Transcript Document

Overall Shell
Mass Balances I
Outline
3.
4.
5.
6.
Molecular Diffusion in Gases
Molecular Diffusion in Liquids
Molecular Diffusion in Solids
Prediction of Diffusivities
7. Overall Shell Mass Balances
1. Concentration Profiles
Overall Shell Mass Balance
Species entering and
leaving the system
by Molecular Transport +
by Convective Transport
* May also be expressed in terms of moles
Mass Generation
by homogeneous Steady-State!
chemical reaction
Overall Shell Mass Balance
* May also be expressed in terms of moles
Common Boundary Conditions:
1.
2.
3.
4.
Concentration is specified at the surface.
The mass flux normal to a surface maybe given.
At solid- fluid interfaces, convection applies: NA = kc∆cA.
The rate of chemical reaction at the surface can be specified.
♪ At interfaces, concentration is not necessarily continuous.
Concentration Profiles
I. Diffusion Through a
Stagnant Gas Film
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
Assumptions:
1.
2.
3.
4.
Steady-state
T and P are constants
Gas A and B are ideal
No dependence of vz on
the radial coordinate
At the gas-liquid interface,
𝑣𝑎𝑝
𝑝𝐴
𝑥𝐴 =
𝑃
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
Mass balance is done in this thin shell
perpendicular to the direction of mass flow
𝑑𝑥𝐴
𝑁𝐴 = −𝑐𝐷𝐴𝐵
+ 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 )
𝑑𝑧
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
𝑑𝑥𝐴
𝑁𝐴 = −𝑐𝐷𝐴𝐵
+ 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 )
𝑑𝑧
Since B is stagnant,
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
𝑁𝐴 = −
(1 − 𝑥𝐴 ) 𝑑𝑧
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
𝑁𝐴 = −
(1 − 𝑥𝐴 ) 𝑑𝑧
Applying the mass balance,
𝑆𝑁𝐴 ǀ𝑧 − 𝑆𝑁𝐴 ǀ𝑧+∆𝑧 = 0
where S = cross-sectional area of
the column
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
𝑆𝑁𝐴 ǀ𝑧 − 𝑆𝑁𝐴 ǀ𝑧+∆𝑧 = 0
Dividing by SΔz and taking the limit as Δz  0,
𝑑𝑁𝐴
−
=0
𝑑𝑧
NA = constant
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
𝑑𝑁𝐴
−
=0
𝑑𝑧
NA = constant
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
But, 𝑁𝐴 = −
(1 − 𝑥𝐴 ) 𝑑𝑧
Substituting,
𝑑
𝑑𝑧
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
=0
1 − 𝑥𝐴 𝑑𝑧
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
𝑑
𝑑𝑧
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
=0
1 − 𝑥𝐴 𝑑𝑧
For ideal gases, P = cRT and so at constant P and T, c = constant
DAB for gases can be assumed independent of concentration
𝑑
𝑑𝑧
1
𝑑𝑥𝐴
=0
1 − 𝑥𝐴 𝑑𝑧
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
𝑑
𝑑𝑧
1
𝑑𝑥𝐴
=0
1 − 𝑥𝐴 𝑑𝑧
Integrating once,
1
𝑑𝑥𝐴
= 𝐶1
1 − 𝑥𝐴 𝑑𝑧
Integrating again,
− ln 1 − 𝑥𝐴 = 𝐶1 𝑧 + 𝐶2
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
− ln 1 − 𝑥𝐴 = 𝐶1 𝑧 + 𝐶2
Let C1 = -ln K1 and C2 = -ln K2,
1 − 𝑥𝐴 = 𝐾1𝑧 𝐾2
B.C.
