Module 4: Tier III Environmental Challenges – Pulp & Paper Industry Created at:

Program for North American Mobility in Higher Education

Introducing Process Integration for Environmental Control in Engineering

Curricula

Module 4: Tier IIIEnvironmental Challenges –

Pulp & Paper IndustryCreated at:

École Polytechnique de Montréal & Texas A&M University, 2003

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Tier III:Open-Ended Problem

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Tier III: Statement of Intent

Tier III: Statement of Intent

The purpose of this is to provide students with an open-ended problem which assimilates the concepts of minimum impact manufacturing including process integration and LCA .

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Problem Statement for Q-1 & 2

You are an environmental engineer in a pulp and paper mill. The head office wants to enhance its competiveness by putting together a technology roadmap with the ultimate goal to be a minimum impact manufacturing mill.

Some information about the mill is given at the following page.

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Mill Description

Conventional pulping technology, ECF bleaching, drying, activated sludge plant

Debarking: dry Lime kiln: normal Lime kiln fuel: heavy fuel oil Lime kiln flue gas: high eff. ESP Bark boiler (HW bark):

Total efficiency 0.87 Fluidized bed boiler Electric power generation from excess heat in mill

condensation turbine Since no information is available concerning the

effluent treatment plant, its efficiency will be consider constant. As a consequence of that, from a relative point of view, the effluent ion loads can be considered proportional to the ones before the effluent treatment.

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Question 1

A few months ago the company ordered a partial LCA study in order to have an idea about its life cycle environmental impacts. As a first step, your boss had asked you to look at this study as well as at the mill simulation and give him your recommendations for environmental improvement. To do this look at unit process contribution to each impacts and perform sensibility analysis. Do not use any normalization or weighting. Without doing calculations, you can also use cost arguments. Also determine, by mass balances by how much fresh water can be theoretically reduced (by recycle).

System boundaries are defined in the LCA study and the main hypothesis are presented next pages.

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Functional Unit

All LCA results are presented relative to the functional unit. The functional unit has been defined as follow: The production of 1 admt of pulp.

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Chemical Production

Chemical production as been included into the system boundaries. Chemicals are considered to be transported an average distance of 100 km using 40 ton diesel trucks and empty trucks return to the supplier. For calculation purpose a weight of 1/10 of the transported chemicals has been assumed for the return of the truck.

No data was available for talc manufacturing. Therefore it has been excluded from the system boundaries. However, its transportation has been considered.

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Birch Growth and Harvesting

Birch growth and harvesting as been included in the boundaries. The wood is transported an average of 100 km. The same assumptions as for chemicals apply.

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Others

By product have been located. A credit has been considered for the

generated energy (but only on the energy).

Pulp is transported an average distance of 200 km to the customer (same assumptions as chemicals).

Industrial landfill is located 5 km from the mill. 16 ton diesel trucks are used to transport the solid wastes, the return of the trucks is considered negligible.

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Necessary documents

LCA Base Case Process Simulation

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Question 2

Your boss is convinced that most of the competitive advantages that can be gained with environmental improvements are related with fresh water reduction.

In this case, recycling the effluent water is the most obvious way to reduce fresh water consumption, but this can result in the build-up of non-process elements and so reduce process performance.

For this reason, he has also mandated a consulting company to perform a water pinch study subject to process constraints.

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Question 2 (Cont’d)

The consultant has first evaluated possibility of direct recycle because it does not implicate major capital costs. Major results are presented in the following table.

Water Consumption 23% reduction

Liquid Effluent Reduction of ion content of 2.3%

Gas Effluent Cl, K: 0.2% increaseNa: 6.8% increase

Energy produced 5% reduction (need more energy to pump)

Dust 13.4% increase

Solid wastes Neglictible difference

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Using the LCA model, discuss if this represents a real environmental improvement. To compare results, normalize against the base case.

A panel of experts has determined that the importance of each impact category can be described by the weights in the following tables. Resources and emissions are weighted separatly.

What is the influence of the weights on the final decision.

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Resource depletion:

Emissions:

Impact Weight

Raw water consumption

0.83

Energy consumption

0.08

Virgin Fiber consumption

0.01

Other resources

0.08

Impact Weight

Carcinogenic substances

0.70

Heavy metals

0.07

Acidification 0.01

Eutrophication

0.01

Summer smog

0.07

Winter smog 0.07

Solid Wastes 0.005

Global Warming

0.065

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Solution – Q1

The process simulation does not give a lot of insights on the environmental impacts of the process. However it is obvious that the bleaching plant consumes a lot on fresh water and rejects a lot in the environment. The following is the solution for potential water reduction

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Solution – Q1 (Cont’d)

Water balances can be summarized by this picture.

The total fresh water consumption is 9.33+0.97+3.83+20.66=34.79 ton/ton of dry pulp.

Only liquid water can be “directly” recycle: 0.967+5.92+0.681+20.66= 28.23 ton/ton of dry pulp.

For mass conservation reasons, only the min of fresh water or liquid effluent can be recycle ie. 28.23 ton.

So the minimum water consumption is 34.79-28.23=6.56 ton (ie a reduction of 81%).

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The following graph show the contribution of each process unit to resource consumption.

Process Contributions - Input Impact Categories

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

Raw WaterConsumption

Virgin FiberConsumption

Natural ResourcesConsumption

Energy Consumption

Landfill

Transport

Chemical Production

Pulp Manufacturing

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The last figure show that the manufacturing activities consumes a lot of resources: water, virgin fiber and other natural resources.

It also shows that chemical production is particularly energy-consuming.

From a first look, reducing chemical and water consumption will result in a significant environmental benefit.

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The following graph show the contribution of each process unit to emission-related environmental impacts.

Process Contributions - Output Impact Categories

0%

20%

40%

60%

80%

100%

Globa

l War

min

g

Acidific

atio

n

Eut

roph

icatio

n

Heavy

Met

als

Carcin

ogen

ic Sub

stanc

es

Summ

er S

mog

Win

ter S

mog

Solid

Was

te

Growth

Harvesting

Landfill

Transport

Chemical Production

Pulp Manufacturing

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From this graph it is possible to note that: Manufacturing activities are a large contributor

to acidification, eutrophication, winter smog and solid wastes;

Chemical production is a large contributor to all impact categories but more specifically eutrophication, heavy metals and summer smog.

Transportation seems also to be a large contributor to several impact categories: global warming, carcinogenic substances and summer smog.

Global warming is due to almost all unit processes.

