COLLEGE OF ENGINEERING

DEPARTMENT OF CHEMICAL ENGINEERING

HEAT AND MASS TRANSFER

PROBSET

TOPIC
INTRODUCTION TO MASS TRANSFER; DIFFUSIONPROBLEM 1
For a mixture of ethanol (Ch3Ch2OH) vapor and methane (CH4), predict the diffusivity using
the method of Fuller et all.
At 1.0132 X 10^5 Pa and 298 and 373K.
At 2.0265 x 10^5 Pa and @298K

Solution:
A = CH4 MA= 16.04 B= CH3CH2OH MB= 46.1 T= 298K P= 1.0 atm

∑va = 1(16.5) 4(1.98) = 24.42

∑vb = 2(16.5) + 6(1.98) 5.48 = 50.36

1
1 1 1 1 1 2
1.00𝑥10−7 𝑇 1.75 ( : ) /2 1.00𝑥10−7 (298)1.75 ( : )
DAB = 𝑀𝐴 𝑀𝐵
2 = 16.04 46.1
2
𝑃∑va 1 ∑vb1 24.421 50.361
( : ) 1.0( : )
3 3 3 3

DAB = 1.43 x 10^-5 m^2/s (298K)

373 1.75
DAB =1.43 x 10^-5 (298) = 2.12 x 10^-5 m^2/s (373K)

DAB = 1.43 x 10^-5 (1/2) = 0.715x 10^-5 m^2/s (298K, 2 atm)

PROBLEM 2

A flat plug 30mm thick having an area of 4.0 x 10^-4 m^2 and made of vulcanized rubber is used
for closing an opening in a container. The gas CO2 at 25C and 2.0 atm pressure is inside the
container. Calculate the total leakage or diffusion of CO2 through the plug to the outside in kg
mol CO2/s at steady state. Assume that the partial pressure of Co2 outside is zero. From
Barrer(B5) the solubility of the CO2 gas is 0.90 m^3 gas (at STP of ) C and 1 atm) per m^3
rubber per atm pressure of CO2. The diffusivity is 0.11 x 10 ^-9 m^2/s.
Solution:
L= 30/1000 = 0.030m
A= 4.0 x 10^-4 m^2
T= 25C
RA1 = 2.0 atm abs
0.90 𝑚3 (𝑆𝑇𝑃)
S=𝑚3 𝑟𝑢𝑏𝑏𝑒𝑟 𝑥 𝑎𝑡𝑚 DAB= 0.11 x10^-9 m^2/s

5 𝑅𝐴1 0.90 (2.0) 10−2 𝐾𝑔𝑚𝑜𝑙

CA1 =2.2414 = = 8.031 𝑋
22.414 𝑚3 𝑟𝑢𝑏𝑏𝑒𝑟

𝐷𝐴𝐵 (𝐶𝐴1;𝐶𝐴2) 𝐴𝐷𝐴𝐵 (𝐶𝐴1;𝐶𝐴2) 4 𝑥 10−4 (0.11𝑥10−9 )(8.031𝑥10−2 ;0)

NA = NA A= = (0.030;0)
=
𝑍2;𝑍1 𝑧2;𝑧1

1.178 𝑥10;13 𝑘𝑔𝑚𝑜𝑙/𝑠

PROBLEM 3
The gas CO2 is diffusing at steady state through a tube 0.20 m long having a diameter of 0.01 m
and containing N2 at 298 K. The total pressure is constant at 101.32 kPa. The partial pressure of
CO2 is 456 mm Hg at one end and 76 mm Hg at the other end. The diffusivity DAB is 1.67 ×
10−5 m2/s at 298 K. Calculate the flux of CO2 in cgs and SI units for equimolar counter
diffusion.
Solution:
The given problem denotes the equimolar counter diffusion of CO2 in a binary gas mixture.
The information given in question are:
Length of tube (z2-z1) = 0.20 m
Total pressure of gases = 101.32 kPa
Temperature of the system (T)= 298 K
Parial pressure of CO2 at one end of tube( @z1) PA1= 456 mmHg = 60.795 KPa
Partial pressure of CO2 at the other end of tube(@z2) PA2= 76 mmHg = 10.1325 KPa
Diffusivity DAB = 1.67 X 10-5 m2 s-1
The equation of molar flux for the steady-state equimolar counter diffusion is given as,
NA = DAB [PA1-PA2] / (RT[Z2-Z1])
where NA is the molar flux of CO2, PA1 and PA2 are the partial pressures of CO2 at the ends of
tube
R is the universal gas constant with a value of 8.314 J mol-1 K-1
T is the temperature and Z2-Z1 is the length of the tube
On substituting the values in the above equation:

Flux of CO2 in SI units:

NA = 1.67 x 10 ^ -5 X ( 60.795 - 10.1325 )/8.314 X 298 X 0.2
NA = 1.71 x 10-6 kgmol m-2 s-1

Flux of CO2 in CGS units:

1 m = 100 cm
1 kg mol = 1000 gmol
using this conversion ,
NA = 1.71 X 10^-6 X1000gmol/100^2cm^2s^2
NA= 1.71 x 10 -7 gmol cm-2 s-1

PROBLEM 4
Hydrogen gas at 2.0 atm and 27°C is flowing in a neoprene tube 3.0 mm inside diameter and 11
mm outside diameter. Calculate the leakage of H2 through a tube 1.0 m long in kg mol H2/s at
steady state.
Given :
r₁ = 3.0/2 x 1000 = 0.0015 m We will take :
r₂ = 11.0/2 x 1000 = 0.0055 m L = 1m
Pₐ₁ = 2.0 atm T = 300 K
Pₐ₂ = 0 Dₐᵦ = 0.180 x 10 ⁻⁹ m²/s
S = 0.053
Now,
Cₐ₁ = S/22.414 (Pₐ₁)
Nₐ = Dₐᵦ ( Cₐ₁ - Cₐ₂ ) ( 2 πL )/In ( r₂/r₁ )
Nₐ = Dₐᵦ ( Pₐ₁ - Pₐ₂ ) 2 πL/22.414 In ( r₂/r₁ )
Nₐ = 0.180 x 10 ⁻⁹ ( 0.053 ) ( 2.0 ) (2 π)/22.414 In ( 0.0055/0.0015 )
Nₐ = 4.12 x 10⁻¹² kg mol H₂/S
At STP, one mole of any gas occupies a volume of 22.414

