DSP: Seminar 2, Binary distillation, McCabe-Thiele method, column with one and multiple feeds

18 kg/kmol 60.5 kg/kmol at 0.4 Mpa 2133.9

Solution:

MH2O Maa

Solution of Example 1: a.

1- Scheme and material balance:

FEED

DESTILAT

BOTTOMS

WWDDFF

WDF

xnxnxnnnn

DFW

WD

WFFD

WDWFDDWDDFF

nnnxxxx

nn

xnxnxnxnDnFxnxn

)(

2- Prepare x-y diagram using equilibrium data:

x-y diagram, Water-Acetic Acid

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1x

y

3- Select on diagram point F, D and W

Δ vhsteam

Seminar 2 Binary distillation, McCabe-Thiele method, column with one and multiple feeds Example 1: Hundred kmol/h of a bubble-point liquid mixture of 75 mol% water and 25 mol% acetic acid is distillated continuously in a distillation tower at a pressure of 1 atmosphere. Distillate contains 95 mol% of light component and Bottoms 95 mol% of heavy component. The reflux ratio is 2 times of minimum reflux ratio.

a- determine the number of theoretical stages and optimum feed stage lacation b- calculate steam requirement in reboiler and requirement of cooling water in total

condenser if the steam pressure is 0.4 Mpa and cooling water is preheated by 20 oC, Only condensation heat of the steam is used and reflux ie returned to the column at boiling point.

Data: Equilibrium data Water Acetic acid at 1 atm (mol frictions) x 0 0.0055 0.053 0.125 0.206 0.297 0.510 0.649 0.803 0.9594 1 y 0 0.0112 0.133 0.240 0.338 0.437 0.630 0.751 0.866 0.9725 1 heat of evaporation at Tw Δvhwater= 2214 kJkg-1 Δvhaa=390 kJkg-1

Solution of Example 1: a.

1- Scheme and material balance:

FEED

DESTILAT

BOTTOMS

WWDDFF

WDF

xnxnxnnnn

DFW

WD

WFFD

WDWFDDWDDFF

nnnxxxx

nn

xnxnxnxnDnFxnxn

)(

2- Prepare x-y diagram using equilibrium data:

x-y diagram, Water-Acetic Acid

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1x

y

3- Select on diagram point F, D and W

x-y diagram, Water-Acetic Acid

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 1x

y

4- Draw q-line q-line is graphical interpretation of material balance of the feed stage, q represents the amount of liquid that accumulates at the feed stage by feeding of 1 kmol of the feed. q-line equation:

11

qx

xq

qy F

For bubble point liquid, q=1, The slop of q-line equation is o

qqtg 90

1

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

5- calculation of minimum reflux ration For calculation minimum reflux ration Rmin the operating line in the rectifying section of the column for at Rmin should be drawn.

11 minmin

min

Rx

xR

Ry D

We have two points of this line one is the intersection of q-line and equilibrium curve and another in the intersection of 45o line and xD line.

x-y diagram, Water-Acetic Acid

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 1x

y

4- Draw q-line q-line is graphical interpretation of material balance of the feed stage, q represents the amount of liquid that accumulates at the feed stage by feeding of 1 kmol of the feed. q-line equation:

11

qx

xq

qy F

For bubble point liquid, q=1, The slop of q-line equation is o

qqtg 90

1

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

5- calculation of minimum reflux ration For calculation minimum reflux ration Rmin the operating line in the rectifying section of the column for at Rmin should be drawn.

11 minmin

min

Rx

xR

Ry D

We have two points of this line one is the intersection of q-line and equilibrium curve and another in the intersection of 45o line and xD line.

