Environmental Challenges – Pulp & Paper Industry

Environmental Challenges – Pulp & Paper Industry
Program for North American Mobility in Higher Education Introducing Process Integration for Environmental Control in Engineering Curricula Module 4: Tier III 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

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 .

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.

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.

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.

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.

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.

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.

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.

Necessary documents LCA Base Case Process Simulation

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.

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% increase Na: 6.8% increase Energy produced 5% reduction (need more energy to pump) Dust 13.4% increase Solid wastes Neglictible difference

Question 2 (Cont’d) 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.

Question 2 (Cont’d) Resource depletion: Emissions: Impact Weight
Raw water consumption 0.83 Energy consumption 0.08 Virgin Fiber consumption 0.01 Other resources Impact Weight Carcinogenic substances 0.70 Heavy metals 0.07 Acidification 0.01 Eutrophication Summer smog Winter smog Solid Wastes 0.005 Global Warming 0.065

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

Solution – Q1 (Cont’d) Water balances can be summarized by this picture. The total fresh water consumption is =34.79 ton/ton of dry pulp. Only liquid water can be “directly” recycle: = ton/ton of dry pulp. For mass conservation reasons, only the min of fresh water or liquid effluent can be recycle ie ton. So the minimum water consumption is =6.56 ton (ie a reduction of 81%).

Solution – Q1 (Cont’d) The following graph show the contribution of each process unit to resource consumption.

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.

Solution – Q1 (Cont’d) The following graph show the contribution of each process unit to emission-related environmental impacts.

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.

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.

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.

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.

Solution - Q2

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.

Solution - Q2 (Cont’d)

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.

Solution - Q2 (Cont’d) 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.

Weight of the raw water consumption
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. Weight of the raw water consumption Aggregated Indicator 0.83 0.76 0.50 0.85 0.30 0.91 0.10 0.97

Solution - Q2 (Cont’d) Weight of the Energy Aggregated Indicator 0.08
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 Aggregated Indicator 0.08 0.76 0.16 0.78 0.32 0.82 0.64 0.90 0.83 0.96

Weight of the Acidification
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. Weight of the Acidification Aggregated Indicator 0.01 1.00 0.10 0.20 0.40 0.99 0.80 0.98

Weight of the Winter Smog
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. Weight of the Winter Smog Aggregated Indicator 0.07 1.00 0.14 0.28 0.56 0.80 1.01

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.

Problem Statement – Q3-7 Consider the following Kraft pulp mill depicted below wash pulp water chips D E D E D screening Brown Stock Washing To papermaking Digester Flue Recovery Boiler Gas concentrator cond. cond. SBL weak smelt ESP black liquor Multiple Effect Evaporators salt cake dust recycle white liquor wash water lime fluegas mud weak white liquor dissolving wash tank mud lime kiln water dregs white liquor filter mud dregs clarifier washer washer & filter green liquor clarifier causticizer grits slaker

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)

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.

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

Question 3 What are the rigorous targets in water discharge and reduction?

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.

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 Freedom Mixer-Splitter Model

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:

Pollutant/Water Load Balance Representation
W1 (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

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).

With this knowledge, the ratio model can be used to relate the pollutant loads in streams 2 and 3: P3new = (P3old/ P2old) * P2new The pollutant load in stream 4 can then be determined by a simple component material balance: P4new = P1new + P1new + P3new

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.