PROBLEM 5
Predict the diffusivity of the enzyme urease in a dilute solution in water at 298 K using the
modified Polson equation and compare the result with the experimental value in Table 6.4-1.
Solution:
Mₐ = 482,700
T = 298 K
DₐB ( experimental value ) = 4.01 × 10 ^−11 m²/s
µ = 0. 8397 x 10 ^-3
DₐB = 9.40 x 10^-15/(µ Mₐ)^⅓
Dₐᵦ = 9.40 x 10^-15/0. 8397 x 10 ^-3(482,700)^⅓
Dₐᵦ = 3.995 × 10^−11 m²/s

PROBLEM 6

The efficacy of pharmaceutical products is reduced by prolonged exposure to high temperature,

light, and humidity. For water vapor–sensitive consumer products that are in tablet or capsule
form, and might be stored in humid environments such as bathroom medicine cabinets, blister
packaging is used to limit the direct exposure of the medicine to humid conditions until
immediately before its ingestion. Consider tablets that are contained in a blister package
composed of a flat lidding sheet and a second, formed sheet that includes troughs to hold each
tablet. The formed sheet is L = 50 µm thick and is fabricated of a polymer material. Each trough
is of diameter D = 5 mm and depth h = 3 mm. The lidding sheet is fabricated of aluminum foil.
The binary diffusion coefficient for water vapor in the polymer is 𝐷𝐴𝐵 = 6 × 10−14 m2 /s while
the aluminum may be assumed to be impermeable to water vapor. For molar concentrations of
water vapor in the polymer at the outer and inner surfaces of 𝐶𝐴 ,s1 = 4.5 × 10−3 kmol/m3 and
𝐶𝐴 ,s2 = 0.5 × 10−3 kmol/m3 , respectively, determine the rate at which water vapor is transferred
through the trough wall to the tablet.

Solution:
Known: Molar concentrations of water vapor at the inner and outer surfaces of a polymer sheet
and trough geometry.
Find: Rate of water vapor molar diffusive transfer through the trough wall.
Schematic:

Assumptions:
1. Steady-state, one-dimensional conditions.
2. Stationary medium.
3. No chemical reactions.
4. Polymer sheet is thin relative to the dimensions of the trough, and diffusion may be analyzed
as though it occurs through a plane wall.
Analysis: The total water vapor transfer rate is the summation of the transfer rate through the
cylindrical walls of the trough and the bottom, circular surface of the trough. From Equation
14.54 we may write
 𝐷𝐴,𝐵 𝐴 𝐷𝐴𝐵 𝜋𝐷 2
𝑁𝐴,𝑋 = (𝐶𝐴,𝑆1 − 𝐶𝐴,𝑆2 ) = ( + 𝜋𝐷𝑕) (𝐶𝐴,𝑆1 − 𝐶𝐴,𝑆2 )
𝐿 𝐿 4

Hence

6✖10−14 𝑚2 /𝑠 𝜋(5✖10−3 𝑚2
𝑁𝐴,𝑋 = * + 𝜋(5✕10;3 𝑚)(3✕10;3 𝑚)+ ✕(4.5✕10;3 )𝑘𝑚𝑜𝑙/𝑚3 =
50✖10−6 𝑚 4
;15
0.32✕10 𝑘𝑚𝑜𝑙/𝑠

PROBLEM 7
Hydrogen gas is maintained at 3 bar and 1 bar on opposite sides of a plastic membrane, which is
0.3 mm thick. The temperature is 25°C, and the binary diffusion coefficient of hydrogen in the
plastic is 8.7 × 10−8 m2 /s. The solubility of hydrogen in the membrane is 1.5 × 10−3 kmol/m3 ⋅
bar. What is the diffusive mass flux of hydrogen through the membrane?
Find: The hydrogen diffusive mass flux n″ A,x (kg/s⋅m2 ).
Schematic:

Assumptions:
1. Steady-state, one-dimensional conditions exist.
2. Membrane is a stationary, nonreacting medium of uniform total molar concentration.
𝑋𝐴,𝑆1 − 𝑋𝐴,𝑆2 𝐷𝐴𝐵
𝑁"𝐴,𝑋 = 𝐶𝐷𝐴𝐵 = (𝐶𝐴,𝑆1 − 𝐶𝐴,𝑆2 )
𝐿 𝐿
The surface molar concentrations of hydrogen may be obtained from Equation 14.62, where
𝐶𝐴,𝑆1 = 1.5✕10;3 𝑘𝑚𝑜𝑙/𝑚3 ⋅ 𝑏𝑎𝑟✕3 𝑏𝑎𝑟 = 4.5✕10;3 𝑘𝑚𝑜𝑙/𝑚3 𝐶𝐴,𝑆2
= 1.5✕10;3 𝑘𝑚𝑜𝑙/𝑚3 ⋅ 𝑏𝑎𝑟✕3 𝑏𝑎𝑟 = 1.5✕10;3 𝑘𝑚𝑜𝑙/𝑚3
Hence,

8.7✕10;8 𝑚2 /𝑠
𝑁"𝐴,𝑋 = (4.5✕10;3 − 1.5✕10;3 )𝑘𝑚𝑜𝑙/𝑚3
0.3✕10;3 𝑚
 𝑁"𝐴,𝑋 = 8.7✕10;7 𝑘𝑚𝑜𝑙/𝑠 ⋅ 𝑚2
On a mass basis,
𝑛"𝐴,𝑋 = 𝑁"𝐴,𝑋 𝜇𝐴
where the molecular weight of hydrogen is 2 kg/kmol. Hence
𝑛"𝐴,𝑋 = 8.7✕10;7 𝑘𝑚𝑜𝑙/𝑠 ⋅ 𝑚2 ✕𝑘𝑔/𝑘𝑚𝑜𝑙 = 1.74✕10;6 𝑘𝑔/𝑠 𝑚2

