24 nov 2011prof. r. shanthini1 course content of mass transfer section lta diffusion theory of...

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


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.


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





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





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





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


= - G (1 – y)(1 – y)2

dy

= - G(1 – y)

dy(88)

Ls

X





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)


(1 – y)

dy

Ls

X

Compare (89) with (79) used for packed columns with dilute solutions. What are the differences?





Inlet gasGin, yin

Gs

YLs

X

Gs

Y+dY

Ls

X+dXz

dz

Z


Rearranging and integrating (89) gives the following:

KyaSGZ =

dy(1 – y)(y – y*)∫

yout

yin

(90)



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)



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



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






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).






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)



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?



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)


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


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)


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


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:


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.)


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.


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


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


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.


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

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