Distillation IVMcCabe thiele method (2)

Mass Transfer for 4th YearChemical Engineering Department

Faculty of EngineeringCairo University

Today isA

Special cases of binary systems distillation using McCabe Thiele method.

•Only one feed•Consisting of two sections•Condenser and Reboiler•No side productsSteps:1. q-line2. Rmin

3. Rop and Top section oper. Line

4. Bottom section oper. line

The Simple Case

XDXFXW

1RxD

(XD,XD)

1Rx

min

D

(XW,XW)

Other Cases

1. Enriching Section2. Stripping Section3. Complex Feed or Multiple feeds4. Open Steam5. Top Side Product6. Bottom Side Product

NOTE: q-line NOT ONLY represents feed, but it represents any stream that changes the flow rates inside the column.

1- Enriching Section

FxF

Vyo

Lxo

DxD

WxW

• Used in case when it’s needed to recover light component from feed containing little amount of it.

• Feed composition is near that of the bottom product (xF is very small)

• Feed is usually saturated vapour• No reboiler is used.

1- Enriching Section

Steps:• Feed is saturated vapour• No. of stages in bottom

section =0• Get point of intersection

of bottom section andq-line directly

• Draw the top section operating line.

XDXW

(XD,XD)

XF

2- Stripping SectionDxD

L’xo’

V’yr’

WxW

FxF

• Used in case when it’s needed to recover heavy component from feed containing little amount of it.

• Feed composition is near that of the top product (xF is very big)

• Feed is usually saturated liquid• No reflux is needed.

2- Stripping Section

Steps:• Feed is saturated liqiud• No. of stages in top

section =0• Get point of intersection

of top section andq-line directly

• Draw the bottom section operating line.

XDXW XF

3- Multiple Feeds

Generally F1 is saturated liquid, and F2 is saturated vapour.The column can now be divided into THREE section: Enriching, Stripping and Middle section.Both operating lines of the top and bottom sections will not be changed.Only we will need to get the middle section line.

F2xF2

Vyo

L’’xo

V”yr

WxW

Lxo

DxD

F1

xF1

3- Multiple Feeds

Since we assumed constant molar flow rates in the column, we can say that:L’=L+F1

V’=V

AndL”=L’V’=V”+F2

To draw the middle section operating line we have to calculate its slope from the above relations

F2

xF2

Vyo

L’’xo

V”yr

WxW

Lxo

DxD

F1

xF1

L V

L’ V’

L” V”

3- Multiple Feeds

Given information:F1, xf1, F2, xf2, xD, xW, RF1 is saturated liqiudF2 is saturated vapour

Steps:Top section operating line as it is

XDXF1XW

1RxD

(XD,XD)

(XW,XW)

XF2

3- Multiple FeedsMiddle line starts from the end of the top section and ends at the beginning of the bottom section.So point of intersection of top section and first q-line is on the middle section line.We need another point or a slope to draw the middle line.

XDXF1XW

1RxD

(XD,XD)

(XW,XW)

XF2

3- Multiple FeedsThe slope of any operating line is L/V in the section it represents.So in the middle section slope is L’/V’

As we knowL’=F1+L and V’=VWhere L=D*R and V=L+D=(R+1)DSo we can now get L’/V’ and draw its line XDXF1XW

1RxD

(XD,XD)

(XW,XW)

XF2

L’/V’

4- Upper Side Product

• Any side product is withdrawn as saturated liquid.

• As this side product will change the flow rates inside the column, there will be a q-line representing the side product.

• The column will be divided into 3 section: Top, middle and bottom• Still the top section not affected, and we

want to draw the middle section line.

Vyo

V’yr

L’xo

Lxo

DxD

F

xF

WxW

SxS

4- Upper Side ProductTo get Operating line of the middle section do MB on the loop:V=L+S+DV.yn+1=L.xn+S.xS+D.xD

(equation of st. line as S,D,xS,xD are constants)

This line starts at the intersection of the top section with the q-line of side product

Vyo

V’yr

L’xo

Lxo

DxD

FxF

WxW

SxS

Lxn

Vyn+1

4- Upper Side ProductSteps:1. Side product is saturated

liquid.2. Feed q-line is drawn

whatever its state3. Draw the top section line

We need to use the equation of the middle section to draw its operating line.

XDXSXW

1RxD

(XD,XD)

(XW,XW)

XF

4- Upper Side ProductBack to equations:V=L+S+D V-L=S+DV.yn+1=L.xn+S.xS+D.xD

We know a point on the line and need to get another point to draw itThe easy point is on 45 lineV.x=L.x+S.xS+D.xD

(V-L).x=S.xS+D.xD

(S+D).x=S.xS+D.xD

XDXSXW

1RxD

(XD,XD)

(XW,XW)

XF

DSxDxSyx DS

..

x=y

5- Bottom Side Product

• The column will be divided into 3 section: Top, middle and bottom

• Still the top section not affected, and we want to draw the middle section line.

