1. Texas A&M University 2 MODULE III Introduction to Process Integration.

Texas A&M UniversityTexas A&M University

2

MODULE III

Introduction to Process Integration


3

1. Introduction

4. Open Ended Problem

2. Foundation Elements

3. Case Study

5. Acknowledgments

Outline

6. References


4

TIER I


5

1. Introduction


6Source : Pollution Prevention through Process Integration, M. M. El-Halwagi

“Do your best; then treat the rest”

1. Introduction


7

Pollution is an ongoing concern that has been addressed in many different ways, from no pollution control, end-of the-pipe treatment

(1970’s), Implementation of Reuse/Recycle (1980’s) up to

Process Integration. The focus of this module is to expose PI tools

for pollution reduction/elimination

1. Introduction


8

What is Process Integration?

“It is a holistic approach to process design, retrofitting and operation which

emphasizes the unity of the process”

Source : Pollution Prevention through Process Integration, M. M. El-Halwagi

1. Introduction


9

The use of PI methods started as early as 1970’s with Pinch Technology (Heat Integration) in order to optimize heat exchanger networks (HEN).

The moving force for mass integration was initially pollution control; El-Halwagi and Manousiouthakis (1989) proposed the use of mass exchange networks (MEN) in analogy to the previously studied HEN.

PI tools can be used in a variety of industries and with approaches as wide as those involving product distribution, life cycle assessment etc (research in these an other areas is currently on their way)

1. Introduction


10



11

2.1. Holistic approach of process integration

2.2. Relationship of process integration to process analysis

2.3. Overview of energy, mass and property integration



12

Holistic: Emphasizing the importance of the whole and the interdependence of its parts. Concerned with wholes rather than analysis or separation into parts

Source : http://dictionary.reference.com

Heuristic: Of or constituting an educational method in which learning takes place through discoveries that result from investigations made by the student

2. Foundation Elements2.1 Holistic Approach of Process Integration


13

Efficient use of resources and raw materials

Efficient use of energy

Pollution reduction

Process debottlenecking

Cost reduction

Other process operation issues

Process Integration can address a wide set of design issues such as:



14

• Traditional process design has been addressed by heuristic methods, based on experience or corporate preferences, in which unit operations equipment have been design individually.

• However little attention has been placed on the relationships with other parts of the process

• Process Integration as a holistic approach, looks at the Big Picture and the relationships among the different operations and equipment alternatives



15

In order to illustrate how Process Integration (PI) can aid in the design process an illustrative example is given we have 3 options for a chemical reactor in order to produce a chemical product, the options to choose from are:

Source : www.aiche.org/cep/ July 2001



16

Using a heuristic approach the “best” option will be a mechanically agitated vessel that produces a yield of 73.9% with a volume of 12m3; however is there any other way to improve the process?



17Source : www.aiche.org/cep/ July 2001

Two designs based on the same solution



18

Using PI tools the following solution was found, 96.9% yield and 9.93m3 of volume.

Two designs based on this solution are shown next; the benefits of using PI tools are evident.

However a thorough analysis of the answer to the problem must be carried out in order to find a feasible design based on the findings obtained using a PI approach

Source : www.aiche.org/cep/ July 2001



19

• In order to find solutions that include the relationship effects among the different options for a given design task, the engineer must use PI in order to find optimum answer to the problems at hand, therefore PI tools should be included in the process design structure. Seider, Seader and Lewin illustrate it as shown in the next slides, for a complete description of the design steps, referred to the above mentioned authors

• Process design is a dynamic process, always making sure that the solutions will agree with the constraints set by the stakeholders (management, governmental agencies, environmentalist groups, general public etc) and the process itself

2. Foundation Elements2.2. Relationship of Process Integration to Process Analysis


20

Process Analysis

“Analysis of the process elements for individual study of performance, by

using mathematical models and computer simulators”




21

Current Situation/Opportunity

(e.g. a new technology is developed etc)

Asses Primitive Problem

(Define the objective of the design task based on the identified opportunity)

Survey Literature

(Identify all sources of useful information for the process design, e.g. Handbooks etc)

Preliminary Data Base Creation

(Thermodynamic data, kinetics, toxicity etc)

Preliminary Process Synthesis, reactions,

Separation, T-P Change Operations,

Task Integration

Source : Product and Process Design Principles : Synthesis, Analysis, and Evaluation W D. Seider J. D. Seader, D.R. Lewin

Equipment Selection

(Assess different options for the given process using process simulators, spreadsheets, in house software etc)

Part I



22

Is the Gross Profit

Favorable?

Yes

No


Equipment Selection

(Assess different options for the given process using process simulators, spreadsheets, in house software etc)

Reject

Part I a Part II Part IV



23


Create Process Flow Sheet

Process Integration

Pilot Plant

Testing Modify Flow Sheet

Create Detailed

Data Base

Prepare Simulation

Model

Heat and Power Integration

Second Law Analysis

Separation Train Synthesis

Dynamic Simulation

Flow Sheet Controllability

Analysis

Qualitative Synthesis

Part I a

Part IIPart VIIs the Process

still Promising? Part IIIGo to I or I a

No Yes



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Detail Design, Equipment Sizing,

Capital Cost Estimation,

Profitability Analysis, Optimization

Part IV

Is the Process still Feasible?

Is the Process still Promising?

Startup Assessment (Additional Equipment, Dynamic Simulation)

Part I or I a

Yes

NoReject Part III

Reliability and Safety Analysis (HAZOP, Pilot

Plant Testing etc)No

YesWritten Report,

Presentation

Part IVFinal Design

(P&ID, Bids etc)Construction Startup

Operation




25

Designing a new plant, retrofitting a existing one, has several operations and for each operation different equipment options and configurations to choose from.

The main problem is that the number of alternatives can be unmanageable. If only heuristics are use for the design, the engineer will risk to miss the true optimal solution to the design problem. Moreover, a design solution for a given problem cannot be use for a different one, since the initial findings are tailored for a specific problem.

Using a PI approach, one can avoid this issue, due to the fact that its methodology can be applied to any problem. The PI methodology is composed of three key components



26

Process Integration

Process Synthesis

Process Analysis

Process Optimization

It defines what process units and how they should be interconnected

Analysis of the process elements for individual study of performance

Minimizing or maximizing a desired function, to find the best option



27

As it has seen, process analysis is a step within the PI methodology.

It is important to emphasize that PI will look at the generalities rather than into the details, and then the designer can analyze the performance of the solutions in order to optimize his/her findings.

The following chart illustrate the impact of the process design steps over the budget

Process Conceptual Detailed Plant Detail Construction Startup &

Develop Design Design Layout Mech. Commission.

Impact

Committed

Spent$

Preliminary equipment selection

Equipment required during

design



28

Mass Integration

“Systematic methodology that provides a fundamental understanding of the global flow of

mass within the process and employs this holistic understanding in identifying performance targets

and optimizing the generation and routing of species throughout the process”


2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration


29

•Mass Exchangers:

A mass exchanger is any direct-contact mass transfer unit that employs a MSA (Mass Separation Agent), to remove selectively certain component (e.g. pollutant) from a rich phase (e.g. waste stream).

The MSA should be partially or totally immiscible in the rich phase

Mass

Exchanger

Outlet Composition

yiout

Rich (Waste) Stream, Flow rate: Gi Inlet Composition

yiin

Lean Stream (MSA) Flow rate: Lj Inlet Composition

xjin

Outlet Composition

xjout



2.3.1 Mass Exchangers


30

When the two phases are in intimate contact the solutes are distributed between the two phases which leads to a depletion of solute in the rich phase and enrichment of the lean phase until equilibrium is reached. The difference in chemical potential for the solute is the moving force for mass transfer (Temperature difference for heat transfer, Pressure difference for fluid movement etc)

Solute Transferred to lean phase

Rich

Phase

Lean

Phase

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.1 Mass Exchangers


31

Mass Exchange involve the following operations: Only counter current operations will be consider because of their higher efficiency

Adsorption

Absorption

Extraction Ion Exchange

Leaching

Stripping


2.3.1. Mass Exchangers


32

Adsorption:Separation of a solute from a liquid or gaseous stream by contacting the carrying phase with a small porous solid particles (adsorbent), usually

arranged in a packed bed. The adsorbent can be regenerated by desorption using inert gas, steam etc

Source : Université d’Ottawa / University of Ottawa - Jules Thibault




33

In order to select an adsorption column the designer must select a suitable adsorbent for the given solute by looking at the appropriate

isotherm data as shown in the plot for a given set of process operation





34

Absorption:A liquid solvent is place in contact with a gas containing a solute to be remove by taking advantage of the preferential solubility of the liquid. Reverse absorption is also know as

stripping (separation of a solute using a gas stream from a liquid phase)





35

Liquid Extraction:It employs a liquid solvent to remove a solute from another liquid by using the

preferential solubility of the solvent to the solute in the MSA





36

Leaching:Selective separation of some constituents

within a solid by contact with a liquid solvent

Solvent Solid

Mixing

SlurryOverflow Solution



Source : University of Ottawa - Jules Thibault


37

Ion Exchange:Cation/anion resins are used to replace undesirable anions

from a liquid phase by non hazardous ions

NaCaRRNaCa 222

Water softeners

Cause of scale forming impurities





38

The mass exchanger is used to provide appropriate contact of the lean and rich phase; there are two principal categories of mass exchange units:

- Multistage (e.g. tray columns, mixer settlers etc), they provide intimate contact follow by phase separation

- Differential (e.g. packed columns, spray towers and mechanically agitated units), continuous contact between phases without intermediate separation and re-contacting




39Heavy Phase Out

Heavy Phase In

Light Phase In

Light Phase Out

Perforated Tray

Shell

Waste In

Waste Out

MSA Out

MSA In

Multiple Mixers / Settlers

Multistage Contactors

Tray Column



40

Heavy Phase In

Light Phase In

Heavy Phase Out

Light Phase Out

Spray Column

Heavy Phase In

Light Phase In

Light Phase Out

Mechanically Agitated Mixer

Mixer

Heavy Phase Out

Differential / Continuous Contactors



41

Equilibrium:

When a rich phase in a solute is put in contact with a lean phase transfer of the solute to the lean phase occurs, also part of the solute In the lean phase also back transfer to the rich phase.