at z = z1,
at z = z2,
xA = xA1
xA = xA2
1 − 𝑥𝐴
1 − 𝑥𝐴2
=
1 − 𝑥𝐴1
1 − 𝑥𝐴1
𝑧−𝑧1
𝑧2 −𝑧1
Concentration Profiles
I. Diffusion Through a Stagnant Gas Film
1 − 𝑥𝐴
1 − 𝑥𝐴2
=
1 − 𝑥𝐴1
1 − 𝑥𝐴1
The molar flux then becomes
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
𝑁𝐴 = −
(1 − 𝑥𝐴 ) 𝑑𝑧
𝑁𝐴
𝑧−𝑧1
𝑧2 −𝑧1
𝑐𝐷𝐴𝐵
1 − 𝑥𝐴2
=
ln(
)
𝑧2 − 𝑧1
1 − 𝑥𝐴1
OR in terms of the driving force ΔxA
*𝑥𝐴1 − 𝑥𝐴2 > 0, i.e. xA1> xA2
ǂ𝑧 − 𝑧
i.e. z2> z1
2
1 > 0,
𝑐𝐷𝐴𝐵
𝑥𝐵2 − 𝑥𝐵1
𝑁𝐴 =
(𝑥𝐴1 − 𝑥𝐴2 )
(𝑥𝐵 )𝑙𝑛 =
(𝑧2 − 𝑧1 )(𝑥𝐵 )𝑙𝑛
𝑥
ln( 𝐵2 )
𝑥𝐵1
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
Two Reaction Types:
1. Homogeneous – occurs
in the entire volume of
the fluid
- appears in the
generation term
2. Heterogeneous – occurs
on a surface (catalyst)
- appears in the
boundary condition
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
Reaction taking place
2A  B
1. Reactant A diffuses to
the surface of the
catalyst
2. Reaction occurs on the
surface
3. Product B diffuses away
from the surface
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
Reaction taking place
2A  B
Assumptions:
1. Isothermal
2. A and B are ideal gases
3. Reaction on the surface
is instantaneous
4. Uni-directional transport
will be considered
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
𝑑𝑁𝐴
=0
𝑑𝑧
𝑑𝑥𝐴
𝑁𝐴 = −𝑐𝐷𝐴𝐵
+ 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 )
𝑑𝑧
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
From stoichiometry,
𝑁𝐵 = −1/2𝑁𝐴
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
𝑁𝐴 = −
1
1 − 𝑥𝐴 𝑑𝑧
2
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
Substitution of NA into the differential equation
𝑑
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
(−
)=0
1
𝑑𝑧
1 − 𝑥𝐴 𝑑𝑧
2
Integration twice with respect to z,
1
−2 ln 1 − 𝑥𝐴 = 𝐶1 𝑧 + 𝐶2 = −(2 ln 𝐾1 )𝑧 − (2 ln 𝐾2 )
2
B.C. 1: at z = 0,
B.C. 2: at z = δ,
xA = xA0
xA = 0
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
The final equation is
𝑧
1
1
(1− )
1 − 𝑥𝐴 = (1 − 𝑥𝐴0 ) 𝛿
2
2
And the molar flux of reactant through the film,
2𝑐𝐷𝐴𝐵
1
𝑁𝐴 =
ln(
)
1
𝛿
1 − 𝑥𝐴0
2
*local rate of reaction per unit of catalytic surface
Concentration Profiles
II. Diffusion With a Heterogeneous Chemical Reaction
Reading Assignment
See analogous problem Example 18.3-1 of
Transport Phenomena by Bird, Stewart and Lightfoot
