Overall Shell Mass Balances I

transcript

Overall ShellMass Balances I

Outline

3. Molecular Diffusion in Gases 4. Molecular Diffusion in Liquids 5. Molecular Diffusion in Solids6. Prediction of Diffusivities

7. Overall Shell Mass Balances1. Concentration Profiles

Overall Shell Mass Balance

Species entering and leaving the system

by Molecular Transport +by Convective Transport

Mass Generationby homogeneous chemical reaction

* May also be expressed in terms of moles

Steady-State!

Overall Shell Mass Balance

* May also be expressed in terms of moles

Common Boundary Conditions:

1. Concentration is specified at the surface.2. The mass flux normal to a surface maybe given.3. At solid- fluid interfaces, convection applies: NA = kc∆cA.4. 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

I. Diffusion Through a Stagnant Gas FilmAssumptions:

1. Steady-state2. T and P are constants3. Gas A and B are ideal4. No dependence of vz on

the radial coordinate

At the gas-liquid interface,

I. Diffusion Through a Stagnant Gas FilmMass balance is done in this thin shell

perpendicular to the direction of mass flow

𝑁 𝐴=−𝑐𝐷𝐴𝐵𝑑𝑥𝐴

𝑑𝑧 +𝑥𝐴(𝑁 𝐴+𝑁𝐵)

I. Diffusion Through a Stagnant Gas Film

𝑑𝑧 +𝑥𝐴(𝑁 𝐴+𝑁𝐵)

Since B is stagnant,

𝑁 𝐴=−𝑐𝐷 𝐴𝐵

(1−𝑥𝐴)𝑑𝑥𝐴

𝑑𝑧

𝑆𝑁 𝐴 ǀ𝑧−𝑆𝑁 𝐴 ǀ𝑧+∆ 𝑧=0

Applying the mass balance,

where S = cross-sectional area of the column

𝑆𝑁 𝐴 ǀ𝑧−𝑆𝑁 𝐴 ǀ𝑧+∆ 𝑧=0

Dividing by SΔz and taking the limit as Δz 0,

−𝑑𝑁 𝐴

𝑑𝑧 =0 NA = constant

−𝑑𝑁 𝐴

𝑑𝑧 =0 NA = constant

𝑑𝑧But,

Substituting,

𝑑𝑑𝑧 ( 𝑐𝐷 𝐴𝐵

(1− 𝑥𝐴 )𝑑𝑥𝐴

𝑑𝑧 )=0

I. Diffusion Through a Stagnant Gas Film𝑑𝑑𝑧 ( 𝑐𝐷 𝐴𝐵

(1− 𝑥𝐴 )𝑑𝑥 𝐴

𝑑𝑧 )=0For ideal gases, P = cRT and so at constant P and T, c = constantDAB for gases can be assumed independent of concentration

𝑑𝑑𝑧 ( 1

𝑑𝑧 )=0

𝑑𝑑𝑧 ( 1

𝑑𝑧 )=0Integrating once,

1(1−𝑥𝐴 )

𝑑𝑥𝐴

𝑑𝑧 =𝐶1

Integrating again,

− ln (1− 𝑥𝐴 )=𝐶1𝑧+𝐶2

I. Diffusion Through a Stagnant Gas Film− ln (1− 𝑥𝐴 )=𝐶1𝑧+𝐶2

Let C1 = -ln K1 and C2 = -ln K2,

1−𝑥𝐴=𝐾 1𝑧𝐾 2

at z = z1, xA = xA1

at z = z2, xA = xA2 ( 1−𝑥𝐴

1− 𝑥𝐴1 )=( 1−𝑥𝐴2

1−𝑥𝐴1 )𝑧− 𝑧 1𝑧 2− 𝑧1

( 1−𝑥𝐴

1− 𝑥𝐴1 )=( 1−𝑥𝐴2

1−𝑥𝐴1 )𝑧− 𝑧 1𝑧 2− 𝑧1

𝑑𝑧𝑁 𝐴=

𝑐𝐷𝐴𝐵

(𝑧 2− 𝑧1 )ln (1− 𝑥𝐴 2

1− 𝑥𝐴1)

*, i.e. xA1> xA2ǂ i.e. z2> z1

𝑁 𝐴=𝑐𝐷𝐴𝐵

( 𝑧2−𝑧1)(𝑥¿¿𝐵)𝑙𝑛(𝑥𝐴1−𝑥𝐴2)¿

The molar flux then becomes

OR in terms of the driving force ΔxA

(𝑥¿¿𝐵)𝑙𝑛=𝑥𝐵 2−𝑥𝐵1

ln (𝑥𝐵2

𝑥𝐵1)

