separation process

BITS Pilani Pilani Campus

PROCESS DESIGN PRINCIPLES I CHE F314

Suresh Gupta Department of Chemical Engineering

BITS-Pilani, Pilani Campus

BITS Pilani, Pilani Campus

CHE F314 Process Design Principles I

• Saturday, 8-8:50 AM

Section-1

Room No. 6107

Mr. Subhajit Majumder

Section-2

Mr. Utkarsh Maheshwari

Room No. 6104

Tutorial Class

2


Lecture-1

Introduction 06-08-2015



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…”

4



5

Process

Design



PROCESS

6

Process

Inputs Outputs

“products” “feed”



DESIGN Definition of Chemical Process Design

7

Raw Material Chemical

Product

Chemical

Process

??

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



• 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

8



• 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

Creative Aspects of Process Design

9



Your role in the chemical process

10

?? Process synthesis vs. Process 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)

Strategy for Process Synthesis and Analysis

11



• 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

Contd..

12



• 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”

Problem Areas Synthesis and Analysis

13



• Synthesis is difficult because there are very

large number (104 – 109) of ways to

accomplish same goal

• Hence design problems are very open-

ended

Contd..

14



• 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.

Objective

15



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

Need of Process Design Principles

16



Hierarchy of Decisions

17

Level 1 • Batch vs Continuous

Level 2 • Input-Output Structure

Level 3 • Recycle Structure of flowsheet

Level 4

• General Structure of Separation System 4a. VRS

4b. LSS

Level 5 • Energy Integration Analysis (EIA)



Decision on Operating mode

Input Information and Batch vs Continuous

18




19




Level 4


4b. LSS





20




Level 4


4b. LSS





21




Level 4


4b. LSS





22




Level 4


4b. LSS




Organization of Course

23

Module-I Strategy for Process

Synthesis and Analysis



24

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

25



Evaluation Scheme

26

EC

No

.

Evaluation

component (EC)

Duratio

n

(Minute

s)

Weightage

(300)

Date and time Nature of

component

1 Mid-Semester Test 90 75 Closed Book

2 Tutorials/Surprise

Tests#

- 70 - Open/Closed

Book

3 Assignment* - 35 To be

announced in

the class in due

course of time

Open Book

4 Comprehensive

Examination

180 120 Closed Book



• Concepts of

• Heat Transfer

• Separation Process I & II

• Chemical Process Calculations

• Chemical Engineering Thermodynamics

• Fluid Mechanics

Prerequisite

27



• Being ONTIME is a good thing!

• Be Interactive!

• Share your idea and views

Points to Remember

28



• 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

Outline

29



Example: Hydrodealkylation of toluene (HDA Process)

– To produce benzene

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 11500F

Hierarchical Approach to Conceptual Design

30

25666

4662356

22 HHCHC

CHHCHCHHC

Rxn 1

Rxn 2



Boiling points

31

Boiling Point (°F)

Diphenyl 491

Toluene 232

Benzene 176.2

Methane −258.68

Hydrogen - 423.182

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Possible flow sheet

HEAT

HEAT

R E C Y C L E

HEAT

Diphenyl (unwanted)

Toluene (C6H5CH3)

H2, CH4 HEAT COMPRESSOR

REACTOR COOLANT FLASH

P R O D U C T

S T A B I L I Z E R

H2, CH4

Recycle H2

Recycle Toluene

1150 – 1300 0F H2, CH4

C6H6

C6H5CH3,Diphenyl

Partial Condenser

Condensed aromatics + Light gases

Light Gases

H2, CH4

Purge

H2, CH4

C6H6, C6H5CH3, C6H5 C6H5CH3, (C6H5)2

C6H6

(Main Product)

Boiling Point (°F)

Diphenyl 491

Toluene 232

Benzene 176.2

Methane −258.68

Hydrogen - 423.182



• 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 Integration

33



Energy Integrated Flow sheet

34

Fig. 2



• 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

Contd..