At first the rate of solute being transfer from the rich phase is bigger than the rate of solute back transfer from the lean phase. However when the concentration of solute in the lean phase increases, the back transfer rate also increases.

Eventually the mass transfer rate and the back transfer rates become equal and an equilibrium is reached

Solute in the rich phase

Equilibrium distribution

function

Maximum attainable composition in the

lean phase


)( *jji xfy (1)




42

In environmental applications the engineer will find very often, diluted systems which can be linearized over the operating range to yield:

Special cases, Raoult’s Law for absorption


Partial pressure at

T

Mol fraction of solute in

gas

Mol fraction of solute in

liquid

jjji bxmy *(2)

*)(j

Total

soluteo

i xP

TPy (3)




43

Henry’s Law for stripping


Mole fraction of solute in

gas

Mol Fraction of solute in

stripping gas

Liquid phase solubility of pollutant at temperature

T

*jji xHy (4)

)(

lub

TP

yPH

Soluteo

ilitySoiTotal

j (5)




44

For solvent extraction

Composition of pollutant in liquid waste

Composition of the

solvent

Distribution Coefficient


*jji xKy (6)




45

The following relationships are used to size multistage mass transfer exchangers:

1 2 NN-1

XJ,0= Xjin

Lj

XJ,2 XJ,N-2 XJ,N-1 XJ,N= XJoutXJ,1

yi,1= yiout yi,N-1yi,3

yi,2 yi,N

yi,N+1= yiin

Gi

Overall Mass Balance:out

jjout

iiin

jjinii xLyGxLyG (7)




46

Rearranging (7):

)(

)(in

jout

j

outi

ini

i

j

xx

yy

G

L

(9)

Eq. (8) represents the operating line in a McCabe-Thiele diagram:

LJ / Giyiin

yiout

xJin xJ

out

Theoretical stages

1

2

Equilibrium Line

Operating Line

)()( inj

outjj

outi

inii xxLyyG (8)




47

•The number of stages for a multistage unit can also be calculated with the following equations, with NTP being the number of theoretical plates

ij

j

jin

jjout

i

jin

jjini

j

ij

Gm

L

bxmy

bxmy

L

Gm

NTP

ln

1ln

(10)

j

ij

ij

iout

jout

i

outj

ini

ij

j

L

Gm

Gm

L

xx

xx

Gm

L

NTP

ln

1ln *

*

(11)

j

jiniout

j m

byx

* (12)





48

(13)

NTP

ij

j

jin

jjout

i

jin

jjini

Gm

L

bxmy

bxmy

oNTPNAP / (14)



When the contact time for each stage is not enough to reach equilibrium, the number of actual plates (NAP) can be calculated using contacting efficiency

Stage efficiency can be define on the rich or lean phase, for the rich phase we have:


49

11ln

1ln

j

ijy

j

ij

jin

jjout

i

jin

jjini

j

ij

L

Gm

L

Gm

bxmy

bxmy

L

Gm

NTP

(15)

xx

yy

NTUHTUH

NTUHTUH

(16)

(17)

Based on rich phase

Based on lean phase



For differential (continuous) mass exchangers, the height is calculated using:


50

For mass exchangers with linear equilibrium:

meanii

outi

ini

yyy

yyNTU

log*)(

(18)

)(

)(ln

)()()( *

jin

jjout

i

jout

jjini

jin

jjout

ijout

jjini

ii

bxmy

bxmy

bxmybxmyyy (19)




51

For mass exchangers with linear equilibrium (cont):

meanjj

outj

inj

xx

xxNTUx

log*)(

(20)

j

jout

iinj

j

jiniout

j

j

jout

iinj

j

jiniout

j

meanjj

m

byx

m

byx

m

byx

m

byx

xx

ln

)( log*

(21)



52

j

ij

j

ij

jin

jjout

i

jin

jjini

j

ij

L

Gm

L

Gm

bxmy

bxmy

L

Gm

NTP

1

1ln(22)

In order to calculate the diameter of the column (m) we have:

(23)

)(

)(4min MASVA

VFRAD




53

In order to calculate the diameter of the column we need volumetric flow rate of air (VFRA), maximum allowable superficial velocity of air (MASVA):

air

airwatersmMASVA

068.0)/(

(24)

AFCAOCTAC (25)



To complete the design of a mass exchange unit, the designer has to look into the costs that the unit will have. The total annual cost (TAC) is given by:


54

Where AOC is the annual operating cost and AFC is the annual fixed cost of the unit. Recall equation (8)

yiin

yiout

xJin,max xJ

outxJin*

JEquilibrium

Line

Operating Line

The number of mass exchange units will be higher for a small , a vanishing driving force. Therefore, it is necessary to assign a minimum driving force between the two lines

Driving Force

Lean End of Exchanger



55

We have:




jjin

jjout

j bxmy )(min. (26)

By using a minimum allowable composition difference, J the designer can identify the minimum practically feasible outlet composition of the waste stream


56

yiin

yiout

xJin xJ

out,max xJout*

J

Equilibrium Line

Operating Line

The number of mass exchange units will be higher for a small , a vanishing driving force. Therefore, it is necessary to assign a minimum driving force between the two lines

Driving Force

Rich End of Exchanger


Remainder :An outlet composition on

the equilibrium line = infinite number of stages



57

We have:j

j

jin

joutj m

byx

max.

(27)

Where, J is the “minimum allowable composition difference” and xJ

out,max is the maximum practically feasible outlet composition of the MSA which satisfies the J driving force

As can be seen from (16 to 19) and (27), there is a trade off between the driving force and the cost/size of the equipment to be use for the separation. To illustrate the use of the previous equations a example is given





58

Example 1

Air stripping is used to remove 95% of the rich trichloroethylene (TCE, molecular weight = 131.4) dissolved in a 200kg/s (3180gpm) waste water stream. The inlet composition of TCE in the waste water is 100ppm. Air (free of TCE) is compressed to 202.6 kPa (2at) and diffused through a packed stripper. The TCE-laden air exiting the stripper is fed to the plant boiler which burns almost all the TCE.

Physical Data:

The stripping operation takes place isothermally at 293K and follows Henry's law. The equilibrium relation for stripping TCE from water is theoretically predicted using:




59

Where yi is the mass fraction of TCE in waste water and xJ is the mass fraction of TCE in air. The air-to-water ratio is recommended by the packing manufacturer to be:

24 m3Air / m3water

Stripper Sizing Criteria:

The maximum allowable superficial velocity of waste water in the column is taken as 0.02m/s (approximately 30 gpm/ft2).The overall height of transfer unit based on the liquid phase is given by:

HTUy = Superficial Velocity of waste water/Kya

jj xy 0063.0 (28)




60

Where ky is the water-phase overall mass transfer coefficient and a is the surface area per unit volume of packing. The value of Kya is provided by the manufacturer to be 0.002s-1

Cost Information:

The operation cost for air compression is basically the electricity utility needed for the isentropic compression. Electric energy needed to compress air may be calculated using: Compression Energy (CE)

11

)/(

1

in

out

isentropicair

in

P

P

M

RTkgkJCE

(29)




61

The isentropic efficiency of the compressor is 60% and the electric energy cost is $0.06/kWhr. The system is operated for 8000hr/y. The fixed cost, $, of the stripper (including installation and auxiliaries, but excluding packing) is given by:

Fixed cost of column = 4700HD0.9

Where H is the height of the column (m) and D is the diameter (m). The cost of packing is $700/m3. The fix cost of the blower, $, is 12000LJ

0.6, where LJ is the flow rate of air (kg/s). Assume negligible salvage value and a five year linear depreciation. (a) estimate the column size, fixed cost and annual operating cost. (b) Due to the potential error in the theoretically predicted value of Henry’s coefficient, it is necessary to asses the sensitivity of your results to variation




62

of the value of Henry’s coefficient. Plot the column height, annualized fixed cost and annual operating cost versus the relative deviation from the nominal value, for 0.5 2.0. The parameter is define by:

= Value of Henry’s Coeffcient/0.0063

(c) Your company is planning to undertake extensive experimentation to obtain accurate values of Henry’s coefficient that can be used in designing and evaluating the cost of this stripper. Based on your results, what would you recommend regarding the undertaking of these experiments?





Waste Water Gi = 200kg/s yi

in = 10-4

yiout = 5*10-6

Air, LJ = ? xJ

in = 0

xJout = ?

StripperBoiler

Exhaust Gas

Stripping of TCE from

Wastewater

Blower



64

Solution: (a)

1. We will first have to calculate the flow and concentrations of the different streams as follows:

33 412.2

293082057.0

29*2

m

kg

KkgmolK

atmmkgmolkg

atm

RT

PM airAir


s

kgAir

kgWater

m

s

kgWater

Waterm

Airm

m

kgLi 06.12

1000

1*200*

125*412.2

3

3

3

3




65

Solution: Continuation

Using the overall mass balance equation we have:

ppmx

airkgmolphenolkgmolx

x

outJ

outJ

outJ

1575

/00157.0

0

10*510*1

200

06.12 64


2. We now will calculate the height and diameter of the column, superficial velocity of waste water (SVWW)

aKSVWWHTU yy /



66


mssm

HTU y 102.0

02.0

1

meanii

yyy

NTUlog

*)(

5100


)0*0063.010*5()00157.0*0063.010*1(

ln

)0*0063.010*5()00157.0*0063.010*1()(

6

4

64

log*

meanii yy




67


ppmyy meanii 43.2910*943.2)( 5log

*

mH 228.3228.3*1

228.343.29

5100

yNTU


mD 568.3)02.0(

)1000/200(4min



68


3. With the equipment dimension we can proceed to calculate the operating and fixed costs


kgkWhrkJ

kWhr

kg

kJ/10*788.1$

1

06.0$*

3600

1*31.107 3

kgkJkgkJCE /31.10711

2

6.0*29

293*314.8

14.1

4.1)/(

4.1

14.1

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.1. Mass Exchangers


69


Annual Operating Cost (AOC):


2.455,53$)06.12(12000

8.592,22$$

700*228.3*)568.3(*4

5.666,47$)568.3*228.3(4700

6.0

32

9.0

Blower

mmmPacking

Stripper

yearyear

hr

hr

s

s

kg

kg

kJAOC /8.234,621$

8000*

13600*06.12*31.107

Equipment Cost (EC):




70

Solution: (b) (c)

Henry’s Law coefficient will affect the FC through the change in the size of the system. By changing one can find different values of Henry’s Law coefficient and use them to calculate the size of the column and then the FC; we will use Excel for this procedure. Since we have a linear 5 year depreciation the FC will be divided by 5


5.714,123$2.455,538.592,225.666,47 FC


Fixed Cost (FC):




71

Alfa Henry H AFC TAC

0.5 0.00315 3.107112 24214.47 645449.3

0.75 0.004725 3.166444 24472.71 645707.5

1 0.0063 3.228434 24742.51 645977.3

1.25 0.007875 3.293275 25024.73 646259.5

1.5 0.00945 3.36118 25320.28 646555.1

1.75 0.011025 3.432384 25630.19 646865

2 0.0126 3.507149 25955.6 647190.4


As the plot and Table 1 show, there is a small change in the TAC and AFC with changing Alfa, meaning that we don’t have appreciable savings by changing the height of the column with more accurate values of Henry’s Law coefficient. Therefore the project is not required; we just saved our company a lot of money!!!!




