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CP302 Separation Process Principles. Mass Transfer - Set 5. Summary: Two Film Theory applied at steady-state. = K G ( p Ab - p A * ) = K L (C A * - C Ab ). N A = k p (p Ab – p Ai ) = k c (C Ai – C Ab ). (52). (51). (59). (62). p Ai = H A C Ai. (53). Liquid film. - PowerPoint PPT Presentation
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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
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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)
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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)
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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
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yAi
xA
yA
yAi = KA xAi
Gas-liquid equilibrium ratio (KA) curve
How to determine KA?
13 Oct 2011 Prof. R. Shanthini 7xAi
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
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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
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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
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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
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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
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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
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(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)
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(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)
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(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)
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(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)
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(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
13 Oct 2011 Prof. R. Shanthini 18
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
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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
13 Oct 2011 Prof. R. Shanthini 20
y = 839.47x
R2 = 0.9588
020406080
100120140160180
0 0.05 0.1 0.15 0.2xA (mole fraction)
p A (
mm
Hg)
Solution to Example on calculating Henry’s constant:
13 Oct 2011 Prof. R. Shanthini 21
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
Solution to Example on calculating Henry’s constant:
13 Oct 2011 Prof. R. Shanthini 22
HA = 591 mm Hg mole fraction of NH3 in water
Solution to Example on calculating Henry’s constant:
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
13 Oct 2011 Prof. R. Shanthini 23
Example 1
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Example 2
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Example 3
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Example 4
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