cp302 masstransfer 02 ok

Prof. R. Shanthini 26 Feb 2013

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CP302 Separation Process PrinciplesMass Transfer - Set 2


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Diffusion of water through stagnant, non-diffusing air: Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed to be dry) is 1 atm and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path is 0.1524 m long. Calculate the rate of evaporation at steady state. The diffusivity of water vapour at 1 atm and 293 K is 0.250 x 10-4 m2/s. Assume that the vapour pressure of water at 293 K is 0.0231 atm.

Answer: 1.595 x 10-7 kmol/m2.s

Example 6.2.2 from Ref. 1 (from Set 1)


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Solution: The set-up of Example 6.2.2 is shown in the figure. Assuming steady state, equation (19) applies.

Water (A)

Air (B)

1

2

z2 – z1

NA = (pA1 - pA2 )

DAB P

RT(z2 – z1) pB,LM

(19)

Data provided are the following: DAB = 0.250 x 10-4 m2/s;

P = 1 atm; T = 293 K; z2 – z1 = 0.1524 m;

pA1 = 0.0231 atm (saturated vapour pressure);

pA2 = 0 atm (water vapour is carried away by air at point 2)

(pA1 – pA2 )

ln[(P - pA2 )/ (P - pA1 )]

pB,LM =

where


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Substituting the data provided in the equations given, we get the following:

NA = (0.0231 - 0) atm (0.250x10-4 m2/s)(1x1.01325x105 Pa)

(0.0231 – 0 )

ln[(1 - 0 )/ (1 – 0.0231 )]

pB,LM = = 0.988 atm

(8314 J/kmol.K) (293 K) (0.1524 m) (0.988 atm)

= 1.595 x 10-7 kmol/m2.s


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Diffusion in a tube with change in path length: Diffusion of water vapour in a narrow tube is occurring as in Example 6.2.2 under the same conditions. However, as shown in the figure, at a given time t, the level is z from the top. As diffusion proceeds the level drops slowly. Drive the equation for the time tF for the level to drop from a starting point z0 m at t = 0 to zF at t = tF s as shown.

Example 6.2.3 from Ref. 1

Water (A)

Air (B)

1

2

ZFz0

z

zF


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Solution: Since the level drops very slowly, we assume pseudo-steady-state condition. Therefore, equation (19) applies in which (z2 – z1) must be replaced by z.

Water (A)

Air (B)

1

2

ZFzF

z0

NA = (pA1 - pA2 ) RTz pB,LM

(22)

DAB P

z


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Water (A)

Air (B)

1

2

ZFzF

z0

Suppose that the level reduce by dz in dt time. Mass balance yields the following:

(NA x A x dt) x MA = ρA x (A x dz) where A is the cross-sectional area of the tube, MA is the molecular mass of water and ρA is the density of water.

(23)

z

dz


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Combining equations (22) and (23), we get

ρA dz = (pA1 - pA2 ) dt RTz pB,LM

(24)MA DAB P

Rearranging (24), we get

ρA z dz = (pA1 - pA2 ) dt RT pB,LM

MA DAB P⌠⌡z0

zF

⌠⌡ 0

tF

tF = (pA1 - pA2 )

ρA RT pB,LM

2 MA DAB P

ρA = (pA1 - pA2 ) tFRT pB,LM

MA DAB P(zF2 – z0

2)2

(zF2 – z0

2)

Is it okay to takepA1 and pA2 as constants?

This above equation is used to experimentally determine diffusivity DAB

(25)


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Evaporation of a Naphthalene sphere: A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and 1.01325 x 105 Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapour pressure at 318 K is 0.555 mm Hg. The DAB of naphthalene in air at 318 K is 6.92 x 10-6 m2/s. Calculate the rate of evaporation of naphthalene from the surface.

If the radius of the sphere decreases slowly with time, drive the equation for the time taken (tF) for the sphere to evaporate completely.


pA2

r1

rpA1

dr


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Therefore, we use the following definition:pA2

rpA1

dr

Solution: All this time the mass transfer of A is taken as NA (in moles/m2.s) and it is assumed to remain constant for systems at steady-state or pseudo steady-state while carrying out the integrations concerned. For the system considered here, we cannot do that since the area across which the mass transfer occurs vary as the radius changes.

