McCabe Thiele Graphical Equilibrium-Stage -...
Transcript of McCabe Thiele Graphical Equilibrium-Stage -...
Dr Saad Al-Shahrani ChE 334: Separation Processes
Limiting condition
McCabe Thiele Graphical
Equilibrium-Stage
When analyzing or designing a process, it is useful to look at limiting
cases to assess the possible values of process parameters. In
distillation analysis, separation of a pair of components can be improved
by increasing the number of stages while holding reflux constant, or by
increasing the reflux flow for a given number of stages. This tradeoff
sets up two limiting cases:
1. Total Reflux (minimum ideal stages)
2. Minimum Reflux (infinite ideal stages)
The design tradeoff between reflux and stages is the standard economic
optimization problem chemical engineers always face -- balancing capital costs
(the number of trays to be built) vs. the operating cost (the amount of reflux to be
recirculated). A good design will operate near a cost optimum reflux ratio.
Dr Saad Al-Shahrani ChE 334: Separation Processes
D
LRD
McCabe Thiele Graphical
Equilibrium-Stage
If the reflux ratio ( ) is increased to very large value, the
operating lines become the 45o line. The infinite reflux ratio occurs
in real life when the column is operated under what are called (total
reflux) condition
a) Minimum number of plates:
Under these conditions, no feed is added to the column (F=0) and
no products are withdrawn (D=0, B=0), but the vapor is raised up
and condensed to the column. So the column is just circulating
vapor and liquid up and down. Most columns are started up under
total reflux conditions.
Dr Saad Al-Shahrani ChE 334: Separation Processes
Distillation of Binary Mixture
Since the liquid flow rate in the column is same as the vapor flow
rate, and
0.1V
L
V
L
The operating line
nD
nn xDL
Dxx
DL
Ly
1
mB
mm xBL
Bxx
BL
Ly
1
,
The composition in the base of the column under total reflux = xB, and
the composition of the liquid in the reflux drum = xD
In this case the number of ideal plates is minimum.
Dr Saad Al-Shahrani ChE 334: Separation Processes
Binary Multistage Distillation
The minimum number of ideal plates can be done by:
a) Graphically as shown in the figure
y
XD XF
XB
XB
x
y1
y2
y3
y4
X1
X2
X3
Operating lines
as total reflux
Composition of liquid
in reflux drum
Composition of liquid
in re-boiler
y1 = xD
y2 = x1
y3 = x2
y4 = x3
xB = xB
Minimum number of
plates = 3+reboiler
Dr Saad Al-Shahrani ChE 334: Separation Processes
Binary Multistage Distillation
b) Analytically (using Fenske Equation)
This equation gives the number of plates required under total reflux at
constant .
It is applicable to multi-component system as well as binary system
(= constant, total reflux, ideal system).
It is very useful for getting quick estimates of the size of a column.
Derivation of Fenske Equation
Consider two component (A,B) forming ideal solution
product bottomin ration mole
product in top ratio mole
/
/
/
/
BA
BA
BB
AA
B
AAB
xx
yy
xy
xy
K
K
(1)
Dr Saad Al-Shahrani ChE 334: Separation Processes
An ideal mixture follows Raoult’s law and = vapor product ratio
Binary Multistage Distillation
P
xP
P
PxPP A
sat
AAA
sat
AA Ay
P
xP
P
PxPP B
sat
BBB
sat
BB By
sat
B
sat
A
BB
sat
B
AA
sat
A
BB
AA
B
AAB
P
P
PxxP
PxxP
xy
xy
K
K
/
/
/
/
sat
B
sat
A PP /
does not change much over the range of temperature
encountered,AB constant
1
, 1 A
A
B
A
A
A
B
A
x
x
x
x
y
y
y
y
(2)
Dr Saad Al-Shahrani ChE 334: Separation Processes
Binary Multistage Distillation
Substitute (2) in (1)
1
1
1
1
1
1
n
n
AB
n
n
x
x
y
y
AB
A
A
A
A
x
x
y
y
11
DL
Dxx
DL
Ly D
nn
1
For plate n+1
Since D = 0 (total reflux), L / V= 1.0 ,
Then yn+1 = xn and 1
1
1
1
n
n
AB
n
n
x
x
x
x
zero
Dr Saad Al-Shahrani ChE 334: Separation Processes
Binary Multistage Distillation
At the top of the column, if a total
condenser is used y1 = xD , n = 0
Substitute in (2)
1
1
1
1 x
x
x
xAB
D
D
For plate (1)
xD
y1
water
Re-boiler
Vb
yb
Lb, xb
steam
x1
x2
x3
xn
xn-1
x0
y2
y3
y4
yn-1
yn
yr
B
BAB
n
n
x
x
x
x
1
1
For plate (n)
. . . .
. . . .