72



0

100000

200000

300000

400000

500000

600000

700000

0 0.5 1 1.5 2 2.5

AFC

TAC

Very slight change



Mass Exchange Networks


2.3.1. Mass Exchange Networks


74



Mass Separation

Agents (MSA)

They are Lean Streams (Ns), LJ, j = 1, 2…Ns

Use to remove pollutants from rich

streams, NR

Process MSA, NSP

Low cost or almost free“In plant”

External MSA, NSE

Must be bought externally

•MSA can be:



75



Flow rates, stream concentration and target concentration of rich streams are known, Gi, yS

S, yit

Inlet compositions of lean streams are also known, xJS flow rate of

lean streams, LJ, is to be determine to minimize network cost

Ns = NSP + NSE(28)

LJ LJC J = 1, 2…NSP

LJC is the flow rate of the Jth MSA available in the plant

(29)



76



Waste streams can be

Disposed

Forwarded to processSinks (equipment)For recycle/reuse

Comply with Environmental

Regulations

Target compositionis the constraint

imposed by processSink



77

• Target composition are assigned by designer based on the following constraints:



Physical (e.g. maximum

Solubility of pollutantIn MSA)

Technical (e.g. avoid corrosion,

Viscosity)

Environmental(e.g. EPA, OSHA

Regulations)

Safety(e.g. stay away of

Flammability limits)

Economic(e.g. optimize cost

Of MSA regeneration)



78

• The following questions will arise:



Which ME operation should we use?

Which MSA should be selected?

How to match MSAs

to the waste streams?

What is the optimum

configuration?


79

• The previous questions will result in a unmanageable number of combinations



• A systematic approach is required

“Targeting Approach”



80

Targeting Approach

“It is based on the identification of

performance targets ahead of design and

without prior commitment to the final network

configuration”



2.3.1.1. Mass Exchangers



Minimum cost of MSA: By combining thermodynamic aspects of the problem with cost data of the MSA, the designer can identify the minimum cost of the separation, without designing the network

Minimum number of mass exchange units: This objective is aim at minimizing fixed cost of the system, by doing so, one can reduce pipe work, foundations, maintenance and instrumentation

GENERALLY

INCOMPATIBLE



82

U = NR + Ni

U = Number of units

Ni = Number of independent synthesis sub-problems in which original synthesis problem can be subdivided

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.1.1. Mass Exchangers

(30)

• In most cases there will be only one independent synthesis problem. In order to avoid the incompatibility of the two targets, one have to use techniques that will identify the MOC solution and then minimize the number of exchangers that satisfy the MOC (Minimum Operating Cost)


83

• In order for the separation to be feasible one have to work in the feasibility area• To relate the different concentrations in one scale, we need to use Equation (27)

yiin

yiout

xJin xJ

out,max xJout*

J

Feasibility area




84


• In order to minimize the cost of external MSA one must maximize the use of in plant MSA

2.3. Overview of Mass, Energy and Property Integration2.3.1.1. Mass Exchangers

• The pinch diagram is a graphical representation that considers the thermodynamic constraints of the system, calculate MR with:

y x1

x2

Pinch Point

Mass Exchanged


R

ti

siii

Ni

yyGMR

,....,2,1

)(

(31)


85

How to construct the pinch diagram?



1. Represent each stream with an arrow

2. Plot mass exchanged versus its composition

3. Tail of the arrow is the supply composition and head is target composition

4. The slope is the flow rate of the stream

5. The vertical distance between the tail and the head represent the amount of pollutant transferred ( MRi ) from the rich stream ( yi ) to the lean stream

y1t y2

t y1s y2

s

MRi

yi

6. Stack the arrows on top of one another starting with the one with the one having the lower composition

R1

R2


86




7. Obtain the composite diagram by using the “diagonal rule”

8. The vertical axis is a relative scale, one can move up and down the curves while maintaining constant the vertical distance

9. Apply the same procedure for the lean streams

y1t y2

t y1s y2

s

MRi

MR2

MR1

yi

10. Plot both composite curves in one graph, slid the lean composite until it touches the rich (waste) composite stream

R1

R2



87



2.3. Overview of Mass, Energy and Property Integration


11. Use the above equation to obtain the horizontal scale and Equation 33 to calculate MS

x1s x1

t

MSiMS2

MS1

yi

x2s x2

t

S1

S2

jj m

byx

1

1

(32)



88





x1

yi

x2

Rich Composite Stream

Lean Composite Stream

SP

sj

tj

cjj

Nj

xxLMS

....,2,1

)(

(33)

Excess Capacity of Process

MSA’s

Load to be removed by

external MSA’s

Mass Exchanged



89





x1

Mass Exchanged

yi

x2


Lean Composite Stream

Pinch Point

Integrated mass exchange:

Maximum amount of pollutant that can be transfer

•The Pinch point is the minimum feasible concentration, it is also a bottleneck, slid up or down the composite curves until they touch, keeping the vertical distance and the concentrations



90

• In order to reduce the excess capacity of process MSA one can either reduce flow rate, or composition. Care must be given when choosing , since it will cause the lean composite curve to move to the right, increasing the load to be removed by external MSAs



Load of pollutant above the pinch to

be removed

)( supplyj

outjjj xxLS (34)


In the case that 2 or more MSAs are overlapped, one have to calculate the composition that will suit the requirements of the plant and compare the costs in order to identify the MSA that will be use in the separation


91

• To calculate cost of recirculation MSA (Cj) and cost of removed pollutant (cj

r) use:



Cost of Make up

MSA ingrecirculat /$ kgCRCMC j (35)

Cost of Regeneration

pollutant removed of kg/$)(

s

jt

j

jrj

xx

Cc (36)




92

• There are cases when there are no process MSAs, therefore a different approach is required in order to construct the pinch diagram



x1

MR

yi

x2


x3

S1

S2

S3

1. Draw the rich composite as before

2. Draw the external MSA as Sj arrows with the tail as the supply composition and the head its target composition

3. Calculate the cj

4. If arrow S2 lies completely to the left of S1 and c2

r < c1r then

eliminate S1



93

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.1.1. Mass Exchangers

x1

MR

yi

x2


x3

S1

S2

S3

5. If arrow S3 lies completely to the left of S2 but c3

r is > c2r then retain

both MSAs6. In order to minimize the

operating cost of the network one should use the cheapest MSA where it is feasible

7. In this case S2 should be used to remove all the rich load to the left and the remaining load is removed by S3

8. Calculate flow rates of S2 and S3 by diving the rich load remove by the composition difference for the MSAs

9. Construct the pinch diagram as shown



94

Example 2A process facility converts scrap tires into fuel via pyrolisis. The discarded tires are fed to a high temperature reactor where heat breaks down the hydrocarbon content of the tires into oils and gaseous fuels. The oils are further processed and separated to yield transportation fuels.

The reactor off gasses are cooled to condense light oils. The condensate is decanted into two layers: organic and aqueous. The organic layer is mixed with the liquid products of the reactor

The aqueous layer is a waste water stream whose organic content must be reduce prior to discharge. The primary pollutant in the waste water is a heavy hydrocarbon. The data for the waste water stream is given in the next slide. A process lean stream is a flare gas (a gaseous stream fed to the flare) which can be used as a process stripping agent. To prevent back propagation of fire from the flare, a seal pot is used.