PROBLEM 8
Biofilms, which are colonies of bacteria that can cling to living or inert surfaces, can cause a
wide array of human infections. Infections caused by bacteria living within biofilms are often
chronic because antibiotics that are applied to the surface of a biofilm have difficulty penetrating
through the film thickness. Consider a biofilm that is associated with a skin infection. An
antibiotic (species A) is applied to the top layer of a biofilm (species B) so that a fixed
concentration of medication, CA,0 = 4 × 10−3 kmol/m3 , exists at the upper surface of the
biofilm. The diffusion coefficient of the medication within the biofilm is DAB = 2 × 10−12 m2
/s. The antibiotic is consumed by biochemical reactions within the film, and the consumption rate
depends on the local concentration of medication expressed as N k A 1CA = − where k1 = 0.1
s−1 . To eradicate the bacteria, the antibiotic must be consumed at a rate of at least 0.2 × 10−3
kmol/s⋅m3 ≤ − × ⋅ − (NA 0.2 10 kmol/s m ) 3 3 since, at smaller absolute consumption rates, the
bacteria will be able to grow back faster than it is destroyed. Determine the maximum thickness
of a biofilm, L, that may be treated successfully by the antibiotic.

Known: Topical antibiotic and biofilm properties, surface concentration of the medication, and
required minimum consumption rate of antibiotic.
Find: Maximum thickness of a bacteria-laden biofilm, L, that may be successfully treated.

Schematic:
Assumptions:
1. Steady-state, one-dimensional conditions.
2. Stationary, homogeneous medium with constant properties.
3. Impermeable bottom of the biofilm.
Analysis: The absolute antibiotic consumption rate will be smallest at x = L, where the antibiotic
concentration is smallest. Thus, we require Ń𝐴 (𝐿) = −0.2✕10;3 𝑘𝑚𝑜𝑙/𝑠. The expression for
the first-order reaction may be combined with Equation 14.74 to write

𝐶
𝐴,𝑂
Ń𝐴 (𝐿) = −𝑘1 𝐶𝐴 (𝐿) = −𝑘1 𝑐𝑜𝑠𝑕 (1)
𝑚𝐿

Where

𝑘1 1/2
0.1𝑠 ;1 1/2
𝑚=( ) =( ) = 2.24✕105 𝑚1
𝐷𝐴𝐵 2✕10;12 𝑚2 /𝑠

Equation 1 may be solved for the maximum allowable thickness:

;𝑘1 𝐶𝐴,𝑂
𝐿 = 𝑚;1 𝑐𝑜𝑠𝑕;1 ( ) (2)
Ń𝐴 (𝐿)

Substituting values into Equation 2

;0.1𝑠−1 ✕4✕10−3 𝑘𝑚𝑜𝑙/𝑚3
yields𝐿 = (2.24✕105 𝑚;1 )𝑐𝑜𝑠𝑕;1 ( ) = 5.9✕10;6 𝑚 = 5.9𝜇𝑚
;0.2✕10−3 𝑘𝑚𝑜𝑙/𝑠．𝑚3

PROBLEM 9
The gas hydrogen at 17°C and 0.010 atm partial pressure is diffusing through a membrane of
vulcanized neoprene rubber 0.5 mm thick. The pressure of H₂ on the other side of the neoprene is
zero. Calculate the steady-state flux, assuming that the only resistance to diffusion is in the
membrane. The solubility S of H₂ gas in neoprene at 17°C is 0.051 m³ (at STP of 0°C and 1
atm)/m³ solid atm and the diffusivity DAB is 1.03 × 10-10 m²/s at 17°C.
Solution: A sketch showing the concentration is shown in Fig. 6.5-1. The equilibrium
concentration cA₁ atthe inside surface of the rubber is,
GIVEN:
TH2O= 170C
P1= 0.010 atm
NeoRUBBER= 0.5 thick
Neosolubility= 0.051 m3
DAB=1.03 x 10-10 m2/s
At STP of 00C and 1 atm /m3

𝑆 0.051(0.010)
CA1 = 22.414 PA1 = = 2.28 X 10-5 Kg mol H2/m3 solid
22.414

Since PA2 at the other side is 0, CA2 = 0.

𝐷𝐴𝐵 (𝐶𝐴1 ;𝐶𝐴2 ) (1.03 𝑋 10−10 )(2.28 𝑋 10−5 ; 0

NA = = = 4.69 X 10– 12 Kg mol H2/s m2
𝑍2 ;𝑍1 (0.5 ; 0)/1000

PROBLEM 10
A polyethylene film 0.00015 m (0.15 mm) thick is being considered for use in packaging a
pharmaceutical product at 300C. If the partial pressure of O2 outside is 0.21 atm and inside the
package it is 0.01 atm, calculate the diffusion flux of O2 at steady state. Use permeability data
from figure (6.5-1). Assume that the resistances to diffusion outside the film and inside are
negligible compared to the resistance of the film.

GIVEN:

P1 = 0.21 atm Molar volume of Gas = 22.414 m3 solid

P2 = 0.01 atm
Tproduct = 300C
(C2H4)n = 0.00015 m => 0.15 mm

SOLUTION:
From Table 6.5-1 PM = 4.17(10-12) m3 solute (STP)/(s m2 atm/m).

𝑷 (𝑷 ;𝑷𝑨𝟐 ) 𝟒.𝟏𝟕(𝟏𝟎−𝟏𝟐 )(𝟎.𝟐𝟏;𝟎.𝟎𝟏)

𝑴 𝑨𝟏
NA = 𝟐𝟐.𝟒𝟏𝟒(𝒁 = = 2.480 x 1010 kg mol/s m2
𝟐 ;𝒁𝟏 ) 𝟐𝟐.𝟒𝟏𝟒(𝟎.𝟎𝟎𝟎𝟏𝟓;𝟎)

Note that a film made of nylon has a much smaller value of permeability PM for O2 and would
make a more suitable barrier.

PROBLEM 11
A sintered solid of silica 2.0 mm thick is porous with a void fraction ε of 0.30 and a tortuosity r
of 4.0. The pores are filled with water at 298 K. At one face the concentration of KCI is held at
0.10 g mol/liter, and fresh water flows rapidly by the other face. Neglecting any other resistances
but that in the porous solid, calculate the diffusion of KCI at steady state.

