13 oct 2011prof. r. shanthini1 course content of mass transfer section lta diffusion theory of...

13 Oct 2011 Prof. R. Shanthini 1

Course content of Mass transfer section

L T A

Diffusion Theory of interface mass transfer Mass transfer coefficients, overall coefficients and transfer units

04 01 03

Application of absorption, extraction and adsorptionConcept of continuous contacting equipment

04 01 04

Simultaneous heat and mass transfer in gas-liquid contacting, and solids drying

04 01 03

CP302 Separation Process PrinciplesMass Transfer - Set 5


Liquid phase

Liquid film

Gas phase

pAb

CAiCAb

pAi

Gas film

Mass transport, NA

NA = kp (pAb – pAi) = kc (CAi – CAb )

pAi = HA CAi (53)

Summary: Two Film Theory applied at steady-state

(52) (59)(51) (62)

1

kp

HA

kc

+1

KG

pA* = HA CAb

pAb = HA CA*

KL

= =HA

(60)

(57)

(58 and 61)

= KG (pAb - pA*) = KL (CA

* - CAb)


Liquid phase

Liquid film

Gas phase

yAb

xAi

xAb

yAi

Gas film

Mass transport, NA

NA = ky (yAb – yAi) = kx (xAi – xAb )

yAi = KA xAi

Summary equations with mole fractions

yA* = KA xAb

yAb = KA xA*

= Ky (yAb - yA*) = Kx (xA

* - xAb)

1

ky

KA

kx

+1

Ky Kx

= =KA

(63)

(65)

(66)

(64)

(67)


Notations used:

xAb : liquid-phase mole fraction of A in the bulk liquid

yAb : gas-phase mole fraction of A in the bulk gas

xAi : liquid-phase mole fraction of A at the interface

yAi : gas-phase mole fraction of A at the interface

xA* : liquid-phase mole fraction of A which would have been in

equilibrium with yAb

yA* : gas-phase mole fraction of A which would have been in

equilibrium with xAb

kx : liquid-phase mass-transfer coefficient

ky : gas-phase mass-transfer coefficient

Kx : overall liquid-phase mass-transfer coefficient

Ky : overall gas-phase mass-transfer coefficient

KA : vapour-liquid equilibrium ratio (or equilibrium distribution coefficient)

13 Oct 2011 Prof. R. Shanthini 5xAi

yAi

yAb

xAb

yA*

xA* xA

yA

yAi = KA xAi

yA* = KA xAb

yAb = KAxA*

Gas-liquid equilibrium ratio (KA) curve


yAi

xA

yA

yAi = KA xAi

Gas-liquid equilibrium ratio (KA) curve

How to determine KA?


yAi

yAb

xAb

yA*

xA* xA

yA

slope my =yAb - yAi

xA * - xAi

slope mx =yAi - yA

*

xAi - xAb

Gas-liquid equilibrium ratio (KA) curve - nonlinear


xAi

yAi

yAb

xAb

yA*

xA* xA

yA

Gas-liquid equilibrium ratio curve (is not linear)

slope mx =yAi - yA

*

xAi - xAb

slope my =yAb - yAi

xA * - xAi

1

kx

1

myky

+1

Kx

= (67)

1

ky

mx

kx

+1

Ky

= (68)

when driving forces for mass transfer are large

Derivation is available on page 109 of Reference 2


Gas & Liquid-side Resistances in Interfacial Mass Transfer

1

KL

1

H kp

= +1

kc

1

KG

1

kp

= +H

kc

fG = fraction of gas-side resistance

=1/KG

1/kp

1/kp

1/kp=+ H/kc kc

kc=+ H kp

fL = fraction of liquid-side resistance

=1/KL

1/kc

1/Hkp

1/kc=+ 1/kc + kc/H

kp=kp


If fG > fL, use the overall gas-side mass transfer coefficient and the overall gas-side driving force.

If fL > fG use the overall liquid-side mass transfer coefficient and the overall liquid-side driving force.

Gas & Liquid-side Resistances in Interfacial Mass Transfer


1

kp

HA

kc

+1

KG KL

= =HA (58 and 61)

The above is also written with the following notations:

1

KOG

1

KG

= +H

KL

=H

KOL


Tutorial discussed.