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Even if it is impossible to talk about the relative importance of each impacts since no weighting has been performed, it is clear from the last two graphs that manufacturing activities, including chemical consumption must be targeted in order to reduce the overall environmental impacts. Transport is also a significant contributor.

The following results show how much a 5% reduction in transportation and chemical consumption will affect the environmental impacts. Manufacturing is more difficult to assess but the impact of an increase of 5% of the yield (from 50% to 52.5%) is also presented. It as been assumed that an increased yield will only impact the quantity of wood required and not the chemical consumption in order to keep both effect separate.

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It is important to note that here only easily manipulable variable have been modified in order to determine which changes will influence the more the environmental impacts.

The most important results are the following: A 5% increase in the yield will result in a:

5.64% reduction in fresh water consumption; 4.70% reduction in virgin fiber consumption; 4.39% reduction in natural resources consumption.

A 5% reduction in transportation will result in a: 4.86% reduction in energy consumption; 4.26 reduction in carcinogenic substances.

A 5% reduction in chemical will not affect significantly the environmental impacts.

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As an environmental engineer, you will propose the followings: Increase the process performance, which

will also reduce costs. Since reducing transportation distance is

not easily realizable, you suggest to find a mode of transportation less pollutant.

Even if a reduction of chemical consumption will necessarily reduce the cost, it is not an environmental priority.

The mass balances have shown that there is a lot of potential for fresh water reduction.

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Solution - Q2

Question 2 - Normalized Results - Resources

0

0.2

0.4

0.6

0.8

1

1.2

Raw Water Consumption Virgin Fiber Consumption Natural ResourcesConsumption

Energy Consumption

Pulp Manufacturing

Chemical Production

Transport

Harvesting

Total

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Solution - Q2 (Cont’d)

The last graph shows the LCA results (resources) for the direct water recycle option. The results have been normalized against the reference case. From this graph, it is possible to say that: Raw water consumption from the

manufacturing process unit has been reduced to 70% of the reference case.

Energy consumption by the manufacturing has been increase by 5%.

Everything else is constant.

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Solution - Q2 (Cont’d)Question 2 - Normalized Results - Resources - Emissions

0

0.2

0.4

0.6

0.8

1

1.2

Global

War

ming

Acidific

ation

Eut

roph

icatio

n

Heavy

Met

als

Carcin

ogen

ic Sub

stanc

es

Summ

er S

mog

Wint

er S

mog

Solid

Was

te

Pulp Manufacturing

Chemical Production

Transport

Landfill

Harvesting

Growth

Total

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The preceding graph shows a reduction in the following impact categories: Acidification from the manufacturing

process unit. It also shows an increase in:

Winter smog from the manufacturing process unit.

All the remaining impact categories are almost constant.

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The aggregated indicators are: Resources: 0.76 Emissions: 1.00

From this it is possible to conclude that the direct water recycle solution has a positive impact on the resource impact categories (almost 25% improvement) and almost no impact on the emissions.

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A lot of importance has been given to the raw water consumption. A sensitivity analysis on the weights has been conducted. First, weight of raw water has been decreased while maintaining the other relative weights constant.

The results are presented in the table. It can be seen than even if the raw water importance passes from 83% to 10%. There is still an environmental benefit.

Weight of the raw water

consumption

Aggregated Indicator

0.83 0.76

0.50 0.85

0.30 0.91

0.10 0.97

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The impact category the most influenced by the direct recycle other than raw water is the energy.

By increasing the weight of energy while maintaining the other ratios constant we obtain the results presented in the table.

The conclusion of the 2 tables is that the environmental improvement is robust to the weights.

Weight of the Energy


0.08 0.76

0.16 0.78

0.32 0.82

0.64 0.90

0.83 0.96

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The same strategy has been applied to the emission impact categories. Sensitivity analysis have been conducted on the acidification and winter smog weights.

Acidification has been reduced so the sensitivity analysis try to determine if more weight on this impact category will reduce significantly the aggregated indicator.

The table shows that even if acidification weight passes from 1% to 80% this will results in only 2% improvement.

Weight of the

Acidification


0.01 1.00

0.10 1.00

0.20 1.00

0.40 0.99

0.80 0.98

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Winter smog has been increased so the sensitivity analysis try to determine if more weight on this impact category will increase significantly the aggregated indicator.

The table shows that even if winter smog weight passes from 7% to 80% this will results in only 1% degradation.

The 2 previous tables show that the emissions indicator is robust to the weights.

Weight of the Winter

Smog


0.07 1.00

0.14 1.00

0.28 1.00

0.56 1.00

0.80 1.01

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Overall conclusion: Direct water recycle results in a

positive resource saving (24%) without compromising the other impact categories.

Furthermore, it is a low cost solution. In consequence, its implementation is

highly recommended.

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Problem Statement – Q3-7

Consider the following Kraft pulp mill depicted below

slaker green liquor clarifier

white liquor clarifier

causticizer

Dig

este

r

conc

entr

ator

Recovery Boiler

ESP

Multiple Effect Evaporators

grits

dregs lime kiln

cond. cond.

screening Brown Stock Washing

chips

white liquor

black liquor weak

pulp wash water

dissolving tank

mud washer

mud filter dregs

washer & filter

weak white liquor

dust recycle

smelt salt

cake

SBL

Flue Gas

wash water

wash water

D E D E D

fluegas lime mud

To papermaking

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Problem Definition

Chips = 6000 tons (wet basis) Moisture = 50% = 0.5*6000 t = 3000 t Pulp Yield = 50 % of Dry = 0.5 * 3000 t Consistency (CY) = 0.12 Dilution Factor (DF)= 2 Wash Water for Pulp = [(1-CY)/CY] +DF Ion Content of Process Water:

Cl = 3.7; K = 1.1; Na = 3.6 (values in ppm)

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Problem Definition

Given this Kraft pulping process, it is desired to develop cost-effective strategies for the reduction of water discharge from the mill. It should be noted that any water reduction objectives will entail the use of recycle; consequently, various species will build up in the process, leading to operation problems.

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Problem Definition

To alleviate the detrimental effect of build-up, comprehensive mass integration strategies are required to provide answers to the following questions: What are the rigorous targets for reduction in water

usage and discharge? Which streams need to be recycled? To which units? Should these streams be mixed or segregated? What interception devices should be added to the

process? To remove what load? What new research needs to be developed to attain the

optimum solutions?

Q3 – 7 will address some of these questions

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Question 3

What are the rigorous targets in water discharge and reduction?

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Species Tracking Model

Before one can begin to tackle the water targeting problem, it is crucial to develop a species tracking model of the system with the right balance in details. A too-simplified model will not adequately

describe the process nor will it capture critical aspects of the process.

