CP302 Separation Process Principles
Mass Transfer - Set 8
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
24 Nov 2011
Prof. R. Shanthini
1
Treated gas
Gout, yout
We have learned to determine
the height of packing in packed
columns with dilute solutions in
the last lecture class.
G
y
Inlet gas
Gin, yin
Prof. R. Shanthini
L
x
Control
volume
Today, we will learn to determine
the height of packing in packed
columns with concentrated
solutions.
24 Nov 2011
Inlet solvent
Lin, xin
Spent solvent
Lout, xout
2
Notations
Gs
Ls
G
L
Y
y
X
x
- inert gas molar flow rate (constant)
- solvent molar flow rate (constant)
- total gas molar flow rate (varies as it looses the solute)
- total liquid molar flow rate (varies as it absorbs the solute)
- mole ratio of solute A in gas
= moles of A / moles of inert gas
- mole fraction of solute A in gas
= moles of A / (moles of A + moles of inert gas)
- mole ratio of solute A in liquid
= moles of A / moles of solvent
- mole fraction of solute A in liquid
= moles of A / (moles of A + moles of solvent)
Solute in the gas phase = Gs Y = G y
Solute in the liquid phase = Ls X = L x
24 Nov 2011
Prof. R. Shanthini
3
Equations for Packed Columns
Mass of solute lost from the
gas over the differential height
of packing dz
= G y - G (y + dy) = - G dy
Treated gas
Gout, yout
was used for packed column
with dilute solution assuming G
can be taken as a constant for
dilute solutions.
For concentrated solution we
ought to use Gs (inert gas molar
flow rate) which is constant
across the column along with Y
(mole ratio of solute A in gas).
24 Nov 2011
Prof. R. Shanthini
Inlet solvent
Lin, xin
G
y
L
x
dz
G
y+dy
L
x+dx
Z
z
Inlet gas
Gin, yin
Spent solvent
Lout, xout
4
Equations for Packed Columns with concentrated solutions
Mass of solute lost from the
gas over the differential height
of packing dz
= Gs Y - Gs (Y + dY) = - Gs dY
Treated gas
Gout, yout
(85)
Relate Gs to G:
Gs = G (1 – y)
Gs
Y
Inlet solvent
Lin, xin
Ls
X
dz
(86)
Gs
Y+dY
Relate Y to y:
Ls
X+dX
Z
z
From y = Y / (Y+1), we get
Y = y / (1 – y)
24 Nov 2011
(87)
Prof. R. Shanthini
Inlet gas
Gin, yin
Spent solvent
Lout, xout
5
Equations for Packed Columns with concentrated solutions
Using (86) and (87), mass of
solute lost from the gas over
the differential height of
packing dz, given by (85), can
be written as follows:
Treated gas
Gout, yout
-Gs dY = - G (1 – y) d[y / (1 – y)]
= - G (1 – y)
=-G
dy
(1 – y)
dy
(1 – y)2
Gs
Y
Ls
X
dz
Gs
Y+dY
Ls
X+dX
Z
z
(88)
Inlet gas
Gin, yin
24 Nov 2011
Inlet solvent
Lin, xin
Prof. R. Shanthini
Spent solvent
Lout, xout
6
Equations for Packed Columns with concentrated solutions
Mass of solute transferred from Treated gas
Gout, yout
the gas to the liquid
Inlet solvent
Lin, xin
= Kya (y – y*) S dz
where S is the inside crosssectional area of the tower.
Gs
Y
Ls
X
dz
Relating (88) to the above at
steady state, we get
-G
dy
(1 – y)
= Kya (y – y*) S dz
Compare (89) with (79) used for
packed columns with dilute
solutions. What are the differences?
