CHAPTER 1 INTRODUCTION TO MASS TRANSFER AND DIFFUSION Chap Ch apte terr 1 ou outl tlin ine e • Simi Simila lari rity ty of Mass Mass,, Heat Heat and and Mome Momen ntu tum m Transfer ransfer Processes • Examp Examples les of Mass-T Mass-Tra ransf nsfer er Proce Processe ssess Law w for Molec Molecul ular ar Diff Diffus usio ion n • Fick’s La • Con Convect vective ive Mass-T Mass-Transf ransfer er Coefficie Coefficients nts Prepared by: Ummi Kalthum Binti Ibrahim Fakulti Kejuruteraan Kimia
OBJECTIVES
Students should be able to : 1. Understand and use Fick’s law and diffusion coefficients (diffusivity) 2. Understand Understand the familiarity of analogies between between correlations correlations for heat, mass and momentum transfer processes 3. Understand Understand the molecular diffusion diffusion in gas, gas, liquid and solid
LECTURE 1
Definition:
• Net movement of mass from one distinct phase to another or through a single phase whether the phase is gas, liquid or solid.
• Mass transfer occurs in many processes, such as absorption, evaporation, adsorption, drying, precipitation, membrane filtration, and distillation.
Mass transfer could occur by the following three ways: 1) Diffusion - the net transport of substances in a stationary solid or fluid under a concentration gradient. 2) Advection - the net transport of substances by the moving fluid , and so
cannot happen in solids. 3) Convection - the net transport of substances caused by both advective transport and diffusive transport in fluids.
1. Momentum transfer ( Newton’s equation) :
d x
dz
= momentum transferred
z
s.m²
= kinematic viscosity (m²/s) = distance (m)
x = momentum/m³ , momentum (kg m/s)
Momentum transfer is
the amount of momentum that one particle gives to another particle. Example: fluid flow, mixing, sedimentation and filtration.
2. Heat transfer (Fourier’s Law)
q A
d c pT dz
q
= heat flux (w/m²)
A
= thermal diffusivity(m²/s)
c pT = J/m³
Heat transfer is
the transfer of energy from high temperature to low temperature (temperature gradient). Example: evaporation, distillation and drying.
3. Mass transfer ( Fick’s Law) :
J A D AB
dc A dz
J A
= molar flux of component A (kg mol A/s.m²)
z = distance due to molecular diffusion (m)
D AB
= molecular diffusivity of A in B (m²/s)
c A = concentration of A (kg mol/m³)
Mass transfer is
the net movement of mass from one location, usually meaning a stream, phase, fraction or component, to another. Example: distillation, absorption, drying, liquid -liquid extraction, adsorption, crystallization, etc.
Momentum transfer
d x
dz
= momentum transferred
s.m² = kinematic viscosity (m²/s) z = distance (m) x = momentum/m³
Heat transfer
q A
q
d c pT
A
dz
z
Mass transfer
J A
D AB
c pT
J A
dc A dz
= heat flux (w/m²) = thermal diffusivity(m²/s) = distance (m) = J/m³
= molar flux of component A (kg mol A/s.m²)
z = distance due to molecular diffusion (m)
D AB = molecular diffusivity of A in B (m²/s)
c
= concentration of A (kg mol/m³)
The similarity: 1) All the fluxes on the left hand-side have units transfer of quantity per unit time per unit area. 2) The transport properties (constant) have units of m2/s. 3) All the concentrations are represented as transfer quantity/m3 .
• Liquid in an open pail of water evaporates into air because of the difference in concentration of water vapor and surrounding air.
• A piece of sugar added to a cup of coffee dissolves by itself and diffuses to the surrounding solution.
• Difference between diffusion and convection mass transfer: Stirring the water with a spoon creates forced convection. That helps the sugar molecules to transfer to the bulk water much faster.
• Transfer or movement of individual molecules through a fluid by means of the random, individual movements of the molecules under concentration gradient.
• If there are a greater number of A molecules near point 1 than point 2, more A molecules will diffuse from 1 to 2.
• The net diffusion of A is from high to low concentration regions.