• We will derive the middle section operating line equation (as in previous case)

Vyo

V’yr

L’xo

Lxo

DxD

FxF

WxW

SxS

Vyo

V’yr

L’xo

Lxo

DxD

F

xF

WxW

SxS

L’Xm+1

V’ym

5- Bottom Side ProductTo get Operating line of the middle section do MB on the loop:L’=V’+S+WL’.x’m+1=V’.y’m+S.xS+W.xW

(equation of st. line as S,W,xS,xw are constants)

This line ends at the intersection of the bottom section with the q-line of side product

5- Bottom Side ProductSteps:1. Side product is saturated

liquid.2. Feed q-line is drawn

whatever its state3. Draw the top section line

We need to use the equation of the middle section to draw its operating line.

XDXSXW

1RxD

(XD,XD)

(XW,XW)

XF

5- Bottom Side ProductBack to equations:L’=V’+S+W L’-V’=S+WL’.x’m+1=V’.y’m+S.xS+W.xw

We know a point on the line and need to get another point to draw itThe easy point is on 45 lineL’.x=V’.x+S.xS+W.xW

(L’-V’).x=S.xS+W.xW

(S+W).x=S.xS+W.xW

XDXFXW

1RxD

(XD,XD)

(XW,XW)

XS

WSxWxSyx WS

..

x=y

6- Open SteamVyo

S

WxW

Lxo

DxD

FxF

• Used in cases when the feed contains water, so heat is added to column in the form of direct heating by steam instead of reboiler.

• The bottom section operating line will change as the equations will be changed.

• Still the intersection of the top section line and the q-line is on the bottom section line.

6- Open SteamVyo

S

WxW

Lxo

DxD

FxF

L’Xm+1

V’ym

Operating line of bottom section:L’+S=V’+W L’-V’=W-SL’.x’m+1+S.yS=V’.y’m+W.xW

To draw this line we need another point, try 45o lineL’.x=V’.x+W.xW

(L’-V’)x=W.xW

(W-S)x=W.xW

SWxWyx W

.

6- Open SteamSteps:1. Draw the top section line2. Draw the q-line3. Locate the point derived

on the 45o line4. Draw the bottom section

line.Here the bottom section will end at a point (xw,0)

Usually S needed to be calculated.XDXFXW

1RxD

(XD,XD)

(XW,ys)

SWxW W

.

4- 100 Kgmol/hr of saturated liquid containing 70 mol% benzene enters a stripping tower at 1 atm. The bottom product flow rate is 15 Kgmol/hr containing only 10% benzene. Saturated steam at 4 atmospheres is available for the reboiler duty. (Hv=2740 KJ/Kg, hL=610 KJ/Kg).Calculate:a- The overhead product flow rate and its composition.b- Number of theoretical plates required.c- Steam consumption in reboiler.Equilibrium data:

lBenzene=7360 Cal/gmol, lToluene=7960 Cal/gmol

T, C 80 85 90 95 100 105 110PB

o, mmHg 760 877 1016 1068 1344 1532 1800PT

o, mmHg 205 345 405 475 557 645 760

Stripping towerF=100 Kgmol/hr (saturated liquid)xF=0.7 P=1 atmW=15 Kgmol/hr xW=0.1Steam: Hv=2740 KJ/Kg, hL=610 KJ/Kglst=2130 KJ/Kg

a)D=F-W=100-15=85 Kmol/hrxD=(F.xF-W.xW)/D=(100*0.7-15*0.1)/85=0.806

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

T, C 80 85 90 95 100 105 110PB

o, mmHg 760 877 1016 1068 1344 1532 1800PT

o, mmHg 205 345 405 475 557 645 760X 1 0.78 0.58 0.48 0.26 0.13 0Y 1 0.90 0.78 0.678 0.46 0.26 0

b)NTS= Reboiler + 4.6

c)Qr=mst.lst=V’.lr

lr is calculated at yr

lr =0.2*7360+0.8*7960lr =7840 Cal/gmol

V’=???Slope of operating line is L’/V’=1.1667And W+V’=L’Get L’ and V’V’=90 kmol/hrL’=105 Kmol/hr

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

L’/V’

L’X’1

V’yr

Wxw

V’y’m

V’y’m

L’x’m+1

L’x’m

FinallyQr=90*7840=705600 Cal/hr= 2949.408 KJ/hrlst=2130 KJ/Kg2949.408 =2130*mst

mst= 1384.7 Kg/hr