35



• We could recover the benzene as overhead

• Remove toluene as the side-stream (below the feed), and

recover the diphenyl as a bottom stream

Distillation Train

Benzene (C6H6) H2, CH4

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

Feed

H2, CH4, C6H6,

C6H5CH3,C6H5

C6H6, C6H5CH3, C6H5

Fig. 3

Diphenyl (C6H5)2

Boiling Point (°F)

Diphenyl 491

Toluene 232

Benzene 176.2

Methane −258.68

Hydrogen - 423.182



Contd..

Benzene (C6H6)

H2, CH4 Toluene (C6H5CH3)

Diphenyl (C6H5)2

Feed

H2, CH4, C6H6,

C6H5CH3,C6H5

C6H5CH3, C6H5

Fig. 4

Boiling Point (°F)

Diphenyl 491

Toluene 232

Benzene 176.2

Methane −258.68

Hydrogen - 423.182



• 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

Contd..

38



• 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

Vapor Recovery System (VRS)

39



• 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..

40



When do we consider designing of VRS?

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

separation system

Contd..

41



• 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 train is finalized)

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

to design the VRS and LRS

– it was decided to consider these design problems just before the

energy integration

Simplified Flowsheet for the Separation Systems

42



• The connections between the VRS and LRS shown in Fig. 5

Contd..

REACTOR SYSTEM

PHASE SPLIT

VAPOR RECOVERY SYSTEM

LIQUID SEPARATION SYSTEM

Benzene

H2, CH4

Liquid

(aromatics)

Diphenyl

Toluene

Toluene

Purge

H2, CH4 Gas Recycle

H2, CH4

Aromatics + Light Gases

Light Gases

Aromatics

Fig. 5



A simplified flow sheet for the process is shown in Fig. 6

Recycle structure of the flow sheet

REACTOR SYSTEM

SEPARATION SYSTEM

Benzene H2, CH4

Diphenyl Toluene

Toluene (Recycle)

Purge

H2, CH4 Gas Recycle

H2, CH4

Aromatics + Light Gases

Fig. 6



• 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.

Contd..

45



Can we still think of simplifying the flowsheet?

46



• 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

Input-Output structure of

the flowsheet

PROCESS

Benzene H2, CH4

Diphenyl Toluene

Purge

H2, CH4 Gas Recycle

H2, CH4

Fig. 7

Liquid Recycle Is this structure of flowsheet correct?



• 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

Contd..

48



• A systematic approach to process design by

reducing the design problem to a hierarchy of

decisions:

1. Batch vs Continuous

2. Input-Output structure of the flow sheet

3. Recycle structure of the flow sheet

4. General structure of the separation system

a) Vapor liquid system

b) Liquid separation system

5. Energy Integration Analysis (HENS)


49



• 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

Contd..

50



• Ethanol is produced by the hydration of ethylene. The primary reactions for

ethanol synthesis are given below:

• 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

51

OHether DiethylEthanol 2

EthanolOHEthylene

2

2



• Draw the following: • General structure of the flow sheet

• Recycle structure of the flow sheet

• Input-output structure of the flow sheet

Problem

52



• Hierarchical Approach to Conceptual Design

IPA Case Study

• Design of a solvent recovery system

Outline

53



Draw the,

1. General structure of the Separation system.

2. Recycle structure of the flowsheet.

3. Input-output structure of the flowsheet.

Contd..

54



55


Economic Decision Making



• Design of A Solvent Recovery System (Ch. 3 of T2)

• Problem Definition

• Economic Potential

• Process alternatives

Outline



• 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)

Problem Definition



• Economic Potential (EP)

• Since stream coming from the same process,

Raw material cost = 0

• Therefore

Economic Potential (EP)

Cost Material Raw - ValueProduct EP

Cr/yr 5.26 Rs.

hr/yr) 50lb/mol)(81 5810.80/lb)( .mol/hr)(Rs (10.3

ValueProduct

0 - ValueProduct

Cost Material Raw - ValueProduct EP

Operating hours



Question 1

How to recover acetone?

60



• Solvent recovery alternatives 1. Condensation

a. High Pressure

b. Low temperature

c. Combination of both

2. Absorption

3. Adsorption

4. A Membrane Separation System

5. A Reaction Process (Acetone as raw material for a new

product)

General Considerations: Process Alternatives



Question 2

Which is the cheapest

alternative?

