separation process

PROCESS DESIGN PRINCIPLES I CHE F314

BITS Pilani Pilani Campus

Suresh Gupta

Department of Chemical Engineering BITS-Pilani, Pilani Campus

C H E F 31 31 4 P r o c e s s D e s i g n P r i n c i p l e s I

Tutorial Class Saturday,, 8-8:50 8-8:50 AM • Saturday Section-1 Room No. 6107 Mr. Subhajit Majumder Section-2 Mr. Mr. Utkarsh Utkarsh Maheshwari Room No. 6104


Lecture-1 Introduction 06-08-2015

C H E F 31 4 P r o c e s s D e s i g n P r i n c i p l e s I

Process Design..

“It is a combination of science and art in a creative activity that helps to make process design such a fascinating challenge to an engineer…”


Process Design


PROCESS

Outputs

Inputs “feed”

Process

“products”


DESIGN Definition of Chemical Process Design

Raw Material

Chemical Process

??

Chemical Product

Chemical process design is about finding a sustainable process that can convert the raw materials to the desired chemical products


Creative Aspects of Process Design • The Purpose of Engineering – to create new material wealth

• This goal in Chemical Engineering is

accomplished – via the chemical transformation – and/or separation of materials


Creative Aspects of Process Design • 50% of the chemical products sold – were developed during the last decade or two (20 – 30

years) – Indication of tremendous success of engineering effort

• Process and Plant Design – Creative activity whereby – Generate ideas, translate them into equation and

processes for producing new materials – or for significantly upgrading the value of existing materials


Your role in the chemical process

Process synthesis vs. Process analysis ??


Strategy for Process Synthesis and Analysis • The goal of a conceptual design is: – To find the best process flow sheet – To estimate the optimum design conditions

• There can be many process alternatives to be

considered • There are many possibilities to consider with only a small chance of success – 104 – 109 alternatives can be generated for a single product plant (since design problems are under-defined)


Contd.. • In some cases it is possible to use design guidelines (rules of thumb or heuristics) – to make some decisions about the structure of the flow

sheet and/or – to set the values of some of the design variables

• In the absence of heuristics - Use shortcut design

methods


Problem Areas Synthesis and Analysis • Design problems are underdefined • To supply this missing information, we must

make assumptions about – What type of process units should be used? – How are they interconnected? – What temperatures, pressures, flow rates are

required? …”Synthesis

Activity”


Contd.. • Synthesis is difficult because there are very large number (104 – 109) of ways to

accomplish same goal • Hence design problems are very open-

ended


Objective • We have to find the process alternative (out of 104 – 109) possibilities – That has the lowest cost – Process is safe – Satisfy environmental constraints – Easy to start up and operate etc.


Need of Process Design Principles

Because of the under defined and open-ended nature of design problems, and because of the lower success rates, it is useful to develop a strategy for solving design problems


Hierarchy of Decisions Level 1

•

Batch vs Continuous

Level 2

•

Input-Output Structure

Level 3

•

Recycle Structure of flowsheet

•

General Structure of Separation System

Level 4

Level 5

4a. VRS 4b. LSS

• Energy

Integration Analysis (EIA)


Input Information and Batch vs Continuous

Decision on Operating mode



•

Batch vs Continuous

Level 2

•


Level 3

•


•


Level 4

Level 5

4a. VRS 4b. LSS

• Energy




•

Batch vs Continuous

Level 2

•


Level 3

•


•


Level 4

Level 5

4a. VRS 4b. LSS

• Energy




•

Batch vs Continuous

Level 2

•


Level 3

•


•


Level 4

Level 5

4a. VRS 4b. LSS

• Energy




•

Batch vs Continuous

Level 2

•


Level 3

•


•


Level 4

Level 5

4a. VRS 4b. LSS

• Energy



Organization of Course

Module-I Strategy for Process Synthesis and Analysis


Module-II Developing a Conceptual Design and Finding the Best Flowsheet

Batch vs Continuous

Input-Output

Recycle

Separation

Heat Exchanger Network

Mass Exchanger Network


Handout


Evaluation Scheme EC Evaluation

Duratio Weightage

No component (EC)

n

.