A too-detailed model cannot be readily incorporated into the process integration and optimization framework and will negatively impact the effectiveness of the optimization computations.

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Species Tracking Model

In order to develop the species tracking module, we will make use of path diagram equations, perform degrees of freedom analysis, and use the mixer splitter models. These topics were covered in Module II, though they are included here as a quick reference

Path Diagram Equation Degrees of FreedomMixer-Splitter Model

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Mathematical Modeling

The modeling techniques covered in module II allow one to describe unit performance without requiring detailed models while still capitalizing on nominal plant data and knowledge about the process. With this information, one can begin to make choices for the selected model and streams/species.

Consider the following unit:

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Pollutant/Water Load Balance Representation

W1 (kg water/s)P1 (kg pollutant/s




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W and P refer to the loads of water and a pollutant, respectively.

Suppose that the load of the water were to change as a result of process improvement (e.g. mass integration). The load of the pollutant will be affected as well; thus, it will be necessary to determine the new load of the pollutant.

Furthermore, suppose that there exists a proportional relationship between the pollutant loads in streams 2 and 3 (much more so than between streams 1 & 3, 1 & 4, etc).

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With this knowledge, the ratio model can be used to relate the pollutant loads in streams 2 and 3:

P3new = (P3

old/ P2old) * P2

new

The pollutant load in stream 4 can then be determined by a simple component material balance:

P4new = P1

new + P1new + P3

new


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Nominal Balance Model

By using these modeling techniques, path equations can be developed for tracking water and targeted NPE’s throughout the process, resulting in a mathematical model for the nominal case study. The nominal case study can then be revised to reflect the impact of mass integration on the process.

48


For this case study, the nominal balance model will be developed with the purpose of tracking water and three nonprocess elements, chloride, potassium, and sodium. These ions were selected because they are among the most important species that cause buildup problems and limit the extent of mass integration

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Using process knowledge, nominal plant data, modeling techniques, initial assumptions, etc., one can begin to develop the nominal balance model unit by unit.

The overall result for the nominal balance model will be provided at this time. However, the full development of the nominal balance is provided at the end of this module for the reader’s understanding.

Nominal Balance

Digester

WasherScreening

MEE Concent .

WhiteLiq Clar

Causticizer

Slaker

LimeKiln

Washers/Filters

RecoveryFurnace

Dissol .Tank

Green LiqClarifier

ESP

smel

tNa2SO4

1

3

5

2

4

6

7

10

9

12

11

18

13

14

15

16

17

20

21

24

25

26

27

28

29

30

31

32

8

23

19

22

W1 = 3000C1 = 1.000K1 = 2.500N1 = 0.973

W2 = 13995C2 = 0.052K2 = 0.015N2 = 0.050

W4= 10995C4 = 0.492K4 = 0.819N4 = 4.347

W6 = 1450C6 = 0.005K6 = 0.002N6 = 0.005

W3 = 5127C3 = 9.278K3 = 39.230N3 = 486.344

W31 = 6143C31 = 11.451K31 = 39.862N31 = 576.226

W30 = 6143C30 = 11.451K30 = 39.862N30 = 576.226

W20 = 6402C20 = 10.623K20 = 44.848N20 = 595.216

W17 = 0C17 = 9.351K17 = 39.479N17 = 499.893

W16 = 0C16 = 4.230K16 = 9.904N16 = 73.033

W18 = 0C18 = 0.182K18 = 0.026N18 = 18.225

W26 = 423C26 = 0.042K26 = 0.042N26 = 12.033

W10 = 8901C10 = 0.000K10 = 0.000N10 = 0.000

W12 = 1024C1 2= 0.000K12 = 0.000N12 = 0.000

W14 = 0C14 = 0.472K14 = 1.146N14 = 0.966

W15 = 1202C15 = 0.197K15 = 0.327N15 = 0.386

W23 = 3.84C23 = 0.038K23 = 0.004N23 = 0.960

W19 = 6402C19 = 1.272K19 = 5.369N19 = 95.323

W5 = 11127C5 = 9.838K5 = 39.230N5 = 483.020

W32 = 1016C32 = 2.173K32 = 0.631N32 = 89.882

W27 = 0C27 = 2.297K27 = 0.604N27 = 0.633

W24 = 5762C24 = 0.021K24 = 0.006N24 = 0.021

W29 = 32C29 = 0.000K29 = 0.000N29 = 0.000

W28 = 8C28 = 0.014K28 = 0.209N28 = 0.576

W7 = 10995C7 = 0.394K7 = 0.655N7 = 3.478

W8 = 1450C8 = 0.104K8 = 0.165N8 = 0.875

W9 = 2225C9 = 9.838K9 = 40.927N9 = 483.020

W22 = 51C22 = 1.455K22 = 5.382N22 = 19.047

W25 = 423C25 = 2.339K25 = 0.647N25 = 12.666

W13 = 1202C13 = 4.899K13 = 11.378N13 = 74.385

W11 = 1202C11 = 9.838K11 = 40.927N11 = 483.020

W21= 6351C21= 9.167K21 = 39.466N21 = 576.169

StripperStripper

Wood chipsPulp

Bleach Plant

W35 = 10995

Bleached pulp to papermaking

33W33 = 30990

37 W37 = 30990C37 = 15.495K37 = 0.155N37 = 15.495

ChemicalsMaterial Balance

51

Back to Question 3

What are the rigorous targets in water discharge and reduction?

The objective here is to minimize the amount of fresh water used in the process and the amount of wastewater discharged from the system.

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Solution – Q3

Beginning with the nominal balance model (figure ), the first step is to identify all possible sources of water entering, leaving or being consumed in the process in order to obtain the overall water balance for the process, as depicted in next figure

53

Bleach Plant = 30990

Chips = 3000

Washer = 13995

Screening = 1450

Washers/Filters = 5762

Screening = 1450

MEE = 8901

Stripper = 1024

ESP = 1202

Washer/Filter dregs = 4

Lime Kiln = 423

Slaker grits = 8

Slaker = 32

BP water = 30990

Pulp leaving Bleach Plant = 10995

OVERALLWater Balance

Water consumedBy reaction = 168

Total Water In =

55197 – 168 = 55029 tpd Total Water Out = 55029 tpd

OVERALL WATER BALANCE

54

Solution – Q3

Next, all streams that use fresh water and all streams that contain potentially recyclable water are identified.

There are four fresh water streams (S2, S6, S24 and S34) giving a total fresh water use of 52,197tpd.

