Module 3: EnvironmentalChallenges – Pulp& PaperIndustry Part2.pdf · In this case, recycling the...
Transcript of Module 3: EnvironmentalChallenges – Pulp& PaperIndustry Part2.pdf · In this case, recycling the...
Program for North American Mobility in Higher Education
Introducing Process Integration for Environmental Control in Engineering Curricula
Module 3: Environmental Challenges –Pulp & Paper Industry
Created 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 plantDebarking: dryLime kiln: normalLime kiln fuel: heavy fuel oilLime kiln flue gas: high eff. ESPBark boiler (HW bark):
Total efficiency 0.87Fluidized bed boilerElectric power generation from excess heat in mill condensation turbineSince no information is available concerning the effluent treatment plant, its efficiency will be consider constant. As aconsequence 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 CaseProcess 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 recyclebecause it does not implicate major capital costs. Major results are presented in the following table.
5% reduction (need more energy to pump)
Energy produced
Neglictible differenceSolid wastes13.4% increaseDust
Cl, K: 0.2% increaseNa: 6.8% increase
Gas Effluent
Reduction of ion content of 2.3%
Liquid Effluent23% reductionWater Consumption
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Question 2 (Cont’d)
Using the LCA model, discuss if thisrepresents a real environmentalimprovement. To compare results, normalizeagainst the base case.A panel of experts has determined that the importance of each impact category can bedescribed by the weights in the followingtables. Resources and emissions are weighted separatly.What is the influence of the weights on the final decision.
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Question 2 (Cont’d)
Resource depletion: Emissions:
WeightImpact
0.08Otherresources
0.01Virgin Fiberconsumption
0.08Energyconsumption
0.83Raw water consumption
0.07Winter smog0.005Solid Wastes
0.70Carcinogenicsubstances
0.07Heavy metals
0.01Eutrophication
0.065Global Warming
0.07Summer smog
0.01Acidification
WeightImpact
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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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Solution – Q1 (Cont’d)
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 W aterConsumption
Virgin FiberConsumption
Natural ResourcesConsumption
Energy Consumption
Landfill TransportChem ical ProductionPulp Manufacturing
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Solution – Q1 (Cont’d)
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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Solution – Q1 (Cont’d)
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%
Global
Warm
ing
Acidific
ation
Eutr
ophic
ation
Heavy
Meta
lsCarc
inoge
nic S
ubsta
nces
Summer
Smog
Wint
er Smog
Solid W
aste
GrowthHarvestingLandfill TransportChemical ProductionPulp Manufacturing
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Solution – Q1 (Cont’d)
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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Solution – Q1 (Cont’d)
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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Solution – Q1 (Cont’d)
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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Solution – Q1 (Cont’d)
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 ManufacturingChemical ProductionTransportHarvestingTotal
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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
Warming
Acidific
ation
Eutrop
hicati
on
Heavy
Meta
lsCarc
inoge
nic Sub
stanc
es
Summer
Smog
Winter
Smog
Solid W
aste
Pulp ManufacturingChemical ProductionTransportLandfill HarvestingGrowthTotal
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Solution - Q2 (Cont’d)
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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Solution - Q2 (Cont’d)
The aggregated indicators are:Resources: 0.76Emissions: 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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Solution - Q2 (Cont’d)
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.
0.910.30
0.970.10
0.850.50
0.760.83
Aggregated Indicator
Weight of the raw water
consumption
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Solution - Q2 (Cont’d)
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.
0.780.16
0.900.64
0.960.83
0.820.32
0.760.08
Aggregated Indicator
Weight of the Energy
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Solution - Q2 (Cont’d)
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.
1.000.10
0.990.40
0.980.80
1.000.20
1.000.01
Aggregated Indicator
Weight of the Acidification
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Solution - Q2 (Cont’d)
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.
1.000.14
1.000.56
1.010.80
1.000.28
1.000.07
Aggregated Indicator
Weight of the Winter Smog
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Solution - Q2 (Cont’d)
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 liquorclarifier
causticizer
Dig
este
r
conc
entra
tor
Recovery Boiler
ESP
Multiple Effect Evaporators
grits
dregslime kiln
cond.cond.
screeningBrown Stock Washing
chips
white liquor
black liquor
weak
pulpwashwater
dissolving tank
mud washer
mud filter dregs
washer& filter
weak white liquor
dust recycle
smeltsalt
cake
SBL
FlueGas
washwater
washwater
D E D E D
fluegaslime mud
To papermaking
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Problem Definition
Chips = 6000 tons (wet basis)Moisture = 50% = 0.5*6000 t = 3000 tPulp Yield = 50 % of Dry = 0.5 * 3000 tConsistency (CY) = 0.12Dilution Factor (DF)= 2Wash Water for Pulp = [(1-CY)/CY] +DFIon 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 EquationDegrees 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 RepresentationW1 (kg water/s)P1 (kg pollutant/s
W2 (kg water/s)P2 (kg pollutant/s
W3 (kg water/s)P3 (kg pollutant/s
W4 (kg water/s)P4 (kg pollutant/s
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Mathematical Modeling
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
Mathematical Modeling
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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.