(Minute

Date and time

(300)

Nature of component

s)

1

Mid-Semester Test

90

75

2

Tutorials/Surprise

-

70

Closed Book -

Tests# 3

Assignment *

Open/Closed Book

-

35

To be

Open Book

announced in the class in due course of time 4

Comprehensive Examination

180

120

Closed Book


Prerequisite • Concepts of • Heat Transfer • Separation Process I & II • Chemical Process Calculations • Chemical Engineering Thermodynamics • Fluid Mechanics


Points to Remember • Being ONTIME is a good thing! • Be Interactive! • Share your idea and views


Outline • Hierarchical Approach to Conceptual Design: HDA

Case Study • Simplified flowsheet for the separation process • Recycle structure of flowsheet • Input-Output Structure of Flowsheet • Hierarchy of Decisions


Hierarchical Approach to Conceptual Design Example: Hydrodealkylation of toluene (HDA Process) – To produce benzene

C 6 H 5CH 3  H 2  C 6 H 6  CH 4

Rxn 1

2C 6 H 6  2C 6 H 5  H 2

Rxn 2

Reaction temperatures for homogeneous reactions: 1150 – 13000F

– If T < 1150 0F – – – –

the reaction rate is very slow If T > 1300 0F a significant amount of hydrocracking takes place Pressure 500 psia ( ≈ 34 atm) Excess hydrogen (H2: aromatics = 5:1) Reactor effluent gas must be rapidly quenched to 1150 0F


Boiling points

Boiling Point (°F)

Diphenyl 491 Toluene 232 Benzene 176.2 Methane −258.68 Hydrogen - 423.182

Possible flow sheet Purge H2, CH4

H2, CH4 HEAT

COMPRESSOR

Recycle H 2

HEAT

Light Gases

1150 – 1300 0F

H2, CH4 C6H6

REACTOR

FLASH

C6H6

HEAT

Diphenyl (unwanted)

H2, CH4

H2, CH4

HEAT

Toluene (C6H5CH3)

COOLANT

Condensed aromatics + Light gases

C6H5CH3, Diphenyl

Recycle Toluene

R E C Y C L E

Partial Condenser

P R O D U C T

(Main Product)

C6H5CH3, (C6H5)2

S T A B I L I Z E R

Boiling Point (°F) Diphenyl 491 Toluene

232

Benzene 176.2 Methane −258.68 Hydrogen - 423.182

C6H6, C6H5CH3, C6H5


Energy Integration • Is the process flow sheet very realistic? • In the last decade (1978), a new design procedure

has been developed – that makes possible to find the minimum heating and

cooling loads for a process – and the Heat Exchanger Network Synthesis (HENS) that gives the ‘Best’ energy integration


Energy Integrated Flow sheet

Fig. 2


Contd.. • Energy Integration flow sheet is more complicated • many more interconnection

• Moreover to apply the Energy Integration (HENS) analysis – we must know the flow rate and composition of every process

stream i.e. all the process heat loads including those of the separation system as well as all the stream temperatures

• Since we need to fix almost all the flow sheet before we can

design the Energy Integration system – since it adds the greatest complication to the process flow sheet – we consider the Energy Integration Analysis (HENS) as last step in

our process design procedure


Distillation Train • We could recover the benzene as overhead • Remove toluene as the side-stream (below the feed), and

recover the diphenyl as a bottom stream H2, CH4

Benzene (C6H6)

Boiling Point (°F) Diphenyl 491 Toluene

232


Feed H2, CH4, C6H6,

Toluene (C6H5CH3) + Small amount of (C6H5)2

C6H5CH3,C6H5

C6H6, C6H5CH3, C6H5

Diphenyl (C6H5)2

Fig. 3


Contd.. H2, CH4

Toluene (C6H5CH3) Benzene (C6H6)

Feed

Boiling Point (°F) Diphenyl 491

H2, CH4, C6H6,

Toluene

C6H5CH3,C6H5

232


C6H5CH3, C6H5

Diphenyl (C6H5)2

Fig. 4


Contd.. • It might be cheaper than using the configuration shown in

the original flow sheet (Fig. 1) • The heurisitics (design guidelines) for separation systems require – A knowledge of the feed composition of the stream entering the

distillation train

• Thus before we consider the decisions associated with the

distillation train, we must specify the remainder of the flow sheet and estimate the process flows • For this reason we consider the design of the distillation train before we consider the design of the heat-exchanger network


Vapor Recovery System (VRS) • Complete separation (of aromatics and light gases) in a

flash drum NOT POSSIBLE! • therefore that some of the aromatics will leave with the flash vapor

(H2 and CH4 lighter gases)

• Moreover some of those aromatics will be lost in the purge

stream • It is possible to recover those aromatics by installing a VRS

either on the flash vapor stream or on the purge stream


Contd.. • As a VRS, one of the following can be used – Condensation (high pressure or low temperature or both) – Absorption – Adsorption – A membrane process

• To find out the economic feasibility of the VRS • we must estimate the flow rates of aromatics lost in the purge as well

as the H2 and CH4 flow in the purge

• Hence before we consider the necessity and / or the design

of a VRS • we must specify the remainder of the flow sheet and we must

estimate the process flows


Contd.. When do we consider designing of VRS?