24 Nov 2011
Prof. R. Shanthini
Gs
Y+dY
Ls
X+dX
Z
z
(89)
Inlet gas
Gin, yin
Spent solvent
Lout, xout
7
Equations for Packed Columns with concentrated solutions
Rearranging and integrating
(89) gives the following:
Treated gas
Gout, yout
Inlet solvent
Lin, xin
yin
Z=
∫
dy
G
KyaS (1 – y)(y – y*)
(90)
Gs
Y
Ls
X
yout
dz
Gs
Y+dY
Inlet gas
Gin, yin
24 Nov 2011
Prof. R. Shanthini
Ls
X+dX
Z
z
Spent solvent
Lout, xout
8
Equations for Packed Columns with concentrated solutions
Multiply the numerator and denominator of (90) by (1 – y)LM:
yin
Z=
∫
(1 – y)LM dy
G
Kya(1 – y)LMS (1 – y)(y – y*)
(91)
yout
where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as
follows:
(1 – y)LM
24 Nov 2011
y* – y
=
ln[(1 – y )/(1 – y* )]
Prof. R. Shanthini
(92)
9
Equations for Packed Columns with concentrated solutions
Even though G and (1 – y)LM are not constant across the
column, we can consider the ratio of the two to be a constant
and take G / [Kya(1 – y)LMS] out of the integral sign in (91)
without incurring errors larger than those inherent in
experimental measurements of Kya. (Usually average values
of G and (1 – y)LM are used.)
Therefore (91) becomes the following:
yin
Z = K a(1G– y) S
y
LM
∫
(1 – y)LM dy
(1 – y)(y – y*)
(93)
yout
NOG
HOG
24 Nov 2011
Prof. R. Shanthini
10
Equations for Packed Columns with concentrated solutions
y* in (93) can be related to the bulk concentration using the
equilibrium relationship as follows:
y* = K x
(94)
x in (94) can be related to y in (93) using the operating line
equation.
We will determine the operating line equation next
24 Nov 2011
Prof. R. Shanthini
11
Equations for Packed Columns with concentrated solutions
The operating equation for the Treated gas
packed column is obtained by Gout, yout
writing a mass balance for
solute over the control volume:
Lin xin + G y = L x + Gout yout
Inlet solvent
Lin, xin
(74)
If dilute solution is assumed, then
Lin = L = Lout and Gin = G = Gout.
G
y
L
x
Control
volume
We somehow have to relate L to
Lin and G to Gout in (74).
Inlet gas
Gin, yin
24 Nov 2011
Prof. R. Shanthini
Spent solvent
Lout, xout
12
Equations for Packed Columns with concentrated solutions
To relate L to Lin, write a mass
balance for solvent over the
control volume:
Treated gas
Gout, yout
Inlet solvent
Lin, xin
Lin (1 – xin) = L (1 – x)
L = Lin (1 – xin) / (1 – x)
(95)
G
y
To relate G to Gout, write the
overall mass balance over the
entire column:
G + Lin = Gout + L
Control
volume
(96)
Inlet gas
Gin, yin
24 Nov 2011
L
x
Prof. R. Shanthini
Spent solvent
Lout, xout
13
Equations for Packed Columns with concentrated solutions
Use (95) to eliminate L from (96):
G + Lin = Gout + Lin (1 – xin) / (1 – x)
G = Gout + Lin (x – xin) / (1 – x)
(97)
Combining (74), (95) and (97), we get the following:
Lin xin + [Gout+Lin(x – xin)/ (1 – x)] y = Lin(1 – xin)x/(1 – x) + Gout yout
Gout yout + [Lin (1 – xin) x / (1 – x)] - Lin xin
y=
Gout + [Lin(x – xin) / (1 – x)]
(98)
Equation (98) is the operating line equation for packed columns
with concentrated solutions. Compare it with equation (76) used
for packed columns with dilute solutions. What are the differences?
24 Nov 2011
Prof. R. Shanthini
14
Equations for Packed Columns with concentrated solutions
If the solvent fed to the column is pure then xin = 0.
Therefore, (98) becomes
Gout yout + [ Lin x / (1 – x) ]
y=
Gout + [ Lin x / (1 – x) ]
24 Nov 2011
Prof. R. Shanthini
(99)
15
Example 1:
Draw the operating curve for a system where 95% of the
ammonia from an air stream containing 40% ammonia by
volume is removed in a packed column. Solvent used in 488
lbmol/h per 100 lbmol/h of entering gas.