There are also four potentially recyclable streams, S8, S10, S12 and S37 giving a total of 42,365 tpd.

The overall water balance diagram has been modified to reflect this information (see figure )

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Washer = 13995

Screening = 1450

Washers/Filters = 5762

Screening = 1450

MEE = 8901

Stripper = 1024

BP water = 30990

Water Balance

Water consumedBy reaction = 168

Total Fresh Water in = 52197 tpd Total recyclable water out = 42365 tpd

Bleach Plant = 30990

FRESH AND RECYCLABLE WATER BALANCE

56

Solution – Q3

If the recyclable water can be intercepted and cleaned up to the point where it is acceptable for use in place of fresh water and if self-recycle is allowed, then one can determine the target for fresh water usage:Minimum water consumption =

52197 – 42365 = 9832 tpd

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Solution – Q3

By adding up the flowrates of the water streams leaving the process except the recyclable streams (S8, s10, s12, s37) and water in the produced pulp, we get a target for wastewater discharge of 1,669 tpd

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Water Consumed 168

WtoBP = 30990

W1 = 3000

W2 = 13995

W6 = 1450

W24 = 5762

Water going out with pulp 10995 tpd

Wastewater Target = 1669 tpd

Fresh WaterTarget = 9832

Chips

Target for minimum water consumption = 52,197 – 42,365 = 9,832 tons per day

OVERALL WATER TARGETING FOR CASE STUDY

59

Solution – Q3

Therefore, the rigorous targets are:Minimum Fresh water target = 9832 tpd Wastewater target = 1669 tpd

60

Limitations on Self-Recycle

Previously, it was permitted to consider recycling the effluent back to the same unit. However, self-recycle may sometimes be forbidden due to numerous reasons such as: To prevent the build-up of impurities in a flow

loop To avoid dynamic instabilities that may arise

due to the high interconnectivity between the input and output

To enhance process reliability by disengaging the dependence of the input from the output.

61

Limitations on Self-Recycle

If self-recycle is not allowed, then it is possible that the targets identified earlier may not be reached even if interception technologies are used to clean up the recyclable water streams. As a result, new targets will need to be determined, which leads to the next question:

62

Question 4a

In the case of no self-recycle with one interceptor, which streams can be intercepted?

63

Solution Q4a

There are four recyclable streams for consideration: W8 – MEE W10 – Concentrator W12 – Screen W37 – Bleach plant effluent

In the development of the nominal balance model, it was assumed that there were no ions in the water leaving the MEE and Concentrator (i.e. it has the same quality as demineralized water); therefore, the only interception candidates are the screen and bleach plant effluents.

64

Question 4b

Choosing the bleach plant effluent for interception and assuming that the quality of the screen effluent is acceptable for direct recycle to the pulping process, what are the new water targets (remember, no self-recycle)?

65

Solution Q4b

The flow of the intercepted bleach plant effluent, along with the screen effluent is more than enough to replace all of the fresh water used in the pulping process. Therefore, the fresh water target for the pulping process is zero.

For the bleach plant, only water meeting dimineralized quality can be used. Thus, the effluents from the multiple effect evaporator and the concentrators can be used, replacing a total of 9925 tpd of BPE.

66

Solution – Q4b

Bleach

PULPING

Consumption by Chemical Reaction

and other losses= 9832 tpd

Fresh water = 21065 tpd

W8 = 1450

W10 = 8901

W12 = 1024

W2 = 13995

W6 = 1450

W24 = 5762

WastewaterTo bio =30990 – 19757= 11233 tpd

13995 + 5762 + 1450 = 21207

interception

67

Solution Q4b

The new targets are now: Pulping process fresh water: 0 tpd Bleach Plant Effluent fresh water

target:30990 – (8901+ 1024) = 21065 tpd

Wastewater target:30990 – 19757 = 11233 tpd

68

Process Integration Strategies

The overall targeting has identified that fresh water consumption can be significantly reduced from 52197 tpd to 9832 tpd.

The next step, then, is to determine how this can be accomplished. What is the optimal strategy for water reduction? How are the streams to be allocated? This cannot be easily perceived simply by looking at the process flowsheet.

Process integration strategies will be employed to determine the optimal ways of reaching the target

69

Why Process Integration?

Process Integration is a holistic approach to the design and operation of complex systems. It is a sound framework that utilizes well-developed and proven mass and energy integration techniques for optimizing the design and operation of a process.

70

Process Integration

It is important to coordinate both process integration and process simulation. The application of process integration provides performance targets, solution strategies, and proposed changes to the process. Process simulation reassesses the process performance as a result of theses changes.

71

Coordination of Process Integration and Simulation

ProcessIntegration

ProcessSimulation

Closing the information loop of integration and simulation ensures that the developed insights and solution strategies are refined and validated.

Process Objectives, Data and constraints

Input/Output relationsNew Processes

Process ModificationsStructural changes

72

Mass Integration Strategies

Now that the rigorous targets have been developed for the minimum feasible water usage and discharge, various cost-effective mass integration strategies should be used to attain the targets. These strategies include Segregation, Low cost/no cost modifications, Direct recycle, Interception high cost process modifications.

The above strategies can be represented as a pyramid (see next slide), where it is desired to begin at the bottom of the pyramid, which represents the lowest cost and perhaps more easily implemented strategies, and work up until the target is achieved.

73

Mixing & Recycling

Interception

Segregation

Target

Chemical Process

Low Cost Process Modifications(LCPM)

HCPM

Mass Integration Strategies

74

Segregation

Segregation refers to avoiding the mixing of streams. In some industrial applications, dilute streams have been mixed with concentrated streams and even different phases have been mixed together unnecessarily. Segregation of streams at the sources can provide several opportunities for cost reduction such as: Generate environmentally benign streams Enhance the opportunities for direct recycle

since dilute streams are easier to recycle. The separate concentrated streams are now

more thermodynamically favorable for interception

75

Low-cost process modifications

In some cases, a change in process conditions (such as temperature, pressure, compositions, etc) may be all this is needed to decrease or eliminate the waste produced in a unit.

Provided that the cost is low, a unit can be replaced with a more environmentally benign one.

76

Recycle

Discharged waste can be reduced by recycling pollutant-laden streams back to the process to be utilized in process or non-process requirements. In some instances, several streams need to be mixed with each other to achieve the desired level of flowrate and composition.

77

Interception

Interception refers to the utilization of separation techniques to selectively remove targeted species from targeted streams. In most industrial applications, inteception is needed to enhance the opportunities of recycling and to generate environmentally benign streams.