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Nominal Balance Model
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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Nominal Balance Model
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
35
2
4
6
7
10
9
12
11
18
13
14
15
16
17
20
21
24
25
26
27
28
29
30
3132
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
Stripper Stripper
Wood chipsPulp
Bleach Plant
W35 = 10995
Bleached pulp to papermaking
33W33 = 30990
37 W37 = 30990C37 = 15.495K37 = 0.155N37 = 15.495
ChemicalsMaterial Balance
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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
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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 tpdTotal Water Out = 55029 tpd
OVERALL WATER BALANCE
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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
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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 Consumed168
WtoBP = 30990
W1 = 3000
W2 = 13995
W6 = 1450
W24 = 5762
Water going out with pulp10995 tpd
Wastewater Target = 1669tpd
Fresh WaterTarget = 9832
Chips
Target for minimum water consumption = 52,197 – 42,365 = 9,832 tons per day
OVERALL WATER TARGETING FOR CASE STUDY
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Solution – Q3
Therefore, the rigorous targets are:Minimum Fresh water target = 9832 tpdWastewater target = 1669 tpd
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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 loopTo avoid dynamic instabilities that may arise due to the high interconnectivity between the input and outputTo enhance process reliability by disengaging the dependence of the input from the output.
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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:
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Question 4a
In the case of no self-recycle with one interceptor, which streams can be intercepted?
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Solution Q4a
There are four recyclable streams for consideration:
W8 – MEEW10 – ConcentratorW12 – ScreenW37 – 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.
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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)?
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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.
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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
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Solution Q4b
The new targets are now:Pulping process fresh water: 0 tpdBleach Plant Effluent fresh water target:30990 – (8901+ 1024) = 21065 tpd
Wastewater target:30990 – 19757 = 11233 tpd
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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
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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.
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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.
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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
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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.
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Mixing & Recycling
Interception
Segregation
Target
Chemical Process
Low Cost Process Modifications(LCPM)
HCPM
Mass Integration Strategies
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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 streamsEnhance the opportunities for direct recycle since dilute streams are easier to recycle. The separate concentrated streams are now more thermodynamically favorable for interception
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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.
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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.
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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.
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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.
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Question 5
What is the optimal water allocation using direct recycle?
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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.
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Source i = 1
Fresh Source
SegregatedSources
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 sinkWj
min ≤ Wj ≤ Wjmax j = 1, 2,….,Nsinks
Wj is the water flowrate entering the jth sinkIon content to each sink
Yion,jmin ≤ Yion,j, ≤ Yion,j
max j = 1, 2,….,NsinksYion,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
WjYion,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 tpdof bleach plant effluentThe 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
273751 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 targetPath equations for tracking water and ClRecycle ModelInterception equationsUnit 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 InterceptionGreen Liquor Interception
Of course, this leads to the next question:
99
Question 7
How much chloride needs to be removed from
Case 1: Green LiquorCase 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 Clremoval 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
273751 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
273751 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
273751 699
9133 30990
30291
Optimum Solution for White Liquor Interception
104
Life Cycle Analysis
But which of the three technologies is the better solution?