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
1. Gas A dissolves in liquid B and
diffuses into the liquid phase
2. An irreversible 1st order
homogeneous reaction takes
place
A + B  AB
Assumption:
AB is negligible in the solution
(pseudobinary assumption)
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
𝑆𝑁𝐴 ǀ𝑧 − 𝑆𝑁𝐴 ǀ𝑧+∆𝑧 − 𝑘1′′′ 𝐶𝐴 𝑆∆𝑧 = 0
𝑘1′′′ first order rate constant
for homogeneous
decomposition of A
S cross sectional area of the liquid
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
𝑆𝑁𝐴 ǀ𝑧 − 𝑆𝑁𝐴 ǀ𝑧+∆𝑧 − 𝑘1′′′ 𝐶𝐴 𝑆∆𝑧 = 0
Dividing by SΔz and taking the limit as Δz  0,
𝑑𝑁𝐴
+ 𝑘1′′′ 𝐶𝐴 = 0
𝑑𝑧
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
𝑑𝑁𝐴
+ 𝑘1′′′ 𝐶𝐴 = 0
𝑑𝑧
If concentration of A is small, then the total c is almost constant and
𝑑𝑐𝐴
𝑁𝐴 = −𝐷𝐴𝐵
𝑑𝑧
Combining the two equations above
𝑑2 𝑐𝐴
′′′
𝐷𝐴𝐵
−
𝑘
1 𝐶𝐴 = 0
2
𝑑𝑧
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
𝑑2 𝑐𝐴
′′′
𝐷𝐴𝐵
−
𝑘
1 𝐶𝐴 = 0
2
𝑑𝑧
𝐵. 𝐶. 1
𝐵. 𝐶. 2
𝑎𝑡 𝑧 = 0,
𝑎𝑡 𝑧 = 𝐿,
𝑐𝐴 = 𝑐𝐴0
𝑑𝑐𝐴
𝑁𝐴 = 0 𝑜𝑟
=0
𝑑𝑧
Multiplying the above equation by
𝐿2
𝑐𝐴0 𝐷𝐴𝐵
gives an equation with dimensionless variables
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
𝑑2 𝑐𝐴
′′′
𝐷𝐴𝐵
−
𝑘
1 𝐶𝐴 = 0
2
𝑑𝑧
𝑑2Γ
2
−
𝜙
Γ=0
2
𝑑𝜁
𝑐𝐴
Γ=
,
𝑐𝐴0
𝑧
𝜁= ,
𝐿
𝜙=
𝑘 ′′′ 𝐿2 /𝐷𝐴𝐵
Thiele Modulus
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
𝑑2Γ
2Γ = 0
−
𝜙
𝑑𝜁 2
𝐵. 𝐶. 1
𝑎𝑡 𝜁 = 0,
𝐵. 𝐶. 2
𝑎𝑡 𝜁 = 1,
Γ=1
𝑑Γ
=0
𝑑𝜁
The general solution is
Γ = 𝐶1 cosh 𝜙𝜁 + 𝐶2 sinh 𝜙𝜁
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
Γ = 𝐶1 cosh 𝜙𝜁 + 𝐶2 sinh 𝜙𝜁
Evaluating the constants,
cosh 𝜙 cosh 𝜙𝜁 − sinh 𝜙 sinh 𝜙𝜁
cosh[ϕ 1 − ζ ]
Γ=
=
cosh 𝜙
cosh 𝜙
Reverting to the
original variables,
𝑐𝐴
=
𝑐𝐴0
𝑘 ′′′ 𝐿2
𝑧
cosh[
1− ]
𝐷𝐴𝐵
𝐿
𝑘 ′′′ 𝐿2
cosh(
)
𝐷𝐴𝐵
Concentration Profiles
III. Diffusion With a Homogeneous Chemical Reaction
Quantities that might be asked for:
1. Average concentration in the liquid phase
𝑐𝐴,𝑎𝑣𝑔
=
𝑐𝐴0
𝐿
(𝑐 /𝑐 )𝑑𝑧
0 𝐴 𝐴0
𝐿
𝑑𝑧
0
tanh 𝜙
=
𝜙
2. Molar flux at the plane z = 0
𝑁𝐴𝑧ǀ𝑧=0
𝑑𝑐𝐴
𝑐𝐴0 𝐷𝐴𝐵
= −𝐷𝐴𝐵
ǀ𝑧=0 =
𝜙 tanh 𝜙
𝑑𝑧
𝐿
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
Assumptions
1. Velocity field is unaffected by
diffusion
2. A is slightly soluble in B
3. Viscosity of the liquid is unaffected
4. The penetration distance of A in B
will be small compared to the film
thickness.
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