II. Diffusion With a Heterogeneous Chemical ReactionTwo 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

II. Diffusion With a Heterogeneous Chemical ReactionReaction taking place

1. Reactant A diffuses to the surface of the catalyst

2. Reaction occurs on the surface

3. Product B diffuses away from the surface

II. Diffusion With a Heterogeneous Chemical ReactionReaction taking place

Assumptions:

1. Isothermal2. A and B are ideal gases3. Reaction on the surface

is instantaneous4. Uni-directional transport

will be considered

II. Diffusion With a Heterogeneous Chemical Reaction

𝑑𝑁 𝐴

𝑑𝑧 =0

𝑑𝑧 +𝑥𝐴(𝑁 𝐴+𝑁 𝐵)

1− 12𝑥𝐴

𝑑𝑥𝐴

𝑑𝑧

From stoichiometry,

II. Diffusion With a Heterogeneous Chemical ReactionSubstitution of NA into the differential equation

𝑑𝑑𝑧 (−

𝑐𝐷𝐴𝐵

1− 12𝑥𝐴

𝑑𝑥𝐴

𝑑𝑧 )=0

Integration twice with respect to z,

−2 ln(1− 12 𝑥𝐴)=𝐶1 𝑧+𝐶2=−¿

B.C. 1: at z = 0, xA = xA0

B.C. 2: at z = δ, xA = 0

II. Diffusion With a Heterogeneous Chemical ReactionThe final equation is

1− 12𝑥𝐴=(1− 1

2𝑥𝐴 0)

(1− 𝑧𝛿 )

And the molar flux of reactant through the film,

𝑁 𝐴=2𝑐𝐷 𝐴𝐵

𝛿 ln( 1

1− 12𝑥𝐴0

*local rate of reaction per unit of catalytic surface

Reading Assignment

See analogous problem Example 18.3-1 of Transport Phenomena by Bird, Stewart and Lightfoot

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)

𝑆𝑁 𝐴 ǀ𝑧−𝑆𝑁 𝐴 ǀ𝑧+∆ 𝑧−𝑘1′ ′ ′𝐶𝐴𝑆 ∆ 𝑧=0

first order rate constant for homogeneous decomposition of AS cross sectional area of the liquid

𝑆𝑁 𝐴 ǀ𝑧−𝑆𝑁 𝐴 ǀ𝑧+∆ 𝑧−𝑘1′ ′ ′𝐶𝐴𝑆 ∆ 𝑧=0

Dividing by SΔz and taking the limit as Δz 0,

𝑑𝑁 𝐴

𝑑𝑧 +𝑘1′ ′ ′𝐶𝐴=0

III. Diffusion With a Homogeneous Chemical Reaction𝑑𝑁 𝐴

𝑑𝑧 +𝑘1′ ′ ′𝐶𝐴=0

If concentration of A is small, then the total c is almost constant and

𝑁 𝐴=−𝐷𝐴𝐵𝑑𝑐𝐴

𝑑𝑧Combining the two equations above

𝐷 𝐴𝐵𝑑2𝑐𝐴

𝑑 𝑧2−𝑘1

′ ′ ′𝐶𝐴=0

𝑑 𝑧2−𝑘1

′ ′ ′𝐶𝐴=0

Multiplying the above equation by gives an equation with dimensionless variables

𝑑 𝑧2−𝑘1

′ ′ ′𝐶𝐴=0

𝑑2Γ𝑑𝜁 2

−𝜙2Γ=0

Γ=𝑐𝐴

𝑐𝐴0,𝜁= 𝑧

𝐿 ,𝜙=√𝑘′ ′ ′𝐿2/𝐷𝐴𝐵

Thiele Modulus

𝑑2Γ𝑑𝜁 2

−𝜙2Γ=0

The general solution is

Γ=𝐶1 cosh (𝜙𝜁 )+𝐶2sinh (𝜙𝜁 )

Γ=cosh (𝜙 ) cosh (𝜙𝜁 )−sinh (𝜙 ) sinh (𝜙𝜁 )

cosh (𝜙 )=cosh [ϕ (1−ζ )]cosh (𝜙 )

Evaluating the constants,

Reverting to the original variables, 𝑐 𝐴

𝑐𝐴0=cosh [√𝑘′ ′ ′ 𝐿2𝐷 𝐴𝐵

(1− 𝑧𝐿 )]

cosh (√𝑘′ ′ ′ 𝐿2𝐷 𝐴𝐵)

III. Diffusion With a Homogeneous Chemical ReactionQuantities that might be asked for:

1. Average concentration in the liquid phase

𝑐𝐴 ,𝑎𝑣𝑔

𝑐 𝐴0=∫0

(𝑐𝐴 ¿𝑐 𝐴0)𝑑𝑧

𝑑𝑧= tanh 𝜙𝜙

2. Molar flux at the plane z = 0

𝑁 𝐴𝑧 ǀ 𝑧=0=−𝐷𝐴𝐵𝑑𝑐 𝐴

𝑑𝑧 ǀ𝑧=0=(𝑐𝐴0𝐷 𝐴𝐵

𝐿 )𝜙 tanh 𝜙

IV. Diffusion into a Falling Liquid Film (Gas Absorption)

Assumptions

1. Velocity field is unaffected by diffusion

2. A is slightly soluble in B3. Viscosity of the liquid is unaffected4. The penetration distance of A in B

will be small compared to the film thickness.

