1.

A liquid mixture containing 50 mole % each of benzene and toluene at

40oC is to be continuously flash vaporized to vaporize 60 mole % of the
feed. The residual liquid product contains 35 mole % benzene. If the
enthalpies per mole of feed, the liquid product and the vapor product are
respectively 2.00*103, 5*103 and 30*103 kJ kmol-1 then:

a. calculate the heat required to be added in kJ kmol-1 of vapour product

b. represent the process on a H-x-y diagram

Solution
(a) Basis: 100 kmol of feed
Feed composition, mole fraction of more volatile component (zF) = 0.5
Feed (F) = 100 kmol
Since 60% of the feed is vaporized,
Vapour (V) = 60 kmol
Liquid (B) = 40 kmol
Mole fraction of more volatile component in the residual liquid (xB) = 0.35
F=V+B (1)
FzF = VyD + BxB (2)
FhF + QH = yDHV + BhB (3)

From equations (2)

100 x 0.5 = 60yD + (40 x 0.35)

Therefore, yD = 0.6

From equation (3),

(100 x 2 x 103) + Q H = (60 x 30 x 103) + (40 x 5 x 103)
Therefore, Q H = 1800 x 103 kJ
QH/V = 1800x 103/60 = 30 x 103kJ kmol-1
(b) The H-x-y diagram for the process is as shown:
2. 1000 kg/hr of a mixture containing 42 mole percent heptane and 58 mole
percent ethyl benzene is to be fractionated to a distillate containing 97 mole
percent heptane and a residue containing 99 mole percent ethyl benzene using a
total condenser and feed at its saturated liquid condition. The enthalpy-
concentration data for the heptane-ethyl benzene at 1 atm pressure are as
follows:

xheptane 0 0.08 0.18 0.25 0.49 0.65 0.79 0.91 1.0

yheptane 0 0.28 0.43 0.51 0.73 0.83 0.90 0.96 1.0
Hl (kJ/kmol) x 10-3 24.3 24.1 23.2 22.8 22.05 21.75 21.7 21.6 21.4

Hv (kJ/kmol) x 10-3 61.2 59.6 58.5 58.1 56.5 55.2 54.4 53.8 53.3

Calculate the following:

a. Minimum reflux ratio;
b. Minimum number of stages at total reflux;
c. Number of stages at a reflux ratio of 2.5;
d. Condenser duty at a reflux ratio of 2.5;
e. Reboiler duty at a reflux ratio of 2.5.
Solution:
Heptane = C7H16 (M.W. = 100)
Ethyl benzene = C6H5C2H5 (M.W. = 106)
Average molecular weight of feed solution
= 0.42 x 100 + 0.58 x 106 = 103.48
Molal flow rate of feed, F = 1000/103.48 = 10.0 kmol/hr
The H-x-y diagram is constructed with the equilibrium data as shown below:

70
60

50
H kJ kmol-1

40
30
20

10
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x,y
Locate the feed, the bottoms and the distillate on the graph and read the
enthalpies:
zF = 0.42 hF = 22.30 x 103 kJ/kmol (from graph)
xD = 0.97 hD = 21.54 x 103 kJ/kmol (from graph)
xB = 0.01 hB = 24.28 x 103 kJ/kmol (from graph)

At minimum reflux ratio, the tie-line passing through F determines P1 and P2

P1 = 98.44 x 103 kJ/kmol P2 = -34.45 x 103 kJ/kmol (from graph)

Enthalpy concentration diagram for minimum reflux

120
100 P1

80
H kJ kmol-1

60
40
20
xB zF yD
0
-0.1
-20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-40 P2
-60
x,y
The enthalpy of the vapor yD which condenses to give liquid xD is
HV1 = 53.70 x 103 kJ/kmol (from graph)
hL0 = hD = 21.54 x 103 kJ/kmol

Minimum Reflux ratio, R = (P1 – HV1) / (HV1 - hL0)

= (98.44 - 53.70) / (53.70- 21.54)
= 1.39

120
100 P1

80 yD, Hv1
60
H kJ kmol-1

40 xD, hD

20
0
-0.1
-20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-40
-60
x,y
Minimum number of stages at total reflux is found from the diagram and = 6.7

70
60

50
H kJ kmol-1

40
30
20

10
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x,y
F = 10.0 kmol/hr
Material balance equations:
F=D+B
F zF = D xD + B xB
i.e.
10.0 = D + B and
10.0 x 0.42 = 0.97 D + 0.01 B
Solving,
0.96 D = 4.0974
D = 4.27 kmol/hr
B = 5.73 kmol/hr
Energy balance equation:
F hF = DP1 + BP2

At reflux ratio of 2.5:

R = (P1 – HV1) / (HV1 - hL0)
(P1 - 53.70) / (53.70 - 21.54)= 2.5

Therefore P1 = 134.1 x 103 kJ/kmol

Substituting for the known quantities in the energy balance equation:

9.99 x 22.30 x 103 = 4.27 x 134.1 x 103 + 5.73 x P2

Hence P2= -61.05 x 103 kJ/kmol

P1= HD + QC / D
P2 = HW - QB / B

Substituting for the known quantities in the above equations,

134.1 x 103 = 21.54 x 103 + QC / 4.27

QC = 480.6 x 103 kJ/hr = 133.5 kW

-61.06 x 103 = 24.28 x 103 - QB / 5.73

QB = 488.9 x 103 kJ/hr = 135.8 kW

Condenser duty = QC = 133.5 kW

Reboiler duty = QB = 135.8 kW

Number of stages is estimated from Ponchon-Savarit method as shown in
the graph, and is equal to 11 (including the reboiler).

150

100
H kJ kmol-1

50

0
-0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-50

-100
x,y
3(a) A binary mixture is separated in an existing tray column with 10 trays. The more
volatile component (A) has a feed stream concentration of xF =0.40; at the top of
the column this component has been enriched to xD=0.90. The reflux ratio is R =
3.0. At the bottom of the column, component A has a concentration of xB=0.10.
The partial reboiler vaporizes 71.0% of the liquid coming out of the column to be
sent back into the column. Determine:
i) the thermodynamic state of the feed stream entering the column;
ii) the overall efficiency E0 of the column.
(b) It is assumed that the column efficiency will decrease to E0 =0.40. The new feed
stream will have the same thermodynamic state and the same composition as
before. The composition of distillate and bottom stream will also be the same. Is it
possible to reach these specifications with the original column without any change
in the column design? Explain your answer. How many theoretical stages are
available now to fulfil the separation task?
(c) Is the reflux ratio, which is needed to fulfil the separation task with 6 theoretical
stages, larger or smaller than in the original column? Explain your answer.
Assumptions:
-the column is adiabatic
-the molar evaporation enthalpies are identical
Equilibrium data in terms of more volatile component A:

x 0.0 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
y 0.0 0.11 0.22 0.37 0.51 0.62 0.7 0.78 0.85 0.9 0.96 1.0
(a) (i) Rectification line : R x
y x D
R 1 R 1

The ROL goes through y=x=0.9 and intersects the y-axis at:

xD 0.9
  0.225
R 1 3 1

Stripping line:
L' Bx
y x B
V' V'

The SOL goes through x = y= 0.1 It is known that 71% of the liquid is
evaporated by the reboiler. Thus V’/L’ = 0.71. The slope of the stripping line is:

L' 1
  1.41
V ' 0.71

The two lines can be drawn on the equilibrium diagram. The line connecting
the two operating lines and the point (xF,y) has an infinite slope (q=1.0) as the
feed is completely liquid at the boiling temperature.
(ii) Step by step construction in the diagram leads to a theoretical number
of stages equal to 7.5. Hence, as 10 stages are in use the overall
efficiency is E0 :

theoretical number of stages 7.5 x100

E0  X 100   75%
actual number of stages 10

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
(b)The reflux ratio at either the top or the bottom of the column can be
increased (for a given feed these two variables are independent of each
other). By increasing the reflux ratio the SOL or ROL will come closer to
the y=x line and less stages are needed to fulfill the separation task.
At total reflux the ROL and SOL are on the y=x line. Step by step
construction suggests 5 stages are required.

1 Now as efficiency is 0.40 and 10

0.9
trays are available then the
0.8
theoretical stages must be 4.
0.7

theoretical number of stages 4 x100

0.6 E0  X 100   40%
actual number of stages 10
0.5

0.4 So theoretical number of stages is

0.3
4 whereas minimum is 5. It is not
0.2
possible to do the separation with
this column.
0.1

0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
(c) The reflux must be higher than 3.0 since the separation is to be fulfilled with
6 theoretical stages. The original column had 7.5 theoretical stages. 5 was
minimum number of stages.
A stream of 100 kmol h-1 contains 0.4 mole fraction of benzene and
0.6 toluene. It is to be separated in a rectifying column, a column
with a condenser, distillate withdrawal and liquid reflux return.
There is no stripping section and no reboiler. The feed is introduced
as a saturated vapour to the bottom stage and a liquid product is
withdrawn from this plate. The distillate product contains 95 mole %
benzene and the bottoms product has a benzene mole fraction xB.
The column operates at 1 atmosphere total pressure.
a) Determine the maximum and minimum values of xB
b) Determine the number of stages required and the fraction
of the feed benzene recovered in the distillate when xB takes the
values 0.375, 0.3 and 0.25.
SOLUTION
a i) The condenser is a total condenser. On the figure the lines x = xD = 0.95
and y = yB = 0.4 are drawn. The operating line connects the diagonal (y = x)
at the point (xD, ya) but since ya= xD this is the point (xD, xD). Its lower end
point will be the point (xB, yB). If the latter point lies on the equilibrium line we
have a pinch point and the value of xB, is at a minimum. For the data, this
pinch point is 0.215 and this will correspond to an infinite number of
equilibrium stages.
The maximum value of xB occurs when the operating line lies along the
diagonal, so xB = 0.4 and number of stages is minimised.
 b) Operating lines are drawn for the three values of xB, 0.375, 0.30 and
0.25. The numbers of ideal stages required are 4, 5 and 7, respectively

Dx D x  x 1 
Recovery = D
 D
  D
 
Vy B
L  D y B  R  1 y B 
xD
 Intercept of the operating line in each case
R 1

yB is 0.4 in each case. So fractional recovery = 2.5 × intercept = (respectively)

12.5%, 37.5% and 51.25%
Alternatively:

Dx 0.95D 0.95
Re cov ery   D
 D
Vy 0.4 *100 40
B

From a benzene mass balance we can find the relationship between the
distillate flowrate, D, and the mole fraction of benzene in the distillate:

0.95D  x (100  D)  40
B

40  100 x
D B

0.95  x B

Combining the two gives

Dx 0.95D 0.95 40  100 x

Re cov ery  D
 .* B

Vy B
0.4 *100 40 0.95  x B

Substituting the three values of xB gives recoveries of 10.3%, 36.5% and 50.9%.
5. An aqueous solution of a volatile component A containing 7.94 mole
percent is preheated to its boiling point and is to be fed to the top of a
continuous stripping column operated at atmospheric pressure. Vapour
from the top of the column is to contain 11.25 mole percent A. No reflux
is to be returned. Two methods are under consideration, both calling for
the same expenditure of heat, namely, a vaporization of 0.562 kmol per
kmol feed in each case. Method 1 is to use a still at the bottom of a plate
column, generating vapour by use of steam condensing inside a closed
coil in the still. In method 2 the still and heating coil are omitted, and live
steam is injected directly below the bottom plate The equilibrium data in
terms of mole fraction A is given below.
(a) Determine:
(i) the composition and flow rate of the bottom product for the two
methods;
(ii) the number of stages required for each method.
(b) Describe the advantages of the two methods.
5.
(a) xF = 0.0794 xD = yD = 0.1125 D = 0.562F
q=1
Method 1
Fxf = Dyd + Bxb and F = D+B
Basis: 1 mole of feed (F = 1)

D xF  xB
From : 
F xD  xB

D 0.0794 - xB
we obtain:  0.562 
F 0.1125  xB

and hence xB = 0.0370

From:
B xD  xF

F xD  xB

we obtain:

B 0.1125  0.0794
  0.437
1.0 0.1125  0.0370

and hence B = 0.437

D = V, F = L, Slope = L/V = F/D =1.779

From material balance, the operating line equation:

F B
y D  xF  xB
D D
With the slope and the intercept are known, the operating line can
be drawn. The start point is the (xb,xb) and the end point is the
intersection with the q line (top stage).

The resulting theoretical stages are 3 including the reboiler.

 Method 2
F+S = D+B but S = D and B = F
B=F=1 D = 0.562
Let S = moles steam/mole feed = 0.562
By a material balance: Fxf + Sxs = Dyd + Bxb
(0.562 x 0.1125) + xB = 0.0794

xB = 0.0162
McCabe-Thiele diagrams use the same operating line for the two
methods, but in Method 2 it extends to the point (0, 0.0162).
(b)
The disadvantage of method 2 is that it requires more stages; 4 theoretical
plates instead of 2 plus a reboiler for Method 1. The advantages of method 2
is that a dirtier liquid can be handled without causing fouling in a reboiler. Also
the liquid leaving the bottom contains a smaller percentage of the more volatile
component (though the total amount is the same by either method). This may
or may not be an advantage.