Mass Transfer Design CCB2053
Dr Sintayehu Mekuria Hailegiorgis
Department of Chemical Engineering
CCB2053 1
Mass Transfer Principles
CCB2053-CHAPTER 2-LECTURE 1
Lesson Outline
Mass transfer principles
Fick’s law of diffusion mass transfer coefficients
Diffusion in gases Equimolar counter diffusion General case of diffusion and convection
Lesson Outcome
At the end of the lesson, the students are able to:
Explain the basics of mass transfer processes and Fick’s law of diffusion.
Apply the principle of diffusion for flux estimation in equimolar counter diffusion processes and general case of diffusion and convection mass transfer in gases.
Separation (recap)
• A physical process in which a mixture is separated into individual
components or group of components.
• Why? Product specification, Recovery, Purification
Mechanical operation
Separation
Homogeneous Heterogeneous
Mass transfer operations
Distillation Extraction Leaching Adsorption
Filtration Centrifugation Settling
Mass transfer principles
• Transport of one component from a region of higher concentration to lower concentration.
• Migration of a substance through another under the influence of concentration gradient.
• Involves the diffusion transport of some component within a single phase or between phases and remain there.
Mass transfer
Dye molecules spread throughout the water in a random fashion
Diffusion • Diffusion (considered in this chapter)
– The movement of a substance from an area of high concentration to an area of low concentration due to concentration difference.
Classification of mass transfer
• Molecular diffusion in stagnate media
• Molecular diffusion in fluids in laminar flow
• Eddy diffusion or turbulent diffusion
• Mass transfer between phases
Fick’s Law of molecular Diffusion
Molecular diffusion: defined as the transfer or movement of individual molecules through a fluid by means of the random, individual movements of the molecules.
This random movement of the molecules is often called a random walk process
Example: If there are a greater number of A molecules near point (1) than at (2), then, since molecules diffuse randomly in both directions, more A molecules will diffuse from point (1) to (2) than from point (2) to (1). The net diffusion of A is from high to low concentration region.
Fick’s Law of molecular Diffusion
• Mass transfer is characterized by the general form of transport equation.
Resistance
forceDrivingprocesstransferofRate
dzJanddcJ AzAz
1
dz
dxcDJ A
ABAZ
The basic law of diffusion called Fick’s Law by Adolf Eugen Fick in 1885
stated “ the molar flux of a species relative to an observer moving the molar
average velocity is proportional to the concentration gradient of the species”
If A diffuses in a binary mixture of A and B then the flux of A is
• The flux is the amount or concentration of solute carried by a fluid past a plane perpendicular to direction of flow or velocity.
• The unit of flux is given kg mol/m2s (SI system) or g mol/cm2s (cgs system)
Flux
Examples of molecular diffusion flux
Fick’s Law of molecular Diffusion
• let c is the concentration of A and B in kmol (A+B)/m3, xA the mole fraction of A in a mixture of A and B. Since c is constant, then
AAA dccxdcdx )(
Substituting in to Eq.(2.3)
dz
dcDJ A
ABA
Where D AB is the proportionality constant called the “diffusion coefficient” or the “ diffusivity” of A in a mixture of A and B.
Equation (2.5) is the mathematical representation of Fick’s law of diffusion in a binary mixture.
It is often called mass transfer equation.
Fick’s Law of molecular Diffusion
AA ccx
Similarities of the three transport equations The three molecular transport process (momentum, heat and mass) are
characterized by the same general type of equation as:
dz
dz
1. Molecular diffusion equation of momentum for constant density, (Newton’s law of viscosity)
2. Fourier’s law of heat conduction for constant and cp
3. Fick’s law of diffusion for constant total concentration, c
dz
dux
zx
dz
TcdAq
p
x
dz
dcDJ A
ABAz
Fick’s Law of molecular Diffusion
Convective mass transfer and mass transfer coefficient
• When a fluid is flowing outside a solid surface in a forced convection motion, mass transfer occurs under the influence of motion in a fluid medium- convective mass transfer.
)( 1 LiLcA cckN
Where:
NA -Convective molar flux in (kmol/m2s)
kc -Mass-transfer coefficient (m/s)
cl1 -Bulk fluid concentration (kg mol A/m3)
cli -Concentration in the fluid next to the surface of the solid (kg mol A/m3)
(2.10)
Convective mass transfer in liquid-solid system
Flowing liquid Flux
CLi .
i,
CL1
.. …………… …..