78

High-Cost Process Modifications

After all other strategies have been exhausted, one may need to employ high cost process modifications. This may include completely new chemistry (such as new solvent or new reaction path), new technology (new plant), etc.

79

Question 5

What is the optimal water allocation using direct recycle?

80

Solution – Q5

To answer this question, a mass allocation representation of the process from the species viewpoint needs to be developed.

For each species, there are sources, those streams that contain the desired species, and sinks, those streams units which can accept the species.

Each sources can be segregated, intercepted to adjust species content, mixed, etc and allocated to the different units or sinks, as depicted in the following figure.

81

Source i = 1

Fresh Source

Segregated Sources

Sinks

SpeciesInterception

Network(SPIN)

j = 1

j = 2

j = Nsinks

Source i = Nsources

Mass & Energy Separating Agents In

Mass & Energy Separating Agents Out

SOURCE-INTERCEPTION-SINK REPRESENTATION

(e.g., El-Halwagi et al., 1996,Spriggs and El-Halwagi, 1998,Dunn and El-Halwagi, 2003)

82

Process Sinks

There are a number of process units, or Nsinks,that employ fresh water and are designated by the index j (j ranges from 1 to Nsinks).

Each jth sink has two sets of contraints on flowrates and composition: Flowrate to each sink

Wjmin Wj Wj

max j = 1, 2,….,Nsinks

Wj is the water flowrate entering the jth sink Ion content to each sink

Yion,jmin Yion,j, Yion,j

max j = 1, 2,….,Nsinks

Yion,j is the compostion of a certain NPE entering the jth sink

83

Process Sinks

Each source, represented by I, is split into Nsink fractions that can be assigned to various sinks. The flowrate of each split is denoted by li,j (see figure)

Each split fraction then has the opportunity to be mixed (or not) and assigned to sinks (see figure)

84

Splitting of the ith source:

where i = 1,2, …, Nsources

ksN

jjii lL

sin

1,

SourceLi

Yion,i

li,j

SPLITTING OF SOURCES TO SINKS

85

Mixing for the jth sink:

where j = 1,2, …, Nsinks

sourcesN

ijijj lFreshWW

1,

jli,jyion,j

Wj

Yion,j

MIXING OF SOURCES BEFORE SINKS

86

Direct Recycle Strategy

For this case study, four sources have been identified: Bleach Plant effluent, Screen effluent, Multiple Effect Evaporator effluent and Concentrator effluent. Fresh Water is included since it is the objective function of the optimization problem (where the objective function is to minimize the flowrate of fresh water via direct recycle).

Four sinks have been identified: Screening, Brown Stock Washer, Washer/Filters, and the Bleach Plant. Waste Treatment is also be included since it is possible that the best allocation for a source may be biotreatment.

The following figure shows the assignment representation for the Direct Recycle/Reuse problem

87

Sinks

Screening

Brown Stock Washer

Washers/Filters

Bleach Plant

Sources

S8 Wastewater from Screening

Fresh water

S10 Condensate from MEE

S12 Condensate from Concenrator

S37 Bleach PlantEffluent

Direct Recycle/Reuse Representation

Biotreatment

88

Direct Recycle Optimization Formulation for Source/Sink Analysis w/Path

Connection

The problem can now be formulated as an optimization problem, where the objective function is the minimization of the flowrate of fresh water. This objective funtion can be represented as:Min. flowrate of fresh water =

Subject to the following constraints

ksN

jjFreshW

sin

1

89

Direct Recycle Optimization Formulation for

Source/Sink Analysis w/Path Connection

maxminjjj WWW j = 1, 2, …, Nsinks

NPE content in feed to each sink:max,jk,

min, Y jkjk YY j = 1, 2, …, Nsinks and k = 1, 2, …, Nk

Splitting for the ith source:

ksN

jjii lL

sin

1, i = 1, 2, …, Nsources

Mixing for the jth sink:

sourcesN

ijijj lFreshWW

1, j = 1, 2, …, Nsinks

Component material balances for the pollutants:

jk

N

ijijkj ylYW

sources

,1

,, **

j = 1, 2, …, Nsinks and k = 1, 2, …, Nk

Flowrate to each sink:

Non-negativity of each fraction of split sources:0, jil i = 1, 2, …, Nsources and j = 1, 2, …, Nsinks

90

Direct Recyce/Reuse Optimization formulation

It should be remember that no self-recycle is permitted and that the bleach plant c,an only accept dimineralized water.

Furthermore, there is an additional issue with respect to the build-up of NPE’s in the recovery furnace which is affected by “sticky temperature”. It is related to Cl, K, and Na through the following constraints where Ci, Ni, and Ki, are the ionic loads of Cl, Na and K, respectively, in the ith source:

23*1.0

1.39181611181611 NNNKKK

)3923

(02.035

181611181611181611 KKKNNNCCC

91

Optimization Solution for Direct Recycle/Reuse

LINGO programming was used to develop and solve the mathematical formulation. The optimal water allocation is depicted in the following slide. The fresh water to screening has been

replaced with 751 tpd of concentrator effluent and 699 tpd of bleach plant effluent

The fresh water to the washers/filters has been replaced with 273 tpd of concentrator effluent and 5489 tpd of MEE effluent.

A portion of the fresh water to the Brown Stock Washers has been replaced with 3412 tpd of MEE effluent and 1450 tpd of screening effluent.

92

WhiteLiq. Clarif.

Causticizer

Screening

MEE Concent.

LimeKiln

Washers/Filters

Slaker

DissolvTank

GreenLiq. Clarif.

RecoveryFurnace

ESP

BSW

DigesterStripper

WoodChips

Stripper

Bleach

5489 3412

1450

273

751 699

9133 30990

30291

Optimum Solution for Direct Recycle/Ruse

93

Results of Direct Material Exchange

The fresh water consumption has been reduced to 40,123 tons per day, a 23% reduction from the nominal fresh water usage of 52,197 tons per day.

This solution is a direct recycle/reuse which requires piping and pumping but involves no capital investment for new processing units.

It should be noted that the mathematical solution can generate alternate solutions that yield the same fresh water consumption but require different piping and allocation alternatives.

94

Question 6

Through direct recycle, fresh water usage went down from 52197 to 40123. However, from water targeting, we know that interception can get the fresh water usage down to 21065 tpd. Consider the interception of the bleach plant effluent. How much Cl must be removed in order to meet meet the fresh water target of 21065 tpd?