We consider the design of the VRS before that for the liquid separation system


Simplified Flowsheet for the Separation Systems Systems • Our goal is to find a way of simplifying flowheets • It is obvious that Fig.1 is much simpler than the figure in

which energy integration (HENS) is included – because of which it was decided that the EIA be carried out at the

end (after distillation distillation train tr ain is finalize f inalized) d)

• Similarly, since we have to know that the process flow rates

to design the VRS and LRS problems just before the – it was decided to consider these design problems energy integration


Contd.. • The connections between between the VRS and LRS shown in Fig. 5 VAPOR RECOVERY SYSTEM

Gas Recycle

Purge H2, CH4

H2, CH4 Light Gases H2, CH4 REACTOR SYSTEM

Aromatics + Light Gases

PHASE SPLIT

Liquid (aromatics)

Toluene Aromatics

Toluene

LIQUID SEPARATION SYSTEM

Fig. 5

Benzene Diphenyl


Recycle Recycle structure of the flow sheet A simplified flow sheet for the process is shown shown in Fig. 6 Purge H2, CH4

Gas Recycle H2, CH4

H2, CH4 REACTOR SYSTEM

Aromatics + Light Gases

Toluene

Toluene (Recycle)

Fig. 6

SEPARATION SYSTEM

Benzene Diphenyl


Contd.. • Use this simple representation – to estimate the recycle flows – their effect on the reactor cost, and – the cost of gas recycle compressor, if any

• For example, we can study: 1. The factors that determine the no. of recycle streams 2. Heat effects in the reactor 3. Equiliblrium limitations in the reactor, etc.


Can we still think of simplifying the flowsheet?


Input-Output structure of the flowsheet • Since raw material costs normally fall in the range from 33-

85% of the total product costs • the overall material balance are the dominant factors in the design

Purge H2, CH4

Gas Recycle H2, CH4

H2, CH4

Benzene PROCESS Diphenyl

Toluene

Liquid Recycle

Fig. 7

Is this structure of flowsheet correct?


Contd.. • Also we do not want to spend any time investigating the

design variables in the ranges • where the products and by products are worth less than the raw

materials

• Thus, we consider the Input-output structure of the flow

sheet and the decisions that affect this structure before we consider any recycle streams • By successively simplifying a flowsheet, we can develop a

general procedure for attacking design problems


Hierarchy of Decisions • A systematic approach to process design by

reducing the design problem to a hierarchy of decisions: 1. 2. 3. 4.

Batch vs Continuous Input-Output structure of the flow sheet Recycle structure of the flow sheet General structure of the separation system a) Vapor liquid system b) Liquid separation system 5. Energy Integration Analysis (HENS)


Contd.. • One great advantage of this approach to design is: – It allows us to calculate equipment cost – to estimate costs

• Then if the potential profit becomes negative at some level • look for a process alternative or , • terminate the design project without having to obtain a complete

solution to the problem

• Another advantage of this procedure: – As we make about the structure of the flow sheet at various levels – We know that if we change these decisions, we will generate process

alternatives

• The goal of a conceptual design is to find the best alternative


Problem •

Ethanol is produced by the hydration of ethylene. The primary reactions for ethanol synthesis are given below: Ethylene  H 2 O  Ethanol 2 Ethanol  Diethylether  H 2 O

•

Initially, the feed (90% ethylene, 8% ethane, and 2% methane) and water are heated by passing through the primary heater. This heated feed is sent to the reactor. The reaction takes place at 560 K and 69 bar. The fractional conversion of ethylene in the reactor is 0.07. The reactants and products are sent to the separator where gaseous and liquid products and reactants are separated. All gaseous products and reactants are scrubbed in a scrubber. Unconverted ethylene and inert gases (ethane and methane) are recycled back. To avoid the accumulation of inert components, some amount of recycled stream is purged. The liquid products and the bottom products of scrubber are sent to the series of distillation columns where side product diethyl ether and water are separated out. The diethyl ether is recycled back and mixed with the feed stream. Ethanol -water azeotrope is produced from the final distillation column.