Diffusion in Gases
Molecular diffusion in gases can be taken place in different ways as:
1. Equimolar counter-diffusion in gases
2. General case for diffusion of gases A and B plus convection
3. Special case for A diffusing through stagnant, non-diffusing B
4. Diffusion through varying cross-sectional area
1. Equimolar counter-diffusion
– Let molecules A diffuse to the right and molecules B diffuse to the left and P constant throughout the system
21 AA pp
12 BB pp
Consider:
◦ Two gases A and B at constant total pressure (P) in two large chambers connected by a tube and molecular diffusion is occurring at steady state.
◦ Partial pressures: pA1 > pA2 and pB2 > pB1
The moles A diffusing to right is eqaual to Moles of B diffusing to the left since total pressure P is constant,
BzAz JJ
Fick’s law for B for constant total concentration c,
Since total pressure P is constant, then
BA ccc BA dcdc
dz
dcDJ
B
BABz
(2.11)
(2.12)
(1.13)
1. Equimolar counter-diffusion…
Equating Eq.(2.5) and (2.12)
dz
dcDJ
dz
dcDJ B
BABzA
ABAz (2.14)
Substituting (2.13) into (2.14) and canceling
BAAB DD (2.15)
i.e. for binary gas mixture of A and B, the diffusion coefficient (diffusivity) DAB for A diffusing into B is the same as DBA for B diffusing into A
1. Equimolar counter-diffusion…
2. General case for diffusion of gases A and B plus convection
•Let us consider what happens when the whole fluid is in motion in a bulk or convective flow to the right as shown in the Figure below
AAdA cvJ (m/s)(kgmol A/m3) (kgmol A/sm2)
Diffusion velocity of A
•For diffusion in stationary fluid, the diffusion flux JA passing a fixed point from left (high concn.) to right (low concn.) can be expressed in terms of velocity of diffusion of A,
•The molar average velocity of the whole fluid relative to the stationary point is vM m/s.
vA
(2.16)
vM vAd
2. General case for…
• Component A is still diffusing to the right with its diffusion velocity Ad,
thus for a stationary observer A is moving faster than the bulk velocity M since its diffusion velocity is added to that of the bulk velocity M
The first term represents the total flux relative to the stationary point, NA (kg mol A/s.m2), the second term the diffusion flux, J relative to the moving fluid and the third term is the convective flux of A relative to the stationary point.
MAdA vvv Convective
velocity of the bulk fluid
Velocity of A relative to a
stationary point
Diffusion velocity of A Multiplying by cA,
(2.17)
MAAdAAA vcvcvc
Hence,
MAAA vcJN (2.18)
2. General case for…
If N = total convective flux of the whole stream relative to the stationary point, then
BAM NNcvN c
NNv BA
M
(2.19)
Substituting equation (2.19) and Fick’s law into (2.18),
BAAA
ABA NNc
c
dz
dxcDN
Convection term
Diffusion term
Equation (2.20) is the general form of equation for diffusion plus convection
with relative to a stationary point. It holds for diffusion in gas, liquid, or solid.
(2.20)
2. General case for …
Note: For Equimolar counter-diffusion, Hence,
BABB
BAB NNc
c
dz
dxcDN (2.21)
A similar equation can be written for NB as shown in eq.(2.21)
The convective term in equation becomes zero. Then,
BA NN (2.22)
dz
dxcDN A
ABA (2.23)
2 Fick’s law of diffusion equation for constant total mass,
dz
dcDJ A
ABAz
3. Convective mass transfer and mass transfer coefficient
)( 1 lilcA cckN
Summary 1 Mass transfer is characterized by the general form of transport equation.
Resistance
forceDrivingprocesstransferofRate
5. General case for diffusion of gases A and B plus convection
4. Equimolar counter-diffusion
BAABBzAz DDandJJ
BAAA
ABA NNc
c
dz
dxcDN
Example-1
1. 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 0.02m 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 if
DAB of the CH4-He mixture is 0.675*10-5 m2/s at 101.32 kpa and
298 K.
Solution
Since total pressure P is constant, where the concentration c is as follows for a gas according to ideal gas law
nRTPV (a)
cRT
P
V
n (b)
Where: n is kg mol A plus B, V is volume, m3 , T is temperature, K, R is 8314.3m3 pa/kg mol K, c is in kg mol/m3
For steady state, the flux J and diffusivity DAB for a gas are constant.