62



• 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

General Considerations: Process Alternatives

63



• 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)

Contd..

64



Design of Gas Recovery System: Flowsheet

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



Alternate Flow sheet: Recycling of solvent




Question 3

Whether discarding the process

water, as shown in Fig. 1 can

ever be justified even when a

pollution treatment facility is

available?

67



• 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..

68



Contd..

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



• This reasoning is the basis for a design heuristic

Design of Gas Absorber: Heuristic

70

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



• Considering the flow sheet shown in Fig. 1

because it is the simplest for further processing

Design of Gas Absorber



• 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

Contd..

72



• 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)

Design of Gas Absorber: Material Balances



Contd..

Acetone 2

1

3




• The gas leaving the absorber (top) will contains

– air, some acetone and some water

– Since water is relatively inexpensive, neglecting this

solvent loss

Contd..



• 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?

Specify Acetone Amount for Material Balance



• 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..

77



Contd..

78

Fig.: %Recovery vs. Cost in Gas Absorber



• 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..

79



Contd..

80

Fig.: %Recovery vs. Cost in Distillation



Design of Gas Absorber: Heuristics

81

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

82

H3: For an isothermal, dilute absorber, choose the

solvent flow rate (L), such that L = 1.4mG

where, m = slope of equilibrium line, and G = gas molar

flow rate (mol/hr)



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

1 atm

– Activity coefficient, γs= 6.7

– Vapour pressure of acetone in air, Pos = 229 mm Hg.

– Air flow rate, G = 687 mol/hr

Material Balances

83

02.2760

)229(7.6

T

o

ss

s

s

vsTs

P

P

x

ym

PxPy

Fugacity

coefficient

Mol

fraction

solute

in gas

Total

pressure

of

system

Activity

coefficient

Mol fraction

of solute in

solvent

Vapor

pressure



• 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

• 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

Contd..

mol/hr 10.0)20.10(99.0

99.01



Material Balances: Flowsheet

Fig.: Stream compositions and flow rates



• 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: Acetone 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

Material Balances: Water Balances



• 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

Stream Cost Calculation: Acetone-Water System



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

Contd..



• For a low pressure absorber, fugacity correction

factors are negligible

• Vapor-liquid equilibrium relationship for the

solvent can be written as

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

Solvent Loss Calculations: Other than Water as Solvent

ssssT xPyP 0



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



• If a solvent is used that is in the homologous

series with the solute, then γs = 1

• Thus, from

Contd..

T

ss

P

Py

0

ssssT xPyP 0



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)

Contd..



Solvent Loss Calculations: Other than Water as Solvent



• Quick way to estimate the solvent loss

– Where ys is the mole fraction of solute in solvent

• Now, the amount of solvent lost

Contd..

syGG 1'

GyGy

y

yy

G

yGm

s

s

s

s

s

ss

1

1

'



• Suppose we consider using MIBK (Methyl Isobutyl

Ketone) as a solvent and we recycle the MIBK

– At 25oC, PT = 1 atm, Ps0 = 0.0237 atm

– ys = Ps0 /PT = 0.0237 / 1 = 0.0237

• Therefore, solvent lost

Cost of Solvent Loss Using MIBK as Solvent

0237.01

0237.00

T

ss

P

Py

mol/hr7169.19mol/hr6870237.0 Gym ss



• 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

Contd..



• Design of Gas Absorber

– Energy Balances

Outline



Material Balances

Acetone 2

1

3




• 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)

Energy Balances for the Acetone Absorber



Contd..

103

Fig.: Stream Temperatures in Gas absorber



• Do not store our product stream (top product from

Distillation column, Acetone) at its boiling point

– so install a product cooler.

– the temperature of the product stream leaving the product

cooler will be 100 0F.

• Acetone product (99 % pure) contains 1 % 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..

104



Contd..

105

Fig.: Stream Temperatures in Still overhead

120oF

90°F

120°F



• Similarly, assume that the bottom stream from the

still is 2120F (i.e. B. P. of water = 1000C)

• Cool this waste stream to 1000F (cooling water

temperature) prior to pollution treatment.