Problem • Draw the following: • • •

General structure of the flow sheet Recycle structure of the flow sheet Input-output structure of the flow sheet


Outline • Hierarchical Approach to Conceptual Design

IPA Case Study • Design of a solvent recovery system


Contd.. Draw the, 1. General structure of the Separation system. 2. Recycle structure of the flowsheet. 3. Input-output structure of the flowsheet.



Economic Decision Making


Outline • Design of A Solvent Recovery System (Ch. 3 of T2) • Problem Definition • Economic Potential • Process alternatives


Problem Definition • As a part of a process design problem – Assume that there is a stream – Containing 10.3 mol/hr of acetone and 687 mol/hr of air – That is being fed to a flare system (to avoid air pollution)


Economic Potential (EP) • Economic Potential (EP)

EP  Product Value - Raw Material Cost • Since stream coming from the same process, Raw material cost = 0 • Therefore EP  Product Value - Raw Material Cost

Operating hours

 Product Value - 0  Product Value  (10.3 mol/hr)(Rs.10.80/lb)( 58 lb/mol)(81 50 hr/yr)  Rs. 5.26 Cr/yr


Question 1 How to recover acetone?


General Considerations: Process Alternatives • Solvent recovery alternatives 1. Condensation a. b. c.

High Pressure Low temperature Combination of both

2. Absorption 3. Adsorption 4. A Membrane Separation System 5. A Reaction Process (Acetone as raw material for a new product)


Question 2 Which is the cheapest alternative?


General Considerations: Process Alternatives • If solute concentration (mole fraction) in a gas < 5 % – Adsorption is the cheapest process – In the present case, it is ≈ 1.5 %

[10.3/(687+10.3) = 0.0147]

• may opt for Adsorption • However, many petroleum companies prefer to use – Condensation or absorption process


Contd.. • Judgment based on:

Concerning the use of technology where we have great deal of experience vs. Using a technology where we have much less experience (Relatively new technology)


Design of Gas Recovery System: Flowsheet Douglas, J. M. Conceptual Design of Chemical Processes, 1988


Alternate Flow sheet: Recycling of solvent Douglas, J. M. Conceptual Design of Chemical Processes, 1988


Question 3 Whether discarding the process water, as shown in Fig. 1 can ever be justified even when a pollution treatment facility is available?


Contd.. • Check the temperature of the process water entering the gas absorber • Cooling water is available from the cooling towers at 90 0F (32 0C) (on the hot summer day) • And that is must be returned to the cooling towers at a temperature less than 120 0F (49 0C)


Contd..

Douglas, J. M. Conceptual Design of Chemical Processes, 1988, pp. 75


Design of Gas Absorber: Heuristic • This reasoning is the basis for a design heuristic H1: If a raw material component is used as the solvent (like water) in a gas absorber, consider feeding the process through the gas absorber


Design of Gas Absorber • Considering the flow sheet shown in Fig. 1

because it is the simplest for further processing


Contd.. • In addition, we must evaluate whether we really

want to use water as the solvent • We arbitrarily choose to consider the flow sheet shown in Fig. 1 because it is the simplest for further processing


Design of Gas Absorber: Material Balances • Identify the components that will appear in every

stream • The inlet gas flow to the absorber – 10.3 mol/hr of acetone + 687 mol/hr of air

• If we use well water as solvent – inlet solvent stream is pure water (100%, solute

concentration is zero)


Contd..

Acetone

2

1

3

Douglas, J. M. Conceptual Design of Chemical Processes, 1988


Contd.. • The gas leaving the absorber (top) will contains – air, some acetone and some water – Since water is relatively inexpensive, neglecting this

solvent loss


Specify Acetone Amount for Material Balance • Specify the product specification in distillate overhead

(2) • Specify amount of acetone leaving in the other two streams (1 and 3) • Recovery of 90, or 99, or 99.9% of the acetone in the gas absorber is possible?


Contd.. • Of course we can recover 90, or 99, or 99.9% or

whatever of the acetone in the gas absorber • Adding more trays to the top of absorber • The cost of the gas absorber will continue to increase as

• Increase the fractional recovery • but the value of the acetone lost to the flare system will

continue to decrease

• There is a trade-off between these two, and • Thus there is an optimal fractional recovery


Contd..