GIVEN:

SiO2 (s) = 2.0 mm thick

KCI = 0.10 mol/liter
ε = 0.30
t= 4.0
T = 293k
SOLUTION:
The diffusivity of KCI in water from table 6.3-1 is DAB = 1.87 x 10-9. m2/s. Also,

CA1 = 0.10/1000
= 1.0 x 10-4 g mol /cm3 => 0.01 kg mol/m3

CA2 = 0. Substituting into Eq. (6.5-13).

 𝜺 𝐷𝐴𝐵 (𝑪𝑨𝟏 ;𝑪𝑨𝟐 ) (0.30)(1.87 𝑥 10−9 )(𝟎.𝟎𝟏;𝟎)
NA = = = 7.01 x 10-9 kg mol KCI/s m2
𝒕(𝒁𝟐 ;𝒁𝟏 ) 𝟒.𝟎(𝟎.𝟎𝟎𝟐;𝟎)

PROBLEM 12
The pressure in a pipeline that transports helium gas at a rate of 2 kg/s is maintained at 1 atm by
venting helium to the atmosphere through a 5-mm- internal-diameter tube that extends 15 m into
the air. Assuming both the helium and the atmospheric air to be at 25°C, determine (a) the mass
flow rate of helium lost to the atmosphere through the tube, (b) the mass flow rate of helium lost
to the atmosphere through the tube

SOLUTION The pressure in a helium pipeline is maintained constant by venting to the

atmosphere through a long tube. The mass flow rates of helium and air through the tube and the
net flow velocity at the bottom are to be determined.
Assumptions
1 Steady operating conditions exist.
2 Helium and atmospheric air are ideal gases.
3 No chemical reactions occur in the tube.
4 Air concentration in the pipeline and helium concentration in the atmosphere are negligible so
that the mole fraction of the helium is 1 in the pipeline and O in the atmosphere (we will check
this assumption later).
Analysis
Typical equimolar counter diffusion process since the problem involves two large reservoirs of
ideal gas mixtures connected to each other by a channel, and the concentrations of species in
each reservoir (the pipeline and the atmosphere) remain constant.
(a) The flow area, which is the cross-sectional area o f the tube, is
𝜋𝐷2 (0.005𝑚)2
𝐴= =𝜋 = 1.963 𝑥 10;5 𝑚2
4 4
Noting that the pressure of helium is 1 atm at the bottom of the tube (x = 0) and O at the top (x =
L), its molar flow rate is determined to be
𝐷𝐴𝐵 𝐴 𝑃𝐴,0 − 𝑃𝐴,𝐿
𝑁𝑕𝑒𝑙𝑖𝑢𝑚 = 𝑁𝑑𝑖𝑓𝑓,𝐴 = ∙
𝑅𝑇 𝐿
 10;5 𝑚2 ;5
(7.20 𝑥
= 𝑠 )(1.963 𝑥 10 ) (1 𝑎𝑡𝑚 − 0)(101.3 𝑘𝑃𝑎 )
𝑘𝑃𝑎 ∙ 𝑚3 15 1 𝑎𝑡𝑚
(8.314 ∙ 𝐾)(298 𝐾)
𝑘𝑚𝑜𝑙

= 3. 85 𝑥 10;12
Therefore,
𝑀𝐻𝑒𝑙𝑖𝑢𝑚 = (𝑁𝑀)𝑕𝑒𝑙𝑖𝑢𝑚 = (3.85 × 10;12 𝑘𝑚𝑜𝑙/𝑠)(4 𝑘𝑔/𝑚𝑜𝑙) = 1.54 𝑥 10;11 Ans

which corresponds to about 0.5 g per year.

(b) Noting that NB = -NA, during an equimolar counter diffusion process, the molar flow rate of
air into the helium pipeline is equal to the molar flow rate of helium. The mass flow rate of air
into the pipeline is

10;12 𝑘𝑚𝑜𝑙 29𝑘𝑔

𝑚𝑎𝑖𝑟 = (𝑁𝑀)𝑎𝑖𝑟 = (−3.85 × )( ) = −122 × 10;12 𝑘𝑔/𝑠
𝑠 𝑚𝑜𝑙

The mass fraction of air in the helium pipeline is

𝑚𝑎𝑖𝑟 112 × 10;12 𝑘𝑔/𝑠

𝑊𝑎𝑖𝑟 = = = 5.6 × 10;11 ≈ 0
𝑚𝑡𝑜𝑡𝑎𝑙 (2 + 112𝑥 10;12 − 1.54 × 10;11 )𝑘𝑔/𝑠
which validates our original assumption of negligible air in the pipeline.

(c) The net mass flow rate through the tube is

𝑚𝑛𝑒𝑡 = 𝑚𝑕𝑒𝑙𝑖𝑢𝑚 + 𝑚𝑎𝑖𝑟 = 1.54 × 10;11 − 112 × 10;12 = −9.66 × 10;11 𝑘𝑔/𝑠

The mass fraction of air at the bottom of the tube is very small, as shown above, and thus the
density of the mixture at x = 0 can simply be taken to be the density of helium, which is
𝑝 101.325 𝑘𝑃𝑎
𝑝 ≅ 𝑝𝑕𝑒𝑙𝑖𝑢𝑚 ≅ = = 0.1637𝑘𝑔/𝑚3
𝑅𝑇 𝑚3
(2.0769 𝑘𝑃𝑎 ∙ ∙ 𝐾) (298 𝐾)
𝑘𝑔

Then the average flow velocity at the bottom part of the tube becomes

𝑚 ;9.66 ×10 −11 𝑘𝑔/𝑠

𝑉 = 𝑝𝐴 = (0.1637 𝑘𝑔/𝑚3 )(1.963×10−5 𝑚2 ) = −3.01 × 10;5 𝑚/𝑠
= −3.01 × 10;5 𝑚/𝑠

PROBLEM 13
Porous alumina spheres, 10 mm diameter, 25% voids, were thoroughly impregnated with an
aqueous potassium chloride, KCl, solution, concentration 0.25 g/cm3. When immersed in pure
running water, the spheres lost 90% of their salt content in 4.75 h. The temperature was 25°C. At
this temperature the average diffusivity of KCl in water over the indicated concentration range is
1.84 x 10-9 m2/s. Estimate the time for removal of 90% of the dissolved solute if the spheres had
been impregnated with potassium chromate, K2CrO4, solution at a concentration 0.28 g/cm3,
when immersed in a running stream of water containing 0.02 g K2CrO4/cm3. The average
diffusivity of K2CrO4, in water at 25°C is 1.14 x 10-9 m2/s.

