Design of packed tower based on overall mass transfer

coefficient

From overall mass transfer equation one can write for packed
tower

where, y* is solute concentration in gas phase that is capable

of remaining in equilibrium with a liquid having a bulk
concentration of x.
• asd

One need to do graphical integration of above eq. Operating

line PQ is drawn in xy plane. Any point (x,y) is taken in
operating line. A vertical line is drawn upto equilibrium line to
get y*.
• asd

Design based on height of a transfer unit (HTU)

Equation above can be written as
• The design equation

can be written as

Where pB,M is called logarithmic mean partial pressure of

species B which is defined as

(𝟏 − 𝒚𝒊 )𝟏 −(𝟏 − 𝒚𝒊 )𝟐
(𝟏 − 𝒚𝒊 )𝑴 =
(𝟏 − 𝒚𝒊 )𝟏
𝒍𝒏
(𝟏 − 𝒚𝒊 )𝟐
• The design equation

can be written as

(𝟏 − 𝒚𝒊 )𝟏 −(𝟏 − 𝒚𝒊 )𝟐
(𝟏 − 𝒚)𝒊𝑴 =
(𝟏 − 𝒚𝒊 )𝟏
𝒍𝒏
(𝟏 − 𝒚𝒊 )𝟐
• It may be noted that

•Remain constant at the packing section though G’ varies.

•This quantity is called ‘height of transfer units’ (HTU) and
designated as HtG.
•It is important to measure the separation effectiveness of the
particular packings for a particular separation process.
•It also describes the mass transfer coefficient. Larger mass
transfer coefficient leads to the smaller value of HTU
• HTG

•The integral part of the above Equation is called number of

gas phase transfer units as NtG.
EXAMPLE 2 (Packed tower design using individual coefficients) A gas
mixture containing 10 mol% SO2 and 90 mol% air at 1 atm total
pressure and 30°C is to be scrubbed with water to remove 97% of
the SO2 in a tower packed with 25 mm ceramic Raschig rings. The
feed gas rate is 1500 kg per hour. Calculate (a) the minimum liquid
rate, (b) Operating line the liquid rate is 1.25 times the minimum and
(c) the packed height.
The Colbum-Drew individual volumetric mass transfer coefficients at
the given conditions are:

The calculated equilibrium data in mole fraction unit (30°C, 1 atm

total pressure) are given below:
Tower cross-section = 0.781 m2
 xy plot
0.120

0.100

0.080

0.060

0.040

0.020

0.000
0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030
Average molecular weight of the feed gas (10% SO2, 90% air) =
(0.10)(64) + (0.90) (28.8) = 32.3 kmol/kg

G1= 46.41 kmol/h; Gs = G1(1-y1) = 46.41(1-.01) = 41.77 kmol/h

y1=0.1 and y2 = 0.00334; G2(1-y2) = 41.77 kmol/h; G2 = 41.91 kmol/h
Ls = 1.25*1655= 2069 kmo/hr, x2=0, x1=0.00217
G1= 46.41 kmol/h; G2 = 41.91 kmol/h
y1=0.1 and y2 = 0.00334
 x y xi yi
0.000000 0.003340 0.000101 0.00166
0.000400 0.022641 0.000632 0.01876
0.000799 0.041210 0.001121 0.03585
0.001199 0.059086 0.001573 0.05285
0.001597 0.076308 0.001994 0.06970
0.001996 0.092911 0.002388 0.09226
0.002168 0.100000 0.002551 0.09927
(𝟏−𝒚𝒊 )𝟏 −(𝟏−𝒚𝒊 )𝟐 (𝟏−𝟎.𝟎𝟎𝟏𝟔𝟔)𝟏 −(𝟏−𝟎.𝟎𝟗𝟗𝟐𝟕)𝟐
(𝟏 − 𝒚)𝒊𝑴 = (𝟏−𝒚𝒊 )𝟏 = (𝟎.𝟗𝟗𝟖𝟑𝟒)𝟏
𝒍𝒏 𝒍𝒏
(𝟏−𝒚𝒊 )𝟐 (𝟎.𝟗𝟎𝟎𝟕𝟑)𝟐
𝟎.𝟎𝟗𝟕𝟔𝟏 𝟎.𝟎𝟗𝟕𝟔𝟏
(𝟏 − 𝒚)𝒊𝑴 = = = 𝟎. 𝟗𝟒𝟗𝟑𝟑
𝒍𝒏(𝟏.𝟏𝟎𝟖𝟑) 𝟎.𝟏𝟎𝟐𝟖𝟐
Tower cross-section = 0.781 m2

G’2 = 41.91/0.781 = 53.661 kmol/h.m2

HtG = G’/kya(1-y)iM = 56.54/(0.94933*270) = 0.213 m

When overall gas phase mass transfer coefficients are used,
the height of the packing is as follows:
 Design Equations based concentration in mole ratio unit