Recall: The velocity of a falling film

𝑣 𝑧 (𝑥 )=𝑣𝑚𝑎𝑥 [1−( 𝑥𝛿 )2]

𝑣 𝑧(𝑥 )=(𝜌 𝑔𝛿2 cos𝛼2𝜇 )[1−(𝑥𝛿 )2]

Concentration ProfilesIV. Diffusion into a Falling Liquid Film (Gas Absorption)

* CA is a function of both x and z

Dividing by WΔxΔz andletting Δx 0 and Δz 0,

𝜕𝑁𝐴𝑧

𝜕 𝑧 +𝜕𝑁 𝐴𝑥

𝜕 𝑥 =0

𝜕𝑁𝐴𝑧

𝜕 𝑥 =0

𝑁 𝐴𝑧=−𝐷𝐴𝐵𝑑𝑐𝐴

𝑑𝑧 +𝑥𝐴(𝑁 𝐴 𝑧+𝑁𝐵 𝑧)

The expressions for ,

Transport of A along the z direction is mainly by convection (bulk motion)

𝑁 𝐴𝑧 ≈𝑐𝐴𝑣𝑀=𝑐 𝐴𝑣𝑧 (𝑥)

𝑁 𝐴= 𝐽 𝐴∗+𝑐𝐴𝑣𝑀Recall: 𝑣𝑀=𝑚𝑜𝑙𝑎𝑟 𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝜕𝑁𝐴𝑧

𝜕 𝑥 =0

𝑁 𝐴𝑥=−𝐷 𝐴𝐵𝑑𝑐 𝐴

𝑑𝑧 +𝑥𝐴(𝑁 𝐴 𝑥+𝑁𝐵𝑥)

The expressions for ,

𝑁 𝐴𝑥 ≈−𝐷 𝐴𝐵𝑑𝑐 𝐴

𝑑𝑧

Transport of A along the x direction is mainly by diffusion

𝜕𝑁𝐴𝑧

𝜕 𝑥 =0

Substituting the expressions for,

𝑣 𝑧(𝜕𝑐𝐴

𝜕 𝑧 )=𝐷 𝐴𝐵𝜕2𝑐 𝐴

𝜕 𝑥2

Substituting the expressions vz,

𝑣𝑚𝑎𝑥 [1−( 𝑥𝛿 )2]( 𝜕𝑐 𝐴

𝜕 𝑧 )=𝐷 𝐴𝐵𝜕2𝑐𝐴

𝜕 𝑥2

𝑣𝑚𝑎𝑥 [1−( 𝑥𝛿 )2]( 𝜕𝑐𝐴

𝜕 𝑥2

Boundary conditions B.C. 1B.C. 2B.C. 3

B.C. 3

BUT we can replace B.C. 3 with

𝑣𝑚𝑎𝑥 [1−( 𝑥𝛿 )2]( 𝜕𝑐 𝐴

𝜕 𝑥2

𝑁 𝐴𝑥 ǀ 𝑥=0=−𝐷𝐴𝐵𝜕𝑐 𝐴

𝜕 𝑥 ǀ𝑥=0=𝑐𝐴0√ 𝐷𝐴𝐵𝑣𝑚𝑎𝑥

𝜋 𝑧

𝑐 𝐴

𝑐𝐴0=1−𝑒𝑟𝑓 𝑥

√ 4𝐷 𝐴𝐵2 𝑧

𝑣𝑚𝑎𝑥

=𝑒𝑟𝑓𝑐 𝑥

√ 4𝐷𝐴𝐵2 𝑧

𝑣𝑚𝑎𝑥

Reading Assignment

See analogous problem Example 4.1-1 of Transport Phenomena by Bird, Stewart and Lightfoot

Quantities that might be asked for:

1. Total molar flow of A across the surface at x = 0

𝑊 𝐴=∫0

𝑁𝐴𝑥 ǀ𝑥=0 𝑑𝑧𝑑𝑦=𝑊 𝑐𝐴 0√ 𝐷𝐴𝐵𝑣𝑚𝑎𝑥

𝜋 ∫0

𝐿 1√𝑧

𝑑𝑧=𝑐𝐴0√𝐷 𝐴𝐵𝑣𝑚𝑎𝑥

𝜋 𝐿

Overall Shell Mass Balances I

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