Solution:
Lin = 488 lbmol/h;
Gin = 100 lbmol/h;
yin = 0.4;
xin = 0
In the inlet air stream: ammonia = 40 lbmol/h; air = 60 lbmol/h
Ammonia removed from the air stream = 0.95 x 40 = 38 lbmol/h
In the outlet air stream: ammonia = 02 lbmol/h; air = 60 lbmol/h
Therefore, Gout = (60+2) = 62 lbmol/h, and
yout = 2 / 62 = 0.0323
24 Nov 2011
Prof. R. Shanthini
16
Example 1:
Using Lin = 488 lbmol/h, Gout = 62 lbmol/h, yout = 0.0323 and
xin = 0 in (98), we get the operating curve as follows:
y
62 x 0.0323 + [488 x / (1 – x)]
y=
62 + [488 x / (1 – x)]
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
yin (bottom of
the tower)
Operating curve
0
0.02
0.04
0.06
0.08
0.1
yout (top of the
tower)
x
24 Nov 2011
Prof. R. Shanthini
17
Example 2:
Draw the equilibrium curve, which is approximately described
by K = 44.223x + 0.4771, on the same plot as in Example 1.
y
Solution:
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
Operating curve
Equilibrium curve
y=Kx
= 44.223x2 + 0.4771x
0
0.02
0.04
0.06
0.08
0.1
x
24 Nov 2011
Prof. R. Shanthini
18
Example 3:
Determine the NOG using the data given in Examples 1 and 2.
Solution:
yin
NOG =
∫
(1 – y)LM dy
(1 – y)(y – y*)
yout
where (1 – y)LM
y* – y
=
ln[(1 – y )/(1 – y* )]
Given x, calculate y from the operating curve and y* from
the equilibrium curve. Using those values, determine the
integral above that gives NOG.
24 Nov 2011
Prof. R. Shanthini
19
Example 3:
The shaded area gives NOG as 3.44
(Numerical integration is better suited
to get the answer.)
35
Integrand
30
25
20
15
10
5
0
0
0.1
yout = 0.0323
24 Nov 2011
0.2
y
Prof. R. Shanthini
0.3
0.4
yin = 0.4
20
Example 4:
Determine the height of packing Z using the data given in
Examples 1 and 2.
Solution:
HOG =
G
Kya(1 – y)LMS
Z = NOG HOG
24 Nov 2011
Need more data to work it out.
Can be calculated once HOG is
known.
Prof. R. Shanthini
21
Summary with overall gas-phase transfer coefficients for
packed column with concentrated solutions
yin
G
Z = K a(1 – y) S
y
LM
∫
(1 – y)LM dy
(1 – y)(y – y*)
(93)
yout
NOG
HOG
where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as
follows:
y* – y
(1 – y)LM =
(92)
ln[(1 – y )/(1 – y* )]
24 Nov 2011
Prof. R. Shanthini
22
Summary with overall liquid-phase transfer coefficients for
packed column with concentrated solutions
L
Z = K a(1 – x) S
x
LM
xout
(1 – x)LM dy
(1 – x)(x* – x)
∫
(93)
xin
NOL
HOL
where (1 – x)LM is the log mean of (1 – x) and (1 – x*) given as
follows:
x* – x
(1 – x)LM =
(92)
ln[(1 – x )/(1 – x* )]
24 Nov 2011
Prof. R. Shanthini
23
Summary: Equations for Packed Columns for dilute solutions
Distributed already:
Photocopy of Table 16.4 Alternative mass transfer
coefficient groupings for gas absorption
from
Henley EJ and Seader JD, 1981, Equilibrium-Stage
Separation Operations in Chemical Engineering, John
Wiley & Sons.
24 Nov 2011
Prof. R. Shanthini
24
Gas absorption, Stripping and Extraction
Gas absorption:
NOG and HOG are used
Stripping:
NOL and HOL are used
Extraction:
NOL and HOL are used
Humidification:
NG and HG are used.
24 Nov 2011
Prof. R. Shanthini
25