Question

Two streams containing a mixture of methanol and water are to be distilled in a distillation column
operating at 1 atm. The first stream enters the column closer to the top as a saturated liquid at 73
°C, with a flow rate of 55 kmol/h. The second stream enters the column at the temperature of 90°C
closer to the bottom, with a flow rate of 40 kmol/h, at the concentration of 30 mol% methanol. The
bottoms stream leaving the reboiler contains 4 mol % methanol and the distillate contains 90 mol
% methanol. The vapour is generated at the bottom of the column in an external reboiler and the
column has a total condenser.+

a) Draw the flowchart of the process and determine the flow rates of the distillate and bottoms
streams.
b) Show all the calculations required to determine all the operating lines for a reflux ratio of 2.
Plot all the operating lines.
c) Determine the minimum reflux ratio.

Answer 1

The flowsheet of the process is shown below

The feed 1 is saturated liquid at 73°C

From Perry handbook

Antonie constants for methanol

A : 5.20409 , B: 1581.341 , C: -33. 50

Antonie comstants for water

A: 5.0768, b: 1659.793 , C: -45. 854

Pas = 10(A-(B/(T+C) ))

T is in K and P is in bar

At saturations conditions

Bubble pressure = Total pressure

Total pressure = 1 atm = 1.013 bar

At T = 73°C= 346 K , the feed 1 is saturated

Solving we get

x1 = 0.6351

Mole fraction of methanol = 0.6351

Feed 1 is saturated liquid

q = 1

Feed 2 enters at 90°C = 363 K

The feed 2 contains 30% methanol

Bubble pressure at 90°C can be calculated by above equation by substituting T = 363 K

P = 1.2499 bar

At dew point

x = 1

Dew pressure

Dew pressure = 0.89109 bar

Bubble pressure = 1.2499 bar

Operating pressure = 1.013 bar

(L/F) = (1.013-0.89109) /(1.2499-0.89109)

(L/F) = 0.3397 = q

q = L/F = 0.3397

Doing overall balance

F1 +F2 = D+B

55+40 = D+B

95 = D+B

Doing overall methanol balance we get

(55)(0.6351)+(40)(0.30) = D(0.90) +B(0.04)

Solving both equations we get

D = 50.151 kmol/h

B = 44.848 kmol/h

Given reflux ratio = 2

L/D = 2

L = 2(50.151) = 100.302 kmol/h

For feed 1

Doing balance around condenser

V = L+ D

V = 100.302+50.151= 150.493 kmol/h

Slope of enriching section operating line for

feed 1 = L/V = 100.302/150.493 = 0.666

Intecept for enriching section operating line =

xd/R+1) = 0.90/3 = 0.30

For stripping section operating line of feed 1 we get

V'+F1 = D+L

Doing component balance

V'(y) +F1(xf) = L(x) + D(xd)

Srripping section operating line equation for feed 1

y = (L'/V')(x) + (D/V')(xd) -(F1/V') (xf)

y = (L'/V') x + (D/V') (0.90) -(F1/V') (0.6351)

For saturated liquid

L' = L+ F1

V'= V

L' = 100.302+55 = 155.302 kmol/h

V' = V = 150.493

Slope of stripping section operating line for feed 1 = L'/V '= (155.302+150.493) = 1.031

Feed 2 is partially vaporized feed

q = 0.3397

Slope of q line = q/(q-1) = 0.3397/(0.3397-1) =

= -0. 5144

From Perry handbook

The equilibrium data for methanol and water at 1 atm is given by

x	0	0.040	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1
y	0	0.23	0.418	0.579	0.655	0.729	0.779	0.825	0.879	0.915	0.958	1

Plotting equilibrium curve and drawing operating line 1 with slope = 0.666

Operating line 2 slope = 1.031

Slope of q line for feed 1 = infinity

Slope of q line for feed 2 = -0. 5144

The three operating line is shown in the graph below

The third operating line is obtained by joining F2 with B

From graph

For feed 1

y' = 0.84

x' = 0.635

xd = 0.90

Rmin = (xd-y') /(y'-x')

Rmin = (0.90-0.6351) /(0.84-0.635) = 1.292

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