If kx, ky are individual gas phase mass transfer coefficients

and KY is overall gas phase mass transfer coefficient, height
of packed tower is expressed as:

; ;
The slope of operating line
• Solute A is to be absorbed from a binary mixture containing
7.5% of A with solvent B in a packed tower. Based on
flooding calculation, a tower diameter of 1.2 m is selected.
Total gas flow rate is 60 kmol/h. The exit gas must not contain
0.2% of solute A. Solute free liquid B enters from the top of the
tower at 40 kmol/h. The gas phase and liquid phase
•Mass transfer coefficients based on mole ratio unit are: kX
=2.05 kmol/m2h (ΔX) and kY = 1.75 kmol/m2h (ΔY). The
equilibrium line Equation is Y = 0.63X. Specific interfacial area
of gas-liquid contact (ā) is 71 m2/m3.
(a) Calculate packing height required for the desired separation.
(b) For 99.5% solute A removal, what % increase in packed height
is needed?
(c) Determine slopes of operating line in each
• Gas flow rate, G1 =60 kmol/h; y1=0.075

𝑑2 1.22
•Area of tower cross-section 𝐴 = 𝜋 =𝜋 4 = 𝟏. 𝟏𝟑𝟏 𝐦𝟐
4

G’1 = 53.05 kmol/m2 .h;

Solute concent ration in exit gas is 0.2 % so
40
•Liquid flow rate L’s = 1.131 =35.37 kmol/m2.h; X2 = 0
Overall mass balance Equation for the solute concentration
in exit liquid as follows:
49.07(0.011-0.00204) = 35.37(X1-0); X1 = 0.1097
Overall gas-phase mass transfer coefficient, KY

KY = 1.138 kmol/m2 h (ΔY)

Y* can be expressed in terms of Y.
The operating line Equation can be expressed as:
= 49.07(Y-0.00204) = 35.37(X-0)
X=1.387(Y-0.00204) We have equilibrium relation
Y* = αX = 0.63×1.387(Y-0.00204) = 0.874Y-0.00178
(a) Packed height, hT=HtoG×NtoG=0.0607×13.9 m=8.46 m.
(b) For 99.5% solute removal, Y2=0.0811×0.0005=4.05×10-4.

49.07(Y-4.05×10-4) = 35.37(X-0)
X = (1.387Y – 5.62×10-4) Hence,
Y* = αX = 0.63×(1.387Y – 5.62×10-4) = 0.874Y-0.000354
Required packed height, hT = HtoG×NtoG=0.0607×22 m=13.35
m

For both the cases the slope of the operating line

35 .37/49.07 = 0.721
 Counter-current multi-stage absorption (Tray absorber)
In tray absorption tower, multi-stage contact between gas and
liquid takes place. In each tray, the liquid is brought into intimate
contact of gas and equilibrium is reached thus making an ideal
stage.
In ideal stage, average composition of liquid leaving the tray is in
equilibrium with liquid leaving that tray. The most important
step in design of tray absorber is the determination of
number of trays.
The schematic of tray tower is presented in figure 4.7. The liquid
enters from top of the column whereas gas is added from the
bottom. The efficiency of the stages can be calculated as:
• Stage efficiency
•Tray type absorber tower
• The following parameters should be known for the
determination of “number of stages”
(1) Gas feed rate Gs
(2) Concentration of gas at inlet and outlet of the tower YN+1
and Y1
(3) Concentration of liquid XN and X0
(4) Minimum liquid rate; actual liquid rate is 1.2 to 2 times
the minimum liquid rate Ls
(4) Equilibrium data for construction of equilibrium curve
Now, the number of theoretic stages can be obtained
graphically or algebraically.
 (A) Graphical Method for the Determination of Number of
Ideal Stages

The overall material balance on tray tower

Gs(YN+1 -Y1)=Ls(XN -X0)
This is the operating line for tray tower.
If the stage (plate) is ideal, (Xn, Yn) must lie on the
equilibrium line, Y*=f(X)
A vertical line is drawn from Q point to D point in equilibrium line at
(XN, YN). From point D in equilibrium line, a horizontal line is
extended up to operating line at E (XN-1, YN).
The region QDE stands for N-th plate (refer Figure 4.8). We may get
fraction of plates. In that situation, the next whole number will be
the actual number of ideal plates.
If the overall stage efficiency is known, the number of real plates can
be obtained from Equation (4.18).
Top plate is located at P(X0, Y1) and bottom plate is marked as Q(XN,
YN+1) in X-Y plane.
Graphical steps
Ideal stages
 Algebraic Determination of Number of Ideal Stages

•If both operating line and equilibrium lines are straight,

number of ideal stages can be calculated algebraically.
•Let solute transfers from gas to liquid (Absorption)
•Equilibrium line, Y=αX
Point (XN, YN) lies on the equilibrium line: YN= α X𝑁
• Where

•On substitution for

•The equation is 1st order difference equation (non

homogeneous)