. . . .
n
n
AB
n
n
x
x
x
x
1
1 1
1
For re-boiler plate
For plate (2) 2
2
1
1
1
1 x
x
x
xAB
Dr Saad Al-Shahrani ChE 334: Separation Processes
Binary Multistage Distillation
If all equations are multiplied together and all the intermediate terms
canceled,
B
Bn
AB
D
D
x
x
x
x
1) (
1
B
BN
AB
D
D
x
x
x
x
1) (
1
1min
AB
BBDD xxxxN
ln
)]1//()1/ln[(1min
AB
BBADBA
AB
BD xxxxN
ln
)//()/ln(
ln
])ration mole/()ration moleln[(1or min
Where n= Nmin + reboiler
Dr Saad Al-Shahrani ChE 334: Separation Processes
McCabe Thiele Graphical
Equilibrium-Stage
Example:
Calculate the minimum number of trays required to achieve a
separate from 5 mole % bottoms to 90 moles % distillate in a binary
column with =2.5
solution xB = 0.05 , xD = 0.9
61.419163.0
14.5 ,
5.2ln
)]05.01/05.0/()9.01/9.0ln[(1 minmin
NN
AB
BBDD xxxxN
ln
)]1//()1/ln[(1min
Dr Saad Al-Shahrani ChE 334: Separation Processes
McCabe Thiele Graphical
Equilibrium-Stage
Example: in a mixture to be fed to a continuous distillation column, the mole
fraction of phenol is 0.35, of o-cresol 0.15, of m-cresol 0.3 and of xylenes
0.2. it is hoped to obtain a product with a mole fraction of phenol 0.952, of
o-cresol 0.0474, of m-cresol 0.0.0006. if p-o= 1.26, m-o=0.7, estimate how
many theoretical plates would be required at total reflux. Assume no phenol in
the bottoms.
Solution:
A light component (o-cresol) B heavy component (m-cresol)
Total balance 100=D + B
For phenol 100*0.35=D*0.952+B*xB,p
= zero
D= 36.8 Kmol, B = 63.2 Kmol
For o – cresol
100*0.15=0.0474*36.8+xB,o*63.2 xB,o=0.21
Dr Saad Al-Shahrani ChE 334: Separation Processes
McCabe Thiele Graphical
Equilibrium-Stage
o-m= 1/0.7=1.43 43.1ln
)]474.0/21.0/()0006.0/0474.0ln[(1min N 5.13min N
component Feed top Bottms
phenol 0.35 0.952 0 p-o= 1.26
o-cresol 0.15 0.0474 0.21 o-o= 1.0
m-cresol 0.3 0.0006 0.474 m-o=0.7
xylenes 0.2 0 0.316
For m – cresol
100*0.3=0.0006*36.8+xB,m*63.2 xB,m=0.474
xB,X=0.316
Dr Saad Al-Shahrani ChE 334: Separation Processes
McCabe Thiele Graphical
Equilibrium-Stage
b) Minimum Reflux Ratio
The next figure shows how changing the reflux ratio affects the
operating lines: the lower the reflux ratio, the closer the operating
line moves toward the equilibrium curve, and the larger the number
of plates.
If the reflux ratio finally reduced to the point where either operating
line intersects or becomes tangent to the VLE curve, an infinite
number of plates will be required and the reflux ratio is minimum.
Dr Saad Al-Shahrani ChE 334: Separation Processes
To obtain the RDmin
McCabe Thiele Graphical
Equilibrium-Stage
111
D
Dn
D
Dn
R
xx
R
Ry
1intercept ab
min
D
D
R
x
1min D
D
R
x
x
a
b x`
(xD,xD)
y
xD
xD
Dr Saad Al-Shahrani ChE 334: Separation Processes
McCabe Thiele Graphical
Equilibrium-Stage
If the equilibrium curve
has a cavity upward, e.g.,
the curve for water-
ethanol shown in the
figure in this case the
minimum reflux ratio
must be computed from
the slope of the operating
line (ac) that is tangent to
the equilibrium
x`
1min D
D
R
x
a
b
y`
Feed line Non-ideal Line VLE
c
Dr Saad Al-Shahrani ChE 334: Separation Processes
McCabe Thiele Graphical
Equilibrium-Stage
Examole. A continuous fractionating column is to be design to separate 30,000 kg/h
of a mixture of 40 percent benzene and 60 percent toluene into an overhead
product containing 97 percent benzene and a bottom product containing 98
percent toluene. These percentages are by weight. A reflux ratio of 3.5 mol to 1
mol of product is to be used. The molal latent heats of benzene and toluene are
7,360 and 7,960 cal/ gmol, respectively. Benzene and toluene form an ideal
system with a relative volatility of about 2.5. The feed has a boiling point of 95 oC at a pressure of 1 atm. (a) Calculate the moles of overhead product and
bottom product per hour. (b) Determine the number of deal plate and the
position of the feed plate (i) if the feed is liquid and at its boiling point. ii)if the
feed is liquid and at 20 oC (specific heat 0.44 cal/ g-oC) (iii) if the feed is a
mixture of two-thirds and one-third liquid. (c) If steam at 20 Ib,/in2(1.36 atm)
gauge is used for heating, how much steam is required per hour for each of the
above three cases, neglecting heat losses and assuming the reflux is a
saturated liquid? (d) If cooling water enters the condenser at 25°C and leaves at
40°C, how much cooling water a required, in gallons per minute?
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
y
.
Feed line
2 1
3
4
5
6
7
8
9
10
R
xD=0.974 xB=0.0235
1D
D
R
x
xF=0.44