SOLUTION For these spheres, a = 0.005 m, and for the KCI diffusion, 𝜃 4.75(3600) = 17 000
s. When the spheres are surrounded by pure water, the ultimate concentration in the spheres CA,∞
= 0.
𝐶𝐴,𝜃 𝐶,𝐴,∞
= 0.1 𝑓𝑜𝑟 90% 𝑟𝑒𝑚𝑜𝑣𝑎𝑙 𝑜𝑓 𝐾𝐶𝐼
𝐶𝐴0 𝐶𝐴,∞

Where Deff is the effective diffusivity.

0.18 𝑎2 0. 18(0.005)2
𝐷𝑒𝑓𝑓 = = = 2.65 × 1010 𝑚2
𝜃 17000

𝐷𝐴𝐵 1.84 × 10;9

⁄𝐷 = = 6.943
𝑒𝑙𝑓 2.65 × 10;10

For the K2 CrO4 diffusion, CAO = 0.28, CA,∞ 0.002, and CA,𝜃 = 0.1(0.028) = 0.028𝑔/𝑐𝑚3

𝐶𝐴,𝜃 𝐶,𝐴,∞ 0.028 − 0.02

𝐸= = = 0.0308 = 𝐸
𝐶𝐴0 𝐶𝐴,∞ 0.28 − 0.02

𝜃
𝐷𝑒𝑓𝑓 = 0.30
𝑎2

𝐷𝐴𝐵 (1.14 × 10);9

𝐷𝑒𝑓𝑓 = = = 1.642 × 10;10 𝑚2/𝑆
6.943 6.943

0.302 0.30(0.005)2
𝜃= = = 45700𝑠 = 12.7 𝑕
𝐷𝑒𝑓𝑓 1.64 × 10;10
=12.7 h
PROBLEM 14
The surface of a mild steel component is commonly hardened by packing the component in a
carbonaceous material in a furnace at a high temperature for a predetermined time. Consider
such a component with a uniform initial carbon concentration of 0.15 percent by mass. The
component is now packed in a carbonaceous material and is placed in a high-temperature
furnace. The diffusion coefficient of carbon in steel at the furnace temperature is given to be 4.8
x 10-10 m/s, and the equilibrium concentration of carbon in the iron at the interface is
determined from equilibrium data to be 1.2 percent by mass. Determine how long the component
should be kept in the furnace for the mass concentration of carbon 0.5 mm below the surface to
reach 1 percent

SOLUTION A steel component is to be surface hardened by packing it in a carbonaceous

material in a furnace. The length of time the component should be kept in the furnace is to be
determined.

Assumptions Carbon penetrates into a very thin layer beneath the surface of the component, and
thus the component can be modeled as a semi-infinite medium regardless of its thickness or
shape

𝑊𝐴 (𝑥, 𝑡) − 𝑊𝐴,𝑖 𝑥
= 𝑒𝑟𝑓𝑐 ( )
𝑊𝐴,𝑠 − 𝑊𝐴,𝐼 2√𝐷𝐴𝐵 𝑡
Substituting the specified quantities gives
0.01 − 0.0015 𝑥
= 0.81 = 𝑒𝑟𝑓𝑐 ( )
0.012 − 0.0015 2√𝐷𝐴𝐷 𝑡

The argument whose complementary error function is 0.81 is determined from Table 4-4 to be
0.17. That is,

𝑥
= 0.17
2√𝐷𝐴𝐵 𝑡
Then solving for the time t gives
𝑥2 (0.0005𝑚)2
𝑡= = = 4505 𝑠 = 1 𝑕 15 𝑚𝑖𝑛
4𝐷𝐴 𝐵(0.17)2 4 × (4.8 × 10;10 𝑚2 /𝑠)(0.17)2

= 1 h 15 mins
PROBLEM 15

Consider a nickel plate that is in contact with hydrogen gas at 358 K and 300 kPa. Determine the
molar and mass density of hydrogen in the nickel at the interface
(Fig. 14–17).
SOLUTION:
A nickel plate is exposed to hydrogen. The molar and mass density of hydrogen in the nickel at
the interface are to be determined.
Assumptions:
Nickel and hydrogen are in thermodynamic equilibrium at the interface.
Properties :
The molar mass of hydrogen is M = 2 kg/kmol (Table A–1). The solubility of hydrogen in nickel
at 358 K is 0.00901 kmol/𝑚3 ⋅bar (Table 14–7).
Analysis
Noting that 300 kPa = 3 bar, the molar density of hydrogen in the nickel at the interface is
determined from Eq. 14–20 to be:

𝐶𝑖 ,𝑠𝑜𝑙𝑖𝑑𝑠𝑡𝑎𝑡𝑒 = 𝑥 𝑃𝑖 ,𝑔𝑎𝑠 𝑠𝑖𝑑𝑒 (𝑘𝑚𝑜𝑙/𝑚3 ) − −→ 𝐸𝑞. 14 − 20.

Where:
is the solubility. Expressing the pressure in bars and noting that the unit of molar concentration is
kmol of species i per 𝑚3 the unit of solubility is kmol/𝑚3 bar. Solubility data for selected gas-
solid combinations are given in Table 14-7. The product of the solubility of a gas and the
diffusion coefficient of the gas in solid is referred to as the permeability , which is measure of
ability of the gas to penetrate a solid. That is ,
= 𝐷𝐴𝐵 where 𝐷𝐴𝐵 is the diffusivity of the gas in the solid. Permeability is inversely
proportional or thickness and has the unit kmol/s bar.