dz
dcDJ A
ABAZ (c)
Rearranging and integrating Eq. (c)
2
1
2
1
A
A
c
c
AAB
z
z
AZ dcDdzJ
21
21
zz
ccDJ AAAB
AZ
(d)
Also, from ideal gas law
nRTPV
V
n
RT
pc AA
A 11
(e)
Substituting Eq. (e) into Eq. (d)
21
21
zzRT
ppDJ AAAB
AZ
(f) Final solution
JAZ = 5.52*10-5 kg mol A/s.m2 or 5.52*10-6 g mol A/s.cm2
EXAMPLE-2
2. Equimolar Counterdiffusion of NH3 and N2 at Steady State.
Ammonia gas (A) and nitrogen gas (B) are diffusing in
counterdiffusion through a straight glass tube 0.610 m long with
an inside diameter of 24.4 mm at 298 K and 101.32 kPa. Both
ends of the tube are connected to large mixed chambers at 101.32
kPa. The partial pressure of NH3 is constant at 20.0 kPa in one
chamber and 6.666 kPa in the other. The diffusivity at 298 K and
101.32 kPa is 2.30 × 10−5 m2/s.
Calculate
a) the diffusion of NH3 in kg mol/s. b) the diffusion of N2. c) the partial pressures at a point 0.305 m in the tube and
plot pA, pB, and P versus distance z.
Solution
T=298K P=101.32KPa pA1=20kpa
T=298K P=101.32KPa pA2=6.666kpa
ø=24.4mm
z=0.61m
Ammonia gas (A) Nitrogen gas (B) Equimolar counterdiffusion DAB=2.3x10-5m2/s
Calculate
a) the diffusion of NH3 in kg mol/s. b) the diffusion of N2. c) the partial pressures at a point 0.305 m in the tube
and plot pA, pB, and P versus distance z.
Similar to example-1 above,
Since total pressure P is constant, where the concentration c is as follows for a
gas according to ideal gas law
nRTPV (a)
cRT
P
V
n
(b)
Where: n is kg mol A plus B, V is volume, m3 , T is temperature, K, R is 8314.3m3 pa/kg mol K, c is in kg mol/m3
For steady state, the flux J and diffusivity DAB for a gas are constant.
dz
dcDJ A
ABAZ (c)
Rearranging and integrating Eq. (c)
2
1
2
1
A
A
c
c
AAB
z
z
AZ dcDdzJ
21
21
zz
ccDJ AAAB
AZ
(d)
Also, from ideal gas law
nRTPV
V
n
RT
pc AA
A 11
(e)
Substituting Eq. (e) into Eq. (d)
21
21
zzRT
ppDJ AAAB
AZ
(f)
Substituting DAB = 8314 m2/s, pA1= 2.0x104 pa, pA2= 6.666x103 pa, R=8314.34 m3 pa /kg
mol K, T=298K and z2-z1=0.61m
JAz = 2.03x10-7 kg mol A/s.m2
Rate of diffusion = JAz S,
Where S is surface area = πr2= π(0.0122m)2=4.68x10-4m2
a) Rate of diffusion = JAz πr2 =9.48x10-11 kg mol A/s
b) JB=?
Similarly JB can be given by;
21
21
zzRT
ppDJ BBBA
BZ
(g)
Where: pB1 = P - pA1 = 101.32kPa-20kpa=81.32kpa
pB2 = P - pA2 =101.32kPa-6.666kpa=94.654kpa
Substituting in to Eq.(g) and calculating
JBz = -2.03x10-7 kg mol B/s.m2
Rate of diffusion = JBz S,
Where S is surface area = πr2= π(0.0122m)2=4.68x10-4m2
Rate of diffusion = JBz πr2 =-9.48x10-11 kg mol A/s
The negative value of JBZ means the flux goes from point 2 to point 1
c) pA1 at 0.05m?
2
211 A
AB
AZA p
D
zzRTJp
(h)
Substituting the calculated value of JAz= 2.03X10-7kg mol/s.m2, z1-z2=0.05m and
the above values , then;
pA1= 1.333x104 pa
From equation (f) solving for pA1
Plots P, pA and pB versus distance z
P
pB1
pA1
z
P, p
A, p
B
pB1
pA2
P
pB1
pA1
P, p
A, p
B
Next Lesson
Diffusion of gas A through stagnant, non-diffusing gas B
Diffusion of gases through varying cross-sectional area
Recommended