95



Stream Description Flowrate

Gi

kg/s

Supply Composition

(ppmw)

yis

Target Composition

(ppmw)

yit

R1 Aqueous layer from decanter

0.2 500 50

Table 1



96

Example 2, ContinuationAn aqueous stream is passed though the seal pot to form a buffer zone between the fire and the source of the flare gas. Therefore, the seal pot can be used as a stripping column in which the flare gas strips the organic pollutant off the waste water while the waste water stream constitutes a buffer solution preventing back propagation of fire. Three external MSAs are considered: a solvent extract S2, an adsorbent S3 and a stripping agent S4. The equilibrium data for the jth MSA and the process MSA are given in the next slide, the equilibrium data is given by

yi = mjxj

Where yi and xj are the mass fractions of the organic pollutant in the waste water and the jth MSA, respectively. Use the pinch diagram to determine the minimum operating cost of the MEN




97


Stream Upper Bound on flow rate

Ljc

kg/s

Supply composition

(ppmw)

xsJ

Target Composition

(ppmw)

xJt

mJ JCJ

$/kg

MSA

S1 0.15 200 900 0.5 200 -

S2 300 1000 1.0 100 0.004

S3 10 200 0.8 50 0.030

S4 20 600 0.2 50 0.050

Table 2

Example 2, Continuation



98



Example 2, Continuation

Condenser

Decanter

Separation Finishing

Seal Pot

Flare

Shredded Tires

Reactor Off Gases

Light oil

Waste water R1

Gaseous Fuel

Water

To atmosphere

To waste water

Liquid Fuel

Flare Gas S1


PyrolisisReactor


99



Solution

Condenser

Decanter

Separation Finishing

MEN

Flare

Shredded Tires

Reactor Off Gases

Light oil

Waste water R1

Gaseous Fuel

To atmosphere

To waste water

Liquid Fuel

Flare Gas, S1

S2 S3 S4

PyrolisisReactor



100


Solution, ContinuationCalculate and plot the pinch diagram, using Equations 31,32,33 and Tables 1 and 2

Pinch Diagram

0

50

100

150

200

0 100 200 300 400 500 600



Mass Exchanged10-6

y. ppmw

MR y

R1

0 50

90 500

S1

0 200

105 550

MR y

R1

0 50

90 500

S1

90 200

195 550

S1

R1


101

Pinch Diagram

0

20

40

60

80

100

120

140

0 100 200 300 400 500 600


Pinch Point

Excess Capacity of Process MSA

Integrated Mass

Exchanged

Mass to be Removed by External MSA

Mass Exchanged 10-6

y. ppmw

New S1 Target Composition

Solution, Continuation


102

• From the pinch diagram the load to be removed by the process MSA is 64 x 10-6 kg/s, the excess capacity is 45 x 10-

6 kg/s; we have to use the whole flare gas flow rate to remove pollutant from the waste water, due to the fire hazard that it represents (we cannot by pass part of it directly to the flare, in order to reduce the excess capacity) from a mass balance or the pinch diagram we find the outlet composition of S1 to be: 400 ppmw



• We now have to evaluate the different external MSAs. The load to be removed by external MSA is approximately 31 x 10-6 kg/s, we need to check the thermodynamic feasibility of each external MSA



103

2. Foundation ElementsSolution, Continuation



104

Pinch Diagram

0

50

100

150

200

0 100 200 300 400 500 600


Mass Exchanged 10-6

y. ppmw


S2S3S4

1000300200

48

60020

10


105

• Calculating the costs of each separation agent, using Equation 36:

c2r = 5.714 $/kg

c3r = 157.89 $/kg

c4r = 86.20 $/kg



Analysis: S2 is not a feasible MSA since its target concentration is higher that the target concentration of the rich stream therefore mass transfer is not possible. S4 is the selected MSA, flow is 31x10-6kg/s annual operating cost is 31x10-6x86.2x3600x24x365 = $84,270.5/yr



106

• Process integration is conformed of mass and energy integration


2.3.1.2. Targeting rules

Process

Energy In

Energy Out

Mass In Mass Out

• In order to achieve a good mass integration, one has to set targeting goals; from an overall mass balance:

Depletion Out Mass Generation In Mass (37)



107

• In order to reduce intake of fresh resources and reduce the discharge of waste streams one need to consider recycle, mixing, segregation and/or interception. In order to identify the recycle (direct or after segregation/interception) strategy that will have a net effect on the system the following procedure follows




1

2

53

4

Fresh Load Terminal Load

FLk,1

FLk,2

FLk,1

TLk,1

TLk,2

TLk,3

TLk,4No Recycle


108

• Identify where recycle of streams will have the biggest net effect




1

2

53

4


FLk,1

FLk,2

FLk,1

TLk,1 + Rk,2 – Rk,1

TLk,2 - Rk,2

TLk,3

TLk,4

No Net effect = Poor Recycle

+ Rk,1

1,1,2,2, kkkk RRRR



109




1

2

53

4


FLk,1 – Rk,2

FLk,2 – Rk,1

FLk,1

TLk,1 – Rk,1

TLk,2 – Rk,2

TLk,3

TLk,4

Effective Recycle from Terminal Streams

1,2, kk RR 1,2, kk RR



110




1

2

53

4


FLk,1 – Rk,2

FLk,2 – Rk,1

FLk,1

TLk,1 – Rk,1

TLk,2 – Rk,2

TLk,3

TLk,4

Effective Recycle from Terminal and Intermediate Streams

1,2, kk RR 1,2, kk RR



111

• Recycle of streams must comply with sink constraints; such as composition and flow rate which a sink can take. In order to take advantage of direct recycle opportunities within a plant one has to identify them by using a graphical technique know and the source/sink mapping diagram




Sink

Source

• Effective recycle should connect fresh intake and out streams

Pollutant Composition

Flo

w R

ate

Lo

ad

,

kg

/s

Acceptable Flow Range

Acceptable Composition

Range


112

• The interception of the two constraints is the area where any source within it can be recycled directly to the sink




Sink

Source

• The maximum amount to be recycle is the minimum between the fresh inlet and outlet load. In order to recycle b and c use the mixing arm rule

• Direct recycling does not require new equipment

• Define equipment constraint from, technical data, operation conditions, physical and chemical properties etc

S

Pollutant Composition

Flo

w R

ate

Lo

ad

, k

g/s

a

b

c



113


• Arm rule:



Pollutant

Composition

Flow Rate

Load, kg/s

Arm c Arm b

yb ys yc

Fs

Fb

Fc c

bSources

Resulting Mixture

bc

bbccs

cbs

FF

yFyFy

FFF

• If a fresh source is mixed with a polluted one, in order to minimize the use of fresh one has to minimize fresh arm

(38)

(39)


114


• Note:



1. The previous method can be simplified for a complex plant since no all equipment will required fresh utilities or discharge waste streams. We will identify those that do and apply the previous method

2. Identifying equipment constraints can reduce fresh and waste streams with little process modifications, by working with minimum requirements


115


• The Composition-Interval Diagram (CID)


2.3.1.3. Synthesis of MEN, Algebraic Approach

The pinch diagram is a very useful tool, however it has accuracy limitations common to any graphical method, therefore an algebraic approach that will overcome these limitations is presented

This diagram shows the mass exchanged between the different streams, thermodynamically feasibility and the location of the pinch point

The number of scales is equal to Nsp + 1, where Nsp is the number of lean streams. Each process is represented by a vertical arrow with supply and target compositions as the tail and head respectively. The horizontal lines are the composition intervals whose number is define as:

1)(2intervals SPR NNN (40)



116



117

• Within each interval it is possible to transfer mass from the rich stream to the lean stream and it is possible to transfer mass from the interval to any MSA that is in an interval below it



Table of Exchangeable Loads (TEL)

• The TEL is used to determine the load of mass exchanged within each interval; for the waste stream the load is:

Wi,kR = Gi(yk-1 – yk) (41)

And the exchangeable load for the lean streams is:

Wj,kS = Lj

c(xj,k-1 – xj,k) (42)


118


• Since one or more streams will pass through one or more intervals we can express the total load of the stream that passes through that interval k; for the waste and lean streams we have



Skjkj

Sk

Rkiki

Rk

WW

WW

, interval through passes

, interval through passes

(44)

(43)



119


• Note that mass can be transferred within each interval from a waste stream to a lean stream, as a result it is possible to transfer mass from a waste stream in a interval to a lean stream in a lower interval, the resulting mass balance is:



interval theleaving and

enteringpollutant of mass residual theare ,1

1

kth

WW

kk

kSkk

Rk

(45)



120


• The graphical representation is:

k



KSkW

RkW

1k

Waste Recovered from Waste Streams

Mass Transferred to

MSA’s

Residual Mass from Preceding Interval

Residual Mass to Next Interval


121

Note:• Initial residual mass for k = 0 is zero• The most negative value of the residual mass load

indicates the excess capacity of MSA’s, in order to reduce it, one can either reduce the flow rate, or the composition of the MSA’s, one this is done one needs to recalculate and apply the previous procedure. The pinch will be represented at the location when the residual mass is zero. This result will be equal to the one given by the pinch diagram

• After reducing flow rate or concentration, the remaining load is the load to be removed but external MSA’s




122

Example 3

3

1)11(2

Intervals

Intervals

N

N



A lean MSA will be used to reduce the composition of a rich stream, the data is give in the table

•Calculate the number of intervals

•Calculate the compositions of each stream for the y and x scales

•Prepare de CID diagram•Calculate a TEL table, using 41, 42•Calculate the cascade diagram, by 43,44


123



Composition Table


124



CID Table


125



TEL Table


126

Cascade Diagram



127

• The excess capacity of the MSA is 0.000027 kg/s of pollutant and the actual flow required for the separation is:

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.1.3. Synthesis of MEN, Algebraic Approach

111.00002.00009.0

0.00002715.0

Capacity Excess

Flow Actual

tFlow Actual

L

xxLL

si

(45)


128

• Recalculating the TEL and cascade diagram



Pinch


129

•The concentrations at which the pinch point is located are:

y = 0.00011x = 0.0002

The quantity leaving the bottom of the cascade diagram is the amount to be removed by external MSA’s, 0.00001 kg/s




130

• In order to minimize the number of mass exchangers to obtain a MOC solution, we will decompose the design problem in to two sub-problems one above and one below the pinch


2.3.1.4. Synthesis of MEN, with Minimum Number of Exchangers

pinch below ,pinch below ,pinch below ,pinch below ,

pinch above ,pinch above ,pinch above ,pinch above ,

pinch below ,pinch above ,

iSRMOC

iSRMOC

MOCMOCMOC

NNNU

NNNU

UUU

(46)



131

• By starting the synthesis of mass exchangers at the pinch point one can ensure that the options will not be compromised at later steps, due to the fact that the pinch point the all streams match at the minimum driving force . The matching of streams will be done in two sections, above and below the pinch, two criteria must be applied to ensure feasibility


2.3.1.5. Feasibility Criteria

RBelowLBelow

AboveLRAbove

NN

NN

(47)

(48)

Stream Population



132

• If the previous inequalities do not hold with the rich and lean streams/branches then splitting of one or more of them is required, as before stream splitting might be required to comply with the following inequalities


2.3.1.5. Feasibility Criteria

Pinch Below

Pinch Above

ij

j

ij

j

Gm

L

Gm

L

(48)

(49)

Thermodynamic Feasibility



133

• The following example will illustrate the procedure for network synthesis; given a process with two waste streams and two process MSA’s


