101011ColumnDist Post

What you should already know:

•  Iden3fy the feed stage, the bo8oms and dis3llate streams, the enriching/rec3fying sec3on, the stripping sec3on

•  Define the external reflux ra3o, the boilup ra3o, constant molal overflow

•  Perform external column balances (mass and energy)

•  For a par3cular stage inside the column, label all streams, perform mass balance

Objec3ves for this lecture

•  Apply the Lewis method to stage-‐by-‐stage analysis of a con3nuous dis3lla3on column

•  Obtain the equa3on of the opera3ng line for a countercurrent rec3fying column

•  For a specified reflux ra3o, deduce the number of equilibrium stages required, using the McCabe-‐Thiele method

Reading assignment: ch. 3 and ch. 4 sec3ons 4.1-‐4.3

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y(MeOH)

x(MeOH)

MeOH-‐H2O rec3fying column

•xD= x0 (x0,y1)

yint•

stage 1 (x1,y1)•

stage 2 (x2,y2)•

stage 3 (x3,y3)•

•(x1,y2)

•(x2,y3)

•xB

1. Draw y=x line

2. Plot xD on y=x

3. Plot yint and draw op. line

4. Step off stages star3ng at xD

5. Stop when you reach xB

Specifica3ons: xD = 0.8, R = 2 Column with total condensor Find N required to achieve xB = 0.1

L/V = R/(R+1) = 2/3 yint = xD(1-‐L/V)= 0.8/3 = 0.26

6. NEVER step over the VLE line.

lowest xB possible for this op. line

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Recap:

•  The external reflux ra3o R is related to, but not the same as, the internal reflux ra3o L/V.

•  The assump3on of constant molal overflow (CMO) allows us to derive a single mass balance equa3on for the rec3fying/enriching sec3on of a dis3lla3on column. When plo8ed on a McCabe-‐Thiele graph, this equa3on is called the opera&ng line.

•  Moving from stage-‐to-‐stage down the column, either algebraically, or graphically, amounts to alterna&ng between the mass balance and the VLE equa&ons.

Objec3ves for this lecture

•  Describe the opera3ng limits for a rec3fying column. •  Find the minimum number of stages, Nmin, required for separa3on by rec3fica3on.

•  Find the minimum reflux ra3o, Rmin, required for separa3on by rec3fica3on.

•  Repeat for a stripping column. •  Assemble the pieces to make a complete dis3lla3on column and iden3fy the feed stage.

Homework: 4D28 (3rd ed; 4D35 in 2nd ed). Assume desired yD = 0.74.

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y(MeOH)

x(MeOH)

Limi3ng cases: rec3fica3on

•xD= x0 (x0,y1)

stage 1 (x1,y1)•

Specifica3ons: xD = 0.8, vary R = L/D

1.  L 0 R = L/D 0 L/V 0

L/V = 0 No reflux!

Max. distance between VLE and op. line Max. separa3on on each equil. stage Corresponds to Nmin. But no dis3llate!

L/V = 1 Total reflux!

2. D 0 R = L/D ∞ L/V = R/(R+1) 1 (L’Hôpital’s Rule)

Opera3ng line is y=x

Column operates like a single equilibrium stage. (Why bother?)

0 ≤ L/V ≤ 1

0 ≤ R ≤ ∞

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y(MeOH)

x(MeOH)

Minimum reflux ra3o

•xD= x0 (x0,y1)

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Specifica3ons: xD = 0.8, vary R

L/V = 0

The number of stages N required to reach the VLE-‐op. line intersec3on point is ∞.

L/V = 1 0 ≤ L/V ≤ 1

0 ≤ R ≤ ∞

Increasing R Decreasing xB (for fixed N) Decreasing D

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• xB ,min for this R

This represents xB,min for a par3cular R.

It also represents Rmin for this value of xB.

Rule-‐of-‐thumb: 1.05 ≤ Ropt/Rmin ≤ 1.25 Can specify Ractual as a mul3ple of Rmin

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y(MeOH)

x(MeOH)

MeOH-‐H2O stripping column (0.7,1) •

•(x1,y2) stage 1

1. Draw y=x line

2. Plot xB = xN+1 on y=x

3. Draw op. line

4. Step off stages star3ng at PR

5. Stop when you reach xD

Specifica3ons: xB = 0.07, Column with par3al reboiler Find N required to achieve xD = 0.55


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V /B = 2

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L /V = B /V +1 = 1.5

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yint = xB(1−L /V ) = 0.1(1−1.5) = −0.05

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y = 1 = 1.5x −0.05

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x = 1.05 /1.5 = 0.7

• xB= xN+1 (xN+1,yN+2)

(xN+1, yN+1)

• (xN,yN+1) • PR

(xN,yN) • • (xN-‐1,yN)

(xN-‐1,yN-‐1) •

• xD

highest xD possible

for this op. line

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stage 2

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y(MeOH)

x(MeOH)

Limi3ng cases: stripping Specifica3ons: xB = 0.07, vary boilup ra3o

Max. distance between VLE and op. line Max. separa3on on each equil. stage Corresponds to Nmin. But no bo8oms product!

Behaves as if the column wasn’t even there. (Why bother?)

2. B 0

Opera3ng line is y=x

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L →V

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L /V →1

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V / B

1. 

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V →0

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L /V →∞

• xB= xN+1

• PR

No boilup!

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L /V = ∞

Total boilup!

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L /V = ∞

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1≤ L /V ≤ ∞

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∞ ≥V / B ≥ 0

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y(MeOH)

x(MeOH)

Minimum boilup ra3o

Specifica3ons: xB = 0.07, vary boilup ra3o

The number of stages N required to reach the VLE-‐op. line intersec3on point is ∞.

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yD ,max for this boilup ra3o

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This represents yD,max for a par3cular boilup ra3o.

It also represents the minimum boilup for this value of xD.

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1≤ L /V ≤ ∞

• xB= xN+1

• PR

No boilup!

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L /V = ∞

Total boilup!

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L /V = 1

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∞ ≥V / B ≥ 0

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y(MeOH)

x(MeOH)

McCabe-‐Thiele analysis of a complete dis3lla3on column

stage 1

1. Draw y=x line

2. Plot xD and xB on y=x

3. Draw both op. lines

4. Step off stages star3ng at either end, using new op. line as you cross their intersec3on 5. Stop when you reach the other end

Specifica3ons: xD = 0.8, R = 2 xB = 0.07, Total condensor, par3al reboiler Find N required Locate feed stage


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V /B = 2

• • PR

• •

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stage 2

• xB

•xD

Feed must enter here

101011ColumnDist Post

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