95

Solution – Q6

In this problem, the objective function has changed from one of minimizing the fresh water consumption to one of minimizing the load of the Cl to be removed from the bleach plant effluent subject to: Desired water target Path equations for tracking water and Cl Recycle Model Interception equations Unit constraints

96

Solution – Q6

Basically, this problem is just like the recycle problem except that the objective function has changed. We know that the fresh water target is now 21065 tpd and that approximately 19750 tpd of intercepted bleach plant effluent is being recycle back to the process. Thus, in order to minimize the load to be intercepted from the BPE, a target has to be set for the maximum recyclable flowrate of the bleach plant effluent (the 19750 tpd).

97

Solution – Q6

Again, LINGO programming was used to solve the mathematical formulation A total of 8.99 tpd of Cl must be removed

from the bleach plant effluent. The fresh water consumption has been

reduced to 21072 tons per day, a 60% reduction from the nominal fresh water usage of 52,197 tons per day.

98

Exploring other interception opportunities

So far, only terminal streams (those streams going directly to waste treatment) have been considered. However, it is possible that other inter-process streams may be intercepted, perhaps providing greater economical and environmental benefits.

A literature search reveals that salt removal technologies exist for other kraft units, among those: White Liquor Interception Green Liquor Interception

Of course, this leads to the next question:

99

Question 7

How much chloride needs to be removed from Case 1: Green Liquor Case 2: White Liquor

in order to meet the fresh water target?

100

Interception Alternatives

This is quick and easy to determine. The objective function will remain the same (minimize chloride removal) as in Q6 but rather than minimizing the Cl removal from the bleach plant effluent, it will be minimized from the white liquor or green liquor streams. Thus, the optimization program only needs to be slightly altered to reflect the stream in question.

Interestingly enough, though it should come as no surprise, the load removal for Green Liquor and White Liquor interception is the same as the case for Bleach Plant interception (approx. 9 tpd of Cl). However, the three solutions are not identical. Each one has a different configuration of optimal water allocation (see figures)

101

WhiteLiq. Clarif.

Causticizer

Screening

MEE Concent.

LimeKiln

Washers/Filters

Slaker

DissolvTank

GreenLiq. Clarif.

RecoveryFurnace

ESP

BSW

DigesterStripper

WoodChips

Stripper

Bleach

5489 3412

1450

273

751 699

9133 30990

30291

Optimum Solution for Bleach Plant Interception

102

WhiteLiq. Clarif.

Causticizer

Screening

MEE Concent.

LimeKiln

Washers/Filters

Slaker

DissolvTank

GreenLiq. Clarif.

RecoveryFurnace

ESP

BSW

DigesterStripper

WoodChips

Stripper

Bleach

5489 3412

1450

273

751 699

9133 30990

30291

Optimum Solution for Green Liquor Interception

103

WhiteLiq. Clarif.

Causticizer

Screening

MEE Concent.

LimeKiln

Washers/Filters

Slaker

DissolvTank

GreenLiq. Clarif.

RecoveryFurnace

ESP

BSW

DigesterStripper

WoodChips

Stripper

Bleach

5489 3412

1450

273

751 699

9133 30990

30291

Optimum Solution for White Liquor Interception

104

Life Cycle Analysis

But which of the three technologies is the better solution?

END

106

Path Diagram Equation

Typically, the Path Diagram Equation defines outlet flows and compositions from key units as functions of inlet flows, inlet compositions and process design and operating conditions

This mass integration tool tracks the targeted species as they propagate through the system and provide the right level of details that will be incorporated into the mass integration analysis

Return to the

flowsheet

107

1. Degrees of Freedom

NV = NS x NC

F= NV - NE = NC (NS - 1)

F: degrees of freedomNV: number of variables

NE: number of equations

NC: number of targeted speciesNS: number of outlet streams

Assumptions:All inlets to a unit are known and it is desired to determine the outputs of the unit.

F must provided as additional modeling equations, assumptions, measurements, or data in order to have an appropriately specified (determined) set of equations that is solvable.

Unit U

Inlet stream(Fresh inputs or outlets from other

units)

Outlet streamsNstreams out

Return to the

flowsheet

108

2. Mixer-Splitter Model

The mixer-splitter model is a modeling technique which relies on nominal data .

The nominal data are those for the plant prior to any changes and can be obtained via simulation, fundamental modeling, direct measurements, or literature data.

There are various of the mixer splitter model: Fixed split model; Flow ratio model ; Species ratio model.

Based on the knowledge of the process, choices can be made for the selected model and streams/species.

Path equations can be developed for water and targeted NPEs throughout the process.

Return to the

flowsheet

109

Fixed Split Flow Model

Fixed SplitModelF

α * F

(1 – α) * F

The Fixed Split model takes a certain split, α, for the flows of streams leaving the unit

Return to the

flowsheet

110

Flow Ratio Model

Flow RatioModel

F G

Gnew = Gold * ( Fnew / Fold )

The Flow Ratio model assumes that streams or components maintain a certain fixed ratio. Thus, if the flow rate of a certain stream increases or decreases, all other related streams adjust according to the same ratio.

Return to the

flowsheet

111

Species Ratio Model

Species RatioModel

F G

IInew = Inew (IIold/ Iold )

Similar to the Flow Ratio Model, the Species Ratio Model maintains a fixed relationship between species in related streams. Thus, if one species changes, the other one adjusts by the fixed ratio. This model is especially useful if one species can be accurately tracked whereas the other one cannot.

I = species 1II = species 2

Return to the

flowsheet

112

Initial Data - Digester

Assumption: all inlet streams are known. Flowrate of wood chips, Chips = 6000 tpd Moisture content of wood chips = 50% Pulp Yield = 50% Pulp = Dry Chips * Yield Mass fraction ions with incoming wood

chips: C1 = 1 * Chips/6000 K1 = 2.50 * Chips/6000 N1 = 0.973 * Chips/6000

Return to the

flowsheet

113

Initial Data – Brown Stock Washer

Composition of ions in incoming wash water: Cl = 3.7 ppm K = 1.1 ppm Na = 3.6 ppm

Consistency of pulp leaving Brown Stock Washer, CY = 0.12

Dilution Factor, DF = 2.0 Ratio of ions in slurry leaving the BSW to the

chloride in the pulp stream leaving the digester Cl = 0.050 K = 0.020 Na = 0.009

Return to the

flowsheet

114

Digester

Digester

BrownStock

Washer

S1

S2

S4

S5

S3

W1 (from moisture contentC1 (from comp of Cl in chips) K1 (from comp of K in chipsN1 (from comp of N in chips