2.3.1.6. Network Synthesis

Example 4


134

• The composition for waste and lean streams are shown in the table

• Number of Intervals = 7

• Calculate the CID

• Calculate TEL

• Revise TEL




135

• CID




136



• TEL


137

2. Foundation Elements• Cascade Diagram


138

• The excess load of the MSA’s is 0.00151kg/s; using Equation 45 and reducing the excess capacity of S2 we have an actual flow of 2.925 kg/s and a revise TEL and cascade diagram can be calculated, with its pinch point at interval 4 and compositions y, x1, x2 = 0.0165, 0.00725, 0.01, respectively




139

2. Foundation Elements• TEL, revised


140

• We will now define the number of mass exchangers• Define feasibility criteria• Match streams



2112

3122

pinch below ,

pinch above ,

MOC

MOC

U

U


141

2. Foundation Elements• Cascade Diagram, revised Pinch Point


142

• The following figure will aid during checking of the feasibility criteria



Pinch Point

R1

R2

S1

S2

G1 = 2.5 kg/s G2 = 1 kg/s L1/m1 = 2.5 kg/s L2/m2 = 1.95 kg/s


143



22

AboveLRAbove NN

Pinch Above ij

j Gm

L

Match:

R1 – S1

R2 – S2


144

Mass Exchanged Loads

R1 = 0.08375 kg/sS1 = 0.03875 kg/sMass exchanged = 0.03875 kg/s

R2 = 0.0135 kg/sS2 = 0.0585 kg/sMass exchanged = 0.0135 kg/s




145

•Remaining load from R1 = 0.045 kg/s•Excess capacity of S2 = 0.045 kg/s

Note that these values are equal, due to the fact that there is no mass transferred trough the pinch. Now we proceed to match exchangers represented by circles with streams; the mass exchanged appears within the circles and composition in arrows. Load to be removed by external MSA is 0.0155kg/s




146

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.1.6. Network Synthesis

R2 S1

S2R1

0.03875 0.03875

0.045

2.5 kg/s 0.05

0.0165

0.0135 0.0135

5 kg/s 0.015

0.00725

0.0451 kg/s 0.03

0.0165 0.01

R2 transfers all its load S1 is depleted

S2 can remove

load

R1 capacity not removed

by S1

x2 **x1 *


147

• In order to calculate the intermediate compositions leaving exchanger R2 – S2 use a material balance using Equation 37:

x2 ** = 0.01 + 0.0135/3 = 0.0145

x1* = 0.05 - 0.045/2.5 = 0.032




148

21 RBelowLBelow NN

•After completing the network design above the pinch we will proceed to do the same below the pinch


Pinch Point

R1 R2 S1 S3 External MSA

G1 = 2.5 kg/s G2 = 1 kg/s L1/m1 = 2.5 kg/s


149

Checking feasibility (Eq. 49) determines that S1 has to be split in two since L1/m > Gi. There are many different combinations in order to achieve it, for this case we will split them arbitrarily and match the streams



Pinch Point

R1 R2S1 S3 External MSA

G1 = 2.5 kg/s G2 = 1 kg/s

L1= 5 kg/sL

12/m

1 =

0.7

25 k

g/s

L1

1/m

1 =

1.7

75 k

g/s


150

Pinch Below ij

j Gm

L

Match:

R1 – S11

R2 – S12



• Mass Exchanged Loads

• R1 = 0.01625 kg/s• S11 = 0.0079875 kg/s• Mass exchanged = 0.0079875 kg/s

• R2 = 0.0105 kg/s• S12 = 0.0032625 kg/s• Mass exchanged = 0.0032625 kg/s


151


•Remaining load from R1 = 0.0082625 kg/s

•Remaining load from R2 = 0.0072375 kg/s

•In order to remove the remaining load from waste streams it is required to use external MSA’s (S3)


152

Pinch Point

G1 = 2.5 kg/s G2 = 1 kg/s

L1= 5 kg/s

2. Foundation ElementsR1 R2

S1S3 External MSA

0.0079875 0.079875

0.0032625 0.0032625

0.0072375

0.00826250.0082625

0.0072375

Calculate the Intermediate Compositions

Can you Suggest another

Configurationfor S3?


153

R2

S1

S2R1

0.03875 0.03875

0.0452.5 kg/s 0.05

0.0165

0.0135 0.0135

5 kg/s 0.015

0.00725

0.045

1 kg/s 0.03

0.01650.01

x2 **

x1 *

L1= 5 kg/s

0.0079875 0.079875

0.0032625 0.0032625

0.0072375

0.00826250.0082625

0.0072375

Pinch Point

Complete Network

S3



Heat Exchange Networks


2.3.2. Heat Integration


155

• Every plant requires energy to be transfer from a hot stream to a cold one; hence the importance a proper heat exchange network in order to have a positive impact in the economics and operation of any process


Heat Exchange Network


Cold Streams In

Cold Streams Out

Hot Streams In

Hot Streams Out


156

• To define the HEN (Heat Exchange Network) problem first we need to define the following:

A number of hot process streams that need to be cooled NH and a number of cold process streams that need to be heated NC, we need to synthesize a network that will achieve the transfer of heat at minimum cost

For hot streams the heat capacity can be expressed as:


tu

su

uP

T

T

FC

eTemperaturTarget

eTemperaturSupply

Capacity Heat ,

(50)

For u = 1,2,…NH



157

In addition for the cold streams we have:


tv

sv

vPf

t eTemperaturTarget

t eTemperaturSupply

c Capacity Heat ,

(51)

For v = 1,2,…NC

A number of cold and hot streams is available whose supply and target temperatures are known but not their flow rates. In order to design a HEN the following questions need to be answered:



158

Which heating/cooling utilities should be used

What is the optimal heat load to be

removed/added by each utility?

What is the

Optimal configurationHow should the hot and

Cold streams be matched?





159

• In order to have heat transfer between two streams the following relationship will established a correspondence between the hot and cold streams temperature:


minTtT (52)



160

• A special case of mass exchanged is the one that compares the heat exchanged problem corresponding T, t, Tmin with yi,xj and j respectively, and having mj, bj equal to zero

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.2. Heat Integration


161


HE

T

NOTE:The order of X and Y axis used here are different from what has been commonly used in the literature. The reason is that there is a strong interactions between mass and energy making the enthalpy expression non linear function of temperature therefore it is easier to have enthalpy in function of temperature, this specially important when combining mass and heat integration

T

H

HE vs. T Approach

T v. H Approach

T min




162

• The procedure use to set up the pinch diagram is exactly the same as the one use for mass integration, by placing the hot and cold streams temperatures in the diagram, starting by their supply temperature as the tail of an arrow and the target temperature as the head of an arrow. The following equation can be used to calculate the vertical distance or heat loss by the hot stream



)(,tu

suuPuu TTCFHH (53)



163


• And for the heat gained by the cold stream we have:

• To construct the pinch diagram we have:

)(,sv

tvvPvv ttcfHC (53)




164


T1t T2

t T1s T2

s T

HE

HH2

HH1

H1

H2

HE

HC2

HC1

C1

C2

t1t t2

t t1s t2

s T

t = T - Tmin




165




Heat Exchanged

Hot Composite Stream

Cold Composite Stream

Thermal Pinch Point

Integrated Heat Exchange



T

t = T - Tmin

Minimum Heating Utility

Minimum Cooling Utility


166

The analysis of the thermal pinch diagram is as follows:

• The cold composite curve cannot be slid down any further otherwise there will not be thermal feasibility, if the cold composite is moved up less heat integration is possible therefore more utilities are required

• Above the pinch there is a surplus of cooling and below the pinch there is a surplus of heating utilities




167

• A similar analysis as the one used for mass integration can be done in order to apply an algebraic cascade diagram, the number z of intervals is:



2)(2int CH NNN (54)

• To construct a Table of Exchangeable Heat Load TEHL we need:

)(

)(

1,,

1,,

zzvpzv

zzuPuzu

ttfcHC

TTCFHH

(55)

(56)



168

• The collective total load for the hot and cold process streams are:


zvNv

vTotalz

zuNu

uTotalz

HCHC

HHHH

C

H

,,...2,1 wherez,

interval through passes

,,...2,1 wherez,

interval through passes

(57)

(58)



169

• As it was mentioned for mass exchanged, it is feasible to transfer heat from a hot process stream to a cold one within each temperature interval, a heat balance around a temperature interval yields:



Z

TotalzHC

TotalzHH

1zr

Heat Added by Process Hot

Stream

Residual Heat from Preceding Interval

Residual Heat to Next Interval zr

Heat Removed by Process Cold

Stream


170


• The resulting heat balance is:


2.3.2. Synthesis of MEN, Algebraic Approach

1 zTotalz

Totalzz rHCHHr

(59)



171

• The resulting TID is:




172

Property Integration:“Functionality based holistic approach to the

allocation and manipulation of streams and processing units which is based on tracking, adjustment, assignment and

matching of functionalities throughout the process”

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.3. Property Integration

Source : Property Integration: Componentless Design Technique and Visualization Tools


173

• Component mass balances are an integral part of process design. There are several design problems in which the designer is interested in a group of properties such as viscosity, corrosion, density etc. Solvent selection is a clear example in which one is interested in its volatility, viscosity, equilibrium distribution, instead of its chemical constituents.

