Direct Sequence vs Indirect Sequence

I V Distillation Sequencing

Outline

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  1. Basic Concepts of Distillation Sequence Design

  3. Performance of Distillation Column (Sieve tray and packed tower)

  4. Separation and Recycle for continues process

BASI C CONCEPT OF DI STI LLATI ON SEQUENCI NG

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY I V.1.

I V.1.1. I ntroduction

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

   Consider: Separation of a homogeneous multi-component fluid mixture into a number of products, whereas all separations are carried out using distillation only.  I f this is the case, how to choice the

  distillation sequence?

   For example: two simple column sequences ( direct and indirect) could be employed in separation of a three-component mixture into three relatively pure products.

Direct Sequence vs I ndirect Sequence

  direct sequence indirect sequence

  the lightest component is taken the heaviest component is taken as overhead in each column bottom product in each column requires less energy for both requires more energy for both reboiling and condensation supplied reboiling and condensation supplied by utilities by utilities component A (light material) is only Component A (light material) is vaporized once vaporized twice can be more energy-efficient if the feed to the sequence has a low flowrate of the light material (A) and a high flowrate of heavy material (C)

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY Alternative sequences for the separation of a four- product mixture.

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY Number of possible distillation sequences using simple columns

   The problem is that there may be significant differences in the capital and operating costs between different distillation sequences that can produce the same products.  I n addition, heat integration may have a significant effect on operating costs ( would be discussed next).

I V.1.2. Practical Constraints

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  1. Remove as early as possible:

  a. A particularly hazardous component  safety consideration

  b. Reactive or heat-sensitive component  to avoid problems of product degradation c. Corrosive component

   to minimize the use expensive material of construction

  2. The main component that difficult to be condensed should be removed as early as possible  using refrigeration system or high pressure system

  3. Don’t take the final product from the bottom of column if:

  a. The component is decomposed in the reboilers (it can contaminates the product) b. Polymerization inhibitors are used to inhibit polymerization of some components when distilled. These polymerization inhibitors tend to be nonvolatile, ending up in the column bottoms

  I V.2. CHOI CE OF SEQUENCE AND

I TS OPERATI NG PRESSURE

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  I V.2.1. Heuristics of Choice of Sequence (Smith, R., 2005)

  1. Component with its relative volatility close to unity or that exhibit azeotropic behavior should be removed last .

  2. The lightest components should be removed alone one by one in column overheads (use direct sequence).

  3. A component composing a large fraction of the feed should be removed first .

  4. Favor splits in which the molar flow between top and

bottom products in individual columns is as near equal as

possible.

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

Example 4.2.1:

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Data for a mixture of alkanes to be separated by distillation are as follows:  Use the heuristics to identify potentially good sequences that are candidates for further evaluation!

   The relative volatilities have been calculated on the basis of the feed composition to the sequence, assuming a pressure of 6 barg using the Peng–Robinson Equation of State with interaction parameters set to zero.

   Different pressures can, in practice, be used for different columns in the sequence

Solution: Alternative- 1

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Heuristic 1 : Do D/ E split last since this separation has the smallest relative volatility.

  Heuristic 2 : Favor the direct sequence: Heuristic 3 : Remove the most plentiful component first:

  Heuristic 4 : Favor near-equimolar splits between top and bottom products:

  All four heuristics are in conflict here: Heuristic 1 suggests doing the D/E split last, and Heuristic 3 suggests it should be done first. Heuristic 2 suggests the A/B split first and Heuristic 4 the C/D split first.

  Solution: Alternative- 2 : Take one of the candidates and accept, say, the A/ B split first.

  Again the heuristics are in conflict Heuristic 1 : Do D/ E split last

• Heuristic 1 again suggests doing the D/E split last, whereas again .

  Heuristic 3 suggests it should be done first.

  : Heuristic 2

• Heuristic 2 suggests the B/C split first and Heuristic 4 the C/D split first.

  Heuristic 3 :

• There are 14 posible sequence
• This process could be continued and possible sequences identified for further consideration.

  Heuristic 4:

• Some possible sequences would _{Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY}

  be eliminated

Quantitative measure as other consideration

  Since heuristics (qualitative procedure) can be in conflict, a

  

  quantitative measure of the relative performance of different sequences would be preferred The vapor flow up the column as a physical measure can be readily

  

  calculated. This provides an indication of both capital and operating cost. More vapor flow up the column, more heat duty required for

  

  reboiler and condenser, increase the operating cost of hot utility (steam) and cold utility (water or refrigerant) A high vapor rate leads to a large diameter column, and also

  

  requires large reboilers and condensers, therefore the capital cost increases Consequently, sequences with lower total vapor load would be

   preferred to those with a high total vapor load.

  But how is the total vapor load predicted?

   _{Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY}

Prediction of the total vapor load

  

V D

  1 R    

  Underwood min min … (4.2.1) Eq. (4.2.1) can also be written at finite reflux.

  Defining R F to be the ratio R/R min (typicaly R/R min = 1.1):

  V D R R

  1   F

    … (4.2.2) min x x

  1   _{DLK DHK}

  R

   

  α R can be calculated: min min

  … (4.2.3)

   

  x x

  1  FLK FHK

  α

   

  = α relative volatility between the key components where: x = _DLK mole fraction of light key in the distillate x = _FLK mole fraction of light key in the feed x = _DHK mole fraction of heavy key in the distillate x = _{FHK Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY} mole fraction of heavy key in the feed

  Assuming a sharp separation:

• only the light key and lighter than LK components in the overhead
• only the heavy key and heavier than HK components in the bottoms

  x F

   

  1 DLK

  1  

  R

     

  min

   

  … (4.2.4) x D

  1

  1

  α  FLK α 

     

  F = feed flow rate where:

  D = distillate flow rate

  R F R _{F F}

     

  V D D F

  Combining (4.2.4) and (4.2.2):

  1    

     

  … (4.2.5) D

  1

  1

  α  α 

     

  thus: R _F

   

  V F F F F F F F F ... ... ...