• The general Fick’s Law equation for a binary mixture A and B: J A
CD AB
C = total concentration of A and B
dx A
x A = mole fraction of A in the mixture
dz
of A and B
• If C is constant : C A Cx A Cdx A J A
d (Cx A ) dC A
D AB
dCA dz
J A = molar flux of component A (kg mol A/s.m²)
z D AB
= distance due to molecular diffusion (m) = molecular diffusivity of A in B (m²/s)
c A = concentration of A (kg mol/m³)
Example 6.1.1 (Molecular Diffusion of Helium in Nitrogen)
• A mixture of He and N2 gas is contained in a pipe at 298 K and 1 atm total pressure which is constant throughout. At one end of the pipe at point 1 the partial pressure p A1 of He is 0.60 atm, and at the other end 0.20m (20cm) p A2 = 0.2 atm. Calculate the flux of He at steady state if DAB of He-N2 mixture is 0.687 x 10-4 m2/s.
Solution: Fick’s law equation:
Integrate:
J A
D AB
dC A dz
PV n V
z2
⌠ ⌡
P
n = kg mol A+B V = volume (m³) T = temperature (K) R = 8314 m³·Pa/kgmol·K C = kgmol A+B/m3
nRT
RT
C
C A2
⌠ ⌡
J A dz = - D AB dC A z1
C A1
J A(z2 – z1) = - D AB (C A2 – C A1) J A = - D AB(C A2 – C A1) z2 – z1 = D AB (P A1 – P A2)
P A1 = 0.6 atm 5 = 0.6 x 1.01325 x 10 Pa = 6.08 x 104 Pa P A2 = 0.2 atm = 0.2 x 1.01325 x 105 Pa = 2.027 x 104 Pa
RT(z2 – z1) = ( 0.687 x 10-4 m2/s) (6.08 x 104 – 2.027 x 104) Pa (298 K) (0.2 - 0)m 8314 m³Pa kg mol K
= 5.63 x 10-6 kg mol A/s.m²
• Equimolar counterdiffusion • Two gases A and B at constant P in two large chambers connected by a tube. • Stirring in each chamber keeps the concentrations uniform. • Molecules A diffuse to the right and B to the left. • Since the total pressure P is constant, the net moles of A diffusing to the right must equal the net moles of B to the left.
pA1 > pA2 pB2 > pB1
J A = - JB
• Equimolar counterdiffusion in gases
J A D AB J B D BA
•
D AB = diffusivity of component A in its
dC A dz
mixture with component B, m 2/s
1
dC A = molar concentration gradient, mol/m 4
dC B
dz
2
dz
Since P = P A + PB = constant,
C = C A + CB = P
This shows that for binary gas mixture of A and B, the diffusivity coefficient D AB for A diffusing into B is the same as DBA for diffusing into A
3
RT
• Differentiating both sides: dC A + dCB = dC = 0
J
D
dC A = - dCB dC A J
4
( )D
dCB
*D AB = DBA
Example 6.2.1 (Equimolar counterdiffusion)
• 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 D AB = 0.230 x 10-4 m2/s a) Calculate the flux J A at steady state b) Repeat for JB
Solution: a)
J A = D AB (P A1 – P A2) RT(z2 – z1) = ( 0.23 x 10-4) (1.013 x 10 4 – 0.507 x 104) 8314(298) (0.1 - 0) = 4.70 x 10-7 kg mol A/s.m² JB = D AB (PB1 – PB2) RT(z2 – z1) = - 4.70 x 10-7 kg mol B/s.m²
PB1 = P - P A1 = 1.0132 x 105 - 1.013 x 104 = 9.119 x 104 Pa PB2 = P – P A2 = 1.0132 x 105 – 0.507 x 104 = 9.625 x 104 Pa
*The negative value for J B means the flux goes fro m point 2 to poi nt 1.