2. Foundation Elements2.3. Overview of Mass, Energy and Property Integration2.3.3. Property Integration

Source : Component less design of recovery and allocation systems: a functionality based approach


174

• Property visualization tools are limited to 3 properties, an algebraic approach is used to deal with more complex cases. The advantage of visualization tools is based on the insides that give of the process, and how the design problem can be addressed. In order to apply this method to a set of properties we need to introduced the concept of cluster

• Properties are not conserved, as a result they cannot be tracked among units without using mass balances, the problem is that often is not possible to identify every single chemical species e.g. Gasoline, Dowtherm


2.3.3. Property Integration


175

Cluster

“Defines as condensed surrogate properties which can be used to characterize the

complex mixture and can be tracked my mapping the raw properties of infinite

compounds onto finite domains”





176

• The problem statement is: given a number of process streams Ns which contain the chemical species of interest, can be used in a number of sinks Nsinks (process units) in order to optimize a a desired objective e.g. minimize usage of fresh resources, maximize use of process resources, minimize cost of external streams etc. Each sink has a set of constraints defined as:



maxsinmin

maxsin,min

RateFlowrateFlowRateFlow

propertypproperty

k

ki



177

• Each stream can be characterized by Nc raw properties with a mixing rule that characterized a given stream



sisii

ths

siisNsii

pofOperatorp

rateflowtotaltheto

streamstheofoncontributiFractionalx

pxp s

,,

,1

)(

)()(


(60)


178

• pi,s can be normalized as:



refi

siisi

p

)( ,

,

• An augmented property index (AUP) for each stream s, is define as the summation of the dimensionless raw property operators:

s

siNis

Ns

AUP c

,...,2,1,1

(61)

(62)


179

• Ci,s is the cluster for property i in stream s



s

sisi AUP

C ,,

• For any stream s, the sum of clusters must be conserved adding up to a constant e.g. unity

s

sNi

Ns

Cc

,...,2,1

11

(63)

(64)

c

sisNsi

Ns

CC s

,...,2,1,1

(65)


180



• The framework for allocation and interception for property integration is:

Property Integration

Network (PIN)

u = 1

u = 2

u = Nsinks

.

.

.

.

.

.

Processed Sources (back to process)

s =1

s =1

Sources Segregated Sources

Sinks


181



• Consider a cluster of stream s to unit u, with three targeted properties i, j, k we have:

sj

sk

sj

sisksjsi

sjsj

si

sk

si

sjsksjsi

sisi

C

C

,

,

,

,,,,

,,

,

,

,

,,,,

,,

1

1

1

1

(66)

(67)


182



1

1

,

,

,

,,,,

,,

sk

sj

sk

sisksjsi

siskC

• In order to obtained an overestimation of the feasibility region we have:

max

min

max

minmax

,

,

,

,

,

1

1

si

sk

si

sj

siC

(68)

(69)


183



min

min

min

minmin

,

,

,

,

,

1

1

si

sk

si

sj

siC

max

min

max

minmax

,

,

,

,

,

1

1

sj

sk

sj

si

sjC

min

min

min

minmin

,

,

,

,

,

1

1

sj

sk

sj

si

sjC

1

1

max

min

max

minmax

,

,

,

,

,

sk

sk

sk

si

skC

(70)

(73)(71)

(72)

• In order to allocate, mix or intercept streams one needs to identify a feasibility region for the sinks, by using the following relationships:


184



1

1

min

min

min

minmin

,

,

,

,

,

sk

sj

sk

si

skC (74)

• These points will now need to be plotted in a ternary diagram will be shown next



185




186


2.3.3. Property Integration Ci

CkCj

Ci,smax

Cj,smin

Cj,smax

Ck,smin Ck,s

max

Cj,smin

Overestimated Region

We need to find the true estimation of the feasibility region (for a more detailed explanation of how to obtained these results, review the references at the end of the module)


187


2.3.3. Property Integration Ci

CkCj

True Region

),,( min,

min,

max, sksjsi

),,( min,

max,

max, sksjsi

),,( min,

max,

min, sksjsi

),,( max,

min,

max, sksjsi

),,( max,

min,

min, sksjsi

),,( max,

max,

min, sksjsi


188


2.3.3. Property Integration• In order to plot these diagrams in a spread sheet, we need to related this

ternary coordinates in a X vs. Y plane as follows:

Ci

Ck

Cj

Y

X

(0.866, 0.50)

(1, 0)

(0, 0)

Ys

Xs

S

Ci,s

siC ,)3

(cos


189



sksjsi

sksisisjsisjs

sksjsi

sisisis

CCCCX

CCY

,.,

,,,,,,

,.,

,,,

5.05.01)

3(cos1

866.0866.0)

3(sin

(75)

• The equations that relate X vs. Y with ternary coordinates are:

(76)



190



•The next step is to set up optimization rules as follows:

Relating cost to fractional contribution of sources

Consider two sources s and s+1 that are mixed to satisfy sinks constraints, let xs and xs+1 denote the fraction contribution of sources s and s+1 to the total flow rate of the mixture. Let s be more expensive than s+1, as Costs>Costs+1, therefore we have:

Costmixture = xs (Costs – Costs+1) + Costs+1

From the previous equation we can conclude that in order to minimize the cost of the mixture xs must be minimized

(77)



191



Rule No. 1

“When two sources (s and s+1) are mixed to satisfy the property constraints of a sink with source s being more expensive than s+1, minimizing Costmixture is achieved by selecting the minimum feasible value of xs”



192



Derivation of relationships between minimum cluster arms (s) and minimum fractional contribution xs

xs cannot be visualized in a ternary diagram, the lever arm on the ternary cluster diagram represents another quantity defined as s, to relate both quantities the AUP is described by equation 62


ssNs

sss

AUPxAUP

AUP

AUPx

s1

(78)

(79)


193



Rearranging we have:


21

1111

])1([

][])1([

ssss

sssssssss

s

s

AUPAUP

AUPAUPAUPAUPAUPAUP

d

dx

Taking the first derivative:

ssss

sss AUPAUP

AUPx

)1(1

1

(80)

(81)


194



Rearranging and simplifying:


21

1

])1([ ssss

ss

s

s

AUPAUP

AUPAUP

d

dx

From the previous development rule 2 is obtained:

(82)


195



Rule No. 2

“On a ternary cluster diagram, minimization of the cluster arm of a source corresponds to minimization of the flow contribution of that source; minimum s corresponds to minimum xs”



196




•Consider the case of fresh external source F, the objective is to minimize its use. A process internal stream W that can be recycled or reused to reduce the use of F. It is desired to mixed them in order to obtain a minimum cost mixture that satisfy sink constraints, the feed to the sink is subject to a number of property constraints that can be mapped in a cluster diagram as follows


197




Ci

CkCj

W

F

Sink

Optimum F

c

b

a

Minimum distance, this is a necessary condition only. For sufficiency AUP and flow rate must be matched as well

Multiple mixtures


198




Ci

CkCj

W1

F

SinkF

For multiple sources the line connecting W1 and W2 represents the possible mixtures, the optimal mixing point is the one that gives the minimum s

Multiple mixtures

Multiple sources case:

W2


199




Ci

CkCj

W

F

Sink

When the process stream W target cannot be met, the stream can be adjusted via an interception device e.g. separation, reaction etc Adjusting properties

will change the cluster value

Adjusting properties

Wintercepted


200




• For a selected mixing point and a desired s, the fresh arm can be drawn to determine the desired location of the desired location of Wintercepted. Moreover, since the values of AUP are known for F and the mixing point of the sink, one can plug the targeted value of xF into Equation 78 to calculate the desired value of AUP for Wintercepted. Once Wintecepted and AUP are known, we can solve the cluster equations backwards to calculate the raw properties of Wintercepted


201




• This is the minimum extent of interception to achieve maximum recycle of W or minimum usage of F since the additional interception will still lead to the same target or minimum usage but will result in a mixing point inside the sink and not just on the surrounding of the sink

• Once the task for interception is define, conventional process synthesis techniques can be apply to develop the design and operating parameters for the interception system. The same procedure can be repeated for multiple mixing points resulting in the task identification of the locus for minimum extend to interception


202




Ci

CkCj

W

F

Sink

Locus for minimum extent of interception

Locus Identification


203



Multiplicity of Optimal values of AUP

A cluster point made of C1sink, C2

sink, C3sink can correspond to multiple combinations

of properties that can give the same cluster values. As a result one can have nMultiple, points within the feasible property domain giving a single cluster value. Three conditions must be satisfied in order to insure feasibility of the sources or mixture of sources going into a sink:

1. The cluster value of the source must be contain within the feasibility region of the sink on the cluster diagram2. The values of AUP for the source and the sink must match3. The flow rate of the source must lie within the acceptable feed flow rate range of the sink



204



From Rule No. 1 minimizing xs will minimize CostMixture, therefore we need to select an AUPm (given for the feasible properties p1,m, p2,m, p3,m) that will be minimized by the following relationship between AUPm and xs.


1

1

1

1

ssm

s

ss

sms

AUPAUPAUP

x

therefore

AUPAUP

AUPAUPx (83)

(84)


205



To minimize xs and as a result the cost we should select:


1

1

max

min

ssmoptimumm

ssmoptimumm

AUPAUPifAUPArgAUP

AUPAUPifAUPArgAUP

If no mixture matches the AUP selected for the sink for the case given by Equation 84 then one has to decrease the value of the sink’s AUP starting with Argmax AUPm till getting the highest value of AUPm within the feasible range of AUP which matches that of the mixture; same procedure is used for Equation 85, by increasing the value of sink’s AUP starting with Argmin AUPm till getting the highest value contained within the feasible range of the sink which matches that of the mixture

(85)

(86)


206




Currently research is being undertaken to design tools that will cover cases for 1, 2 and more than three properties. This is a very dynamic and changing field of research


207

TIER II


208

• A tire to fuel processing plant flow sheet is shown in the next slide which is a more complete description for the one given in Example 2. Tire shredding is achieved by using high pressure water jets. The shredded tires are fed to the process while the spent water is filtered. The wet cake collected from the filtration system is forwarded to solid waste handling.