Recall: The velocity of a falling film
𝜌𝑔𝛿 2 cos 𝛼
𝑣𝑧 (𝑥) =
2𝜇
𝑣𝑧 𝑥 = 𝑣𝑚𝑎𝑥
𝑥 2
1−( )
𝛿
𝑥 2
1−( )
𝛿
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
* CA is a function of both x and z
𝑁𝐴𝑧ǀ𝑧 𝑊∆𝑥 − 𝑁𝐴𝑧ǀ𝑧+∆𝑧 𝑊∆𝑥
+𝑁𝐴𝑥ǀ𝑥 𝑊∆𝑧 − 𝑁𝐴𝑥ǀ𝑥+∆𝑥 𝑊∆𝑧 = 0
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝑁𝐴𝑧ǀ𝑧 𝑊∆𝑥 − 𝑁𝐴𝑧ǀ𝑧+∆𝑧 𝑊∆𝑥
+𝑁𝐴𝑥ǀ𝑥 𝑊∆𝑧 − 𝑁𝐴𝑥ǀ𝑥+∆𝑥 𝑊∆𝑧 = 0
Dividing by WΔxΔz and
letting Δx  0 and Δz  0,
𝜕𝑁𝐴𝑧 𝜕𝑁𝐴𝑥
+
=0
𝜕𝑧
𝜕𝑥
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝜕𝑁𝐴𝑧 𝜕𝑁𝐴𝑥
+
=0
𝜕𝑧
𝜕𝑥
The expressions for 𝑁𝐴𝑧 ,
𝑁𝐴𝑧
𝑑𝑐𝐴
= −𝐷𝐴𝐵
+ 𝑥𝐴 (𝑁𝐴𝑧 + 𝑁𝐵𝑧 )
𝑑𝑧
Transport of A along the z direction is mainly by convection (bulk motion)
Recall:
𝑁𝐴 = 𝐽𝐴∗ + 𝑐𝐴 𝑣𝑀
𝑣𝑀 = 𝑚𝑜𝑙𝑎𝑟 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑁𝐴𝑧 ≈ 𝑐𝐴 𝑣𝑀 = 𝑐𝐴 𝑣𝑧 (𝑥)
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝜕𝑁𝐴𝑧 𝜕𝑁𝐴𝑥
+
=0
𝜕𝑧
𝜕𝑥
The expressions for 𝑁𝐴𝑥 ,
𝑁𝐴𝑥
𝑑𝑐𝐴
= −𝐷𝐴𝐵
+ 𝑥𝐴 (𝑁𝐴𝑥 + 𝑁𝐵𝑥 )
𝑑𝑧
Transport of A along the x direction is mainly by diffusion
𝑁𝐴𝑥
𝑑𝑐𝐴
≈ −𝐷𝐴𝐵
𝑑𝑧
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝜕𝑁𝐴𝑧 𝜕𝑁𝐴𝑥
+
=0
𝜕𝑧
𝜕𝑥
Substituting the expressions for𝑁𝐴𝑥 𝑎𝑛𝑑 𝑁𝐴𝑧 ,
𝑣𝑧
𝜕𝑐𝐴
𝜕 2 𝑐𝐴
= 𝐷𝐴𝐵
𝜕𝑧
𝜕𝑥 2
Substituting the expressions vz,
𝑣𝑚𝑎𝑥
𝑥 2
1−( )
𝛿
𝜕𝑐𝐴
𝜕 2 𝑐𝐴
= 𝐷𝐴𝐵
𝜕𝑧
𝜕𝑥 2
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝑣𝑚𝑎𝑥
𝑥 2
1−( )
𝛿
𝜕𝑐𝐴
𝜕 2 𝑐𝐴
= 𝐷𝐴𝐵
𝜕𝑧
𝜕𝑥 2
Boundary conditions
B.C. 1 𝑎𝑡 𝑧 = 0,
B.C. 2 𝑎𝑡 𝑥 = 0,
B.C. 3 𝑎𝑡 𝑥 = 𝛿,
𝑐𝐴 = 0
𝑐𝐴 = 𝑐𝐴0
𝜕𝑐𝐴
𝜕𝑥
=0
BUT we can replace B.C. 3 with
B.C. 3 𝑎𝑡 𝑥 = ∞, 𝑐𝐴 = 0
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝑣𝑚𝑎𝑥
𝑥 2
1−( )
𝛿
𝑐𝐴
2
=1−
𝑐𝐴0
𝜋
or
where
2 𝑧/𝑣
𝑥/ 4𝐷𝐴𝐵
𝑚𝑎𝑥
exp −𝜉 2 𝑑𝜉
0
𝑐𝐴
= 1 − 𝑒𝑟𝑓
𝑐𝐴0
erf 𝑥 =
𝜕𝑐𝐴
𝜕 2 𝑐𝐴
= 𝐷𝐴𝐵
𝜕𝑧
𝜕𝑥 2
𝑥
= 𝑒𝑟𝑓𝑐
2
4𝐷𝐴𝐵
𝑧
𝑣𝑚𝑎𝑥
𝑥
2
0 exp(−𝑥 )𝑑 𝑥
∞
2
0 exp(−𝑥 )𝑑 𝑥
=
𝑥
2
4𝐷𝐴𝐵
𝑧
𝑣𝑚𝑎𝑥
2 𝑥
2
exp(−
𝑥
) 𝑑𝑥
𝑥 0
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
𝑐𝐴
= 1 − 𝑒𝑟𝑓
𝑐𝐴0
𝑁𝐴𝑥ǀ𝑥=0
𝑥
2
4𝐷𝐴𝐵
𝑧
𝑣𝑚𝑎𝑥
= 𝑒𝑟𝑓𝑐
𝑥
2
4𝐷𝐴𝐵
𝑧
𝑣𝑚𝑎𝑥
𝜕𝑐𝐴
𝐷𝐴𝐵 𝑣𝑚𝑎𝑥
= −𝐷𝐴𝐵
ǀ𝑥=0 = 𝑐𝐴0
𝜕𝑥
𝜋𝑧
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
Reading Assignment
See analogous problem Example 4.1-1 of Transport
Phenomena by Bird, Stewart and Lightfoot
Concentration Profiles
IV. Diffusion into a Falling Liquid Film (Gas Absorption)
Quantities that might be asked for:
1. Total molar flow of A across the surface at x = 0
𝑊
𝑊𝐴 =
0
= 𝑊𝑐𝐴0
= 𝑐𝐴0
𝐿
0
𝑁𝐴𝑥ǀ𝑥=0 𝑑𝑧𝑑𝑦
𝐷𝐴𝐵 𝑣𝑚𝑎𝑥
𝜋
𝐷𝐴𝐵 𝑣𝑚𝑎𝑥
𝜋𝐿
𝐿
0
1
𝑑𝑧
𝑧