Fig.: %Recovery vs. Cost in Gas Absorber


Contd.. • There is optimum fractional recovery of bottoms

in the distillation column • As we add more & more plates in stripping section (bottom of distillation) of this column, • the still cost increases, but the value of the acetone

lost to the sewer decreases


Contd..

Fig.: %Recovery vs. Cost in Distillation


Design of Gas Absorber: Heuristics

H2: It is desirable to recover more than 99% of all valuable materials (we normally use 99.5% recovery as a first guess)


Design of Gas Absorber: Heuristics H3: For an isothermal, dilute absorber, choose the solvent flow rate (L ), such that L = 1.4m G where, m = slope of equilibrium line, and G = gas molar flow rate (mol/hr)


Material Balances • For the acetone-water system at 77 oF (25oC) and

1 atm – Activity coefficient, γ s= 6.7 – Vapour pressure of acetone in air, P os = 229 mm Hg. – Air flow rate, G = 687 mol/hr

 y s P T   x s P v



y s

Vapor pressure

o



P  s Total s



6.Mol 7(229 ) fraction

 2.02

Fugacitym Mol of760 solute in Activity x s pressure P coefficient fraction T ofcoefficient solvent solute system in gas


Contd.. • Solvent flow rate ( L) = 1.4 mG = 1.4 x 2.02 x 687

= 1943 mol/hr • For a 99.5% recovery of acetone in the gas absorber, – The acetone lost from top of absorber = 0.005 (10.3) =

0.05 mol/hr

• And the acetone flow to the distillation column, 0.995 (10.3) = 10.25 mol/hr

• If 99.5% of acetone entering the still is recovered

overhead, – Then acetone as distillate = 0.995 (10.25) =10.20 mol/hr


Contd.. • Also if the product composition of acetone is

specified to be 99%, – Then the amount of water in the product stream

(distillate) will be

 1  0.99   (10.20)  0.10 mol/hr  0.99 

• Then the bottom flows of acetone and water are – Acetone: 0.005 (10.25) = 0.05 mol/hr – Water: 1943 - 0.1 = 1942.9 mol/hr


Material Balances: Flowsheet

Fig.: Stream compositions and flow rates


Material Balances: Acetone Balances • Acetone entering Absorber = Acetone leaving

absorber (bottom) + Acetone lost from absorber (top) – 10.3 mol/hr = (10.25 + 0.05) mol/hr

• Acetone leaving absorber (entering distillation

column) = Acetone in distillate + Acetone in bottom – 10.25 mol/hr = (10.2 + 0.05 ) mol/hr


Material Balances: Water Balances • Water entering Absorber = Water leaving absorber

= Water entering distillation 1943 mol/hr = 1943 mol/hr

• Water entering distillation column = Water in

distillate + Water in bottom 1943 mol/hr = 10.2 (1-0.99)/0.99 + (1943 – x) x

= 0.10 + 1942.9 = 1943 mol/hr


Stream Cost Calculation: Acetone-Water System • For acetone water system, with no recycling and

99.5 % recoveries Acetone loss in absorber overhead (assume $0.27/lb of Acetone = ($0.27/lb)×(58 lb/mol)×(0.0515 mol/hr)×(8150 hr/yr) = $6600/yr Acetone loss in still bottom = ($0.27/lb)×(58 lb/mol)×(0.05 mol/hr)×(8150 hr/yr) = $6600/yr


Contd.. Pollution treatment cost (assume $0.25/lb BOD and 1 lb acetone/lb BOD) = ($0.25/lb BOD)×(1 lb BOD/1 lb acetone)×(58 lb/mol) ×(0.05 mol/hr)×(8150 hr/yr) = $6100/yr Sewer charges (assume $0.20/1000 gal) = ($0.20/1000 gal)×(1 gal/8.34 lb)×(18 lb/mol)× (1942.9 mol/hr)×(8150 hr/yr) = $6800/yr


Contd.. Solvent water (assume $0.75/1000 gal) =($0.75/1000 gal) (1 gal/8.34 lb) (18 lb/mol) (1943 mol/hr) (8150 hr/yr) = $25,600/yr

• Each of these costs all together

is essentially negligible compared to economic potential of $1.315×106 /yr, – We want to continue developing the design


Solvent Loss Calculations: Other than Water as Solvent • For a low pressure absorber, fugacity correction

factors are negligible • Vapor-liquid equilibrium relationship for the solvent can be written as P T y s   P x