W2 (from consistency)C2 (from comp of Cl in wash water) K2 (from comp of K in wash water)N2 (from comp of N in wash water)

All species data will be calculated as anoutput stream from white liquor clarifier

W4 (from dilution factorC4 (from ratio to C5) = 0.05*C5 K4 (from ratio to K5) = 0.02*K5

N4 (from ratio to N5) = 0.009*N5

W5 C5 K5 N5

115

Digester

W1 = Moisture*Chips=0.5*6000=3000

W4 = [(1-CY)/CY]*Pulp=[(1-.82)/(0.82)]*3000

DF = (W2 - W4 )*Pulp; DF is given as 2

W2 can be determined from after W4 has been calculated.Ion Content in streams 2 and 4:

C2 = (3.7*10-6) *W2; C4 = 0.05*C5

K2 = (1.1*10-6) *W2; K4 = 0.02*K5

N2 = (3.6*10-6) *W2; N4 = 0.009*N5

Recalling the assumption that all inlet streams are known, thenstream 5 will need to be determined. The number of unknowns is our (flowrate of water and the three ions in S5); these can be obtainedVia the four material balances for the 4 species

W5 = W1 + W2 + W3 – W4

C5 = C1 + C2 + C3 – C4

K5 = K1 + C2 + K3 – K4

N5 = N1 + C2 + N3 – N4

116

Multiple Effect Evaporator

80% of the water in the weak black liquor is evaporated (water recovery ratio is 0.8).

It is assumed that no ions are in the condensate of the multiple effect evaporators

The material balances can be used to calculated the concentrated stream leaving the multiple effect evaporators

W10 = Water recovery in evaporator * W5

W9 = W5 - W10

C9 = C5 - C10

K9 = K5 - K10

N9 = N5 - N10

117

Multiple Effect Evaporators

MultipleEffect

Evaporators

W10 = Water Recovery * W5

C10 = 0 K10 = 0N10 = 0

S5

W5 C5 K5 N5

Black Liquor

S10

Evaporator Condensate

S9

Evaporator Concentrate

W9 C9 K9 N9

118

Multiple Effect Evaporator

46% of the water in the black liquor entering the concentrators is evaporated (water recovery ratio is 0.46).

Again, it is assumed that no ions are in the condensate of the multiple effect evaporators

The material balances can be used to calculated the concentrated stream leaving the multiple effect evaporators

W12 = Water recovery in concentrator * W9

W11 = W9 - W12

C11 = C9 - C12

K11 = K9 - K12

N11 = N9 - N12

119

Concentrator

Concentrator

W11 C11

K11

N11

S9

Evaporator Concentrate

W9 C9 K9 N9

S10

Concentrator Condensate

W12 = Water Recovery * W9C12 = 0

K12 = 0N12 = 0

S11

Strong Black Liquor

120

Recovery Furnace and Electrostatic Precipitator (ESP)

It is assumed that all the water in the strong black liquor leaves with the ESP off-gas so W15= W11.

The ions in the solids return, ESP dust and off-gass are related to the ions in the strong black liquor stream:C13 = 0.278*C11; C14 = 0.048*C11; C15 = 0.02*C11

K13 = 0.498*K2; K14 = 0.028*K11; K15 = 0.008*K11

N13 = 0.154*N2; N14 = 0.002*N11; N15 = 0.0008*N11

121

Recovery Furnace and ESP

The component material balance around the ESP is:W13 - W14 - W15 - W16 = 0.0

C13 - C14 - C15 - C16 = 0.0

K13 - K14 - K15 - K16 = 0.0

N13 - N14 - N15 - N16 = 0.0 It is assumed that the saltcake has a makeup flow of 0.0375 *

Pulp. Knowing this and the molecular formula for saltcake, N18 = 2*23/142 * Saltcake

The content of Cl and K in the saltcake is obtained by assuming ratios to Na in the saltcake. In addition, there is virtually no water in saltcake.W17 = 0.0

W18 = 0.0

C18 = 0.01*N18

K18 = 0.0014*N18

122

Recovery Furnace and ESP

The ion content in the smelt is determined via component material balance around the Recovery Furnace and ESPC11 + C18 - C15 - C14 - C17 = 0.0

K11 + K18 - K15 - K14 - K17 = 0.0

N11 + N18 - N15 - N14 - N17 = 0.0

123

Smelt

The smelt flowrate consists of the saltcake + solids in strong black liquor (SBL) – solids lost with the purge streams (S14 and S15) – solids volatilized in the furnace. Assuming that 5% of the solids in the SBL leave the ESP in the flue gas and that 47% of the SBL solids are volatized in the furnace:Smelt = Saltcake + SBL – 0.05*SBL – 0.47*SBL

OrSmelt = Saltcake + 0.48*SBL

124

Recovery Furnace and Electrostatic Precipitator (ESP

Recovery Furnace

ESP

W15 = W11

C15 (from ratio to C11) K15 (from ratio to K11)N15 (from ratio to N11)

W18 = 0C18 (from ratio to N18) K18 (from ratio to N18)

N18 = 0.324 * Salt Cake

W14 = 0C14 (from ratio to C11)

K14 (from ratio to K11)N14 (from ratio to N11)

W17 C17

K17

N17

S17

S14

S18

S15

S11

Strong Black Liquor

W11 C11

K11

N11

Smelt

Dust PurgeOff-gas

Salt Cake = 0.0375*Pulp

125

Dissolving Tank

The dissolving water-to-smelt ratio used in the dissolving tank is typically 85/15W19 = (85/15)*Smelt

The ionic content of S19 is determined by assuming ratios of Cl and K to those in the smelt and Na to the white liquorC19=0.136*C17

K19=0.136*K17

N19=0.196*N3

Component material balances around the dissolving tank are used to evaluate the ionic content of the feed to the green liquor clarifierW20 - W17 - W19 = 0.0

C20 - C17 - C19 = 0.0

K20 - K17 - K19 = 0.0

N20 - N17 - N19 = 0.0

126

Dissolving Tank

DissolvingTank

W17 C17

K17

N17

S17

Smelt

W19 = 5.67 * SmeltC19 (from ratio to C17)


S19

Dissolving Water = (85/15)*Smelt

S20

Feed to green liquor clarifierW20 C20

K20

N20

127

Green Liquor Clarifier

Typical overflow ratios were used to obtain the flows and ion concentrations in the overflow and underflow stream. W21=0.992*W20

C21=0.863*C20

K21=0.880*K20

N21=0.968*N20

Component material balance can then be written around the green liquor clarifier:W22 - W21 - W20 = 0.0