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

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

Material Balance Bleach Plant Screening Washer Digester ESP MEE
Chemicals C6 = 0.005 W33 = 30990 K2 = 0.015 2 C4 = 0.492 6 K6 = 0.002 W7 = 10995 33 N2 = 0.050 K4 = 0.819 N6 = 0.005 C7 = 0.394 N4 = 4.347 K7 = 0.655 Bleach Plant Screening 7 N7 = 3.478 W35 = 10995 Washer 8 Pulp Bleached pulp 4 W8 = 1450 Wood chips C8 = 0.104 to papermaking K8 = 0.165 N8 = 0.875 37 W37 = 30990 W1 = 3000 10 C37 = C1 = 1.000 W10 = 8901 C10 = 0.000 W12 = 1024 K37 = 0.155 W15 = 1202 K1 = 2.500 C1 2= 0.000 N37 = C15 = 0.197 N1 = 0.973 Digester K10 = 0.000 N10 = 0.000 12 K12 = 0.000 W16 = 0 K15 = 0.327 1 Stripper N12 = 0.000 C16 = 4.230 N15 = 0.386 Stripper 15 K16 = 9.904 W5 = 11127 N16 = C5 = 9.838 W9 = 2225 K5 = ESP W14 = 0 C9 = 9.838 14 C14 = 0.472 W3 = 5127 N5 = 16 K9 = K14 = 1.146 C3 = 9.278 W11 = 1202 5 K3 = MEE N9 = Concent . C11 = 9.838 N14 = 0.966 13 W13 = 1202 N3 = K11 = C13 = 4.899 3 9 N11 = K13 = W32 = 1016 11 Recovery N13 = White C32 = 2.173 W26 = 423 W18 = 0 K32 = 0.631 Liq Clar C26 = 0.042 C18 = 0.182 18 Furnace N32 = K26 = 0.042 26 K18 = 0.026 N26 = N18 = Na SO Lime 2 4 32 smelt W17 = 0 W31 = 6143 27 17 C17 = 9.351 C31 = 31 Kiln K17 = K31 = N17 = W27 = 0 N31 = C27 = 2.297 W25 = 423 C25 = 2.339 W19 = 6402 K27 = 0.604 Dissol . K25 = 0.647 C19 = 1.272 Causticizer N27 = 0.633 25 N25 = K19 = 5.369 N19 = Tank Washers/ 19 W29 = 32 24 W23 = 3.84 C29 = 0.000 W30 = 6143 Filters C23 = 0.038 W20 = 6402 20 K29 = 0.000 30 C30 = 23 K23 = 0.004 C20 = N29 = 0.000 K30 = K20 = W24 = 5762 N23 = 0.960 N30 = N20 = C24 = 0.021 W22 = 51 29 K24 = 0.006 N24 = 0.021 C22 = 1.455 W21= 6351 K22 = 5.382 Green Liq Slaker N22 = C21= 9.167 W28 = 8 K21 = C28 = 0.014 28 N21 = 21 Clarifier K28 = 0.209 N28 = 0.576 22

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.

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

OVERALL WATER BALANCE OVERALL Water Balance Screening = 1450
MEE = 8901 Chips = 3000 OVERALL Water Balance Stripper = 1024 Washer = 13995 ESP = 1202 Screening = 1450 Washer/Filter dregs = 4 Lime Kiln = 423 Washers/Filters = 5762 Slaker grits = 8 Bleach Plant = 30990 Slaker = 32 Water consumed By reaction = 168 BP water = 30990 Pulp leaving Bleach Plant = 10995 Total Water In = 55197 – 168 = tpd Total Water Out = tpd

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 )

FRESH AND RECYCLABLE WATER BALANCE
Screening = 1450 MEE = 8901 Water Balance Washer = 13995 Stripper = 1024 Screening = 1450 Washers/Filters = 5762 Bleach Plant = 30990 Water consumed By reaction = 168 BP water = 30990 Total Fresh Water in = tpd Total recyclable water out = tpd

Minimum water consumption =
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 – = 9832 tpd

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

OVERALL WATER TARGETING FOR CASE STUDY
Wastewater Chips W1 = 3000 Target = 1669 tpd W2 = 13995 Water Consumed 168 Fresh Water W6 = 1450 Target = 9832 Water going out with pulp 10995 tpd W24 = 5762 WtoBP = 30990 Target for minimum water consumption = 52,197 – 42,365 = 9,832 tons per day

Solution – Q3 Therefore, the rigorous targets are:
Minimum Fresh water target = 9832 tpd Wastewater target = 1669 tpd

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.

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:

Question 4a In the case of no self-recycle with one interceptor, which streams can be intercepted?

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.

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)?

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.

