mass transfer - engineers · pdf fileper unit time is called mass transfer flux of a species....

CHEMICAL ENGINEERING GATE-2019 MASS TRANSFER

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MASS TRANSFER-MT

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T A B L E O F C O N T E N T

S. No. Title Page no.

1. Introduction 3

2. Diffusion 10

3. Drying and Humidification 24

4. Absorption and Stripping 34

5. Distillation 40

6. Extraction and Leaching 57

7. Adsorption 64

Annexure - A Important Formulae used in Mass Transfer 66

Annexure - B Practice Set 72

Annexure - C Gate Based Questions 86



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1. Introduction 3

2. Diffusion 10



5. Distillation 40


7. Adsorption 64






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1. Introduction 3

2. Diffusion 10



5. Distillation 40


7. Adsorption 64






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Chapter-1Introduction

The process of mass transfer occurs due to concentration difference of mixture components. Itoccurs from a region of higher concentration to a region of lower concentration. Concentration differenceis the pillar of mass transfer. Various separation techniques such as distillation, gas adsorption, liquidextraction, drying, crystallization, etc. are studied in Mass transfer, where operation may occurisothermally or non-isothermally. Mass transfer operation may occur simultaneously with heat transfer,for example; Drying, Humidification, Distillation, crystallization etc. Mass transfer operation may occurin one direction example, gas absorption. The contacting can be done in six ways namely, Gas-Gas, Gas-Liquid, Gas-solid, Liquid-Liquid and Liquid-Solid and Solid-Solid.(a) Liquid-Vapour Distillation.(b) Liquid-gas Gas absorption, stripping, Humidification and Dehumidification.(c) Liquid-Solid Crystallization, Leaching, Adsorption.(d) Liquid- Liquid Extraction.(e) Solid-Gas Adsorption, Drying.

The equillibrium between phases is attained after sufficiently long time. Since Mass transfer occursdue to both molecular diffusion and turbulence, the detailed study of factors affecting mass transfer is

studied. At the phase interface there is no resistance due to thermodynamic equillibrium (T, P, i ) at

interface. Rate of mass transfer is measured by deviation from equillibrium i.e. higher the deviation,higher is the driving force.

1.1. Some Important Definitions:

1) Distillation:Distillation is a vapour-liquid operation in

which the mixture components are separatedby use of thermal energy. When liquidmixture is heated, different components exertdifferent vapour pressure, expressed in termsof relative volatility. This pressure differenceresults in separation of components in such away that the top product contains higheramount of light component and bottomproducts contains higher amounts of heaviercomponent, as shown in figure 1.1. Adistillation example is separation of crudepetroleum into gasoline, kerosene, etc. Fig. 1.1. Schematic diagram of a distillation

column



3









column



3









column



4

2) Absorption and Stripping:

Gas absorption is a gas-liquid operation inwhich one or more constituents of a gasmixture are separated by using a suitable liquidsolvent i.e. component moves from gas phaseto liquid phase as shown in figure 1.2. Exampleof gas absorption methods is ammonia washingfrom ammonia-air mixture by means of water.Stripping is opposite of absorption i.e. acomponent moves from liquid phase to gasphase.

3) Liquid-Liquid Extraction:It is a liquid-liquid operation, also called

as solvent extraction, in which components of aliquid mixture are separated by treating it withsuitable solvent which dissolves one or moreconstituents of mixture more preferably. It is anefficient separation process in cases whereseparation is either not possible or noteconomical by using distillation eg. Separationof components of an azeotropic mixture.

4) Crystallization:It is liquid-solid operation in which we

obtain uniform crystals of good purity. Thesaturated liquid is subjected to changes intemperature and pressure in such a way thatcrystals get separated from the feed liquor asshown in the figure 1.4.

5) Drying:Drying is gas-solid operation in which a

relatively small amount of water is removedfrom solid material, by contacting it with acontinuous stream of gas (air) as shown infigure 1.5.

1.2. Important concepts to remember:

1) Ideal gas law :

PV = nRT

Where, P = Pressure (kPa), V = Volume (m3)

Fig. 1.2. Schematic diagram of an absorptioncolumn

Fig. 1.3. Schematic diagram of a Liquid-LiquidExtractor

Fig. 1.4. Schematic diagram of a crystallizer

Fig. 1.5. Schematic diagram of a dryer



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1) Ideal gas law :

PV = nRT








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1) Ideal gas law :

PV = nRT








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n = number of moles (kmol), T = Temperature (K)

R = 8.314512m kPa

kmol K

= Universal gas constant

2) Vapour Pressure v sA A(p or p ) and partial pressure A B(p , p ) : The vapour pressure of a liquid

is defined as the absolute pressure at which the liquid and its vapour are in equillibrium at giventemperature in a closed vessel. The partial pressure of a gas component that is present in a gaseousmixture, is a pressure that would be exerted by that component if it alone were present in the samevolume and at the same temperature.

vA A Ap x p

3) Dalton’s law : Dalton’s law mathematically is given by: A BP=p +p , Where P is the total pressure

exerted by gaseous mixture. A Bp and p are the partial pressures of component gases A and B

respectively.

