20 Oct 2011Prof. R. Shanthini1 Course content of Mass transfer section LTA Diffusion Theory of...

20 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 6


Plate columns are used mainly in distillation

You probably know how to calculate the number of plates required for the desired separation.


Packed columns are used in absorption, stripping and adsorption

There are no plates here.

We need to calculate the height of packing in a packed column where the separation process takes place.

How to do that?


Equations for Packed ColumnsInlet solventLs,in, Lin, Xin, xin

Treated gasGs,out, Gout, Yout, yout

Spent solventLs,out, Lout, Xout, xout

Inlet gasGs,in, Gin, Yin, yin

Counter-current packed column used for absorption

Gs

GYy

Ls

LXx


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


Relating Y to y and X to x:

Therefore y = Y / (Y+1)

y =

Similarly, it can be shown x = X / (X+1)

moles of solute Amoles of solute A + moles of inert gas

=1

1 + (moles of inert gas / moles of solute A )

=Y

Y + 1 =

1

1 + (1/Y)


For dilute mixers

Y - mole ratio of solute A in gas= moles o fA / 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)

For dilute mixtures, y ≈ Y x ≈ X


For dilute mixers

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)

For dilute mixtures, Gs ≈ G Ls ≈ L


Equations for Packed Columns for dilute solutionsInlet solventLin, xin

Treated gasGout, yout

Spent solventLout, xout

Inlet gasGin, yin

Gy

Lx

The operating equation for the packed column is obtained by writing the 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. Therefore, the above becomes

L xin + G y = L x + G yout

y = (L / G) x + yout - (L / G) xin

The operating line therefore becomes

(74)

(75)

(76)


Mass transfer of solute from gas phase to liquid phase is assumed to be explained by the Two-film Theory:

Equations for Packed Columns for dilute solutions


Rate of mass transfer from vapour phase to liquid phase is therefore given by the following:

ky a (y - yi) = kx a (xi - x)

ky and kx are film mass transfer coefficients based on unit area.

But the area for mass transfer in a packed column in difficult to determine.

Therefore we use a factor ‘a’ which gives the area of mass transfer per unit volume of packed bed.

In packed columns, we use ‘kya’ and ‘kxa’ which are the film mass transfer coefficients based on unit volume of packed bed.

Molar fractions yi and xi are the interface properties that are related by the equilibrium ratio K (= yi / xi).

(77)

(78)



Relating equilibrium mole fractions to operating mole fractions:

Equation (77) gives

Equilibrium line: K = yi / xi

y = (L / G) x + yout - (L / G) xinOperating line:

y - yi

ky akx a

x - xi = -

Equilibrium line(slope = K)

x

y

Operating line (slope = L/G)

yout - (L / G) xin (xi,yi)

Slope = - ky akx a

(x,y)

(76)

(78)


20 Oct 2011 Prof. R. Shanthini 13x

y

(x,y)

(xi,yi)

(x,y*)

(x*,y)

y* is the gas-phase mole fraction that would have been in equilibrium with the liquid-phase mole fraction x. y* = K x

x* is the liquid-phase mole fraction that would have been in equilibrium with the gas-phase mole fraction y. y = K x*

y - yi

x* - xi yi – y*xi - x

K = =

yout - (L / G) xin



1

kxa

1

kya +

1

Kxa= (77)y - yi

x* - xi

1

kya

1

kxa +

1

Kya=

yi – y*xi - x

Since the liquid usually has strong affinity for the solute, mass transfer resistance is mostly in the gas. Therefore, determination of the packed height of a column most commonly involves the overall gas-phase mass transfer coefficient based on unit volume of packed base, which is Kya.

1

kxa

1/K

kya +=

1

kya

K

kxa += (78)



Inlet solventLin, xin



Inlet gasGin, yin

Gy

LxLet us now obtain equations

required to determine the height of packing required for a specified separation in a packed column operated with dilute solution.

Gy+dy

Lx+dx

z

dz

Z






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

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.

Gy+dy

Lx+dx

z

dz

Z

Therefore, mass balance for solute gives the following:

-G dy = Kya (y – y*) S dz (79)



Rearranging and integrating (79) gives the following:

KyaSG dz

dyy – y*




Inlet gasGin, yin

Gy

Lx

Gy+dy

Lx+dx

z

dz

Z

=∫yout

yin

∫0

Z

The packed height is therefore given by:

G

KyaSZ

dyy – y*

= ∫yout

yin

HOG NOG



G

KyaS

dyy – y*∫

yout

yin

Definition of HOG and NOG:

HOG ≡

NOG ≡

which is termed as the overall height of a transfer unit (HTU) based on the gas phase which has the dimension of length

which is termed as the overall number of transfer units (NTU) based on the has phase which is dimensionless

Z = HOG NOG

Height of the packed column is written in terms of transfer units as follows:

(80)

(81)

(82)



G

KyaS

dyy – y*∫

yout

yin

HTU: HOG =

NTU: NOG =

The smaller the HTU, the more efficient is the contacting.

It represents the overall change in solute mole fraction divided by the average mole-fraction driving force. The larger the NTU, the greater is the extent of contacting required.



dyy – y*∫

yout

yin

Integration of NOG = (81)

Equilibrium relationship gives: y* = K x

Operating line gives: x = (G/L) y - (G/L) yout + xin

Using these the denominator in (81) can be written as follows:

y - y* = y - K x

= y - K(G/L) y + K(G/L) yout - K xin

= (1 - KG/L) y + (KG/L) yout - K xin(82)



dy

∫yout

yin

NOG =(1 - KG/L) y + (KG/L) yout - K xin

Combining (81) and (82), we get

= ln[(1 - KG/L) y + (KG/L) yout - K xin]

(1 - KG/L)yout

yin

= (1 - KG/L) yin + (KG/L) yout - K xin

(1 - KG/L)

1

yout - K xin

ln



NOG =yin - K xin + KG/L (yout - yin)

(1 - KG/L)

1

yout - K xin

ln

yin - K xin + KG/L (yout - K xin + K xin - yin)

(1 - KG/L)

1

yout - K xin

ln=

(1 - KG/L) (yin - K xin)

(1 - KG/L)

1

yout - K xin

ln= + KG/L (83)



Example 6.10 of Ref 2

Experimental data have been obtained for air containing 1.6% by volume of SO2 being scrubbed with pure water in a packed column of 1.5 m2 in cross-sectional area and 3.5 m in packed height. Entering gas and liquid flow rates are 0.062 and 2.2 kmol/s, respectively. If the outlet mole fraction of SO2 in the gas is 0.004 and column temperature is near ambient with KSO2 = 40, calculate the following:

a) The NOG for absorption of SO2

b) The HOG in meters

c) The volumetric, overall mass-transfer coefficient, Kya for SO2 in kmol/m3.s


Example 6.11 of Ref 2 (modified)

A gaseous reactor effluent consisting of 2 mol% ethylene oxide in an inert gas is scrubbed with water at 30oC and 20 atm. The total gas feed rate is 2500 lbmol/h, and the water rate entering the scrubber is 3500 lbmol/h. The column, with a diameter of 4 ft, is packed in two 12-ft-high sections with 1.5 in metal Pall rings. A liquid redistributer is located between the two packed sections. Under the operating conditions for the scrubber, the K-value for ethylene oxide is 0.85 and estimated values of kya and kxa are 200 lbmol/h.ft3 and 2643 lbmol/h.ft3 , respectively. Calculate the following:

a) Kya

b) HOG and NOG

c) Yout and xout

20 Oct 2011Prof. R. Shanthini1 Course content of Mass transfer section LTA Diffusion Theory of...

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