Drive the relationship between the following under ideal conditions:

(i) kp and ky

(ii) kc and kx

(iii) KG and Ky

(iv) KL and Kx

(v) HA and KA


(i) Drive the relationship between kp and ky

(yAb – yAi)Therefore, kp = ky (pAb – pAi)

Start from NA = kp (pAb – pAi) = ky (yAb – yAi)

Since, the partial pressure (pA) can be related to the mole fraction in vapour phase (yA) and the total pressure in the vapour phase (P) by pA = yA P for an ideal gas, the above expression can be rewritten as follows:

kp = ky

(yAb – yAi)(yAb P – yAi P)

= ky / P

ky = kp PTherefore (69)


(ii) Drive the relationship between kc and kx

(xAi – xAb)Therefore, kc = kx (CAi – CAb)

Start from NA = kc (CAi – CAb) = kx (xAi – xAb)

Since, the concentration (CA) can be related to the mole fraction in liquid phase phase (xA) and the total concentration in the liquid phase (CT) by CA = xA CT, the above expression can be rewritten as follows:

kc = kx(xAi – xAb)

(xAi CT – xAb CT)= kx / CT

kx = kc CTTherefore (70)


(iii) Drive the relationship between KG and Ky

(yAb – yA*)

Therefore, KG = Ky (pAb – pA*)

Start from NA = KG (pAb – pA*) = Ky (yAb – yA

*)

using pA = yA P for an ideal gas, the above expression can be rewritten as follows:

KG = Ky

(yAb – yA*)

(yAb P – yA* P)

= Ky / P

Ky = KG PTherefore (71)


(iv) Drive the relationship between KL and Kx

(xA* – xAb)Therefore, KL = Kx (CA

* – CAb)

Start from NA = KL (CA* – CAb) = Kx (xA

*– xAb)

Since CA = xA CT, the above expression can be rewritten as follows:

KL = Kx(xA

* CT – xAb CT)

= Kx / CT

Kx = KL CTTherefore (72)

(xA* – xAb)


(v) Drive the relationship between HA and KA

Start from the equilibrium relationship pAi = HA CAi

Since pAi = yAi P and CAi = xAi CT, the above expression can be written as follows:

Therefore

yAi P = HA xAi CT

We know yAi = KA xAi

Combining the above, we get the following:

KA xAi P = HA xAi CT

KA = HA CT / P (73)True ONLY for dilute system


Example on calculating Henry’s constant:

Use the NH3-H2O data at 293 K given in the table below to calculate the Henry’s law constant (HA = pA / xA) at low concentrations of NH3, where pA is the equilibrium partial pressure of ammonia over aqueous solution having xA mole fraction of ammonia.

Wt NH3 per 100 wts. H2O

20.0 15.0 10.0 7.5 5.0 4.0 3.0 2.5 2.0

pA (mm Hg) 166 114 69.6 50 31.7 24.9 18.2 15.0 12.0


Solution to Example on calculating Henry’s constant:

The mass concentration data given in the table must be converted to mole fraction of ammonia in the liquid. It is done as follows: xA = (moles A) / (moles A + moles water)

= (mA / MA) / [(mA / MA) + mH2O / MH2O)]

= (20 / 17) / [(20 / 17) + 100 / 18)] for the first data point = 0.175

Wt NH3 per 100 wts. H2O

20.0 15.0 10.0 7.5 5.0 4.0 3.0 2.5 2.0

xA (mm Hg) 0.175 0.137 0.095 0.0735 0.0503 0.0401 0.0301 0.0258 0.0208

pA (mm Hg) 166 114 69.6 50 31.7 24.9 18.2 15.0 12.0

HA (mm Hg) 949 832 732 680 630 621 605 581 576


y = 839.47x

R2 = 0.9588

020406080

100120140160180

0 0.05 0.1 0.15 0.2xA (mole fraction)

p A (

mm

Hg)



y = 590.94x

R2 = 0.9836

02468

101214161820

0 0.01 0.02 0.03 0.04xA (mole fraction)

p A (

mm

Hg)

HA = 591 mm Hg per mol fraction of NH3 in water



HA = 591 mm Hg mole fraction of NH3 in water


HA = 591 mm Hg per mol fraction of NH3 in water

Henry’s law constant determined above is HA = pA / xA.

Use the relationship pA = yA PT, where pA is the partial pressure of NH3 in air, yA is the mol fraction NH3 in air and PT is the total pressure.

Henry’s law constant therefore becomes HA = yA PT / xA, from which we get KA = yA / xA = HA/PT

At 1 atm total pressure,

KA = 591 / 760

= 0.778 mol fraction NH3 in air / mol fraction NH3 in water


Example 1


Example 2


Example 3


Example 4

13 oct 2011prof. r. shanthini1 course content of mass transfer section lta diffusion theory of...

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