• The filtrate is mixed with 0.20 kg/s of fresh water makeup to compensate for water losses with the wet cake, 0.08 kg water/s and the shredded tires 0.12 kg water/s. The mixture of filtrate and water make up is fed to a high pressure compression station for recycling the shredding unit. Due to the pyrolisis reactions, 0.08kg water is generated

3. Case Study3.1. Tire to Fuel Processing Plant


209

• The plant has two primary sources for waste water, the decanter (0.20 kg water/s and the seal pot 0.15 kg/s. The plant has been shipping the waste water for off-site treatment. The cost of wastewater transportation and treatment is $0.02/kg leading to a wastewater treatment cost of approximately $129,000/yr



210

3. Case Study

Filtration

Compression

Water JetShredding

PyrolisisReactor Separation Finishing

Condenser

Decanter

Flare

SealPot

Fresh Water 0.20 kg/s 0 ppmw

Wet Cake to Solid Handling 0.08 kg/s, 0 ppmw

Tires

Shredded Tires

Reactor Off-Gases

Gaseous Fuel

Waste water to treatment 0.20 kg/s

500 ppmw

Fresh water

0.15 kg/s 0 ppmw

Light Oil Flare Gas, 0.15 kg/s

200 ppmw

Waste water to treatment, 0.15 kg/s 0 ppmw

Liquid Fuel

To Atmosphere

Tire to Fuel Plant Flow Sheet


211

• The plant wishes to stop off site treatment of wastewater to avoid the cost ($129,000/yr) and alleviate legal liability concerns in case of transportation accident or inadequate treatment of wastewater treatment. For capital budget authorization, the plant has the following economic criteria:

3. Case Study

years 4Savings Annual

investment capital Fixed periodPayback

system site-on

treatmentsite-off

cost operating Annual

- cost avoided Annual Savings Annual

3.1. Tire to Fuel Processing Plant


212

Economic Data• Fixed cost of extraction system associated with S2. $ = 130,000

(flow rate of wastewater, kg/s)0.60

• Fixed cost of adsorption system associated with S3, $ = 800,000 (flow rate of wastewater, kg/s)0.72

• Fixed cost of stripping system associated with S4, $ = 280,000 (flow rate of wastewater, kg/s)0.66

• A biotreatment facility that can handle 0.35kg/s waste water has a fixed cost of $260,000 and an annual operating cost of $72,000/yr

Technical Data• Water may be recycle to two sinks: the seal pot and the water-jet



213

Compression station. The following constrains on flow rate and composition of the pollutant (heavy organic) should be satisfied:

Seal Pot• 0.10 Flow rate of feed water (kg/s) 0.20• 0 Pollutant content of feed water (ppmw) 500

Make up to water-jet compression station• 0.18 Flow rate of make up water (kg/s) 0.20• 0 Pollutant content of make up water (ppmw) 50



214

Solution

We will start with an overall mass balance, note that 0.12 kg/s of water are lost in the process and cannot be re used

3. Case Study

Water Generation 0.08kg/s

0.2 kg/s to Compression Station

0.15 kg/s to Seal Pot

0.08 kg/s from Wet Cake

0.15 kg/s from Seal Pot

0.2 kg/s from Decanter



215

Solution

From the overall mass balance we can set the targets for fresh use and wastewater production

3. Case Study

Water Generation 0.08kg/s

0.2 kg/s

0.15 kg/s

0.08 kg/s

0.35 kg/s

No Fresh Water

0.08 kg/s

Wastewater

The source diagram is shown in the next slide



216

Source/Sink Diagram

00.020.040.060.080.1

0.120.140.160.180.2

0.22

0 50 100 150 200 250 300 350 400 450 500 550

ppmw

kg

/s

3. Case Study

Seal Pot

WW from Decanter

Compression Station

WW from Seal Pot

WW from Wet Cake


217

• From the source/sink diagram we can see that wastewater from the decanter can be accepted by the seal pot only; the outlet composition of the wastewater coming from the seal pot is 400 ppmw (from the pinch diagram) as shown in Example 2



218

Pinch Diagram

0

50

100

150

200

0 100 200 300 400 500 600

Mass Exchanged 10-6

3. Case Study

Composition from Seal Pot



219

Source/Sink Diagram

00.020.040.060.080.1

0.120.140.160.180.2

0.22

0 50 100 150 200 250 300 350 400 450 500 550

ppmw

kg

/s

Seal Pot

W W from Decanter

Compression Station

W W from Seal Pot

W W from Wet Cake

3. Case Study


220

• Wastewater coming from the seal pot cannot be recycled directly to the compression station due to its high pollutant composition, therefore it is required to treat it using an external MSA as shown in Example 2; for this case S4 is the best stripping agent, which will bring down the composition to 50ppmw



221

Source/Sink Diagram

00.020.040.060.080.1

0.120.140.160.180.2

0.22

0 50 100 150 200 250 300 350 400 450 500 550

ppmw

kg

/s

3. Case Study

Seal Pot

WW from Decanter

Compression Station

WW from Stripper

WW from Wet Cake

WW from Seal Pot



222

3. Case Study

Filtration

Compression

Water JetShredding

PyrolisisReactor Separation Finishing

Condenser

Decanter

Flare

SealPot

Wet Cake to Solid Handling

Tires

Shredded Tires

Reactor Off-Gases

Gaseous Fuel

Light Oil

Flare Gas

Liquid Fuel

To Atmosphere

Tire to Fuel Plant Flow Sheet (Revised)

Stripper


223

• Now we will proceed to compare the different alternatives in order to make a decision. For the bio-treatment plant we have:

3. Case Study

Annualized Saving Cost = $129,000/yr - $72,000/yr = $57,000/yr

Pay Back = $260,000 / $57,000/yr = 4.56 years

• For the recycling/stripping system:

Annualized Saving Cost = $129,000/yr - $84,270.5/yr = $44,729.5/yr

Pay Back = $96,791.6 / $44,729.5/yr = 2.16 years

Fixed Cost of Stripping = $280,000(0.2)0.66 = $96,791.6



224

• From the results we can conclude that the recycling/stripping alternative is the best economical and technical option. We need to point out that the water contained in the wet cake will not be recovered or treated



225

3. Case Study3.2. Pulp and Paper Process Plant

• Wood chips are chemically cooked in a Kraft digester using a white liquor (mainly NaOH and Na2S). Black liquor (spent white liquor) is converted back to white liquor by a recovery cycle. The digested pulp is then bleached to obtain bleached pulp (fiber I). The plant also buys pulp from another plant (fiber II), the pulp is then sent to two different paper machines (Sink I and Sink II). Paper machine I uses 200 tons/hr of fiber I. A mix of fiber I and II (20 ton/hr and 30 ton/hr, respectively) is fed to paper machine II. Due to interruptions and other disturbances, a certain amount of partly and completely manufactured paper is rejected


226

3. Case Study• The rejected fiber is referred as broke, which is passed

through a hydro-pulper and a hydro-sieve resulting in two streams, an underflow which is burnt and an overflow which goes to waste treatment. Part of the broke contains fiber which can be recycle for paper making.

• The properties that are important for the process are:– Objectionable material (OM), undesirable material in the fiber – Reflectivity (R), reflectance of an infinite thick material

compared to a standard– Absorption coefficient (k), measure of absorptivity of light into

the fibers

3.2. Pulp and Paper Process Plant


227

3. Case Study

• The mixing rules are:

61

6

2

1

2

1

s

s

s

s

RxR

g

mkx

g

mk

OMxOM

sNs

ssNs

ssNs



228

3. Case Study

Kraft Digester

Chemical Recovery

Cycle

Bleaching Paper Machine I

PaperMachine II

Hydro-Pulper

Hydro-Sieve

Fiber II

Fiber I

Reject

Reject

Paper II

Paper I

Underflow

Broke (Overflow)

Pulp

Black Liquor

White Liquor

Wood Chips

OM =0.085

k = 0.0013

R = 0.95

OM =0.0

k = 0.0012

R = 0.85

OM =0.0

k = 0.00065

R = 0.95

20 t/hr

30 t/hr

200 t/hr


229


Property Lower Bound Upper BoundOM (mass fraction) 0 0.03

k (m2 / gm) 0.00115 0.00125R 0.85 0.95

Flowrate (ton/hr) 95 100

Property Lower Bound Upper BoundOM (mass fraction) 0 0

k (m2 / gm) 0.0007 0.00125R 0.9 0.95

Flowrate (ton/hr) 45 45

Constraints for Paper Machine I, (Sink I)

Constraints for Paper Machine II, (Sink II)


230


Source

OM

(mass

fraction)

k (m2 / gm) R

Maximum Available Flowrate (ton/hr)

Cost

($/ton)Broke 0.085 0.0013 0.95 35 0Fiber I 0 0.0012 0.85 230Fiber II 0 0.00065 0.95 395

Properties of Fiber Sources


231

1. Determine the optimal allocation of the three sources, fiber I, II and broke for a direct recycle reuse without new equipment

2. In order to maximize use of process resources and minimize wasteful discharge (broke) how should the designer change the properties of the broke as to achieve maximum recycle?



232

SolutionIn order to translate the data from property domain to cluster domain we will select arbitrarily reference values as:


0.1

/001.0

02.02

R

gmmk

OMref

ref


233

We will proceed to calculate the cluster values for the sources as follows:


38.173.065.00

577.1377.02.10

28.673.03.125.4

II

I

RkOMFiber

RkOMFiber

RkOMBroke

AUP

AUP

AUP

Source OM

k

R

Broke 0.085/0.02 0.0013/0.001 0.956/16

Fiber I 0 0.0012/0.001 0.856/16

Fiber II 0 0.00065/0.001 0.956/16


234


12.028.6

735.0

21.028.6

3.1

677.028.6

25.4

,

,

,

BrokeR

Brokek

BrokeOM

C

C

C

Similarly for Fiber I and II we obtain:


235


239.0577.1

377.0

761.0577.1

2.1

0577.1

0

,

,

,

IFiberR

IFiberk

IFiberOM

C

C

C

533.038.1

735.0

471.038.1

65.0

038.1

0

,

,

,

IIFiberR

IIFiberk

IIFiberOM

C

C

C

Now we can proceed to transform the ternary points to X vs. Y plot


236


452.05.01

586.0866.0

,,

,

BrokeOMBrokekBroke

BrokeOMBroke

CCX

CY

239.05.01

0866.0

I ,I ,I

I ,I

FiberOMFiberkFiber

FiberOMFiber

CCX

CY

530.05.01

0866.0

II ,II ,II

II ,II

FiberOMFiberkFiber

FiberOMFiber

CCX

CY


237

Ternary / X-Y Diagram

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

Y

Broke

Fiber I Fiber II

COM

Ck

CR


238

• Now we need to proceed to locate sinks in the diagram by using the point illustrated in slide 187