NA = nA

AArea of mass transfer

Mass transfer per timeMass transfer per area per time (26)

For the given system, equation (26) gives

NA = nA

4πr2(27)

r1


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Setting NB = 0 and rearranging the above gives:pA2

rpA1

dr

Since the given system is a case of A diffusing through stagnant, non-diffusing B, we could start from equation (15).

which can be rearranged to give

(NA + NB)NA = - + (15)dpA

dz

DAB

RT

pA

P

NA = - dpA

(1 - pA/P) dz

DAB

RT

Using equation (26) and replacing dz by dr, the above could be written as follows:

NA = = - dpA

(1 - pA/P) dr

DAB

RT

nA

4πr2r1


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pA2

rpA1

dr

Integrating the above between r1 and some point r2 (which is a large distance away) gives the following:

(28)

= - dpA

(1 - pA/P)

DAB

RT

nA

4π

dr

r2

= - dpA

(1 - pA/P)

DAB

RT

nA

4π

dr

r2 ⌠⌡

r1

r2

⌠⌡pA1

pA2

- =DABP

RT

nA

4π1r1

1r2

P - pA2

P – pA1

r1


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pA2

rpA1

dr

Since r2 >> r1, 1/ r2 is much smaller than 1/ r1 in equation (28), and therefore neglected. So, equation (28) gives the following:

(29)

=DABP

RT

nA

4πr1

P - pA2

P – pA1

= DABP

RTr1

nA

4πr12

pA1 - pA2

pB,LMNA =

Using equation (27) and the definition of pB,LM, the above could be written as

Equation (29) gives the rate of evaporation of naphthalene from the surface at r1 radius. r1


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Data provided are the following: DAB = 6.92 x 10-6 m2/s;

P = 1 atm; T = 318 K; r1 = 2 mm = 2/1000 m;

pA1 = 0.555/760 atm (saturated vapour pressure);

pA2 = 0 atm (no naphthalene vapour far away from the sphere)

Substituting the data provided, we get the following:

NA = (0.555/760 - 0) atm (6.92x10-6 m2/s)(1.01325x105 Pa)

(0.555/760 – 0 )

ln[(1 - 0 )/ (1 – 0.555/760 )]

pB,LM = = 0.9996 atm

(8314 J/kmol.K) (318 K) (2/1000 m) (0.9996 atm)

= 9.686 x 10-8 kmol/m2.s


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pA2

pA1

d(30) =

DABPRT

nA

4π2

pA1 - pA2

pB,LMNA =

Since the radius of the sphere is said to decrease slowly, we assume pseudo-steady-state condition. Therefore, equation (29) applies in which r1 is replaced by as follows:

Sphere is enlarged.

r1

Now, we need to drive the equation for the time taken (tF) for the sphere to evaporate completely. That is, the radius of the sphere is r1 at t = 0 and it is zero at t = tF.


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pA2

pA1

d

Sphere is enlarged.

r1

Suppose that the surface reduce by d in dt time. Mass balance yields the following:

where A is the surface area of the sphere at radius , MA is the molecular mass of the sphere and ρA is the density of the sphere.

(31)(NA x A x dt) x MA = ρA x (A x (-d))


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Combining equations (30) and (31), we get

- ρA d = (pA1 - pA2 ) dt RT pB,LM

MA DAB P

pA2

r1

pA1

d

Sphere is enlarged.

Rearranging and integrating, we get

- ρA d = (pA1 - pA2 ) dt RT pB,LM

MA DAB P⌠⌡r1

0

⌠⌡ 0

tF

Is it okay to assume that pA1 and pA2 remain constants as the radius of the sphere () reduces from r1 to 0?


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Integration gives the following results:

pA2

r1

pA1

d

Sphere is enlarged.

tF = (pA1 - pA2 )

ρA r12 RT pB,LM

2 MA DAB P

ρA = (pA1 - pA2 ) tFRT pB,LM

MA DAB P(r12 – 0)2

(32)

The above gives the time taken for the sphere to vanish completely at a slow rate of evaporation.


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Objectives of the slides that follow:

Estimating diffusivities


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Diffusivities for different systems could be estimated using the empirical equations provided in the following slides as well as those provided in other reference texts available in the library and other sources.

Estimating Diffusivity


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Diffusivity of gases

An example at 1 atm and 298 K:

System Diffusivity (cm2/s)

H2-NH3 0.783

H2-CH4 0.726

Ar-CH4 0.202

He-CH4 0.675

He-N2 0.687

Air-H2O 0.260

Air-C2H5OH 0.135

Air-benzene 0.0962


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DAB - diffusivity in cm2/s

P - absolute pressure in atmMi - molecular weight

T - temperature in KVi - sum of the diffusion volume for component i

DAB is proportional to 1/P and T1.75

Binary Gas Diffusivity


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Binary Gas Diffusivity


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T - temperature in Kμ - viscosity of solution in kg/m sVA - solute molar volume at its normal boiling point

in m3/kmol

DAB is proportional to 1/μ and T

DAB =9.96 x 10-12 T

μ VA1/3

For very large spherical molecules (A) of 1000 molecular weight or greater diffusing in a liquid solvent (B) of small molecules:

applicable for biological

solutes such as proteins

Diffusivity in Liquids


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MB - molecular weight of solvent B

T - temperature in Kμ - viscosity of solvent B in kg/m sVA - solute molar volume at its normal boiling point in m3/kmol

Φ - association parameter of the solvent, which 2.6 for water, 1.9 for methanol, 1.5 for ethanol, and so on