Given from the table ;

= 0.00901 kmol/𝑚3 𝑏𝑎𝑟
Density Hydrogen= 2kg/kmol
𝑃𝐻 2 = 300 kPa = 3 bar, the molar density of hydrogen in the nickel at the interface is determined

Using Eq. 14-20

𝐶𝐻2 ,𝑠𝑜𝑙𝑖𝑑𝑠𝑡𝑎𝑡𝑒 = 𝑥 𝑃𝐻2 ,𝑔𝑎𝑠 𝑠𝑖𝑑𝑒

𝑘𝑚𝑜𝑙
= (0.00901 ) 𝑥 3𝑏𝑎𝑟 = 0.027 𝑘𝑚𝑜𝑙/𝑚3
𝑚3 𝑏𝑎𝑟

It corresponds to mass density of:

𝑃𝐻2 ,𝑠𝑜𝑙𝑖𝑑𝑠𝑡𝑎𝑡𝑒 = 𝐶𝐻 2 ,𝑠𝑜𝑙𝑖𝑑 𝑠𝑖𝑑𝑒 𝑀𝐻 2

= (0.026𝑘𝑚𝑜𝑙/𝑚3 )(2𝑘𝑔/𝑚𝑜𝑙) = 0.0054 𝑘𝑔/𝑚3

Hence, there will be 0.027kmol or 0.054 kg of 𝐻2 gas in each 𝑚3 volume of nickel adjacent to
the interface.

PROBLEM 16
Pressurized hydrogen gas is stored at 358 K in a 4.8-m-outer-diameter spherical container made
of nickel (Fig. 14-23). The shell of the container is 6 cm thick. The molar concentration of
hydrogen in the nickel at the inner surface is determined to be
0.087 kmol/𝑚3 . The concentration of hydrogen in the nickel at the
outer surface is negligible. Determine the mass flow rate of hydrogen by diffusion through the
nickel container.

SOLUTION Pressurized hydrogen gas is stored in a spherical container. The diffusion rate of
hydrogen through the container is to be determined.

Assumptions : 1 Mass diffusion is steady and one-dimensional since the hydrogen concentration
in the tank and thus at the inner surface of the container is practically constant, and the hydrogen
concentration in the atmosphere and thus at the outer surface is practically zero. Also, there is
thermal symmetry about the center. 2 There are no chemical reactions in the nickel shell that
result in the generation or depletion of hydrogen.

Properties The binary diffusion coefficient for hydrogen in the nickel at the specified
temperature is 1.2𝑥10;12 𝑚2 /𝑠 (Table 14-36).

Analysis : We can consider the total molar concentration to be constant (𝐶 = 𝐶𝐴 + 𝐶𝐵 ≅ 𝐶𝐵 =

.
𝑐𝑖𝑛𝑠𝑡𝑎𝑛𝑡) and the container to be a stationary medium since there is no diffusion of nickel
molecules (𝑁𝐵 = 0 ) and the concentration of the hydrogen in the container is extremely low (𝐶𝐴
<< 1). Then the molar flow rate of hydrogen through this spherical shell by diffusion can readily

be determined from Eq. 14-28 to be

𝐶𝐴,1 − 𝐶𝐴 ,2
𝑁𝑑𝑖𝑓𝑓 = 4𝜋𝑟1 𝑟2 𝐷𝐴𝐵
𝑟2 − 𝑟1
(0.087 − 0)𝑘𝑚𝑜𝑙/𝑚3
= 4𝜋(2.34𝑚)(2.40𝑚)(1.2𝑥10;12 𝑚2 /𝑠 )
(2.40 − 2.34)𝑚

= 1.228𝑥10;10 𝑘𝑚𝑜𝑙/𝑠
The mass flow rate is determined by multiplying the molar flow rate by the molar mass of
hydrogen, which is M = 2 kg/kmol,
. .

𝑚𝑑𝑖𝑓𝑓 = 𝑀𝑁𝑑𝑖𝑓𝑓 = (2𝑘𝑔/𝑘𝑚𝑜𝑙)(1.228𝑥10;10 𝑘𝑚𝑜𝑙/𝑠)

= 2.46𝑥10;10 𝑘𝑔/𝑠

Therefore, hydrogen will leak out through the shell of the container by diffusion at a rate of 2.46
× 10;10 kg/s or 7.8 g/year. Note that the concentration of hydrogen in the nickel at the inner
surface depends on the temperature and pressure of the hydrogen in the tank and can be
determined as explained in Example 14-3. Also, the assumption of zero hydrogen concentration
in nickel at the outer surface is reasonable since there is only a trace amount of hydrogen in the
atmosphere (0.5 part per million by mole numbers).

PROBLEM 17

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. The diagram is similar to Fig. 6.2-1, 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.
(a) Calculate the flux 𝐽𝐴 at steady state.
(b) Repeat for 𝐽𝐵

Equation (6.1-13) can be used, where 𝑃 = 1.0132 𝑥 105 𝑃𝑎, 𝑧2 − 𝑧2 = 0.10 𝑚, 𝑎𝑛𝑑 𝑇 = 298 𝐾.
Substituting into Eq. (6.1-13) for part (a).
𝐷𝐴𝐵 (𝑃𝐴1− 𝑃𝐴2 ) (0.23 𝑥 10−4 )(1.013 𝑥 104 ;0.507 𝑥 104 )
𝐽𝐴 = =
𝑅𝑇(𝑧2 ; 𝑧1 ) 8314(298)(0.10;0)

= 4.70 𝑥 10;7 𝑘𝑔. 𝑚𝑜𝑙. 𝐴/𝑠. 𝑚2

Rewriting Eq. (6.1-13) for component B for part (b) and noting that 𝑃𝐵1 = 𝑃 − 𝑃𝐴1 =
1.0132 𝑥 105 − 1.013 𝑥 104 = 9.119 𝑥 104 𝑃𝑎 𝑎𝑛𝑑 𝑃𝐵2 = 𝑃 − 𝑃𝐴2 = 1.0132𝑥 105 −
0.507 𝑥 104 = 9.625 𝑥 104 𝑃𝑎.
 𝐷𝐴𝐵 (𝑃𝐵1 − 𝑃𝐵2 ) (0.23 𝑥 10;4 )(9.119 𝑥 104 − 9.625 𝑥 104 )
𝐽𝐵 = =
𝑅𝑇(𝑧2 − 𝑧1 ) 8314(298)(0.10 − 0)
= 4.70 𝑥 10;7 𝑘𝑔. 𝑚𝑜𝑙 𝐵/𝑠. 𝑚2
The negative value for 𝐽𝐵 means the flux goes from point 2 to point 1.

