Post on 26-Dec-2015
24 Nov 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 8
24 Nov 2011 Prof. R. Shanthini 2
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gy
Lx
Control volume
We have learned to determine the height of packing in packed columns with dilute solutions in the last lecture class.
Today, we will learn to determine the height of packing in packed columns with concentrated solutions.
24 Nov 2011 Prof. R. Shanthini 3
Notations
Gs - inert gas molar flow rate (constant)
Ls - solvent molar flow rate (constant)
G - total gas molar flow rate (varies as it looses the solute) L - total liquid molar flow rate (varies as it absorbs the solute) Y - mole ratio of solute A in gas
= moles of A / moles of inert gas y - mole fraction of solute A in gas
= moles of A / (moles of A + moles of inert gas) X - mole ratio of solute A in liquid
= moles of A / moles of solvent x - 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 4
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gy
Lx
Mass of solute lost from the gas over the differential height of packing dz = G y - G (y + dy) = - G dy
was used for packed column with dilute solution assuming G can be taken as a constant for dilute solutions.
Gy+dy
Lx+dx
z
dz
Z
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).
Equations for Packed Columns
24 Nov 2011 Prof. R. Shanthini 5
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gs
YLs
X
Mass of solute lost from the gas over the differential height of packing dz
= Gs Y - Gs (Y + dY) = - Gs dY
Gs
Y+dY
Ls
X+dXz
dz
Z
Relate Gs to G:
Gs = G (1 – y)
Relate Y to y:
From y = Y / (Y+1), we get
Y = y / (1 – y)
(85)
(86)
(87)
Equations for Packed Columns with concentrated solutions
24 Nov 2011 Prof. R. Shanthini 6
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gs
Y
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:
-Gs dY = - G (1 – y) d[y / (1 – y)]
Gs
Y+dY
Ls
X+dXz
dz
Z
Equations for Packed Columns with concentrated solutions
= - G (1 – y)(1 – y)2
dy
= - G(1 – y)
dy(88)
Ls
X
24 Nov 2011 Prof. R. Shanthini 7
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gs
Y
Mass of solute transferred from the gas to the liquid
= Kya (y – y*) S dz
where S is the inside cross-sectional area of the tower.
Gs
Y+dY
Ls
X+dXz
dz
ZRelating (88) to the above at steady state, we get
-G = Kya (y – y*) S dz (89)
Equations for Packed Columns with concentrated solutions
(1 – y)
dy
Ls
X
Compare (89) with (79) used for packed columns with dilute solutions. What are the differences?
24 Nov 2011 Prof. R. Shanthini 8
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gs
YLs
X
Gs
Y+dY
Ls
X+dXz
dz
Z
Equations for Packed Columns with concentrated solutions
Rearranging and integrating (89) gives the following:
KyaSGZ =
dy(1 – y)(y – y*)∫
yout
yin
(90)
24 Nov 2011 Prof. R. Shanthini 9
Equations for Packed Columns with concentrated solutions
Kya(1 – y)LMSGZ =
(1 – y)LM dy(1 – y)(y – y*)∫
yout
yin
(91)
Multiply the numerator and denominator of (90) by (1 – y)LM:
where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as follows:
(1 – y)LM
y* – y
ln[(1 – y )/(1 – y* )]
= (92)
24 Nov 2011 Prof. R. Shanthini 10
Equations for Packed Columns with concentrated solutions
Kya(1 – y)LMSGZ =
(1 – y)LM dy(1 – y)(y – y*)∫
yout
yin
(93)
Therefore (91) becomes the following:
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.)
HOGNOG
24 Nov 2011 Prof. R. Shanthini 11
Equations for Packed Columns with concentrated solutions
x in (94) can be related to y in (93) using the operating line equation.
y* in (93) can be related to the bulk concentration using the equilibrium relationship as follows:
y* = K x (94)
We will determine the operating line equation next
24 Nov 2011 Prof. R. Shanthini 12
Equations for Packed Columns with concentrated solutions
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gy
Lx
The operating equation for the packed column is obtained by writing a mass balance for solute over the control volume:
Control volume
Lin xin + G y = L x + Gout yout
If dilute solution is assumed, then Lin = L = Lout and Gin = G = Gout.