0 s s s

• With greater than 99% recovery of the solute, x s ≈ 1


Alternate Flow sheet: Using Solvent Recycle (Other than Water as solvent)

Fig.: Solvent Recycle to Gas-Absorber Douglas, J. M. Conceptual Design of Chemical Processes, 1988, pp. 75


Contd.. • If a solvent is used that is in the homologous series with the solute, then γ s = 1 • Thus, from

P T y s   P x

0 s s s

y s 

0 s

P

P T


Contd.. Homologous series

• A series of chemical compounds of (1) Uniform chemical type (2) Showing a regular graduation in physical properties and (3) Capable of being represented by a general molecular formula – e.g. alkanes: CnH2n+2 (CH4, C2H6, C3H8, etc.) – Ketone: CnH2nO (acetone, C3H6O, MIBK, C6H12O)


Solvent Loss Calculations: Other than Water as Solvent


Contd.. • Quick way to estimate the solvent loss



G  G 1  y s '



– Where y s is the mole fraction of solute in solvent

• Now, the amount of solvent lost m s  G ' y s

 G   y s    1  y s   y s  G  y s G    1  y s 


Cost of Solvent Loss Using MIBK as Solvent • Suppose we consider using MIBK (Methyl Isobutyl

Ketone) as a solvent and we recycle the MIBK – At 25oC, P T = 1 atm, P s0 = 0.0237 atm – y s = P s0 /P T = 0.0237 / 1 = 0.0237

y s 

P s0 P T



0.0237 1

 0.0237

• Therefore, solvent lost m s  y s G  0.0237  687 mol/hr  19.7169 mol/hr


Contd.. • MIBK lost (assume $0.35/lb of MIBK = $35/mol of

MIBK) – ($35/mol) (19.7169 mol/hr) (8150 hr/yr)

= $4.464×106/yr

• This value is much higher than E.P. ( = $1.315×106) – So we drop any idea of using MIBK as solvent


Outline • Design of Gas Absorber – Energy Balances


Material Balances

Acetone

2

1

3

Douglas, J. M. Conceptual Design of Chemical Processes, 1988


Energy Balances for the Acetone Absorber • Since the inlet composition to the gas absorber is

quite dilute (10.3/687) (i.e. Acetone/Air) – assume that the absorber will operate isothermally

(constant temperature)


Contd..

Fig.: Stream Temperatures in Gas absorber


Contd.. • Do not store our product stream (top product from

Distillation column, Acetone) at its boiling point – so install a product cooler. cooler. – the temperature of the product stream leaving the product

cooler will be 100 0F.

• Acetone product (99 % pure) pure) contains 1 % water. water. – guess that the temperature of the overhead is essentially

the same as the boiling point of acetone (56.5 0C or 135 0F)


Contd.. 120oF

120°F

90°F

Fig.: Stream Temperatures in Still overhead


Contd.. • Similarly, assume that the bottom stream from the

still is 2120F (i.e. B. P. of water = 100 0C) • Cool this waste stream to 100 0F (cooling water

temperature) prior to pollution treatment.


Contd..

Fig.: Stream Temperatures at Still bottom


Energy Integration Flowsheet


Energy Balances for the Acetone Absorber • Must specify the temperature of the stream

entering the distillation column


Contd.. • If we do not preheat the feed stream entering the

distillation column to close the saturated liquid condition,

What will be consequences? – 7


Contd.. • Energy Balances – With the specified stream temperatures and estimated

stream flows, heat loads of various streams can be calculated





Qi  F i C Pi T in  T out

– Thus we can decide on HEN and calculate • The H.E. areas • Annualized H.E. capital costs and • The utility costs


Process Alternative • We noted that the still bottom was almost pure

water (0.05 mol acetone and 1943 mol of water) • For this case, the column reboiler uses 25-psia

(lps) at 276 degree F • As a process alternative, we could eliminate the reboiler and feed live steam to the column (alternative)


Do we have to face any problem in this case?


Next Lecture • Design of Absorber – Determination of number of plates – Cause-and-effect relationship of design variables – Opportunities for simplification of unit operation


Outline • Equipment Design Consideration – Number of plates in gas absorber – Cause-and-effect relationship of process design

variables – Simplifying unit-operation models (Back-of-the-Envelop design equation)

• Rules of thumb: Liquid flow rate to absorber


Equipment Design Consideration • Calculate the size & cost of the absorber and

distillation column • Need to understand the cause-and-effect

relationships (Input-output models) of the design variables • System vs. Unit Approach


Gas Absorber • For isothermal dilute system, the

Kremser’s Eqn.