C22 - C21 - C20 = 0.0

K22 – K21 - K20 = 0.0

N20 - N21 - N20 = 0.0

128

Green Liquor Clarifier

Green LiquorClarifier

S20

Feed to green liquor clarifierW20 C20

K20

N20

C21 (from ratio to C20)


S21

Dust OverflowW21 (from ratio to

W20)

S22

UnderflowW22 C22

K22

N22

129

Washer/Filter System

The dregs leaving the washer/filter system contains very little water and is determined by relating it to the water content in the underflow from the GLC. The ion content in the dregs is determined by assuming ratios of the ions to the water in the dregsW23 = 0.075*W22

C23 = 0.010*C22

K23 = 0.001*K22

N23 = 0.250*N22

The overflow from the white liquor clarifier is determined by assuming a ration to Green Liquor overflowW32 = 0.160*W21

C32 = 0.237*C21

K32 = 0.016*K21

N32 = 0.156*N21

130

Washer/Filter System

The wash water is assumed to be 90% of the smelt dissolution water and the ionic content for C, K and N is based on the typical values of 3.7, 1.1, and 3.6 ppm, respectively

W24 = 0.9*W19

C24 = (3.7*10-6) *W24

K24 = (1.1*10-6) *W24

N24 = (3.6*10-6) *W24

Component material balances can then be written for the washer/filter systemW22 + W24 + W32 - W19 - W23 - W25 = 0.0

C22 + C24 + C32 - C19 - C23 - C25 = 0.0

K22 + K24 + K32 - K19 - K23 - K25 = 0.0

N22 + N24 + N32 - N19 - N23 - N25 = 0.0

131

Washer/Filter

Washer/Filter

S22

UnderflowW22 C22

K22

N22

S23

Dregs

S19

To dissolving tank

S25

Feed to lime kilnW25 C2

5K2

5N25



S32

Overflow from WLCW32 (from ratio to

W21)

S24

Wash Water

Ion content of Cl, K and Na is assumed to be 3.7, 1.1 and 3.6 ppm

132

Lime Kiln

The lime leaving the lime kiln is assumed to have no water. It is assumed that 95% of the sodium entering the lime kiln leaves in the off-gas. The ratio of C and K to water in the off-gas is assumed to be 0.001W27 = 0.0

W25 = W26

C26 = 0.001 *W26

K26 = 0.001 *W26

N24 = 0.05 *N25

A material balance can then be written around the kilnC25 – C26 – C27 = 0

K25 – K26 – K27 = 0

N25 – N26 – N27 = 0

133

Lime Kiln

Lime Kiln

S27

To SlakerW27 C2

7K2

7N27 S25

Lime mud from washer/filterW25 C2

5K2

5N25

S26

Kiln Off-gasW26 C2

6K2

6N26

134

Slaker The slaking reaction is given byCaO + H2O = Ca(OH)2

The amount of water consumed by the reaction is 0.32 of the consumed limeWATERSLK = 0.32*Lime

The amount of lime fed to the slaker is 35% of the pulpLime = 0.35 * Pulp

The vapor leaving the slaker makes up %5 of the water in the green liquor overflow and is ion-freeW29 = 0.005 * W21

C296 = 0.0

K296 = 0.0

N294 = 0.0

135

Slaker

The ion content in the grits is related to the green liquor overflowW28 = 0.0013 * W21

C28 = 0.0015 *C21

K28 = 0.0053 *K21

N28 = 0.0015 *N215

The component material balance can then be written around the slaker:W21 + W27 - W28 - W29 - W30 - Waterslk = 0.0

C21 + C27 - C28 - C29 - C30 = 0.0

K21 + K27 - K28 - K29 - K30 = 0.0N21 + N27 - N28 - N29 - N30 = 0.0

136

Slaker

Slaker

S30

To Causticizer

S27

From Lime Kiln

S21

Green Liquor Overflow

S29

Slaker Vapor

S28

Grits

C29 = 0

K29 = 0

N29 = 0

W29 = 0.005 *W21



W21 (from ratio to W21)

Water consumed by Rxn 0.032*Lime = 0.35*Pulp

137

Causticer/White Liquor Clarifier(WLC)

The causticing system provides an addition residence time for the causticizing reaction to take place so it can be assumed that the water and ionic content of the stream entering and exiting the system is the same (the chemical forms may change)W31 = W30

C31 = C30

K31 = K30

N31 = N30

The material balance around the WLC is thenW21 - W32 - W3 = 0.0

C21 - C32 - C2 = 0.0

K21 – K32 - K3 = 0.0

N21 - N32 - N3 = 0.0

138

Causticizer and White Liquor Clarifier

WhiteLiquor

Clarifier

Causticizer

S30

From Slaker

S31

S31

White Liquorto digester

S32

To washer/filter

139

Bleaching

Bleaching

140

Degrees of Freedom Analysis

As stated earlier, a too-detailed model can hinder optimization. Consequently, a degree of freedom analysis should be conducted to determine the number of unknowns that can be specified before the remaining variables can be solved.

141

Degree of Freedom Analysis

Unit U

INLET STREAMS

(fresh inputs or outlets from other units)

OUTLET STREAMS

NS

Given the following generic unit

andF = Number of Degrees of FreedomNV = number of unknown variables (NS*NC)NE = number of equations in each variableNS = number of outlet streamsNC = number of targeted species

142

Degree of Freedom Analysis

Then the number of assumptions, additional modeling equations, measurements, data, etc that must be provided in order to have a properly specified and solvable set of equations is:

F = NV – NE = NC(NS – 1)

143

Flowrate

Load

Sink CompositeCurve

WasteDischarged

MaterialRecyclePinch

FreshUsage

SourceCompositeCurve

(El-Halwagi et al., 2003)

PINCH-BASED VISUALIZATION TOOL FOR BPX PROBLEM

144

Screen

Brown StockWasher

Washer/Filter

Bleach Plant

BiotreatmentFresh H2O

MEE Cond.

Conc. Cond.

Screen Cond

BPE

F10, z10

F12, z12

F8, z8

F37, z37

Fresh, z5

G6

G2

G24

G37

G2

y6

y24

y37

y2

y2

f101

u = 1

u = 2

u = 3

u = 4

u = 5

f102

fresh4

f121

f103f105

f125

f85

f37bio

f371 f104

fresh1

f82

f83

fresh3

fresh2

f122f123

f124

f81

f83

f84

f373

f372

Optimization of Source/Sink Analysis With Path Connections

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Module 4: Tier III Environmental Challenges – Pulp & Paper Industry Created at: