Assignment #6 Chemical Engineering Separations

Due 10/21/11

Please complete the following problems: 1) 7.8 2) 7.12 3) 7.15 4) 7.16 5) 7.19 6) A distillation system is to be designed to separate propane from a binary propane-butane mixture at 101.43 kPa. The system is designed to process 100 kgmol/hour of a feed containing 50 mole percent of propane which contains 25 % saturated liquid and 75% saturated vapor. The composition of the bottoms product is 92 mole% butane. The distillation system employs a total condenser and a partial reboiler. 91.8% of the propane is recovered in the distillate product. The actual vapor flow rate inside the top section of the column is 170.5 kgmol/hr. Assuming constant molal overflow for the distillation tower, calculate the following after constructing a flow chart of this process; 1) All external flow rates and compositions. 2) Determine all appropriate operating lines and Q lines and construct these lines on Figure 2. 3) Determine the number of stages and optimum feed location. 4) What are the compositions of the vapor and liquid passing streams between the second and third stages from the top of the column?

Equilibrium Data for Propane-Butane at 101.43 kPa

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 (liquid phase mole fraction)

y (v

apo

r p

has

e m

ole

fra

ctio

n)

Exercise 7.8

Subject: Separation of ethyl alcohol and water at 1 atm. with two countercurrent cascades.

Given: One cascade (a) with given liquid feed to top stage and given vapor feed to bottom

stage. Another cascade (b) with total condenser and reflux, and given vapor feed to bottom

stage. Vapor-liquid equilibrium data for 1 atm.

Assumptions: Constant molar overflow to give straight operating lines on a y-x diagram.

Find: (a) Compositions of V4 and L1 for 4 stages in cascade (a).

(b) Number of equilibrium stages for 85 mol% alcohol in exit vapor of cascade (a).

(c) Compositions of D and L1 for 4 stages in cascade (b).

(d) Number of equilibrium stages for 50 mol alcohol in D of cascade (b).

Analysis: From the given vapor-liquid equilibrium data, in the composition range of interest,

ethyl alcohol is more volatile than water. Therefore, the y and x coordinates in a y-x plot pertain

to ethyl alcohol.

(a) Since L = 100 mol and V = 100 mol, the slope of the operating line from Eqs. (7-6) or (7-11)

= L/V = 100/100 = 1. The terminal points on the operating line as (y, x) are: (?, 0.7) at the top

and (0.3, ?) at the bottom. To determine the compositions of V4 and L1 for 4 stages, this

operating line is located so that exactly 4 stages are stepped off in a y-x diagram, as shown below.

From the diagram, the ethanol compositions are 76 mol% in V4 and 24 mol% in L1.

Exercise 7.8 (continued) Analysis: (continued)

(b) It is impossible to obtain an overhead vapor with 85 mol% ethanol. With an infinite

number of stages, the highest concentration of ethanol in the overhead vapor corresponds to that

in equilibrium with the top liquid feed containing 70 mol% ethanol. From the given vapor-liquid

equilibrium data, the highest concentration is an ethanol mole fraction of 0.82.

(c) Since the bottom vapor feed, V0 = 100 mol and D = 50 mol, by overall material

balance, L1 = V0 - D = 100 - 50 = 50 mol. Because of the assumption of constant molar

overflow, L = LR = L1 = 50 mol. By material balance around the condenser or because of

constant molar overflow, V = V4 = LR + D = 50 + 50 = 100 mol. The slope of the operating

line from Eqs. (7-6) or (7-11) = L/V = 50/100 = 0.5. To determine the compositions of D and L1

for 4 stages, an operating line of this slope is located so that exactly 4 stages are stepped off in a

y-x diagram, as shown below. From the diagram, the ethanol compositions are 45 mol% in D

and 16 mol% in L1.

(d) Since the distillate is 50 mol% ethanol, 25 moles of ethanol and 25 moles of water

leave in the distillate. Because the feed is 30 moles of ethanol and 70 moles of water, L1 , the

leaving liquid, contains 5 moles of ethanol and 45 moles of water. Thus, the terminal points on

the operating line, because of the total condenser, as (y, x), are: (0.5, 0.5) at the top and

(0.3, 0.1 ) at the bottom. However, point (0.3, 0.1) above the equilibrium line is impossible.

Exercise 7.12

Subject: Distillation of a mixture of benzene (A) and toluene (B) at 101 kPa.

Given: Column consisting of a partial reboiler, one theoretical plate, and a total condenser.

Produce a distillate of 75 mol% benzene from a saturated liquid feed of 50 mol% benzene.

Assumptions: Constant molar overflow. Constant relative volatility = aA,B = 2.5.

Find: Number of moles of distillate per 100 moles of feed for:

(a) Feed to the reboiler and no reflux.

(b) Feed to the reboiler and a reflux ratio, L/D = 3.

(c) Feed to the plate and a reflux ratio of 3.

(d) Same as (c) except a partial condenser.

(e) Feed to the reboiler with minimum reflux.

(f) Feed to the reboiler with total reflux.

Analysis: Either a graphical or analytical method can be used. Because the relative volatility is

assumed constant, use an analytical method. For each part, the theoretical plate and the partial

reboiler are equilibrium stages. Benzene is the more volatile component, so the y-x diagram is

based on benzene. Because the relative volatility = constant = 2.5, the equilibrium relationship is

given by Eq. (7-3),

yx

x

x

x=

+ -=

+

a

a1 1

2 5

1 15( )

.

. (1)

Take as a basis, 100 moles of feed. Therefore, the feed is 50 moles of A and 50 moles of B.

(a) With no reflux, separation occurs only in the reboiler. The vapor leaving the reboiler

is totally condensed to become the distillate with yD = xD = 0.75. Solve Eq. (1) for equilibrium x,

xy

y yB

D

D D

=+ -

=+ -

=a( )

.

. . ( . ).

1

0 750

0 750 2 5 1 0 7500545 (2)

Because the distillate and bottoms have benzene mole fractions greater than the mole fraction of

the feed (0.5), it is impossible to obtain a distillate with a benzene mole fraction of 0.75.

(b) From the reflux ratio, L = 3D, V = L + D = 4D. Therefore, D/V = 1/4 and L/V =3/4.

Use a subscript of D for distillate, R for reflux, B for streams leaving the reboiler, and 1 for the

theoretical plate, when used. With 1 theoretical plate, from part (a),

y1 = 0.75 xD = 0.75 x1 = 0.545

Benzene material balance around plate 1,

y V x L y V x LB D+ = +1 1 (3)

Solving for yB

y y x xL

VB D= + -

FHGIKJ = + -

FHGIKJ =1 1 0 750 0545 0 750

3

40596( ) . ( . . ) .

Exercise 7.12 (continued) Analysis: (b) (continued)

The mole fraction of benzene in the bottoms product is in equilibrium with yB =0.596.

Therefore,

the form of Eq. (2) applies,

xy

y yB

B

B B

=+ -

=+ -

=a( )

.

. . ( . ).

1

0596

0596 2 5 1 05960 371

Overall total material balance, F = 100 = D + B (4)

Overall benzene material balance, xFF = xDD + xBB or 50 = 0.75D + 0.371B (5)

Solving Eqs. (4) and (5), D = 34.2 moles or 34.2 mol/100 mol feed, and B = 65.8 moles.

(c) With the feed to the theoretical plate, the following results apply from part (b), y1 = 0.75

xD = 0.75 x1 = 0.545

Benzene material balance around Stage 1, which now includes the feed,

xFF + y V x L y V x LB D+ = +1 1 (6)

Solving for yB,

y yV

Vx

L

Vx

L

Vx

F

V

V

V

L

V

L

V VB D F=FHGIKJ +FHGIKJ -FHGIKJ -FHGIKJ =

FHGIKJ +

FHGIKJ -

FHGIKJ -FHGIKJ1 1 0 750 0545 0 750 050

100. . . . (7)

Because the feed is a saturated liquid, , V V L L= = + and 100

From above, V = 4D and L/V = 3/4. Also, L V L V V V/ / / / /= + = +100 3 4 100

Therefore, Eq. (7) becomes,

yV V V D

B = + +FHG

IKJ -

FHGIKJ -FHGIKJ = - = -0 750 0545

3

4

1000 750

3

4050

1000596

4 50596

1125. . . . .

..

. (8)

The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the left-

hand part of Eq. (2),

xy

y yB

B

B B

=+ -2 5 1. ( )

(9)

Overall total material balance, F = 100 = D + B (10)

Overall benzene material balance, xFF = xDD + xBB or 50 = 0.75D + xBB (11)

Solving Eqs. (8), (9), (10), and (11),

yB = 0.647, xB = 0.423, D = 23.5 moles or 23.5 mol/100 mol feed, B = 76.5 moles

(d) With a partial condenser, the mole fraction of the liquid reflux is that in equilibrium

with the vapor distillate. Therefore, from the above results,

yD = 0.75 xR = 0.545 y1 = 0.596 x1 = 0.371

Benzene material balance around the theoretical plate, which includes the feed,

xFF + y V x L y V x LB R+ = +1 1 (12)

Solving for yB,

y yV

Vx

L

Vx

L

Vx

F

V

V

V

L

V

L

V VB R F=FHGIKJ +FHGIKJ -FHGIKJ -FHGIKJ =

FHGIKJ +

FHGIKJ -

FHGIKJ -FHGIKJ1 1 0596 0 371 0545 050

100. . . . (13)

Exercise 7.12 (continued) Analysis: (d) (continued)

Because the feed is a saturated liquid, , V V L L= = + and 100

From above, V = 4D and L/V = 3/4. Also, L V L V V V/ / / / /= + = +100 3 4 100

Therefore, Eq. (13) becomes,

yV V V D

B = + +FHG

IKJ -

FHGIKJ -FHGIKJ = - = -0596 0 371

3

4

1000545

3

4050

1000 466

12 90 466

323. . . . .

..

. (14)

The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the left-

hand part of Eq. (2),

xy

y yB

B

B B

=+ -2 5 1. ( )

(15)

Overall total material balance, F = 100 = D + B (16)

Overall benzene material balance, xFF = yDD + xBB or 50 = 0.75D + xBB (17)

Solving Eqs. (14), (15), (16), and (17),

yB = 0.405, xB = 0.214, D = 53.4 moles or 53.4/100 mol feed, B = 46.6 moles

(e) At minimum reflux, with the feed sent to the still pot (partial reboiler), an infinite number of

theoretical plates is needed between the condenser and reboiler. This part is not completely

specified. In order to compute the distillate, we must assume a bottoms benzene mole fraction

less than that in the feed. Suppose we choose that mole fraction to be 0.45. Then the operating

line will intersect the equilibrium line at x = 0.45, creating the pinch zone of infinite stages. The

value of y at the intersection is given by Eq. (1):

y =+

=2 5 0 45

1 15 0 450 672

. ( . )

. ( . ).

Therefore the operating line passes through the two points, as {y, x}, of {0.75, 0.75} and

{0.672, 0.450}. Therefore, the slope = L/V = (0.75 - 0.672)/(0.75 - 0.45) = 0.260. Now compute

the overall material balances:

Overall total material balance, F = 100 = D + B (18)

Overall benzene material balance, xFF = yDD + xBB or 50 = 0.75D + 0.45B (19)

Solving Eqs. (18) and (19), D = 16.67 moles or 16.67 mol/100 mol feed, and B = 83.33 moles

Calculations for other values of the benzene mole fraction in the bottoms can be made in the

same manner.

(f) At total reflux, there is no distillate, but there is a boilup. The moles of distillate per 100

moles of feed = 0.

Exercise 7.15

Subject: Effect of loss of plates in a distillation column separating a benzene-toluene mixture.

Given: Saturated vapor feed of 13,600 kg/h of 40 wt% benzene and 60 wt% toluene. Column

with 14 plates above the feed location. Plate efficiency is 50%. Reflux ratio is 3.5. Previously,

with 10 plates in the stripping section, column could achieve a distillate of 97 wt% benzene and a

bottoms of 98 wt% toluene. Vapor-liquid equilibrium data in Exercise 7.13.

Assumptions: Constant molar overflow. Total condenser and partial reboiler.

Find: (a) If column with 10 inoperable plates can yield a distillate of 97 wt% benzene,

assuming that we no longer can achieve the 98 wt% bottoms product.

(b) The distillate flow rate.

(c) The composition of the bottoms.

Analysis: (a) First convert the feed to kmol/h and mole fractions, using molecular weights of

78.11 for benzene and 92.13. The result is:

Component kmol/h Mole fraction

Benzene 69.65 0.44

Toluene 88.57 0.56

Total: 158.22 1.00

For a distillate of 97 wt% benzene, the mole fraction for benzene, the more volatile of the two

components, is,

xD =

+

=

97

781197

7811

3

9213

0 974.

. .

.

With a reflux ratio of 3.5, from Eq. (7-7), the slope of the rectifying section operating line is,

L/V = R/(1 + R) = 3.5/4.5 = 0.778

The q-line is a horizontal line at y = 0.44. For 14 plates with 50% efficiency, the column has the

equivalent of 7 equilibrium stages + 1 for the partial reboiler.

The McCabe-Thiele construction is shown on the next page, where it is seen that it is possible to

obtain the desired distillate composition.

(b) and (c) From the McCabe-Thiele diagram, the mole fraction of benzene in the bottoms is

xB = 0.24. As a weight fraction, this corresponds to,

0 24 7811

0 24 7811 0 76 92130 211

. ( . )

. ( . ) . ( . ).

+= weight fraction or 21.1 wt% benzene

Compute the distillate rate by overall molar material balances.

Overall total mass balance: 158.22 = D + B (1)

Overall benzene mass balance: 69.65 = 0.974D + 0.240B (2)

Solving Eqs. (1) and (2): D = 43.16 kmol/h B = 115.06 kmol/h

By weight, D = 43.16[0.974(78.11) + 0.026(92.13)] = 3,387 kg/h

Exercise 7.15 (continued) Analysis: (a) (continued)

McCabe-Thiele Diagram

Exercise 7.16

Subject: Effect on the separation of A from B by distillation when 3 of 7 theoretical plates

rust and drop to the bottom of the column.

Given: . Column has 7 theoretical plates + partial reboiler. Saturated liquid feed of 100 kmol/h

of 50 mol% A is sent to plate 5 from the top. Distillate contains 90 mol% A. The L/V = 0.75 in

the rectifying section. Vapor-liquid equilibrium data.

Assumptions: Constant molar overflow. Total condenser.

Find: Case 1: Column before the 3 plates rust and drop.

(a) Composition of the bottoms product.

(b) The L/V in the stripping section.

(c) The kmol/h of bottoms product.

Case 2: If plates 5, 6, and 7 counted down from the top are lost:

(a) Composition of bottoms product.

Case 3: Same as Case 2, except replace reflux with the same molar flow rate of product

containing 80 mol% A:

(a) Composition of distillate.

(b) Composition of bottoms.

Analysis: Case 1: Apply the McCabe-Thiele method in terms of component A, which is

more volatile than B. The rectifying section operating line passes through [0.90, 0.90] with a

slope of 0.75. The q-line is vertical through x = 0.50. Step off 4 stages in the rectifying section.

Then, by trial and error, find an xB with a corresponding stripping section operating line that

gives 4 equilibrium stages in the stripping section. The result is shown on the following page,

where:

(a) Bottoms contains 7 mol% A and 93 mol% B.

(b) The slope of the stripping section operating line from the coordinates of the line is:

{[0.90 0.75(0.90 0.50] 0.07}/(0.50/ 1.20 07) 3.= - - - - =L V

(c) By material balances, F = D + B and FxF = 50 = 0.9D + 0.07B. Solving these two

equations, distillate flow rate = 51.8 kmol/h and bottoms flow rate = 48.2 kmol/h

Case 2: We now have 4 equilibrium stages and a partial reboiler, with the feed being

sent to the reboiler. Assume that utility rates are such that L/V and L V/ are the same as in Case

1. Then, on the McCabe-Thiele diagram, the values of xD and xB must shift so that 5 stages are

stepped off, with the fifth, which is the reboiler, intersecting the stripping section operating line

at the 45o line. This is shown on the McCabe-Thiele diagram on the following page.

(a) From this diagram, the mole fractions of benzene in the distillate and bottoms are

0.80 and 0.21, respectively.

Case 3: Since the distillate composition in Case 2 is 80 mol% benzene, the results would be the

same as Case 2 if an 80 mol% benzene stream from another column were used as reflux.

Exercise 7.16 (continued)

Analysis: Case 1 (continued)

Exercise 7.16 (continued)

Analysis: Case 2 (continued)

Exercise 7.19

Subject: Normal and abnormal operation of a distillation column separating a methanol -

water mixture at 101 kPa.

Given: Column with 7 theoretical plates, a total condenser, and a partial reboiler. A feed of

100 kmol/h of 50 mol% methanol in water is sent to plate 3 from the bottom. During normal

operation, distillate is 90 mol% methanol and bottoms is 5 mol% methanol, with a reflux rate of

1 mole per mole distillate. During abnormal operation, the following data are obtained:

Stream kmol/h mol% methanol

Feed 100 51

Bottoms 62 12

Distillate 53 80

Reflux 94 -

Vapor-liquid equilibrium data are given at 101 kPa, where methanol is the more volatile species.

Assumptions: Constant molar overflow.

Find: Most probable cause for abnormal operation.

Recommended further tests.

If 90 mol% methanol distillate could be obtained by increasing the reflux ratio for a

constant vapor rate.

Analysis:

First determine whether the normal operation can be verified by the McCabe-Thiele

method. With L/D = R = 1, from Eq. (7-7), the slope of the rectifying operating line = L/V =

R/(1 + R) = 1/2 = 0.5. Also, xF = 0.5, xD = 0.90, and xB = 0.05.

What is not known is the phase condition of the feed. If a saturated liquid feed is

assumed, giving a vertical q-line as shown in the plot below, stepping stages up from the bottom,

with the feed stage to plate 3 from the bottom, less than 2 theoretical plates are needed in the

rectifying section, while 4 are present. The construction is shown on the next page.

Therefore, it appears that the feed is not a saturated liquid, but is partially vaporized.

Exercise 7.19 (continued) Analysis: Normal Operation (continued)

By trial and error, using q-lines of various slopes, the following McCabe-Thiele diagram is

consistent with the given data. It shows a q-line with a slope of -0.34.

slope = /( 1)

Therefore, slope/(slope-1)=-0

m

.34/(-0.34-1.0)=0.25

From Eq. (7-19 olar fraction vaporize), = 1d 0.7. 51 0 25

-

=

- = - =

q q

q

q

Exercise 7.19 (continued) Analysis: Normal operation (continued)

The material balance for the normal operation is as follows, using the overall balances,

F = 100 = D + B and 0.5F = 0.5(100) = 50 = xDD + xBB = 0.90D + 0.05B.

Stream kmol/h mol% methanol

Feed 100 50

Bottoms 47.06 5

Distillate 52.94 90

Reflux 52.94 90

Exercise 7.19 (continued) Analysis: Abnormal operation For the abnormal operation, first check the overall total material balance using the given

data. F = 100 kmol/h. D + B = 53 + 62 = 115 kmol/h. Therefore, it appears that we have

115 - 100 = 15 kmol/h more flow out of the distillation system. Now check the methanol overall

material balance using the given data. Methanol flow rate in = 0.51(100) = 51 kmol/h. Methanol

flow rate out = 0.80(53) + 0.12(62) = 49.84 kmol/h. Therefore, the methanol balance is close,

with only about a 2% discrepancy. Now check the water overall material balance using the given

data. Water flow in = 0.49(100) = 49 kmol/h. Water flow out = 0.20(53) + 0.88(62) = 65.16

kmol/h. Therefore, we have 65.16 - 49 = 16.16 kmol/h more water out than in. This is a

significant discrepancy. It appears certain that water is leaking into the distillation system. Two

possibilities are: (1) leakage of condenser cooling water into the condensate, or (2) leakage of

reboiler steam into the boilup vapor. A reboiler steam leak may not be serious because the steam

might not get to the top of the column to dilute the methanol product. A condenser cooling water

leak could be very serious because part of it would end up in the distillate, thereby diluting the

methanol product. Because of the impure methanol distillate for the abnormal operation, it

appears that a condenser cooling water leak is the fault. Check this next. We note that the distillate flow rate for the abnormal operation is almost exactly the same

as that for the normal operation. A flow rate equal to that of he leakage passes out the bottom of

the column. In normal operation, the water passing out in the distillate = 0.1(53) = 5.3 kmol/h,

while for the abnormal operation, the water passing out in the distillate = 0.2(53) = 10.6 kmol/h.

Thus, an additional 5.3 kmol/h of water leaves in the distillate. For the abnormal operation, the

overhead vapor rate = 53 + 94 = 147 kmol/h and, therefore, 53/147 x 100% = 36% of the

overhead vapor (total condensate) is distillate. Thus, if 15 kmol/h of water leaked into the

overhead vapor, then, we would expect 0.36(15) = 5.4 kmol/h would be expected to leave with

the distillate. This compares very well with the 5.3 kmol/h additional water calculated above by

material balance. If the degree of fractionation within the column is about the same as for the

normal operation, it could be concluded that a condenser cooling water leak is to blame.

To check the cooling water leak, could meter the cooling water in and out of the

condenser and see if there is a difference.

If the vapor rate is kept constant and the reflux rate is increased, then the distillate rate

must be decreased. Assume a vapor rate of 147 kmol/h, with 30 kmol/h to distillate and 117

kmol/h to reflux. Then, 30/117 x 100% = 25.6% of the overhead vapor is distillate. Therefore,

the water leak to the distillate would be 0.256(15) = 3.84 kmol/h. If the fractionation were

otherwise the same as for normal operation so that the overhead vapor was 90 mol% methanol,

the dilution with leakage would result in 0.1(30 - 3.84) + 3.84 = 6.46 kmol/h of water in 30

kmol/h. Thus, methanol purity = (30 - 6.46)/30 x 100% = 78.5 mol%. However, the higher

reflux ratio would increase the fractionation, so as to increase the purity above this value. A

further increase in fractionation could be achieved, if the feed were condensed to a saturated

liquid and additional heat was transferred in the reboiler. But, even if a pure methanol overhead

vapor were achieved, the methanol purity after dilution with the water leakage would be :

(30 - 3.84)/30 x 100% = 87.2 mol% methanol. Must eliminate the leak.