Contd..

106



Contd..

107

Fig.: Stream Temperatures at Still bottom



Energy Integration Flowsheet

108



• Must specify the temperature of the stream

entering the distillation column

Energy Balances for the Acetone Absorber

109



• If we do not preheat the feed stream entering the

distillation column to close the saturated liquid

condition,

What will be consequences? – 7

Contd..

110



• Energy Balances

– With the specified stream temperatures and estimated

stream flows, heat loads of various streams can be

calculated

– Thus we can decide on HEN and calculate

• The H.E. areas

• Annualized H.E. capital costs and

• The utility costs

Contd..

111

outinPiii TTCFQ



• 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)

Process Alternative

112



Do we have to face any problem in

this case?

113



• Design of Absorber

– Determination of number of plates

– Cause-and-effect relationship of design variables

– Opportunities for simplification of unit operation

Next Lecture

114



• 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

Outline



• 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

Equipment Design Consideration

116



• For isothermal dilute system, the Kremser’s Eqn.

• Pure water as the solvent,

Gas Absorber

117

mG

L

mxy

mxy

mG

L

Ninout

inin

ln

11ln

1

0in

x



• From the rules of thumb, discussed earlier,

Contd..

118

inout yy 99.01

GP

PmGL

y

y

T

in

out

4.14.1

01.099.01



Effect of Design Variables: Column Pressure

119

• If we double the column pressure (PT), • L decreases by a factor of 2, • but since L/mG = 1.4, i.e. constant • both L and m are = f(PT), decreases • The number of plates required in gas absorber does

not change.

GP

PmGL

T

4.14.1

mG

L

mxy

mxy

mG

L

Ninout

inin

ln

11ln

1



Contd..

120

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?

121



• 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 Design Variables: Solvent MIBK

122



• If we change the inlet water (solvent to absorber)

temperature to 400C (112 0F),

γ = 7.8 & Po = 421 mm Hg, Po ∝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)

Effect of Design Variables: Operating Temperatures



• Also called as Back-of-the-Envelope Design

equation

• Significance and order of magnitude of various

terms in Kremser’s Eqn.

Simplifying unit-operation models

mG

L

mxy

mxy

mG

L

Ninout

inin

ln

11ln

1



Contd..

125

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

mG

L

mxy

mxy

mG

L

Ninout

inin

ln

11ln

1



Contd..

126

R.H.S. For pure solvents, xin = 0 (Solute concentration in pure solvent = 0) Numerator of R.H.S. of Kremser Eqn. =

mG

L

mxy

mxy

mG

L

Ninout

inin

ln

11ln

1

11ln

out

in

y

y

mG

L



Contd..

127

Rules of thumb indicate

100&4.1 out

in

y

y

mG

L

Thus

14011

out

in

y

y

mG

L

1<<40



• Applying the order of magnitude criteria

( 1 << 40)

• The denominator of R.H.S. Kremser Eqn.

• From Taylor series expansion,

Contd..

128

out

in

out

in

y

y

mG

L

y

y

mG

L1ln11ln

1lnln

mG

L

1ln



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

we get

Contd..

129

4.0ln4.014.11 mG

L

mG

L



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



• 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

Rules of Thumb: Liquid Flow Rate to gas Absorbers

131

mG

L

mxy

mxy

mG

L

Ninout

inin

ln

11ln

1



Contd..

132

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



L/mG < 1

INFINITE No. of trays are required for near

complete recovery (Infinite capital cost)

L/mG = 2

5 plates are required for complete recovery

(≈100 %)

– Large L correspond to dilute feeds to the distillation

column

Contd..

133



L/mG > 2

We obtain tiny, inexpensive absorbers but very expensive D.C.

• Based on above arguments, 1 < L/mG < 2

L/mG = 1.5: Observe the shape of the curves near L/mG • Better trade off (with high recoveries)

• Decreasing no. of trays in absorber (capital cost Vs. Increasing capital cost & operating cost of D.C.)

• Common Rule of Thumb: By decreasing L ( such that L/mG=1.4) almost 100 % recovery

Contd..

134

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