Solution – Q4b 13995 + 5762 + 1450 = 21207 Wastewater To bio
=30990 – 19757 = tpd Fresh water = tpd Bleach interception W2 = 13995 PULPING W8 = 1450 W6 = 1450 Consumption by Chemical Reaction and other losses = 9832 tpd W24 = 5762 W10 = 8901 W12 = 1024 = 21207

Solution Q4b The new targets are now:
Pulping process fresh water: 0 tpd Bleach Plant Effluent fresh water target: 30990 – ( ) = tpd Wastewater target: 30990 – = tpd

Process Integration Strategies
The overall targeting has identified that fresh water consumption can be significantly reduced from 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

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.

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.

Coordination of Process Integration and Simulation
Process Objectives, Data and constraints Process Modifications Structural changes Process Simulation Process Integration Input/Output relations New Processes Closing the information loop of integration and simulation ensures that the developed insights and solution strategies are refined and validated.

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.

Mass Integration Strategies
Target Chemical Process HCPM Interception Mixing & Recycling Low Cost Process Modifications (LCPM) Segregation

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

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.

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.

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.

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.

Question 5 What is the optimal water allocation using direct recycle?

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.

Spriggs and El-Halwagi, 1998,
SOURCE-INTERCEPTION-SINK REPRESENTATION Mass & Energy Separating Agents In Sinks Segregated Sources Source i = 1 j = 1 Species Interception Network (SPIN) j = 2 Source i = Nsources Fresh Source j = Nsinks (e.g., El-Halwagi et al., 1996, Spriggs and El-Halwagi, 1998, Dunn and El-Halwagi, 2003) Mass & Energy Separating Agents Out

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  Wjmax j = 1, 2,….,Nsinks Wj is the water flowrate entering the jth sink Ion content to each sink Yion,jmin  Yion,j,  Yion,jmax j = 1, 2,….,Nsinks Yion,j is the compostion of a certain NPE entering the jth sink

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)

SPLITTING OF SOURCES TO SINKS
Yion,i li,j Splitting of the ith source: where i = 1,2, …, Nsources

MIXING OF SOURCES BEFORE SINKS
Wj Yion,j li,j yion,j j Mixing for the jth sink: where j = 1,2, …, Nsinks

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

Direct Recycle/Reuse Representation
Sinks Sources S Wastewater 8 Screening from Screening S Condensate 10 from MEE Brown Stock Washer S Condensate 12 from Concenrator Washers/Filters S Bleach Plant 37 Effluent Bleach Plant Fresh water Biotreatment

Min. flowrate of fresh water =
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

Direct Recycle Optimization Formulation for Source/Sink Analysis w/Path Connection
Flowrate to each sink: j = 1, 2, …, Nsinks NPE content in feed to each sink: j = 1, 2, …, Nsinks and k = 1, 2, …, Nk Splitting for the ith source: i = 1, 2, …, Nsources Mixing for the jth sink: j = 1, 2, …, Nsinks Component material balances for the pollutants: j = 1, 2, …, Nsinks and k = 1, 2, …, Nk Non-negativity of each fraction of split sources: i = 1, 2, …, Nsources and j = 1, 2, …, Nsinks

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:

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.

Optimum Solution for Direct Recycle/Ruse
9133 30990 BSW Screening Bleach 1450 Wood Chips 751 699 30291 3412 Digester 5489 273 ESP Stripper Stripper White Liq. Clarif. MEE Concent. Recovery Furnace Lime Kiln Dissolv Tank Causticizer Washers/ Filters Green Liq. Clarif. Slaker

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.

Question 6 Through direct recycle, fresh water usage went down from to However, from water targeting, we know that interception can get the fresh water usage down to 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 tpd?

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

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 tpd and that approximately 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 tpd).

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

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:

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?

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)

Optimum Solution for Bleach Plant Interception

Optimum Solution for Green Liquor Interception

Optimum Solution for White Liquor Interception

Life Cycle Analysis But which of the three technologies is the better solution?