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 conditionsThis 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 variablesNE: number of equationsNC: 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 RatioModelF 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 RatioModelF 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 tpdMoisture content of wood chips = 50%Pulp Yield = 50%Pulp = Dry Chips * YieldMass fraction ions with incoming wood chips:
C1 = 1 * Chips/6000K1 = 2.50 * Chips/6000N1 = 0.973 * Chips/6000
Return to the flowsheet
113
Initial Data – Brown Stock Washer
Composition of ions in incoming wash water:Cl = 3.7 ppmK = 1.1 ppmNa = 3.6 ppm
Consistency of pulp leaving Brown Stock Washer, CY = 0.12Dilution Factor, DF = 2.0Ratio of ions in slurry leaving the BSW to the chloride in the pulp stream leaving the digester
Cl = 0.050K = 0.020Na = 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*C5K4 (from ratio to K5) = 0.02*K5N4 (from ratio to N5) = 0.009*N5
W5C5K5N5
115
DigesterW1 = Moisture*Chips=0.5*6000=3000W4 = [(1-CY)/CY]*Pulp=[(1-.82)/(0.82)]*3000DF = (W2 - W4 )*Pulp; DF is given as 2W2 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*C5K2 = (1.1*10-6) *W2; K4 = 0.02*K5N2 = (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 isour (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 – W4C5 = C1 + C2 + C3 – C4K5 = K1 + C2 + K3 – K4N5 = 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 evaporatorsThe material balances can be used to calculated the concentrated stream leaving the multiple effect evaporators
W10 = Water recovery in evaporator * W5W9 = W5 - W10C9 = C5 - C10K9 = K5 - K10N9 = N5 - N10
117
Multiple Effect Evaporators
MultipleEffect
Evaporators
W10 = Water Recovery * W5C10 = 0 K10 = 0N10 = 0
S5
W5C5K5N5
Black Liquor
S10 Evaporator Condensate
S9 Evaporator Concentrate
W9C9K9N9
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 evaporatorsThe material balances can be used to calculated the concentrated stream leaving the multiple effect evaporators
W12 = Water recovery in concentrator * W9W11 = W9 - W12C11 = C9 - C12K11 = K9 - K12N11 = N9 - N12
119
Concentrator
Concentrator
W11C11K11N11
S9 Evaporator Concentrate
W9C9K9N9
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 = 0.0C13 - C14 - C15 - C16 = 0.0K13 - K14 - K15 - K16 = 0.0N13 - N14 - N15 - N16 = 0.0It 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 * SaltcakeThe 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 = 0.0W18 = 0.0C18 = 0.01*N18K18 = 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 = 0.0K11 + K18 - K15 - K14 - K17 = 0.0N11 + 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 = W11C15 (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)
W17C17K17N17
S17
S14
S18
S15
S11
Strong Black LiquorW11C11K11N11
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 Cland K to those in the smelt and Na to the white liquorC19=0.136*C17K19=0.136*K17N19=0.196*N3Component material balances around the dissolving tank are used to evaluate the ionic content of the feed to the green liquor clarifierW20 - W17 - W19 = 0.0C20 - C17 - C19 = 0.0K20 - K17 - K19 = 0.0N20 - N17 - N19 = 0.0
126
Dissolving Tank
DissolvingTank
W17C17K17N17
S17
Smelt
W19 = 5.67 * SmeltC19 (from ratio to C17) K19 (from ratio to K17)
N19 (from ratio to N17)
S19 Dissolving Water = (85/15)*Smelt
S20 Feed to green liquor clarifier
W20C20K20N20
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*W20C21=0.863*C20K21=0.880*K20N21=0.968*N20
Component material balance can then be written around the green liquor clarifier:W22 - W21 - W20 = 0.0C22 - C21 - C20 = 0.0K22 – K21 - K20 = 0.0N20 - N21 - N20 = 0.0
128
Green Liquor Clarifier
Green LiquorClarifier
S20 Feed to green liquor clarifier
W20C20K20N20
C21 (from ratio to C20)
K21 (from ratio to K20)N21 (from ratio to N20)
S21 Dust Overflow
W21 (from ratio to W20)
S22 Underflow
W22C22K22N22
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*W22C23 = 0.010*C22K23 = 0.001*K22N23 = 0.250*N22The overflow from the white liquor clarifier is determined by assuming a ration to Green Liquor overflowW32 = 0.160*W21C32 = 0.237*C21K32 = 0.016*K21N32 = 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, respectivelyW24 = 0.9*W19C24 = (3.7*10-6) *W24K24 = (1.1*10-6) *W24N24 = (3.6*10-6) *W24Component material balances can then be written for the washer/filter systemW22 + W24 + W32 - W19 - W23 - W25 = 0.0C22 + C24 + C32 - C19 - C23 - C25 = 0.0K22 + K24 + K32 - K19 - K23 - K25 = 0.0N22 + N24 + N32 - N19 - N23 - N25 = 0.0
131
Washer/Filter
Washer/Filter
S22 Underflow
W22C22K22N22
S23
Dregs
S19
To dissolving tank
S25 Feed to lime kiln
W25
C25
K25
N25
C32 (from ratio to C21)
K32 (from ratio to K21)N32 (from ratio to N21)
S32 Overflow from WLC
W32 (from ratio to W21)
S24
Wash WaterIon content of Cl, K and Na is assumed to be 3.7, 1.1 and 3.6 ppm
132
Lime KilnThe 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 = 0.0W25 = W26C26 = 0.001 *W26K26 = 0.001 *W26N24 = 0.05 *N25A material balance can then be written around the kilnC25 – C26 – C27 = 0K25 – K26 – K27 = 0N25 – N26 – N27 = 0
133
Lime Kiln
Lime Kiln
S27 To Slaker
W27
C27
K27
N27 S25 Lime mud from washer/filterW25
C25
K25
N25
S26 Kiln Off-gas
W26
C26
K26
N26
134
SlakerThe slaking reaction is given by
CaO + H2O = Ca(OH)2The amount of water consumed by the reaction is 0.32 of the consumed limeWATERSLK = 0.32*LimeThe amount of lime fed to the slaker is 35% of the pulpLime = 0.35 * PulpThe vapor leaving the slaker makes up %5 of the water in the green liquor overflow and is ion-freeW29 = 0.005 * W21C296 = 0.0K296 = 0.0N294 = 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 = 0.0C21 + 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 = 0N29 = 0
W29 = 0.005 *W21
C21 (from ratio to C21)
K21 (from ratio to K21)N21 (from ratio to N21)
W21 (from ratio to W21)
Water consumed by Rxn0.032*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 = W30C31 = C30K31 = K30N31 = N30The material balance around the WLC is thenW21 - W32 - W3 = 0.0C21 - C32 - C2 = 0.0K21 – K32 - K3 = 0.0N21 - 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 outletsfrom 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