•
 First Order Difference Equations
Differential equation are great for modeling situations where
there is a continually changing population or value. If the
change happens incrementally rather than continuously then
differential equations have their shortcomings. Instead we
will use difference equations which are recursively defined
sequences. Examples of incrementally changes include
salmon population where the salmon spawn once a year,
interest that is compound monthly, and seasonal businesses
such as ski resorts.
 First Order Difference Equation A first order difference
equation is a recursively defined sequence in the form
yn+1 = f(n,yn) n = 0,1,2,...
• What makes this first order is that we only need to know
the most recent previous value to find the next value. It
also comes from the differential equation
y' = g(n,y(n))
Recalling the limit definition of the derivative this can be
written as
 if we think of h and n as integers, then the smallest that h
can become without being 0 is 1. The differential equation
becomes, h= 1
, h=1

y(n+1) - y(n) = g(n,y(n))

y(n+1) = y(n) +g(n,y(n))
 Now letting f(n,y(n)) = y(n) +g(n, y(n))
and putting into sequence notation gives
yn+1 = f(n, yn)
If the first order difference depends only on yn (autonomous
in Diff EQ language), then we can write
y1 = f(y0), y2 = f(y1) = f(f(y0))= f 2(y0);
y3 = f(y2) = f(f(f(y0))) = f 3(y0)
In general, yn = K1f n(y0)
 Solutions to a finite difference equation with
yn+1 = yn ; Are called equilibrium solutions.
We find them by setting yn = f(n,yn)
An finite difference equation is called linear if f(n,yn) is a
linear function of yn.
 Example
Each year, 1000 salmon are stocked in a creak and the salmon
have a 30% chance of surviving and returning to the creak the
next year. How many salmon will be in the creak each year
and what will be population in the very far future?
Solution
This is a linear finite difference equation with
yn+1 = 0.3yn + 1000
We have
y0 = 1000, y1 = 0.3y0 + 1000,
y2 = 0.3y1 + 1000 = 0.3(0.3y0 + 1000) + 1000
 y3=0.3y2+1000 = 0.3(0.3(0.3y0 + 1000) + 1000) +1000
= 1000 + .3(1000) + .32(1000) + .33y0
In general,
yn = 1000(1 + 0.3 + 0.32 + 0.33 + ... + 0.3n-1) + 0.3ny0
The first term is a geometric series, so the equation can be
written as
• Notice that the limiting population will be 1000/0.7
= 1429 salmon.

More generally for the linear first order difference equation

yn+1 = r yn + b
The solution is
• Where

•On substitution for

•The equation is 1st order difference equation (non

homogeneous)

•
•Solution by finite difference method
•Corresponding homogeneous Equation

•Non-homogeneous Equation has a particular solution,

which is constant.
•Assuming YN=YN+1, we have, Y=K2
• Particular

•The complete solution is as follows:

Initial conditions: N=0; Y0=αX0

• Solve for K1

•Su substitute it

When N=N+1;

or
• asd

•Taking logarithm of both the sides

• when
 Let solute is transferred from liquid to gas (stripping).

When
• The equations

•And for stripping

•The equation are called “Kremser Equations”.

•Example Problem 4.2. It is desired to absorb 95% of acetone by
water from a mixture of acetone and nitrogen containing 1.5%
of the component in a counter current tray tower. Total gas
input is 30 kmol/hr and water enters the tower at a rate of 90
kmol/hr. The tower operates at 27oC and 1 atm. The
equilibrium relation is Y=2.53X.
•Determine the number of ideal stages necessary for the
separation using (a) graphical method as well as (b) Kremser
analysis method
Solution 4.2: Basis: 1 hour
GN+1=30 kmol yN+1=0.015; L0=90 kmol
Moles acetone in = 30×0.015 moles=0.45 moles
Moles nitrogen in = (30-0.45) moles=29.55 moles
Moles acetone leaving (95% absorbed) = 0.45×(1-0.95)
=0.0225 moles
Gs=29.55 moles, Ls=90 moles; α=2.53 [as, Y=2.53X]

YN+1 = 0.015
• Rewriting Equation for operating line as

•29.55 × (0.015 − 7.61 × 10−4 ) = 90(X𝑁 − 0)

•XN=4.68×10-3

(a) Solution by graphical method

Construction of operating line PQ:
P(X0, Y1)=P(0, 7.61×10-4) and Q(XN, YN+1)=Q(4.68×10-3, 0.015)
Construction of equilibrium line (Y=2.53X):

X 0 0.001 0.002 0.003 0.004 0.005

Y 0 0.00253 0.00506 0.00759 0.01012 0.01265

From graphical construction (Figure 4.9), the number of

triangles obtained is more than 7. Hence number of ideal
stages is 8.
• Graphical
(b) Solution by Kremser analysis method
As Ā≠1, according to Kremser analysis method:

Number of ideal stages is 8

• Gr