Return to the flowsheet
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

(Fresh inputs or outlets from other units)
Degrees of Freedom 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. NV = NS x NC F= NV - NE = NC (NS - 1) F: degrees of freedom NV: number of variables NE: number of equations NC: number of targeted species NS: number of outlet streams Unit U Inlet stream (Fresh inputs or outlets from other units) Outlet streams Nstreams out Return to the flowsheet

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

Fixed Split Flow Model Fixed Split Model α * F 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

Gnew = Gold * ( Fnew / Fold ) Return to the flowsheet
Flow Ratio Model Flow Ratio Model G F 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

IInew = Inew (IIold/ Iold ) Return to the flowsheet
Species Ratio Model Species Ratio Model G F I = species 1 II = species 2 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. Return to the flowsheet

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 = * Chips/6000 Return to the flowsheet

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

Digester S2 Brown Stock Washer S4 S1 Digester S5 S3
W2 (from consistency) S2 C2 (from comp of Cl in wash water) K2 (from comp of K in wash water) N2 (from comp of N in wash water) Brown Stock Washer S4 W4 (from dilution factor C4 (from ratio to C5) = 0.05*C5 K4 (from ratio to K5) = 0.02*K5 S1 N4 (from ratio to N5) = 0.009*N5 Digester W1 (from moisture content C1 (from comp of Cl in chips) S5 K1 (from comp of K in chips N1 (from comp of N in chips W5 C5 S3 K5 N5 All species data will be calculated as an output stream from white liquor clarifier

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, then stream 5 will need to be determined. The number of unknowns is our (flowrate of water and the three ions in S5); these can be obtained Via 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

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

Multiple Effect Evaporators
W10 = Water Recovery * W5 C10 = 0 Evaporator Condensate K10 = 0 N10 = 0 S5 Black Liquor S9 Multiple Effect Evaporators Evaporator Concentrate W5 C5 K5 N5 W9 C9 K9 N9

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

Concentrator Condensate Evaporator Concentrate
W12 = Water Recovery * W9 C12 = 0 Concentrator Condensate K12 = 0 N12 = 0 S9 S11 Evaporator Concentrate Concentrator Strong Black Liquor W9 C9 K9 N9 W11 C11 K11 N11

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 = *N11

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 * 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 = *N18

Recovery Furnace and ESP
The ion content in the smelt is determined via component material balance around the Recovery Furnace and ESP C11 + C18 - C15 - C14 - C17 = 0.0 K11 + K18 - K15 - K14 - K17 = 0.0 N11 + N18 - N15 - N14 - N17 = 0.0

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 Or Smelt = Saltcake *SBL

Recovery Furnace and Electrostatic Precipitator (ESP
W15 = W11 S15 C15 (from ratio to C11) K15 (from ratio to K11) Off-gas S14 N15 (from ratio to N11) Dust Purge W14 = 0 S11 ESP C14 (from ratio to C11) K14 (from ratio to K11) Strong Black Liquor N14 (from ratio to N11) W11 C11 K11 N11 Recovery Furnace S17 Smelt S18 Salt Cake = *Pulp W17 C17 W18 = 0 K17 C18 (from ratio to N18) K18 (from ratio to N18) N17 N18 = * Salt Cake

Dissolving Tank The dissolving water-to-smelt ratio used in the dissolving tank is typically 85/15 W19 = (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 liquor C19=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 clarifier W20 - W17 - W19 = 0.0 C20 - C17 - C19 = 0.0 K20 - K17 - K19 = 0.0 N20 - N17 - N19 = 0.0

Dissolving Water = (85/15)*Smelt Feed to green liquor clarifier
Dissolving Tank S17 Smelt W17 C17 K17 N17 S19 Dissolving Tank Dissolving Water = (85/15)*Smelt W19 = 5.67 * Smelt C19 (from ratio to C17) K19 (from ratio to K17) S20 N19 (from ratio to N17) Feed to green liquor clarifier W20 C20 K20 N20