(74)
We somehow have to relate L to Lin and G to Gout in (74).
24 Nov 2011 Prof. R. Shanthini 13
Equations for Packed Columns with concentrated solutions
Inlet solventLin, xin
Treated gasGout, yout
Spent solventLout, xout
Inlet gasGin, yin
Gy
Lx
Control volume
Lin (1 – xin) = L (1 – x)
To relate L to Lin, write a mass balance for solvent over the control volume:
To relate G to Gout, write the overall mass balance over the entire column:
L = Lin (1 – xin) / (1 – x) (95)
G + Lin = Gout + L (96)
24 Nov 2011 Prof. R. Shanthini 14
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
y = Gout + [Lin(x – xin) / (1 – x)]
Gout yout + [Lin (1 – xin) x / (1 – x)] - Lin xin (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 15
Equations for Packed Columns with concentrated solutions
If the solvent fed to the column is pure then xin = 0.
Therefore, (98) becomes
y = Gout + [ Lin x / (1 – x) ]
Gout yout + [ Lin x / (1 – x) ] (99)
24 Nov 2011 Prof. R. Shanthini 16
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
Ammonia removed from the air stream = 0.95 x 40 = 38 lbmol/h
In the inlet air stream: ammonia = 40 lbmol/h; air = 60 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 17
Example 1:
y = 62 + [488 x / (1 – x)]
62 x 0.0323 + [488 x / (1 – x)]
Using Lin = 488 lbmol/h, Gout = 62 lbmol/h, yout = 0.0323 and xin = 0 in (98), we get the operating curve as follows:
00.05
0.10.15
0.20.25
0.30.35
0.40.45
0.5
0 0.02 0.04 0.06 0.08 0.1
x
y
Operating curve
yin (bottom of the tower)
yout (top of the tower)
24 Nov 2011 Prof. R. Shanthini 18
Example 2:
Draw the equilibrium curve, which is approximately described by K = 44.223x + 0.4771, on the same plot as in Example 1.
Solution:
00.05
0.10.15
0.20.25
0.30.35
0.40.45
0.5
0 0.02 0.04 0.06 0.08 0.1
x
y Equilibrium curve
y = K x = 44.223x2 + 0.4771x
Operating curve
24 Nov 2011 Prof. R. Shanthini 19
Example 3:
Determine the NOG using the data given in Examples 1 and 2.
(1 – y)LM dy(1 – y)(y – y*)∫
yout
yin
NOG =
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.
Solution:
24 Nov 2011 Prof. R. Shanthini 20
Example 3:
0
5
10
15
20
25
30
35
0 0.1 0.2 0.3 0.4
y
Inte
gran
d
yin = 0.4yout = 0.0323
The shaded area gives NOG as 3.44(Numerical integration is better suited
to get the answer.)
24 Nov 2011 Prof. R. Shanthini 21
Example 4:
Determine the height of packing Z using the data given in Examples 1 and 2.
HOG =
Solution:
Kya(1 – y)LMSG
Z = NOG HOG
Need more data to work it out.
Can be calculated once HOG is known.
24 Nov 2011 Prof. R. Shanthini 22
Summary with overall gas-phase transfer coefficients for packed column with concentrated solutions
where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as follows:
(1 – y)LM
y* – y
ln[(1 – y )/(1 – y* )]
= (92)
Kya(1 – y)LMSGZ =
(1 – y)LM dy(1 – y)(y – y*)∫
yout
yin
(93)
HOGNOG
24 Nov 2011 Prof. R. Shanthini 23
where (1 – x)LM is the log mean of (1 – x) and (1 – x*) given as follows:
(1 – x)LM
x* – x
ln[(1 – x )/(1 – x* )]
= (92)
Kxa(1 – x)LMSLZ =
(1 – x)LM dy(1 – x)(x* – x)∫
xin
xout
(93)
HOLNOL
Summary with overall liquid-phase transfer coefficients for packed column with concentrated solutions
24 Nov 2011 Prof. R. Shanthini 24
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 25
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.