Transcript C H A P T E R 13 The Transfer of Heat

CP302 Separation Process Principles
Mass Transfer - Set 5
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
adsorption
Concept of continuous contacting equipment
04
01
04
Simultaneous heat and mass transfer in gasliquid contacting, and solids drying
04
01
03
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Summary: Two Film Theory applied at steady-state
NA = kp (pAb – pAi) = kc (CAi – CAb ) = KG (pAb - pA*) = KL (CA* - CAb)
(52)
(51)
pAi = HA CAi
(59)
(53)
pAb = HA CA*
(60)
pA* = HA CAb
(57)
1
KG
=
HA
KL
=
(62)
1
kp
+
HA
kc
Liquid
film
Gas
film
Liquid
phase
pAb
Gas
phase
pAi
CAi
CAb
(58 and 61)
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Mass transport, NA
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Summary equations with mole fractions
NA = ky (yAb – yAi) = kx (xAi – xAb )
= Ky (yAb - yA*) = Kx (xA* - xAb)
(63)
yAi = KA xAi
(64)
yAb = KA xA*
(65)
yA* = KA xAb
(66)
1
Ky
=
KA
Kx
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=
1
ky
+
KA
kx
Liquid
film
Gas
film
Liquid
phase
yAb
(67)
Gas
phase
yAi
xAi
xAb
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Mass transport, NA
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Notations used:
xAb
yAb
xAi
yAi
xA*
yA*
kx
ky
Kx
Ky
KA
: liquid-phase mole fraction of A in the bulk liquid
: gas-phase mole fraction of A in the bulk gas
: liquid-phase mole fraction of A at the interface
: gas-phase mole fraction of A at the interface
: liquid-phase mole fraction of A which would have been in
equilibrium with yAb
: gas-phase mole fraction of A which would have been in
equilibrium with xAb
: liquid-phase mass-transfer coefficient
: gas-phase mass-transfer coefficient
: overall liquid-phase mass-transfer coefficient
: overall gas-phase mass-transfer coefficient
: vapour-liquid equilibrium ratio (or equilibrium distribution
coefficient)
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Gas-liquid equilibrium ratio (KA) curve
yA
yAb
yAb = KAxA*
yAi
yA*
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yAi = KA xAi
yA* = KA xAb
xAb
xAi
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x A*
xA
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Gas-liquid equilibrium ratio (KA) curve
yA
How to determine KA?
yAi
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yAi = KA xAi
xAi
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xA
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Gas-liquid equilibrium ratio (KA) curve - nonlinear
yA
yAb
slope my =
yAb - yAi
xA * - xAi
yAi
yAi - yA*
slope mx = x - x
Ai
Ab
*
yA
xAb
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xAi
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x A*
xA
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Gas-liquid equilibrium ratio curve (is not linear)
yA
when driving forces for
mass
transfer are large
y
Ab
slope my =
1
1
=
+
Kx
kx
yAi
1
myky
yAb - yAi
xA * - xAi
(67)
yAi - yA*
slope mx = x - x
Ai
Ab
*
yA
xAi
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1
1
mx
xAbKy = ky +xA* kx
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xA
Derivation is available on page 109 of Reference 2
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Gas & Liquid-side Resistances in Interfacial Mass Transfer
1
KG
=
1
kp
+
H
kc
fG = fraction of gas-side resistance
1/kp
=
1/KG
1
KL
=
1
H kp
1/kp
=
1/kp + H/kc
+
=
kc
kc + H kp
1
kc
fL = fraction of liquid-side resistance
1/kc
=
1/KL
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1/kc
=
=
1/Hkp + 1/kc
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kp
kp + kc/H
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Gas & Liquid-side Resistances in Interfacial Mass Transfer
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.
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1
=
KG
HA
=
KL
1
+
kp
HA
kc
(58 and 61)
The above is also written with the following notations:
1
KOG
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=
H
KOL
=
1
KG
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+
H
KL
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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
Start from NA = kp (pAb – pAi) = ky (yAb – yAi)
Therefore, kp = ky
(yAb – yAi)
(pAb – pAi)
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:
(yAb – yAi)
kp = ky
= ky / P
(yAb P – yAi P)
Therefore
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ky = kp P
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(69)
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(ii) Drive the relationship between kc and kx
Start from NA = kc (CAi – CAb) = kx (xAi – xAb)
(xAi – xAb)
Therefore, kc = kx
(CAi – CAb)
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
Therefore
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(xAi – xAb)
= kx / CT
(xAi CT – xAb CT)
kx = kc CT
Prof. R. Shanthini
(70)
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(iii) Drive the relationship between KG and Ky
Start from NA = KG (pAb – pA*) = Ky (yAb – yA*)
(yAb – yA*)
Therefore, KG = Ky
(pAb – pA*)
using pA = yA P for an ideal gas, the above expression
can be rewritten as follows:
(yAb – yA*)
KG = Ky
(yAb P – yA* P)
Therefore
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Ky = KG P
Prof. R. Shanthini
= Ky / P
(71)
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(iv) Drive the relationship between KL and Kx
Start from NA = KL (CA* – CAb) = Kx (xA*– xAb)
(xA* – xAb)
Therefore, KL = Kx
(CA* – CAb)
Since CA = xA CT, the above expression can be rewritten
as follows:
KL = Kx
Therefore
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(xA* – xAb)
(xA* CT
– xAb CT)
Kx = KL CT
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= Kx / CT
(72)
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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:
yAi P = HA xAi CT
We know
yAi = KA xAi
Combining the above, we get the following:
KA xAi P = HA xAi CT
Therefore
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KA = HA CT / P
(73)
True ONLY for dilute system
Prof. R. Shanthini
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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
20.0
100 wts. H2O
15.0
10.0
7.5
5.0
4.0
pA (mm Hg)
114
69.6
50
31.7
24.9 18.2
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3.0
2.5
2.0
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
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pA (mm Hg)
Solution to Example on calculating Henry’s constant:
180
160
140
120
100
80
60
40
20
0
y = 839.47x
2
R = 0.9588
0
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0.05
0.1
0.15
xA (mole fraction)
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0.2
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Solution to Example on calculating Henry’s constant:
pA (mm Hg)
HA = 591 mm Hg per mol fraction of NH3 in water
20
18
16
14
12
10
8
6
4
2
0
y = 590.94x
2
R = 0.9836
0
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0.01
0.02
0.03
xA (mole fraction)
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0.04
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Solution to Example on calculating Henry’s constant:
HA591
= 591
mm
mole
ofof
NH
water
HA =
mm
HgHg
per
molfraction
fraction
NH
water
3 in
3 in
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
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Example 1
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Example 2
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Example 3
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Example 4
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