PROBLEM 18
Predict the diffusion coefficient of acetone 𝐶𝐻3 𝐶𝑂𝐶𝐻3 ) in water at 25℃ and 50℃ using Wilke-
𝑚
Chang equation. The experimental value is 1.28 𝑥 10;9 𝑠 𝑎𝑡 25℃ (298 𝐾).

Solution: From Appendix A.2 the viscosity of water at 25.0℃ is

;3 ;3
𝜇𝐵 = 0.8937 𝑥 10 𝑃𝑎. 𝑠 𝑎𝑛𝑑 𝑎𝑡 50℃, 0.5494 𝑥 10 . From table 6.3-2 for 𝐶𝐻3 𝐶𝑂𝐶𝐻3 with 3
carbons + 6 hydrogens +1 oxygen.
𝑉𝐴 = 3(0.0148) + 6(0.0037) + 1(0.0074) = 0.0740𝑚3 𝑘𝑔. 𝑚𝑜𝑙
For water the association parameter ∅ = 2.6 𝑎𝑛𝑑 𝑀𝐵 = 18.02 𝑘𝑔. 𝑚𝑎𝑠𝑠 /𝑘𝑔𝑚𝑜𝑙

For 25℃, 𝑇 = 298𝐾. Substituting into Eq. (6.3-9)

𝑇
𝐷𝐴𝐵 = (1.173 𝑥 10;16 )(𝜙𝑀𝐵 )1/2
𝜇𝐵 𝑉𝐴0.6
1
(1.173 𝑥 10;16 )(2.6 𝑥 18.02)2 (298)
=
(0.8937 𝑥 10;3 )(0.0740)0.6
= 1.277 𝑥 10;9 𝑚2 /𝑠
Principles of Transport Processes and Separation Processes 4th Edition by Christie John
Geankoplis

PROBLEM 19
Nitrogen is diffusing under steady condition through a mixture of 2%𝑁2 , 20% 𝐶2 𝐻6 , 30% 𝐶2 𝐻4
and 48% 𝐶4 𝐻10 at 298 K and 100 kPa. The partial pressure of nitrogen at two planes (1 mm)
apart are 13.3 and 6.67 kPa, respectively. Calculate the rate of 𝑁2 across the two planes. The
diffusivity of 𝑁2 through 𝐶4 𝐻10 , 𝐶2 𝐻6 and 𝐶2 𝐻4 may be taken as
9.6 𝑥 10;6 𝑚2 /𝑠, 14.8 𝑥 10;6 𝑚2 /𝑠 𝑎𝑛𝑑 16.3 𝑥 10;6 𝑚2 /𝑠 , respectively.
Solution:
Since Stagnant diffusion:
 𝐷𝐴𝑚 𝑃𝑇 𝑃𝑇 − 𝑃𝐴2
𝑁𝐴 = ln * +
𝑅𝑇 Δ𝑧 𝑃𝑇 − 𝑃𝐴1
1 − 𝑦𝐴 𝑦𝐵 𝑦𝐶 𝑦𝐷
= + +
𝐷𝐴𝑚 𝐷𝐴𝐵 𝐷𝐴𝐶 𝐷𝐴𝐷
1 − 0.02 0.48 0.2 0.3
= ;6
+ ;6
+
𝐷𝐴𝑚 9.6 𝑥 10 14.8 𝑥 10 16.3 𝑥 10;6
𝐷𝐴𝑚 = 1.22 𝑥 10;5 𝑚2 /𝑠
1.22 𝑥 10;5 100 100 − 6.67
𝑁𝐴 = ln [ ] = 0.0492 𝑘𝑚𝑜𝑙/𝑚2 . 𝑠
8.314 𝑥 298 0.001 100 − 13.3

PROBLEM 20

Oxygen (A) is diffusing through carbon monoxide (B) under steady-state conditions. with the
carbon monoxide nondiffusing. The total pressure is 1 x 10^5 N/m², and the temperature 0° C.
The partial pressure of oxygen at two planes 2.0 mm apart is. respectively, 13 000 and 6500
N/m². The diffusivity for the mixture is 1.87 x 10-5 m2/s. Calculate the rate of diffusion of
oxygen in kmol/s through each square meter of the two planes.
Solution Equation (2.30) applies. DAB = 1.87 x 10-5 m²/s, P1= 10^5 N/m² z = 0.002 m, R =
8314 N • m/kmol • K, T= 273 K; PA1=13 x 10³, Pb1 = 10^5 - 13 x 10³ - = 87 x 10³, PA2 = 6500,
PB2 = 10^5 – 6500 = 93.5 x 10³, All in N/m².
PB, m = PB1 – PB2 = (87 – 93.5 )(10³) = 90 200 N/m²
ln(PB1/PB2) ln (87/93.5)
NA = DABt__(PA1 – PA2) = (1.87 x 10-^5)(10^5)(13 – 6.5)(10³)
RT2pb,m 8314(273)(0.002)(90.2 x 10³)

= 2.97 x 10^-5 kmol / m² • s

PROBLEM 21
Recalculate the rate of diffusion of oxygen (A) in illustration 2.1, Assuming that non diffusing
gas is a mixture of methane (B) and hydrogen (C) in the volume ratio 2: 1, The diffusivities are
estimated to be DO2 – H2 = 6.99 x 10-5, DO2- CH4 = 1.86 x 10^ -5 m²/s.
SOLUTION Equation (2.25) will become Eq. (2.30) for this case, P1 = 10^5 N/m², T=273 K,
PA1 = 13 x 10³, PA2 = 6500, P1M = 90.2 x 10³, All in N/m²; z = 0.002 m, R = 8314 N•m/kmol •
k, as in illustration 2.1. In eq ( 2.36) Y’B = 2/(2+1) = 0.667, Y’C = 1-0.667 = 0.333

DA,m = ___ 1_________ =____________1_______________________

Y’B/DAB + Y’C /DA,C 0.667/(1.86 x 10^-5 ) + 0.333/ (6.99 x 10^-5)

= 2.46 x 10^-5 m²/s

Therefore eq.(2.30) becomes,
NA = (2.46 x 10^-5)(13 000 – 6500) = 3.91 x 10^-5 kmol / m² • s
8314(273)(0.002)(90 200)

PROBLEM 22
Calculate the rate of diffusion of acetic acid (A) across a film of nondiffusing water (B) solution
1mm thick at 17° C when the concentrations on opposites sides of the film are, respectively, 9
and 3 wt % acid. The diffusivity of acetic acid in the solution is 0.95 x 10^-9 m²/s.