DAB is proportional to 1/μB and T

DAB =1.173 x 10-12 (Φ MB)1/2 T

μB VA0.6

For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B):

applicable for biological solutes

Diffusivity in Liquids


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DoAB is diffusivity in cm2/s

n+ is the valence of cation

n- is the valence of anion

λ+ and λ- are the limiting ionic conductances in very dilute

solutions T is 298.2 when using the above at 25oC

DAB is proportional to T

DoAB =

8.928 x 10-10 T (1/n+ + 1/n-)

For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B):

(1/λ+ + 1/ λ-)

Diffusivity of Electrolytes in Liquids


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Objectives of the slides that follow:

Mathematical modelling of steady-state one dimensional diffusive mass transfer in solids


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Diffusion in solids are occurring at a very slow rate.

However, mass transfer in solids are very important.

Examples:

Leaching of metal oresDrying of timber, and foodsDiffusion and catalytic reaction in solid catalystsSeparation of fluids by membranes

Treatment of metal at high temperature by gases.

Diffusion in solids


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Diffusion in solids

Diffusion in solids occur in two different ways:

- Diffusion following Fick’s law (does not depend on the structure of the solid)

- Diffusion in porous solids where the actual structure and void channels are important


30

Diffusion in solids following Fick’s Law

NA = + (13)dCA

dz-DAB (NA + NB)

CT

CA

Start with equation (13) from Set 1:

Bulk term is set to zero in solids

NA = dCA

dz-DAB

Therefore, the following equation will be used to describe the process:

(33)


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Diffusion through a slab

NA = (CA1 - CA2)

Applying equation (33) for steady-state diffusion through a solid slab, we get

(34)

z2-z1

CA2

CA1

z2 - z1

where NA and DAB are taken as constants.

Drive (34) and compare it with heat conduction equivalent.

DAB


32

Relating the concentration and solubilityThe solubility of a solute gas in a solid is usually expressed by the notation S.

m3 solute at STP

m3 solid . atm partial pressure of solute

CA = kmol solute /m3 solid S pA

22.414

STP of 0oC and 1 atm

Unit used in general is the following:

Relationship between concentration and solubility:

where pA is in atm


33

Relating the concentration and permeabilityThe permeability of a solute gas (A) in a solid is usually expressed by the notation PM. in m3 solute at STP (0oC and 1 atm) diffusing per second per m2 cross-sectional area through a solid 1 m thick under a pressure difference of 1 atm.

m3 solute at STP . 1 m thick solid

s . m2 cross-sectional area . atm pressure difference

PM = DAB S

Unit used in general is the following:

Relationship between concentration and permeability:

where DAB is in m2/s and S is in m3/m3.atm


34

Diffusion of H2 through Neoprene membrane: The gas hydrogen at 17oC and 0.010 atm partial pressure is diffusing through a membrane on vulcanized neoprene rubber 0.5 mm thick. The pressure of H2 on the other side of 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 17oC is 0.051 m3 (at STP of 0oC and 1 atm)/m3 solid. atm and the diffusivity DAB is 1.03 x 10-10 m2/s at 17oC.

Answer: 4.69 x 10-12 kmol H2/m2.s



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Diffusion through a packging film using permeability: A polythene film 0.00015 m (0.15 mm) thick is being considered for use in packaging a pharmaceutical product at 30oC. If the partial pressure of O2 outside the package is 0.21 atm and inside it is 0.01 atm, calculate the diffusion flux of O2 at steady state. Assume that the resistances to diffusion outside the film and inside are negligible compared to the resistance of the film. Permeability of O2 in polythene at 303 K is 4.17 x 10-12 m3 solute (STP)/(s.m2.atm.m).

Answer: 2.480 x 10-12 kmol O2/m2.s

Would you prefer nylon to polythene? Permeability of O2 in nylon at 303 K is 0.029 x 10-12 m3 solute (STP)/(s.m2.atm.m). Support your answer.



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Diffusion through a cylinder wall

CA2

Applying equation (33) for steady-state diffusion through a cylinder wall of inner radius r1 and outer radius r2 and length L in the radial direction outward, we get

r1r2

r

NA = = dCA

dr-DAB

(35)nA

2 π r L

Area of mass transfer

Mass transfer per time

CA1

Mass transfer per area per time

nA = 2πL DAB(CA1 - CA2) (36)ln(r2 / r1)



37

Diffusion through a spherical shell

CA2

Applying equation (33) for steady-state diffusion through a spherical shell of inner radius r1 and outer radius r2 in the radial direction outward, we get

r1r2

r

NA = = dCA

dr-DAB

(37)nA

4 π r2

Area of mass transfer

Mass transfer per time

CA1

Mass transfer per area per time

nA = 4πr1r2 DAB(CA1 - CA2) (38)(r2 - r1)



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Diffusion in porous solids

Read Section 6.5C (page 445 to 446) of Ref. 1

cp302 masstransfer 02 ok

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