterdiffusion
o Ammonia gas (A) is diffusing through a uniform tube 0.10 m long
containing N2 gas (B) at 1.0132 x 105
Pa pressure and 298 K. At point 1,
pA1 = 1.013 x 104
Pa , and at point 2, pA2 = 0.507 x 104
Pa. The diffusivity
DAB = 0.230 x 10-4
m2
/s.
1. 1. Calculate the flux J*A at steady state
2. 2. Repeat for J*B
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Numerical
value
Units
82.057 cm3
.atm/kg mol . K
82.057x 10-3
m3
.atm/kg mol . K
8314.34 J/kg mol . K
8314.34 m3
.Pa / kg mol .K
8314.34 kg . m2
/s2
. kg mol.K
o.7302 ft3
.atm/lb mol.0
R

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BKF 2432: MASS TRANSFER FKKSA, UMPProblem 6.2-1 (pg 452) Equimolar Counterdiffusion of a Binary Gas
Mixture
• Helium and nitrogen gas are contained in a conduit 5 mm in diameter
and 0.1 m long at 298 K and a uniform constant pressure of 1.0 atm
abs. The partial pressure of He at one end of the tube is 0.060 atm and
the other end is 0.020 atm. Calculate the following for steady-state
equimolar counterdiffusion:
1. Flux of He in kg mol/s.m2
1. 2. Flux of N2
2. 3. Partial pressure of He at a point 0.05 m from either end.
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Solution:
DAB = 0.687 x 10-4
m2
/s (Table 6.2-1)
z2-z1 = 0.1m
pA1 = 0.060 atm
pA2 = 0.020 atm
R = 82.06 x 10-3
cm3
.atm/g mol.K (Table A.1-1)
a) Eqn. (6.1-13)
b)

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Diffusion of Gases A and B Plus
Convection (General Case) (pg 416)
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Convection is the concerted, collective
movement of ensembles of molecules
within fluids (e.g., liquids, gases)

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For equimolar counterdiffussion, NA=-NB ,
then NA=J*A=-NB=-J*B
Eq 6.17

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‘A’ Diffusing Through
Stagnant, Nondiffusing ‘B’
(Special Case)
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Example 6.2-2 (pg 419) Diffusion of Water
Through Stagnant, Nondiffusing Air
Water in the bottom of a narrow metal tube is held at a constant
temperature of 20o
C. The total pressure of air (assumed dry) is1.0 atm
and the temperature is 20o
C. Water evaporates and diffuses through
the air in the tube, and the diffusion path z2– z1 is 0.1524 m (0.5 ft)
long. The diagram is similar to Fig 6.2-2a. Calculate the rate of
evaporation at steady state in lb mol/hr.ft2
and kg mol/s.m2
. The
diffusivity of water vapor at 20o
C and 1 atm pressure is 0.250x10-4
m2
/s.
Assume that the system is isothermal.
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Problem 6.2-3 (pg 453) Diffusion of A Through
Stagnant B and Effect of Type of Boundary on Flux
Ammonia gas is diffusing through N2 under steady state conditions
with N2nondiffusing since it is insoluble in one boundary. The total
pressure is 1.013 x 105
Pa and the temperature is 298 K. The partial
pressure of NH3 at one point is 1.333 x 104
Pa, and at the other point 20
mm away it is 6.666 x 103
Pa. The DAB for the mixture at 1.013 X 105
Pa
and 298 K is 2.30 x 10-5
m2/s.
a) calculate the flux of NH3 in kg mol/s.m2
b) do the same as (a) but assume that N2 also diffuses, both boundaries
are permeable to both gases and the flux is equimolar
counterdiffusion. In which case is the flux greater?
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Diffusion Through Cross
Sectional Area (Sphere)
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Example 6.2-4 (pg 421) Evaporation of
Naphthalene Sphere
 A sphere of naphthalene having a radius of 2.0 mm is suspended in a
large volume of still air at 318 K and 1.01325 x 105
Pa. The surface
temperature of the naphthalene can be assumed to be at 318 K and its
vapor pressure at 318 K is 0.555 mm Hg. The DAB of naphthalene in air
at 318 K is 6.92 x 10-6
m2
/s. Calculate the rate of evaporation of
naphthalene from the surface.
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Problem 6.2-5 (pg 453) Mass Transfer from a
Naphthalene Sphere to Air
Mass transfer is occurring from a sphere of naphthalene having radius
of 10 mm. The sphere is in large volume of still air at 52.6°C and 1 atm
abs pressure. The vapor pressure of naphthalene at 52.6°C is 1.0
mmHg. The diffusitivity of naphthalene in air at 0°C is 5.16 x 10-6
m2
/s.
Calculate the rate of evaporation of naphthalene from the surface in
kg mol/s.m2
.
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Diffusion Coefficient
for Gases
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Example 6.2-5 (pg 427) Estimation of
Diffusivity of a Gas Mixture
• Normal butanol (A) is diffusing through air (B) at 1 atm abs. Using
the Fuller et al. method, estimate the diffusivity DAB for the following
temperatures and compare with the experimental data.
• Given MA (butanol) = 74.1 kg (mass)/kg mol,
• MB (air) = 29 kg (mass)/kg mol]
a) For 0o
C.
b) For 25.9o
C
c) For 0o
C and 2.0 atm abs
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Numerical
value
Units
82.057 cm3
.atm/kg mol . K
82.057 x 10-3
m3
.atm/kg mol . K
8314.34 J/kg mol . K
8314.34 m3
.Pa / kg mol .K
8314.34 kg . m2
/s2
. kg mol . K
o.7302 ft3
.atm/lb mol.0
R
Gas Law Constant R (pg 955)

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TEST 1 EP 203

Chap 1(a) molecular-diffusion_in_gas(2)

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