4) More Volatile Component: More volatile component is lower boiling point component or thecomponent with higher vapour pressure at a given temperature. It is also called as the lightercomponent.

5) Less Volatile Component: In a binary system, it is the component with higher boiling point orlower vapour pressure at a given temperature. It is also called as the heavier component.

6) Concentration:6.1. Mass Concentration (Also known as mascon): It is defined as the mass of constituent divided bythe volume of mixture. It is denoted by i or i.

ii

m

V where, im mass of constituent i , V = volume of mixture.

For a pure chemical the mass concentration equals its density.

For Binary system having components A and B, mass density ( ρ ) of solution is given by-

A B A Bρ + ρ = ρ, where ρ and ρ are mass concentrations of A and B respectively.

6.2. Molar concentration (Molarity):Molar concentration is defined as the number of moles of species A per unit volume of the solution.

Mathematically, It is given as 3 A AA

A

n ρc (kmol/m )

V M

where, An = moles of species A, MA= molecular weight of component A

For a binary system of A and B, the total molar concentration of the solution is given by

A BC = c +c

6.3. Mass Fraction A(x ') :



5


R = 8.314512m kPa

kmol K




vA A Ap x p



respectively.




ii

m







A

n ρc (kmol/m )

V M



A BC = c +c




5


R = 8.314512m kPa

kmol K




vA A Ap x p



respectively.




ii

m







A

n ρc (kmol/m )

V M



A BC = c +c




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AA

A B

Wx '

W + W , where A BW and W are given weights of A and B respectively

6.4. Mole Fraction A(x ) : The mole fraction (xA) of species A can be defined as the ratio of number of

moles of component to the total no. of moles.

Mathematically, for a binary system , AA

A B

nx =

n + n

Note: A B A Bx + x =1 and x' + x' =1

In gas phase the concentrations are expressed in terms of partial pressures. In case of an ideal gas

A Ap V = n RT or A AA

pc = =

V RT

n

Where, Ap = partial pressure of species A in the mixture

An = number of moles of A

V= molar volume of mixture

T= absolute temperature (K)

R= universal gas constant

xA in terms of Ap is given as:

A A AA

c p /RT px = = =

C P/RT P

Where P is the total pressure exerted by the gas mixture.

A BA B

p p PC = c + c = + =

RT RT RT

7) Velocity of Species:Different chemical species are moving at different velocity in a diffusing mixture. The bulk velocityof the mixture would be some sort of an average velocity.For a mixture of n species local mass average velocity u is defined as

n

i ii = 1

n

ii =1

ρ uu =

ρ

Where ui is velocity of ith species.

In case of binary system, A A B Bρ u +ρ uu =

ρ

Local molar average velocity U of mixture is given by

n

i ii = 1

n

ii = 1

c UU =

c



6

AA

A B

Wx '





A B

nx =

n + n




pc = =

V RT

n







A A AA

c p /RT px = = =

C P/RT P


A BA B

p p PC = c + c = + =

RT RT RT


n

i ii = 1

n

ii =1

ρ uu =

ρ



ρ


n

i ii = 1

n

ii = 1

c UU =

c



6

AA

A B

Wx '





A B

nx =

n + n




pc = =

V RT

n







A A AA

c p /RT px = = =

C P/RT P


A BA B

p p PC = c + c = + =

RT RT RT


n

i ii = 1

n

ii =1

ρ uu =

ρ



ρ


n

i ii = 1

n

ii = 1

c UU =

c



7

In a binary system, A A B Bc U +c UU =

C8) Flux: Flux is a vector quantity. The amount of species (mass or molar units) that crosses a unit area

per unit time is called flux (mass transfer flux or molar flux respectively) of a species.

8.1. Mass Flux: The mass flux of species i is defined as the mass of species i that passes through a unit

area per unit time. It is also defined as mass flow rate per unit area.