),,(),,(

),,(),,(

),,(),,(

minmaxminminmaxmin

minmaxmaxminmaxmax

minminmaxminminmax

,I ,I ,,,,

,I ,I ,,,,

,I ,I ,,,,

ISinkRSinkkSinkOMsksjsi



For Sink I:


239


229.0677.1/377.0

681.0677.1/15.1

09.0677.1/15.0

677.1

377.00.1/85.0

15.1001.0/00115.0

15.02.0/03.0

max ,

min ,

max ,

66min

min

max

,

I ,

I ,

ISinkOM

ISinkk

ISinkOM

C

C

C

AUPISinkR

Sinkk

SinkOM


240


213.0777.1/377.0

703.0777.1/25.1

084.0777.1/15.0

777.1

377.00.1/85.0

25.1001.0/00125.0

15.02.0/03.0

max ,

min ,

max ,

66min

max

max

,

I ,

I ,

ISinkOM

ISinkk

ISinkOM

C

C

C

AUPISinkR

Sinkk

SinkOM


241


23.0627.1/377.0

77.0627.1/25.1

0627.1/0

627.1

377.00.1/85.0

25.1001.0/00125.0

02.0/0

max ,

min ,

max ,

66min

max

min

,

I ,

I ,

ISinkOM

ISinkk

ISinkOM

C

C

C

AUPISinkR

Sinkk

SinkOM


242


),,(),,(

),,(),,(

),,(),,(

maxmaxminmaxmaxmin

maxminminmaxminmin

maxminmaxmaxminmax

,I ,I ,,,,

,I ,I ,,,,

,I ,I ,,,,




For Sink I, continuation:


243


361.0035.2/735.0

565.0035.2/15.1

074.0035.2/15.0

035.2

735.01/95.0

15.1001.0/00115.0

15.02.0/03.0

max ,

min ,

max ,

66max

min

max

,

I ,

I ,

ISinkR

ISinkk

ISinkOM

C

C

C

AUPISinkR

Sinkk

SinkOM


244


39.0886.1/736.0

61.0886.1/15.1

0886.1/0

886.1

736.01/95.0

15.1001.0/00115.0

02.0/0

min ,

min ,

min ,

66max

min

min

,

I ,

I ,

ISinkR

ISinkk

ISinkOM

C

C

C

AUPISinkR

Sinkk

SinkOM


245


37.0986.1/736.0

63.0986.1/25.1

0986.1/0

986.1

736.01/95.0

25.1001.0/00125.0

02.0/0

min ,

min ,

min ,

66max

max

min

,

I ,

I ,

ISinkR

ISinkk

ISinkOM

C

C

C

AUPISinkR

Sinkk

SinkOM


246

Sink I and Sources

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

Y

Broke

Fiber I

Fiber II

COM

CRCk

COM Ck Xsink I Ysink I

0 01 0

0.5 0.8660.677 0.210 0.452 0.5860.000 0.761 0.239 0.0000.000 0.471 0.529 0.0000.090 0.681 0.274 0.0780.084 0.703 0.255 0.0730.000 0.770 0.230 0.0000.074 0.565 0.398 0.0640.000 0.610 0.390 0.0000.000 0.630 0.370 0.000

Sink I


247

• Similarly for Sink II we have:


Sink II Low High RefOM 0 0 0.02k 0.0007 0.001 0.001R 0.9 0.95 1F 45 45

COM Ck Xsink I Ysink I

0 01 0

0.5 0.8660.677 0.210 0.452 0.5860.000 0.761 0.239 0.0000.000 0.471 0.529 0.0000.090 0.681 0.274 0.0780.084 0.703 0.255 0.0730.000 0.770 0.230 0.0000.074 0.565 0.398 0.0640.000 0.610 0.390 0.0000.000 0.630 0.370 0.0000.000 0.568 0.432 0.0000.000 0.702 0.298 0.0000.000 0.702 0.298 0.0000.000 0.488 0.512 0.0000.000 0.488 0.512 0.0000.000 0.630 0.370 0.000

OM

Min OM

Max k

Min k

Max R

Min R

Max

0 0 0.7 1.25 0.531441 0.735092


248

Sink I - II and Sources

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

Y

Broke

Fiber I Fiber II

Sink II

COM

CRCk

Sink I


249

Now we proceed to identify the minimum distance for Sink I, that will minimize the use of fresh sources


Sink I and Sources

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

Y

Broke

Fiber I

Fiber II

COM

CRCk

Sink I


250

Sink I - II and Sources

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

X

Y

(0.27, 0.85)

COM

CkCR

In order to get the length of the arm to obtain s one can measure it from the graph or:

221

221 )()( yyxxd

or

By Equation 65


251

The distance between mixture and broke is:


533.0

)586.0085.0()452.027.0(

)()(

22

221

221

d

d

yyxxd

The Total distance is:

623.0

)0586.0()239.0452.0( 22

d

d


252

Therefore s is:


855.0623.0

533.0 IFiber

Using Equation 65:

855.0677.00

677.0098.0

, ,

,

, ,

BrokeOMIFiberOM

BrokeOMMixtureOM

IFiber

IFiberOMIFiberBrokeOMBrokeMixtureOM

CC

CC

CCC


253

From Equation 86, AUPmoptimum = 2.035


103.1577.1

035.2855.0 IFiberX

855.0677.00

677.0098.0

, ,

,

, ,

BrokeOMIFiberOM

BrokeOMMixtureOM

IFiber

IFiberOMIFiberBrokeOMBrokeMixtureOM

CC

CC

CCC

Therefore xs is:


254

TIER III


255

4. Open Ended ProblemAn ethylene/ethyl benzene plant is shown in the next flow sheet. Gas oil is being cracked with steam in a pyrolysis furnace to form ethylene, low BTU gases, hexane, heptanes, and heavier hydrocarbons. The ethylene is then reacted with benzene to form ethyl benzene. Two waste water streams are formed one of the streams is the quench water recycle for the cooling tower and the second one is the waste water from the ethyl benzene portion of the plant. The primary pollutant present in the two waste water streams in benzene. Benzene must be removed from the waste water that will be use to quench the cooling tower, coming from the settling unit to a concentration of 180ppm before it can be recycled back to the cooling tower and the boiler water treatment process. Benzene must also be removed from the waste water stream coming from the lower separation unit down to a composition of 380ppm before the waste water stream can be sent to biotreatment


256

PyrolysisFurnace

BoilerWater

Treatment

Cooling Tower

Settling

UpperSeparation

Ethyl benzeneReactor

Lower Separation

Gas Oil

Steam

FreshWater

Refuse

Waste water 150kg/s 1100ppm

FreshWater

Benzene

Waste water 70kg/s 2100ppm

Vent Fuel

Recycle Quenched Water

To Biotreatment

Low BTU gasesHexane 0.8kg/s 10ppmw

Heptane 0.4kg/s 17ppmw

Heavy Hydrocarbons

Ethylbenzene


257

4. Open Ended ProblemThe heptane and hexane streams will be used to recover part of the benzene, the desired final composition of them is unknown and has to be determined by the engineer, after which they are sent to finishing and storage. The mass transfer driving forces 1 and 2, should be at least 25,000 and 29,000ppmw respectively. The equilibrium data for benzene transfer from waste water to hexane (1) and heptane (2) are:

y = 0.012x1

y = 0.009x2

Where y, x1 and x2 are in mass fractions. Two external MSA are being considered for removing of benzene; air and activated carbon. Air is compressed to 2 atmospheres before stripping. Following stripping, benzene is separated from air using condensation.


258

4. Open Ended Problem• Henry’s law can be used to predict equilibrium for the

stripping process. Activated carbon is regenerated using steam in a ratio of 2kg steam : 1 kg of benzene adsorbed on activated carbon. Make up at a rate of 1.2% of recirculating activated carbon is needed to compensate for losses due to regeneration and deactivation. Over the operating range, the equilibrium relation for the transfer of benzene from waste water onto activated carbon can be described by:

y = 6.8x10-4x4


259

4. Open Ended Problem1. Label the rich and lean streams2. Construct a pinch diagram, identify pinch location,

minimum load of benzene to be removed by external and excess capacity of MSA’s Consider the four MSA’s to choose from and find the MOC needed to remove benzene. Use the cost data found in slide 97

3. Apply the algebraic approach4. Design the network for the plant and draw a modified

flow sheet5. Comment on your results, what limitations do you think

have the methods used in the calculations if any, what conclusions can you draw based on your results?


260

I wish to thank for their cooperation and guidance:

• Dr. Mahmoud M. El-Halwagi Professor Texas A&M• Dr. Jules Thibault Professor University of Ottawa• Dr John T. Baldwin Professor Texas A&M• Dr. Dustin and Georgina Harrel Texas A&M• Vasiliki Kazantzi PhD student Texas A&M• Qin Xiaoyun Researcher Candidate Texas A&M• Daniel Grooms PhD student Texas A&M

William Acevedo, April 2004

5. Acknowledgments


261

• El-Halwagi M. Mahmoud, Pollution Prevention through Process Integration Systematic Design Tool, Academic Press, 1997

• El-Halwagi M. Mahmoud, Glasgow M. Ian, Eden R. Mario, Qin Xiaoyun, Property Integration: Componentless Design Techniques and Visualization Tools, Texas A&M

• Kazantzi V., Harell D., Gabriel F., Qin X., El-Halwagi M.M., Property Based Integration For Sustainable Design, AIChE Annual Meeting, 2003

• Seider D. Warren, Seader J.D., Lewin Daniel R., Product and Process Design Principles, Wiley International, 2004, 2d ed

• Shelley, M.D. and El-Halwagi M.M., Component-less Design of Recovery and Allocation Systems: A Functionality based Clustering Approach, Computers and Chemical Engineering, 24, 2081-2091, 2000

• Qin X., Gabriel F., Harell D., El-Halwagi M.M., Algebraic Techniques for Property Integration Via Componentless Design, Texas A&M

References

1. Texas A&M University 2 MODULE III Introduction to Process Integration.

Documents

Transcript of 1. Texas A&M University 2 MODULE III Introduction to Process Integration.