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

Green Liquor Clarifier
S20 Feed to green liquor clarifier W20 C20 K20 N20 S21 Green Liquor Clarifier Dust Overflow W21 (from ratio to W20) C21 (from ratio to C20) K21 (from ratio to K20) S22 N21 (from ratio to N20) Underflow W22 C22 K22 N22

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 dregs W23 = 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 overflow W32 = 0.160*W21 C32 = 0.237*C21 K32 = 0.016*K21 N32 = 0.156*N21

Washer/Filter System W24 = 0.9*W19
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 system W22 + 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

Washer/Filter S25 S32 Feed to lime kiln Overflow from WLC S19
W32 (from ratio to W21) W25 C25 C32 (from ratio to C21) K25 K32 (from ratio to K21) N25 N32 (from ratio to N21) S19 Washer/Filter To dissolving tank S23 S24 Dregs Wash Water S22 Ion content of Cl, K and Na is assumed to be 3.7, 1.1 and 3.6 ppm Underflow W22 C22 K22 N22

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 0.001 W27 = 0.0 W25 = W26 C26 = *W26 K26 = *W26 N24 = 0.05 *N25 A material balance can then be written around the kiln C25 – C26 – C27 = 0 K25 – K26 – K27 = 0 N25 – N26 – N27 = 0

Lime mud from washer/filter
Lime Kiln S26 Kiln Off-gas W26 C26 K26 S27 N26 To Slaker Lime Kiln W27 C27 K27 N27 S25 Lime mud from washer/filter W25 C25 K25 N25

Slaker The slaking reaction is given by CaO + H2O = Ca(OH)2
The amount of water consumed by the reaction is 0.32 of the consumed lime WATERSLK = 0.32*Lime The amount of lime fed to the slaker is 35% of the pulp Lime = 0.35 * Pulp The vapor leaving the slaker makes up %5 of the water in the green liquor overflow and is ion-free W29 = * W21 C296 = 0.0 K296 = 0.0 N294 = 0.0

Slaker The ion content in the grits is related to the green liquor overflow W28 = * W21 C28 = *C21 K28 = *K21 N28 = *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 = 0.0 N21 + N27 - N28 - N29 - N30 = 0.0

Slaker S30 S29 To Causticizer Slaker Vapor S27 From Lime Kiln Slaker
W29 = *W21 C29 = 0 S27 K29 = 0 Slaker From Lime Kiln N29 = 0 S21 S28 Green Liquor Overflow Grits Water consumed by Rxn 0.032*Lime = 0.35*Pulp W21 (from ratio to W21) C21 (from ratio to C21) K21 (from ratio to K21) N21 (from ratio to N21)

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 then W21 - W32 - W3 = 0.0 C21 - C32 - C2 = 0.0 K21 – K32 - K3 = 0.0 N21 - N32 - N3 = 0.0

Causticizer and White Liquor Clarifier
to digester White Liquor Clarifier S32 To washer/filter S31 Causticizer S30 From Slaker

Bleaching Bleaching

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.

Degree of Freedom Analysis
Given the following generic unit INLET STREAMS OUTLET STREAMS (fresh inputs or outlets from other units) Unit U NS and F = Number of Degrees of Freedom NV = number of unknown variables (NS*NC) NE = number of equations in each variable NS = number of outlet streams NC = number of targeted species

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)

PINCH-BASED VISUALIZATION TOOL FOR BPX PROBLEM
(El-Halwagi et al., 2003)

Optimization of Source/Sink Analysis With Path Connections
MEE Cond. G6 f101 Screen F10, z10 y6 u = 1 f102 f105 f103 Conc. Cond. G2 f121 Brown Stock Washer f122 F12, z12 f123 y2 u = 2 f125 f124 fresh2 Screen Cond f81 f82 G24 F8, z8 f83 Washer/Filter f84 y24 f85 u = 3 fresh3 BPE f372 f371 f104 G37 f373 F37, z37 f83 Bleach Plant f37bio y37 u = 4 fresh1 Fresh H2O G2 fresh4 Biotreatment Fresh, z5 y2 u = 5

Environmental Challenges – Pulp & Paper Industry

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