(i) Mass flux relative to fixed coordinate is given by, i i i=ρ ui

For a binary system, the mass flux of A and B relative to stationary coordinate are

A A A B B B= u and = ρ u respectivelyi i

(ii) Mass flux relative to the mass average velocity u is given by, i ( )i ij u u For a binary system, the mass flux of A and B relative to mass average velocity are-

A A A B B Bj = ρ (u – u) and j = ρ (u – u) respectively

8.2. Molar Flux: Molar flux is defined as moles of species i that passes through a unit area per unit time.

It is also defined as molar flow rate per unit area.

(i) Molar flux relative to the stationary coordinate is given by, i i iN = c U

For a binary system, the molar fluxes of A and B with respect to stationary coordinates are-

A A A B B BN = c U and N =c U respectively

(ii) Molar flux relative to molar average velocity U (also known as bulk velocity) is given by,

i i iJ = c (U – U)

For a binary system, the molar fluxes of A and B with respect to an observer moving with bulk

velocity, A A AJ c (U U) and J =c (U –U) respectivelyB B B

1.3. Dimensionless numbers used in mass transfer:

(1) Sherwood number (Sh)

AB

k'L convective mass transportSh = , k' = Mass transfer coefficient

D molecular mass transport

Where L is the characteristics length

(2) Schmidt Number (Sc)

AB

ν 1 Momentum diffusivitySc = = =

D Mass diffusivityABD

(3) Prandtl Number (Pr)



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i i iJ = c (U – U)





AB





AB






7

















i i iJ = c (U – U)





AB





AB






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Momentum diffusivityPr =

α ThermaldiffusivitypC

K

(4) Lewis Number (Le)

AB P AB

Sc α K 1 Momentum diffusivityLe = =

Pr D ρC D Mass diffusivity

(5) Reynolds Number (Re)ρvD Inertial force

Reμ Viscous force

(6) Stanton number (St)k' Sh

Stυ ReSc

(7) Peclet number (Pe)Pe ReSc

1. Calculate the equillibrium composition of the liquid and vapour phase for a mixture of methyl alcoholand water at a temperature of 400 K and under a pressure of 45 kPa.Assume that both liquid and vapour behave ideallyData :Vapour pressure of water at 400 K = 15 kPaVapour pressure of methanol at 400 K = 60 kPa

Solution:Let x1 and y1 be mole fraction of methyl alcohol in liquid and vapour respectively.

1p partial pressure of methyl alcohol

v1 1 1 1p =p x =60x

2p water partial pressure

v2 2 2 1p =p x =15 1–x

and we know total pressure 1 2P p p

45 = 60x1 +15(1–x1)

45 – 15 = 60x1–15x1

30 = 45x1

x1 = 0.67

1 1 11

p p x 60 0.67y 0.89

P P 45

Hence at equillibrium:

Liquid phase = 0.67 mole fraction of methyl alcohol



8



K


AB P AB




Reμ Viscous force


Stυ ReSc





v1 1 1 1p =p x =60x


v2 2 2 1p =p x =15 1–x


45 = 60x1 +15(1–x1)

45 – 15 = 60x1–15x1

30 = 45x1

x1 = 0.67

1 1 11

p p x 60 0.67y 0.89

P P 45





8



K


AB P AB




Reμ Viscous force


Stυ ReSc





v1 1 1 1p =p x =60x


v2 2 2 1p =p x =15 1–x


45 = 60x1 +15(1–x1)

45 – 15 = 60x1–15x1

30 = 45x1

x1 = 0.67

1 1 11

p p x 60 0.67y 0.89

P P 45





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Vapour phase = 0.89 mole fraction of methyl alcohol.

2. A mixture of benzene and toluene boils at 380 K under a pressure of 101.325 kPa. Determine thecomposition of boiling liquid assuming that mixture obeys Raoult’s law. At 380 K the vapourpressure of benzene is 160 kPa and the toluene is 70 kPa.

Solution:

Let mole fraction of benzene in liquid xA

P = 100.325 kPa (given data)

Ap =160kPa

Bp =70 kPa

A B

vB

A v v

P– px =

p – p

A

101.325 70x = =0.348

160 70

Mole fraction of benzene in boiling liquid

= 0.348× 100 = 34.8

Hence composition of the boiling liquid = 34.8 mole%



9



Solution:



Ap =160kPa

Bp =70 kPa

A B

vB

A v v

P– px =

p – p

A

101.325 70x = =0.348

160 70


= 0.348× 100 = 34.8




9



Solution:



Ap =160kPa

Bp =70 kPa

A B

vB

A v v

P– px =

p – p

A

101.325 70x = =0.348

160 70


= 0.348× 100 = 34.8




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mass transfer - engineers · pdf fileper unit time is called mass transfer flux of a species....

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