The component subscript on xa indicates mole fraction A, to distinguish it from x meaning

distance in the x direction.

Density of the 9% solution is 1012 kg/m³. Therefore,

Xa1 = ____0.09/60.03________ = 0.0015 = 0.0288 mole fraction acetic acid
0.09/60.03 + 0.91/ 18.02 0.0520

Xb1 = 1 – 0.0288 = 0. 9712 mole fraction water

M = _____1_____ = 19.21 kg/kmol _p_ = 1012_ = 52.7 kmol/m³
0.0520 m 19.21
Similarly the density of the 3% solution is 1003.2 kg/m³, xa2 = 0.0092, xb2 = 0.9908, M = 18.40,
and p/M 54.5.
(P) = 52.7 + 54.5 = 53.6 kmol/m³ Xbm = 0.9908 – 0.9712__ = 0.980
Mav 2 ln(0.9908/0.9712)
:
Na = 0.95 x 10^-9__ 53.6 (0.0288 – 0.0092) = 1.018 x 10^-6 kmol/m² • s
0.001 (0.980)

PROBLEM 23

The solute HCI (A) is diffusing through a thin film of water (B) 2.0 mm thick at 283 K. The
concentration of HCl at point 1 at one boundary of the film is 12.0 wt % HCI (𝑑𝑒𝑛𝑠𝑖𝑡𝑦 1 =
1060.7 𝑘𝑔/𝑚2 ). and at the other boundary at point 2 it is 6.0 wt % HCI ρ2 = 1030.3 𝑘𝑔/𝑚2 ).
The diffusion coefficient of HCl in water is 2.5 × 10-9 ms. Assuming steady state and one
boundary impermeable to water, calculate the flux of HCl in kg mol/s • m2.

𝐷𝐴𝐵 = 2.5 𝑥10;9 𝑚2 /𝑠

12.0/36.47 0.329
𝑋𝐴1 = = = 0.0632
12.0 88.0 5.21
+
36.47 8.02

𝑋𝐵1 = 1 − 0.0632 = 0.9368

100
𝑀1 = 5.21 = 19.18 kg/kg mol solution

6.0/36.47 0.1645
𝑋𝐴2 = 6.0 84.0 = =0.03055
: 5.385
36.47 8.02

𝑋𝐵1 = 1.0 − 0.0355 = 0.9694

100
𝑀2 = = 18.60
5.385

𝑋𝐵1 − 𝑋𝐵1 0.9694 − 0.9368

𝑋𝐵𝑀 = = = 0.953
𝑋𝐵1 0.9694
ln(𝑋𝐵1) ln(0.9368)
𝑝1 𝑝2 1060.7
𝑀1 + 𝑀2 + 1030.3/8.60
𝐶𝐴 = = 19.18 = 55.4 𝑘𝑔 𝑚𝑜𝑙/𝑚3
2 2

𝐷𝐴𝐵 𝐶𝐴 (𝑋𝐴1 − 𝑋𝐴2 ) (2.50 𝑋 10;9 ) (55.4)(0.0632 − 0.03055)

𝑁𝐴 = =
(𝑍2 − 𝑍1 )𝑋𝐵𝑀 (0.002)(0.953)
 𝑁𝐴 = 2.372 𝑥 10;6 kg mol A/s m2

PROBLEM 24
Estimation of Diffusivity of Methanol in H2O. The diffusivity of dilute methanolin water has
been determined experimentally to be 1.26 x 10-9 m2 /s at 288 K. (a) Estimate the diffusivity at
293 K using the Wilke-Chang equation. (b) Estimate the diffusivity at 293 K by correcting the
experimental value.
𝑉𝐴 = 0.0148 + 4(0.0037) + 1(0.0074)

1.173 𝑋 10−16 ( 𝐵 𝑀𝐵)1/2 𝑇

𝐷𝐴𝐵 = 𝐵 = 1.6
𝜇𝐵 𝑉 0.6 𝐴 𝐴

1
1.173 𝑥 106 (2.6 𝑥 18.02)2 (293)
=
1.005 𝑥 10;3 (0. 0370)0.6
𝐷𝐴𝐵 = 1.692 𝑥 10;9 𝑚2 /𝑠

B.)

𝐷𝐴𝐵𝑒𝑥𝑝 = 1.26 𝑥 10;9 𝑚2 /𝑠

298 1.404
𝐷𝐴𝐵 = 1.26 𝑥 10;9 ( )( )
288 1.005
𝐷𝐴𝐵 = 1.45 𝑥 10;9 𝑚2 /𝑠

PROBLEM 25

A layer of pulverized cork 6 in. (152 mm) thick is used as a layer of thermal insulation in a flat
wall. The temperature the cold side of the cork is 40°F (4.4°C), and that of the warm side is
180°F (82.2°C). The thermal conductivity of the cork at 32°F (0°C) is 0.021 Btuft-h-°F (0.036
W/m-°C), and that at 200°F (93.3°C) is 0.032 (0.055). The arca of the wall is 25 ft* (2.32 m?).
What is the rate of heat flow through the wall in Btu per hour (watts)?
Solution The arithmetic average temperature of the cork layer is (40 + 180)/2 = 110°F. By linear
interpolation the thermal
conductivity at 110°F is

(110 − 32)(0.032 − 0.021)

𝑘 = 0.021 +
200 − 32
= 0.021 + 0.005 = 0.026 𝐵𝑡𝑢/𝑓𝑡 − 𝑕 − ℉

6
𝐴 = 25 𝑓𝑡 2 ∆𝑇 = 180 − 40 = 140℉ 𝐵= = 0.5 𝑓𝑡
12
0.026 𝑥 25 𝑥 140 𝐵𝑡𝑢
𝑞= = 182 (53.3)
0.5 𝑕