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

The minimum reflux ratio can be calculated from the slope of this line

'''

min xyyx

R D

or from the section 1min R

xD on the y axis for x=0

6- calculate reflux ratio (R) as R=1.2 Rmin

7- calculate the section 1R

xD on the y axis for x=0

8- draw the operating lineof the rectifying section of the column by connecting points (0,

1RxD ) and (xD, yD)

9- draw the operating line of the striping section of the column, by connecting intersection point of q-line and operating line of rectifying section with point (xw, yw)

y‘

x‘

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

The minimum reflux ratio can be calculated from the slope of this line

'''

min xyyx

R D

or from the section 1min R

xD on the y axis for x=0

6- calculate reflux ratio (R) as R=1.2 Rmin

7- calculate the section 1R

xD on the y axis for x=0

8- draw the operating lineof the rectifying section of the column by connecting points (0,

1RxD ) and (xD, yD)

9- draw the operating line of the striping section of the column, by connecting intersection point of q-line and operating line of rectifying section with point (xw, yw)

y‘

x‘

x-y diagram, Water-Acetic Acid

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 1x

y

10- draw steps between equilibrium curve and operating lines

x-y diagram, Water-Acetic Acid

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 1x

y

Number of theoretical stages= number of steps – 1 (reboiler) Optimal feed stage = intersection of q-line and operating lines

x-y diagram, Water-Acetic Acid

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 1x

y

10- draw steps between equilibrium curve and operating lines

x-y diagram, Water-Acetic Acid

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 1x

y

Number of theoretical stages= number of steps – 1 (reboiler) Optimal feed stage = intersection of q-line and operating lines

b. Enthalpy balance of reboiler:

vvvmsteamsteam hnvhm

)1( waavwwatervvv yhyhh

)1()1( qnRnn FDvm Where msteam is the mass flow of steam, steamvh is heat of condensation of water at pressure of the steam, nvm is mol flow of boil-up and vvh is heat of evaporation of boil-up at reboiler temperature.

water

vvvmsteam vh

hnm

Enthalpy balance of total condenser:

DvDDvvnOHOHOH hRnhntCpm )1(222

)1( DaavDwatervDv xhxhh Where mH2O is the mass flow of cooling water, nvn is mol flow of overhead vapor and Dv h is heat of evaporation of distillate at condenser temperature.

b. Enthalpy balance of reboiler:

vvvmsteamsteam hnvhm

)1( waavwwatervvv yhyhh

)1()1( qnRnn FDvm Where msteam is the mass flow of steam, steamvh is heat of condensation of water at pressure of the steam, nvm is mol flow of boil-up and vvh is heat of evaporation of boil-up at reboiler temperature.

water

vvvmsteam vh

hnm

Enthalpy balance of total condenser:

DvDDvvnOHOHOH hRnhntCpm )1(222

)1( DaavDwatervDv xhxhh Where mH2O is the mass flow of cooling water, nvn is mol flow of overhead vapor and Dv h is heat of evaporation of distillate at condenser temperature.

DSP: Seminar 2, Binary distillation, McCabe-Thiele method, column with one and multiple feeds

kJkg-1

Solution of Example 1: a.

1- Scheme and material balance:

FEED

DESTILAT

BOTTOMS

WWDDFF

WDF

xnxnxnnnn

DFW

WD

WFFD

WDWFDDWDDFF

nnnxxxx

nn

xnxnxnxnDnFxnxn

)(

2- Prepare x-y diagram using equilibrium data:

x-y diagram, Water-Acetic Acid

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1x

y

3- Select on diagram point F, D and W

Example 2: Determine the number of theoretical stages and optimum feed stages locations in column in example 1, if we have two feed streams, one the same as in example 1 and another 50 kmol/h 50 mol% vaporized, which is consist of 50 mol% water and 50 mol% acetic acid. All other data are same as in example 1.

Solution of Example 2: Scheme and mass balance:

F1

F2

D

W

nL1 nv1

nL3

nL2 nv2

nv3

1

2

3

WWDDFFFF xnxnxnxnnWnDnFnF

2211

21

WD

WFFFFFFD xx

xnnxnxnn

)( 212211

nW =nF1+nF2-nD q1=1 q2=0.5 Section 1: Same as rectifying section in previous example nL1=nD.R nV1=nD(R+1) Equation of operating line:

11

Rx

xR

Ry D

q-line equation:

11 11

1

qx

xq

qy F

Solution of Example 1: a.