   A  B   LK  A  B   LK  HK   NC … (4.2.6)

       

  1 

   α 

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

Example 4.2.2:

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Using the data (below) for a ternary separation of benzene, toluene, and ethyl benzene. Based on the vapor flow-up, determine whether the direct or indirect sequence should be used!

  Symbol Component Flowrate

  (kmole/h) Relative Volatility

  Relative Volatility betwwen adjacent components

  A Benzene 269

  3.53

  1.96 B Toluene 282

  1.80

  1.80 C Ethyl Benzene

  57

  1.00 Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

Solution of Example 4.2.2:

    

    

  Direct Sequence: A/ BC and B/ C         

  Hence we should use the direct sequence

   900 4 . 1387  kmole/h  2287 4 .

  V

  1 269 282 269 57 282

  1 1 .

  1 269 282 296 1 80 .

  1 1 .

  1 96 .

  

       

    

    

     

   

     

  I ndirect Sequence: AB/ C and A/ B

        

  1713 8 . 

  7 . 965   kmole/h

  V 748 1 .

  1 269 269 57 282

  1 1 .

   

  1

  1 1 .

  1 80 .

  

      

    

    

  57 82 282 1 96 .

Direct and I ndirect Sequence ( example 4.2.2) 269 kmol/h

  269 kmol/h 282 kmol/h 269 kmol/h 282 kmol/h 282 kmol/h 57 kmol/h 57 kmol/h

  V =2287.4 kmol/h

  ∑

  V =1713.8 kmol/h _{Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY} ∑

Example 4.2.3:

  Using the Underwood Equations, determine the best distillation sequence, in terms of overall vapor load, to separate the mixture of alkanes in Example 4.2.1 into relatively pure products. The recoveries are to be assumed to be 100% . Assume the ratio of actual to minimum reflux ratio to be 1.1 and all columns are fed with a saturated liquid. Neglect pressure drop across each column. Relative volatilities can be calculated from the Peng–Robinson Equation of State with interaction parameters assumed to be zero (see Chapter 4).

  Determine the rank order of the distillation sequences on the basis of total vapor load for all column pressures fixed to 6 barg w ith relative volatility calculated from the feed to the sequence .

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

Solution of Example 4.2.3:

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY I V.3.

  PERFORMANCE OF DI STI LLATI ON COLUMN Plate/ Tray Column and Packed Column ( )

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

Distillation Tray and Packing

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Distillation Tray Distillation Packing

Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

Plate/ Tray Column vs Packed Column Plate/ Tray Column Packed Column

  Contact of vapor-liquid relatively good

  chanelling and backmixing could be

  happen More liquid hold-up

• Easy to be cleaned
• Small Pressure drop, prefer to vacuum operation
• Cheaper for corrosive fluid
• Prefer to small diameter Can be used for liquid that contains solid particles

  Solid particle plugs the packed

• Foaming liquid
• Lighter Products can be taken from the

  side-stream

I V.4. SEPARATI ON AND RECYCLE SYSTEM FOR CONTI NUES PROCESS

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I V.4.1. I ntroduction

  Do separation for some reasons: 

  1. to achieve product specification 2. to meet environment law Material to be separated:

   1. reactants 2. main product 3. byproduct (could be sold 4. waste (could not be sold)

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  I V.4.2. Function of Process Recycles

  1. Reactor conversion : Consider FEED

   PRODUCT with conversion of about 95%

I ncomplete conversion in the reactor requires a recycle for

unconverted feed material.

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  2. Byproduct formation Consider:

  1 st

  : FEED  PRODUCT + BYPRODUCT or

  1 st

  : FEED  PRODUCT

  2 nd

  : PRODUCT  BYPRODUCT

  Using a purge saves the cost of a separator but incurs raw material losses, and possibly waste treatment and disposal costs. This might be worthwhile if the FEED- BYPRODUCT separation is expensive.

  3. Recycling byproducts for improved selectivity Consider:

  If a byproduct is formed via a reversible secondary reaction then recycling the byproduct can inhibit its formation at source.

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  4. Recycling byproducts or contaminants that damage the reactor When recycling unconverted feed material, it is possible

   that some byproducts or contaminants, such as products of corrosion, can poison the catalyst in the reactor.

  I t is clearly desirable to remove such damaging  components from the recycle stream.

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  5. Feed impurities If the impurity has an adverse effect on the reaction or poisons the catalyst

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  5. Feed impurities ( continued) I if the impurity does not have a significant effect on the reaction, then it could perhaps be passed through the reactor and be separated

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  As with its use to

  

  separate byproducts, the purge saves the cost of a separation, but incurs raw material losses. This might be

  

  worthwhile if the

  FEED-I MPURI TY separation is

  expensive. Care should be taken to ensure that the resulting increase in

  

  concentration of

  I MPURI TY in the reactor does not have an adverse effect on reactor performance.

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY 6. Reactor diluents and solvents.

  An inert diluent such as steam is sometimes needed in the

  

  reactor to lower the partial pressure of reactants in the vapor phase. Diluents and solvents are normally recycled.

   Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY 7. Reactor heat carrier.

  The introduction of an extraneous component as a heat carrier effects the recycle structure of the flowsheet.

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

   This simplifies the recycle structure of the flowsheet and removes the need for one of the separators.  The use of the product as heat carrier is obviously restricted to situations where the product does not undergo secondary reactions to unwanted byproducts.