 L  yin  mxin     1  1 ln   mG  y  mx    N  1   L  ln    mG  o u t

• Pure water as the solvent,

xin  0

in


Contd.. • From the rules of thumb, discussed earlier,





yout  1  0.99 yin  y   y  1  0.99   0.01      P      L  1.4mG  1.4  G   P    o u t

in

T


Effect of Design Variables: Column Pressure • • • • •

If we double the column pressure ( PT), L decreases by a factor of 2,   P   L  1.4mG  1.4 G but since L/mG = 1.4, i.e. constant  P  both L and m are = f( PT), decreases The number of plates required in gas absorber does not change.  L   y  mx    T

  1  1  mG  y  mx    L  ln    mG 

ln 

N  1 

in

ou t

in

in


Contd.. Lower values of L means 1. D.C. feed will be more concentrated 2. The reflux ratio decreases 3. The vapor rate in the still decreases 4. The column diameter decreases 5. Sizes of condenser and reboiler decreases (load decreases) 6. Steam and cooling water requirement decreases


Is there any consequence of increasing the pressure?


Effect Design Variables: Solvent MIBK • For MIBK • ᵞ = 1 in place of 6.7 in case of water • Liquid rate could be decreased as m will decrease • Decreases the D.C. cost • No. of plates in absorber will not change as L/mG is

constant


Effect of Design Variables: Operating Temperatures • If we change the inlet water (solvent to absorber)

temperature to 400C (112 0F), γ = 7.8 & P o = 421 mm Hg, P o

T , γ ∝T

∝

• Thus ‘m’ increases , L increases ( So the D. C.

Cost increases) • But number of trays in absorber does not change (L/mG = Const)


Simplifying Simplifying unit-operation models called as Back-of-the-Envelope Back-of-the-Envelope Design • Also called equation • Significance and order of magnitude of various terms in Kremser’s Eqn.  L   yin  m xin      1  1 ln   m G  yo u t  m xin    N  1   L  ln   m G 


Contd..  L   y  mx  1 ln   mG  y  mx  N  1   L  ln   mG  in

o u t

in

   1

in

 

 

L.H.S. of Kremser Eqn. = N +1 Assuming N @ =15-20 =15- 20 trays trays and 10% error is allowed N + 1 ≈N L.H.S of Kremser Eqn. = N


Contd..  L   y  mx      1  1 ln   mG  y  mx    N  1   L  ln  mG   in

o ut

in

in

R.H.S. For pure solvents, x in = 0 (Solute concentration in pure solvent = 0) Numerator of R.H.S. of Kremser Eqn. =  L  y ln

  1  mG  y

in

ou t

  1 


Contd.. Rules of thumb indicate L mG

 1.4 &

yin you t

 100

Thus  L   yin  1  1  40  1   mG  yout  1<<40


Contd.. • Applying the order of magnitude criteria

( 1 << 40)  L  L  yin  yin       1  ln   ln   1  1  mG  yout     mG  yout 

• The denominator of R.H.S. Kremser Eqn.  L  ln   ln1   mG  • From Taylor series expansion,

ln 1  


Contd.. 

L mG

 1  1.4  1  0.4  ln

L mG

 0.4

• With these, simplification and replacing ‘ln’ by ‘log’

we get


Contd..

• Approximation for a recovery of 99% gives 10 trays

instead of actual value of 10.1 •

For recovery of 99.9% gives 16 trays which is a very good estimation


Rules of Thumb: Liquid Flow Rate to gas Absorbers • For Isothermal, dilute gas absorbers – Kremser Eqn. can be used for calculating No. of trays reqd. (N ) for a specified recovery as a function of L/mG

 L   y  mx      1  1 ln   mG  y  mx    N  1   L  ln   mG  in

ou t

in

in


Contd..

Fig.: Liquid Flow Rate vs. Fractional Recovery : Kremser Eqn. Douglas, J. M. Conceptual Design of Chemical Processes, 1988, pp. 86


Contd.. L /m G < 1

INFINITE No. of trays are required for near complete recovery (Infinite capital cost) L /m G = 2

5 plates are required for complete recovery (≈100 %) – Large L correspond to dilute feeds to the distillation

column

separation process

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