1- Scheme and material balance:

FEED

DESTILAT

BOTTOMS

WWDDFF

WDF

xnxnxnnnn

DFW

WD

WFFD

WDWFDDWDDFF

nnnxxxx

nn

xnxnxnxnDnFxnxn

)(

2- Prepare x-y diagram using equilibrium data:

x-y diagram, Water-Acetic Acid

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1x

y

3- Select on diagram point F, D and W

x-y diagram, Water-Acetic Acid

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 1x

y

4- Draw q-line q-line is graphical interpretation of material balance of the feed stage, q represents the amount of liquid that accumulates at the feed stage by feeding of 1 kmol of the feed. q-line equation:

11

qx

xq

qy F

For bubble point liquid, q=1, The slop of q-line equation is o

qqtg 90

1

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

5- calculation of minimum reflux ration For calculation minimum reflux ration Rmin the operating line in the rectifying section of the column for at Rmin should be drawn.

11 minmin

min

Rx

xR

Ry D

We have two points of this line one is the intersection of q-line and equilibrium curve and another in the intersection of 45o line and xD line.

Solution of Example 2: Scheme and mass balance:

F1

F2

D

W

nL1 nv1

nL3

nL2 nv2

nv3

1

2

3

WWDDFFFF xnxnxnxnnWnDnFnF

2211

21

WD

WFFFFFFD xx

xnnxnxnn

)( 212211

nW =nF1+nF2-nD q1=1 q2=0.5 Section 1: Same as rectifying section in previous example nL1=nD.R nV1=nD(R+1) Equation of operating line:

11

Rx

xR

Ry D

q-line equation:

11 11

1

qx

xq

qy F

Section 2: nL2=nL1+nF1q1 nV2=nV1+nF1(q1-1) Equation of operating line:

2

11

2

2

V

FFDD

V

L

nxnxn

xnn

y

q-line equation:

11 22

2

qx

xq

qy F

Section 3: nL3=nL2+nF2q2 nV3=nL3-nW Equation of operating line:

WL

Ww

WL

L

nnxn

xnn

ny

33

3

Calculation steps:

1- Follow steps 1-8 as in example 1, considering new equation for mass balance 2- Calculate nL1, nV1, nL2, nV2, nL3, nV3 3- Plot q2-line using q2-line equation, one point of the line is (xF2, yF2), for obtaining

another point choose an x and calculate a y. 4- Plot the operating line in section 3, using operating line equation. One point is (xW,

yW), for obtaining another point choose an x and calculate a y. Plot operating line in section 2 by connecting two intersections, intersection of operating line in section 1 and q1-line and intersection of operating line in section 3 and q2-line.

x-y diagram, Water-Acetic Acid

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 1x

y

4- Draw q-line q-line is graphical interpretation of material balance of the feed stage, q represents the amount of liquid that accumulates at the feed stage by feeding of 1 kmol of the feed. q-line equation:

11

qx

xq

qy F

For bubble point liquid, q=1, The slop of q-line equation is o

qqtg 90

1

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

5- calculation of minimum reflux ration For calculation minimum reflux ration Rmin the operating line in the rectifying section of the column for at Rmin should be drawn.

11 minmin

min

Rx

xR

Ry D

We have two points of this line one is the intersection of q-line and equilibrium curve and another in the intersection of 45o line and xD line.

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

The minimum reflux ratio can be calculated from the slope of this line

'''

min xyyx

R D

or from the section 1min R

xD on the y axis for x=0

6- calculate reflux ratio (R) as R=1.2 Rmin

7- calculate the section 1R

xD on the y axis for x=0

8- draw the operating lineof the rectifying section of the column by connecting points (0,

1RxD ) and (xD, yD)

9- draw the operating line of the striping section of the column, by connecting intersection point of q-line and operating line of rectifying section with point (xw, yw)

y‘

x‘

x-y diagram, Water-Acetic Acid

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 1x

y

q-line

The minimum reflux ratio can be calculated from the slope of this line

'''

min xyyx

R D

or from the section 1min R

xD on the y axis for x=0

6- calculate reflux ratio (R) as R=1.2 Rmin

7- calculate the section 1R

xD on the y axis for x=0

8- draw the operating lineof the rectifying section of the column by connecting points (0,

1RxD ) and (xD, yD)

9- draw the operating line of the striping section of the column, by connecting intersection point of q-line and operating line of rectifying section with point (xw, yw)

y‘

x‘

x-y diagram, Water-Acetic Acid

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 1x

y

10- draw steps between equilibrium curve and operating lines

x-y diagram, Water-Acetic Acid

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 1x

y

Number of theoretical stages= number of steps – 1 (reboiler) Optimal feed stage = intersection of q-line and operating lines

x-y diagram, Water-Acetic Acid

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 1x

y

10- draw steps between equilibrium curve and operating lines

x-y diagram, Water-Acetic Acid

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 1x

y

Number of theoretical stages= number of steps – 1 (reboiler) Optimal feed stage = intersection of q-line and operating lines

b. Enthalpy balance of reboiler:

vvvmsteamsteam hnvhm

)1( waavwwatervvv yhyhh

)1()1( qnRnn FDvm Where msteam is the mass flow of steam, steamvh is heat of condensation of water at pressure of the steam, nvm is mol flow of boil-up and vvh is heat of evaporation of boil-up at reboiler temperature.

water

vvvmsteam vh

hnm

Enthalpy balance of total condenser:

DvDDvvnOHOHOH hRnhntCpm )1(222

)1( DaavDwatervDv xhxhh Where mH2O is the mass flow of cooling water, nvn is mol flow of overhead vapor and Dv h is heat of evaporation of distillate at condenser temperature.

b. Enthalpy balance of reboiler:

vvvmsteamsteam hnvhm

)1( waavwwatervvv yhyhh

)1()1( qnRnn FDvm Where msteam is the mass flow of steam, steamvh is heat of condensation of water at pressure of the steam, nvm is mol flow of boil-up and vvh is heat of evaporation of boil-up at reboiler temperature.

water

vvvmsteam vh

hnm

Enthalpy balance of total condenser:

DvDDvvnOHOHOH hRnhntCpm )1(222

)1( DaavDwatervDv xhxhh Where mH2O is the mass flow of cooling water, nvn is mol flow of overhead vapor and Dv h is heat of evaporation of distillate at condenser temperature.

Example 2: Determine the number of theoretical stages and optimum feed stages locations in column in example 1, if we have two feed streams, one the same as in example 1 and another 50 kmol/h 50 mol% vaporized, which is consist of 50 mol% water and 50 mol% acetic acid. All other data are same as in example 1.

Solution of Example 2: Scheme and mass balance:

F1

F2

D

W

nL1 nv1

nL3

nL2 nv2

nv3

1

2

3

WWDDFFFF xnxnxnxnnWnDnFnF

2211

21

WD

WFFFFFFD xx

xnnxnxnn

)( 212211

nW =nF1+nF2-nD q1=1 q2=0.5 Section 1: Same as rectifying section in previous example nL1=nD.R nV1=nD(R+1) Equation of operating line:

11

Rx

xR

Ry D

q-line equation:

11 11

1

qx

xq

qy F

Solution of Example 2: Scheme and mass balance:

F1

F2

D

W

nL1 nv1

nL3

nL2 nv2

nv3

1

2

3

WWDDFFFF xnxnxnxnnWnDnFnF

2211

21

WD

WFFFFFFD xx

xnnxnxnn

)( 212211

nW =nF1+nF2-nD q1=1 q2=0.5 Section 1: Same as rectifying section in previous example nL1=nD.R nV1=nD(R+1) Equation of operating line:

11

Rx

xR

Ry D

q-line equation:

11 11

1

qx

xq

qy F

Section 2: nL2=nL1+nF1q1 nV2=nV1+nF1(q1-1) Equation of operating line:

2

11

2

2

V

FFDD

V

L

nxnxn

xnn

y

q-line equation:

11 22

2

qx

xq

qy F

Section 3: nL3=nL2+nF2q2 nV3=nL3-nW Equation of operating line:

WL

Ww

WL

L

nnxn

xnn

ny

33

3

Calculation steps:

1- Follow steps 1-8 as in example 1, considering new equation for mass balance 2- Calculate nL1, nV1, nL2, nV2, nL3, nV3 3- Plot q2-line using q2-line equation, one point of the line is (xF2, yF2), for obtaining

another point choose an x and calculate a y. 4- Plot the operating line in section 3, using operating line equation. One point is (xW,

yW), for obtaining another point choose an x and calculate a y. Plot operating line in section 2 by connecting two intersections, intersection of operating line in section 1 and q1-line and intersection of operating line in section 3 and q2-line.

DSP: Seminar 2, Binary distillation, McCabe-Thiele method, column with one feed

Data to PictureEqulibrium 0 0

0.0055 0.0112 nF= 100 kmol/h nD= 77.77778

0.053 0.133 xF= 0.75 nW= 22.222220.125 0.24 q= 10.206 0.338 xD/Rmin+1 0.3550.297 0.437 Rmin= 1.676056

0.51 0.63 R= 3.3521130.649 0.7510.803 0.866 xD= 0.95 xD/R+1= 0.218285

0.9594 0.9725 xw= 0.05 Steps 13.51 1 P= 101.325 kPa NTS 12.50

line

0 01 1

q-line0.75 0.750.75 0.83

Operating line in rectifying section0.95 0.95

0 0.218285F

0.75 00.75 0.75

D0.95 00.95 0.95

W0.05 00.05 0.05

45 o

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

x-y diagram, Water-Acetic Acid

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 1x

y

x-y diagram, Water-Acetic Acid

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 1x

y

DSP: Seminar 2, Binary distillation, McCabe-Thiele method, column with one feed

2214 kJkg-1K-1 MH2O 18 Kg/kmol

390 kJkg-1K-1 Maa 60.05 Kg/kmolPsteam 0.4 Mpa

2133.9 kJkg-1 338.4977

yw 0.1325555.73 kJ/kmol/K

4053.87 kg/h

cpwater 4.18 kJ/kg/K39030.37 kJ/kmol/K

dtH2O 20 K

158034.6 kg/h

Δvhwater= Δvhaa=

nvm

msteam

mH2O

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Δ vhsteam

Δv hv

Δv hD

x-y diagram, Water-Acetic Acid

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 1x

y

x-y diagram, Water-Acetic Acid

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DSP: Seminar 2, Binary distillation, McCabe-Thiele method, column with multiple feeds

Data to PictureEqulibrium 0 0

0.0055 0.0112 nF1= 100 kmol/h nD=nF1*xF1+nF2*xF2-(nF1+nF2)*xW/(xD-xW)=0.053 0.133 xF1= 0.75 nW= nF1+nF2-nD=0.125 0.24 q1= 1 nL1= 344.52270.206 0.338 nF2= 50 kmol/h nV1= 447.30050.297 0.437 xF2= 0.5 nL2= 444.5227

0.51 0.63 q2= 0.5 nV2= 447.30050.649 0.751 nL3= 469.52270.803 0.866 xD= 0.95 nV3= 422.3005

0.9594 0.9725 xw= 0.051 1 P= 101.325 kPa

steps 14.3line xD/Rmin+1 0.355 N 13.3

0 0 Rmin= 1.6760561 1 R= 3.352113

q1-line0.75 0.75 xD/R+1= 0.2182850.75 0.83

q2-line0.5 0.50.2 0.8

Operating line in section10.95 0.95

0 0.218285F1 F2

0.75 0 0.5 00.75 0.75 0.5 0.5

D0.95 00.95 0.95

Operating line in section 3 Operaiting line in section 20.05 0.050.5 0.55032

W0.05 00.05 0.05

45 o

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DSP: Seminar 2, Binary distillation, McCabe-Thiele method, column with multiple feeds

nD=nF1*xF1+nF2*xF2-(nF1+nF2)*xW/(xD-xW)= 102.7778nF1+nF2-nD= 47.22222

kmol/h

Optimal Feed stage 1 6Optimal Feed stage 2 10

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0 0.0055 0.053 0.125 0.206 0.297 0.51 0.649 0.803 0.95940 0.0112 0.133 0.24 0.338 0.437 0.63 0.751 0.866 0.9725

xd/Rmin+1 0.35Rmin 1.714286R= 3.428571

xd/R+1 0.214516

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

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1x-y diagram, Water-Acetic Acid

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x'

0 0.2 0.4 0.6 0.8 1 1.20

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11

0 01 1

0.75 00.75 0.75

0.95 00.95 0.95

0.05 00.05 0.05

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

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1x-y diagram, Water-Acetic Acid

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0 0.2 0.4 0.6 0.8 1 1.20

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