• The rate at which moles of A passed a fixed point to the right, which will be taken as a possible flux, JA kg mol A/s.m²
• This flux can be converted to a velocity of diffusion of A to the right by JA(kg mol A/s.m²) = C Ad A
m kg mol A Ad = diffusion velocity of s m³ , A in m/s
molar average velocity = (m/s) m
• The velocity of A relative to the stationary point is the sum of the diffusion velocity and the average/convective velocity: A
Ad m
A Ad
m
• Multiply by C A
C A A C A Ad C A m NA (kg mol A/s.m²) --- total flux C A Ad C A m ❖ N A
J A = diffusion flux
convective flux
J A
C A m
Let N be the total convective flux of the whole stream relative to the stationary point:
N
m ❖ N A
C m
N A
N A
N B
N B
C J A
C A
N
A
N B
• Similarly: N A N B
CD AB
CD BA
dx A dz dx B dz
C A
N
N B
N
N B
A
C C B
A
C
• For equimolar counterdiffusion: N A
N B
convective term become zero N A N B
CD AB
CD BA
dx A dz
dx B dz
C A C
C B C
0
N
A
N B
0
N
A
N B
• In this case, one boundary at the end of the diffusion path is in impermeable to component B, so it cannot pass through.
• Example (Figure a): the evaporation of a pure liquid benzene (A) at the bottom • • • •
of tube, where nondiffusing air (B) is pass over the top. The benzene vapor (A) diffuses through the air (B) in the tube. The boundary at the liquid surface at point 1 is impermeable to air, since air is insoluble in benzene. Hence, air(B) cannot diffuse into or away from the surface. At point 2, the partial pressure P = 0 since a large volume of air is passing by.
• To derive the case for A diffusing in stagnant, nondiffusing B , NB = 0 N A
CD AB
dx A dz
C A C
N
• The convective flux of A: C A C
N
A
• At P = constant,
N A
N A
0
0
A
C A
C A C
C
P RT
D AB
,
dC A dz
P A
D AB dP A RT dz
C A C
N B
X AC X A
x A P ,
N A
P A P
N A
P A P
C A C
PA P
• Rearranging and integrating
N A 1
P A
D P RT
z2
AB
N A dz
D AB RT
z1
N A z2 z1
N A
D AB RT
D AB P RT D AB P RT
dP A dz P A 2
P A 1
dP A 1 PA P
P A 2
P
P P dP
A
P A 1
P A 2
A P A 1
P P A 2
D ABP
RT z2
z1
In
P P
Log mean value of B PBM
P P A1 P A 2 PA1
PB1 = P – PA1 PB2 = P – PA2
= PA2 + PB2
A
InP P In
✓P = PA1 + PB1
* N A
PB 2
PB1
In PB 2 / PB1
P A1
D AB P
RT z2
In P
z1 PBM
P
A1
P A 2
P A 2 / P
P A 2
P A1
Example 6.2.2 (Diffusion of water through stagnant, nondiffusing air) Water in the bottom of a narrow metal tube is held at a constant temp. of 293 K. The total pressure of air is 1.01325 x 10⁵ Pa (1.0 atm) and the temp. is 293 K (20°C). Water evaporates and diffuses through the air in the tube and the diffusion path z₂ - z₁ is 0.1524 m long. Calculate the rate of evaporation at steady state. The diffusivity of water vapor at 293 K and 1 atm pressure is 0.250 x 10-4 m2/s. Solution:
• DAB = 0.250 x 10-4 m2/s • From appendix A.2.2, vapor pressure of water at 20°C : PA1 = 17.54 mmHg = 2.338 kPa = 2.338 x 10³ Pa
z₂ - z₁
PA2 = 0 (pure air)
PB1 =P-PA1 = 1.01325 x 10⁵ - 2.338 x 10³ = 9.899 x 10⁴ Pa PB2 =P-PA2 = 1.01325 x 10⁵ - 0 = 1.01325 x 10⁵ Pa
water
For A diffusing in stagnant, nondiffusing B: N A
N A
CD AB
CD AB
dx A dz dx A dz
C A
A
C C A C
N N
A
At P = constant:
N A Integrate:
D AB dP A RT dz
P A P
N A
NB
0
NB
PBM = PB2 – PB1 = 1.01325 x 10⁵ Pa - 9.899 x 10⁴ Pa In(PB2/PB1 )
1.01325 10 In Pa 9 . 899 10 5
4
= 1.001 x 10⁵ Pa
N A
D ABP RT z2 z1 PBM
P P A1
A 2
0.25 10 m / s 1.01325 10 Pa2.338 10 Pa 8314J / kgmol.K 293K 0.1524m1.001 10 Pa 4
2
5
3
5
1.595 10
7
kgmol / s.m2
